Solutions to Chapter 1 Problems 11
Some typical applications of engineering economy are listed below. Different organi...
3288 downloads
12969 Views
6MB Size
Report
This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form
Solutions to Chapter 1 Problems 11
Some typical applications of engineering economy are listed below. Different organizations have been assumed, but many applications do not vary significantly by the type of organization. Manufacturing Company • • • • •
Addition (expansion) to an existing plant. Improvements to a material handling system. Selecting between numerical controlled milling machines produced by different manufacturers. Selecting the preferred conceptual (preliminary) design for a new consumer product. Selecting the preferred detailed design for an application (e.g. a precompressor option for a present series of large gasoline engines).
Service Company • • • • •
Selecting the best type (and model) of light truck for use in a service fleet. Determining the preferred location of a distribution center. Selecting an upgraded computer system to integrate and improve the accomplishment of a number of present functions in an organization. Selecting the preferred conceptual (preliminary) design for a new consumer service and its basic delivery process. Selecting the preferred detailed design for a service (e.g., a home health care service involving professional nursing care to meet the periodic needs of patients).
Government Organization • • • • •
12
Evaluating the contracting of garbage pickup and disposal versus continued accomplishment by a city work force. Analyzing the significant repair and upgrading of a highway bridge versus its replacement with a new structure. Selecting the preferred conceptual (preliminary) design for a new elementary school. Evaluating the life cycle cost for a new major weapon system. Determining the preferred replacement cycle for medium sized backhoes in the highway department equipment fleet.
The economic aspects of an engineering project are of equal importance to its physical aspects. It is very important to the engineering profession and the public that the designs for products, structures, systems, and services result in economic consequences acceptable to the user(s). Otherwise, the basic social needs (or wants) will not be satisfied.
3
Chapter 1 Solutions
13
Some nonmonetary factors (attributes) that might be important are: • • • • •
Safety Reliability (from the viewpoint of user service) Quality in terms of consumer expectations Aesthetics (how it looks, and so on) Patent considerations
14
The increased use of automation in the activities of all organizations should increase the importance of engineering economy studies. Some reasons for this viewpoint are: increased investment costs; the tradeoff between increased investment costs and other costs such as personnel; the impact on revenue of improved service time, better flexibility to meet changing demand, and so on; the consideration of new costs associated with software, maintenance, and changed skill requirements in the work force; and other factors that impact the economic and other consequences of increasing automation.
15
The results of engineering economic analyses are used to assist decision making. A decision involves making a choice among two or more alternatives; thus, fundamentally the decision is among alternatives.
16
A potential alternative, in order to be selected as a feasible alternative for detailed analysis, must be considered capable of achieving the outcomes for the project based on preliminary analysis. The planned outcomes for a project are the goals, objectives, performance criteria, and other results that have been established.
17
Uncertainty refers to the variation of actual values that will occur in the future from the estimated values developed at the time of the study associated with a project. Engineering economic analysis is prospective in that it deals with the analysis of the estimated future consequences of alternative courses of action. Thus, uncertainty is inherently involved. Some causes of uncertainty are: • • • • • •
Rapid changes in the demand for products and services. Inflation and price changes. Changes in technology. Competition in the world marketplace. Changes in regulatory requirements. Lack of an adequate costing base (data) for a new project.
18
The perspective related to cash flow of $1,000 between two individuals is different for the person paying the amount than it is for the one receiving it; that is, the first person sees a negative cash flow of $1,000 and the second person sees a positive cash flow of the same amount. Thus, the viewpoint or perspective related to the future consequences of a course of action is in terms of an individual organization, constituency group, and so on. Normally, the perspective used is that of the owners of the organization.
19
There is not a single "correct" plan for accomplishing a postevaluation. However, important activities to be included in a technical plan for this purpose are:
Chapter 1 Solutions
4
110
111
•
Establishing the original baseline for the selected alternative. This involves explicitly defining each of the original goals, objectives, performance criteria, and other results established for the project; and the estimates or projections originally developed for each of these results.
•
Determine which results can be compared in monetary terms, quantified in other measures, or made explicit with a selected description technique.
•
Obtaining (and developing) the economic data and other information needed about actual results achieved.
•
Accomplishing the comparison of the original baseline for the selected alternative with the actual results achieved.
•
Identifying the differences in the comparison, and determining the probable cause for these differences.
•
Deciding upon improvements that could be made in engineering economic analyses or the decision making process within the organization to achieve a better outcome for similar projects in the future.
(a)
Assume that the MTBF and the average cost per repair are known or can be developed from historical data for the existing equipment. Then, if no changes were made in the replacement item that affects the average cost per repair the impact of the improvement in the MTBF can be estimated in the monetary unit (e.g. dollars) directly based on the 40% change in the MTBF only. Specifically, the maintenance cost under these circumstances would be estimated to be 40% less.
(b)
Estimating the economic consequences of this improvement in the product will take close cooperation between engineering design, marketing, purchasing, production and other functions within the organization. The impact of these changes can be developed by estimating the increase in production cost, deciding upon any changes in product prices, estimating changes in sales (revenues) that will result from the improved product, and then estimating the change in the net income of the organization.
(c)
This situation deals with two basic tasks. These tasks are developing feasible alternatives for improved discharge levels that the company believes will maintain a "good neighbor" policy and estimating the costs associated with each alternative. Once these tasks are completed, a typical decision situation exists and the monetary consequences will be defined by the alternative selected.
The relationship between engineering economic analysis and engineering design is characterized by its integrated nature. As indicated in Figure 11, each of the first six activities of the analysis procedure has information transfer with one or two of the six activities in the design process. Information from the design process activities is used in doing the steps of the analysis procedure, particularly for procedure Steps 1 and 2.
5
Chapter 1 Solutions
Similarly, the economic results of the procedure Steps 3, 4 and 5 are an integral part of the design process Activity 4. Likewise, the preferred alternative based on the economic aspects of the situation (Activity 6) is critical to the selection and specification of the preferred design alternative (Activity 5). Iteration within the analysis procedure steps occurs in conjunction with any iteration in the design process activities. 112
(a)
Problem:
To find the least expensive method for setting up capacity to produce drill bits.
(b)
Assumptions: The revenue per unit will be the same for either machine; startup costs are negligible; breakdowns are not frequent; previous employee’s data are correct; drill bits are manufactured the same way regardless of the alternative chosen; inhouse technicians can modify the old machine so its life span will match that of the new machine; neither machine has any resale value; there is no union to lobby for inhouse work; etc.
(c)
Alternatives:
(1) Modify the old machine for producing the new drill bit (using inhouse technicians); (2) Buy a new machine for $450,000; (3) Get McDonald Inc. to modify the machine; (4) Outsource the work to another company.
(d)
Criterion:
Least cost in dollars for the anticipated production runs, given that quality and delivery time are essentially unaffected (i.e., not compromised).
(e)
Risks:
The old machine could be less reliable than a new one; the old machine could cause environmental hazards; fixing the old machine inhouse could prove to be unsatisfactory; the old machine could be less safe than a new one; etc.
(f) Nonmonetary Considerations: Safety; environmental concerns; quality/ reliability differences; “flexibility” of a new machine; job security for inhouse work; image to outside companies by having a new technology (machine); etc. (g) Post Audit: Did either machine (or outsourcing) fail to deliver high quality product on time? Were maintenance costs of the machines acceptable? Did the total production costs allow an acceptable profit to be made?
Chapter 1 Solutions
6
113
(a)
Problem A: Subject to time, grade point average and energy that Mary is willing/able to exert, Problem A might be "How can Mary survive the senior year and graduate during the coming year (earn a college degree)?" Problem B: Subject to knowledge of the job market, mobility and professional ambition, Mary's Problem B could be "How can I use my brother's entrylevel job as a spring board into a higherpaying position with a career advancement opportunity (maybe no college degree)?"
(b) Problem A  Some feasible solutions for Problem A would include: (1)
Get a loan from her brother and take fewer courses per term, possibly graduating in the summer.
(2)
Quit partying and devote her extra time and limited funds to the task of graduating in the spring term (maybe Mary could get a scholarship to help with tuition, room and board).
Problem B  Some feasible solutions for Problem B would include:
114
(1)
Work for her brother and take over the company to enable him to start another entrepreneurial venture.
(2)
Work parttime for her brother and continue to take courses over the next couple of years in order to graduate.
(3)
Work for her brother for one or two semesters to build up funds for her senior year. While interviewing, bring up the real life working experience and request a higher starting salary.
A Typical Discussion/Solution: (a)
One problem involves how to satisfy the hunger of three students  assume a piping hot delicious pizza will satisfy this need. (Another problem is to learn enough about Engineering Economy to pass  or better yet earn an “A” or a “B”  on the final examination and ace the course. Maybe a pizza will solve this problem too?) Let’s use “hunger satisfaction with a pizza” as the problem/need definition.
(b)
Principle 1  Develop the Alternatives i) Alternative A is to order a pizza from “PickUp Sticks” ii) Alternative B is to order a pizza from “Fred’s” Other options probably exist but we’ll stick to these two alternatives Principle 2  Focus on the Differences Difference in delivery time could be an issue. A perceived difference in the quality of the ingredients used to make the pizza could be another factor to consider. We’ll concentrate our attention on cost differences in part (c) to follow. Principle 3  Use a Consistent Viewpoint 7
Chapter 1 Solutions
Consider your problem from the perspective of three customers wanting to get a good deal. Does it make sense to buy a pizza having a crust that your dog enjoys, or ordering a pizza from a shop that employs only college students? Use the customer’s point of view in this situation rather than that of the owner of the pizza shop or the driver of the delivery vehicle. Principle 4  Use a Common Unit of Measure Most people use “dollars” as one of the most important measures for examining differences between alternatives. In deciding which pizza to order, we’ll use a costbased metric in part (c). Principle 5  Consider All Relevant Criteria Factors other than cost may affect the decision about which pizza to order. For example, variety and quality of toppings and delivery time may be extremely important to your choice. Dynamics of group decision making may also introduce various “political” considerations into the final selection (can you name a couple?) Principle 6  Make Uncertainty Explicit The variability in quality of the pizza, its delivery time and even its price should be carefully examined in making your selection. (Advertised prices are often valid under special conditions  call first to check on this!) Principle 7  Revisit Your Decision After you’ve consumed your pizza and returned to studying for the final exam, were you pleased with the taste of the toppings? On the downside, was the crust like cardboard? You’ll keep these sorts of things in mind (good and bad) when you order your next pizza! (c) Finally some numbers to crunch  don’t forget to list any key assumptions that underpin your analysis to minimize the cost per unit volume (Principles 1, 2, 3, 4 and 6 are integral to this comparison) Assumptions: (i) weight is directly proportional to volume (to avoid a “meringue” pizza with lots of fluff but meager substance), (ii) you and your study companions will eat the entire pizza (avoids variable amounts of discarded leftovers and hence difficulttopredict cost of cubic inch consumed) and (iii) data provided in the Example Problem are accurate (the numbers have been confirmed by phone calls). Analysis:
Alternative A “PickUpSticks” Volume = 20″ x 20″ x 1 ¼″ = 500 in.3 Total Cost = $15 (1.05) + $1.50 = $17.25 Cost per in.3 = $0.035 Alternative B “Fred’s” Volume = (3.1416)(10″)2 (1.75″) = 550 in.3 Total Cost = $17.25 (1.05) = $18.11 Cost per in.3 = $0.033
Chapter 1 Solutions
8
Therefore, order the pizza from “Fred’s” to minimize total cost per cubic inch. (d)
115
Typical other criteria you and your friends could consider are: (i) cost per square inch of pizza (select “PickUpSticks”), (ii) minimize total cost regardless of area or volume (select “PickUpSticks”), and (iii) “Fred’s” can deliver in 30 minutes but “PickUpSticks” cannot deliver for one hour because one of their ovens is not working properly (select “Fred’s”).
Definition of Need Some homeowners need to determine (confirm) whether a storm door could fix their problem. If yes, install a storm door. If it will not basically solve the problem, proceed with the problem formulation activity. Problem Formulation The homeowner’s problem seems to be one of heat loss and/or aesthetic appearance of their house. Hence, one problem formulation could be: “To find different alternatives to prevent heat loss from the house.” Alternatives • Caulking of windows • Weather stripping • Better heating equipment • Install a storm door • More insulation in the walls, ceiling, etc. of the house • Various combinations of the above
116 STEP 1—Define the Problem Your basic problem is that you need transportation. Further evaluation leads to the elimination of walking, riding a bicycle, and taking a bus as feasible alternatives. STEP 2—Develop Your Alternatives (Principle 1 is used here.) Your problem has been reduced to either replacing or repairing your automobile. The alternatives would appear to be 1. Sell the wrecked car for $2,000 to the wholesaler and spend this money, the $1,000 insurance check, and all of your $7,000 savings account on a newer car. The total amount paid out of your savings account is $7,000, and the car will have 28,000 miles of prior use. 2. Spend the $1,000 insurance check and $1,000 of savings to fix the car. The total amount paid out of your savings is $1,000, and the car will have 58,000 miles of prior use. 3. Spend the $1,000 insurance check and $1,000 of your savings to fix the car and then sell the car for $4,500. Spend the $4,500 plus $5,500 of additional savings to buy the newer car. The total amount paid out of savings is $6,500, and the car will have 28,000 miles. 4. Give the car to a parttime mechanic, who will repair it for $1,100 ($1,000 insurance and $100 of your savings), but will take an additional month of repair time. You will also have to rent a car for that time at $400/month (paid out of savings). The total amount paid out of savings is $500, and the car will have 58,000 miles on the odometer. 5. Same as Alternative 4, but you then sell the car for $4,500 and use this money plus $5,500 of additional savings to buy the newer car. The total amount paid out of savings is $6,000, and
9
Chapter 1 Solutions
the newer car will have 28,000 miles of prior use. ASSUMPTIONS: 1. The less reliable repair shop in Alternatives 4 and 5 will not take longer than an additional month to repair the car. 2. Each car will perform at a satisfactory operating condition (as it was originally intended) and will provide the same total mileage before being sold or salvaged. 3. Interest earned on money remaining in savings is negligible. STEP 3—Estimate the Cash Flows for Each Alternative (Principle 2 should be adhered to in this step.) 1. Alternative 1 varies from all others because the car is not to be repaired at all but merely sold. This eliminates the benefit of the $500 increase in the value of the car when it is repaired and then sold. Also this alternative leaves no money in your savings account. There is a cash flow of −$8,000 to gain a newer car valued at $10,000. 2. Alternative 2 varies from Alternative 1 because it allows the old car to be repaired. Alternative 2 differs from Alternatives 4 and 5 because it utilizes a more expensive ($500 more) and less risky repair facility. It also varies from Alternatives 3 and 5 because the car will be kept. The cash flow is −$2,000 and the repaired car can be sold for $4,500. 3. Alternative 3 gains an additional $500 by repairing the car and selling it to buy the same car as in Alternative 1. The cash flow is −$7,500 to gain the newer car valued at $10,000. 4. Alternative 4 uses the same idea as Alternative 2, but involves a less expensive repair shop. The repair shop is more risky in the quality of its end product, but will only cost $1,100 in repairs and $400 in an additional month’s rental of a car. The cash flow is −$1,500 to keep the older car valued at $4,500. 5. Alternative 5 is the same as Alternative 4, but gains an additional $500 by selling the repaired car and purchasing a newer car as in Alternatives 1 and 3. The cash flow is −$7,000 to obtain the newer car valued at $10,000. STEP 4—Select a Criterion It is very important to use a consistent viewpoint (Principle 3) and a common unit of measure (Principle 4) in performing this step. The viewpoint in this situation is yours (the owner of the wrecked car). The value of the car to the owner is its market value (i.e., $10,000 for the newer car and $4,500 for the repaired car). Hence, the dollar is used as the consistent value against which everything is measured. This reduces all decisions to a quantitative level, which can then be reviewed later with qualitative factors that may carry their own dollar value (e.g., how much is low mileage or a reliable repair shop worth?). STEP 5—Analyze and Compare the Alternatives Make sure you consider all relevant criteria (Principle 5). 1. Alternative 1 is eliminated, because Alternative 3 gains the same end result and would also provide the car owner with $500 more cash. This is experienced with no change in the risk to the owner. (Car value = $10,000, savings = 0, total worth = $10,000.)
Chapter 1 Solutions
10
2. Alternative 2 is a good alternative to consider, because it spends the least amount of cash, leaving $6,000 in the bank. Alternative 2 provides the same end result as Alternative 4, but costs $500 more to repair. Therefore, Alternative 2 is eliminated. (Car value = $4,500, savings = $6,000, total worth = $10,500.) 3. Alternative 3 is eliminated, because Alternative 5 also repairs the car but at a lower outofsavings cost ($500 difference), and both Alternatives 3 and 5 have the same end result of buying the newer car. (Car value = $10,000, savings = $500, total worth = $10,500.) 4. Alternative 4 is a good alternative, because it saves $500 by using a cheaper repair facility, provided that the risk of a poor repair job is judged to be small. (Car value = $4,500, savings = $6,500, total worth = $11,000.) 5. Alternative 5 repairs the car at a lower cost ($500 cheaper) and eliminates the risk of breakdown by selling the car to someone else at an additional $500 gain. (Car value = $10,000, savings = $1,000, total worth = $11,000.) STEP 6—Select the Best Alternative When performing this step of the procedure, you should make uncertainty explicit (Principle 6). Among the uncertainties that can be found in this problem, the following are the most relevant to the decision. If the original car is repaired and kept, there is a possibility that it would have a higher frequency of breakdowns (based on personal experience). If a cheaper repair facility is used, the chance of a later breakdown is even greater (based on personal experience). Buying a newer car will use up most of your savings. Also, the newer car purchased may be too expensive, based on the additional price paid (which is at least $6,000/30,000 miles = 20 cents per mile). Finally, the newer car may also have been in an accident and could have a worse repair history than the presently owned car. Based on the information in all previous steps, Alternative 5 was actually chosen. STEP 7—Monitor the Performance of Your Choice This step goes handinhand with Principle 7 (revisit your decisions). The newer car turned out after being “test driven” for 20,000 miles to be a real beauty. Mileage was great, and no repairs were needed. The systematic process of identifying and analyzing alternative solutions to this problem really paid off! 117 An inclass team exercise! Left to the creativity of the students and the insights/guidance of the instructor!
11
Chapter 1 Solutions
Solutions To Chapter 3 Problems 31
A representative WBS for a lawnmower (home use) through level 3 is shown below. (Suggestion to instructor: Copy part of an owners manual and distribute it to the class.)
Level 1
Mower (Walk behind, self propelled)
2
1.1
1.2
Chassis
3
1.3
Drive Train
Controls
Cutting & 1.4 Collection Mechanism
1.1.1 Deck
1.2.1 Engine
1.3.1 Handle
1.4.1 Blade Assembly
1.1.2 Wheels
1.2.2 Starter Assembly
1.3.2 Engine Control Linkage
1.4.2 Side Chute
1.1.3 Axles
1.2.3 Transmission
1.3.3 Blade Control Linkage
1.4.3 Grass Bag Adapter
1.2.4 Drive Disc Assembly
1.3.4 Speed Control Linkage
1.4.4 Bag
1.2.5 Driven Disc Assembly
1.3.5 Drive Control Assembly
1.2.6 Clutch Linkage
1.3.6 Cutting Height Adjustment
1.2.7 Belt Assemblies
32
A representative cost and revenue structure for construction, 10years of ownership and use, and the sale of a home is: Cost or Revenue Category Captial Investment
Annual Operating and Maintenance Costs
Major Repair or Replacement Costs Real Estate Fees Asset Sales
Chapter 3 Solutions
Typical Cost and Revenue Elements Real estate (lot) cost; architect/engineering fees; construction costs (labor,material, other); working capital (tools, initial operating supplies, etc.); landscaping costs. Utilities (electricity, water, gas, telephone, garbage); cable TV; painting (interior and exterior); yard upkeep (labor and materials); routine maintenance (furnace, air conditioner, hot water heater, etc.); insurance; taxes. Roof; furnace; air conditioner; plumbing fixtures; garage door opener; driveway and sidewalks; patio; and so on. Acquisition; selling. Sale of home (year 10).
49
33
34
(a)
Relatively easy to estimate. Contact a real estate firm specializing in commercial property.
(b)
Accurate construction cost estimate can be obtained by soliciting bids from local construction contractors engaged in this type of construction.
(c)
Working capital costs can be reasonably estimated by carefully planning and identifying the specific needs, and then contacting equipment distributors and other suppliers; estimating local labor rates and company staffing; establishing essential inventory levels; and so on.
(d)
Total capital investment can be reasonably estimated from a, b, and c plus the cost of other fixed assets such as major equipment, plus the cost of any consulting or engineering services.
(e)
It is not easy to estimate total annual labor and material costs due to the initial uncertainty of sales demand for the product. However, these costs can be controlled since they occur over time and the expenditure can be matched to the demand.
(f)
First year revenues for a new product are difficult to estimate. A marketing consultant could be helpful.
⎛I C 2009 = C 2004 ⎜⎜ 2009 ⎝ I 2004
35 I 2008
36
37
(a)
⎞ ⎟⎟ ⎠
⎛ 293⎞ = $200,000⎜ ⎟ = $262,780.27 ⎝ 223⎠
⎛ 53 ⎞ ⎛ 57 ⎞ ⎛ 62 ⎞ 0.70⎜ ⎟ + 0.05⎜ ⎟ + 0.25⎜ ⎟ ⎝ 33 ⎠ × 100 = 153.5 ⎝ 38 ⎠ ⎝ 41 ⎠ = 0.70 + 0.05 + 0.25
I R = 0.13(131) + 0.20(150) + L + 0.06(140) = 154.9 I C = 0.13(176) + 0.20( 210) + L + 0.06(172) = 203.4
(b)
⎛ 203.4 ⎞ CC = $314,300 ⎜ ⎟ = $412,710 ⎝ 154.9 ⎠
(a)
I 2008 = 0.30(200) + 0.20(175) + 0.50(162) = 176
(b)
I 2004 = 0.30(160) + 0.20(145) + 0.50(135) = 144.5 ⎛ 176 ⎞ C2008 = $650,000 ⎜ ⎟ = $791,696 ⎝ 144.5⎠
38
Let
CA = cost of new boiler, CB = cost of old boiler, today
50
SA = 1.42X SB = X
Chapter 3 Solutions
⎛ 221⎞ CB = $181,000 ⎜ ⎟ = $246,920 ⎝ 162 ⎠ 0.8
. X⎞ ⎛ 142 CA = $246,920 ⎜ ⎟ = $326,879 ⎝ X ⎠ Total cost with options = $326,879 + $28,000 = $354,879 39
Intrabuilding connctions: ($20 to $50 per foot) x (3,000 ft) = $60,000 to $150,000 Cable installation: ($20 per foot) x (3,000 ft) = $60,000 Broadband Taps: ($17 to $20 per tap) x (50 taps) = $850 to $1,000 Broadband NIUs: ($500 to $1,000 per port) x (6 ports) = $3,000 to $6,000 Broadband Modem: ($1,000 per modem) x (2 modems) = $2,000 Network manager/analyzer: ($30,000 per unit) x (1 unit) = $30,000 ____________ Total Cost: $155,850 to $249,000 Difference between high and low estimates is 59.8%. Therefore, the estimate should not be expected to be more accurate than about +50%.
310
CB(8 years ago) = $2,500,000 SA = 1,500,000 lbs/yr X = 0.65 ⎛S ⎞ CA= CB ⎜ A ⎟ ⎝ SB ⎠
311
X
CB(now) = $2,500,000(1.12)8 = $6,189,908 SB = 500,000 lbs/yr
⎛ 1,500,000 ⎞ = $6,189,908 ⎜ ⎟ ⎝ 500,000 ⎠
0.65
= $12,641,919
(a)
Cost = (15,000 miles/year)($0.42/mile) = $6,300 / year
(b)
Several of the cost elements are not simple linear functions of miles driven. For example, while insurance costs are likely to be higher if the vehicle is to be driven 30,000 instead of 15,000 miles per year, it is not likely that the insurance cost will double. Also, taxes, license, and registration fees (as well as finance charges) are independent of the miles driven per year. These elements are fixed in that the same cost is incurred whether the vehicle is driven one mile or 50,000 miles per year.
312
The average compound rate of growth is 2.4% per year. C2008 = C2006[(1+0.24)2] or C2008 = $10.2(1.0486) = $10.7 million
313
(a)
⎛ 194 ⎞ Cnow(80kW) = $160,000 ⎜ ⎟ = $165,989 ⎝ 187 ⎠ ⎛ 120 ⎞ Cnow(120kW) = $165,989 ⎜ ⎟ ⎝ 80 ⎠
0.6
= $211,707
Total Cost = $211,707 + $18,000 = $229,707
Chapter 3 Solutions
51
(b)
⎛ 40 ⎞ Cnow(40kW) = $165,989 ⎜ ⎟ ⎝ 80 ⎠
0.6
= $109,512
Total Cost = $109,512 + $18,000 = $127,512 314 1 mile = 5,280 feet; 25 miles = 132,000 feet; 50 yards = 150 feet A light pole is to be installed every 50 yards or every 150 feet. Thus, 132,000 feet/150 feet = 880 light poles need to be installed. Cost of installing light poles = 880($2,500) = $2,200,000 Total cost = $15,000/mile(25 miles) + $2,200,000 = $2,575,000 315
⎛ 964 ⎞ C2006(250 ft2) = $13,500 ⎜ ⎟ = $15,680 ⎝ 830 ⎠ ⎛ 150 ⎞ C2006(150 ft2) = $15,680 ⎜ ⎟ ⎝ 250 ⎠
316
0.6
= $11,541
K = 126 hours; s = 0.95 (95% learning curve); n = (log 0.95)/(log 2) = 0.074 (a)
Z8 = 126(8)0.074 = 108 hours; Z50 = 126(50)0.074 = 94.3 hours 5
(b)
C5 = T5/5;
T5 = 126
∑u
0.074
= 587.4 hrs; C5 = 587.4 / 5 = 117.5 hrs
u=1
317
K = 400 hours; s = 0.91 (91% learning curve); n = (log 0.91)/(log 2) = 0.136 24
C24 = T24/24;
T24 = 400
∑u
0.136
= 7,210.9 hrs; C5 = 7,210.9 / 24 = 300.4 hrs
u =1
318
K = 1.15X; s = 0.90 (90% learning curve); n = (log 0.90)/(log 2) = 0.152 Z30 = 1.15X(30)0.152 = 0.686X After 30 months, a (1  0.686) = 31.4% reduction in overhead costs is expected (with respect to the present cost of $X).
52
Chapter 3 Solutions
319
(a)
HOME
Framing Site Work/ Foundation 1.1 1.1.1 Site Clearing and Grading 1.1.2 Footing 1.1.3 Floor Slab 1.1.4 Sidewalks 1.1.5 Driveway 1.1.6 Patio 1.1.7 Underground Utilities
1.2
1.2.1 Studs 1.2.2 Windows & Door Frames 1.2.3 Rafters/Joists 1.2.4 Sheathing 1.2.5 Subflooring (House Only)
Exterior Walls 1.3
1.3.1 Siding/Brick Veneer 1.3.2 Insulation 1.3.3 Windows 1.3.4 Doors 1.3.5 Vehicle Door (Garage Only)
Roofing
Interior
1.4
1.5
1.4.1 Shingles/Building Paper 1.4.2 Flashing 1.4.3 Gutters/Downspouts 1.4.4 Attic Insulation
Specialties
Mechanical
Electrical
Landscaping
1.6
1.7
1.8
1.9
1.6.1 Cabinets 1.6.2 Appliances 1.6.3 Miscellaneous
1.7.1 Bathroom Fixtures 1.7.2 Sinks/Water Heater 1.7.3 Heating & A/C 1.7.4 Miscellaneous
1.8.1 Wiring/Breaker Box 1.8.2 Switches/Receptacles 1.8.3 Lighting Fixtures 1.8.4 Intercom System 1.8.5 Miscellaneous
1.9.1 Trees 1.9.2 Shrubbery 1.9.3 Lawn
1.5.1 Drywall 1.5.2 Baseboard/Trim 1.5.3 Painting/Finish 1.5.4 Flooring (House Only) 1.5.5 Doors (House Only)
Architect/ Engineer 1.10
Overhead/ Profits 1.11
Land (lot) 1.12
Chapter 3 Solutions
53
(b)
Capital Investment. The information provided below will be needed by the student to develop a semidetailed estimate of the Capital Investment. 1) General Assumptions • Lot Size: 100 ft by 120 ft (approximately 0.54 acres). • House is a single story (1level) construction; assume that the house plus garage is a rectangle 36 ft wide. • Driveway, patio and sidewalks are concrete construction. The dimensions are: Driveway: 12 ft wide; 80 ft long Patio: 15 ft by 22 ft Sidewalks: 3.25 ft wide; 110 linear ft required • Underground Utilities required are: Sewer: 82 linear ft Gas: 76 linear ft Water: 80 linear ft 2) Cost Factors ITEM COST ($/UNIT) Land $ 22,000 Sidewalks 2.35/ft2 Driveway 2.75/ft2 Patio 2.60/ft2 Site Clearing/Grading 2,460/acre (1) Footing 6.40/acre Floor Slab (including Garage)(2) 2.40/ft2 Sewer 9.35/ft Gas 5.70/ft Water 8.24/ft Framing: House 12.25/ft2 Garage 11.30/ft2 Exterior Walls 10.95/ft2 Vehicle Door (Garage) 670 Roofing 4.60/ft2 Interior: House 11.90/ft2 Garage 5.15/ft2 Specialties (House Only) 3.90/ft2 Mechanical: House 6.90/ft2 Garage 1.15/ft2 Electrical: House 1.68/ft2 Garage 0.54/ft2 1 $/ft means cost per linear foot 2 $/ft2 means cost per square foot of building floor space (except for landscaping, sidewalks, driveway, and patio)
54
Chapter 3 Solutions
2)
Cost Factors (continued) ITEM COST ($/UNIT) Landscaping (excluding building area) 0.30/ft2 (3) Architect/Engineer 8% Overhead/Profit(3) 16% Working Capital(4) 1% 3 Percentage does not apply to costs of land or landscaping. 4 One percent of total construction costs.
1.12 1.1.1 1.1.2 1.1.3 1.1.4 1.1.5 1.1.6 1.1.7 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11
320
(a)
ESTIMATED COST AREA/ELEMENTS CAPITAL INVESTMENT Land (lot) $22,000 Construction Costs: Site Clearing/Grading $ 1,328 Footing 1,551 Floor Slab 7,356 Sidewalks 840 Driveway 2,640 Patio 858 Underground Utilities 1,859 Framing 35,098 Exterior Walls 34,232 (Including Garage Door) Roofing 14,099 Interior 31,473 Specialties 9,555 Mechanical 17,423 Electrical 4,359 Landscaping 2,681 Architect/Engineer 13,253 Overhead/Profit (Contractor) 26,027 Subtotal 204,632 Working Capital 2,046 TOTAL $228,678
Find I B1 and I B2 , the weighted cost indexes. I B1 (for equipment and installation costs): I B1,C Weighted Current Year Index = Weighted Reference Year Index I B1,R I B1,C = 0.41(201) + 0.22(212) + 0.09(200) + 0.28(184) = 198.6 I B1 =
I B1,R = 0.41(122) + 0.22(131) + 0.09(118) + 0.28(135) = 127.3
I B1 =
Chapter 3 Solutions
198.6 = 1.56 or 156% 127.3 55
I B2 (for other project and support costs): I B2 =
I B2,C Weighted Current Year Index = Weighted Reference Year Index I B2,R
I B2,C = 0.38(206) + 0.31(194) + 0.11(162) + 0.20(179) = 192.0 I B2,R = 0.38(136) + 0.31(128) + 0.11(105) + 0.20(113) = 125.5
I B2 = (b)
192.0 = 1.53 or 153% 1255 .
Find: CA = estimated cost of new automated warehousing system. ⎛S ⎞ C A = C B1 ⎜ A ⎟ ⎝ SB ⎠
X
( I B1 ) + C B2 ( I B2 )
⎛ 11,000⎞ = $1,226,000⎜ ⎟ ⎝ 5,800 ⎠
0.7
. ) + $234,000(153 . ) (156
= $3,351,600 321
The following table facilitates the intermediate calculations needed to compute the values of b0 and b1 using Equations 38 and 39. i 1 2 3 4 5 6 7 8 9 10 Totals
b1 =
xi 14,500 15,000 17,000 18,500 20,400 21,000 25,000 26,750 28,000 30,000 216,150
xi2 210,250,000 225,000,000 289,000,000 342,250,000 416,160,000 441,000,000 625,000,000 715,562,500 784,000,000 900,000,000 4,948,222,500
yi 800,000 825,000 875,000 972,000 1,074,000 1,250,000 1,307,000 1,534,000 1,475,500 1,525,000 11,637,500
x iy i 11,600,000,000 12,375,000,000 14,875,000,000 17,982,000,000 21,909,600,000 26,250,000,000 32,675,000,000 41,034,500,000 41,314,000,000 45,750,000,000 265,765,100,000
(10)(265,765,100,000) − (216,150)(11,637,500) = 51.5 (10)(4,948,222,500) − (216,150) 2
56
Chapter 3 Solutions
b0 =
11,637,500 − (51.5)(216,150) = 50,631 10
(a) The resulting CER relating supermarket building cost to building area (x) is:
Cost = 50,631 + 51.5x So the estimated cost for the 23,000 ft2 store is: Cost = $50,631 + ($51.5/ft2)(23,000 ft2) = $1,235,131 (b) The CER developed in part (a) relates the cost of building a supermarket to its planned area using the following equation:
Cost = 50,631 + 51.5x Using this equation, we can predict the cost of the ten buildings given their areas. i
xi
yi
( x i − x )( y i − y )
(yi  Costi)2
Costi
( xi − x ) 2
( yi − y ) 2
1
14,500
800,000
797,345
7,048,179
2,588,081,250
50,623,225
132,314,062,500
2
15,000
825,000
823,094
3,633,147
2,240,831,250
43,758,225
114,751,562,500
3
17,000
875,000
926,089
2,610,081,256
1,332,581,250
21,298,225
83,376,562,500
4
18,500
972,000
1,003,335
981,896,725
597,301,250
9,703,225
36,768,062,500
5
20,400
1,074,000
1,101,181
738,780,429
109,046,250
1,476,225
8,055,062,500
6
21,000
1,250,000
1,132,079
13,905,356,010
53,043,750
378,225
7,439,062,500
7
25,000
1,307,000
1,338,069
965,288,881
484,901,250
11,458,225
20,520,562,500
8
26,750
1,534,000
1,428,190
11,195,807,942
1,901,233,750
26,368,225
137,085,062,500
9
28,000
1,475,500
1,492,562
291,099,988
1,990,523,750
40,768,225
97,188,062,500
10
30,000
1,525,000
1,595,557
4,978,246,304
3,029,081,250
70,308,225
130,501,562,500
Totals
216,150
11,637,500 11,637,500
35,677,238,861
14,220,537,500 276,140,250
767,999,625,000
x=
1 (216,150) = 21,615 10
y=
1 (11,637,500) = 1,163,750 10
Using Equations 310 and 311, we can compute the standard error and correlation coefficient for the CER. SE =
R=
322
35,677,238,861 = 59,730 10 14,220,537,500 ( 276,140,250)( 767,999,625,000)
xi = weight of order (lbs) yi = packaging and processing costs ($)
Chapter 3 Solutions
57
= 0.9765
(a) y = b0 + b1x
∑x
i
∑y
= 2530 x = 253
y = 102.4
= 1024
i
∑x y i
b1 =
i
∑x
2 i
∑y
= 658,900
2 i
= 106,348
= 264,320
264,320 − (253)(1024) = 0.279 658,900 − (253)(2530)
b0 = 102.4 – (0.279)(253) = 31.813; (b) R =
y = 31.813 + 0.279x
S xy S xx S yy
Sxy = 264,320 – (2530)(1024)/10 = 5,248 Sxx = 658,900 – (2530)2/10 = 18,810 Syy = 106,348 – (1024)2/10 = 1,490.4 R=
5248 (18,810)(1490.4)
= 0.99
(c) y = 31.813 + (0.279)(250) = $101.56 323
Using the spreadsheet template of Figure 38, the total manufacturing cost and unit selling price are estimated to be $221.43 and $248.00, respectively.
58
Chapter 3 Solutions
MANUFACTURING COST ELEMENTS A: B: C: D: E: F: G: H: I: J: K: L: M: N: O: P: Q: R:
*
324
Factory Labor Planning Labor Quality Control TOTAL LABOR Factory Overhead General & Admin. Expense Production Material Outside Manufacture SUBTOTAL Packing Costs TOTAL DIRECT CHARGE Other Direct Charge Facility Rental TOTAL MANUFACTURING COST Quantity (lot size) MANUFACTURING COST / UNIT Profit/Fee UNIT SELLING PRICE
4.20
$
Factor Estimate Factor of Row
Direct Estimate
11.15
Row Total $
150%
A $ 26.20 74.87
7%
A
12%
P
46.83
46.83 70.25 26.20 74.87 218.15 3.28 221.43 221.43 1.00 221.43 26.57 $ 248.00
1 unit = 1 lot of 100 wire cutters
The estimate of direct labor hours is based on the time to produce the 50th unit. K = 1.76 hours s = 0.8 (80% learning curve) n = (log 0.80)/(log 2) = 0.322 Z50 = 1.76(50)0.322 = 0.5 hours Factory Labor Production Material Factory Overhead Packing Costs Total Manufacturing Cost Desired Profit Unit Selling Price
325
Unit Estimate Units Cost/Unit
= ($15/hr)(0.5 hr/widget) = $375 / 100 widgets = (1.25)($7.50 / widget) = (0.75)($7.50 / widget) = (0.20)($26.25 / widget)
= $7.50 / widget = $3.75 / widget = $9.375 / widget = $5.625 / widget = $26.25 / widget = $5.25 / widget = $31.50 / widget
K = 5.26 hours s = 0.85 (85% learning curve) n = (log 0.85)/(log 2) = 0.2345 Z20 = (5.26)(20)0.2345 = 2.61 hours Direct labor = ($15/hour)(2.61 hours/unit) = $39.15/unit Factory overhead = (1.2)($39.15/unit) = $46.98/unit Production material = $300/unit Packing costs = (0.2)($39.15/unit) = $ 7.83/unit _____________ Total manufacturing cost = $393.96/unit (a) Maximum profit that company can have to remain competitive = $420 $393.96 = $26.04/unit. Profit margin = $26.04/$393.96 = 0.0661 = 6.61%
Chapter 3 Solutions
59
(b) At 15% profit margin, the target cost is given by (Eq. 313): TC = $420/(1.15) = $365.22/unit Hence, the target cost cannot be achieved. Two possible ways to achieve target cost: (1) Reduce profit margin to less than or equal to 6.61% (2) Try to obtain production material at a lower cost. Since the TMC exceeds TC by ($393.96  $365.22) = $28.74/unit, therefore, the material cost should be reduced by at least $28.74/unit. 326
Profit = Revenue – Cost $25,000 = ($20.00/unit)(x) – [($21.00/unit)(.2 hours/unit)(x) + ($4.00/unit)(x) + (1.2)($21.00/unit)(.2 hours/unit)(x) + ($1.20/unit)(x)] $25,000 = 5.56x;
327
x = 4,497 units
Before examining potential cost reductions, the current manufacturing cost must be estimated. Total Direct Labor Time = Assembly Time + Handling Time A 95% learning curve applies to the assembly time. K = 1 hr,
s = 0.95,
n = log 0.95/log 2 = 0.074
Time to assemble all 20,000 units = (1)
20, 000
∑ (u )
−0.074
u =1
Using a computer program, T20,000 = 10,378.26 hours. Adding in handling time as 10% of assembly time: Total Factory (direct) Labor = (10,378.26)(1.1) = 11,416.09 hours Using the spreadsheet template of Figure 38 (below), the total manufacturing cost and selling price are $2,284.94 and $2,741.92, respectively.
60
Chapter 3 Solutions
M ANUFACTURING CO ST ELEM ENTS A: B: C: D: E: F: G: H: I: J: K: L: M: N: O: P: Q: R:
Units
Unit Estim ate Cost/Unit
Factory Labor 11416.09 $ Planning Labor Q uality Control TO TAL LABO R Factory O v erhead G eneral & Adm in. Expense Production M aterial 20000.00 $ O utside M anufacture 20000.00 $ SUBTO TAL Packing Costs TO TAL DIRECT CHARG E O ther Direct Charge Facility Rental TO TAL M ANUFACTURING CO ST Q uantity (lot size) M ANUFACTURING CO ST / UNIT Profit/Fee UNIT SELLING PRICE
Factor Estim ate Factor of Row
Direct Estim ate
15.00
Row Total $
10% 50%
A A
200% 300%
D D
10%
D
10%
D
20%
P
171,241.35 17,124.14 85,620.68 273,986.16 547,972.32 821,958.48 4,000,000.00 40,000,000.00 45,643,916.96 27,398.62 45,671,315.58 27,398.62 45,698,714.19 20,000.00 2,284.94 456.99 $ 2,741.92
200.00 2,000.00
Given the competitor's selling price of $2,500 and assuming a desired return on sales (ROS) of 20%, the target cost is given by: Target Cost = $2,500/1.20 = $2,083 It will be difficult for the company to assemble the computer for less than $2,083. The estimated outside manufacture cost alone is $2,000 per computer. This cost must be reduced if the target cost of $2,083 is to be achieved. As shown in the following spreadsheet, the target cost of $2,083 can be reached if the outside manufacture cost can be reduced to $1,715 (a 14.25 % reduction). MANUFACTURING COST ELEMENTS A: B: C: D: E: F: G: H: I: J: K: L: M: N: O: P: Q: R:
Units
Unit Estim ate Cost/Unit
Factory Labor 11416.09 $ Planning Labor Quality Control TOTAL LABOR Factory Ov erhead General & Adm in. Expense Production Material 20000.00 $ Outside Manufacture 20000.00 $ SUBTOTAL Packing Costs TOTAL DIRECT CHARGE Other Direct Charge Facility Rental TOTAL MANUFACTURING COST Quantity (lot size) MANUFACTURING COST / UNIT Profit/Fee UNIT SELLING PRICE
Factor Estim ate Factor of Row
15.00
Direct Estim ate
Row Total $
10% 50%
A A
200% 300%
D D
10%
D
10%
D
20%
P
200.00 1,715.06
171,241.35 17,124.14 85,620.68 273,986.16 547,972.32 821,958.48 4,000,000.00 34,301,285.81 39,945,202.77 27,398.62 39,972,601.38 27,398.62 40,000,000.00 20,000.00 2,000.00 400.00 $ 2,400.00
328 There is not a unique answer to this problem. In Problem 319, the information provided permitted a specific answer to be calculated. In this problem it is suggested that the class be organized into groups for solution (preferably after Problem 319 has been individually worked and the “school solution” discussed). In solving this some important points are:
Chapter 3 Solutions
61
• • • 329
The capital investment requirements at time 0 were calculated in Problem 319. Annual recurring costs should be calculated based on the local utility, repair, and maintenance costs found to be representative. Cost for yard maintenance should be included. At the end of Year 10, net sale value should be estimated.
mW 0.8 Boiler Cost = $300,000 (106 mW ) = $451,440 9 mW 0.6 Generator Cost = $400,000 ( 6 mW ) = $510,170
Tank Cost = $106,000
(
91, 500 gal 80 , 000 gal
) 0.66 = $115,826
Total Cost = (2)($451,440) + (2)($510,170) + $115,826 + $200,000 = $2,239,046 330
Y3 = 846.2 = Y1(3)n Y5 = 783.0 = Y1(5)n 846.2 = (3/5)n or 1.0807 = (0.6)n 783.0 log 1.0807 = n log (0.6) 0.03371 = n, thus n = 0.1520 − 0.22185 and n = log s/log 2, where s = learning curve as decimal 0.152 = log s/0.30103 0.04576 = log s, s = 0.9 (90% learning curve)
331
The following spreadsheet was used to calculate a 2005 estimate of $320,274,240 for the plant.
62
Chapter 3 Solutions
Element Code
Units/Factors
1.1.12
Price/Unit
600
Subtotal
$2,000
1.1.3
$
1,200,000
$
3,000,000
1.1
Totals
$
4,200,000
1.23
2.3969
a
$ 110,000,000
$ 263,659,000
$ 263,659,000
1.4
4.4727
b
$
5,000,000
$
$
22,363,500
80,390
c
$
60
$
4,823,400
$
15,000,000
$
900,000 $
20,723,400
$
9,328,340
22,363,500
1.5.13 labor materials 1.5.4
600
$
1,500
1.5 1.9
3%
$ 310,944,672
$ 320,274,240
TOTAL ESTIMATED COST IN 2005
a
Factor value for boiler and support system (WBS elements 1.2 and 1.3): ⎛ 492 ⎞ ⎛ 1 ⎞ ⎜ ⎟⎜ ⎟ ⎝ 110 ⎠ ⎝ 2 ⎠
b
0.9
= 2.3969
Factor value for the coal storage facility (WBS element 1.4):
⎛ 492 ⎞ ⎜ ⎟ = 4.4727 ⎝ 110 ⎠ c
Labor time estimate for the 3rd facility (WBS elements 1.5.1, 1.5.2, and 1.5.3): K = 95,000 hours, s = 0.9, n = log(0.9)/log(2) = 0.152 Z3 = 95,000(3)0.152 = 80,390 hours
Chapter 3 Solutions
63
Solutions to Spreadsheet Exercises 332 A 1 K= 2 s= 3 Labor Rate/hr = $ 4 5 n= Unit #
6 7 1 8 2 9 3 10 4 11 5 12 6 7 13 8 14 9 15 16 10 17 11 12 18 13 19 14 20 21 15 22 16 17 23 18 24 19 25 20 26 27 21 22 28 23 29 24 30 25 31 32 26 33 27 28 34 29 35 30 36 37 31 38 32 33 39 34 40 35 41 42 36 43 37 38 44 39 45 40 46 47 41 48 42 43 49 44 50 45 51 52 46 53 47 54 48 55 49 56 50 57 58 Total Labor Cost
B
C
D
E
100 90.0% 15.00 0.152
Unit Time 100.00 90.00 84.62 81.00 78.30 76.16 74.39 72.90 71.61 70.47 69.46 68.54 67.71 66.96 66.26 65.61 65.01 64.45 63.92 63.42 62.95 62.51 62.09 61.69 61.31 60.94 60.59 60.26 59.94 59.63 59.33 59.05 58.77 58.51 58.25 58.00 57.76 57.53 57.30 57.08 56.87 56.66 56.46 56.26 56.07 55.88 55.70 55.52 55.35 55.18
Cumulative Cumulative Total Average 100.00 100.00 190.00 95.00 274.62 91.54 355.62 88.91 433.92 86.78 510.08 85.01 584.47 83.50 657.37 82.17 728.98 81.00 799.45 79.94 868.90 78.99 937.45 78.12 1005.16 77.32 1072.11 76.58 1138.37 75.89 1203.98 75.25 1268.99 74.65 1333.44 74.08 1397.35 73.54 1460.78 73.04 1523.73 72.56 1586.24 72.10 1648.33 71.67 1710.02 71.25 1771.32 70.85 1832.27 70.47 1892.86 70.11 1953.12 69.75 2013.06 69.42 2072.69 69.09 2132.02 68.77 2191.07 68.47 2249.85 68.18 2308.35 67.89 2366.60 67.62 2424.61 67.35 2482.37 67.09 2539.89 66.84 2597.19 66.59 2654.27 66.36 2711.14 66.13 2767.80 65.90 2824.25 65.68 2880.51 65.47 2936.58 65.26 2992.46 65.05 3048.15 64.85 3103.67 64.66 3159.02 64.47 3214.20 64.28 $ 48,212.93
64
Chapter 3 Solutions
333 A Spacecraft
B Weight 0 1 2 3 4 5 6
C Cost
100 400 530 750 900 1,130 1,200
$ $ $ $ $ $ $
D
E
F
G
H
I
J
K
600 278 414 557 689 740 851
SUMMARY OUTPUT Regression Statistics Multiple R 0.705850276 R Square 0.498224612 Adjusted R Square 0.397869535 Standard Error 151.9036833 Observations 7 ANOVA df Regression Residual Total
1 5 6
SS MS F Significance F 114557.2121 114557.212 4.9646179 0.076342997 115373.645 23074.729 229930.8571
Coefficients Standard Error t Stat Pvalue 341.9170907 125.2152843 2.73063383 0.0412491 0.346423227 0.155476261 2.22814225 0.076343
Intercept Weight
Lower 95% Upper 95% Lower 95.0% Upper 95.0% 20.04148132 663.7927001 20.04148132 663.7927001 0.053240574 0.746087027 0.053240574 0.746087027
Observed vs. Predicted Values
RESIDUAL OUTPUT Observation 1 2 3 4 5 6 7
Predicted Cost 376.5594133 480.4863813 525.5214008 601.7345106 653.6979946 733.3753367 757.6249626
Chapter 3 Solutions
Residuals 223.4405867 202.4863813 111.5214008 44.73451063 35.30200539 6.624663274 93.37503741
Cost
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56
$900 $800 $700 $600 $500 $400 $300 $200 $100 $
10
0
20
0
30
0
40
0
50
0
60
0
70
0
80
Weight
65
0
90
0
1 ,0
00 1 0 0 2 00 3 00 1, 1, 1,
334 Column A MANUFACTURING COST ELEMENTS A: Factory Labor B: Planning Labor C: Quality Control D: TOTAL LABOR E: Factory Overhead F: General & Admin. Expense G: Production Material H: Outside Manufacture I: SUBTOTAL J: Packing Costs K: TOTAL DIRECT CHARGE L: Other Direct Charge M: Facility Rental N: TOTAL MANUFACTURING COST O: Quantity (lot size) P: MANUFACTURING COST / UNIT Competitor's Selling Price Desired Return on Sales Target Cost
Column B Unit Estimate Units Cost/Unit 36.48
Column C Column D Column E Factor Estimate Direct Row Factor of Row Estimate Total
$ 10.54 12% 11%
A A
105% 15%
D D
5%
I
1%
K
$ 384.50 46.14 42.29 472.93 496.58 70.94 $ 110.23 110.23 28.00 28.00 1,178.69 58.93 1,237.62 12.38 1,250.00 50 25.00
$ 27.50 10% $ 25.00
Solutions to Case Study Exercises 335 Other cost factors include maintenance, packaging, supervision, materials, among others. Also, the case solution presents a beforetax economic analysis. 336 Left as an exercise for the student. However, by observation, it appears that the factory overhead and factory labor are good candidates since they comprise the largest percentage contributions to the per unit demanufacuring cost. 337 A 50% increase in labor costs equates to a factor of 15%; a 90% increase in Transportation equates to a factor of 38%. The corresponding demnaufacturing cost per unit is $5.19. The per unit cost of using the outside contractor (i.e., the target cost) is $11.70. Should the proposed demanufacturing method be adopted, the revised per unit cost savings is $6.51 for a 55.6% reduction over the per unit cost for the outside contractor.
66
Chapter 3 Solutions
DEMANUFACTURING COST ELEMENTS
A: B: C: D: E: F: G: H: I: J:
Unit Elements Factor Estimates Units Cost/Unit Factor of Row
Factory Labor 24.5 hrs $ 12.00/hr Quality Costs  Training 15% TOTAL LABOR Factory Overhead  Set up Costs 150% Transportation Cost 38% TOTAL DIRECT CHARGE Facitily Rental TOTAL DEMANUFACTURING COST Quantity  Lot Size Demanufacturing Cost/Unit Outside Cost/Unit  Target Cost
A C C
Row Total $ $ $ $ $ $
294.00 44.10 338.10 507.15 192.72 699.87 $ 1,037.97 200 $ 5.19
$ 11.70
Solutions to FE Practice Problems 338
K = 460 hours; s = 0.92 (91% learning curve); n = (log 0.92)/(log 2) = 0.120 30
C30 = T30/30; T30 = 460
∑u
0.120
= 10,419.63 hrs;
u =1
C30 = 10,419.63 / 30 = 347.3211 Select (d) 339
1,500 + 800 + (.07.05)(1.85)(10)x = 0 700 + 0.37x = 0 x = 700/0.37 = 1,892 miles/year Select (c)
340
ACcurrent = $4,000 Proposed: N = 13 years, SV = 11% of first cost $4,000 = I (A/P,12%,13) – (0.11)I (A/F,12%,13) $4,000 = I(0.1557) – (0.003927)I $4,000 = I (0.1517) Select (c) I = $26,358
341
Let X = average time spent supervising the average employee. Then the time spent supervising employee A = 2X and the time spent supervising employee B = 0.5X. The total time units spent by the supervisor is then 2X + 0.5X + (8)X = 10.5X. The monthly cost of the supervisor is $3,800 and can be allocated among the employees in the following manner: $3,800/10.5X = $361.90 / X time units. Employee A (when compared to employee B) costs (2X – 0.5X)($361.90/X) = $542.85 more for the same units of production. If employee B is compensated accordingly, the monthly salary for employee B should be $3,000 + $542.85 = $3,542.85. Select (a)
342
Type X filter: cost = $5, changed every 7,000 miles along with 5 quarts oil;
Chapter 3 Solutions
67
between each oil change, 1 quart of oil must be added after each 1,000 miles Type Y filter: cost = ?, changed every 5,000 miles along with 5 quarts of oil no additional oil between filter changes oil = $1.08 / quart Common multiple = 35,000 miles For filter X = 5 oil changes; 5($5 + 5($1.08) + 6($1.08)) = (5)$16.88 = $84.40 For filter Y = 7 oil changes; 7CY + 7(5)($1.08) = 7X + $37.8 $84.40 = 7CY + $37.8 Select (d) $46.60 = 7CY; CY = $6.66
68
Chapter 3 Solutions
Solutions to Chapter 4 Problems 41
I = P(N)(i) = $10,000 (4.25 yrs.) (0.10/yr) = $4,250 F = P + I = $18,060
42 i = 12% / yr (simple interest) 0 1
2
3
4
5
6
End of Year P = $10,500 F = $10,500 + $10,500 (0.12) (6) = $10,500 + $7,560 = $18,060 I = $7,560 43
I = $1,000 (2.5 yrs.) (0.08) = $200; F = P + I = $1,000 + $200 = $1,200 Select c.
44 Year 1 2 3 4 5 6 7 8
Amount Owed at Beginning of Year $2,000 2,000 2,000 2,000 1,000 1,000 1,000 1,000 Total Interest =
Interest Accrued for Year $ 200 200 200 200 100 100 100 100 $ 1,200
Total Amount Owed at End of Year $2,200 2,200 2,200 2,200 1,100 1,100 1,100 1,100
Principal Payment $ 0 0 0 1,000 0 0 0 1,000
Total End of Year Payment $ 200 200 200 1,200 100 100 100 1,100
$200 in interest is payable each year for the first four years. $100 in interest is payable each year for the second four years. The total interest paid over the eight year period is $1,200.
71
Chapter 4 Solutions
45 Year 1 2 3 4 5 6 7 8
Amount Owed Interest at Beginning Accrued of Year for Year $2,000.00 $ 200.00 2,200.00 220.00 2,420.00 242.00 2,662.00 266.20 1,928.20 192.82 2,121.02 212.10 2,333.12 233.31 2,566.43 256.64 Total Interest = $1,823.07
Total Amount Owed at End of Year $2,200.00 2,420.00 2,662.00 2,928.20 2,121.02 2,333.12 2,566.43 2,823.07
Principal Payment $ 0 0 0 1,000 0 0 0 1,000
Total End of Year Payment $ 0.00 0.00 0.00 1,000.00 0.00 0.00 0.00 2,823.07
The total interest paid in this problem is $1,823.07. The difference between the total interest paid in this problem and the interest paid in Problem 44 ($1,823.07  $1,200 = $623.07) is due to the compounding effect. That is, in this problem, you are paying interest on the upaid interest.
46
(a)
Amount Owed at Month BOM 1 $17,000 2 17,000 3 8,500 4 8,500 Total Interest =
Interest Accrued for Month $ 170 170 85 85 $510
Total Amount Owed at End of Month $17,170 17,170 8,585 8,585
Principal Payment $ 0 8,500 0 8,500
Total EOM Payment $ 170 8,670 85 8,585
Total interest paid over the four month loan period = $510. (b)
A = $17,000 (A/P, 1.5%, 4) = $17,000 (0.259445) = $4,410.57 Amount Owed at Month BOM 1 $17,000.00 2 12,844.43 3 8,626.53 4 4,345.36 Total Interest =
Interest Accrued for Month $ 255.00 192.67 129.40 65.18 $ 642.25
Total Amount Owed at End of Month $17,255.00 13,037.10 8,755.93 4,410.54
From the above table: EOM 3 Principal Payment = $4,281.17 Total Interest Paid = $642.25
Chapter 4 Solutions
72
Principal Payment $ 4,155.57 4,217.90 4,281.17 4,345.39
Total EOM Payment $ 4,410.57 4,410.57 4,410.57 4,410.57
47
(a)
Interest Paid = (0.06) (Beginning of Year Principal) For Year 2:
$411.54 = (0.06)(BOY2) BOY2 = $6,859
BOYk = BOYk1  Principal Repayment at EOYk1 = BOYk1  Pk1 For Year 1:
$6,859 = $10,000  P1 P1 = $3,141
For Year 3:
BOY3 = BOY2  P2 = $6,859  $3,329.46 = $3,529.54 I3 = BOY3 (0.06) = $3,529.54 (0.06) = $211.77 Σ Principal Repayment = $10,000 = P1 + P2 + P3 P3 = $10,000  $3,141  $3,329.46 = $3,529.54
(b)
The amount to be repaid at the beginning of year 3 equals the remaining principal which equals the principal repayment at the end of year 3 (since the entire loan must be paid off at the end of year 3). Thus, the amount left to be repaid at the beginning of year 3 = $3,529.54.
(c)
The amount is less because part of the principal is repaid each year.
48
F = $10,000 (F/P, 6%, 5) = $10,000 (1.3382) = $13,382
49
P = $50,000 (P/F, 12%, 4) = $50,000 (0.6355) = $31,775
410
$50 = $25 (F/P, 10%, N) 2 = (1.10)N log (2) = N log (1.1) log(2) N= log(1.1) N = 7.27 years
411
$240,000 = $189,500 (1+i)11 1.26649 = (1+i)11 11 1.26649  1 = i, or i = 2.17% per year
412
(a) Years of growth = 2005 – 1709 = 296 104 = (1) (F/P, I, 296) 104 = (1+i)296
73
Chapter 4 Solutions
296
104  1 = i, or i = 1.58% per year $187 (U .S ) $1.80 (b) = 104 BritishPou nds BritishPou nd
413
F = $5,000 (F/A, 8%,5) = $5,000 (5.8666) = $29,333
414
A = $150,000 (A/F, 9%, 20) = $150,000 (0.0195) = $2,925
415
A = $5,000 (A/F, 8%, 15) = $5,000 (0.0368) = $184.00 F = $5,000 i = 8% / yr 0
1
2
3
14
15
A = $184 416
A = $25,000 (A/P, 10%, 8) = $25,000 (0.1874) = $4,685
417
P = $2,500 (P/A, 5%, 4) = $2,500) (3.5460) = $8,865
418
(a) Cash flow in year k = ($14/ton)(10,000 tons)(1.05)k1 $187,613 $178,679 i = 10% / yr
$170,171 $162,068
$154,350 $140,000
0
1
$147,000
2
3
4
End of Year
Chapter 4 Solutions
74
5
6
7
(b)
Solution based on Section 4.8: F7 = $140,000(F/P,15%,6) + $147,000(F/P,15%,5) + $154,350(F/P,15%,4) + $162,068(F/P,15%,3) + $170,171(F/P,15%,2) + $178,679(F/P,15%,1) + $187,613 = $140,000(2.3131) + $147,000(2.0114) + $154,350(1.7490) + $162,068(1.5209) + $170,171(1.3225) + $178,679(1.15) + $187,613 = $1,754,102.16 Solution based on Section 4.14: $140,000[1 − ( P / F ,15%,7)( F / P,5%,7)] (F/P,15%,7) 0.15 − 0.05 $140,000[1 − (0.3759)(1.4071)] = (2.6600) 0.10 = $1,754,268.81
F7 = P0 (F/P,15%,7) =
The difference in F7 amounts is due to rounding the interest factor values. 419
Mrs. Green’s Point of View: F = $7,000 End of Month 0
1
2
3
4
34
35
36
A = $415.90 / month P = (0.30)($20,000) = $6,000 420
A = $25,000 (A/P,6%,10) = $25,000 (0.1359) = $3,397.50
75
Chapter 4 Solutions
A = $22,000
421
0
1
2
3
4
5
EOY
i = 15% / yr
P=?
P = $22,000 (P/A, 15%, 5) = $22,000 (3.3522) = $73,748.40 The company can justify spending up to $73,748.40 for this piece of equipment. 422
P ≤ $350 (P/A, 8%, 18) ≤ 350 (9.3719) ≤ $3,280.16 Thus, the homeowner can afford to spend no more than $3,280.16 for the thermal windows.
423
The annual savings required to justify the $14,000 investment are: A = $14,000 (A/P, 10%, 4) = $14,000 (0.3155) = $4,417
424
F4 = ? A = $10,000
0
1
2
3
4
EOY
i = 15% / yr
$100,000 Equivalent receipts = Equivalent expenditures F4 + $10,000 (F/A,15%, 4) = $100,000 (F/P, 15%, 4) so, F4 = $100,000 (F/P, 15%, 4)  $10,000 (F/A,15%, 4) = $100,000 (1.7490)  $10,000 (4.9934) = $174,900  $49,934 = $124,966 425
(a)
P = A ( P/A, i%, N); $1,000 = $200(P/A, 12%,N); (P/A, 12%, N) = 5.0000
Chapter 4 Solutions
76
By looking at the 12% interest table in Appendix C under the P/A column, N ≈ 8 years. (b)
P = A ( P/A, i%, N); $1,000 = $200 (P/A, i%,10); (P/A, i%, 10) = 5.0000 (P/A, 15%, 10) = 5.0188 and (P/A, 18%, 10) = 4.4941, thus 15% < i < 18% 18%  15% i%  15% By using linear interpolation = 5.0188  4.4941 5.0188  5.00 ∴ i = 15.11%
426
(c)
P = A (P/A, i%, N) = $200 (P/A, 12%, 5) = 200 (3.6048) = $720.96
(d)
A = P (A/P, i%, N) = $1,000 (A/P, 12%, 5) = $1,000 (0.2774) = $277.40
72 = 7.2 years. 10 Compare to the exact solution:
Solution by Rule of 72: N ≈
$10,000 = $5,000 (F/P, 10%, N) = $5,000 (1 + 0.10)N 2 = 1.1 N ; ln(2) = N ln(1.1); and N = 7.2725 years 427
F6 = $2,000 (F/A, 5%, 6) = $2,000 (6.8019) = $13,603.80 and F10 = F6 (F/P, 5%, 4) = $13,603.80 (1.2155) = $16,535.42
428 F12 = $24,000 (F/P, ½%, 12) = $24,000 (1.0617) = $25,480.80 and A = F12 (A/P, ½%, 36) = $25,480.80 (0.0304) = $774.62 per month 429
F4 = $10,000 (F/P, 15%, 4)  $4,000 = $10,000 (1.7490)  4,000 = $13,490 Select d.
77
Chapter 4 Solutions
$309 / yr
430 i = 8% / yr
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19
EOY
A=?
P12 (of deposits) = A (F/A, 8%,12), and P13 = P12 (F/P, 8%, 1) P13' (of withdrawals) = $309 (P/A, 8%, 5) By letting P13 = P13', we have [A (F/A, 8%,12)](F/P, 8%, 1) = $309 (P/A, 8%, 5) [A(18.9771)] (1.08) = $309 (3.9927) A = $60.20 431
The value of N (years) at which you become a millionaire is found as follows: $1,000,000 = $10,000 (F/P, 10%, N); 100 = (F/P, 10%, N); 100 = (1.1)N By logarithms, or trial and error, N = 48.317 years = 49 years (to nearest whole year).
432
(a)
$1,000 = $263.80 (P/A,10%, N) (P/A,10%, N) = $1,000/$263.80 = 3.7905 From Appendix C, (P/A,10%,5) = 3.7908 Thus N ≈ 5 years
(b) 433
P2 = $263.80 (P/A, 10%, 3) = $263.80 (2.4869) = $656.04
P0 = $1,000 (P/A, 10%, 5)  $10,000 (P/F, 10%, 15)  $10,000 (P/F, 10%, 30) = $1,000 (3.7908)  $10,000 (0.2394)  $10,000 (0.0573) = $6,757.80 A = $6,757.80 (A/P, 10%, 50) = $6,757.80 (0.1009) = $681.86
Chapter 4 Solutions
78
434
P0 = $2,000  $200 (P/F, 8%, 1)  $800 (P/F, 8%, 2) + $1,000 (P/A, 8%, 10)(P/F, 8%, 3) =  $2,000  $200 (0.9259)  $800 (0.8573) + $1,000 (6.7101) (0.7938) = $2,455.46 (lumpsum receipt now)
435
P = $3,000 (P/A, 20%, 10) + $7,000 (P/A, 20%, 5) = $3,000 (4.1925) + $7,000 (2.9906) = $33,511.70 You can afford to pay as much as $33,511.70 for the equipment.
436
$200,000 = $14,480 (F/A, i%, 10) (F/A, i%, 10) = 13.8122 From Appendix C, i ≈ 7% per year
437
This is a deferred annuity, the time periods are months, and i = 3/4 % per month: P71 = $500 (P/A, 3/4%, 60) = $500 (48.1733) = $24,086.65 P0 = $24,086.65 (P/F, 3/4% ,71) = 24,086.65 (0.58836) = $14,171.62
438
Using time = 0 as the reference point, set P0(LHS) = P0(RHS): $2,000(P/F,8%,2) + $5,000(P/F,8%,6)
= Z(P/F,8%,4)  2Z(P/F,8%,5) + 3Z(P/F,8%,6) $2,000(0.8573) + $5,000(0.6302) = Z(0.7350)  2Z(0.6806) + 3Z(0.6302) $4,865.6 = 1.2644 Z Z = $3,848.15
439
Original Payments = A = $100,000 (A/P, 8%, 30) = $100,000 (0.0888) = $8,880 Balance at EOY 8 = $100,000 (F/P, 8%, 8)  $8,880 (F/A, 8%, 8) = $100,000 (1.8509)  $8,880 (10.6366) = $90,636.99 or P8 = $8,880 (P/F, 8%, 22) = 8,880 (10.2007) = $90,636.99 New Payment = $8,880(3) = $26,640 Balance at EOY 12 = $90,636.99 (F/P, 8% , 4)  $26,640 (F/A, 8% , 4) = $90,636.99 (1.3605)  $26,640 (4.5061) = $3,269.12
79
Chapter 4 Solutions
440
Original Payments = A1 = $1,000 (A/P, 10%, 10) = $1,000 (0.1627) = $162.70 Balance at EOY 3 = $162.70 (P/A, 10%, 7) + $500 = $162.70 (4.8684) + $500 = $1,292.09 New Payments = A2 = $1,292.09 (A/P, 10%, 12) = $1,292.09 (0.1468) = $189.68
441
A1 = [$10,000(F/P,9%,4)  $5,000](A/F,9%,4) = [$10,000(1.4116)  $5,000](0.2187) = $1993.67 A2 = $5,000(A/P,9%,4) = $5,000(0.3087) = $1543.50
442
Number of monthly deposits = (5 years)(12 months/yr) = 60 $400,000 = $200,000(F/P, i' / month, 60) + $676(F/A, i' / month, 60) Try i' / month = 0.75%: Try i' / month = 1%:
$400,000 > $364,126.69, ∴ i' / month > 0.75% $400,000 < $418,548.72, ∴ i' / month < 1%
Using linear interpolation: 1% − 0.75% i' /month − 0.75% ; = $400,000 − $364,126.69 $418,548.72 − $364,126.69
i' / month = 0.9148%
Therefore, i' / year = (1.009148)12  1 = 0.1155 or 11.55% per year 443
P0 = $100 (P/A, 7%, 10)  $200 (P/F, 7%, 4)  $200 (P/F, 7%, 8) P0 = $100(7.0236)  $200(0.7629)  $200(0.5820) = $433.28
444
F5 = $100 (F/A, 8%, 4) (F/P, 8%, 1) + $100 (P/A, 8%, 2) = 100 (4.5061)(1.08) + 100 (1.7833) = $664.99
445
Using time = 0 as the reference point, set P0(LHS) = P0(RHS) $1,000(P/F,12%,1) + $1,000(P/F,12%,3)  $1,000(P/F,12%,5) = W + W(P/F,12%,7) $1,000(0.8929) + $1,000(0.7118)  $1,000(0.5674) = W + W(0.4523) $1,037.30 = 1.4523 W W = $714.25
446
Left Side : P0 =  Z (P/A, 10%, 4) = Z(3.1699) Right Side: P0 = $1,000 (P/F, 10%, 2)  $2,000 (P/F, 10%, 7) = $1,000 (0.8264)  $2,000 (0.5732) = $200 Set P0 (Left) = P0 (Right):
Z(3.1699)= $200 Z = $63.09
Remember that the cash flow is negative but expressed as a positive cost.
Chapter 4 Solutions
80
447
P0 = $800 + $1,000(P/A,10%,10) + $100(P/G,10%,9)(P/F,10%,1) = $800 + $1,000(6.1446) + $100(19.422)(0.9091) = $8,710.25 A = $8,710.25(A/P,10%,10) = $8,710.25(0.1627) = $1,417.16
448
P0 = $100,000  $10,000 (P/A, 15%,10)  $30,000 (P/F, 15%, 5) = $100,000  $10,000 (5.0188)  $30,000 (0.4972) = $165,104
449
Equivalent cash inflows = Equivalent cash outflows Using time 0 as the equivalence point and N = total life of the system: $2,000(P/F,18%,1) + $4,000(P/F,18%,2) + $5,000(P/A,18%,N2)(P/F,18%,2) = $20,000 $2,000(0.8475) + $4,000(0.7182) + $5,000(P/A, 18%, N2)(0.7182) = $20,000 $5,000 (P/A,18%,N2)(0.7182) = $15432.20 (P/A, 18%, N2) = 4.297 From Table C17, (P/A,18%,8) = 4.078 and (P/A,18%,9) = 4.303 Thus, (N2) = 9 and N = 11 years . Note that if the system lasts for only 10 years, the present equivalent of the cash inflows < present equivalent of the cash outflows.
450
Left side: PL = 10 (P/F, 15%, 1) + H (P/A, 15%, 164) (P/F, 15%, 4) + 0.7H (P/A, 15%, 6) (P/F, 15%, 7) = 10 (0.8696) + H(5.4206) (0.5718) + 0.7H (3.7945) (0.3759) = 4.0953H – 8.696 Right side: PR = P0 + 2P0 (P/F, 15%, 10) = P0 + 2P0 (0.2472) = 0.5056 P0 set PR = PL: 0.5056 P0 – 4.0953H – 8.696 P0 = 17.199 – 8.1H
451
A = $500 + $100(A/G,8%,20) = $500 + $100(7.0369) = $1,203.69
81
Chapter 4 Solutions
452
P0(rental income) = $1,300(P/A,9%,15)  $50(P/G,9%,15) = $1,300(8.0607)  $50(43.807) = $8,288.56 The present equivalent of the rental income is greater than the present equivalent of the $8,000 investment, so the rental property appears to be a good investment.
453
P0 (loan amount) = P0 (loan repayment) $10,000 = Z(P/G,7%,8)(P/F,7%,2) $10,000 = Z(18.825)(0.8734) $10,000 = 16.4417 Z Z = $608.21
454
4Z
4Z $10,000
3Z 2Z Z
⇔ 0
1
⇔ 0
1
2
3
4
5
0
1
i = 8% / yr Select time 1 as the reference point, set P1 (LHS) = P1 (RHS): $10,000(F/P,8%,1) $10,000(1.0800) $10,800 Z 455
(a)
= 4Z (P/A,8%,4)  Z (P/G,8%,4) = 4Z (3.3121)  Z (4.650) = 8.5984 Z = $1,256.05
F = $600 (P/G, i%, 6)(F/P i%, 6) = $10,000 If i = 7%, F = $9,884.81 < $10,000, ∴ i > 7% If i = 8%, F= $10,019.37 > $10,000, ∴ i < 8% Thus, 7% < i < 8%. Using linear interpolation, i%  7% 8%  7% = $10,000  $9,884.81 $10,019.37  $9,884.81 i = 7.86%
(b)
F = $600 (P/G, 5%, N) (F/P, 5%, N) = $10,000 If N = 6, F = $9,622.99 < $10,000, ∴ N > 6 If N = 7, F= $13,704.03 > $10,000, ∴ N < 7
Chapter 4 Solutions
82
2
3
4
5
Z 2Z 3Z
Thus, 6 < N < 7. Using linear interpolation, N  6 7  6 = $10,000  $9,622.99 $13,704.03  $9,622.99 N = 6.1 periods. If an integer value of N is desired, choose N = 7 periods. (c)
F = $1,000(P/G,10%,12)(F/P,10%,12) = $1,000(29.901)(3.1384) = $93,841.30
(d)
G = F(P / F,10%,6)
1 1 = $8,000(0.5645) = $466.34 (P / G,10%,6) 9.684
456
P0 = $100(P/G,15%,7)  $400(P/F,15%,5) = $100(10.19)  $400(0.4972) = $820.12
457
Using time 1 as the reference point, set P1(LHS) = P1(RHS) K(P/A,12%,2) (P/F,12%,2) K(1.6901) (0.7972) 1.3473 K K
= $100(P/A,12%,6) + $110(P/G,12%,6) = $100(4.1114) + $110(8.93) = $1,393.44 = $1,034.25
458
P0 = $100 (P/A, 10%, 4) + $100 (P/G, 10%, 8)
459
F = $1,000 (F/A, 8%, 4) (F/P, 8%, 1)  $200 (P/G, 8%, 4) (F/P, 8%, 5) = $1,000 (4.5061) (1.08)  $200 (4.65) (1.4693) = $3,500.14
460
P0 = $100 + $100 (P/A, 9%, 6) + $100 (P/F, 9%, 3) = $100 + $100 (4.486) + $100 (0.7722) = $625.82 A = $625.82 (A/P, 9%, 7) = $625.82 (0.19869) = $124.34
461
A = [$2,000 (P/A,8%,4) + $400 (P/G,8%,4)] (P/F,8%,2) (A/F,8%,11) = [$2,000 (3.3121) + $400 (4.650)] (0.8573) (0.0601) = $437.14
462
Equivalent cash outflows = Equivalent cash inflows Using time 9 as the reference point, set F9(outflows) = F9(inflows) A(F/A,8%,5) = [$400(P/A,8%,4)  $100(P/G,8%,4)](F/P,8%,10) + $500(F/A,8%,3) A(5.8666) = [$400(3.3121)  $100(4.65)](2.1589) + $500(3.2464)
83
Chapter 4 Solutions
A(5.8666) = $3,479.51 A = $593.10 463
Savings in heat loss: Savings will increase by:
$3,000(0.8) = $2,400 next year $200(0.8) = $160 each year (gradient)
P0(savings) = $2,400(P/A,10%,15) + $160 (P/G,10%,15) = $2,400 (7.6061) + 160 (40.15) = $24,678.64 The present equivalent value of the savings is greater than the installation cost of $18,000. Therefore recommend installing the insulation. 464
Using time 3 as the equivalence point, $500 (F/A,18%,2)(F/P,18%,2) $500 (2.18)(1.3924) $1,517.72 Q
465
= Q(P/G,18%,4) = Q(3.483) = 3.483Q = $435.75
Using time 1 as the equivalence point:  R (P/G,15%,N  1) = 27R (P/F,15%,3) + 27R (P/F,15%,9)  R (P/G,15%,N  1) = 27R (0.6575) + 27R (0.2843) (P/G,15%,N  1) = 10.0764 From Table C15, (P/G,15%,7) = 10.192, thus N  1 = 7 and N = 8 years.
466
Use the EOY 5 as the reference point and equate P0 equivalents as follows: B (P/F,10%,5) B (0.6209) 0.6209B B
= $4,000 (P/A,10%,4)  $1,000 (P/G,10%,4) = $4,000(3.1699)  $1,000(4.378) = $8,301.60 = $13,370.26
467 Note N = 6 $175,000[1 − ( P / F ,18%,5)( F / P,8%,6)] 0.18 − 0.08 $175,000[1 − (0.3704)(1.58690] = 0.10 = $721,371
P0 =
You can afford to spend as much as $721,371 for a higher quality heat exchanger.
Chapter 4 Solutions
84
468
P=
$500[1 − P / F ,12%,15)( F / P,6%,15)] 0.12 − 0.06
= $4,684.51 469
$1,304.35 (1 + f )10 = $5,276.82; Solving yields f = 15% $1,304.35[1 − ( P / F ,20%,11)( F / P,15%,11)] P1 = 0.20 − 0.15 $1,304.35[1 − (0.1346)(4.6524)] 0.05 = $9,750.98
=
∴ P0 = $9,750.98 (F/P,20%,1) = $9,750.98 (1.20) = $11,701.18 ∴ A = $11,701.18 (A/P,20%,10) = $11,701.18 (0.2385) = $2,790.73 470
(a) P0 =
$10,000[1 − ( P / F ,12%,8)( F / P,7%,8)] 0.12 − 0.07
$10,000[1 − (0.4039)(1.7182)] 0.05 = $61,204
=
We can justify spending up to $61,204 for the device (b) Using the result of part (a): A = $61,204 (A/P, 12%, 8) = $61,204 (0.2013) = $12,320 471
P2 =
$1,000[1 − ( P / F ,15.5%,23)( F / P,10%,23)] 0.155 − 0.10
1 (8.9543)] (1.155) 23 = 0.055 $1,000[1 − (0.03636)(8.9543)] = 0.055 $1,000[1 −
= $12,262.21 ⎡ 1 ⎤ P0 = P2 (P/F, 15.5%, 2) = $12,262.21 ⎢ 2 ⎣1.155 ⎥⎦
85
Chapter 4 Solutions
= $12,262.21 (0.7496) = $9,191.75 472
P2 =
$8,140[1 − ( P / F ,15.5%,23)( F / P,10%,23)] 0.155 − (−0.10) $8,140[1 −
=
1 (1 − 0.1) 23 ] 23 (1.155) 0.255
$8,140[1 − (0.03636)(0.08863)] 0.255 = $31,818.70
=
⎡ 1 ⎤ P0 =P2 (P/F, 15.5%, 2) = $31,818.70 ⎢ 2 ⎣1.155 ⎥⎦
= $31,818.70 (0.7496) = $23,851.30 473
$25,000 =
X [1 − ( P / F ,15%,6)( F / P,8%,6)] 0.15 − 0.08
$25,000 (0.07) = X [1  (0.4323)(1.5869)] $1.750 X= 0.314 = $5,573.25 474
(a) P0 =
$5,000[1 − ( P / F ,8%,10)( F / P,6%,10)] (P/F, 8%, 2) 00.08 − 0.06
$5,000[1 − (0.4632)(1.7908) (0.8573) 0.02 = $36,542.72
=
(b)
P0′ = $4,000 (P/A, 8%, 12) + G (P/G. 8%, 12)
= $4,000 (7.5361) + G (34.634) = $30,144,40 + 34.634G (c) set P0 = P0′ $36,542.72 = $30,144.40 + 34.634G G = $184.74
Chapter 4 Solutions
86
475
Left Hand Side: P2 = $1,500 (P/A,30%,6)  $100 (P/G,30%,6) P0 = P2 (P/F,30%,2) = [$1,500 (P/A,30%,6)  $100 (P/G,30%,6)] (P/F,30%,2) Right Hand Side: Z(1+ f )1 = 1.2Z, ∴ f = 0.20 or 20% per year P0 =
Z [1 − ( P / F ,30%,8)( F / P,20%,8)] 0.30 − 0.20
Set P0(Left Hand Side) = P0(Right Hand Side) and solve for Z. F=?
476 i = 4% / yr 1
2
3
4
5
i = 7% / yr 6 7
8 9 10 11 12 13 14 15 16
17 18
$2,000 F6 = $2000(F/A,4%,6) = $2000(6.6330) = $13,266 F8 = F6(F/P,4%,2) = $13,266(1.0816) = $14,348.51 F18 = F8(F/P,7%,10) = $14,348.51(1.9672) = $28,226.38 Answer: $28,226.38 is withdrawn 12 years after last deposit. 477
r ⎤ ⎡ (a) r = 10%, M = 2/yr ; i = ⎢1 + ⎥ ⎣ M⎦
M
⎡ 0.1⎤ − 1 = ⎢1 + − 1 = 0.1025 =10.25% 2 ⎥⎦ ⎣
r ⎤ ⎡ (b) r = 10%, M = 4/yr ; i = ⎢1 + ⎥ ⎣ M⎦
M
⎡ 0.1⎤ − 1 = ⎢1 + − 1 = 0.1038 = 10.38% 4 ⎥⎦ ⎣
r ⎤ ⎡ (c) r = 10%, M = 52/yr ; i = ⎢1 + ⎥ ⎣ M⎦
478
2
M
4
⎡ 0.1⎤ − 1 = ⎢1 + ⎣ 52 ⎥⎦
52
− 1 = 0.1051 = 10.51%
(a) ieff/yr = (1+.016)12 – 1 = 4.936 or 493.6% (b) It is unlikely that this rate of growth could be sustained for very long.
87
Chapter 4 Solutions
EOY
479
imonthly =
6.00% / year = 1/2% / month 12 months / year
A = $100,000 (A/F, 1/2%, 60) = 100,000 (0.0143) = $1,430 Select d. 480
(a)
i / 6 months = 8% / 2 = 4%; N = 5 years = 10 6month periods A = $3,000 (A/F, 4%,10) = $3,000 (0.0833) = $249.99
(b)
⎡ 0.12 ⎤ r = 12%; M = 12; ieff/yr = ⎢1 + 12 ⎥⎦ ⎣
12
− 1 = 0.1268 or 12.68%
A = $125,000 (P/A,12.68%,10) = $125,000 (0.18194) =$22,742.33 481
r = 12%; M = 12; i / yr = (1 +
012 . 12 )  1 = 0.1268 or 12.68% 12
$10,000
0
11 12
13
14
15
16
17
End of Year
P0 = ? P0 = $10,000(P/A,12.68%,6)(P/F,12.68%,11) = $10,000 (4.034)(0.2689) =$10,847.43
482
i / mo. = 15% / 12 = 1.25% ⎡ (1 + 0.0125 )72 − 1 ⎤ P = $100(P/A,1.25%,72) = $100 ⎢ = $100(47.2987) 72 ⎥ ⎣ (0.0125 )(1 + 0.0125 ) ⎦
= $4,729.87 483
i / qtr = 12% / 4 = 3%; N = 10 years = 40 quarters F = $7,500 (F/P,3%,40) = $7,500 (3.2620) = $24,465 OR
Chapter 4 Solutions
88
4
⎛ 0.12 ⎞ r = 12%, M =4, ieff/yr = ⎜1 + ⎟ − 1 = 0.1255 or 12.55% 4 ⎠ ⎝ F = $7,500 (F/P,12.55%,10) = $7,500 (3.2620) = $24,465
484
(a) A = $10,000 (A/P, 1% per month, 36 months) = $10,000 (0.0332) = $332 (b) 0 = $9,800  $332 (P/A, i' per month, 36 months) By trial and error, i' = 1.115% per month so the true APR is 12 (1.115%) = 13.38% per year
485
i / month = 3/4 %; $3,350 = $100 (F/A, 3/4%, N), ∴(F/A, 3/4%, N) = 33.5 From Table C3, N = 30 months.
486
i / month = 16.5% / 12 = 1.375% F = $340 (F/P,1.375%,11) = 340 (4.1253) = $1402.63
487
(a) (1 + i/mo)12 – 1 = 0.0616778 (1 + i/mo)12 = 1.0616778 i/mo = 12 1.0616778  1 = 0.00499 or ≈ 0.5% / mo. r = (12) (0.5%) = 6% per year (b) A = $9,500 (A/P, 0.5% per month, 36 months) = $9,500 (0.0304) = $288.80
488
A = $20,000 (A/P,1%,60) = $444 P (of remaining 40 payments) = $444 (P/A,1%,40) or P = $14,579 Select c. 4
489
⎡ 0.08 ⎤ r = 8%, M = 4, ∴i / yr = ⎢1 + − 1 = 0.0824 = 8.24% 4 ⎥⎦ ⎣
F = $5,000(F/P,8.24%,3) = $5,000(1.2681) = $6,340.50 490
Effective rate = (1 + 0.01)4 – 1 = 4.06% F = $1,000 (F/A, 4.06%, 5) (1 + 0.0406) 5 − 1 ] = $5,423 = $1,000 [ 0.0406 or A = $1,000 (A/F, 1%, 4) = $1,000 (0.2463) = $246.30 F20 = $246.30 (F/A, 1%, 20) $246.30 (22.0190) = $5,423 89
Chapter 4 Solutions
491
5 years x 12 month/year = 60 months; i/mo. = 9%/12 = 3/4% A = $15,000 (A/P, 3/4%, 60) = $15,000 (0.0208) = $312 Select c.
492
ieff = er  1; 0.192 = er  1; 1.192 = er; ln(1.192) = r = 17.56%
493
(a) P = $100,000 (P/F, 15%, 10) = $100,000 (0.2472) = $24,720 (b) P = $100,000 (P/F, 15%, 10) = $100,000 (0.2231) = $22,310
494
F = $10,000 (F/P, 14%, 1) (F/P, 12%, 1)(F/P, 10%,2)(F/P,12%,1) = $10,000 (1.14) (1.12)(1.21)(1.12) = $17,303.19
495
P = $1,000 (P/F, 8%, 1) + $2,000 (P/F, 10%, 1)(P/F, 8%, 1) + $1,000 (P/F,6%, 1)(P/F, 8%, 1)(P/F, 10%, 1)(P/F, 8%, 1) + $2,000 (P/F, 6%,3)(P/F, 8%, 1)(P/F, 10%, 1)(P/F, 8%, 1) = $1,000 (0.9259) + $2,000 (0.9091)(0.9259) + $1,000 (0.9434)(0.9259)(0.9091)(0.9259) + $2,000 (0.8396)(0.9259)(0.9091)(0.9259) = $4,653.33
496
F1 = $1,000 (F/P,15%,1)  $1,000 = $1,000 (1.15)  $1,000 = $2,150 F2 = F1 (F/P,15%,1) + $12,000 = $2,150 (1.15) + $12,000 = $9527.50 F4 = F2 (F/P,10%,1)(F/P,6%,1) = $9527.50(1.10)(1.06) = $11,109.06
497
(a) False; (b) False; (c) False; (d) True; (e) False; (f) True; (g) False; (h) False; (i) False
498
(a)
A = $8,000(A/F,8%,10) =
(b)
P = $1,000(P/A,8%,12) = $1,000(7.4094) = $7,409.40
(c)
r = 8%/2% = 4%
$8,000 $8,000 = = $543.67 (F / A,8%,10) 14.7147
⎡ e (0.04)(12) − 1⎤ F = $243(F/A,4%,12) = $243 ⎢ 0.04 ⎥ = $243(15.0959) = $3,668.30 −1 ⎦ ⎣ e (d) 499
F = $1,000(F/P,8%,9) = $1,000(2.0544) = $2,054.40
Set Equivalent cash outflows = Equivalent cash inflows Using time 9 as the equivalence point, Z (F/P,20%,9) = $500(F/A,20%,5) + Z (F/P,20%,6) 6.0496 Z = $500(7.7609) + 3.3201 Z
Chapter 4 Solutions
90
2.7295 Z = $3,880.45 Z = $1,421.67 4100 F18 = $10,000 (F/P,8%,18) = $10,000 (4.2207) = $42,207 4101 r = 10%/2 = 5% per 6 months, compounded continuously P = $100 (P/A,5%,10) = $100(7.6743) = $767.43 4102 $45,000 = $3,000 (F/A, r %, 10) 15.000 = (F/A, r %, 10) From Appendix D, (F/A,8%,10) = 14.7147 and (F/A,10%,10) = 16.3380 Thus, 8% < r < 10%. Using linear interpolation, r = 8.35%. 4103 r = 10% / 2 = 5% every six months, compounded continuously A = $18,000(A/P, 5%,24) =
$18,000 $18,000 = = $1,320.66 (P/A, 5%,24) 13.6296
4104 (a) P = $3,500 (P/A,10%,5) = $3,500 (3.7412) = $13,094.20 (b) $16,000 = $7,000 (F/P, r%, 9) $16,000 = $7,000 e9r; 2.2857 = e9r ; 9r = ln(2.2857) = 0.8267 r = 0.0919 or 9.19% 4105 r = 10% per year, compounded continuously (a)
F = $2000 (F/A,10%,30] = $2000 (181.472) = $362,944
(b)
A = $362,944(A/P,10%,10)=
4106 (a) (b)
$362,944 $362,944 = = $60,386 (P / A,10%,10) 6.0104
True; i/yr = er 1 = e0.1 1 = 0.1052 or 10.52% > r = 10% True; In fact, more than half of the principal is still owed after the tenth monthly payment is made. r = 10% / 12 = 0.833% per month compounded continuously. ⎛
⎞ ⎟ ⎟ ⎝ e 0.00833(24) ( e 0.00833 − 1) ⎠
P0 = $185(P/A,0.833%,24) = $185 ⎜⎜
e 0.00833(24) − 1
= $185(21.6627) = $4,008 Amount still owed immediately following = $185(P/A,0.833%,10) = $185(13.1595) = $2,435 tenth payment
91
Chapter 4 Solutions
$2,435 * 100 = 61% of principal is still owed after tenth payment. $4,008 2
(c)
⎡ 0.08 ⎤ False; r = 8%, M = 2, i/yr = ⎢1 + − 1 = 0.0816 = 8.16% 2 ⎥⎦ ⎣ F = P(F/P, i%, N) ?
$1,791 = $900 (F/P,8.16%,10) ?
$1,791 = $900 (1.0816)10 $1,791 ≠ $1,972 (d)
False; (A/P, i%, N) + i ≠ (A/F, i%, N). However, if we subtract i from the capital recovery factor, the sinking fund factor is obtained. N
(e)
False; (P/A, i%, N) =
∑ (P / F, i%, k) ≠ 1 + i N
k=1
(f)
Part i)
(P/A,i%,N) (F/P,i%,N) = (F/A,i%,N)
Note:
⎡ (1 + i) N − 1⎤ ⎡ i (1 + i) N ⎤ ( F / A , i%, N ) = (F/A,i%,N)(A/P,i%,N) = ⎢ ⎥⎢ ⎥ N ( P / A , i%, N ) i ⎣ ⎦ ⎣ (1 + i) − 1⎦ = (1+i)N = (F/P,i%,N)
Part ii) (A/G,i%,N)(P/A,i%,N) = (P/G,i%,N) Note:
⎤ ⎡ (1 + i) N − 1⎤ ⎡1 N (A/G,i%,N)(P/A,i%,N) = ⎢ − ⎥⎢ N N ⎥ ⎣ i (1 + i) − 1⎦ ⎣ i(1 + i) ⎦ 1 ⎡ (1 + i) N − 1 N ⎤ − = ⎢ ⎥ i ⎣ i(1 + i) N (1 + i) N ⎦ = (P/G,i%,N)
4107 Assumptions: 1. Retirement deposits are made at the end of the year and are based on the beginning of year salary. 2. A deposit is made into the retirement account just before 10% of balance is withdrawn. 3. First withdrawal from retirement account occurs at age 67 to cover expenses for upcoming year. Cash flow series: 1. Annual deposits from first job (age 23 to 45, years 1 to 22) A1 = 0.05($43,000) = $2,150
Chapter 4 Solutions
92
f = 4% PW(8%) =
$2,150[1 − ( P / F ,8%,22)( F / P,4%,22)] = $30,324.43 0.08 − 0.04
2. Lottery deposit = $5,000 (age 30, year 7) PW(8%) = $5,000(P/F,8%,7) = $2,917.50 3. PWwithdrawal(8%) = 0.10($30,324.43 + $2,917.50) = $3,324.19 Actual amount of withdrawal = $3,324.19(F/P,8%,22) = $18,071.96 4. Annual deposits from second job (age 47 to 66, years 25 to 44) A = (0.05)($85,000) = $4,250 G = (0.05)($4,500) = $225 PW(8%) = $4,250(P/A,8%,20)(P/F,8%,24) + $225(P/G,8%,20)(P/F,8%,24) = $12,284.88 PW of Account Balance at Retirement (age 23, year 0): = $30.324.43 + $2,917.50  $3,324.19 + $12.284.88 = $42,202.62 Actual Account Balance at Retirement (age 67, year 44): FW(8%) = $42,202.62(F/P,8%,44) = $42,202.62 (1.08)44 = $1,247,339.44 (a) Desired final account balance at age 87 (one year after last withdrawal) = (10)($100,000) = $1,000,000 $1,247,339
67
$1,000,000
68
69
70
85
86
87
⇔
87
A=? A = [$1,247,339(P/F,8%,1)  $1,000,000(P/F,8%,21)](A/P,8%,20) = $97,438 (b) FWage37(8%) of withdrawal = $3,324.19(F/P,8%,64) = $457,937
93
Chapter 4 Solutions
Each grandchild could have received an additional $457,937/10 = $45,794
Chapter 4 Solutions
94
Solutions to Spreadsheet Exercises 4108 EOY 0 1 2 3 4 5 6 7 8 P A F
Cash Flow $ (15,000) $ 2,000 $ 2,500 $ 3,000 $ 3,500 $ 4,000 $ 4,000 $ 4,000 $ 4,000 $ 2,189 $ 410 $ 4,692
Notice that a cell is not designated for the interest rate. Therefore, the interest rate must be entered specifically in the NPV, PMT, and FV financial functions. The following entries in the designated cells will yield the results for P, A, and F: B11 = NPV (0.1, B3:B10) + B2 B12 = PMT (0.1, A10, B11) or = PMT (0.1, 8, B11) B13 = FV (0.1, A10, B12) or = FV (0.1, 9, B12)
P = $2,189.02 A = $410.32 F = $4,692.38
4109 i / yr Years in series
18% 8
EOY 0 1 2 3 4 5 6 7 8
P0 =
$
1,294
F8 =
$
4,864
A=
$
317
Cash Flow $ $ $ $ $ $ $ $ $
100 200 500 400 400 400 400 400
In Example 412, the project was acceptable at the original MARR = 20%. A MARR of 18% serves to make the project even more attractive.
95
Chapter 4 Solutions
4110 i / yr = f= A1 =
25% 20% $ 1,000
N=
10
EOY 1 2 3 4 5 6 7 8 9 10 P= A= F=
$ $ $ $ $ $ $ $ $ $
1,000 1,200 1,440 1,728 2,074 2,488 2,986 3,583 4,300 5,160
$ 6,703 $ 1,877 $ 62,430
Solutions to Case Study Exercises 4111 Amount set aside each month = $311.40 Interest rate = 3%/12 = ¼% per month First payment occurs at the end of the first month of year 6 Amount available at end of 10year time frame: F = $311.40 (F/A, ¼%, 60) = $311.40 (64.6467) = $20,130.98 4112 Assume the utility cost remains at $150 per month. Assume a property tax and insurance rate equal to 25% of the principal and interest (P&I) payment. Total Monthly Payment = $800 = 1.25 (P&I) P&I = $640 Assume a 30year mortgage at 6% compounded monthly. Mortgage amount = $640 (P/A, ½%, 360) = $640 (166.7916) = $106,747 Maximum purchase price = $106,747 + $40,000 = $146,747 4113
A = $320; i = 10%/12 = 0.833% per month; N = 120 months F = $320 (F/A, 0.833%, 120) = $320 (204.7981) = $65,535.39
Chapter 4 Solutions
96
Solutions to FE Practice Problems 4114
I = (P)(N)(i) = $3,000 (7) (0.06) = $1,260 F = P + I = $3,000 + $1,260 = $4,260 Select (e)
4115 P= $100,000 (P/F, 10%, 25) = $100,000 (0.0923) = $9,230 Select (b) 4116 F = $2,000 (F/A,2%,30) = $2,000 (40.5681) = $81,136 Select (d) 4117
i / mo. = 6% / 12 = 0.5%, N = 30 x 12 = 360 months P = $1,500 (P/A, 0.5%, 360) ⎡ (1.005) 360 − 1 ⎤ = $1,500 ⎢ 360 ⎥ ⎣ (0.005)(1.005) ⎦ = $1,500 (166.7916) = $250,187 Select (c)
4118
A = $8,000; G = $7,000 Atotal = $8,000 + $7,000 (A/G, 12%, 5) = $8,000 + $7,000 (1.7746) = $20,422 Select (a)
4119
$9,982 = $2,500 (P/A, i', 5) 3.9928 = (P/A, i', 5); i' = 8% / yr $9,982 = G (P/G. 8%, 5) $9,982 = G (7.372) G = $1,354 Select (d)
4120
i / mo. = 9% / 12 = ¾% per month; N = 7 x 12 = 94 months A = $30,000 (A/P, ¾% per month, 84 months) ⎡ (0.0075)(1.0075) 84 ⎤ A = $30,000 ⎢ ⎥ 84 ⎣ (1.0075) −1 ⎦ = $30,000 (0.0161) = $483 / month 97
Chapter 4 Solutions
Select (c) 4121
r = 12%; i = e0.12 – 1 = 0.1275 or 12.75% compounded annually Select (c)
4122
i / mo. = 12%/12 = 1% per month; N = 4 x 12 = 48 months A = $5,000 (A/P, 1% per month, 48 months) = $5,000 (0.0263) = $131.50 Select (a)
4123
F = $7,000 (F/P, 12%, 3) = $7,000 e(0.12)(3) = $7,000e0.36 = $7,000 (1.4333) = $10,033 Select (c)
Chapter 4 Solutions
98
Solutions to Chapter 5 Problems 51
No. A higher MARR reduces the present worth of future cash inflows created by savings (reductions) in annual operating costs. The initial investment (at time 0) is unaffected, so higher MARRs reduce the price that a company should be willing to pay for this equipment.
52
Monthly gasoline savings = 10 gallons x $2.00 per gallon = $20 per month. $1,200 = $20 per month x (P/A, 0.5% per month, N months); N = 72 months to recover the investment plus interest
53
(a)
PW (12%) =  $640,000 + $180,000 (P/A,12%,8)  $42,000 (P/A,12%,8) + $4,000 (P/G, 12%,6)(P/F,12%,2) + $20,000 (P/F,12%,8) =  $640,000 + $180,000 (4.9676)  $42,000 (4.9676) + $4,000 (8.930)(0.7972) + $20,000 (0.4039) = $82,082.78 > 0, therefore the proposal is acceptable
(b) Based on PW and IRR, the proposal is acceptable. 54
PW(12%) = $13,000 + $3,000 (P/F,12%,15)  $100 (P/A,12%,15)  $200 (P/F,12%,5)  $550 (P/F,12%,10) = $13,000 + $3,000(0.1827)  $100(6.8109)  $200(0.5674)  $550(0.3220) = $13,423.57
55
PW(15%) = $10,000 + ($8,000  $4000)(P/A,15%,5)  $1,000 (P/F,15%,5) = $10,000 + $4000(3.3522)  $1,000 (0.4972) = $2,911.60 FW(15%) = $10,000 (F/P,15%,5) + ($8,000  $4000)(F/A,15%,5)  $1,000 = $10,000 (2.0114) + $4000(6.7424)  $1,000 = $5,855.60 AW(15%) = $8,000  $4000  [$10,000(A/P,15%,5)  ( $1,000)(A/F,15%,5)] = $4000  [$10,000(0.2983) + $1,000(0.1483)] = $868.70
56
Assume miles driven each year is roughly constant at 12,000. A fourspeed transmission car will consume 400 gallons of gasoline each year, and the fuel cost is $800 per year. The annual savings due to a six speed transmission is 0.04 ($800) = $32. Over a ten year life, the present worth of savings will be $32 (P/A, 6%, 10) = $235.52. This is how much extra the motorist should be willing to pay for a sixspeed transmission.
101
Chapter 5 Solutions
57
Your Point of View:
$1,300
0
1
$1,400 $1,500
$1,800 $1,600 $1,700 +
2
3
4
5
$12k
6
EOY
$10,000 (RE) = $1,300 (yr. 1) = $1,300 + (k1)($100) for 1 ≤ k ≤ 6
58
PW(15%) = $10,000 + $1,300 (P/A, 15%, 6) + $100 (P/G, 15%, 6) + $1,200 (P/F, 15%, 6) = $10,000 + $1,300 (3.7845) + $100 (7.937) + $12,000 (0.4323) = $901.15 > 0; the investment appears to be a good one if the risk is low Capital Investment = $300,000  $600,000  $250,000  $100,000 = $1,250,000 Revenue = $750,000 per year Market Value = $400,000 + $350,000 + $50,000 = $800,000 Expenses = $475,000 per year PW(15%) = $1,250,000 + ($750,000  $475,000)(P/A,15%,10) + $900,000(P/F,15%,10) = $1,250,000 + $275,000(5.0188) + $800,000(0.2472) = $327,930 > 0 Therefore, they should invest in the new product line.
59
Both cars last for 6 years. Cost of gasoline stays constant. Assume that fuel expense is the only expense to consider (ignore tire wear, etc.). Assume a standard auto gets 25 mpg. Composite auto annual fuel expense (20,000 mi./yr.)(1 gal./75 mi.)($2.00/gal.) = $533.33/yr.
Chapter 5 Solutions
102
Traditional auto annual fuel expense (20,000 mi./yr.)(1 gal/.25 mi.)($2.00 gal.) = $1,660/yr. So spending an extra P = ($1,600  $533.33)(P/A, 10%, 6) will be justified by the annual fuel savings. P = $1,066.67 (4.3553) = $4,645.67 If the composite auto costs more than $4,646 compared to the “traditional” auto, the purchase should not be made. (This ignores “prestige value” of being the first on the block with a composite car.) Reduced fuel expense per year is being traded off against extra investment for the composite auto. 510 Desired yield per 6 months = 3%; N = 3 x 2 = 6 periods i = [(1.13)1/2 – 1)]100 = 6.3% per semiannual period VN = $1,000 (P/F, 6.3%,6) + 0.03 ($1,000) (P/A,6.3%,6) = $1,000 (0.693) + $30 (4.876) = $69.30 + $146.28 = $836.58 511 Desired yield per year = 10% VN = $1,000 (P/F, 10%, 10) + 0.14 ($1,000) (P/A,10%,10) = $1,000 (.3855) + $140 (6.1446) = $1,245.74 512
Desired yield per quarter = 12%/4 = 3%; N = 4(10) = 40 quarters (a)
VN = $10,000 (P/F,3%,40) + 0.02 ($10,000) (P/A,3%,40) = $10,000 (0.3066) + $200 (23.1148) = $7,688.96
(b)
Quarterly interest payment = (0.02)($1,000,000) = $20,000
⎛ 0.08 ⎞ i/qtr = ⎜ 1 + ⎟ ⎝ 2 ⎠
1/ 2
− 1 = 0.0198 or 1.98%
Equivalent value of interest payments at EOY10: F40 = $20,000(F/A,1.98%,40) = $20,000(60.1408) = $1,202,816 F40 redemption cost = $1,000,000 2
⎛ 0.08 ⎞ i/yr = ⎜ 1 + ⎟ − 1 = 0.0816 or 8.16% ⎝ 2 ⎠
103
Chapter 5 Solutions
A = [$1,000,000 + $1,202,816] (A/F,8.16%,10) = $2,202,816 (0.0685) = $150,892.90 513 True interest rate in the IRR associated with the equation PW(i') = 0. ($1,000,000 – $50,000) – $40,000 (P/A,i',15%)  $70,256 (P/A,i',15) – $1,000,000 (P/F,i',15) = 0 $950,000  $110,526 (P/A,i',15%)  $1,000,000 (P/F,i',15) = 0 at 10% $950,000  $110256 (7.6061)  $1,000,000 (0.2394) = $128,018 at 12% $950,000  $110,256 (6.8109)  $1,000,000 (0.1827) = +$16,357 by interpolation, i' = 11.773% 514
Interest payments from bond = $100,000 (0.0725) = $7,250 every six months P0 = 0 = $100,000 + $7,250 [P/A,i'%,10(2)] + $110,000 [P/F,i'%,10(2)] PW(7%) = $5,230.50 > 0, ∴ i'% > 7% PW(8%) = $5,223.78 < 0, ∴ i'% < 8% Linear interpolation between 7% and 8% yields: i'% = 7.5% per six months.
515
Original Mortgage Payment: i / mo = 10%/12 = 0.833% A = $100,000(A/P,0.833%,360) A = $100,000(0.0087757) = $877.57 Remaining principal after two years of payments: P24 = $877.57(P/A,0.833%,336) P24 = $877.57(112.61763) = $98,829.86 New payment if refinanced over remaining 28 years: 0.583%
i / mo = 7%/12 =
A = $98,829.86(A/P,0.583%,336) A = $98,829.86(0.0067961) = $671.66 Savings per month if refinanced = $877.57  $671.66 = $205.91 Number of months Susie should remain in her townhouse to make this a worthwhile venture: $4,500 = $205.91(P/A,0.5%,N) (P/A,0.5%,N) = $4,500/$205.91 = 21.8542 From Table C2, N ≈ 24 months.
Chapter 5 Solutions
104
516
PW
1
2
3
4
5
6
7
8
9
10
11
2000 2600 3200 4100
4700
5000
PW of the base ($2,000) = $2,000 (P/A,10%,11) = $2,000 (P/A,10%,11) ⎡ (1.10)11 − 1 ⎤ = $2,000 ⎢ = $12,990.12 11 ⎥ ⎣ (0.10)(1.10) ⎦ PW of the gradient ($300) = $300 (P/G, 10%,11) =
⎤⎡ 1 ⎤ 300 ⎡ (1.10)11 − 1 − 11⎥ ⎢ ⎢ 11 ⎥ 0.10 ⎣ 0.10 ⎦ ⎣ (1.10) ⎦
= $7,818.88 Total PW = $12,990.12 + $7,818.88 = $20,909.00 517 PW of 5 yr lining: PW = [ $2,500 (A/P, 10%,5)] (P/A,10%,20) = $2,500 (0.2638)(8.5136) = $5,614.72
PW 10 yr lining: PW = [$6,500 (A/P,10%,10)] (P/A,10%,20) = $6,500 (0.1627) (8.5136) = $9,003.56
518 Principal contributed Future worth at 6% per year a b
Janet $10,000 $56,571a
Bill $25,000 $54,865b
$1,000 (F/A,6%,10) (F/P,6%,25) $1,000 (F/A,6%,25) Janet will have accumulated more money in her retirement account than Bill by age 65 even though her contribution of principal was less than half of what Bill invested. The reason for this is that Janet had compounding of interest working for her longer than Bill had. The lesson is – start saving early!
105
Chapter 5 Solutions
519 $2,900 $2,900 $2,800 $2,700 $2,500 $2,400 $2,300 $2,200 $2,100
$500
1
2 3 4 5 6 7 8 9 10
(EOY)
Let A = $2,900, G = $100 (delayed 1 year) F6 = $2,000 P0 = $2,900 (P/A,6$,10) – 100(P/G,6%,9)(P/F,6%,1)  $2,000(P/F,6%,6) = $2,900 (7.3601)  $100 (24.5768) (0.9434)  $2,000 (0.7050) = $17,615.71 520
0.1 P = P (A/P,8%,N) so N ≈ 21 years A payout duration table can be constructed for selected payout percentages and compound interest rates as follows: Interest Rate/Year 4% 6% 8% 10% 13.0* 15.7 20.9 never Payout/Yr 10% (% of principal) 20% 5.7 6.1 6.6 7.3 30% 3.7 3.8 4.0 4.3 * Note: Table entries are years.
521
F14 = A(F/A,4%,17) = 23.6975A F11 = 23.6975A (P/F,4%,3) = 21.067A so then 21.067A = $32,500(P/F,4%,1)+ $34,125(P/F,4%,2) + $35,750 (P/F,4%,3) and A = $4,490/year
522
(a) CW(10%) =
$1,500 ⎡ $10,000 (A F ,10%,4) ⎤ +⎢ ⎥( P / F,10%,1) = $34,591 010 . 010 . ⎦ ⎣
(b) Find the value for N for which (A/P,10%,N) = 0.10 From Table C13, N = 80 years 523
Opportunity Cost = Investment at BOY x (P/F, 15%, 1) = BOY (0.15) Capital Recovery Amount = Opportunity Cost + Loss in Value During Year
Chapter 5 Solutions
106
Investment Opportunity at Beginning Cost of Interest of Year (i = 15%) Year 1 $10,000 $10,000 (0.15) = 1,500 2 7,000 (0.15) = 1,050 10,0003000 = 7,000 3 5,000 (0.15) = 750 7,000  2,000 = 5,000 4 3,000 (0.15) = 450 5,000  2,000 = 3,000
Loss in Capital Value Recovery During Year Amount $3,000 1,500 + 3,000 = 4,500 2,000 1,050 + 2,000 = 3,050 2,000 750 + 2,000 = 2,750 1,000 450 + 1,000 = 1,450
P0 = $4,500 (P/F,15%,1) + $3,050 (P/F,15%,2) + $2,750 (P/F,15%,3) +$1,450 (P/F, 15%, 4) = $4,500 (0.8696) + $3,050 (0.7561) + $2,750 (0.6575) + $1,450 (0.5718) = $ 8,856.54 A = $8,856.54 (A/P,15%,4) = $8,856.49 (0.3503) = $3,102.45 This same value can be obtained and confirmed with Equation 55: CR(i%) = I (A/P, i%, N)  S (A/F, i%, N) = $10,000 (A/P,15%,4)  $2,000 (A/F,15%,4) = $10,000 (0.3503)  $2,000 (0.2003) = $3,102.12 Note: The Annual Worth from the table and the CR amount from Equation 55 are the same. 524
FW(18%) = $8,000 (F/P,18%,10) + $2,400 (F/A,18%,10) = $8,000 (5.2338) + $2,400(23.5213) = $14,580.72 Therefore, recommend process R since FW ≥ 0. AW(18%) = $8,000 (A/P,18%,10) + $2,400 = $8,000 (0.2225) + $2,400 = $620 Therefore, recommend process R since AW ≥ 0.
525 AW (of costs) = $125,000 (A/P,8%,10) + $1,000 + $1,000 (A/G,8%,10) – $40,000 (A/F,8%,10) = $125,000 (0.14900) + $1,000 + $1,000 (3.8713)  $40,000 (0.0690) = $18,625 + $1,000 + $3,871.30  $2,760 = $20,736.30 per year
107
Chapter 5 Solutions
526
Year 1 2 3 4
527
Investment Opportunity at Beginning Cost of Interest of Year (i = 15%) $1,000 $50 1,000  200 = 800 800(0.05) = 40 600 30 600  200 = 400 20
Loss in Value During Year $250  $50 = $200 200 200 400  300 = 100
Capital Recovery Amount $250 240 230 20 + 100 = 120
(a) (b) (c) (d) (e) (f)
Loss in Value = Capital Recovery Amount  Opportunity Cost Investment at BOY2 = Investment at BOY1  Loss in Value during Year 1 Opportunity Cost = Investment at BOY * (0.05) Investment at BOY4 = Investment at BOY3  Loss in Value during Year 3 Loss in Value during year 4 = Investment at BOY4  Salvage Value at EOY4 Capital Recovery Amount = Opportunity Cost + Loss in Value during Year
(a)
⎡ 1 ⎤ lim (P / F,i%, N) = lim ⎢ ⎥=0 i→∞ i→∞ (1 + i) N ⎣ ⎦
Therefore, as i → ∞ , the PW approaches $3,000. (b)
By inspection, θ = 5 years. Therefore θ' ≥ 5 years PW05(12%) = $3,000  $1,000(P/F,12%,1) + $1,000(P/A,12%,3)(P/F,12%,1) + $2,000(P/F,12%,5) = $3,000  $1,000(0.8929) + $1,000(2.4018)(0.8929) + $2,000(0.5674) = $613.53 < 0 , ∴ θ' > 5 years PW06(12%) = PW05(12%) + $2,000(P/F,12%,6) = $613.53 + $2,000(0.5066) = $399.67 > 0, ∴ θ' = 6 years
(c)
PW(0%) = $3,000  $1,000(P/F,0%,1) + $1,000(P/A,0%,3)(P/F,0%,1) + $2,000(P/F,0%,5)  $2,000(P/F,0%,6) =  $3,000  $1,000(1) + $1,000(3)(1)+ $2,000(1)  $2,000(1) =  $1,000 AW(0%) = PW(0%)(A/P,0%,6) =  $1,000 (0.1667) =  $166.70
Chapter 5 Solutions
108
528
(a)
(Maintenance, taxes, insurance) = $2,000 + 0.05 ($75,000) = $5,750/year cV =
$1.00 / unit direct labor + $0.75 / unit new material + $ 0.30 / unit scrap + $ 0.60 / unit overhead $ 2.65 / unit
X = annual production in units
[Unit costs (not including CR)] = $2.65 / unit +
$5,750 = $2.94 / unit 20,000 units
CR = ($75,000+$15,000)(A/P,20%,5)  ($3,000+$15,000)(A/F,20%,5) = $90,000 (0.3344)  $18,000 (0.1344) 108 = $27,676.80 CR / unit =
$27,676.80 = $1.38 / unit 20,000 units
Total Cost / unit = CT = $2.94 + $1.38 = $4.32 / unit Thus the unit price for the bid should be greater than $4.32. (b)
Selling Price = (1 + 0.20) CT = (1.20) ($4.32 / unit) = $5.18 /unit
529
$10,000 = $200 (F/A,i,45) (F/P,i,3) or i per month is approximately equal to 0.4165% which equates to i per year of (1.004165)12 – 1 = 0.0511 (5.11% per year). This is a conservative investment when Stan makes the assumption that the investment firm will pay him $10,000 when he leaves the service at the end of 4 years (i.e., there is little risk involved). Stan should probably take this opportunity to invest money while he is in the service. It beats U.S. savnings bonds which pay about 4% per year.
530
PW(i'%) of outflows = PW(i'%) of inflows $308.57 (P/A,i'%,35) = $7,800 (P/A,i'%,35) = 25.2779 (P/A,1%,35) = 29.4085 and (P/A,2%,35) = 24.9986, Therefore, 1% < i'% < 2%. Linear interpolation yields: i'% = 1.9% per month
109
Chapter 5 Solutions
A.P.R = (1.9% per month)(12 months/year) = 22.8% compounded monthly
531 $1,000
i=?
1
2
3
23
24
EOM
$56.44
$1,000 = 56.44 (P/A,i',24) i (P/A,i,24) 2 18.9139 i' 17.7174 16.9355 3
i'−2 1 = 18.9139 − 17.7174 18.9139 − 16.9355 i '−2 1 = 1.1960 1.9784 i' – 2 =
1.196 1.9784
i' = 2.604% / mo.; i/yr = (1.026)12 – 1 = 0.3607 or 36.1% / yr. 532
$1,000 = $58.50 (P/A,i'%,24) (P/A,i'%,24) = $1,000/$58.50 = 17.094 From Appendix C: (P/A,2%,24) = 18.9139 and (P/A,3%,24) = 16.9355 By linear interpolation: i'% = 2.9% per month Thus, ieffective = (1.029)12  1 = 0.409 or 40.9% per year.
533
PW = 0 = $1,000  $117 (P/A,i'%,2)  $58.50(P/A,i'%,20) (P/F,i'%,2) By trial and error: PW(3%) = $44 and PW(4%) = $44 By linear interpolation: i'% = 3.5% per month Thus, ieffective = (1.035)12  1 = 0.511 or 51.1% per year.
Chapter 5 Solutions
110
534
PWcost = PWbenefit $17,000 = (0.8) ($4,000) (P/A,i%,10) + (0.8)($300)(P/G,i%,10) i = 14%
535 Year 0 1
Plan 2 $850 0
Plan 1 $90 $90
Plan 2 – Plan 1 $760 +$90
A = P x i (as a decimal) i = A/P, = $90 / $760 = 11.84% Choose Plan 1 if MARR > 11.84% 536
Assume that the task is to produce 40,000 pieces / yr. Old Machine (No Brake): Production time/year = (40,000 pieces/yr)(2.5 min/pc)(1hr/60 min) = 1,666.67 hr/ yr Operator salary = $16.50/hr (1,666.67 hr / yr) = $27,500.06 / yr Overhead = $4.00/hr (1,666.67 hr/ yr) = $ 6,666.68 / yr Total Cost = $34,166.74 New Machine (With Brake): Production time/year = (40,000 pieces/yr)(2.05 min/pc)(1hr/60 min) = 1,366.67 hr/yr Operator salary = $16.50 / hr( 1,366.67 hr / yr) = $22,550.06 / yr Overhead = $4.00 / hr (1,366.67 hr/ yr) = $ 5,466.68 / yr Maintenance = $ 250.00 / yr Total Cost = $28,266.74/ yr Savings per year (with brake) = $34,166.74  $28,266.74 = $5,900 / yr Thus, PW = 0 = X + $5,900 (P/A,15%,5) X =  $5,900 (3.3522) X =  $19,778 Thus the company could afford to pay up to $19,778 for a new brake.
537
General Equation: PW(i'%) = 0 = $450,000  $42,500(P/F,i'%,1) + $92,800(P/F,i'%,2) + $386,000(P/F, i'%, 3) + $614,600(P/F, i'%, 4)  $202,200(P/F, i'%, 5) PW(20%) = $17,561 > 0, ∴i'% > 20% PW(25%) = $41,497 < 0, ∴i'% < 25% Linear interpolation between 20% and 25% yields: i'% = 21.5% > 10%, so the new product line appears to be profitable.
111
Chapter 5 Solutions
However, due to the multiple sign changes in the cash flow pattern, the possibility of multiple IRRs exists. The following graph of PW versus i indicates that multiple IRRs do not exist for this problem.
PW(i%) $400,000 $300,000
IRR = 21.5%
$200,000 $100,000 $0 10%
$100,000
20%
30%
40%
50%
60%
70%
80%
90%
100%
$200,000 $300,000 $400,000
538
(a) 15.2% (b) 18.8% (c) 21.5% (d) 20%
539
The following cash flow diagram summarizes the known information in this problem. F = $54,000
3251993 1302010 i'% = ? per month
P = $13,500 The value of N is the number of time periods separating P and F. If monthly compounding is assumed, N equals (9 months in 1993) + (192 months from 1994 through 2009) + (1 month in 2010) = 202 months (to the nearest integer). Therefore, we can determine the unknown interest rate using:
Chapter 5 Solutions
112
F = P (1+ i'%)N; $54,000 = $13,500 (1+ i'%)202 Using logarithms, i'% per month = 0.69%. The nominal rate of interest = (12)(0.69%) = 8.28% per year and the effective annual interest rate = [(1.0069)12  1](100%) = 8.6% per year. 540
(a)
PW(i'%) = 0 =  $23,000  $1,200 (P/A,i'%,4)  $8,000 (P/F,i'%,4) + $5,500 (P/A,i'%,11)(P/F,i'%,4) +$33,000 (P/F,i'%,15) By linear interpolation, i'% = IRR = 10%
(b)
FW (12%) =  $23,000(F/P,12%,15)  $1,200(F/A,12%,4)(F/P,12%,11)  $8,000(F/P,12%,11) + $5,500 (F/A,12%,11) + $33,000 =  $23,000(5.4736)  $1,200(4.7793)(3.4785)  $8,000(3.4785) + $5,500 (20.6546) + $33,000 = $27,070.25
(c)
⏐$23,000  $1,200(P/A,12%,4)  $8,000(P/F,12%,4)⏐ (F/P,i'%,15) = $5,500(F/A,12%,11) + $33,000 [$23,000 + $1,200(3.0373) + $8,000(0.6355)] (F/P,i'%,15) = $5,500(20.6546) + $33,000 15 $31,728.76 (1 + i') = $146,600.30 i' = ERR = 0.1074 or 10.74%
541
$20,000 = $250 [F/A,i'%,(7621)] (F/P,i'%,1) By linear interpolation, i = 1.24 % per year
542
PW(15%) =  $1,500 + $200(P/F,15%,1) + $400(P/F,15%,2) + $450(P/F,15%,3) + $450(P/F,15%,4) + $600(P/F,15%,5) + $900(P/F,15%,6) + $1,100 (P/F,15%,7) PW(15%) =  $1,500 + $200(0.8696) + $400(0.7561) + $450(0.6575) + $450(0.5718) + $600(0.4972) + $900(0.4323) + $1,100(0.3759) = $630.43 ≥ 0, therefore the project is acceptable.
113
Chapter 5 Solutions
FW(15%)
=  $1,500 (F/P,15%,7) + $200 (F/P,15%,6) + $400 (F/P,15%,5) + $450 (F/P,15%,4) + $450 (F/P,15%,3) + $600 (F/P,15%,2) + $900 (F/P,15%,1) + $1,100 FW(15%) =  $1,500 (2.66) + $200 (2.3131) + $400 (2.0114) + $450 (1.749) + $450 (1.5209) + $600 (1.3225) + $900 (1.15) + $1,100 = $1,677.14 ≥ 0, therefore the project is acceptable. AW(15%) = PW (A/P,15%,7) = $630.43 (0.2404) = $151.56 ≥ 0, therefore the project is acceptable. IRR: PW(i'%) = 0 =  $1,500 + $200 (P/F,i'%,1) + $400 (P/F,i'%,2) + $450 (P/F,i'%,3) + $450 (P/F, i'%, 4) + $600 (P/F, i'%, 5) + $900 (P/F, i'%, 6) + $1,100 (P/F, i'%, 7) By linear interpolation, IRR = 24.9% ≥ 15%, therefore the project is acceptable. θ
Simple Payback Period:
∑ (cash flow)
k
≥ $1,500 when θ = 4 years < 5 years,
k=1
Therefore, the project is acceptable. θ'
Discounted Payback Period:
∑ (cash flow)
k
(P/F,15%,k) ≥ $1,500
k=1
We must find the PW of all cash flows up to year k. Because θ' ≥ θ, the Discounted Payback Period is at least 4 years. PW04(15%) =  $1,500 + $200(P/F,15%,1) + $400(P/F,15%,2) + $450(P/F,15%,3) + $450 (P/F,15%,4) =  $1,500 + $200(0.8696) + $400(0.7561) + $450(0.6575) + $450(0.5718) = $470.46 < 0 thus θ' > 4 years PW05(15%) = $470.46 + $600 (P/F,15%,5) = $470.46 + $600 (0.4972) = $172.14 < 0 thus θ' > 5 years PW06(15%) = $172.14 + $900 (P/F,15%,6) = $172.14 + $900 (0.4323) = $216.93 > 0 thus θ' = 6 years ≤ 6 years Therefore the project is acceptable. ERR: ⏐$1,500(F/P,i'%,7)⏐= $200 (F/P,15%,6) + $400 (F/P,15%,5) + $450(F/P,15%,4) + $450 (F/P,15%,3)
Chapter 5 Solutions
114
+ $600 (F/P,15%,2) + $900 (F/P,15%,1) + $1,100 $1,500(F/P,i'%,7) = $200(2.3131) + $400(2.0114) + $450(1.749) + $450(1.5209) + $600(1.3225) + $900(1.15) + $1,100 7 i' = 0.209 or 20.9% $1,500(1 + i') = $5,667.14 Since ERR = 20.9% ≥ 15%, the project is acceptable. 543
(a)
(b)
544
θ = 4 years thus θ' ≥ 4 years PW04(25%) = $500  $200(P/F,25%,1) + $500(P/F,25%,3) + $500(P/F,25%,4) = $199.2 < 0, thus θ' > 4 years PW05 (25%) = $199.2 + $500 (P/F,25%,5) = $35.35 < 0, thus θ' > 5 years PW06(25%) = $35.35 + $500 (P/F,25%,6) = $95.70 > 0, thus θ' = 6 years 0 = $500  $200 (P/F,i'%,1) + $500 (P/A,i'%,4)(P/F,i'%,2) By linear interpolation, i'% = 29.4% per year
(a) PW(12%) = P + $90,000 (P/A, 12%, N) + $7,000 (P/F, 12%, N) = 0
Affordable Price (P)
N years 5 6 7 8 9 10
P (= affordable price) $328,403.80 $373,572.20 $413,908.10 $449,911.30 $482,062.20 $510,772.00
$550,000 $500,000 $450,000 $400,000 $350,000 $300,000 5
545
6
7 8 Useful Life (N)
9
10
(b)
θ ≈ $344,000 / $90,000 ≈ 4 years
(a)
PW(15%) = $100  $50 (P/F, 15%, 1) + [$20 (P/A, 15%, 4) + $100 (P/G, 15%, 4)](P/F, 15%, 2) = $100  $50(0.8696) + [$20(2.8550) + $100 (3.786)](0.7561)
115
Chapter 5 Solutions
= $185.95 ≥ 0 Yes, this project is financially profitable. EOY 0 1 2 3 4 5
(b)
Cash Flow  $100  50 0 20 120 220
Balance  $100  150  150  130  10 210
Balance becomes positive at the end of year 5. Thus, θ = 5 years. (c)
Since θ = 5 years, θ' ≥ 5 years PW05(15%) = $100  $50(P/F,15%,1) + [$20 (P/A,15%,3) +$100 (P/G,15%,3)] (P/F,15%,2) PW05(15%) = $100  $50(0.8696) + [$20(2.2832) + $100(2.071)](0.7561) = $47.63 > 0, thus θ' = 5 years
546
(a)
PW(i'%) = 0 = $100,000 + $20,000 (P/A,i'%,5) + $10,000 (P/G,i'%,5) + $10,000 (P/F,i'%,5) PW(20%) = $12,891 > 0, ∴ i'% > 20% PW(25%) = $897 < 0, ∴ i'% < 25% By linear interpolation, i'% = IRR = 24.7%
(b)
(c)
547
EOY 1 2 3 4
Cumulative Cash Flow $100,000 + $20,000 = $80,000 < 0 80,000 + 30,000 = 50,000 < 0 50,000 + 40,000 = 10,000 < 0 10,000 + 50,000 = 40,000 > 0 ∴θ = 4 years
Although this project is profitable (IRR > MARR), it is not acceptable since θ = 4 years is greater than the maximum allowable simple payback period of 3 years.
IRR method: PW(i'%) = 0 = $500,000(P/F,i'%,1) + $300,000(P/F,i'%,2) + [$100,000 + $100,000(P/A,i'%,7) + $50,000(P/G,i'%,7)](P/F,i'%,3)  $2,500,000 (P/F,i'%,4)
Chapter 5 Solutions
116
i'% 1 2 3 4 5
Present Worth $103,331.55 63,694.68 30,228.14 2,175.18 21,130.28
i'% 30 31 32
Present Worth $12,186.78 5,479.09 1,182.76
There are two internal rates of return: 4.9% and 31.2% per year. ERR Method: ⏐$2,400,000⏐(P/F,8%,4)(F/P,i'%,10) = $500,000(F/P,8%,9) +$300,000(F/P,8%,8) + $100,000(F/P,8%,7) + $150,000(P/A,8%,6)(F/P,8%,6) + $50,000(P/G,8%,6)(F/P,8%,6) After solving, the external rate of return is 7.6% per year. 548 (a)
2 sign changes → 0,1 or 2 internal rates of return PW(i) = $1,000  $5,000 (1+i)1 + $6,000 (1+i)2 Let x = 1/(1+i) PW(i%) = 1 − 5x − 6x 2 1,000 From the quadratic formula: x = 1/2 and 1/3 If x = 1/2; 1/2 = 1/(1+i); i = 1.00, or 100% per year If x = 1/3; 1/3 = 1/(1+i); i = 2.00, or 200% per year
(b)
549
⏐$5,000(P/F,10%,1)⏐(F/P,i'%,2) = $1,000(F/P,10%,2) + $6,000 $5,000(0.9091)(F/P,i'%,2) = $1,000(1.21) + $6,000 i' = 0.259 or 25.9% per year $4,545.50 (1+i')2 = $7,210
The general equation to find the internal rate of return is: PW(i'%)= 0 = $520,000 + $200,000 (P/A, i'%, 10)  $1,500,000 (P/F, i'%, 10) (a)
i'%
PW(i'%)
i'%
PW(i'%)
0 0.5 1 4 10
 $ 20,000 0 10,250 89,000 131,000
15 20 25 30 40
$113,000 76,000 33,000  10,350  89,100
i'% = 1/2% and 28.8% per year.
117
Chapter 5 Solutions
$150,000 $100,000 PW(i%)
IRR = 28.8%
$50,000 $0 5%
$50,000 $100,000
10%
15%
20%
25%
30%
35%
IRR = 0.5%
i%
(b) Assume ε = 20% per year.
⏐$520,000$1,500,000(P/F,20%,10)⏐(F/P,i'%,10)=$200,000(F/A,20%,10) [$520,000 + $1,500,000(0.1615)](F/P,i'%,10) = $200,000(25.9587) i' = 0.2115 or 21.15% $762,250 (1+i')10 = $5,191,740 ERR > 20%, therefore the project is economically acceptable. 550
(a)
In all three cases, IRR = 15.3%. This is true for EOY 0 as a reference point in time, and also for EOY 4 as a reference point in time.
(b)
PW1(10%) = $137.24 at EOY 0 PW2(10%) = $137.24 at EOY 4 at EOY 0 PW2(10%) = $93.73 PW3(10%) = $686.18 at EOY 4 PW3(10%) = $468.67 at EOY 0 Select (3) to maximize PW(10%). However, the PW(IRR=15.3%) would be zero for all three situations.
551 (a) PW = 16X (P/F,15%,16) + 16Y (P/A, 15%,11) (P/F,15%,4) – 2Y (P/G,15%,11) (P/F,15%,4) = 16X (0.1069) + 16Y (5.2337) (0.5718) – 2Y (19.129) (0.5718) = 1.710X + 26.006Y (b) AW = PW (A/P,15%,20) = 1.710X + 26.006Y (0.1598) = 0.2732X + $4.1558Y 552
(a) PW = 23Y (P/A,15%,102) (P/F,15%,2) – 3Y (P/G,15%, 102) (P/F,14%,2)
Chapter 5 Solutions
118
+ 5Y (P/A,15%, 1710) (P/F,15%,10) + 2Y (P/G,15%,1710) (P/F,15%,10) PW = 23Y (4.4873) (0.7561) – 3Y (12.481) (0.7561) + 5Y (4.1604) (0.2472) + 2Y (10.192) (0.2472) = 59.91Y (b) A = 59.91 Y (A/P,15%,17) = 59.91Y (0.1654) = 9.908Y 553
(a)
Let Q = monthly payment at age 65 0.8Q = monthly payment at age 62 (20% penalty for early retirement) Let i = 0.5% per month (6% per year) How long does it take for the “deferred” full payment plan to catch up to the “early” retirement plan in terms of the future wealth of your uncle? 0.8Q (F/A,1/2%,N) = Q(F/A,1/2%,N36) ( F / A,1/ 2%, N  36) 0.8 = ( F / A,1/ 2%, N)
(Basic Equation)
By trial and error, it takes N ≈ 348 months (past age 62) before the “deferred” plan overtakes the “early” plan. So, it takes 29 years past age 62 before Social Security retirement at age 65 is preferred. If your uncle's health is good and he has longevity in his family tree, he should probably start drawing social security payments at age 65. (b)
At 1.5% per month, you can't live long enough (practically speaking) to catch up to the early retirement option. So take Social Security starting at age 62! The basic calculation is shown below: F / A,1.5%, N  36) ( (Basic Equation) 0.8 = ( F / A,1.5%, N) By trial and error: age 85, 0.8 > 0.578; age 100, 0.8 > 0.584; and age 120, 0.8 > 0.585.
(c)
At i = 0% per month, the basic equation is: ( F / A,0%, N  36) 0.8 = ( F / A,0%, N) By trial and error, at age 77 0.8 ≈ 0.795, or basic breakeven. But at age 85, FWearly = $220,800 and FWdeferred = $240,800. Thus, if your uncle expects to live past age 77, deferring social security payments to the regular age of 65 years is preferred at i = 0% per month.
119
Chapter 5 Solutions
554
$10,000,000
2000
2004 2005 2006
2009
2014 ~ Forever
$250,000/yr
X X Amount at July 2004: $100,000(1.05)4  $3,000,000 = $12,155,000 – $3,000,000 = $9,155,000 $9,155,000
2004 2005
2009
2014
$250,000/yr X
X
$9,155,000 = $250,000(P/A,5%, ∞ ) + X(A/F,5%,5)(P/A,5%, ∞ ) = $250,000 (20) + 0.181X (20) or X = $1,147,790 every 5 years 555
(a)
AW(15%) = $710,000 (A/P,15%,5) + $198,000 + $100,000 (A/F,15%,5) = $1,828 per year. Yes, it’s a good investment opportunity.
(b)
IRR: 0 = $710,000 (A/P, i',5) +$198,000 +$100,000(A/F, i',5) so i' = 15.3% θ = 4 years; θ' = 5 years
(c) Other factors include sales price of reworked units, life of the machine, the company’s reputation, and demand for the product.
Chapter 5 Solutions
120
Solutions to Spreadsheet Exercises 556 Desired Ending Balance Current Age Age at Retirement
$ 250,000 25 60 Interest Rate per Year 8% $ 2,207 $ 3,420 $ 5,463 $ 9,207
4% N
5 10 15 20
$ $ $ $
4,458 6,003 8,395 12,485
$ $ $ $
12% 1,036 1,875 3,470 6,706
The formula used in cell C7 is: =PMT(C$6,$D$3$D$2$B7,,$D$1) Note that this uses the fv parameter of the PMT function instead of the pv parameter. The formula in cell C7 was copied over the range C7:E10. The trend in the table shows that as the interest rate increases, less has to be saved each year. Also, the longer Jane delays the start of her annual savings, the larger the annual deposit will have to be.
121
Chapter 5 Solutions
557 MARR = Capital Investment = Market Value = Useful Life = Annual Savings = Annual Expense = EOY 0 1 2 3 4 5 5
11% $ $ $
10,000 2,000 5
$
5,311
$
3,000
$ $ $ $ $ $ $
Cash Flow (10,000) 2,311 2,311 2,311 2,311 2,311 2,000
EOY 0 1 2 3 4 5
Present Value = Annual Worth = Future Worth = Internal Rate of Return =
$ $ $ $ $ $
Cash Flow (10,000) 2,311 2,311 2,311 2,311 4,311
$ $ $
(271.88) (73.56) (458.13) 10.0%
From Example 511, the IRR = 10%. Therefore, any MARR > 10% will result in negative equivalent worth values. The following table displays the results of different MARRs on the profitability measures.
Present Worth Annual Worth Future Worth
Chapter 5 Solutions
MARR = 15% MARR = 5% $ (1,258.82) $ 1,572.47 $ (375.52) $ 363.20 $ (2,531.93) $ 2,006.92
122
558 MARR = Reinvestment rate = Capital Investment = Market Value = Useful Life = Annual Savings = Annual Expense = EOY 0 1 2 3 4 5 5
20% 20% $ $ $ $
25,000 5,000 5 8,000
$
$ $ $ $ $ $ $
Cash Flow (25,000) 8,000 8,000 8,000 8,000 8,000 5,000
EOY 0 1 2 3 4 5
Present Value = Annual Worth = Future Worth = Internal Rate of Return = External Rate of Return =
$ $ $ $ $ $
Cash Flow (25,000) 8,000 8,000 8,000 8,000 13,000
$ $ $
934.28 312.41 2,324.80 21.58% 20.88%
This spreadsheet was created using the spreadsheet for Example 511. The following table displays the results of different MARRs on the profitability measures.
Present Worth Annual Worth Future Worth IRR ERR
MARR = 18% MARR = 22% $ 2,202.91 $(240.89) $ 704.44 $ (84.12) $ 5,039.73 $(651.04) 21.58% 21.58% 20.88% 20.88%
The original recommendation is unchanged for a MARR = 18%. However, the recommendation does change for MARR = 22% (which is greater than the IRR of the project's cash flows). Note that the ERR is unaffected by changes in the MARR. This is because 1) the reinvestment rate was assumed to remain at 20%, and 2) there is only a single net cash outflow occurring at t=0.
123
Chapter 5 Solutions
559 MARR = Capital Investment = Market Value = Useful Life = Net Annual Savings = EOY 0 1 2 3 4 5 5
20% $ $ $
28,750 5,000 5
$
8,000
$ $ $ $ $ $ $
Cash Flow (28,750) 8,000 8,000 8,000 8,000 8,000 5,000
EOY 0 1 2 3 4 5
$ $ $ $ $ $
Cash Flow (28,750) 8,000 8,000 8,000 8,000 13,000
IRR = The following IRR values were determined by changing the values of the capital investment (cell B2) and market value (cell (B3) in the spreadsheet. Capital Investment = (1+0.15)*25000 = (10.15)*25000
IRR 15.76% 29.03%
Market Value = (1+0.15)*5000 = (10.15)*5000
IRR 22.06% 21.08%
A change in the capital investment amount has a larger impact on IRR. This is a result of both the magnitude of the cash flow and its timing.
Chapter 5 Solutions
124
15.76%
Solutions to Case Study Exercises 560
Average number of wafers per week: (10 wafers/hr)(168 hr/wk)(0.90) = 1,512 Added profit per month: (1,512 wafers/wk)(4.333 wk/month)($150/wafer) = $982,724 PW(1%) = $250,000  $25,000(P/A, 1%, 60) + $982,724(P/A, 1%, 60) = $42,804,482 The increase in CVD utilization serves to make the project even more attractive.
561
Average number of wafers per week: (15 wafers/hr)(168 hr/wk)(0.80) = 2,016 New breakeven point: X=
$1,373,875 = $3.50 / wafer (2,016)(4.333)(44.955)
$3.50/$100 = 0.035 extra microprocessors per wafer. The retrofitted CVD tool would significantly reduce the breakeven point. 562
Average number of wafers per week: (10 wafers/hr)(150 hr/wk)(0.90) = 1,350 Added profit per month: (1,350 wafers/wk)(4.333 wk/month)($150/wafer) = $877,433 PW(1%) = $250,000  $25,000(P/A, 1%, 60) + $877,433(P/A, 1%, 60) = $38,071,126 New breakeven point: X=
$1,373,875 = $5.225 / wafer (1,350)(4.333)(44.955)
$5.225/$100 = 0.05225 extra microprocessors per wafer.
125
Chapter 5 Solutions
Solutions to FE Practice Problems 563
i / mo. = 9%/12 = ¾% per month; N = 4 x 12 = 48 months A = $8,000 (A/P, ¾% per month, 48 months) = $8,000 (.0249) = $199.20 Select (d)
564
EOY 0 1 2 3 4
Cumulative PW (i = 12%) $300,000 $300,000 + $111,837.50(P/F,12%,1) = $200,140.30 200,140.30 + $111,837.50(P/F,12%,2) = $110,983.45 $110,983.45 + $111,837.50(P/F,12%,3) = $31,377.52 $31,377.52 + $111,837.50(P/F,12%,4) = $39,695.21 > 0 ∴ Ø' = 4
Select (a) 565
$14,316 = X(A/P,8%, ∞ ) $14,316 = X (0.08); X = $178,950 Select (a)
566
Find i% such that PW(i%) = 0 0 = $3,345 + $1,100(P/A,i'%,4) PW(10%) = $141.89 tells us that i'% > 10% PW (12%) = $3.97 tells us that i'% < 12% (but close!)
∴ IRR = 11.95% Select (b)
567
VN = C(P/F,i%,N) + rZ(P/A,i%,N) = $981 N = 8 periods r = 10% per period ($1,000/$100 = 10%) C = Z = $1,000 $981= $1,000(P/F,i%,8) + (.10)($1,000)(P/A,i%,8) $981 = $1,000(P/F,i%,8) + $100(P/A,i%,8) try i = 10%; $466.50 + $533.49 = $999.99 try i = 12%; $403.90 + $496.76 = $900.66 by observation, i% is > 10% but very close to 10% ∴ rate of return = 10.35%
Chapter 5 Solutions
126
Select (c) 568
AW = $5,123 + $1,100 (P/A,10%,20) (P/A,10%,20) = 4.6573 Using the interest tables, (P/A,10%,6) = 4.3553 and (P/A,10%,7) = 4.8684. Thus, 6 < N < 7. Using linear interpolation we find that N ≈ 6.5. ∴ N = Ø' = 7 Select (a)
569 AW = ($5,000 + $500 (A/G.1%24))*(F/P,1%,12) = ($5,000 + $500*11.02337)*1.0100*12.6825 = $134,649
Select (e) 570
i = 12%
f = 5% $7000[1 − (P/F, 12%, 6)(F/P, 5%, 6)] 0.12 − 0.05 $7000[1 − (0.5066)(1.3401)] = $9,000 + 0.07 = $41,110.53 AW(12%) = $41,110.53 (A/P, 12%, 6) = $41,110.53(0.2432) = $9,998.08 PW (12%) = $9,000 +
Select (c)
571
For simplicity, assume a loan amount of $1,000. The yearly payment due on this loan is $1,000 (A/P, 6%, 5) = $1,000 (0.2374) = $237.40. The application fee due is $1,000 (0.1367) = $136.70. Thus the borrower walks away with $1,000  $136.70 = $863.30. The effective interest rate being charged can be found by solving the following equivalence equation for i': $863.30 = $237.40 (P/A,i'%,5) (P/A,i'%,5) = 3.6365 Using the interest tables, (P/A,10%,5) = 3.7908 and (P/A,12%,5) = 3.6048. Thus, 127
Chapter 5 Solutions
10% < i'% < 12%. Using linear interpolation we find that i' = 11.65%. Select (a) 572
P = $8,000 (P/A,12%,4)(P/F,12%,16) = $8,000 (3.0373) (0.1631) = $3,963.07 Select (d)
Chapter 5 Solutions
128
Solutions to Chapter 6 Problems 61 (a) Acceptable alternatives are those having a PW(15%) ≥ 0. Alt I: PW (15%) = $100,000 + $15,200(P/A,15%,10) + $10,000(P/F,15%,10) = $21,242.24 Alt II: PW(15%) = $152,000 + $31,900 (P/A,15%,10) = $8,099.72 Alt III: PW(15%) = $184,000 + $35,900(P/A,15%,10) + $15,000(P/F,15%,10) = $117.08 Alt IV: PW(15%) = $220,000 + $41,500(P/A,15%,10) + $20,000(P/F,15%,10) = $6,775.80 Thus, alternatives II, III and IV are economically infeasible, and Alternative II should be selected because it has the only positive PW value.
62
(b)
If total investment capital is limited to $200,000, Alternative II should be selected since it is within the budget and is also economically feasible.
(c)
Rule 1; the net annual revenues vary among the alternatives.
Present Worth Method, MARR = 10% per year PWD1 (10%) = $600,000  ($68,000 + $40,000 + $660,000 + $12,000)(P/A,10%,10) = $5,392,788 PWD2 (10%) = $760,000  ($68,000 + $45,000 + $600,000 + $15,000)(P/A,10%,10) = $5,233,268.80 PWD3 (10%) = $1,240,000  ($120,000 + $65,000 + $420,000 + $25,000)(P/A,10%,10) =  $5,111,098.00 PWD4 (10%) = $1,600,000  ($126,000 + $50,000 + $370,000 + $28,000)(P/A,10%,10) =  $5,127,000.40 Select Design D3 to minimize the present worth of costs. Future Worth Method, MARR = 10% per year FWD1 (10%) = $600,000 (F/P, 10%,10) – ($68,000 + $40,000 + $660,000 + $12,000) (F/A,10%,10) = $13,987,392 FWD2 (10%) = $760,000 (F/P,10%,10) = ($68,000 + $45,000 + $600,000 + $15,000) (F/A,10%,10) = $13,573,639 FWD3 (10%) = $1,240,000 (F/P,10%,10) – ($120,000 + $65,000 + $420,000 + $25,000) (F/A,10%,10) = $13,256,750 FWD4 (10%) = $1,600,000 (F/P,10%,10) – ($126,000 + $50,000 + $370,000 + $28,000) (F/A,10%,10) = $13,297,988 Select Design D3 to minimize the future worth of costs. Annual Worth Method, MARR = 10% per year
131
Chapter 6 Solutions
AWD1 (10%) = $600,000 (A/P,10%,10) – ($68,000 + $40,000 + $660,000 + $12,000) = $877,620 AWD2 (10%) = $760,000 (A/P,10%,10) – ($68,000 + $45,000 + $600,000 + $15,000) = $851,652 AWD3 (10%) = $1,240,000 (A/P,10%,10) – ($120,000 + $65,000 + $420,000 + $25,000) = $831,748 AWD4 (10%) = $1,600,000 (A/P,10%,10) – ($126,000 + $50,000 + $370,000 + $28,000) = $834,320 Select Design D3 to minimize the annual worth of costs. 63
64
(a)
AW1 (12%) = $2,500(A/P,12%,5) + $750 + $2,000 (A/F,12%,5) = $2,500(0.2774) + $750 + $2,000(0.1574) = $371.30
(b)
AW2 (12%) = $4,000 (A/P,12%,5) + $1,200 + $2,000 (A/F,12%,5) = $405.20
(c)
AW∆ (12%) = [$4,000  ($2,500)](A/P,12%,5) + ($1,200  $750)
(d)
Alternative 2 should be selected. In parts (a) and (b), Alternative 2 was shown to have the largest AW. In part (c), the annual worth of the incremental investment needed for Alternative 2 was shown to be positive meaning that it earns more than the required 12% per year.
Assume all units are produced and sold each year. AWA(20%) = $30,000 (A/P,20%,10) + 15,000 ($3.10  $1.00)  $15,000 + $10,000 (A/F,20%,10) = $9,730 AWB(20%) = $6,000 (A/P,20%,10) + 20,000 ($4.40  $1.40)  $30,000 + $10,000 (A/F,20%,10) = $16,075 AWC(20%) = $40,000 (A/P,20%,10) + 18,000 ($3.70  $0.90)  $25,000 + $10,000 (A/F,20%,10) = $16,245 Select Design C to minimize the annual worth.
65
FW Method: FWA = $170,000 (F/P,10%,10) + ($114,000  $70,000) (F/A,10%,10) = $260,317 FWB = $330,000 (F/P,10%,10) + ($147,000  $79,000) (F/A,10%,10) = $227,822 FWC = $300,000 (F/P,10%,10) + ($130,000  64,000) (F/A,10%,10) = $273,758 Choose C
PW Method:
Chapter 6 Solutions
132
PWA = $170,000 + ($114,000  $70,000) (P/A,10%,10) = $100,362 PWB = $330,000 + ($147,000  $79,000) (P/A,10%,10) = $87,833 PWC = $300,000 + ($130,000  $64,000) (P/A,10%,10) = $105,544 Choose C AW Method: AWA = $170,000 (A/P,10%,10) + $114,000  $70,000 = $16,341 AWB = $330,000 (A/P,10%,10) + $147,000  $79,000 = $14,309 AWC = $300,000 (A/P,10%,10) + $130,000  $64,000 = $17,190 Choose C 66
Wet Tower, Mechanical Draft AW(12%) = $3,000,000 (A/P,12%,30) ⎛ 200hp ⎞ ⎛ 0.746kw ⎞ ⎟⎟ (8,760 hr/yr) ($0.022/kWh)  40 ⎜ ⎟ ⎜⎜ ⎝ 0.9 ⎠ ⎝ hp ⎠ ⎛ 150hp ⎞ ⎛ 0.746kw ⎞ ⎟⎟ (8,760 hr/yr) ($0.022/kWh)  $150,000  40 ⎜ ⎟ ⎜⎜ ⎝ 0.9 ⎠ ⎝ hp ⎠ = $372,300  $1,277,948  $479,230  $150,000 = $2,229,478/yr.
Wet Tower, Natural Draft AW(12%) = $8,700,00(A/P,12%,30) ⎛ 150hp ⎞ ⎛ 0.746kw ⎞ ⎟⎟ (8,760 hr/yr) ($0.022/kWh)  20 ⎜ ⎟ ⎜⎜ ⎝ 0.9 ⎠ ⎝ hp ⎠  $1,076,670  $479,230 = $1,558,900/yr. Dry Tower, Mechanical Draft AW(12%) = $5,100,000(A/P,12%,30)  $1,277,948/2 ⎛ 100hp ⎞ ⎛ 0.746kw ⎞ ⎟⎟ (8,760 hr/yr) ($0.022/kWh)  $170,000  40 ⎜ ⎟ ⎜⎜ ⎝ 0.9 ⎠ ⎝ hp ⎠ = $2,080,858/yr. Dry Tower, Natural Draft AW(12%) = $9,000,000(A/P,12%,30)  $638,974  $120,000 = $1,875,874/yr. The wet cooling tower with natural draft heat removal from the condenser water is the mose economical (i.e., least costly) alternative. Noneconomic factors include operating considerations and licensing the plant in a given location with its unique environmental characteristics. 67
PWA(20%) = $28,000 + ($23,000  $15,000)(P/A,20%,10) + $6,000(P/F,20%,10) = $6,509 133
Chapter 6 Solutions
PWB(20%) = $55,000 + ($28,000  $13,000)(P/A,20%,10) + $8,000(P/F,20%,10) = $9,180 PWC(20%) = $40,000 + ($32,000  $22,000)(P/A,20%,10) + $10,000(P/F,20%,10) = $3,540 Select Alternative B to maximize present worth. Note: If you were to pick the alternative with the highest total IRR, you would have incorrectly selected Alternative A. 68
Assume zero salvage value for both alternatives. A: Incandescent lighting system 10 kW (8,760 hr/yr) = 87,600 kWh/yr.($0.045/kWh) = $3,942/yr. B: Fluorescent lighting system 4.5 kW (8,760 hr/yr) = 39,420 kWh/yr.($0.045/kWh) = $1,773/yr. Net PWA(15%) = $3,942(P/A,15%,5) = $13,214.20 Net PWB(15%) = $11,000  $1,773(P/A,15%,5) = $16,943.45 Therefore, keep existing incadescent lighting system.
69
FWA(10%) = $780,000(F/P,10%,10) + $138,060(F/A,10%,10) = $177,231 FWB(10%) = $1,840,000(F/P,10%,10) + $311,000(F/A,10%,10) = $184,123 Select Alternative B to maximize future worth.
610
Rank alternatives by increasing capital investment: A,B,C,D and E. Since the IRR on Equipment A > MARR, it is an acceptable base alternative. Therefore, IRR∆ on ∆(BA) must be examined. ∆(BA) AW∆ (i'∆%) = ($50,000 + $38,000)(A/P, i'∆%,10) + ($14,100  $11,000) = 0 = $12,000 (A/P, i'∆%,10) + $3,100 = 0 i'∆ = 22.4% > MARR, therefore select B. ∆(CB) AW∆ (i'∆%) = $5,000 (A/P, i'∆%,10) + $2,200 = 0 i'∆ = 42.7% > MARR, therefore select C.
∆(DC) AW∆ (i'∆%) = $5,000 (A/P, i'∆%,10) + $500 = 0
Chapter 6 Solutions
134
Since Σ cash outflows = Σ cash inflows, i'∆ = 0 % < MARR, therefore keep C. ∆(EC) AW∆ (i'∆%) = $15,000 (A/P, i'∆%,10) + $2,900 = 0 i'∆ = 14.2% < MARR, therefore keep C as best. Answer: Select C. 611
The ranking of the alternatives by increasing capital investment is A, C, and B. Is Alternative A an acceptable base alternative? PWA (i'%) = $200,000 + $90,000 (P/A, i'%,6) = 0 By trial and error, i' = IRRA = 38.7% > MARR, ∴ acceptable as base alternative. Or, PWA (15%) = $200,000 + $90,000 (P/A,15%,6)= $140,605 which is greater than zero, ∴acceptable as base alternative. ∆(CA)  $12,500  105,000 32,500
∆(BA) $30,000 18,000 18,000
IRR∆ Is increment justified? Current best design
11.4% No A
55.8% Yes B
PW∆(15%) Is increment justified? Current best design
$9,068 No A
$38,121 Yes B
∆ Capital Investment ∆ Net cash flow (year 1) ∆ Net cash flow (years 26)
Both methods result in the decision to invest in design B. 612
Capacity (units/year) A1:
(7,400 units/month)(12 months/year)[1 + 4(0.007)] = 91,286 units/year
A2:
7,400(12)[1 + (6.5)(0.007)] = 92,840 units/year
Therefore, either alternative will produce the maximum 91,000 units per year that are estimated to be sold. Alternative A1 AW(18%) = $260,000 (A/P,18%,5)  $9,400 + (91,000  88,800)($48.20) = $13,492 Alternative A2 AW(18%) = $505,000 (A/P,18%,5) + $6,200 + (91,000  88,800)($48.20) 135
Chapter 6 Solutions
= $49,259 Yes (the project should be implemented); select A1; because it maximizes the AW value [and A2 is not economically justified at (demand) = 91,000 units/year]. 613
(a) AW Method AWA(12%) = $85,600 (A/P,12%,7)  $7,400 = $26,155 AWB(12%) = $63,200 (A/P,12%,7)  $12,100 = $25,947 AWC(12%) = $71,800 (A/P,12%,7)  $10,050 = $25,781 Select Design C to minimize annual equivalent cost. (b) ERR Method The ranking of alternatives based on increasing capital investment is B, C, and A. Since these are cost alternatives, the one with the least capital investment (B) is the base alternative. ∆ (C  B)  $8,600 2,050 13.4% Yes C
∆ Capital investment ∆ Annual expense savings ERR∆ Is increment justified? Current best alternative
∆ (A  C) $13,800 2,650 9.9% No C
Decision: Select Design C. Note: ERR∆ was found by solving the following equation for i'%: ∆ Capital investment (F/P, i'%,7) = ∆ Annual expense savings (F/A,12%,7) 614
Design A: All components have a 20 year life. Capital Investment Concrete pavement: ($90/ft)(5,280 ft/mi) = $475,200 /mile Paved ditches: 2×($3/ft)(5,280 ft/mi) = $31,680 /mile Box culverts: (3 culverts/mile)($9,000/culvert) = $27,000 /mile Total Capital Investment = $533,880 /mile Maintenance Annual maintenance: $1,800 /mile Periodic cleaning of culverts*: (3 culverts/mile)($450/culvert) = $1,350 /mile every 5 years AWA(6%) = $533,880(A/P,6%,20)  $1,800  $1,350(A/F,6%,5)* = $48,594 /mile
PWA(6%) = $533,880  [$1,800 + $1,350(A/F,6%,5)](P/A,6%,20) = $557,273 /mile
Chapter 6 Solutions
136
* assumes a cleaning also occurs at the end of year 20. Design B: All components have a 10 year life. Capital Investment (Year 0) Bituminous pavement: ($45/ft)(5,280 ft/mi) Sodded ditches: 2×($1.50/ft)(5,280 ft/mi) Pipe culverts: (3 culverts/mile)($2,250/culvert) Total
= $237,600 /mile = $15,840 /mile = $6,750 /mile = $260,190 /mile
Capital Investment (EOY 10) Bituminous pavement: ($45/ft)(5,280 ft/mi) = $237,600 /mile Sodded ditches: 2×($1.50/ft)(5,280 ft/mi) = $15,840 /mile Replacement culverts: (3 culverts/mile)($2,400/culvert) = $7,200 /mile Total = $260,640 /mile Maintenance Annual pavement maintenance: = $2,700 /mile Annual cleaning of culverts: (3 culverts/mile)($225/culvert) = $675 /mile Annual ditch maintenance: 2×($1.50/ft)(5280 ft/mi) = $15,840 /mile Total = $19,215 /mile AWB(6%) = [$260,190 + $260,640(P/F,6%,10)](A/P,6%,20)  $19,215 = $54,595 /mile PWB(6%) = $260,190  $260,640(P/F,6%,10)  $19,215(P/A,6%,20) = $626,126 /mile Select Design A (concrete pavement) to minimize costs. 615
Operating hours/year = (250 days/year) × (6 hrs/day) = 1,500 hours/year (a) Motor A: Annual taxes and insurance = 0.025 ($850) Annual energy expense:
= $21.25 /year
⎛ 10 hp ⎞ ⎛ 0.746 kW ⎞ ⎛ $0.051 ⎞ ⎛ 1,500 hr ⎞ = ⎜ ⎟⎜ ⎟ = $671.40 /year ⎟⎜ ⎟⎜ ⎝ 0.85 ⎠ ⎝ hp ⎠ ⎝ kW  hr ⎠ ⎝ year ⎠ Motor B: Annual taxes and insurance = 0.025 ($700) Annual energy expense:
137
= $17.50 /year
Chapter 6 Solutions
⎛ 10 hp ⎞ ⎛ 0.746 kW ⎞ ⎛ $0.051 ⎞ ⎛ 1,500 hr ⎞ = ⎜ ⎟⎜ ⎟ = $713.36 /year ⎟⎜ ⎟⎜ ⎝ 0.80 ⎠ ⎝ hp ⎠ ⎝ kW  hr ⎠ ⎝ year ⎠ These are cost alternatives. Motor B is the base alternative since it requires the least capital investment.
∆ Capital investment ∆ Taxes and insurance ∆ Annual energy savings
∆(AB) $150.00  3.75 + 41.96
PW(i'∆%) = 0 = $150 + ($41.96  $3.75)(P/A, i'∆%,5) By trial and error, i'∆% = 8.6%. The additional investment required by Motor A is not justified (earns less than the MARR). Select Motor B. (b) PWA(10%) = $850  ($21.25 + $671.40)(P/A, 10%,5) = $3,475.70 PWB(10%) = $700  ($17.50 + $713.36)(P/A, 10%,5) = $3,470.54 Select Motor B to minimize equivalent costs. 616
Compare future worths of the two alternatives at the end of year 5: FW1(10%) = $80,000(F/P,10%,5) + $20,000(F/A,10%,5) + $4,000(F/P,10%,1) + $6,000 = $3,662 FW2(10%) = $80,000(F/P,10%,3) + $33,000(F/A,10%,3) = $2,750 Select Alternative 1 to maximize the future worth measure of profitability. Note: We could have computed the present worth at 10% of ∆ (12). The PW on the difference is $562, so again Alternative 1 would be recommended.
617
(a) Use AW to deal with different useful lives AWx(5%) = $6,000(A/P,5%,12)  $2,500 = $3,176.80 AWy(5%) = $14,000(A/P,5%,18) + $2,800(A/F,5%,18)  $2,400 = $3,497.60 Select Alternative X (can also calculate PW over 36 years and compare) (b) PWx(5%) = $6,000  $2,500(P/A,5%,12)  $8,000(P/A,5%,6)(P/F,5%,12) = $50,767.45 PWy(5%) = $3,497.60(P/A,5%,18) = $40,885.54
Chapter 6 Solutions
138
Select Alternative Y (can also calculate AW over 18 years and compare) 618
(a) Assume repeatability so that AWs can be directly compared (over a 15year study period). AWA(8%) = $1,200(A/P,8%,3)  $160 60hp (0.746 kW/hp)(800 hrs/yr.)($0.07/kWh) 0.90 = $3,410.67 AWB(8%) = $1,000(A/P,8%,5)  $100 60hp (0.746 kW/hp)(800 hrs/yr.)($0.07/kWh) 0.80 = $3,483.70 By a slim margin, select Motor A (b) Increased capital investment of Motor A (relative to Motor B) is being traded off for improved electrical efficiency and lower annual energy expenses.
619 Number of machines needed: D1:
3,450 = 2.40 (3 machines) 2,000(0.8)(0.9)
D2:
2,350 = 1.96 (2 machines) 2,000(0.75)(0.8)
Annual equivalent cost of ownership: D1: $16,000(3)(A/P,15%,6)  $3,000(3)(A/F,15%,6) = $11,653.80 D2: $24,000(2)(A/P,15%,8)  $4,000(2)(A/F,15%,8) = $10,116.00 Annual operating expenses (assume repeatability): D1: $5,000(3) = $15,000 D2: $7,500(2) = $15,000 Total equivalent annual cost: D1: $11,653.60 + $15,000 = $26,653.80 D2: $10,116.00 + $15,000 = $25,116.00
139
Chapter 6 Solutions
Select Machine D2 to minimize total equivalent annual cost. 620
(a) PWA(15%) = $272,000  $28,800(P/A,15%,9) + $25,000(P/F,15%,6)  $66,000(P/A,15%,3)(P/F,15%,6) = $463,758 PWB(15%) = $346,000  $19,300(P/A,15%,9) + $40,000(P/F,15%,9) = $426,720 Select B to minimize equivalent cost. (b) Alternative A is the base alternative because it requires the least capital investment. Year 0 15 6 78 9
∆ (B  A) cash flow $346,000  ($272,000) = $74,000  19,300  ($28,800) = 9,500  19,300  ($28,800 + $25,000) =  15,500  19,300  ($94,800) = 75,500  19,300 + $40,000  ($94,800) = 115,500
PW∆(i'∆%) = 0 = $74,000 + $9,500(P/A, i'∆%,5)  $15,500(P/F, i'∆%,6) + $75,500(P/A, i'∆%,2)(P/F, i'∆%,6) + $115,500(P/F, i'∆%,9) By trial and error, i'∆% = 22.5% > MARR. Therefore the incremental investment is justified and Alternative B should be selected. Note that there are multiple sign changes. It is possible that there are multiple IRR∆s. (c) Again, Alternative A would be the base alternative. Using the incremental cash flows computed in part (b), the ERR∆ is found by solving the following equation: ⏐$74,000  $15,500(P/F,15%,6)⏐ (F/P, i'∆%,9) = $9,500(F/A,15%,5)(F/P,15%,4) + $75,500(F/A,15%,3) + $40,000 $80,701(1 + i'∆)9 = $404,202 ERR∆ = 19.9% > MARR. Therefore, select Alternative B (d) PWL(15%) = $94,800 (P/A,15%,9) = $452,348 Thus, leasing crane A is not preferred to the selected alternative (B), but would be preferred to the purchase of crane A.
Chapter 6 Solutions
140
621 (a) Each longlife light bulb will last 20,000 hrs/5,000 hrs/yr. = 4 years. The annual equivalent cost for each bulb is: $15.95/6(A/P,12%,4)+(0.06kw/0.85)($0.10/kwh)(5,000 hr/yr.) = $0.266(0.3292) + $35.29 = $35.38/year To provide the same level of lighting for one year, five standard light bulbs will be required. The annual equivalent cost for the standard bulbs is: $0.60(5)(A/P,12%,1)+(0.06kw/0.95)($0.10/kwh)(5,000 hr/yr.) = $3.36+$31.58+$34.94 The standard light bulb is less expensive by $0.44 per year. Based only on the economic data given, the standard light bulb should be chosen (especially if several hunderd bulbs are involved). (b) What’s been omitted from the analysis is the expense and the inconvenience of changing a light bulb. If this expense is more than 9 cents per bulb, the longlife bulb is the better choice. Furthermore, if the light bulbs are in a difficult to reach location, the safety associated with changing the longlife bulbs twenty times less frequently than the standard bulbs would quickly make the longlife bulbs the better choice. 622
Assume a common 40 yr. life and use the AW method (PW could be used if 2 life cycles of Boiler A are explicitly considered over a 40 year study period.) AWA(10%) = $50,000(A/P,10%,20) + $10,000(A/F,10%,20)  $9,000 = $14,704 AWB(10%) = $120,000(A/P,10%,40) + $20,000(A/F,10%,40)  $3,000  $300(A/G,10%,40) = $17,962 or PWA(10%) = $143,735 over 40 years; PWB(10%)  $175,580 over 40 years.
623
Assume repeatability. AWA(20%) = $2,000(A/P,20%,5) + ($3,200  $2,100) + $100(A/F,20%,5) = $444.64 AWB(20%) = $4,200(A/P,20%,10) + ($6,000  $4,000) + $420(A/F,20%,10) = $1,014.47 AWC(20%) = $7,000(A/P,20%,10) + ($8,000  $5,100) + $600(A/F,20%,10) = $1,253.60 Select Alternative C to maximize annual worth
141
Chapter 6 Solutions
624
Assume repeatability. (a) Machine A: CR cost/yr = $35,000(A/P,10%,10)  $3,500(A/F,10%,10) = $5,475.05 Maint. Cost/yr = $1,000 $5,475.05 + $1,000 CR and maintenance cost/part = = $0.6475 10,000 parts $16 / hr Labor cost/part = = $5.3333 3 parts / hr Total cost/part (to nearest cent) = $5.98 Machine B: CR cost/yr = $150,000(A/P,10%,8)  $15,000(A/F,10%,8) = $26,799 Maint. Cost/yr = $3,000 $26,799 + $3,000 CR and maintenance cost/part = = $2.98 10,000 parts $20 / hr Labor cost/part = = $3.33 6 parts / hr Total cost/part (to nearest cent) = $6.31 Select Machine A to minimize total cost per part. (b) Machine A: $8 / hr = $2.67 3 parts / hr. (other costs remain unchanged) Total cost/part (to nearest cent) = $3.31
Labor cost/part =
Machine B: $10 / hr = $1.67 6 parts / hr. (other costs remain unchanged) Total cost/part (to nearest cent) = $4.65
Labor cost/part =
Select Machine A to minimize total cost per part.
Chapter 6 Solutions
142
625
(a) Assume repeatability. AWA = $20,000(A/P,20%,5)  $5,500 + $1,000(A/F,20%,5) = $12,053.60 AWB = $38,000(A/P,20%,10)  $4,000 + $4,200(A/F,20%,10) = $12,901.30 Select A (b) AWA(20%) = $12,053.60 AWB(20%) = $38,000(A/P,20%,5)  $4,000 +$15,000(A/F,20%,5) = $14,691.20 Select A
626
(a) How many machines will be needed? 6,150/(3,000*0.85*0.95) = 2.54 (3 lathes of type L1 or 4,400/(3,000*0.90*0.90) = 1.81 (2 lathes of type L2) (b) The annual cost of ownership, based on capital investment, is $18,000*3*(A/P,18%,7) = $14,169.60 for lathe L1 and $25,000*2*(A/P,18%,11) = $10,740.00 for L2. Repeatability is assumed. (c) Since the workers are paid for idle time we cannot make any reductions in annual expenses. Thus, annual expenses for 3 L1s equals: 3*$5,000 = $15,000/yr. For 2 L2s, the annual expense equals 2*$9,500 = $19,000/yr. (d) The total equivalent annual cost for the two options equals: $14,169.60 + $15,000 = $29,169.60 for lathe L1 and $10,740.00 + $19,000 = $29,740.00 for lathe L2. Hence, lathe L1 is the preferred choice to minimize equivalent annual choice.
627
(a) Repeatability assumption AWE1(15%) =  $14,000 (A/P,15%,5)  $14,000 + $8,000 (A/F,15%,5) = $16,990 AWE2(15%) = $65,000(A/P,15%,20)  $9,000 + $13,000 (A/F,15%,20) = $19,260 Select Alternative E1 to minimize costs. (b) Coterminated assumption (5year study period) AWE1(15%) =  $16,990; unchanged from Part (a) Imputed market value (MV5) for Alternative E2: MV5 = [$65,000(A/P,15%,20)  $13,000 (A/F,15%,20)](P/A,15%,15) + $13,000 (P/F,15%,15) = $61,590 AWE2(15%) =  $65,000(A/P,15%,5)  $9,000 + $61,590(A/F,15%,5) = $19,256 (slight difference from Part (a) is due to rounding)
143
Chapter 6 Solutions
Select Alternative E1 to minimize costs. The reason AWE2 in Part b is the same as in Part a is the annual expenses are the same over the 20year period in Part a as they are over the 5year period in Part b. 628
[ −$50,000(A P ,10%,25) + $5,000(A F,10%,25)  $1,200] 0.10 = $66,590
CWA(10%) =
CWB(10%) =
−$90,000( A P ,10%,50)  $5,000(P A ,10%,15)(A P ,10%,50)  $1,000 0.10
= $139,183 Select Plan A to minimize costs. 629
[ −$50,000(A P ,10%,20) + $10,000(A F,10%,20)  $9,000] 0.10 = $147,000
CWD1(10%) =
[ −$120,000(A P ,10%,50) + $20,000(A F,10%,50)  $5,000] 0.10 = $170,900 Select Design D1 to minimize costs.
CWD2(10%) =
630
[ −$274,000(A P ,15%,83)  $10,000  $50,000(A F,15%,6)] 0.15 = $378,733
CWL(15%) =
[ −$326,000(A P ,15%,92)  $8,000  $42,000(A F,15%,7)] 0.15 = $404,645
CWH(15%) =
Select Bridge Design L to minimize cost. 631
Alternative E1 PWE1(15%) = $210,000  $31,000 (P/A,15%,6)  $2,000 (P/G,15%,6) + $21,000 (P/F,15%,6) =  $334,118 Alternative E2 Calculate an imputed MV at EOY 6: Capital Recovery Amount = $264,000 (A/P,15%,10) + $38,000 (A/F,15%,10) = $50,742 MV6 =  $50,742  (P/A,15%,4) + $38,000 (P/F,15%,4) = $166,597
Chapter 6 Solutions
144
(1088 . )6 −1 015 . − 0.057 = 0.088 or, 8.8%; (P/A,8.8%,6) = = 4.5128 1057 . 0.088(1088 . )6 $19,000 PWE2(15%) = $264,000 (P/A,8.8%,6) + $166,597 (P/F,15%,6) 1057 . = $273,100
iCR =
Select Alternative E2. 632
Method: Incremental PW Order alternatives by increasing capital investment: ER3, ER1, ER2. Is ER3 an acceptable base alternative? PWER3(12%) = $81,200 + $19,750(P/A,12%,6) = $0.15 ≈ 0. Since PW(MARR=12%) ≥ 0, ER3 is an acceptable base alternative. Analyze ∆ (ER1  ER3) $25,800[1 − ( P / F ,12%,6)( F / P,6%,6)] 0.12 − 0.06 $25,800(0.2814)  $19,750(4.1114) = $17,400 + 0.06 = $22,402 > 0
PW∆(12%)= ($98,600  $81,200) +
The additional capital investment earns more than the MARR. Therefore, design ER1 is preferred to design ER3. Analyze ∆ (ER2  ER1) PW∆(12%) = ($115,000  $98,600) + $29,000(P/A,12%,6) + $150(P/G,12%,6) $25,800[1 − ( P / F ,12%,6)( F / P,6%,6)] 0.12 − 0.06 $25,800(0.2814) = $16,400 + $29,000(4.1114) + $150(8.93) 0.06 = $16,832 < 0 The additional capital investment required by design ER2 has a negative PW (earns less than the MARR). Therefore, design ER1 is preferred to design ER2. Decision: Recommend Design ER1
633
Assume repeatability CWS1(10%) =
−$72,000(A P ,10%,9)  $2,200  $300( A G ,10%,9) + $8,400( A F ,10%,9) 0.10
= $150,927 145
Chapter 6 Solutions
For Alternative S2, the PW of the annual expenses over a single life are: $2,100[1 − (P / F,10%,12)(F / P,5%,12] 0.10 − 0.05 = $17,969
PW(10%) =
CWS2(10%) =
− $90,000 ( A P ,10%,12)  $17,969(A/P,10%,12) + $13,000( A F ,10%,12) 0.10
= $152,414 Alternative S1 is preferred to minimize costs. 634
Assume repeatability. Let: NA = life of blower costing $42,000 NB = life of the more efficient blower H = hours of operation per year C = cost of electricity per kilowatthour X = cost of the more efficient blower Operation Cost of A = OCA =
90 hp (0.746 kW / hp)(H hrs / yr)(C $ / kW  hr) 0.72
AWA(20%) = $42,000 (A/P,20%,NA)  (OCA) Operation Cost of B = OCB =
90 hp (0.746 kW / hp)(H hrs / yr)(C $ / kW  hr) 0.81
AWB(20%) =  X (A/P,20%,NB)  (OCB) Set AWA(20%) = AWB(20%) and solve for X. $42,000 (A/P,20%,NA)  X (A/P,20%, NB) = OCB  OCA 635
This study points out two important conditions that may arise in investment studies. First, the investment of an amount beyond the minimum may be necessary to make a project sufficiently effective to provide enough return to be financially sound. Second, it is necessary to know how far the investing of additional capital should be carried. From a practical viewpoint, if no better opportunity for investment of additional amounts of capital exists than is offered by the increase above the point of a satisfactory rate of return, it is logical to invest the additional increment and obtain this contentment return. This process could continue as long as the return from each additional increment is greater than the contentment (minimum attractive) rate. In this problem a building with 50 stories should be recommended when the MARR is 15%.
Chapter 6 Solutions
146
636
The following example program is written in FORTRAN. REAL REAL REAL REAL REAL REAL REAL REAL REAL REAL REAL
MARR, HP, HPD, DPY, CPKWH, ELECCOST, LIFE ICA, ICB, ICC SVA, SVB, SVC EFFA, EFFB, EFFC LIFEA, LIFEB, LIFEC AWA, AWB, AWC TLIFEA, TLIFEB, TLIFEC TSVA, TSVB, TSVC TAWA, TAWB, TAWC IRRBA, IRRCA, IRRCB, IRRAB, IRRAC, IRRBC EXTERNAL AW, IRR
WRITE (*,'(/A\)') 'Enter MARR (as an integer): ' READ (*,*) MARR MARR = MARR / 100.0 WRITE (*,'(A\)') 'Enter horsepower: ' READ (*,*) HP WRITE (*,'(A\)') 'Enter # of hours/day that motor will be used: ' READ (*,*) HPD WRITE (*,'(A\)') 'Enter # of days/year that motor will be used: ' READ (*,*) DPY WRITE (*,'(A\)') 'Enter cost per kilowatthour (in dollars): ' READ (*,*) CPKWH WRITE (*,'(//A/)') 'MOTOR A:' WRITE (*,'(A\)') ' Enter READ (*,*) LIFEA WRITE (*,'(A\)') ' Enter READ (*,*) ICA WRITE (*,'(A\)') ' Enter READ (*,*) SVA WRITE (*,'(A\)') ' Enter READ (*,*) EFFA WRITE (*,'(//A/)') 'MOTOR B:' WRITE (*,'(A\)') ' Enter READ (*,*) LIFEB WRITE (*,'(A\)') ' Enter READ (*,*) ICB WRITE (*,'(A\)') ' Enter READ (*,*) SVB WRITE (*,'(A\)') ' Enter READ (*,*) EFFB WRITE (*,'(//A/)') 'MOTOR C:' WRITE (*,'(A\)') ' Enter READ (*,*) LIFEC WRITE (*,'(A\)') ' Enter READ (*,*) ICC WRITE (*,'(A\)') ' Enter READ (*,*) SVC WRITE (*,'(A\)') ' Enter READ (*,*) EFFC
life of motor: ' initial cost of motor: ' salvage value of motor: ' efficiency of motor ( P4 > P3 > P1.
Chapter 6 Solutions
152
(b) MARR = Useful Life =
Annual Output Capacity Selling price =
10% 5 P1
P2
$
P3
120,000 0.500
Expenses Capital Investment Power Labor Maintenance Tax & Insurance
$ $ $ $
24,000 2,720 26,400 1,600
$ $ $ $
30,400 2,720 24,000 1,800
$ $ $ $
49,600 4,800 16,800 2,600
$ $ $ $
52,000 5,040 14,800 2,000
$
480
$
608
$
992
$
1,040
Reject Rate Revenue
$
8.4% 54,960 $
0.3% 59,820 $
P4
2.6% 58,440 $
5.6% 56,640
EOY 0 1 2 3 4 5
$ $ $ $ $ $
P1 (24,000) 23,760 23,760 23,760 23,760 23,760
$ $ $ $ $ $
P2 (30,400) 30,692 30,692 30,692 30,692 30,692
$ $ $ $ $ $
P3 (49,600) 33,248 33,248 33,248 33,248 33,248
$ $ $ $ $ $
P4 (52,000) 33,760 33,760 33,760 33,760 33,760
PW = AW = FW =
$ $ $
66,069 17,429 106,405
$ $ $
85,947 22,673 138,418
$ $ $
76,436 20,164 123,101
$ $ $
75,977 20,043 122,362
For a selling price of $0.50, P2 is still the recommended alternative. However, the overall ranking of the alternatives is now P2 > P3 > P4 > P1.
153
Chapter 6 Solutions
(c) MARR =
10%
Useful Life =
Expenses Capital Investment Power Labor Maintenance Tax & Insurance
Annual Output Capacity Selling price = Scrap Price =
5
120,000 $ $
0.375 0.100
$
P1 24,000
$
P2 30,400
$
P3 49,600
$
P4 52,000
$ $ $ $
2,720 26,400 1,600 480
$ $ $ $
2,720 24,000 1,800 608
$ $ $ $
4,800 16,800 2,600 992
$ $ $ $
5,040 14,800 2,000 1,040
$
8.4% 42,228
$
0.3% 44,901
$
2.6% 44,142
$
5.6% 43,152
EOY 0 1 2 3 4 5
$ $ $ $ $ $
P1 (24,000) 11,028 11,028 11,028 11,028 11,028
$ $ $ $ $ $
P3 (49,600) $ 18,950 $ 18,950 $ 18,950 $ 18,950 $ 18,950 $
P4 (52,000) 20,272 20,272 20,272 20,272 20,272
PW = AW = FW =
$ $ $
17,805 4,697 28,675
$ $ $
Reject Rate Revenue
P2 (30,400) $ 15,773 $ 15,773 $ 15,773 $ 15,773 $ 15,773 $ 29,392 7,754 47,336
$ $ $
22,235 5,866 35,810
$ $ $
24,847 6,555 40,016
Including a scrap price of $0.10, P2 is still the recommended alternative. The overall ranking of the alternatives remains P2 > P4 > P3 > P1.
Chapter 6 Solutions
154
(d) MARR = Useful Life =
Annual Output Capacity Selling price = Scrap Price =
15% 5
P1
P2
$ $
P3
120,000 0.375 0.100
Expenses Capital Investment Power Labor Maintenance Tax & Insurance
$ $ $ $
24,000 2,720 26,400 1,600
$ $ $ $
30,400 2,720 24,000 1,800
$ $ $ $
49,600 4,800 16,800 2,600
$ $ $ $
52,000 5,040 14,800 2,000
$
480
$
608
$
992
$
1,040
Reject Rate Revenue
$
8.4% 42,228 $
0.3% 44,901 $
P4
2.6% 44,142 $
5.6% 43,152
EOY 0 1 2 3 4 5
$ $ $ $ $ $
P1 (24,000) 11,028 11,028 11,028 11,028 11,028
$ $ $ $ $ $
P2 (30,400) 15,773 15,773 15,773 15,773 15,773
$ $ $ $ $ $
P3 (49,600) 18,950 18,950 18,950 18,950 18,950
$ $ $ $ $ $
P4 (52,000) 20,272 20,272 20,272 20,272 20,272
PW = AW = FW =
$ $ $
12,968 3,868 26,082
$ $ $
22,474 6,704 45,202
$ $ $
13,923 4,154 28,005
$ $ $
15,955 4,760 32,091
When all changes occur simultaneously, P2 is still the recommended alternative. The overall ranking of the alternatives remains P2 > P4 > P3 > P1.
155
Chapter 6 Solutions
642 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
A MARR = Useful Life =
B
C
D1 100,000 29,000 10,000
Capital Investment Annual Expense Market Value
$ $ $
EOY 0 1 2 3 4 5 5
$ $ $ $ $ $ $
EOY 0 1 2 3 4 5
∆(D4D1) $ (22,000) $ 7,000 $ 7,000 $ 7,000 $ 7,000 $ 11,000
∆IRR ∆PW(20%)
D
E
20% 5
$
D2 140,600 16,900 14,000
$ $ $
∆(D4D1) (22,000) 7,000 7,000 7,000 7,000 7,000 4,000
$ $ $
D3 148,200 14,800 25,600
∆(D2D4) (18,600) 5,100 5,100 5,100 5,100 5,100 
∆(D3D4) $ (26,200) $ 7,200 $ 7,200 $ 7,200 $ 7,200 $ 7,200 $ 11,600
∆(D2D4) $ (18,600) $ 5,100 $ 5,100 $ 5,100 $ 5,100 $ 5,100
∆(D3D4) $ (26,200) $ 7,200 $ 7,200 $ 7,200 $ 7,200 $ 18,800
$ $ $ $ $ $ $
21.0% 542 $
11.5% (3,348) $
D4 $ 122,000 $ 22,000 $ 14,000
20.0% (6)
If the annual expenses for D4 can be reduced to $22,000, it would be preferred to D3. This solution can be found either by systematically changing the value of cell E6 or by using the solver tool (set value of cell D27 equal to 0 by changing the value of cell E6).
Chapter 6 Solutions
156
643 (a) MARR
20%
Capital investment Annual energy expense Annual maintenance: Starting year First year Increase per year Useful life (years) Market Value
EOY 4 5 6 7 8 9 AW(20%)
SP240 $ 33,200 $ 2,165
$ $ $
1 1,100 500 5 
HEPS9 $ 47,600 $ 1,720
$ $ $
4 500 100 9 5,000
$ $
SP240 (4,765) (5,265)
HEPS9 $ (2,220) $ (2,320) $ (2,420) $ (2,520) $ (2,620) $ 2,280
$
(15,187)
$ (13,621)
157
Chapter 6 Solutions
(b)
MARR
20%
Capital investment Annual energy expense Annual maintenance: Starting year First year Increase per year Useful life (years) Market Value EOY 0 1 2 3 4 5 6 7 8 9 AW(20%)
SP240 $ 28,519 $ 2,165
$ $ $
1 1,100 500 5 
HEPS9 $ 47,600 $ 1,720
$ $ $
4 500 100 9 5,000
$ $ $ $ $ $
SP240 (28,519) (3,265) (3,765) (4,265) (4,765) (5,265)
HEPS9 $ (47,600) $ (1,720) $ (1,720) $ (1,720) $ (2,220) $ (2,320) $ (2,420) $ (2,520) $ (2,620) $ 2,280
$
(13,621)
$ (13,621)
Difference $ 
If the capital investment for SP240 was $28,519, we would be indifferent as to which pump was implemented.
644 MARR = Useful Life =
10% 10
Design A $ 170,000 $ 114,000 $ 70,000 22.5% ∆IRR Answer: Select Design C Capital Investment Annual Receipts Annual Expenses
Chapter 6 Solutions
Design B Design C $ 330,000 $ 300,000 $ 147,000 $ 130,000 $ 79,000 $ 64,000 10.9% 6.8%
158
Solutions to Case Study Exercises 645 Other factors that could be included are distribution costs to grocers, storage costs at grocery stores and projected revenues in the likely event that half gallons are sold for less than twice the price of a quart, and quarts are sold for less than twice the price of a pint. Such nonproportional pricing is common for consumer food products. 646 If landfill costs double, the recommendation for the assumptions given in the case study will be to "produce ice cream and yogurt in halfgallon containers." 647 At your grocery store you will probably discover that the profit associated with a smaller container of ice cream is greater than that of a larger container. For a fixed amount of consumption (e.g. 20,000,000 gallons per year), it is likely that packaging in pint containers is the way to go. Ned and Larry's Ice Cream Company apparently already knows this to be true.
Solutions to FE Practice Problems 648
Set PWall now (i') = PW3 checks (i') and solve for i' $125,000 = $50,000 + $50,000 (P/A,i',2) (P/A,i',2) = 1.5; i' ≈ 21.6% Select (c) If the MARR > 21.6%, select $125,000 now; if MARR < 21.6%, select $50,000 at time 0, 1, and 2. If MARR = 21.6%, Bill would be indifferent.
649
Savings per plan = $40,000  $30,000 = $10,000 / year Let X = number of planes operated per year. $500,000 = $10,000 (X) (P/A,10%,15) + $100,000 (P/F,10%,15) X=
$500,000 − $100,000(0.2394) $10,000(7.6061)
X = 6.26 or 7 planes per year Select (a)
159
Chapter 6 Solutions
650
PWB(15%) = $16,000 + $57,000 (P/A,14%,3) + $6,150 (P/F,15%,3) = $1,058 Select (c)  Alternative B to maximize PW
651
FWC(12%) = $13,000 (F/P,12%,10)  $400 (F/A,12%,10) + $1,750 = $45,645 Select (a) – Alternative A to minimize costs.
652
IRR on ∆ (B – C): 0 = $3,000 + ($460  $100) (P/A,i',6) + $3,350 (P/F,i',6) i' = 13.4% > 10% Select (c) – Alternative B
653
∆ PWW→X(15%) = $550 + $15 (P/A,15%,8) + $200 (P/F,15%,8) = $417.31 < 0
Select (a) – Alternative W 654
Rank Order: DN→D→C→A→E→B Assuming the MARR ≤ 42.5%, Alternative D is the base alternative. The first Comparison to be made based on the tank ordering would be D→C. Select (e)
655
PWA→B(15%) = [$90,000  ($60,000)] + ($12,000  $20,000) (P/A,15%,10) + ($15,000  $10,000) (P/F,15%,10) = $68,914 Select (a)
656
PWC(i') = 0 = $4,000 + $13,000 (P/A,i',10) i' = 30.8% Select (a)
657
Eliminate Alt. B and Alt. E (IRR < 15%) PWA(15%) = $60,000 + $20,000 (P/A,15%,10) + $10,000 (P/F,15%,10) = $42,848 PWC(15%) = $40,000 + $13,000 (P/A,15%,10) + $10,000 (P/F,15%,10) = $27,716 PWD(15%) = $30,000 + $13,000 (P/A,15%,10) + $10,000 (P/F,15%,10) = $37,716 Select (b) – Alternative A to maximize PW.
Chapter 6 Solutions
160
Solutions to Chapter 7 Problems 71
The actual magnitude of depreciation cannot be determined until the asset is retired from service (it is always paid or committed in advance). Also, throughout the life of the asset we can only estimate what the annual or periodic depreciation cost is. Another difference is that relatively little can be done to control depreciation cost once an asset has been acquired, except through maintenance expenditures. Usually much can be done to control the ordinary outofpocket expenses such as labor and material.
72
To be considered depreciable, a property must be: 1) used in a business to produce income; 2) have a determinable life of greater than one year; and 3) lose value through wearing out, becoming obsolete, etc.
73
Personal property is generally any property that can be moved from one location to another, such as equipment or furniture. Real property is land and anything erected or growing on it.
74
Tangible property can be seen or touched, such as equipment and land. Intangible property is personal property that has no physical substance (other than the paper documenting the property), such as a copyright or a patent.
75
The cost basis is usually the purchase price of the property, plus any sales taxes, transportation costs, and the cost of installation or improving the property to make it fit for intended use. Salvage value is not considered, nor is the cost of the land the property is on.
76
(a)
d2 =
(b)
GDS recovery period = 5 years (from Table 74) d2 = 0.32 ($35,000) = $11,200 Assuming the ADS recovery period is 7 years (that is, equal to the class life):
(c)
d2 = 77
2 ⎡⎛ 5 ⎞ ⎤ ⎜ ⎟ ($35,000)⎥ = $7,142.86 ⎢ 7 ⎣⎝ 7 ⎠ ⎦
1 ($35,000) = $5,000 7
From Table 72, the GDS recovery period is 3 years. (a) Basis = $195,000 d*3 = $195,000 (0.3333 + 0.4445 + 0.1481) = $180,550.50 (b)
d4 = 0.0741 ($195,000) = $14,449.50
(c)
BV2 = $195,000 (1  0.3333  0.4445) = $43,329 163
Chapter 7 Solutions
78
79
Under MACRS, the ADS might be preferred to the GDS in several cases. If profits are expected to be relatively low in the near future, but were going to increase to a fairly constant level after that, the ADS would be a way to “save up” depreciation for when it is needed later. In essence, income taxes would be deferred until a later time when the firm is financially more able to pay them. Basis = $120,000 (a)
d2 = $120,000(1 – 0.2)(0.2) = $19,200
(b)
BV1 = $120,000(1 – 0.2) = $96,000
(c)
BV10 = $120,000(1 – 0.2)10 = $12,885
(d)
From Table 72, the recovery period is 7 years (asset class 00.11).
(e)
r1 = $120,000(0.1429) = $17,148
(f)
BV3 = $120,000(1 – 0.1429 – 0.2449 – 0.1749) = $52,476
(g)
Applying the halfyear convention, d4 = (0.5)($120,000)(0.1249) = $7,494
710
(a)
Assumption: cost (“unadjusted”) basis = cash purchase price of machine + cost of shipping, sales taxes, and nonrecurring installation (startup costs) = total capital investment $15,000 $1,000 $1,200 B = $17,200
(b)
711
Purchase Price Shipping Cost + Sales Taxes Installation Unadjusted Basis (Total Capital Investment)
d = (B  S) / 5 = ($17,200  $1,000) / 5 = $3,240/yr BV3 = $17,200  3($3,240) = $7,480 Net selling price = $1,500  $500 = $1,000 They failed to recover $7,480  $1,000 = $6,480
Basis = $60,000 and SVN = $12,000. Find d3 and BV5. (a)
d3 = dk =
B − SVN $60,000 − $12,000 = $3,428.57 = N 14
BV5 = $60,000  (5)($3,428.57) = $42,857.15
Chapter 7 Solutions
164
(b)
(1)
Year, k 1 2 3 4 5 a b c d
(2) (3) (4) Depreciation Method Beginning of 200% Declining StraightLine Depreciation Year BVa Balance Methodb Method c Amount Selectedd $ 60,000.00 $8,571.43 $3,428.57 $8,571.43 51,428.57 7,346.94 3,032.97 7,346.94 44,081.63 6,297.38 2,673.47 6,297.38 37,784.25 5,397.75 2,344.02 5,397.75 32,386.50 4,626.64 2,038.65 4,626.64
Column 1 for year k  column 4 for year k = the entry in column 1 for year k+1 Column 1 x (2 / 14) Column 1 minus estimated SVN divided by remaining years from the beginning of the current year through the fourteenth year. Select the larger amount of column 2 or column 3. From the above table, d3 = $6,297.38 and BV5 = $32,386.50  $4,626.64 = $27,759.86
(c) From Table 72, the GDS recovery period is 7 years.
d3
= $60,000 (0.1749) = $10,494
BV5 = $60,000  $60,000 (0.1429 + 0.2449 + 0.1749 + 0.1249 + 0.0893) = $13,386
712
(d)
From Table 72, the ADS recovery period is 14 years. ⎛ $60,000 ⎞ d1 = d15 = (0.5)⎜ ⎟ = $2,142.86 ⎝ 14 ⎠ ⎛ $60,000 ⎞ d 2 = d 3 =L = d14 = ⎜ ⎟ = $4,285.71 ⎝ 14 ⎠ BV5 = $60,000  [$2,142.86 + 4($4,285.71)] = $40,714.30
(a) (b)
From Table 74, the GDS Recovery period is 5 years. In year 4 there was a $17,280 depreciation deduction. Year 1 2 3 4 5 6
Depreciation Calculation $150,000 (0.2) 150,000 (0.32) 150,000 (0.192) 150,000 (0.1152) 150,000 (0.1152) 150,000 (0.0576)
165
Depreciation Deduction $30,000 48,000 28,800 17,280 17,280 8,640
EndofYear Book Value $120,000 72,000 43,200 25,920 8,640 0
Chapter 7 Solutions
713
(c)
From (b), the book value at beginning of year 5 (end of year 4) is $25,920
(a)
From Table 72, the GDS recovery period is 5 years and the ADS recovery period is 6 years. GDS depreciation deductions: d1 = 0.2000 ($300,000) = $60,000 d2 = 0.3200 ($300,000) = $96,000 d3 = 0.1920 ($300,000) = $57,600
d4 = 0.1152 ($300,000) = $34,560 d5 = 0.1152 ($300,000) = $34,560 d6 = 0.0576 ($300,000) = $17,280
ADS depreciation deductions: ⎛ $300,000 ⎞ d 1 = d 7 = (0.5)⎜ ⎟ = $25,000 ⎝ ⎠ 6 $300,000 d2 = d3 = L = d6 = = $50,000 6 (b)
Assume calculation of PW at time of purchase and the depreciation deduction is taken at the end of the year. PWGDS = $60,000 (P/F,12%,1) + $96,000(P/F,12%,2) + $57,600(P/F,12%,3) + $34,560(P/F, 12%,4) + $34,560(P/F,12%,5) + $17,280(P/F,12%,6) = $221,431.15 PWADS = $25,000 (P/F,12%,1) + $50,000 (P/A,12%,5)(P/F,12%,1) + $25,000(P/F,12%,7) = $194,566.30 Difference = PW∆ = $221,431.15  $194,566.30 = $26,864.85
714
From Table 74, the GDS recovery period is 5 years. Cost basis = $99,500 + $15,000 trade in = $114,500 (a)
d3 = 0.192 ($114,500) = $21,984
(b)
BV4 = $114,500  $114,500 (0.2 + 0.32 + 0.192 +0.1152) = $19,786
(c)
R = 2/9.5 = 0.2105 d4* = $114,500 [1  (1  0.2105)4] = $70,015
715
$25,000 − $5,000 = $0.20/unit 100,000 units = (10,000 units) ($0.2/unit) = $2,000
Depreciation per unit of production = d4
BV4 = $25,000  (60,000 units + 10,000 units)($0.20/unit) = $11,000
Chapter 7 Solutions
166
716
(a)
Income taxes = $50,000 (0.15) + $25,000 (0.25) + $15,000 (0.34) = $18,850
(b)
Depreciation + Expenses = $220,000  $90,000 = $130,000
717
EOY 0 1 2 3 4a 4b
(A)
(B)
BTCF $10M $4M $4M $4M $4M 0
Depr. $2.5M $2.5M $2.5M $2.5M
(C) = (A)  (B)
TI $1.5M $1.5M $1.5M $1.5M 0
(D) = t (C)
T (40%) $0.6M $0.6M $0.6M $0.6M 0
(E) = (A) + (D)
ATCF  $10M $3.4M $3.4M $3.4M $3.4M
M ≡ millions of dollars PW(15%) = $10M + $3.4M (P/A,i%,4) = $.293M IRR: 0 = $10M + $3.4M (P/A,i%,4); i' = 13.54% Because PW(10%) < 0 and IRR < 15%, this project should not be recommended. From Table 74, d4 = 0.0741 ($195,000) = $14,450 BV2 = $195,000  $195,000 (0.3333 + 0.4445) = $43,329
718
(a) (b)
719
t = 0.06 + 0.34(1  0.06) = 0.3796, or 37.96%; t = 0.12 + 0.34(1  0.12) = 0.4192, or 41.92%
720
Income taxes owed (with venture) = $50,000(0.15) + ($7,000 + $8,000)(0.25) = $11,250 Income taxes owed (without venture) = $50,000(0.15) + $7,000(0.25) = $9,250
721
PW(i) = 0 = $9,000 + ($10,000)(0.10/2)(10.28)(P/A,i%,15) + [$10,000 – ($10,000  $9,000)(0.28)](P/F,i%,15) = $9,000 + $360(P/A,i%,15) + $9,720(P/F,i%,15) i% PW(i) 4% $400.14 i% 0 5% $587.99 i% = 4.4%/6 months, r = 8.8%/year, ieff. = 8.99%/year
722
Assume repeatability.
167
Chapter 7 Solutions
Alternative A: Plastic d = ($5000$1000)/5 = $800 EOY 0 15 5
(A)
(B)
(C) = (A)  (B)
BTCF $5,000 $300 0
Depr. $800 
TI $1,100  $1,000
(D) = t (C)
(E) = (A) + (D)
T (40%) $440 $400
ATCF  $5,000 $140 $400
AWA(12%) = $5,000(A/P,12%,5) + $140 + $400(A/F,12%,5) = $1,184 Alternative B: Copper d = ($10,000  $5,000)/10 = $500 EOY 0 110 10
(A)
(B)
(C) = (A)  (B)
BTCF $10,000 $100 0
Depr. $500 
TI $600  $5,000
(D) = t (C)
(E) = (A) + (D)
T (40%) $240 $2,000
ATCF  $10,000 $140 $2,000
AWB(12%) = $10,000(A/P,12%,10) + $140 + $2,000(A/F,12%,5) = $1,516 Select Alternative A: Plastic. 723
Assume repeatability. Purchase Option: From Table 72, the ADS recovery period is 6 years (asset class 36.0). Applying the half year convention, depreciation deductions can be claimed over a 7year period. d1 = d7 = (0.5)($30,000/6) = $2,500 d2 = d3 = .... = d6 = ($30,000/6) = $5,000 EOY 0 1 2 3 4 5 6 7 8
(A)
(B)
BTCF $30,000 0 0 0  10,000 0 0 0 0
Depr. $2,500 5,000 5,000 5,000 5,000 5,000 2,500 
(C) = (A)  (B)
TI $2,500  5,000  5,000 15,000  5,000  5,000  2,500 0
(D) = t (C)
T (40%) $1,000 2,000 2,000 6,000 2,000 2,000 1,000 0
(E) = (A) + (D)
ATCF  $30,000 1,000 2,000 2,000  4,000 =  10,000 + 6,000 2,000 2,000 1,000 0
PW(12%) =  $30,000 + $1,000(P/A,12%,7) + $1,000(P/A,12%,5)(P/F,12%,1)  $6,000(P/F,12%,4) Chapter 7 Solutions
168
=  $26,030.47 AW(12%) =  $26,030.47 (A/P,12%,8) =  $5,240 Leasing Option ATCF =  (1  0.40)(Leasing Cost) = AW(12%) For the leasing option to be more economical than the purchase option,  (0.6) (Leasing Cost) <  $5,240 Leasing Cost < $8,733 If Leasing Cost < $8,733 per year, lease the tanks; otherwise, purchase the tanks. 724
Assume repeatability. Fixture X (A)
(B)
(C) = (A)  (B)
EOY BTCF Depr. TI 0 $30,000 15 $3,000 $6,000 $9,000 6 $3,000 0 $3,000 6* $6,000 $6,000** *Market Value **Depreciation Recapture AWX(8%) = $4,989
(D) = t (C)
(E) = (A) + (D)
T (50%) $4,500 $1,500 $3,000
ATCF  $30,000 $1,500 $1,500 $3,000
Fixture Y EOY 0 1 2 3 4 5 6 7 8 8
(A)
(B)
BTCF $40,000 $2,500 $2,500 $2,500 $2,500 $2,500 $2,500 $2,500 $2,500 $4,000*
Depr. $8,000 $12,800 $7,680 $4,608 $4,608 $2,304 0 0 
(C) = (A)  (B)
TI $10,500 $15,300 $10,180 $7,108 $7,108 $4,804 $2,500 $2,500 $4,000
(D) = t (C)
(E) = (A) + (D)
T (50%) $5,250 $7,650 $5,090 $3,554 $3,554 $2,402 $1,250 $1,250 $2,000
ATCF  $40,000 $2,750 $5,150 $2,590 $1,054 $1,054 $98 $1,250 $1,250 $2,000
* Market Value AWB(8%) = $5,199 Select Fixture X.
169
Chapter 7 Solutions
725
The sales revenue is $40 per ticket x 60,000 tickets per year = $2,400,000 per year and the investment in working capital is (1/12)($2,4000,000) = $200,000. The the total investment is $800,000 = $200,000 = $1,000,000 and depreciation is $800,000 / 4 = $200,000 per year. Total annual operating costs are $24 (60,000 tickets) + $400,000 = $1,840,000. (A)
EOY 0 1 2 3 4a 4b
(B)
(C) = (A)  (B)
BTCF Depr. $1,000,000 $560,000 $200,000 $560,000 $200,000 $560,000 $200,000 $560,000 $200,000 $200,000 
TI $360,000 $360,000 $360,000 $360,000 
(D) = t (C)
(E) = (A) + (D)
T (50%) ATCF  $1,000,000 $122,400 $437,600 $122,400 $437,600 $122,400 $437,600 $122,400 $437,600 $200,000
PW(15%) = $1,000,000 + $437,600 (P/A,15%,4) + $200,000 (P/F,15%,4) = $363,690 >> 0; the investment should be made. 726
Machine A: (A)
EOY 0 112 12
BTCF $20,000 $12,000 $4,000
(B)
(C) = (A)  (B)
(D) = t (C)
Depr $1,333.33 
TI $10,666.67 0
T (40%) $4,266.67 0
0.1468
(E) = (A) + (D)
ATCF $20,000 $7,733.33 $4,000
0.0468
(a) AW = $20,000(A/P,10%,12) + $7733.33 +$4,000(A/F,10%,12) = $4,984.53 8.9847
(b) PW = $4,984.53(P/A,10%,24) = $44,784.51 (c) IRR see part (c) below
Machine B: (A)
EOY 0 18
(B)
BTCF $30,000 $18,000
(C) = (A)  (B)
Depr $3,750
TI $14,250
0.1874
(a) AW = $30,000(A/P,10%,8) + $12,300 = $6,678 8.9847
Chapter 7 Solutions
170
(D) = t (C)
T (40%) $5,700
(E) = (A) + (D)
ATCF $30,000 $12,300
(b) PW = $6,678(P/A,10%,24) = $60,000 Choose Machine B. (c) IRR; use qualitative reasoning and ∆ (BA); by observation from differences in PWs, both IRRs > MARRAT = 10% and i'%B > i'%A ∴ Choose Machine B. 727
Assume repeatability of investments over a common multiple of lives.
Design S1 (A)
(B)
EOY BTCF Depr 0  $100,000 1 20,000 $20,000 2 20,000 32,000 3 20,000 19,200 4 20,000 11,520 5 20,000 11,520 6 20,000 5,760 7 20,000 0 7 30,000 
(C) = (A)  (B)
TI 
(D) = t (C)
(E) = (A) + (D)
T (40%) $0 4,800  320  3,392  3,392  5,696  8,000 12,000
ATCF PW(10%) $100,000 $100,000 $0 20,000 18,182  12,000 24,800 20,495 800 19,680 14,786 8,480 16,608 11,343 8,480 16,608 10,312 14,240 14,304 8,075 20,000 12,000 6,158 30,000 18,000 9,238 PW (10%) = $ 1,411 AW(10%) = PW (A/P,10%,7) = $1,411 (0.2054) = $290
Design S2 (A)
EOY BTCF 0 $200,000 1 40,000 2 40,000 3 40,000 4 40,000 5 40,000 6 40,000 6 50,000
(B)
Depr $40,000 64,000 38,400 23,040 23,040 11,520 
(C) = (A)  (B)
TI $0 24,000 1,600 16,960 16,960 28,480 50,000
(D) = t (C)
(E) = (A) + (D)
T (40%) $0 9,600  640  6,784  6,784 11,392 20,000
ATCF PW(10%) $200,000 $200,000 40,000 36,364 49,600 40,989 39,360 29,571 33,216 22,687 33,216 20,624 28,608 16,149 30,000 16,935 PW(10%) = $16,681
AW(10%) = PW(A/P,10%,6) =  $16,681 (0.2296) =  $3,830
Therefore, choose Design S1 since it has the greater annual worth (Note: Neither design makes money  so if a system is not required, do not buy either S1 or S2.) Repeatability is assumed.
171
Chapter 7 Solutions
728
(a)
Straightline depreciation:
Method I (A)
(B)
(C) = (A)  (B)
(D) = t (C)
(E) = (A) + (D)
EOY BTCF Depr TI T (40%) ATCF 0 $10,000 $10,000 15 $14,150 $1,800 $15,950 $6,380 $7,770 5 $1,000 0 0 $1,000 PW0(12%) = $10,000  $7,770(P/A,12%,5) +$1,000(P/F,12%,5) = $37,441,68 To have a basis for computation, assume that Method I is duplicated for years 610. Transform the additional PW5(12%) = $37,449.68 to the present and get: PW(12%) = PW0(12%) + (P/F,12%,5)PW5(12%) = $58,867.10 Method II (A)
EOY 0 110 10
BTCF $40,000 $7,000 $5,000
(B)
(C) = (A)  (B)
Depr $3,500 
TI $10,500 0
(D) = t (C)
T (40%) $4,200 0
(E) = (A) + (D)
ATCF $40,000 $2,800 $5,000
PW(12%) = $40,000  $2,800(P/A,12%,10) +$5,000(P/F,12%,10) = $54,210.76 Thus, Method II is the better alternative. (b) Method I Assume that MV5 is $1,000 The MACRS property class is 5 years. This means that the taxlife is 6 years which is greater than the useful life of 5 years.
Chapter 7 Solutions
172
(A)
(B)
(C) = (A)  (B)
(D) = t (C)
(E) = (A) + (D)
EOY BTCF Depr TI T (40%) ATCF 0 $10,000 $10,000 1 $14,150 $2,000 $16,150 $6,460 $7,690 2 $14,150 $3,200 $17,350 $6,940 $7,210 3 $14,150 $1,920 $16,070 $6,428 $7,722 4 $14,150 $1,152 $15,302 $6,120.80 $8,029.20 5 $14,150 $1,152 $15,302 $6,120.80 $8,029.20 5 $1,000 $1,000 $400 $600 6 0 $576 $576 $230.40 $230.40 PW (12%) = $37,311.71 PW(12%) over 10 years = $37,311[1+(P/F,12%,5)] = $58,482 To get a figure for comparison, convert PW(12%) to annual worth over the useful life of 5 years: AW(12%) = $37,311.71(A/P,12%,5) = $10,350.63 With straight line depreciation the annual worth is AWSL(12%) = $37,441.68(A/P,12%,5) = $10,368.69 We note that the annual worths are basically the same. This is due to the fact that, whatever depreciation method we use, the depreciation deductions are small relative to the annual expenses. Method II The MACRS class life is 7 years. Assume that MV5 is $5,000 (A)
EOY BTCF 0 $40,000 1  $7,000 2  $7,000 3  $7,000 4  $7,000 5  $7,000 6 $7,000 7 $7,000 8 $7,000 9 $7,000 10 $7,000 10 $5,000 PW(12%) = $51,869.57
(B)
(C) = (A)  (B)
Depr $5,716 $9,796 $6,996 $4,996 $3,572 $3,568 $3,572 $1,784 0 0 
TI $12,716 $16,796 $13,996 $11,996 $10,572 $10,568 $10,572 $8,784 $7,000 $7,000 $5,000
(D) = t (C)
(E) = (A) + (D)
T (40%) $5,086.40 $6,718.40 $5,598.40 $4,798.40 $4,228.80 $4,227.20 $4,228.80 $3,513.60 $2,800 $2,800 $2,000
ATCF $40,000 $1,913.60 $281.60 $1,401.60 $2,201.60 $2,771.20 $2,772.80 $2,771.20 $3,486.40 $4,200 $4,200 $3,000
AW over the useful life of 10 years: AW(12%) $51,869.67(A/P,12%,10) = $9,180.11 Thus, Method II is chosen also in this case. AWSL(12%) = $54,210.76(A/P,12%,10) = $9,594.45 173
Chapter 7 Solutions
We notice a more significant difference in this case. Here, the timing of the depreciation deductions is of greater importance. 729
Let X = Annual operating expenses Note: The $4,000 shown in the BTCF column (A) represents the savings in rent. Thus, Column A is the incremental BTCF between leasing and purchasing the equipment. (A)
(B)
EOY BTCF 0  $ 12,000 1 4,000  X 2 4,000  X 3 4,000  X 4 4,000  X 5 4,000  X 6 4,000  X 7 4,000  X 8 4,000  X 8 5,000 Solve for X:
(C) = (A)  (B)
Depr $1,200 2,400 2,400 2,400 2,400 1,200 0 0 
(D) = t (C)
(E) = (A) + (D)
TI T (40%) ATCF $ 12,000 $2,800  X  $1,120 + 0.4X 2,880  0.6X 1,600  X  640 + 0.4X 3,360  0.6X 1,600  X  640 + 0.4X 3,360  0.6X 1,600  X  640 + 0.4X 3,360  0.6X 1,600  X  640 + 0.4X 3,360  0.6X 2,800  X  1,120 + 0.4X 2,880  0.6X 4,000  X  1,600 + 0.4X 2,400  0.6X 4,000  X  1,600 + 0.4X 2,400  0.6X 5,000 2,000 3,000
PW(10%) = 0 =  $12,000 + $2,880(P/F,10%,1) + $3,360(P/A,10%,4)(P/F,10%,1) + $2,880(P/F,10%,6) + $2,400(P/F,10%,7) + ($2,400 + $3,000)(P/F, 10%,8)  (0.6X)(P/A,10%,8) X = $1,773.60 The added annual expense can be as high as $1,774 and the IRR will still exceed 10%. 730
After  tax MARR 015 . = = 0.25 , or 25% 1 t 1 − 0.40
(a)
Beforetax MARR =
(b)
Year 1 2 3 4
(c)
BV8 = 0, therefore TI8 = $10,000 (Property having a 7year recovery period is fully depreciated after N+1=8 years.)
Depreciation $12,861 22,041 15,741 11,241
Chapter 7 Solutions
Year 5 6 7 8
174
Depreciation $8,037 8,028 8,037 4,014
(d)
EOY 0 1 2 3 4 5 6 7 8 8 (e)
(A)
(B)
(C) = (A)  (B)
(D) = t (C)
(E) = (A) + (D)
BTCF
Depr
TI
T (40%)
ATCF
$90,000 15,000 $12,861 15,000 22,041 15,000 15,741 15,000 11,241 15,000 8,037 15,000 8,028 15,000 8,037 15,000 4,014 10,000 
$2,139 7,041 741 3,759 6,963 6,972 6,963 10,986 10,000
$856 2,816 296 1,504 2,785 2,789 2,785 4,394 4,000
PW (15%)
$90,000 14,144 17,816 15,296 13,496 12,215 12,211 12,215 10,606 6,000 PW(15%) =
$90,000 12,299 13,472 10,058 7,717 6,073 5,279 4,592 3,467 1,961 $25,082
No, reject the project because PW(ATCF) < 0 at MARR = 15%. (A)
731
EOY 0 1 2 3 4 5 6 . . N
BTCF $50,000 14,000 14,000 14,000 14,000 14,000 14,000 14,000 14,000 14,000
(B)
(C) = (A)  (B)
Depr $10,000 10,000 10,000 10,000 10,000 0 0 0 0
TI $4,000 4,000 4,000 4,000 4,000 14,000 14,000 14,000 14,000
(D) = t (C)
T (40%) $1,600 1,600 1,600 1,600 1,600 5,600 5,600 5,600 5,600
(E) = (A) + (D)
ATCF $50,000 12,400 12,400 12,400 12,400 12,400 8,400 8,400 8,400 8,400
Let X = N  5 years. Set PW(10%) = 0 and solve for X: 0 = $50,000 + $12,400(P/A,10%,5) + $8,400(P/A,10%,X)(P/F,10%,5) (P/A,10%,X) = 0.5741 X ≈ 1 year. Thus, N = 5 + X = 6 years
175
Chapter 7 Solutions
732 Manufacturing designed for varying degrees of automation:
Degrees Investment Cost Annual Labor Cost Annual Power & Maintenance Cost A $10,000 $9,000 $ 500 B 14,000 7,500 800 C 20,000 5,000 1,000 D 30,000 3,000 1,500 (A) (A)
EOY 0 15
BTCF $10,000 $9,500
(B)
(C) = (A)  (B)
Depr $2,000
TI $11,500
(D) = t (C)
T (40%) $4,600
(E) = (A) + (D)
ATCF $10,000 $4,900
Straight Line Depreciation: ($10,0000)/5 = $2,000 (B) (A)
EOY 0 15
BTCF $14,000 $8,300
(B)
(C) = (A)  (B)
Depr $2,800
TI $11,100
(D) = t (C)
T (40%) $4,440
(E) = (A) + (D)
ATCF $14,000 $3,860
Depreciation: ($14,0000)/5 = $2,800 (C) (A)
EOY 0 15
BTCF $20,000 $6,000
(B)
(C) = (A)  (B)
Depr $4,000
TI $10,000
(D) = t (C)
T (40%) $4,000
(E) = (A) + (D)
ATCF $20,000 $2,000
Depreciation: ($20,0000)/5 = $4,000 (D) (A)
EOY 0 15
BTCF $30,000 $4,500
(B)
(C) = (A)  (B)
Depr $6,000
TI $10,500
Depreciation: ($30,0000)/5 = $6,000
Chapter 7 Solutions
176
(D) = t (C)
T (40%) $4,200
(E) = (A) + (D)
ATCF $30,000 $300
(a) Annual Worth
AWA = $10,000(A/P,15%,5)  $4,900 = $7,883 AWB = $14,000(A/P,15%,5)  $3,860 = $8,036.20 AWC = $20,000(A/P,15%,5)  $2,000 = $7,966 AWD= $30,000(A/P,15%,5)  $300 = $9,249 Select to automate to Degree A. (b) Present Worth
PWA = $10,000  $4,900(P/A,15%,5) = $26,425.78 PWB = $14,000  $3,860(P/A,15%,5) = $26,939.49 PWC = $20,000  $2,000(P/A,15%,5) = $26,704.40 PWD= $30,000  $300(P/A,15%,5) = $31,005.66 Select to automate to Degree A. (c) IRR
Degree Investment Annual Cost
A B C D $10,000 $14,000 $20,000 $30,000 $4,900 3,860 2,000 300
Because all numbers are costs, we cannot compare with the alternative of do nothing, thus select A as our base comparison. BA $4,000 + $1,040(P/A, i'%,5) = 0, i'% ≅ 9.431%; since i'% < 15%, reject B CA $10,000 + $2,900(P/A, i'%,5) = 0, i'% ≅ 13.8%; since i'% < 15%, reject C DA $20,000 + $4,600(P/A, i'%,5) = 0, i'% ≅ 4.8%; since i'% < 15%, reject D and select A.
733
t = s + f (1  s) = 0.04 + 0.34(1  0.04) = 0.3664, or 36.64%
177
Chapter 7 Solutions
(A)
(B)
(C)
EOY Investment Revenues Expenses 0 $1,000,000 1 X $636,000 2 X 674,160 3 X 714,610 3 280,000c d 3 580,000 
(D)=(A)+(B)+(C)
(E)
BTCF  $1,000,000 X  636,000 X  674,160 X  714,610 280,000 580,000
Depra $139,986 186,690 31,101b 
a
Cost basis for depreciation calculations = $420,000 Only a halfyear of depreciation is allowable c Market value of depreciable investment d Assumed value of nondepreciable investment (land and working capital) b
(F)=(D)(E)
EOY 0 1 2 3 3 3 e
TI X  775,986 X  860,850 X  745,711 217,777e 
(G)= t (F)
(H)=(D)+(G)
T(36.64%)
ATCF $1,000,000 0.6336X  351,679 0.6336X  358,745 0.6336X  441,381 200,207 580,000
0 0.3664X + 284,321 0.3664X + 315,415 0.3664X + 273,229 79,793 
MV  BV3 = $280,000  $62,223 = $217,777
PW(10%) = 0 = $1,000,000 + (0.6336 X)(P/A,10%,3)  $351,679(P/F,10%,1)  $358,745(P/F,10%,2)  $441,381(P/F,10%,3) + $780,207(P/F,10%,3) X=
734
$1,361,618 = $864,135 / year (2.4869)(0.6336)
(a) (A)
EOY 0 1 2 3 4 5 6
(B)
Lease Other Expense Expenses $80,000  $4,000  60,000  4,000  50,000  4,000  50,000  4,000  50,000  4,000  50,000  4,000
Chapter 7 Solutions
(C) = (A) + (B)
BTCF $84,000 64,000 54,000 54,000 54,000 54,000
178
(D) = t (C)
T (40%) $33,600 25,600 21,600 21,600 21,600 21,600
(E) = (C) + (D)
ATCF PW (8%) $50,400 $46,667 38,400 32,922 32,400 25,720 32,400 23,815 32,400 22,051 32,400 20,417 PW = $171,592
(b) 735
AW(8%) = PW (A/P,8%,6) = $171,592 (0.2163) = $37,115
Depreciation schedule (MACRS 5year property class). The cost basis (B) is assumed to be $345,000. Year 1 2 3 4 5 6
BVk1 $345,000 276,000 165,600 99,360 59,616 19,872
rk 0.2000 0.3200 0.1920 0.1152 0.1152 0.0576
dk $ 69,000 110,400 66,240 39,744 39,744 19,872
BVk $276,000 165,600 99,360 59,616 19,872 0
(a) Economic Value Added (EVA):
t = 0.5 T(50%)  $21,500  800  22,880  36,128  36,128  46,064  60,000
EOY BTCF Depr TI NOPAT 1 $112,000a $ 69,000 $ 43,000 $21,500 2 110,400 1,600 800 3 66,240 45,760 22,880 4 39,744 72,256 36,128 5 39,744 72,256 36,128 6 112,000 19,872 92,128 46,064 6 120,000b 120,000c 60,000 a BTCFk = $120,000  $8,000 = $112,000 b MV6 = $120,000 c Gain on disposal = MV6  BV6 = $120,000  0 = $120,000 EOY 1 2 3 4 5 6 6
$ 21,500 800 22,880 36,128 36,128 46,064
EVA  0.10 ( $345,000)  0.10 ( 276,000)  0.10 ( 165,600)  0.10 ( 99,360)  0.10 ( 59,616)  0.10 ( 19,872)
=  $13,000 =  26,800 = 6,320 = 26,192 = 30,166 = 44,077 60,000
PW(10%)  $11,818  22,148 4,748 17,888 18,730 24,880 33,870 Total: $66,150
The present equivalent of the EVA = $66,150.
179
Chapter 7 Solutions
Aftertax cash flow (ATCF):
(b)
EOY 0 1 2 3 4 5 6 6
BTCF $345,000 112,000
T(50%) 0  $21,500  800  22,880  36,128  36,128  46,064  60,000
112,000 120,000
ATCF $345,000 90,500 111,200 89,120 75,872 75,872 65,936 60,000
PW(10%) $345,000 82,274 91,896 66,957 51,822 47,109 37,222 33,870 Total: $66,150
Yes; the PW(10%) of the ATCF ($66,150) is the same as the present equivalent of the annual EVA amounts.
736
Beginningofyear book values: Year 1 2 3 4 5 6
BVk1 $ 180,000 144,000 86,400 51,840 31,104 10,368
dk $ 36,000 57,600 34,560 20,736 20,736 10,368
BVk $ 144,000 86,400 51,840 31,104 10,368 0
EOY NOPATa (0.10) BVk1 EVAb 1 0 $ 18,000  $ 18,000 2  $ 13,392 14,400  27,720 3 893 8,640  7,747 4 9,464 5,184 4,280 5 9,464 3,110 6,354 6 15,892 1,037 14,855 7 22,320 0 22,320 8 22,320 0 22,320 9 22,320 0 22,320 10 22,320 0 22,320 10 18,600 18,600 a From Table 76: Column (C) algebraically added to Column (D). b Equation 723: EVAk = NOPATk  i • BVk1 PW(10%) =  $18,000 (P/F,10%,1)  L + $14,855 (P/F,10%,6)
Chapter 7 Solutions
180
+ $22,320 (P/A,10%,4) (P/F,10%,6) + $18,600 (P/F,10%,10) = $17,208
737
Depreciation schedule (3year property class). The cost basis (B) is assumed to be $84,000. rk dk BVk Year BVk1 1 $ 84,000 0.3333 $ 28,000 $ 56,000 2 56,000 0.4445 37,338 18,662 3 18,662 0.1481 12,440 6,222 4 6,222 0.0741 6,222 0 Aftertax cash flow (ATCF): EOY 0 1 2 3 4
BTCF  $ 84,000 40,000 40,000 40,000 40,000
Depr 0 $ 28,000 37,338 12,440 6,222
TI 0 $ 12,000 2,662 27,560 33,778
 t =  0.5 T (50%) 0  $ 6,000  1,331  13,780  16,889
ATCF  $ 84,000 34,000 38,669 26,220 23,111
PW(12%) = $84,000 + $34,000 (P/F,12%,1) + L + $23,111 (P/F,12%,4) = $10,535 AW(12%) = $10,535 (A/P,12%,4) = $3,468 Economic Value Added (EVA): EVAb EOY NOPATa (0.12) BVk1 1 $ 6,000 $ 10,080  $ 4,080 2 1,331 6,720  5,389 3 13,780 2,239 11,541 4 16,889 747 16,142 a From ATCF analysis above: NOPATk = (TI)k  [T(50%)]k b Equation 723: EVAk = NOPATk  i • BVk1 PW(12%) =  $4,080 (P/F,12%,1)  L + $16,142 (P/F,12%,4) = $10,535 * AW(12%) = $10,535 (A/P,12%,4) = $3,468 *
Annual equivalent EVA.
181
Chapter 7 Solutions
738 (A)
EOY 0 1 2 3 4 5 5
BTCF $160,000 $35,000 $35,000 $35,000 $35,000 $35,000 $30,000**
(B)
(C) = (A)  (B)
Depr* $26,000 $41,600 $24,960 $14,976 $7,488 
TI $9,000 $6,600 $10,040 $20,024 $27,512 0$14,976***
(D) = t (C)
T (50%) $4,500 $3,300 $5,020 $10,012 $13,756 $7,488
(E) = (A) + (D)
ATCF $160,000 $30,500 $38,300 $29,980 $24,988 $21,244 $37,488
*d = $130,000 · rk(p) (see Table 73) **Assume land recovered at original cost of $30,000 ***MV – BV; Market Value of equipment (purchased) is negligible at the end of year 5. Assume 1/2 year on year 5 depreciation (recapture = $14,976) PW(5%) = $160,000 + $30,500(P/F,5%,1) + $38,300(P/F,5%,2) + $29,980(P/F,5%,3) + $24,988(P/F5%,4) + ($21,244+$37,488)(P/F,5%,5) = $160,000 + $30,500(0.9524) + $38,300(0.9070) + $29,980(0.8638) + $24,988(0.8227) + $58,732(0.7835) = $3,742.83 PW(MARRAT) < 0, not a profitable investment
Chapter 7 Solutions
182
739 Use a study period of 3 years
Quotation I EOY BTCF BTCF Depr. Depr. Book Gain (Loss) Capital Operating Fact. Value On Disp. 0 $(180,000) $180,000 1 $(28,000) 0.2000 $(36,000) $144,000 2 $(28,000) 0.3200 $(57,600) $ 86,400 3 $50,000 $(28,000) 0.0960 $(17,280) $ 69,120 $ (19,120) PW of ATCF, Quotation I: $(143,174) IT = Income Taxes Quotation II EOY BTCF BTCF Depr. Depr. Book Gain (Loss) Capital Operating Fact. Value On Disp. 0 $(200,000) $200,000 1 $(17,000) 0.2000 $(40,000) $160,000 2 $(17,000) 0.3200 $(64,000) $ 96,000 3 $60,000 $(17,000) 0.0960 $(19,200) $ 76,800 $ (16,800) PW of ATCF, Quotation II: $(136,848) IT = Income Taxes
∆ in Ord. Inc
∆ Cash Flow ∆ Cash Flow for IT (Cap) for IT (Oper)
$(64,000) $(85,600) $(45,280) $ 7,648
∆ in Ord. Inc
$ 25,600 $ 34,240 $ 18,112
∆ Cash Flow ∆ Cash Flow for IT (Cap) for IT (Oper)
$(57,000) $(81,000) $(36,200) $ 6,720
$ 22,800 $ 32,400 $ 14,480
ATCF $(180,000) $ (2,400) $ 6,240 $ 47,760
ATCF $(200,000) $ 5,800 $ 15,400 $ 64,200
Accept Quotation II.
183
Chapter 7 Solutions
740 Machine 1: Year
BTCF
Depr
0 1 2 3 4 5 5
$60,000 $16,000 $16,000 $16,000 $16,000 $16,000 0
$12,000 $19,200 $11,520 $6,912 $3,456
TI $28,000 $35,200 $27,520 $22,912 $19,456 $6,912
Income Tax (40%) $11,200 $14,080 $11,008 $9,164.80 $7,782.40 $2,764.80
ATCF $60,000 $4,800 $1,920 $4,992 $6,835.20 $8,217.60 $2,764.80 AW1(12%) = $21,306.41
Machine 2: Year
BTCF
Depr
0 1 2 3 4 5 6 7 8
$66,000 $18,000 $18,000 $18,000 $18,000 $18,000 $18,000 $18,000 $18,000
$13,200 $21,120 $12,672 $7,603.20 $7,603.20 $3,801.60 0 0
TI $31,200 $39,120 $30,672 $25,603.20 $25,603.20 $21,801.60 $18,000 $18,000
Income Tax (40%) $12,480 $15,648 $12,268.80 $10,241.28 $10,241.28 $8,720.64 $7,200 $7,200
ATCF $66,000 $5,520 $2,352 $5,731.20 $7,758.72 $7,758.72 $9,279.36 $10,800 $10,800
AW2(12%) = $20,163.04
Therefore, select Machine 2. ATMARR = 0.20(10.40) = 0.12 or 12%
741
Year
BTCF
0 1 2 3 4 5 6 7 8
$25,000 $8,000 $8,000 $8,000 $8,000 $8,000 $8,000 $8,000 $10,500
Depr $5,000 $8,000 $4,800 $2,880 $2,880 $1,440 0 0
TI
Income Tax
$3,000 0 $3,200 $5,120 $5,120 $6,560 $8,000 $10,500
$1,200 0 $1,280 $2,048 $2,048 $2,624 $3,200 $4,200
PW = $6,831.37 > 0 Purchase the van.
Chapter 7 Solutions
184
ATCF $25,000 $6,800 $8,000 $6,720 $5,952 $5,952 $5,376 $4,800 $6,300
PW of ATCF $25,000 $6,071.72 $6,377.60 $4,783.30 $3,782.50 $3,377.16 $2,723.48 $2,171.04 $2,544.57
742
Leasing Option: PW(10%) = $166,036; AW(10%) = $43,800 EOY
BTCF (A)
Depr
0 0 1 2 3 4 5
$75,000 $55,000 $55,000 $55,000 $55,000 $55,000 +$75,000

Interest
0 0 0 0 0
Principal Payment (B)
0 0 0 0 0
0 0 0 0 0
TI $55,000 $55,000 $55,000 $55,000 $55,000
Taxes Payable (40%) (C) $22,000 $22,000 $22,000 $22,000 $22,000
ATCF (A) + (B) + (C) $75,000 $33,000 $33,000 $33,000 $33,000 $33,000 $75,000
Purchasing Option: PW(10%) = $191,197; AW(10%) = $50,438 EOY
BTCF (A)
Depr
Interest
0 1 2 3 4 5a 5b
$350,000 $20,000 $20,000 $20,000 $20,000 $20,000 +$150,000
$116,655 $155,575 $51,835 $25,935 0 
$28,000 $19,376 $10,062 0 0 
Principal Payment (B) $350,000 $107,800 $116,424 $125,738 0 0 
TI
$164,655 $194,951 $81,897 $45,935 $20,000 +$150,000
Taxes Payable (40%) (C) $65,862 $77,980 $32,759 $18,374 $8,000 $60,000
ATCF (A) + (B) + (C) $0 $89,938 $77,820 $123,041 $1,626 $12,000 +$90,000
Payment = $350,000 (A/P,8%,3) = $135,800 Select the lease.
185
Chapter 7 Solutions
Solutions to Spreadsheet Exercises 743 MARR Cost Basis Useful Life Market Value
10% $500,000 10 $ 20,000
DB Rate MACRS Recovery Period EOY
200% 7
SL Method DB Method
1 2 3 4 5 6 7 8 9 10 PW(10%)
$ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $ 48,000 $294,939
$100,000 $ 80,000 $ 64,000 $ 51,200 $ 40,960 $ 32,768 $ 26,214 $ 20,972 $ 16,777 $ 13,422 $319,534
MACRS Method $ 71,429 $122,449 $ 87,464 $ 62,474 $ 44,624 $ 44,624 $ 44,624 $ 22,312
$360,721
The MACRS method results in the largest PW of the depreciation deductions.
744 (a) Aftertax MARR = effective tax rate =
15%
Capital Investment =
40.00%
Market Value = Annual Savings = Useful Life =
EOY 0 1 2 3 4 4
BTCF $(10,000,000) $ 4,000,000 $ 4,000,000 $ 4,000,000 $ 4,000,000 $ 
$ 10,000,000
Depreciation Deduction $ $ $ $
2,500,000 2,500,000 2,500,000 2,500,000
Chapter 7 Solutions
Taxable Income
$ $ $ $ $
1,500,000 1,500,000 1,500,000 1,500,000 
186
$ $
4,000,000 4
Cash Flow for Income Taxes $ $ $ $
(600,000) (600,000) (600,000) (600,000) $ 
ATCF
Adjusted ATCF
$ (10,000,000) $ 3,400,000 $ 3,400,000 $ 3,400,000 $ 3,400,000 $ 
$(10,000,000) $ 3,400,000 $ 3,400,000 $ 3,400,000 $ 3,400,000
PW = IRR =
(293,073.57) 13.54%
(b) Aftertax MARR = effective tax rate =
EOY 0
15% 40.00%
BTCF
Depreciation Deduction
Capital Investment =
$ 10,000,000
Market Value =
$
1,024,000
Annual Savings = Useful Life =
$
4,000,000 4
Taxable Income
Cash Flow for Income Taxes
ATCF
$(10,000,000)
Adjusted ATCF
$ (10,000,000)
1
$ 4,000,000
$
2,244,000
$
1,756,000
$
(702,400)
$
3,297,600
2
$ 4,000,000
$
2,244,000
$
1,756,000
$
(702,400)
$
3,297,600
3
$ 4,000,000
$
2,244,000
$
1,756,000
$
(702,400)
$
3,297,600
4
$ 4,000,000
$
2,244,000
$
1,756,000
$
(702,400)
$
3,297,600
4
$ 1,024,000

$
1,024,000
$

$
$(10,000,00 0) $ 3,297,600 $ 3,297,600 $ 3,297,600 $ 4,321,600
PW = IRR = If a market value of at least $1,024,000 could be expected at the end of year 4, this investment would be acceptable.
187
Chapter 7 Solutions
51.97 15.00%
745 Aftertax MARR =
10%
Class Life = State tax rate =
5 6%
Federal tax rate =
34%
effective tax rate =
$
180,000
Market Value = Annual Savings =
$ $
30,000 36,000
Useful Life =
10
38.00%
EOY
0
Capital Investment =
BTCF
$
Depreciation Deduction
Taxable Income Cash Flow for Income Taxes
(180,000)
ATCF
$
Adjusted ATCF
(180,000)
(180,000)
$
36,000
$
15,000 $
21,000 $
(7,980)
$
28,020
$
28,020
$
36,000
$
36,000
$
15,000 $
21,000 $
(7,980)
$
28,020
$
28,020
$
36,000
3
$
36,000
$
15,000 $
21,000 $
(7,980)
$
28,020
$
28,020
$
36,000
4
$
36,000
$
15,000 $
21,000 $
(7,980)
$
28,020
$
28,020
$
36,000
5
$
36,000
$
15,000 $
21,000 $
(7,980)
$
28,020
$
28,020
$
36,000
6
$
36,000
$
15,000 $
21,000 $
(7,980)
$
28,020
$
28,020
$
36,000
7
$
36,000
$
15,000 $
21,000 $
(7,980)
$
28,020
$
28,020
$
36,000
8
$
36,000
$
15,000 $
21,000 $
(7,980)
$
28,020
$
28,020
$
36,000
9
$
36,000
$
15,000 $
21,000 $
(7,980)
$
28,020
$
28,020
$
36,000
10
$
36,000
$
15,000 $
21,000 $
(7,980)
$
28,020
$
58,020
$
66,000
10
$
30,000

$
30,000
PW =
$
3,737
$
52,771
AW =
$
608
$
8,588
$

$
Solutions to FE Practice Problems From Table 72, the GDS recovery period for woord products equipment (Asset Class 24.4) is 7 years. Select (b) Cost Basis = $100,000 + $10,000 = $110,000 $110,000 − $5,000 dk = = $10,500 10 Select (a) 748
$
2
IRR =
747
(180,000)
1
The project is still acceptable when straightline depreciation is used.
746
$
Adjusted BTCF
BV10 = $110,000 – (10)($10,500) = $5,000 Select (c)
Chapter 7 Solutions
188
10.4%
16.1%
749
d6 = $110,000 (0.0892) = $9,812 Select (a)
750
Using the half year convention: d5* = $110,000 [0.1429 +0.2449 +0.1749 + 0.0893(0.5)] = $80,547.50 BV5 = $110,000  $80,547.50 = $29.452.50 Select (a)
751
dk =
$12,000 − $2,000 = $1,250 8
Select (d) 752
dA*= $12,000 (0.1429 + 0.2449 + 0.1749 + 0.1249) = $8,251.20 BV4 = $12,000  $8,251.20 = $3,748.80 Select (a)
753
MACRS: d3 = $50,000 (0.1920) = $9,600 Straight Line: d3 =
$50,000 − $5,000 = $9,000 5
Difference = $9,600  $9,000 = $600 754
Select (d)
Using Equation (715): 0.40 = 0.20 + federal rate (1 0.20) federal rate = 0.25 or 25% Select (b)
755
BTCF5 = (RE) + MV = [$40,000 + $30,000] + $40,000 = $110,000 Select (e)
189
Chapter 7 Solutions
756
TI3 = $70,000  $135,000(0.1481) = $41,120 Select (d)
757
After Tax MARR = (10.40)(20%) = 12% PW(12%) = $70,000(10.40)(P/A,12%,5) = $42,000(3.6048) = $151,402 Select (c)
758
ATCFk = ($110,000  $65,000) – 0.40($110,000  $65,000  $25,000) = $37,000 Select (e)
759
0.1481 ) = $28,889 2 Deprecitaion Recapture = $50,000  $28,889 = $21,111 Taxes = 0.40($21,111) = $8,444 BV3 = $195,000  $195,000(0.3333 + 0.4445 +
Select (a) 760 Income Tax = $300,000(0.40) = $120,000
Select (c) 761 ∆ Income Tax = ∆ Taxable Income (0.40) = $150,000(0.40) = $60,000
Select (e)
Chapter 7 Solutions
190
Solutions to Chapter 8 Problems 81
A = $1,000; N = 10 (a)
f = 6% per year; ir = 4% per year In Part (a), the $1,000 is an A$ uniform cash flow (annuity) ic = 0.04 + 0.06 + (0.04)(0.06) = 0.1024, or 10.24% per year PW(ic) = $1,000 (P/A,10.24%,10) = $1,000(6.0817) = $6,082
(b)
In Part (b), the $1,000 is a R$ uniform cash flow (annuity) because the A$ cash flow is $1,000 (1.06)k where 1 ≤ k ≤ 10; i.e., ⎛ 1 ⎞ (R$)k = (A$)k ⎜ ⎟ ⎝1+ f ⎠
kb
k⎛
1 ⎞ = $1,000 (1.06) ⎜ ⎟ ⎝ 1 + 0.06 ⎠
k0
= $1,000; 1 ≤ k ≤ 10
PW(ir) = $1,000 (P/A,4%,10) = $8,111
82
f = 4% per year; N = ? (when does $1 equal $0.50 in today's purchasing power) From Equation 81, with k = 0, we have
⎛ 1 ⎞ ⎟ (R$)N = $0.50 = ($1) ⎜⎜ N ⎟ ⎝ (1.04) ⎠ (1.04)N = 2 N (ln (1.04)) = ln (2) N=
ln(2) ≅ 18 years ln(1.04)
In 18 years, the dollar's purchasing power will be onehalf of what it is now if the general price inflation rate is 4% per year.
83
Situation 1: FW5 (A$) = $2,500 (F/P, 8%, 5) = $2,500 (1.4693) = $3,673 Situation 2: FW5 (A$) = $4,000 (given) Choose situation 2. (Note: The general inflation rate, 5%, is a distractor not needed in the solution.)
193
Chapter 8 Solutions
84
f = 6% per year; ir = 9% per year; b = 0 Alternative A: Estimates are in actual dollars, so the combined (market) interest rate must be used to compute the present worth (PW). ic = ir + f + (ir)(f) = 0.09 + 0.06 + (0.09)(0.06) = 0.1554 or 15.54% per year PW(15.54%) =  $120,000(P/F,15.54%,1)  $132,000(P/F,15.54%,2)  $148,000(P/F,15.54%,3)  $160,000(P/F,15.54%,4) =  $120,000(0.8655)  $132,000(0.7491)  $148,000(0.6483)  $160,000(0.5611) =  $388,466 Alternative B: Estimates are in real dollars, so the real interest rate must be used to compute the present worth (PW). PW(9%) =  $100,000(P/A,9%,4)  $10,000(P/G,9%,4) =  $100,000(3.2397)  $10,000(4.511) =  $369,080 Alternative B has the least negative equivalent worth in the base time period (a PW value in this case since b = 0).
85
Alternative I II III
Present Worth of Costs $10,000 = $10,000 $7,000 + $5,000(P/F,8%,6) = $10,151 $5,000 + $5,000(P/F,8%,3) + $5,000(P/F,8%,6) = $12,120
Therefore, Alternative I is most economical with respect to the objective of minimizing equivalent cost. 86
The engineer's salary has increased by 6.47%, 7.18%, and 6.96% in years 2, 3, and 4, respectively. These are annual rates of change. By using Equation 81, but with each year's general price inflation taken into account separately, the R$ equivalents in year 0 dollars are calculated as follows. EOY
Salary (R$ in Year 0)
1
$34,000 (P/F,7.1%,1)
= $ 31,746
2 3
$36,200 (P/F,7.1%,1)(P/F,5.4%,1) $38,800 (P/F,7.1%,1)(P/F,5.4%,1)(P/F,8.9%,1)
= 32,069 = 31,564
4
$41,500 (P/F,7.1%,1)(P/F,5.4%,1)(P/F,8.9%,1)(P/F,11.2%,1)
=
Chapter 8 Solutions
194
30,361
87
( 0) R$ 10 = $400M(1.75) = $700M A$10 = $920M = $700M(1+f)10 1.314 = (1+f)10 f = 10 1.314  1 = 0.0277 or 2.77%
88
ic = (1.02)(1.05)  1 = 0.071 or 7.1% Amount FW of Deposit @ EOY10 EOY Deposited (A$) (A$) ___________________________________________________________ 1 $1,000 $ 1,838 2 1,080 1,870 3 1,166.40 1,885 4 1,259.71 1,890 5 1,360.49 1,917 6 1,469.33 1,933 7 1,586.87 1,949 8 1,713.82 1,966 9 1,850.93 1,982 10 1,999.00 1,999 $19,231 FW10 = $19,231
89
Unit cost (8 years ago) = $89 / ft2 X = 0.92 eC = 5.4% per year eAE = 5.66% per year (a)
SB = 80,000 ft2 SA = 125,000 ft2 ic = MARRc = 12% per year f = 7.69% per year
Using the power sizing technique (exponential cost estimating model) from Section 3.4.1, with an adjustment for the price increase in construction costs, we have:
⎛S ⎞ CA = CB ⎜ A ⎟ ⎝ SB ⎠
X
(1 + eC)8
⎛ 125,000⎞ = ($89/ft2)(80,000 ft2) ⎜ ⎟ ⎝ 80,000 ⎠ = $16,350,060
0.92
(1.054)8
Total Capital Investment = $16,350,060 (1 + 0.05 + 0.042 + 0.08 + 0.31) = $24,230,790
195
Chapter 8 Solutions
(b)
Note: The building is not being sold at the end of the 10 years. Therefore, working capital is not considered to be recovered at that time.
PW(12%) = $24,230,790 – ($5)(125,000 ft2) = $24,230,790 –
[1 − ( P / F ,12%,10)( F / P,5.66%,10)] 0.12 − 0.0566
625,000(0.4416) 0.0634
= $28,584,102 (c)
ir =
0.12  0.0769 = 0.04 or 4% per year 1.0769
Assuming the base year to be the present (b = 0), we have: AW(4%) = $28,584,440 (A/P,4%,10) = $3,524,460 810
In this problem, b = 2007 (i.e., the purchasing power of a real dollar is defined by the fiscal year 2007 dollar), and f = 5.6% per year. For years 2005 and 2006, more 2007 dollars are required than the actual dollar amounts spent in those two years. This is because the 2007 dollar has less purchasing power than actual dollars in 2005 and 2006. Also, the entries in Column 4 indicate that in real dollars, the annual budget amounts decrease between 2006 and 2009. This contrasts with the actual dollar amounts in Column 2 which increase during this period. This difference reflects that the actual dollar amounts after 2006 increase at an annual rate less than the general price inflation rate and purchasing power is decreasing each year. (1)
811
Fiscal Year
(2) Budget Amount (A$)
2005 2006 2007 2008 2009
$1,615,000 1,728,000 1,780,000 1,858,300 1,912,200
(3)
(4)
(P/F,f %, kb) [1/(1.056)k  b]
Estimate (R$), b=2007
1.1151 1.0560 1.0000 0.9470 0.8968
$1,800,887 1,824,768 1,780,000 1,759,810 1,714,861
f = 7% per year; ic = 5% per year; N = 20 years F20 (A$) = $30,000 (F/P,7%,20) = $30,000(3.8697) = $116,091
Chapter 8 Solutions
196
But, since the account will only earn interest at the rate of 5% compounded per year (versus the general inflation rate of f = 7% per year) we have: P0 (A$) = $116,091 (P/F,5%,20) = $116,091(0.3769) = $43,755 812
MARR = ic = 25% per year; Assume f = 8% per year; Let k = 0 Note that the estimated cash flows are in R$ except for the contract maintenance agreement ($3,000 / year). However, the PW of $3,000 per year at ic = 25% per year is equal to the PW of the R$ equivalent at ir. Therefore, the PW of the cash flows as a function of N is: PW = $50,000 + $18,000 (P/A, ir , N)  $3,000 (P/A, 25%, N) and, ir =
0.25  0.08 = 0.1574 or 15.74% per year 1.08
By trial and error we have:
N 3 4 5
PW $15,257 6,455 1,230
The life of the computer system must be at least 5 years for it to be economically justified. 813
In 10 years, the investor will receive the original $10,000 plus interest that has accumulated at 10% per year, in actual dollars. Therefore, the market rate of return (IRRc) is 10%. Then, based on Equation 85, the real rate of return (IRRr) is: ir' =
814
815
ic  f = 0.0185, or 1.85% per year 1+ f
(a)
FW(A$) = $2,000 (F/A, ic = 12%, 20) = $144,105
(b)
FW(R$) = FW(A$) (P/F, f %, 20) = $144,105 (P/F, 6%, 20) = $44,932
(Equation 81) (b = 1991)
(a) Cost in year 2020 = $15,000 (F/P,6%,15) = $35,949 Cost in year 2021 = $38,106 Cost in year 2022 = $40,392 Cost in year 2023 = $42,816 Total (undiscounted dollars) = $157,263
197
Chapter 8 Solutions
(b) Because ic = f , P2020 = 4($35,949) = $143,796 so A = $143,796 (A/P,0.5%,180 months) = $143,796 (0.00844) = $1,214 per month 816
$20,000 = $10,000(F/P,ic,11) or ic = 0.065 (6.5%) per year. ir= (icf)/(1+f) = (0.0650.03)/(1.03) = 0.03398 or 3.4% per year. This is a failrly good return in real terms. Historically, real returns have been in the 23% per year ball park.
817
Assume neither machine has a salvage value. Incremental analysis ∆(BA) (A)
(B)
(C)
(D)
(E)
(F)
(G)
Year
BTCF (R$)
Adjustment (1.10)Year1
BTCF (A$)
Depreciation
Taxable Income: (C) – (D)
Cash Flow for Income Taxes t(E)
ATCF (A$) CD+F
0
$50,000 500 500 500 500 500 500 500 500 500 500
1.000 1.100 1.210 1.331 1.464 1.611 1.772 1.949 2.144 2.358
500 550 605 666 732 805 886 974 1,072 1,179
$10,000 16,000 9,600 5,760 5,760 5,760 0 0 0 0
$4,750 7,725 4,498 2,547 2,514 1,037  443  487  536  589
$50,000 5,250 8,275 5,103 3,213 3,246 1,843  443  487  536  590
1 2 3 4 5 6 7 8 9 10
∑
cash outflows > $50,000
>
∑
 $9,500  15,450  8,995  5,095  5,028  2,075  886  974 1,072 1,179
cash inflows $28,986
Therefore, IRR∆ < 0%. The company should choose Machine A. 818 Option 1: Software with 3 year upgrade agreement. Year
(A) BTCF (A$)
(B) Depreciation
0 1 2 3
$X 0 0 0
$X/3 $X/3 $X/3
Taxable Cash Flow Income: for Income Taxes C=A+B D = t(C) $X/3 $0.1133X $X/3 $0.1133X $X/3 $0.1133X
PW1(20%) = $X + $0.1133X(P/A,20%,3) Chapter 8 Solutions
198
ATCF (A$) A+D $X $0.1133X $0.1133X $0.1133X
Year
(A) BTCF (A$)
(B) Depreciation
1 2 3
$20,000  22,000  24,200

Cash Flow for Income Taxes D = t(C)  $20,000 $6,800  22,000 7,480  24,200 8,228 Taxable Income: C=A+B
ATCF (A) A+D $13,200  14,520  15,972
PW2(20%) = $13,200(P/F,20%,1)  $14,520(P/F,20%,2)  $15,972(P/F,20%,3) = $30,326 Set PW1 = PW2 and solve for X. $X + $0.1133X(P/A,20%,3) = $30,326 $0.761X = $30,326 X = $39,836 Therefore, $39,836 could be spent for software with a 3 year upgrade agreement (i.e., Option 1). 819
(a) In two years: $1 (1.026)2 = 6.4X or, $1 = 6.4X / (1.026)2 = 6.08 units of X. (b)
820
In three years: $1 = (6.4X) (1.026)3 = 6.91 units of X.
iUS = 26% per year (a)
fe = 8% per year ifc = 0.26 + 0.08 + (0.26)(0.08) = 0.3608, or 36.08% per year (IRR on project in Country A currency)
(b)
fe = 6% per year ifc = 0.26 + ( 0.06) + (0.26)( 0.06) = 0.1844, or 18.44% per year (IRR on project in Country B currency)
199
Chapter 8 Solutions
821 (a) NCF EOY (Tmarks) 0 $3,600,000 1 450,000 2 1,500,000 3 1,500,000 4 1,500,000 5 1,500,000 6 1,500,000 7 1,500,000
Exchange Rate (Tmarks/$) 20.000 22.400 25.088 28.099 31.470 35.247 39.476 44.214
NCF ($) $180,000 20,089 59,790 53,383 47,664 42,557 37,998 33,926 PW (18%) =
PW(18%)  $180,000 17,025 42,941 32,489 24,585 18,602 14,074 10,649  $19,635
Project is not economically acceptable. (b)
IRRfc in terms of Tmarks: PW(i'%) = 3,600,000 + 450,000 (P/F, i'%, 1) + 1,500,000 (P/A, i'%, 6) (P/F, i'%, 1) By linear interpolation, i'% = IRRfc = 0.2798, or 28.0% per year.
(c)
From Equation 87, we have: (IRR)US =
IRR fc  f e 0.28  0.12 = = 0.1429, or 14.29% < 18% 1.12 1 + fe
Note: This confirms our recommendation in part (a). 822
ifc = 20% per year; fe = 2.2% per year Current exchange rate = $1 per 92 ZKrons i US =
0.20  (0.022) = 22.7% 1  0.022
PW(22.7%) = $168,000,000  $32,000,000(P/F,22.7%,1) + $69,000,000(P/A,22.7%,10) = $168,000,000  $32,000,000(0.8150) + $69,000,000(3.8357) = $70,583,300 > 0 Yes, this project will meet the company's economic decision criteria.
Chapter 8 Solutions
200
823 (a) (A)
(B)
(C)
Year
BTCF (R$)
Adjustment (1.10)Year
BTCF (A$)
1 2 3 4 5 6
 $4,000  4,000  4,000  4,000  4,000  4,000
1.100 1.210 1.331 1.464 1.611 1.772
(D)
$4,400  4,840  5,324  5,856  6,442  7,086
(E)
(F)
(G)
Taxable Cash Flow Lease Payment Income: for Income Taxes (C) – (D) (A$) t(E) $80,000 60,000 50,000 50,000 50,000 50,000
$84,000  64,840  55,324  55,856  56,442  57,086
$33,760 25,936 22,130 22,342 22,577 22,834
ATCF (A$) CD+F $50,640  38,904  33,194  33,514  33,865 34,252
(b) ic = (1.05)(1.09524) – 1 = 0.15 = 15% per year PW = $50,640(P/F,15%,1)  $38,904(P/F,15%,2)  $33,194(P/F,15%,3)  $33,514(P/F,15%,4)  $33,865(P/F,15%,5)  $34,252(P/F,15%,6) = $146,084 EUAC = $146,084(A/P,15%,6) = $38,595 824 EOY
Savings
O&M Costs
Net CF
PW(Net CF)
1
15,000.00
3,500.00
11,500.00
10,454.55
2
16,050.00
3,750.00
12,300.00
10,165.29
3
17,173.00
4,000.00
13,173.50
9,897.45
4
18,375.00
4.250.00
14,125.00
9,648.01
5
19,661.94
4,500.00
15,161.94
9,414.37
6
21,038.28
4,750.00
16,288.28
9,194.31
7
22,510.96
5,000.00
17,510.96
8,985.89
8
24,086.72
5,250.00
518,836.72
8,787.47
9
25,772.79
5,500.00
20,272.79
8,597.64
10
27,576.89
5,750.00
21,826.89
8,415.21 PW = $93,560.18
201
Chapter 8 Solutions
Alternate Solution: Savings as geometric gradient, costs as uniform gradient PW = $15,000/(1.07)[P/A,(0.100.07)/1.07,10] – 3,500(P/A,10%,10) – 250(P/G,10%,10) i(r) = (0.100.07)/1.07 = 2.8037% PW = $93,560.51
825
Demand = 500 million BTU/year; Efficiency = 80%, N = 12 years f = 10% per year; b = 0 MARR = ic = 18% per year 3 ⎛ 500 million Btu ⎞ ⎛ 1,000 ft of gas ⎞ Annual gas demand = ⎜ ⎟ = 625,000 ft3 of gas ⎟⎜ ⎝ ⎠ ⎝ million Btu ⎠ 0.8
A1 =
$2.50(1.1) $2.75 = 3 1000ft 1000ft 3
⎛ $2.75 ⎞ [1 − (P / F,18%,12)(F / P,10%,12)]] PW(18%) =  (625,000 ft3) ⎜ 3 ⎟ 0.18 − 0.10 ⎝ 1000 ft ⎠ ⎛ $2.75 ⎞ =  (625,000 ft3) ⎜ (7.1176) 3 ⎟ ⎝ 1000 ft ⎠
= $12,233 826
(a) Cost of compressor replacement at EOY 8 (A$) = $500(1 + 0.06)8 = $797 Annual maintenance expense: A1(A$) = $100 (1.06) = $106 Annual electricity expense: A1(A$) = $680(1.10) = $748
PW(15%) = $2,500  $797 (P/F,15%,8) $106 [1 − ( P / F ,15%,15)( F / P,6%,15)] 0.15 − 0.06
$748 [1 − ( P / F ,15%,15)( F / P,10%,15)] 0.15 − 0.10 = $2,500  $797 (0.3269)  $106(0.70546)  $748(0.48662) 0.09 0.05 = $10,871
Chapter 8 Solutions
202
A$: AW(15%) = $10.871(A/P,15%,15) = $1,859 (b) ir = 0.15 − 0.06 = 0.085or 8.5% per year 1.06 = $10,871 (0.1204) = $1,309 827
f = 4.5% per year; ic (aftertax) = 12% per year; t = 40%; b = 0 increase rate = 6 % per year (applies to annual expenses, replacement costs, and market value) Analysis period = 20 years; Useful life = 10 years MACRS (GDS) 5year property class Capital investment (and cost basis, B) Market value (at end of year 10) in year 0 dollars
= $260,000 = $50,000
Annual expenses (in year 0 dollars) = $6,000 Annual property tax = 4% of capital investment (does not inflate) Assume like replacement at end of year 10. EOY 0 1 2 3 4 5 6 7 8 9 10 10 10 11 12 13 14 15 16 17 18 19 20 20
Annual Property Expenses Taxes $6,360 $10,400 6,742 10,400 7,146 10,400 7,575 10,400 8,029 10,400 8,511 10,400 9,022 10,400 9,563 10,400 10,137 10,400 10,745 10,400
11,390 12,073 12,798 13,565 14,379 15,242 16,157 17,126 18,154 19,243
18,625 18,625 18,625 18,625 18,625 18,625 18,625 18,625 18,625 18,625
ATCF ATCF* BTCF Depr. TI T(40%) (A$) (R$) $260,000 $260,000 $260,000 16,760 52,000 68,760 27,504 10,744 10,281 17,142 83,200 100,342 40,137 22,995 21,057 17,546 49,920 67,466 26,986 9,440 8,272 17,975 29,952 47,927 19,171 1,196 1,003 18,429 29,952 48,381 19,352 923 741 18,911 14,976 33,887 13,555 5,356 4,113 19,422 0 19,422 7,769 11,653 8,563 19,963 0 19,963 7,985 11,978 8,423 20,537 0 20,537 8,215 12,322 8,292 21,145 0 21,145 8,458 12,687 8,170 89,542 89,542 35,817 53,725 34,595 465,620 465,620 299,826 30,015 93,124 123,139 49,256 19,241 11,856 30,698 148,998 179,696 71,878 41,180 24,282 31,423 89,399 120,822 48,329 16,906 9,540 32,190 53,639 85,829 34,332 2,142 1,157 33,004 53,639 86,643 34,657 1,653 854 33,867 26,820 60,687 24,275 9,592 4,743 34,782 0 34,782 13,913 20,869 9,875 35,751 0 35,751 14,300 21,451 9,713 36,779 0 36,779 14,712 22,067 9,562 37,868 0 37,868 15,147 22,721 9,421 160,357 160,357 64,143 96,214 39,894
203
Chapter 8 Solutions
* ATCF(R$) = ATCF(A$) × 1/(1.045)k ir =
012 . − 0.045 = 0.0718 or 7.18% per year 1.045 20
PW =
20
∑ ATCF (A$) (P/F, 12%, k) = ∑ ATCF (R$) (P/F, 7.18%, k) = $359,665 k
k
k=0
828
k=0
dk = ($150,0000)/3 = $50,000 (A)
(B)
Year Revenues Expenses (A$) (A$)
0 1 2 3
$84,000 88,200 92,610
$150,000 21,800 23,762 25,900
(C)
(D)
(E)
(F)
(G)
BTCF (A$) A+B
Depreciation
Taxable Income: CD
Cash Flow for Income Taxes t(E)
ATCF (A$) C+F
$6,100 7,219 8,355
$150,000 56,100 57,219 58,354
$150,000 62,200 64,438 66,709
$50,000 50,000 50,000
$12,200 14,438 16,709
For discounting purposes, ic = 26% would be used since the ATCFs are expressed in actual dollars. 829
Annual revenues in year k (A$) = $360,000(1.025)k Annual expenses in year k (A$) = $239,000(1.056)k (a)
The values in the following table are expressed in A$.
Annual Annual EOY Revenues Expenses BTCF 0 $220,000 1 $369,000  $252,384 116,616 2 378,225  266,518 111,707 3 387,681  281,442 106,239 4 397,373  297,203 100,170 5 407,307  313,847 93,460 6 417,490  331,422 86,068 6 40,000
Chapter 8 Solutions
Depr TI $44,000 $72,616 70,400 41,307 42,240 63,999 25,344 74,826 25,344 68,116 12,672 73,396 40,000
204
T(39%)  $28,320  16,110  24,960  29,182  26,565  28,624  15,600
ATCF (A$) $220,000 88,296 95,597 81,279 70,988 66,895 57,444 24,400
6
PW(10%) =
∑ ATCF (P/F,10%,k) = $136,557 k
k=0
Total investment that can be afforded (including new equipment) = $136,557 + $220,000 = $356,557 (b)
ATCFk (R$) = ATCFk (A$)(P/F,4.9%,k) ATCFk (A$)  $220,000 88,296 95,597 81,279 70,988 66,895 57,444 24,400
Year, k 0 1 2 3 4 5 6 6
830
(P/F,4.9%,k) 1.0000 0.9533 0.9088 0.8663 0.8258 0.7873 0.7505 0.7505
ATCFk (R$)  $220,000 84,173 86,879 70,412 58,622 52,666 43,112 18,312
Purchase (A$ Analysis): Oper., Ins. & Investment / Other Expenses (O,I,OE) EOY Market Value 0  $600,000 1  $27,560b 2  29,214 3  30,966 4  32,824 5  34,794 6  36,881 6 101,355a EOY 0 1 2 3 4 5 6 6
BTCF  $600,000  62,440  67,233  72,407  77,995  84,030  90,549 101,355
Maintenance Expense  $34,880c  38,019  41,441  45,171  49,236  53,667
Deprd
TI
T(34%)
$120,000 192,000 115,200 69,120 69,120 34,560
 $182,440  259,233  187,607  147,115  153,150  125,108 101,355
$62,030 88,139 63,786 50,019 52,071 42,537  34,461
205
BTCF  $600,000  62,440  67,233  72,407  77,995  84,030  90,549 101,355 ATCF (A$)  $600,000 410 20,906 8,621  27,976  31,959  48,012 66,894
Chapter 8 Solutions
Notes:
a
(MV)6 = $90,000 (1.02)6 = $101,355 b (O,I,OE)k = $26,000 (1.06)k c (Maint)k = $32,000 (1.09)k d Cost Basis = $600,000 ic = 0.13208 + 0.06 + (0.13208)(0.06) = 0.20, or 20% per year 6
FW6 (A$) =
∑ ATCF (F/P, 20%, 6  k) =  $1,823,920 k
k=0
Lease (A$ Analysis): EOY 1 2 3 4 5 6
EOY 1 2 3 4 5 6
Leasing Costs  $300,000  200,000  200,000  200,000  200,000  200,000
Oper., Ins. & Other Expenses  $27,560  29,214  30,966  32,824  34,794  36,881
BTCF  $362,440  267,233  272,407  277,995  284,030  290,549
Depr 0 0 0 0 0 0
Maint. Expense  $34,880  38,019  41,441  45,171  49,236  53,667
TI  $362,440  267,233  272,407  277,995  284,030  290,549
BTCF  $362,440  267,233  272,407  277,995  284,030  290,549
T(34%) $123,230 90,859 92,618 94,518 96,570 98,787
6
FW6 (A$) =
∑ ATCF (F/P, 20%, 6  k) =  $1,952,551 k
k=0
Therefore choose Purchase Alternative due to smaller FW of costs.
Chapter 8 Solutions
206
ATCF (A$)  $239,210  176,374  179,789  183,477  187,460  191,762
831
Assuming that EOY 1 cost for purchased components is $85,000,000 EOY 0 1 2 3 4 5
Cash Flow (w) 20,000,000 85,000,000 80,750,000 76,712,500 72,876,875 69,233,031
Cash Flow (w/o) 0 85,000,000 85,000,000 85,000,000 85,000,000 85,000,000
PW
300,467,957
306,405,977
i(r) = 0.178947368 (P/A,i(r),5) = 3.134641881 PW(w) = 20,000,000 – (85,000,000/(1.05))[P/A,((0.12+0.05)/(10.05)),5] = 300,467,957.81 PW(w/o) = 85,000,000(P/A,12%,5) = 306,45,977.20 PW(Difference) = $5,938,019.39 AW(Difference) = $1,647,264.37 Assuming that EOY 1 cost for purchased components is $85,000,000 (10.05) EOY 0 1 2 3 4 5
Cash Flow (w) 20,000,000.00 80,750,000.00 76,712,500.00 72,876,875.00 69,233,031.25 65,771,379.69
Cash Flow (w/o) 0.00 85,000,000.00 85,000,000.00 85,000,000.00 85,000,000.00 85,000,000.00
PW
286,444,559.92
306,405,977.20
i(r) = 0.178947368 (P/A,i(r),5) = 3.134641881 PW(w) = 20,000,000 – 85,000,000[P/A,((0.12+0.05)/(10.05)),5] = 286,444,559.92 PW(w/o) = 85,000,000(P/A,12%,5) = 306,405,977.20 PW(Difference) = $19,961,417.28 AW(Difference) = $5,537,491.42
207
Chapter 8 Solutions
832
This is intended to be a tailormade exercise (at the discretion of the instructor). Assumptions: • Salary and fringe benefits for the new analyst will be $28,000 (1.3) = $36,400 in year 1 purchasing power. This increases 6% per year thereafter. • Staff retirements occur at the end of the year. Therefore, there are no realized savings in year 1. Savings of $16,200 in year 2, $32,400 in year 3, and $48,600 each year thereafter are expressed in real purchasing power keyed to year 0. • Firstyear savings on purchases are 3% of $1,000,000 (1.10) = $33,000 and this increases by 10% per year thereafter. • Contingency costs will not be considered as cash flows until they are spent (we assume they won’t be spent). • The effective income tax rate is = 38%. • There is no market value at the end of the 6year project life. Capital EOY Investment 0  $80,000 1 2 3 4 5 6
EOY 0 1 2 3 4 5 6
BTCF $80,000 9,400 9,918 31,619 55,926 61,398 67,376
Service Contract  $6,000  6,000  6,000  6,000  6,000  6,000
Depr. $16,000 25,600 15,360 9,216 9,216 4,608
New Manpower Savings on Analyst Savings Purchases $36,400  38,584  40,899  43,353  45,954  48,711
TI $25,400 15,682 16,259 46,710 52,182 62,768
6
PW (15%) =
∑ ATCF (P/F, 15%, k) = $10,263 k
k=0
Chapter 8 Solutions
$ 0 18,202 38,588 61,356 65,037 68,940
208
$33,000 36,300 39,930 43,923 48,315 53,147
T(38%) $9,652 5,959 6,178 17,750 19,829 23,852
Total BTCF $80,000  9,400 9,918 31,619 55,926 61,398 67,376
ATCF (A$) $80,000 252 15,877 25,441 38,176 41,569 43,524
In view of MARR of 15%, this investment should be undertaken. The instructor may wish to ask the class to explore various “what if” questions involving changes in the assumptions listed above. For example, how much change would occur in the PW value if we assume staff retirements occur at the beginning of the year?)
209
Chapter 8 Solutions
Solutions to Spreadsheet Exercises 833 3 4 5 6 7 8
A Savings Interest Rate = Average Inflation Rate = Desired amount in 2022 (R$) = $ Desired Amount in 2022 (A$) =
B
C
D
4.00% 4.00% 500,000
$ 1,332,918
Savings Bank Balance Salary (A$) (A$) (A$) 9 Year 10 1997 $ 60,000 $ 11,532 $ 11,532 11 1998 $ 64,800 $ 12,455 $ 24,448 12 1999 $ 69,984 $ 13,451 $ 38,877 13 2000 $ 75,583 $ 14,527 $ 54,959 14 2001 $ 81,629 $ 15,689 $ 72,846 15 2002 $ 88,160 $ 16,944 $ 92,704 16 2003 $ 95,212 $ 18,300 $ 114,712 17 2004 $ 102,829 $ 19,764 $ 139,065 18 2005 $ 111,056 $ 21,345 $ 165,972 19 2006 $ 119,940 $ 23,053 $ 195,664 228,387 20 2007 $ 129,535 $ 24,897 $ 2008 $ 139,898 $ 26,888 $ 264,411 21 2009 $ 151,090 $ 29,040 $ 304,027 22 2010 $ 163,177 $ 31,363 $ 347,551 23 2011 $ 176,232 $ 33,872 $ 395,324 24 2012 $ 190,330 $ 36,581 $ 447,719 25 2013 $ 205,557 $ 39,508 $ 505,135 26 2014 $ 222,001 $ 42,669 $ 568,010 27 2015 $ 239,761 $ 46,082 $ 636,812 28 2016 $ 258,942 $ 49,769 $ 712,053 29 2017 $ 279,657 $ 53,750 $ 794,285 30 2018 $ 302,030 $ 58,050 $ 884,107 31 2019 $ 326,192 $ 62,694 $ 982,165 32 2020 $ 352,288 $ 67,710 $ 1,089,162 33 2021 $ 380,471 $ 73,126 $ 1,205,855 34 2022 $ 410,909 $ 78,977 $ 1,333,066 35 36 $ 147 37 Difference between Desired and Actual = 38 39 40 41 42 43 44 45 46 47 48 49
Chapter 8 Solutions
210
E
F
G
834 A 1 Starting Salary = Annual Salary 2 Increase = Savings Interest 3 Rate = Average Inflation 4 Rate = Desired amount in 5 2027 (R$) = 6 Desired Amount in 7 2027 (A$) = 8
B $
C
60,000
D % Salary to Save Annually
E 8.64%
F
G 864
8.00% 7.50% 3.75% $
500,000
$
1,508,736
Salary (A$) Savings (A$) Year 9 1997 $ 60,000 $ 5,184 10 1998 $ 64,800 $ 5,599 11 1999 $ 69,984 $ 6,047 12 2000 $ 75,583 $ 6,530 13 2001 $ 81,629 $ 7,053 14 2002 $ 88,160 $ 7,617 15 2003 $ 95,212 $ 8,226 16 2004 $ 102,829 $ 8,884 17 2005 $ 111,056 $ 9,595 18 2006 $ 119,940 $ 10,363 19 2007 $ 129,535 $ 11,192 20 2008 $ 139,898 $ 12,087 21 2009 $ 151,090 $ 13,054 22 2010 $ 163,177 $ 14,099 23 2011 $ 176,232 $ 15,226 24 2012 $ 190,330 $ 16,445 25 2013 $ 205,557 $ 17,760 26 2014 $ 222,001 $ 19,181 27 2015 $ 239,761 $ 20,715 28 29 2016 $ 258,942 $ 22,373 30 2017 $ 279,657 $ 24,162 31 2018 $ 302,030 $ 26,095 32 2019 $ 326,192 $ 28,183 33 2020 $ 352,288 $ 30,438 34 2021 $ 380,471 $ 32,873 35 2022 $ 410,909 $ 35,502 36 2023 $ 443,781 $ 38,343 37 2024 $ 479,284 $ 41,410 38 2025 $ 517,626 $ 44,723 39 2026 $ 559,036 $ 48,301 40 2027 $ 603,759 $ 52,165 41 42 Difference between Desired and Actual = 43 44 45 46 47 48 49 50 51 52
Bank Balance (A$) $ 5,184 $ 11,172 $ 18,056 $ 25,941 $ 34,939 $ 45,176 $ 56,791 $ 69,935 $ 84,775 $ 101,496 $ 120,300 $ 141,410 $ 165,070 $ 191,548 $ 221,141 $ 254,171 $ 290,994 $ 331,999 377,615 $ $ 428,308 $ 484,594 $ 547,034 $ 616,244 $ 692,900 $ 777,741 $ 871,574 $ 975,284 $ 1,089,841 $ 1,216,302 $ 1,355,825 $ 1,509,677 $
941
211
Chapter 8 Solutions
Solutions to Case Study Exercises 835
PW(maintenance costs) =
A1[1 − ( P / F ,8%,5)( F / P,4%,5)] 0.08 − 0.04
= A1 (4.29785)
where A1 = $1,000 for the induction motor and A1 = $1,250 for the synchronous motor. PW(electricity costs) =
A1[1 − ( P / F ,8%,5)( F / P,5%,5)] 0.08 − 0.05
= A1 ( 4.37834 )
where A1 = $50,552 for the induction motor and A1 = $55,950 for the synchronous motor. PWTC(induction motor) = $17,640 + $1,000(4.29785) + $50,552(4.37834) = $17,640 + $4,298 + $221,334 = $243,272 PWTC(synchronous motor) = $24,500 + $1,250(4.29785) + $55,950(4.37834) = $24,500 + $5,372 + $244,968 = $274,840 PWTC(A) = (3)($243,272) + [$17,640 + $4,298 + (350/400)($221,334)] = $945,421 PWTC(B) = (3)($274,840) + [$24,500 + $5,372 + (50/500)($244,968)] = $878,889 PWTC(C) = (3)($243,272) + [$24,500 + $5,372 + (350/500)($244,968)] = $931,166 PWTC(D) = (3)($274,840) + [$17,640 + $4,298 + (50/400)($221,334)] = $874,125 Option (D) has the lowest present worth of total costs. Thus, the recommendation is still to power the assembly line using three 500 hp synchronous motors operated at a power factor of 1.0 and one 400 hp induction motor. 836
PW(electricity costs) =
Chapter 8 Solutions
A1[1 − ( P / F ,8%,8)( F / P,6%,8)] 0.08 − 0.06
212
= A1 (6.9435)
where A1 = $50,552 for the induction motor and A1 = $55,950 for the synchronous motor. PWTC(induction motor) = $17,640 + $1,000(6.5136) + $50,552(6.9435) = $17,640 + $6,514 + $351,008 = $375,162 PWTC(synchronous motor) = $24,500 + $1,250(6.5136) + $55,950(6.9435) = $24,500 + $8,142 + $388,489 = $421,131 PWTC(A) = (3)($375,162) + [$17,640 + $6,514 + (350/400)($351,008)] = $1,456,772 PWTC(B) = (3)($421,131) + [$24,500 + $8,142 + (50/500)($388,489)] = $1,334,884 PWTC(C) = (3)($375,162) + [$24,500 + $8,142 + (350/500)($388,489)] = $1,430,070 PWTC(D) = (3)($421,131) + [$17,640 + $6,514 + (50/400)($351,008)] = $1,331,423 Option (D) has the lowest present worth of total costs. Thus, the recommendation is still to power the assembly line using three 500 hp synchronous motors operated at a power factor of 1.0 and one 400 hp induction motor. 837
The following four options will be considered: (A) (B) (C) (D)
Four induction motors (three at 400 hp, one at 100 hp) Three synchronous motors (two at 500 hp, one at 300 hp) Three induction motors at 400 hp plus one synchronous motor at 100 hp Two synchronous motors at 500 hp plus one induction motor at 300 hp.
PWTC(A) = (3)($364,045) + [$17,640 + $6,514 + (100/400)($339,891)] = $1,201,262 PWTC(B) = (2)($408,827) + [$24,500 + $8,142 + (300/500)($376,185)] = $1,076,007 PWTC(C) = (3)($364,045) + [$24,500 + $8,142 + (100/500)($376,185)] = $1,200,014 PWTC(D) = (2)($408,827) + [$17,640 + $6,514 + (300/400)($339,891)] 213
Chapter 8 Solutions
= $1,096,726 Option (B) has the lowest present worth of total costs. Thus, the recommendation is to power the assembly line using three 500 hp synchronous motors: two operated at 500 hp and one at 300 hp.
Solutions to FE Practice Problems 838
Expected cost of Machine in 2004 = $2,550(1.07)4 = $3,342.53 True percentage increzse in cost =
$3,930 − $3.342.53 x 100% = 17.58% $3,343.53
Select (d) 839 ir = 0.07; f = 0.09 ic = 0.07 + 0.09 + (0.07)(0.09) = 0.1663 Select (a) 840 A$ Analysis: ic – 0.098 + 0.02 + (0.098)(0.02) = 0.12 or 12% PWA(12%) = $27,000 + $4,000(P/A,12%,5) = $2,581 PWB(12%) = $19,000 + $5,000(P/A,12%,5) = $976 Neither alternative is acceptable. Select (c) 841
A1 = $25,000 (1.04) = $26,000 PW =
=
$26,000[1 − ( P / F ,7%, ∞)( F / P,4%, ∞ ] 0.07 − 0.04
$26,000 0.03
⎡ (1.04) ∞ ⎤ ⎢1 − ∞ ⎥ ⎣ (1.07) ⎦ 0
= $866,667 Select (a)
Chapter 8 Solutions
214
842 ($1.60 Canadian/Euro)($0.75 US/$1.00 Canadian) = $1.20 US/Euro Select (c)
215
Chapter 8 Solutions
Solutions to Chapter 10 Problems 101
Advantages of taking uncertainty into account include the following: • A more comprehensive analysis of strengths and weaknesses of alternatives is possible when uncertainty in project outcomes is made explicit. • Variability in economic attractiveness of a project can be quantified and compared with variability associated with other projects. • Consideration of uncertainty usually forces estimators to be more detailed and deliberate in preparing their estimates and communicating them to other persons. • The need for more information from possible new sources is often identified. • Attempting to deal with uncertainty may formalize a rational framework for resolving misconceptions concerning alternatives under study and for establishing a continuing evaluation of future investment opportunities. Some likely sources of uncertainty in engineering economy studies are: • Insufficient experience and data related to similar engineering situations. • Bias in the data, or incorrect data. • A changing external economic and technical environment. • Misinterpretation of data (incorrect assessment). • Errors in economic analysis techniques. • Lack of managerial talent in terms of having a “partner” in preparation and interpretation of findings.
102
Individual assignment. (T&I)
103
$12,500 (A/P,15%,10) 
100(0.746)($0,05)(1,000)
ηα
 500 – 187
= $16,000 (A/P,15%,10) (T&I)
100 3,730 (0.746)(0.05)(1,000) – 250 – 240  2,490  687 0.92 ηα = 3,190 – 4,054.35 – 490 3,730 = 7,734.35 + 3,177 
ηα 3,730 = ηα ; 4,557.35
ηα = 0.818
249
Chapter 10 Solutions
104
If Extra Cost = $390,000: (+ 30%) $1,400,000  $850,000(P/F,10%, ˆt ) = $1,250,000  $1,240,000(P/F,10%, ˆt ) $390,000(P/F,10%, ˆt ) = $150,000 (P/F,10%, ˆt ) = 0.385 ˆt = 10 years If Extra Cost = $210,000: (30%) $210,000(P/F,10%, ˆt ) = $150,000 (P/F,10%, ˆt ) = 0.714 4−3 4−X = ; X = 3.5 0.683 − 0.7513 0.683  0.714 ˆt = 3.5 years Y Select Design 2 Deferral Time,
Tˆ years)
 30%
105
(a)
9 8 7 6 5 4 3 2 1
Select Design 1
X + 30%
% Change in Deferred Cost Savings
AW1(15%) = $4,500 (A/P, 15%, 8) + $1,600  $400 + $800 (A/F, 15%, 8) = $4,500(0.2229) + $1,200 + $800 (0.0729) = $255 AW2(15%) = $6,000(A/P, 15%,10) + $1,850  $500 + $1,200(A/F,15%,10) = $6,000(0.1993) + $1,350 + $1,200 (0.0493) = $213 Therefore, the initial decision is to select Alternative 1. To determine the market value of Alternative 2 (MV2) so that the initial decision would be reversed, equate the AWs: AW1(15%) = AW2(15%) $255 = $6,000 (A/P, 15%, 10) + $1,350 + MV2 (A/F, 15%, 10) $255 = $6,000 (0.1993) + $1,350 + MV2 (0.0493) MV2 = ($255.27  $154.2)/0.0493 = $2,050 The market value of Alternative 2 would have to be $2,050 or greater for the initial decision to be reversed.
Chapter 10 Solutions
250
(b)
Set AW1(15%) = AW2(15%) and solve for N assuming market values remain constant. $4,500 (A/P,15%,N) + $1,200 + $800 (A/F,15%,N) = $213 $4,500 (A/P, 15%, N) + $986.64 + $800 (A/F, 15%, N) = 0 By trial and error, N = 7.3 years.
106
Let H = annual hours of operation. (a)
⎛ 100 hp ⎞ ⎛ 0.746 kW ⎞ ⎛ $0.10 ⎞ AWABC(12%) =  $170  ⎜ ⎟⎜ ⎟ (H) ⎟⎜ ⎝ 0.8 ⎠ ⎝ hp ⎠ ⎝ kW  hr ⎠
 $1,900(A/P,12%,10) AWABC(12%) =  $506.3  $9.33H ⎛ 100 hp ⎞ ⎛ 0.746 kW ⎞ ⎛ $0.10 ⎞ AWXYZ(12%) =  $310  ⎜ ⎟ (H) ⎟⎜ ⎟⎜ ⎝ 0.9 ⎠ ⎝ hp ⎠ ⎝ kW  hr ⎠
 $6,200(A/P,12%,10) AWXYZ(12%) =  $1,407.40  $8.29H To find the breakeven value of H, set AWABC = AWXYZ. $506.3  $9.33H =  $1,407.40  $8.29H $1.04H = $901.10 H = 866.4 hrs/year The XYZ brand motor is preferred if H = 867 hours per year. (b)
XYZ brand is preferred for H ≥ 867 hours per year. Therefore, if the expected annual hours of operation is 2,000, motor XYZ should be selected. Check: AWABC(12%) = $506.30  $9.33(2,000) = $19,166 AWXYZ(12%) = $1,407.40  $8.29(2,000) = $17,987 Motor XYZ has the minimum annual equivalent cost at H = 2,000.
107
Assume repeatability. Set AWA(10%) = AWB(10%) and solve for breakeven value of X. AWA(10%) = $5,000 (A/P,10%,5) + $1,500 + $1,900 (A/F,10%,5) = $492,22 AWB(10%) = X((A/P,10%,7) + $1,400 + $4,00 (A/F,10%,7) = 0.2054X + $1,821.60 $492.22 = 0.2054X + $1,821.60 X = ($1,821.60  $492.22) / (0.2054) = $6,472.15
251
Chapter 10 Solutions
At X = 0, AWB(10%) = $1,821.60 > $492.22, so Alt. B is preferred. At X = $6,500, AWB(10%) = $486.50 < $492.22, so Alt. A is preferred. Select Alt. B for 0 ≤ X ≤ $6,472.15 108 Investment Annual Savings Market Value Study Period i
Machine A $50,000 $12,000 $5,000 8 years 5%, 10%, 15%, 20%
Machine B $75,000 $18,000 0 8 years 5%, 10%, 15%, 20%
∆ (BA): 0 = 25k + 6k (P/A,i',7) + 1k (P/F,i',8)
Polygon Chart of PW vs. MARR
20%
5% 40k 30 20 10 0 10
B
A 10 0 10 20 30 40
40 30 20 10 0 10
10%
10 0 10 20 30 40 15%
MARR 5% 10% 15% 20%
Chapter 10 Solutions
PWA(MARR) 31k 16.3k 5.5k 2.8k
252
PWB(MARR) 41.3k 21k 5.8k 6k
109
Aftertax, A$ Analysis: Let X = annual beforetax revenue requirement. Annual EOY Investment Revenue 0 $1,166,0001 1 X 2 X 3 X 4 X 2 4 441,741
Annual Expenses  $519,7503  545,738  573,024  601,676
BTCF (A$) $1,166,000 519,750+X 545,738+X 573,024+X 601,676+X 441,741
1
Capital Investment = (55 trucks)($21,000/truck) = $1,166,000 Market Value = MV4 = 0.35($1,166,000)(1.02)4 = $441,741 3 Annual Expenses in year k = (55 trucks)(20,000 mi/truck)($0.45/mi)(1.05)k = $495,000(1.05)k 2
EOY 0 1 2 3 4 4
BTCF (A$)  $1,166,000 519,750 + X 545,738 + X 573,024 + X 601,676 + X 441,741
Depr. $388,628 518,287 172,685 86,401 
TI  908,378 + X 1,064,025 + X  745,709 + X  688,077 + X 441,741
T(38%) 345,1840.38X 404,3300.38X 283,3690.38X 261,4690.38X 167,862
ATCF (A$)  $1,166,000 174,566+0.62X 141,408+0.62X 289,655+0.62X 340,207+0.62X 273,879
PW(15%) = 0 =  $1,166,000  $174,566(P/F,15%,1)  $141,408 (P/F,15%,2)  $289,655(P/F,15%,3)  $340,207(P/F,15%,4) + $273,879(P/F,15%,4) + 0.62X(P/A,15%,4) 0 =  $1,653,096 + 1.77X Thus, X = $1,653,096/1.77 = $933,953 in annual revenues per year. Breakeven Point Interpretation: The equivalent uniform annual revenue of $933,953 per year is the breakeven point between signing the contract (and purchasing the trucks, etc.), and not signing the contract (and making no change in current operations). 1010 (a)
⎛ $45 ⎞ ⎛ 365 days⎞ Annual Revenues = (150 rooms)(0.6) ⎜ ⎟ = $1,478,250 ⎟⎜ ⎝ room  day ⎠ ⎝ year ⎠ AW(10%) = $1,478,250  $125,000  $5,000,000(A/P,10%,15) + (0.2)($5,000,000)(A/F,10%,15)  $1,875,000 (A/P,10%,5) = $232,625 > 0 Yes, the project is economically feasible.
253
Chapter 10 Solutions
(b)
Sensitivity with respect to Decision Reversal Capital Investment: $232,625*  $5,000,000(A/P,10%,15)X = 0 X = 0.3538 or 35.38% * Market value is assumed to remain constant at $1,000,000. Occupancy Rate:
$232,625 + 45(150)(365)(0.6)X = 0 X = 0.1574 or 15.74%
MARR: (find the IRR and calculate % change) By trial and error, IRR = 14.3%. Therefore, the MARR must increase by 14.3/10  1 = 43 % The decision is most sensitive to changes in occupancy rate (requires the smallest percent change to reverse the decision). (c)
Annual worth is a linear function with respect to capital investment and occupancy rate  we can construct the plot using points from parts (a) and (b). Annual worth is nonlinear with respect to the MARR, therefore additional data points are necessary. i 6% 8% 10% 12% 14%
% change 40% 20% 0 20% 40%
AW $557,200 378,559 232,625 112,679 12,808
Occupancy Rate
$1,000,000 $800,000
Capital Investment
Annual Worth
$600,000 $400,000 $200,000
MARR $0 $200,000 $400,000 40
30
20
10
0
$600,000 % Change
Chapter 10 Solutions
254
10
20
30
40
1011
Cash Flow Diagram
$26,000 $18,000 $22,000 $14,000
$10,000
1
0
1
2
3
4
5
$20,000
$30,000 AW = [$20,000  $30,000(P/F,10%,1) + $5,000(P/F,10%,6)](A/P,10%,6) + [$10,000(1+x)(P/A,10%,5) + 4,000(1+y)(P/G,10%,5)](P/F,10%,1)(A/P10%,6)
Sensitivity of AW to estimation errors in both year 1 savings and gradient amount: Gradient Amount
First Year Savings
% change 50 40 30 20 10 0 10 20 30 40 50
50 40 30 3,386 2,813 2,240 2,595 2,022 1,449 1,803 1,231 658 1.012 439 133 221 352 925 570 1,143 1,716 1,362 1,934 2,507 2,153 2,726 3,298 2,944 3,517 4,090 3,735 4,308 4,881 4,527 5,099 5,672
20 10 0 10 20 30 40 50 1,668 1,095 522 51 623 1,196 1,769 2,342 876 304 269 842 1,415 1,987 2,560 3,133 85 488 1,060 1,633 2,206 2,779 3,351 3,924 706 1,279 1,852 2,424 2,997 3,570 4,143 4,715 1,497 2,070 2,643 3,216 3,788 4,361 4,934 5,507 2,289 2,861 3,434 4,007 4,580 5,152 5,725 6,298 3,080 3,653 4,225 4,798 5,371 5,944 6,516 7,089 3,871 4,444 5,017 5,589 6,162 6,735 7,308 7,880 4,662 5,235 5,808 6,381 6,953 7,526 8,099 8,672 5,454 6,026 6,599 7,172 7,745 8,317 8,890 9,463 6,245 6,818 7,390 7,963 8,536 9,109 9,681 10,254
255
Chapter 10 Solutions
1012 (a) AWpump M(10%) = $1,700 (A/P,10%,10)  $40(X/100) 2.5 gal/hr ($1.50/gal)X = 276.59 – 4.15X
AWpump N(10%) = $2,220 (A/P,10%,10)  $25(X/100) 2 gal/hr ($1.50/gal)X 276.59 + 4.15X = 361.19 + 3.25X 0.9X = 84.6 or X = 94 hrs/yr (b) Suppose X = 200 hrs/yr, AWM = $1,106.59; AWN = $1,011.19
0
AW
M Select N if X > 94 Select M if X < 94 N Neg 94
200 Hrs/yr
1013 (a) Analysis bsed on most likely estimates. Assume repeatability.
AWA(20%) = $57,000 (A/P,20%,12)  $30,000 + $20,000 ($4.40  $1.40) + $9,000 (A/F,20%, 12) = $17,385.60 AWB(20%) = $36,000 (A/P,20%,10)  $25,000 + $18,000 ($3.70  $0.90) + $9,000 (A/F,20%,10) = $17,160.50 At the most likely estimates, Design A is marginally preferred to Design B. (b) Breakeven sales for Design A:
AWA(20%) = $0 = $57,000 (A/P,20%,12)  $30,000 + X ($4.40  $1.40) + $9,000 (A/F,20%,12); X = 14,205 units / year
Chapter 10 Solutions
256
Breakeven sales for Design B: AWB(20%) = $0 = $36,000 (A/P,20%10)  $25,000 + X($3.70  $0,90) + $9,000 (A/F,20%,10); X = 11,872 units / year 1014 (a) Analysis at most likely estimates:
PW(15%) = $30,000 + ($20,000 = $5,000) (P/A,15%,5) + $1,000 (P/F,15%,5) = $20,780.20
Sensitivity to changes in capital investment: +5%: PW(15%) = $30,000(1.05) _ ($20,000  $5,000)(P/A,15%,5) + $1,000(P/F,15%,5) = $19,280.20 5%: PW*15%) = $30,000(0.95) + ($20,000  $5,000)(P/A,15%,5) + $1,000(P/F,15%,5) = $22,280.20 Breakeven percent change: PW(15%) = 0 = $30,000(1 + x%) + ($20,000  $5,000)(P/A,15%,5) + $1,000(P/F,15%,5) x = 0.693 or +69.3% increase in capital investment cost Sensitivy to changes in annual expenses: +10%: PW(15%) = $30,000 + [$20,000  $5,000(1.1)](P/A,15%,5) + $1,000(P/F,15%,5) = $19,104.10 10%: PW(15%) = $30,000 + [$20,000  $5,000)(0.9)](P/A,15%,5) + $1,000(P/F,15%,5) = $22,456.30 Breakeven percent change: PW(15%) = 0 = $30,000 + [$20,000  $5,000(1 + x%)](P/A,15%,5) + $1,000(P/F,15%,5) x = 1.24 or +124% increase in annual expenses Sensitivity to changes in annual revenue: +20%: PW(15%) = $30,000 + [$20,000(1.2)  $5,000](P/A,15%,5) + $1,000(P/F,15%,5) = $34,189
257
Chapter 10 Solutions
20%: PW(15%) = $30,000 + [$20,000(0.8)  $5,000](P/A,15%,5) + $1,000(P/F,15%,5) = $7,371.40 Breakeven percent change: PW(15%) = 0 = $30,000 + [$20,000(1 + x%) = $5,000](P/A,15%,5) + $1,000(P/F,15%,5) x = 0.31 or – 31% (decrease in annual revenues) Sensitivity to changes in market value: +20%: PW(15%) = $30,000 + ($20,000  $5,000)(P/A,15%,5) + $1,000(1.2)(P/F,15%,5) = $20,879.64 20%: PW(15%) = $30,000 + ($20,000  $5,000)(P/A,15%,5) + $1,000(0.8)(P/F,15%,5) = $20,680.76 Breakeven percent change: PW(15%) = 0 = $30,000 + ($20,000  $5,000)(P/A,15%,5) + $1,000(1 + x%)(P/F,15%,5) X = 41.79 or 417.9% (decrease in market value) Sensitivity to changes in useful life: At +20%, n = 6 PW(15%) = $30,000 + ($20,000  $5,000)(P/A,15%,6) + $1,000(P/F,15%,6) = $27,199.80 At 20%, n = 4 PW(15%) = $30,000 + ($20,000  $5,000)(P/A,15%,4) + $1,000(P/F,15%,4) = $13,396.80 Breakeven percent change: At n = 3 (40%): PW(15%) = $30,000 + ($20,000  $5,000)(P/A,15%,3) + $1,000(P/F,15%,3) = $4,905.50 At n =2 (60%): PW(15%) = $30,000 + ($20,000  $5,000)(P/A,15%,2) + $1,000(P/F,15%,2) = $4,858.40 Breakeven life ≈ 2.5years, a 50% change
Chapter 10 Solutions
258
Recommendation: Proceed with the project. PW is positive for values of factors (changed individually) within stated accuracy ranges. However, if all factors are at their worst values, PW(15%) = $1,065 [capital investment at +5%, annual expenses at +10%, annual revenues, market value, and useful life at 20%].
Capital Investment
35 30 PW (Thousands of Dollars)
Annual Revenues Useful Life
$20,780 25
Market Value
20 Annual Expenses
15 10 5
60 50 40 30 20 10 0 10 20 30 40 50 60 70
Percent Devation Changes from Most Likely Estimate
(b) Factors can be ranked base on the breakeven percent change.
Highest need
Lowest need
Annual Revenues Useful Life Capital Investment Annual Expense Market Value
31% 50% +69.3% +124% 418%
1015 Repair cost = $5,000: PW(i'%) = 0 = $10,000 + $4,000(P/A,i'%,5)  $5,000(P/F,i'%,3) By trial and error, IRR = 15.5%
Repair cost = $7,000: PW(i'%) = 0 = $10,000 + $4,000(P/A,i'%,5)  $7,000(P/F,i'%,3) By trial and error, IRR = 9.6% Repair cost = $3,000 PW(i'%) = 0 = $10,000 + $4,000(P/A,i'%,5)  $3,000(P/F,i'%,3) By trial and error, IRR = 21%
259
Chapter 10 Solutions
The sensitivity analysis indicates that if the repairs at the end of year 3 cost $5,000 or less, it will be economical to invest in the machine. However, if the repairs cost $7,000, the IRR of the purchase is less than the MARR. A followup analysis would be to determine the maximum repair cost that would still result in the desired return of 10%. Let R = repair cost at the end of year 3. PW(10%) = 0 =  $10,000 + $4,000(P/A,10%,5)  R (P/F,10%,3) 0.7513(R) = $4,163.20 R = $6,872 As long as the repair cost at the end of year 3 does not exceed $6,872 (which represents a 37.44% increase over the estimated cost of $5,000), the IRR of the purchase will be ≥ 10%. 1016 PW(Land Cost) = $50,000+$50,000(P/F,i%,40) = $50,000[1  (P/F,i%,40)]
The following equations were used to generate a spreadsheet analyzing the sensitivity of the decision due to changes in estimates of the MARR over the range 10% to 20%. No. of Floors 2 3 4 5
PW Equation $200,000 + ($ 40,000  $15,000)(P/A,i%,40)  $50,000[1  (P/F,i%,40)] $250,000 + ($ 60,000  $25,000)(P/A,i%,40)  $50,000[1  (P/F,i%,40)] $320,000 + ($ 85,000  $25,000)(P/A,i%,40)  $50,000[1  (P/F,i%,40)] $400,000 + ($100,000  $45,000)(P/A,i%,40)  $50,000[1  (P/F,i%,40)]
The following table shows that the optimal height of the proposed building is 4 floors over the specified range of the MARR. However, unless the MARR is less than 17%, the proposed building would not be profitable.
Chapter 10 Solutions
260
PW MARR 10% 11% 12% 13% 14% 15% 16% 17% 18% 19% 20%
2  $4,419  25,455  43,368  58,764  72,109  83,769  94,031  103,123  111,230  118,499  125,051
Number of Floors 3 4 $43,372 $217,848 * 14,056 167,832 *  10,930 125,164 *  32,420 88,439 *  51,059 56,567 *  67,351 28,693 *  81,696 4,142 *  94,410  17,626 *  105,748  37,044 *  115,917  54,463 *  125,085  70,170 *
5 $88,953 43,077 3,945  29,733  58,958  84,516  107,026  126,983  144,785  160,754  175,153
* Indicates optimal number of floors for the corresponding MARR value. 1017
1) Assume salvage value for each system = 0 2) The difference in user cost will be projected as a savings for system #2. Savings = $0.02/vehicle 3) There are approximately 8,760 hours/year
System 1: AW method CR = $32,000(A/P,10%,10) = $5,206.40 Annual Maintenance = $75 Operation Cost = (28 KW) (8760 hours/year) = 245,280 KW – hours
(245,280 KW − hour )(0.02) = $6,289.23 0.78 AC#1 = $5,206.40 + $75 + $6,289.23 = $11,570.23 System #2: CR = $45,000(A/P,10%,10) = $7,321.50 Annual Maintenance = $100 (245,280 KW − hour )(0.02) Operation Cost = = $5,450.67 0.90 Let N = # vehicles using intersection Savings per vehicle = 0.2 N AC#2 = $7,321.50 + $100 + $5,450.67 – 0.2N = $12,872.17  $0.02N
261
Chapter 10 Solutions
For Breakeven point: $11,570.63 = $12,872.17  0.2N; N = 65,076.83/year For ADT = Average Daily Traffic N=
65,076.83 = 179 vehicles/day 365
1018 Let X = annual Btu requirement (in 103 Btu)
AWoil (10%) = $80,000(A/P,10%,20) + $4,000  ($1.10/140)(X) = $5,400  $0.008X AWgas (10%) = $60,000(A/P,10%,20) + $6,000  ($0.02/1)(X) = $1,050  $0.02X To find breakeven value of X, set AWoil (10%) = AWgas (10%) and solve for X. $5,400  $0.008X = $1,050  $0.02X X = 362,500 or 362,500,000 Btu per year Now let's examine the sensitivity of the decision to changes in the annual Btu requirement. The following table and graph indicate that the conversion to natural gas is preferred if the annual Btu requirement is less than the breakeven point, else the conversion to oil is preferred.
% change in X 30 20 10 0 10 20 30
Chapter 10 Solutions
Annual Worth Oil Gas $7,430 $6,125 7,720 6,850 8,010 7,575 8,300 8,300 8,590 9,025 8,880 9,750 9,170 10,475
262
$0
Annual Worth
 $2,000  $4,000
Gas  $6,000
Oil
 $8,000  $10,000  $12,000 30%
20%
10%
0%
10%
20%
30%
Percent Change in Annual Btu Requirement
AW(optimistic) =  $90,000(A/P,10%,12) + $35,000 + $30,000(A/F,10%,12) = $23,192
1019 (a)
AW(most likely) = $100,000(A/P,10%,10) + $30,000 + $20,000(A/F,10%,10) = $14,984 AW(pessimistic) = $120,000(A/P,10%,6) + $20,000 = $7,552 (b)
Useful Life
O M P
Net Annual Cash Flow O M P $21,256 $16,256 $6,256 19,984 14,984 4,984 14,632 9,632  368
1020 Build 4lane bridge now: PW(12%) = $350,000
Build twolane bridge now: Optimistic: widen bridge to four lanes in 7 years PW(12%) = $200,000  [$200,000 + (7)($25,000)](P/F,12%,7) = $369,613 Most Likely: widen bridge to four lanes in 5 years PW(12%) = $200,000  [$200,000 + (5)($25,000)](P/F,12%,5) = $384,405 Pessimistic: widen bridge to four lanes in 4 years PW(12%) = $200,000  [$200,000 + (4)($25,000)](P/F,12%,4) = $390,650
263
Chapter 10 Solutions
Recommend building the 4lane bridge now. In this problem, there is no difficulty in interpreting the results since building the 4lane bridge now is preferred to a delay in widening the bridge for 7 years (optimistic estimate). The advantage of pessimistic, most likely, and optimistic estimates is that the uncertainty involved is made explicit. Therefore, the information should be more useful in decisionmaking. However, when mixed results are obtained, significant judgement is required in reaching a decision. Mixed results would occur in this problem, for example, if the PW of a 7year delay in widening the bridge were less than $350,000. ⎛ 100 hp ⎞ ⎛ 0.746 kW ⎞ ⎛ $0.10 ⎞ 1021 AWABC(12%) = $170  ⎜ ⎟⎜ ⎟ (1000 hr ) ⎟⎜ ⎝ 0.8 ⎠ ⎝ hp ⎠ ⎝ kW  hr ⎠  $1,900(A/P,12%,10) = $9,831
⎛ 100 hp ⎞ ⎛ 0.746 kW ⎞ ⎛ $0.10 ⎞ AWXYZ(20%) = $310  ⎜ ⎟⎜ ⎟ (1000 hr ) ⎟⎜ ⎝ 0.9 ⎠ ⎝ hp ⎠ ⎝ kW  hr ⎠  $6,200(A/P,20%,10) = $10,078 Select the ABC brand motor. A problem with using a riskadjusted MARR is that the risk associated with one or more factors is not made explicit. In this problem, it would be preferable to make pessimistic, most likely, and optimistic estimates of "less reliability" in terms of maintenance expenses for Motor XYZ. Then, each of these situations could be examined using a MARR = 12% and compared with Motor ABC.
Chapter 10 Solutions
264
1022 (a) EOY 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 20
Alternative A: BTCF  $108,000,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000  3,460,000 43,200,000
Depr $5,400,000 10,260,000 9,234,000 8,316,000 7,484,400 6,728,400 6,372,000 6,372,000 6,382,800 6,372,000 6,382,800 6,372,000 6,382,800 6,372,000 6,382,800 3,186,000 0 0 0 0 
TI  $8,860,000  13,720,000  12,694,000  11,776,000  10,944,400  10,188,400  9,832,000  9,832,000  9,842,800  9,832,000  9,842,800  9,832,000  9,842,800  9,832,000  9,842,800  6,646,000  3,460,000  3,460,000  3,460,000  3,460,000 43,200,000
T(40%) $3,544,000 5,488,000 5,077,600 4,710,400 4,377,760 4,075,360 3,932,800 3,932,800 3,937,120 3,932,800 3,937,120 3,932,800 3,937,120 3,932,800 3,937,120 2,658,400 1,384,000 1,384,000 1,384,000 1,384,000  17,280,000
ATCF  $108,000,000 84,000 2,028,000 1,617,600 1,250,400 917,760 615,360 472,800 472,800 477,120 472,800 477,120 472,800 477,120 472,800 477,120  801,600  2,076,000  2,076,000  2,076,000  2,076,000 25,920,000
20
PW(10%) =
∑
ATCFk (P/F,10%, k) =  $99,472,154
k=0
Alternative B: Compute ATCFs for current estimate of capital investment. Using the ATCFs shown in the following table: 20
PW(10%) =
∑
ATCFk (P/F,10%, k) =  $79,065,532
k=0
265
Chapter 10 Solutions
ATCFs for Alternative B given original capital investment amount: EOY 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 20
BTCF  $17,000,000  12,400,000  12,400,000  12,400,000  12,400,000  15,400,000  12,400,000  12,400,000  12,400,000  12,400,000  15,400,000  12,400,000  12,400,000  12,400,000  12,400,000  15,400,000  12,400,000  12,400,000  12,400,000  12,400,000  12,400,000 0
Depr $ 850,000 1,615,000 1,453,500 1,309,000 1,178,100 1,059,100 1,003,000 1,003,000 1,004,700 1,003,000 1,004,700 1,003,000 1,004,700 1,003,000 1,004,700 501,500 0 0 0 0 
TI  $13,250,000  14,015,000  13,853,500  13,709,000  16,578,100  13,459,100  13,403,000  13,403,000  13,404,700  16,403,000  13,404,700  13,403,000  13,404,700  13,403,000  16,404,700  12,901,500  12,400,000  12,400,000  12,400,000  12,400,000 0
T(40%) ATCF  $17,000,000 $5,300,000  7,100,000 5,606,000  6,794,000 5,541,400  6,858,600 5,483,600  6,916,400 6,631,240  8,768,760 5,383,640  7,016,360 5,361,200  7,038,800 5,361,200  7,038,800 5,361,880  7,038,120 6,561,200  8,838,800 5,361,880  7,038,120 5,361,200  7,038,800 5,361,880  7,038,120 5,361,200  7,038,800 6,561,880  8,838,120 5,160,600  7,239,400 4,960,000  7,440,000 4,960,000  7,440,000 4,960,000  7,440,000 4,960,000  7,440,000 0 0
Alternative B (revised to include extra investment permissible to breakeven) EOY BTCF 0  $42,731,490 1  12,400,000 2  12,400,000 3  12,400,000 4  12,400,000 5  15,400,000 6  12,400,000 7  12,400,000 8  12,400,000 9  12,400,000 10  15,400,000 11  12,400,000 12  12,400,000 13  12,400,000 14  12,400,000 15  15,400,000 16  12,400,000 17  12,400,000 18  12,400,000 19  12,400,000 20  12,400,000 20 0
Chapter 10 Solutions
Depr TI $2,136,574  $14,536,574 4,059,492  16,459,492 3,653,542  16,053,542 3,290,325  15,690,325 2,961,292  18,361,292 2,662,172  15,062,172 2,521,158  14,921,158 2,521,158  14,921,158 2,525,431  14,925,431 2,521,158  17,921,158 2,525,431  14,925,431 2,521,158  14,921,158 2,525,431  14,925,431 2,521,158  14,921,158 2,525,431  17,925,431 1,260,579  13,660,579 0  12,400,000 0  12,400,000 0  12,400,000 0  12,400,000 0
266
T(40%) ATCF  $42,731,490 $5,814,630  6,585,370 6,583,797  5,816,203 6,421,417  5,978,583 6,276,130  6,123,870 7,344,517  8,055,483 6,024,869  6,375,131 5,968,463  6,431,537 5,968,463  6,431,537 5,970,172  6,429,828 7,168,463  8,231,537 5,970,172  6,429,828 5,968,463  6,431,537 5,970,172  6,429,828 5,968,463  6,431,537 7,170,172  8,229,828 5,464,232  6,935,768 4,960,000  7,440,000 4,960,000  7,440,000 4,960,000  7,440,000 4,960,000  7,440,000 0 0
The above solution for Alternative B is the result of a trial and error procedure for obtaining identical present worths of ATCFs. Extra Capital = $42,731,490  $17,000,000 = $25,731,490 This solution takes into account depreciation credits arising from the extra capital that can be invested in Alternative B to breakeven with Alternative A. (b)
Coterminate both alternatives at the end of year 10. Alternative A: EOY BTCF 0  $108,000,000 1  3,460,000 2  3,460,000 3  3,460,000 4  3,460,000 5  3,460,000 6  3,460,000 7  3,460,000 8  3,460,000 9  3,460,000 10  3,460,000 10 43,200,000
Depr $5,400,000 10,260,000 9,234,000 8,316,000 7,484,400 6,728,400 6,372,000 6,372,000 6,382,800 3,186,000 
TI  $8,860,000  13,720,000  12,694,000  11,776,000  10,944,400  10,188,400  9,832,000  9,832,000  9,842,800  6,646,000 4,935,600
T(40%) ATCF  $108,000,000 $3,544,000 84,000 5,488,000 2,028,000 5,077,600 1,617,600 4,710,400 1,250,400 4,377,760 917,760 4,075,360 615,360 3,932,800 472,800 3,932,800 472,800 3,937,120 477,120 2,658,400  801,600  1,974,240 41,225,760
*MV10 = $43,200,000; BV10 = $38,264,400 10
PW(10%) =
∑
ATCFk (P/F,10%, k) =  $87,010,230
k =0
Alternative B: EOY BTCF 0  $17,000,000 1  12,400,000 2  12,400,000 3  12,400,000 4  12,400,000 5  15,400,000 6  12,400,000 7  12,400,000 8  12,400,000 9  12,400,000 10  15,400,000 10 0
Depr TI $850,000  $13,250,000 1,615,000  14,015,000 1,453,500  13,853,500 1,309,000  13,709,000 1,178,100  16,578,100 1,059,100  13,459,100 1,003,000  13,403,000 1,003,000  13,403,000 1,004,700  13,404,700 501,500  15,901,500  6,023,100
T(40%) ATCF  $17,000,000 $5,300,000  7,100,000 5,606,000  6,794,000 5,541,400  6,858,600 5,483,600  6,916,400 6,631,240  8,768,760 5,383,640  7,016,360 5,361,200  7,038,800 5,361,200  7,038,800 5,361,880  7,038,120 6,360,600  9,039,400 2,409,240 2,409,240
*MV10 = 0; BV10 = $6,023,100 10
PW(10%) =
∑
ATCFk (P/F,10%, k) =  $60,788,379
k =0
267
Chapter 10 Solutions
If the study period is reduced to 10 years, Alternative B would still be recommended. Sensitivity of Alternative B to cotermination at EOY 10: ⎛ − $79,065,532 + $60,788,379 ⎞ ⎜ ⎟ ×100% = 23.1% less expensive. − $79,065,532 ⎝ ⎠ (c)
If our annual operating expenses of Alternative B double, the extra present worth of cost in part (a) equals: ($2.1 million)(1  0.4) • (P/A, 10%, 20) = $10,727,090. This makes the total present worth of Alternative B equal to: $79,065,532  $10,727,090 = $89,792,622. Because $89,792,622 > $99,472,154, the initial decision to adopt Alternative B is not reversed. EOY 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 20
Chapter 10 Solutions
BTCF  $17,000,000  14,500,000  14,500,000  14,500,000  14,500,000  17,500,000  14,500,000  14,500,000  14,500,000  14,500,000  17,500,000  14,500,000  14,500,000  14,500,000  14,500,000  17,500,000  14,500,000  14,500,000  14,500,000  14,500,000  14,500,000 0
Depr $ 850,000 1,615,000 1,453,500 1,309,000 1,178,100 1,059,100 1,003,000 1,003,000 1,004,700 1,003,000 1,004,700 1,003,000 1,004,700 1,003,000 1,004,700 501,500 0 0 0 0 
268
TI  $15,350,000  16,115,000  15,953,500  15,809,000  18,678,100  15,559,100  15,503,000  15,503,000  15,504,700  18,503,000  15,504,700  15,503,000  15,504,700  15,503,000  18,504,700  15,001,500  14,500,000  14,500,000  14,500,000  14,500,000 0
T (40%) $6,140,000 6,446,000 6,381,400 6,323,600 7,471,240 6,223,640 6,201,200 6,201,200 6,201,880 7,401,200 6,201,880 6,201,200 6,201,880 6,201,200 7,401,880 6,000,600 5,800,000 5,800,000 5,800,000 5,800,000 0
ATCF  $17,000,000  8,360,000  8,054,000  8,118,600  8,176,400  10,028,760  8,276,360  8,298,800  8,298,800  8,298,120  10,098,800  8,298,120  8,298,800  8,298,120  8,298,800  10,098,120  8,499,400  8,700,000  8,700,000  8,700,000  8,700,000 0
Solutions to Spreadsheet Exercises 1023 Capital Investment = Efficiency = Useful Life = Annual Maintenance =
CR amount = Maintenance = Taxes = Total = Power Expense =
$
Alpha 12,500 $ 68% 10 500 $
Beta 16,000 92% 10 250
$ $ $ $ $
Alpha 2,491 500 188 3,178 5.48
Beta 3,188 250 240 3,678 4.05
$
$ $ $ $ $
MARR = Electricity = Annual Taxes =
Hours 350 $
Alpha 5,097
$
$
15% 0.05 1.5%
Beta Difference 5,097 $ (0)
Hours Scroll Bar Hours 0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 1000
EUAC: Alpha $ 3,178 $ 3,452 $ 3,726 $ 4,001 $ 4,275 $ 4,549 $ 4,823 $ 5,097 $ 5,371 $ 5,645 $ 5,919 $ 6,194 $ 6,468 $ 6,742 $ 7,016 $ 7,290 $ 7,564 $ 7,838 $ 8,112 $ 8,387 $ 8,661
EUAC: Beta $ 3,678 $ 3,881 $ 4,083 $ 4,286 $ 4,489 $ 4,692 $ 4,894 $ 5,097 $ 5,300 $ 5,502 $ 5,705 $ 5,908 $ 6,111 $ 6,313 $ 6,516 $ 6,719 $ 6,922 $ 7,124 $ 7,327 $ 7,530 $ 7,732
Difference $ (500) $ (428) $ (357) $ (286) $ (214) $ (143) $ (71) $ (0) $ 71 $ 143 $ 214 $ 286 $ 357 $ 428 $ 500 $ 571 $ 643 $ 714 $ 786 $ 857 $ 928
At an efficiency of 68% for the Alpha motor, the breakeven annual hours of operation for the motors would be 350.
269
Chapter 10 Solutions
1024 A Effective tax 3 rate = 4 5 6 7 8 9 10 11 12
B
EOY 1 0 1 2 3 4 5 6 7 8
D
9.40%
Total 13 Investment = $ 14 DB rate = 15 16 17 18 19 20 21 22 23 24 25
C
0 1 2 3 4 5 6 7 8
$ $ $ $ $ $ $ $ $
1.10 
Depreciation Deduction
(900,000) (1,100,000) 1,000,000 1,100,000 1,200,000 1,000,000 1,000,000 1,000,000 900,000 800,000
$ $ $ $ $ $ $ $
400,000 640,000 384,000 230,400 230,400 115,200 
Taxable Income
$ $ $ $ $ $ $ $
600,000 460,000 816,000 769,600 769,600 884,800 900,000 800,000
Cash Flow for Income Tax
$ $ $ $ $ $ $ $
26 27 28 For an effective tax rate of 9.4%, all cumulative IRR amounts are positive. 29 30 31 32 33 34 35 36 37 38 39 40 41
Chapter 10 Solutions
F $ $ $ $ $ $ $ $ $
G 1.80 2.00 2.10 1.90 1.80 1.80 1.70 1.50
$ $ $ $ $ $ $ $ $
H 0.80 0.90 0.90 0.90 0.80 0.80 0.80 0.70
2,000,000 200% BTCF
$ $ $ $ $ $ $ $ $ $
E
270
(56,400) (43,240) (76,704) (72,342) (72,342) (83,171) (84,600) (75,200)
ATCF $ (900,000) $ (1,100,000) $ 943,600 $ 1,056,760 $ 1,123,296 $ 927,658 $ 927,658 $ 916,829 $ 815,400 $ 724,800
"Cumulative" IRR
0.01% 19.78% 28.24% 33.13% 36.06% 37.70% 38.66%
I
1025 MARR Installation Expense ($/in.) Annual Tax and Insurance Rate
20% $
5%
Insulation Thickness (in.) 3 4 5 6 7 8
50% 40% 30% 20% 10% 0% 10% 20% 30% 40% 50%
150
Heat Loss (Btu/hr) 4,400 3,400 2,800 2,400 2,000 1,800
$ $ $ $ $ $ $ $ $ $ $
3 (500.35) (577.44) (654.53) (731.61) (808.70) (885.79) (962.88) (1,039.97) (1,117.05) (1,194.14) (1,271.23)
Installation Expense $ (450) $ (600) $ (750) $ (900) $ (1,050) $ (1,200)
$ $ $ $ $ $ $ $ $ $ $
4 (451.05) (510.62) (570.19) (629.76) (689.33) (748.89) (808.46) (868.03) (927.60) (987.17) (1,046.73)
Useful Life Operating Hours per Yr. Cost of Heat Loss ($/Btu)
20
$
Annual Taxes and Insurance $ (22.50) $ (30.00) $ (37.50) $ (45.00) $ (52.50) $ (60.00)
Cost of Heat Removal ($/yr) $ (770.88) $ (595.68) $ (490.56) $ (420.48) $ (350.40) $ (315.36)
8,760 0.00002
Equivalent Annual Worth 5 6 $ (436.80) $ (440.06) $ (485.85) $ (482.11) $ (534.91) $ (524.16) $ (583.97) $ (566.20) $ (633.02) $ (608.25) $ (682.08) $ (650.30) $ (731.13) $ (692.35) $ (780.19) $ (734.40) $ (829.25) $ (776.44) $ (878.30) $ (818.49) $ (927.36) $ (860.54)
Equivalent Annual Worth $ (885.79) $ (748.89) $ (682.08) $ (650.30) $ (618.52) $ (621.79)
7 (443.32) (478.36) (513.40) (548.44) (583.48) (618.52) (653.56) (688.60) (723.64) (758.68) (793.72)
$ $ $ $ $ $ $ $ $ $ $
$ $ $ $ $ $ $ $ $ $ $
8 (464.11) (495.64) (527.18) (558.72) (590.25) (621.79) (653.32) (684.86) (716.40) (747.93) (779.47)
1026 A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
B Optimistic $ 80,000 12 $ 30,000 $ 35,000 12%
Capital Investment Useful Life Market Value Net Annual Cash Flow MARR
AW = $ Investment = Market Value = MARR =
Net Annual Cash Flow
$ $
$ $ $
C Most Likely $ 95,000 10 $ 20,000 $ 30,000 12%
23,328 $ 95,000 20,000 12.00%
35,000 $ 30,000 $ 20,000 $
D Pessimistic $ 120,000 6 $ $ 20,000 12%
14,326 $
E
F
G
H
I
J
K
(9,187)
Table values are AW
Useful Life 12 10 6 20,492 $ 19,326 $ 14,358 15,492 $ 14,326 $ 9,358 5,492 $ 4,326 $ (642)
271
Chapter 10 Solutions
Solutions to Case Study Exercises 1027 The following sensitivity graph was created using the results from the case study:
$400,000 $300,000
Material Cost
Capacity Utilization
Useful Life
$200,000
AW(15%)
Selling Price
$100,000 $0 $100,000 $200,000 40
30
20
10
0
10
20
30
40
$300,000
Percent Change in Factor
Raw Material Costs: Assume, because of competition, that only 50% of any material cost increase (above $27/yd3) can be recovered through an increase in the selling price ($45/yd3). The resulting AW(15%) values with 10, 20, and 30% material cost increases are:
AW(15%)
Percent Increase in Material Cost (M) 10 20 30 $35,397 $17,172  $1,053
1028 Selling price and capacity utilization are the two factors that appear to impact the AW of the project the most when changes from their best estimates occur. The following information is needed:
Project Factor (variable) Selling Price, S Capacity Utilization, U
Deviation Range  10% to + 0%
Best Estimate $45/yd3
 33% to + 20%
75%
Range Estimate Minimum Maximum $40.50/yd3 $45/yd3 50%
90%
In the following calculations and twodimensional graphical plot, the best estimates of useful life (10 yrs) and material costs ($27/yd3) are used. Also, in the twodimensional plot, capacity utilization (U) is represented as a
Chapter 10 Solutions
272
variable on the abscissa and selling price (S) is represented by a set of curves. The two plotted curves are based on the maximum and minimum values of S and are the boundaries of the set of curves representing this variable. When the selling price is $40.50/yd3, we have (U in decimal form): AW(15%) =  $28,190 + $40.50 (72) (250) U (Capital Recovery) (Revenue)  72 (250) ($27) U  (1 + U)($9,143)  $145,188 (Material Costs) (O&M expenses) (other annual expenses) = $233,857 U  $182,521 And, when selling price is $45/yd3, we have based on a similar formulation: AW(15%) = $314,857 U  $182,521 The following twodimensional graph shows the combined impact of changes in these two factors (capacity utilization and selling price) on the AW(15%) of the project: AW(15%)
$120,000 Maximum: $100,850
$100,000 $80,000 Selling Price (S) = $45/yd3
Annual Worth (15%)
$60,000 $40,000 $20,000 U (%)
$0 50
60
70
80
90 Capacity Utilization
$20,000
Selling Price (S) = $40.50/yd3
$40,000 $60,000 $80,000
Minimum: $65,590
273
Chapter 10 Solutions
1029 The three most sensitive factors are: capacity utilization (U), selling price (S), and material costs (M). Two approaches are used in this solution. The first involves a modified O  ML  P approach to develop scenarios. The second uses selected changes in the three factors, in combination, for the development of scenarios.
Modified O  ML  P scenarios
Factor Selling Price (S) Capacity Utilization (U) Material Costs (M) *
O* $45 90% $27
Estimated Value ML* P $45 $40.50 (10%) 75% 50% $27 $35.10 (+30%)
For S and M the optimistic and most likely estimates are the same. The three factors have 2 × 2 × 3 = 12 combinations or scenarios.
M − 27 ⎞ ⎛ AW(15%) = $72(250)(U) ⎜ S + ⎟ ⎝ 2 ⎠  $72(250)(U)(M)  $9,143(1 + U)  $182,521 M − 27 ⎞ ⎡⎛ ⎤ = $18,000(U) ⎢⎜ S + ⎟ − M⎥ 2 ⎠ ⎣⎝ ⎦ *
Annual Revenue Material costs O&M expenses Other expenses  $9,143(U)  $182,521
Assumes that 50% of any material cost increase above $27/yd3 can be recovered through an increase in the selling price ($45/yd3 or $40.50/yd3).
Capacity Utilization (U) O: 90% ML: 75% P: 50% *
Selling Price (S) O, ML: $45 P: $40.50 Material Cost (M) O, ML: $27 P: $35.10 O, ML: $27 P: $35.10 $101* $35 $28  $38 54 1 7  62  25  62  66  102
AW(15%) of scenario (in thousands) based on calculation above.
This modified O  ML  P analysis highlights the importance of U > 75% and material costs (M) remaining close to $27/yd3. The project results are sensitive to combinations of changes in these factors.
Chapter 10 Solutions
274
Other Selected Scenarios Factor Capacity Utilization Material Cost Selling Price AW(15%)
Percent Deviation for Combination: A B C 10% 15% 25% +10% +10% + 5% 0% + 3% 0% $2,800 $2,850 $18,940
These additional scenarios further emphasize the sensitivity of the project to the combined effect of modest changes in the three factors.
Solutions to FE Practice Problems 1030 Existing Bridge: PWE(12%) = $1,6000,000  $20,000(P/A,12%,20)  $70,000[(P/F,12%,5) +(P/F,12%,10) + (P/F,12%,15)] = $1,824,435
New Bridge: PWN(12%) = X – [$24,000 + (5)($10,000)](P/A,12%,20) + ($0.25)(4000,000)(P/A,12%,20) = X + $194,204 Set PWE(12%) = PWN(12%) and solve for X. $1,824,435 = X + $194,204 X = $2,018,639 Select (b) 1031 X = average number of vehicles per day
AW(12%) = 0 = $117,000/mile(A/P,12%,25) 
(0.03)($117,000) mile
⎛ 1,250 − 710accidents ⎞ + (X vehicles/day)(365 days/yr)($1,200/accident) ⎜ ⎟ ⎝ 1,000,000vehicle − miles ⎠ X = 77.91 vehicles/day Select (a)
275
Chapter 10 Solutions
1032 AW(12%) = $8,000(A/P,12%,7) + X($0.50  $0.26)  $2,000 = 0
X = 15,637 Select (e) 1033 AW(12%) = $16,000(A/P.12%,7) + X($0.50  $0.16)  $4,000 = 0
X = 22,076 Select (c) 1034 AWA(12%)  $8,000(A/P,12%,7) + 35,000($0.50  $0.26)  $2,000 = $4,647
AWB(12%) = $16,000(A/F,12%,7) + 35,000($0.50  $0.16)  $4,000 = $4,394 Select (b) – Install Machine A 1035 False 1036 False 1037 False 1038 True 1039 False
Chapter 10 Solutions
276