Solutions Manual for Engineering Economy 17th Global Edition By William Sullivan Elin Wicks Patrick by digitaldownload87

Solutions Manual for Engineering Economy 17th Global Edition By William Sullivan Elin Wicks Patrick Koelling (All Chapters 1-14)

Solutions to Chapter 1 Problems A Note To Instructors: Because of volatile energy prices in today's world, the instructor is encouraged to vary energy prices in affected problems (e.g. the price of a gallon of gasoline) plus and minus 50 percent and ask students to determine whether this range of prices changes the recommendation in the problem. This should make for stimulating inclass discussion of the results.

1-1

(a)

66 million tons per year × (0.05) = 3.3 million tons per year of reduced greenhouse gas $2.1 billion ÷ 3.3 million tons per year = $636.36 per ton

(b)

3 billion tons per year × (0.02) = 60 million tons per year of reduced greenhouse gas $2.1 billion/3.3 million tons = $X billion/60 million tons X = $40.65 billion

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1-2

The primary disadvantage is the installed cost of $8,000 (rather expensive!). A key advantage of synthetic turf is the annual savings due to lawn maintenance. This may be worth up to $1,000 per year when periodic lawn mowing, watering, and feeding are avoided. See what other monetized advantages and disadvantages members of your class can identify. Be creative!

1-3

Cost per Watt-hour = $0.90/1.5 Watt-hours = $0.60 per Watt-hour At a cost of $0.60 per Watt-hour, it would cost (1,000)($0.60 per Watt-hour) = $600 per kilo Watt-hour for power from a single AAA battery. This is 3,000 times more costly than energy from your local utility. No wonder we turn off our battery operated devices when we're not using them!

1-4

At first glance, Tyler’s options seem to be: (1) immediately pay $803 to the owner of the other person’s car or (2) submit a claim to the insurance company. If Tyler keeps his Nissan for five more years (an assumption), the cost of option 2 is $500 + ($60 × 2 payments/year) × 5 years = $1,100. This amount is more than paying $803 out-of-pocket, so Tyler appears to have made the most economical choice. What we don’t know in this problem is the age and condition of the other person’s car. If we assume it’s a clunker, another option for Tyler is to offer to buy the other person’s car and fix it himself and then sell it over the internet. Or Tyler could donate the unrepaired (or repaired) car to his favorite charity.

1-5

(a)

18,000 miles per year/25 mpg = 720 gallons per year of E20 Savings = 720 gallons per year ($4.40 ‒ $3.73) = $482.4 per year

(b)

Gasoline saved = 0.20 (720 gal/yr)(1,000,000 people) = 144 million gallons per year

1-6

An economic tradeoff is trading (exchanging) one economic outcome for another. Examples include (a) whether to ride the bus or drive your car to campus today, (b) whether to pack your lunch or eat at a local fast food restaurant and (c) whether to carry an umbrella (an inconvenience) or hope for no rain if you don’t take your umbrella.

1-7

The cost of a wood-framed home is 2,500 ft2 × $140/ft2 = $350,000. The ICF home will cost $350,000 (1.10) = $385,000. So, the cost difference to construct an ICF home is $35,000. The energy savings per month is 5% × $200 = $100 per month. Therefore, it will take 350 months to pay back the extra ICF construction cost through monthly energy savings.

1-8

Increased earnings with new technology = $2,000,000(0.05) = $100,000

1-9

Strategy 1: Change oil every 4,000 miles. Cost = (16,000/4,000)($35) = $140/year Strategy 2: Change oil every 8,000 miles. Cost = (16,000/8,000)($35) = $70/year Savings = $70 per year

1-10

The halogen bulb will use 60 × 30 = 1,800 kWh over 30,000 hours. Electricity costs with a halogen bulb will be (1,800 − 300)($0.2) = $300 higher than with an LED bulb. Additionally, the halogen bulb will have to be replaced every 1,000 hours, which will cost (30)($5) = $150, which is $120 more than the cost of a single LED bulb. Therefore, over the span of 30,000 usage hours, the LED bulb saves $420.

1-11

100 gallons × $4.50 per gallon = $450 saved over 45,000 miles of driving. This comes down to $450 ÷ 45,000 = $0.01 per mile driven. So, Brand A saves a penny for each mile driven.

1-12

(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; in-house 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 in-house 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 in-house could prove to be unsatisfactory; the old machine could be less safe than a new one; etc.

(f)

Non-monetary Considerations: Safety; environmental concerns; quality/reliability differences; “flexibility” of a new machine; job security for in-house 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?

1-13

(a)

Problem A: Subject to energy that Farah is willing/able to exert, her ability to do well in the Masters program, and expectations about the future state of the job market, Problem A might be "How can Farah motivate herself to complete the Masters program and get a job after two years, which is at least as good as the one in hand?” Problem B: Subject to knowledge of the job market, mobility, and professional ambition, as well as the probability of winning another scholarship a few years from now, Problem B could be "How can Farah use the new job as a springboard to better pay and career advancement, such that she can either pay for a Masters degree, increase her chances of winning a scholarship, or dispense with the degree altogether?”

(b) Problem A – Some feasible solutions could include 1) Spend some time finding out about future job opportunities after the Masters program. Speak to the employer who has offered her a job and see if they would be willing to take her on after two years. Use university resources to establish employment prospects after the Masters. 2) Take a holiday or some time off to motivate herself to perform well on a second degree and in future exams. Problem B – Some feasible solutions could include 1) Explore the possibility of working part-time with the company and finishing the Masters degree simultaneously; or of working for the company for a few years and then being given a sabbatical to finish her Master’s degree. 2) Find out about scholarship chances for working professionals who are re-entering school after taking some time off to work. 3) Project future earnings to see when she would be able to save enough to either pay for the degree herself or get access to a student loan. 4) Develop a career path that can give her the experience and the skills that she would have got on the Masters program by working on different assignments or getting on-the-job training, for example.

1-14

A Typical Discussion/Solution: (a)

One way to phrase this problem is: How do the three students satisfy their hunger in the most economical way possible, both in terms of time and money?

(b)

Principle 1 - Develop the Alternatives i) Alternative A is to sign up for college meals ii) Alternative B is to cook on their own Other options probably exist but we’ll stick to these two alternatives Principle 2 - Focus on the Differences The biggest difference is likely to be in explicit money costs. Another difference is in the time taken to satisfy hunger in either case. A third difference could be quality of the food and of the ingredients. We’ll concentrate our attention on cost differences in part (c) to follow. Principle 3 - Use a Consistent Viewpoint Consider your problem from the perspective of three students as customers wanting to get a good deal. Use the customer’s point of view in this situation rather than that of the principal of the college, or the owners of the local supermarket, or the neighbors of the students. 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 meal system to select, we’ll use a cost-based metric in part (c). Principle 5 - Consider All Relevant Criteria Factors other than cost may affect the decision about which meal system to commit to. For example, students may not end up eating every meal at college even if they have signed up for it. Another factor is how well prepared these students are for the final exams. 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 Students should check the timings of college meals and whether there is a holiday period when the college kitchen would be closed. Another area of uncertainty is about the shopping and cooking time associated with preparing your own meals. Are all students equally efficient cooks? Since the time taken by each student will vary, how would the three friends allocate cooking time between themselves? Principle 7 - Revisit Your Decision After you’ve completed the semester, how did you do on the final exam? Did your decision have an impact on study times? How good was the food that you consumed? How much money did you save? You will keep these sorts of things in mind (good and bad) when you plan for your next semester!

1-14

continued c)

Finally some numbers to crunch – don’t forget to list any key assumptions that underpin your analysis to minimize the costs of time and money (Principles 1, 2, 3, 4 and 6 are integral to this comparison). Assumptions: (i) the costs given in the example are all accurate, (ii) cooking and meal preparation time will be equally divided between all three, (iii) students will not miss a single college meal if they have signed up for the system, (iv) students never eat out at a restaurant or a friend’s place. Analysis: Alternative A Cost of meals on weekdays: $500 per term per student, or $50 per week or $5 per meal. Cost of meals on weekends: (1/2)($5) × 4 = $2.50 per meal × 4 = $10 Cost of meals in the week = $50 + $10 = $60. Time taken to prepare home meals: 1 student-hour per meal or 20 minutes a student for all three. Value of time per meal = $15/hr × 20 mins = $5 per meal. Value of time spent cooking per week = $5 × 4 = $20 Total cost per week per student: $60 + $20 = $80. Alternative B Cost of meals for the week: $2.50 per meal × 7 × 2 = $35 per student Time taken to prepare home meals: 20 minutes × 14 = 280 mins per student. Value of time spent cooking per week = $15/hr × 280 mins = $70. Total cost per week per student: $35 + $70 = $105. Therefore, sign up for the college meal system to minimize total cost. Typical other criteria you and your friends could consider are: (i) minimize explicit money costs paid out, ignoring the value of time (pick cooking yourselves), (ii) minimize total time spent, ignoring money costs (pick college meal system), and (iii) minimize wasted funds (pick cooking yourself because you will invariably miss a few college meals because of your social calendar).

1-15

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

1-16

STEP 1—Define the Problem: Your basic problem is that you need transportation. STEP 2—Develop Your Alternatives (Principle 1 is used here.): Your problem has been reduced to buying a new car or a used one. The alternatives would appear to be 1. Buy a new car for $20,000. Spend $10,000 of your savings and borrow $10,000. You will have to repay $10,000 after two years, with 10% in interest over two years, generating an interest cost of $1,000. The car will be brand new with 0 miles of prior use. Your savings account will be run down to 0. You can resell the car after two years for $10,000 (50% less than the original cost). 2. Buy a used car for $9,000. Spend the money out of your savings. You will have $1,000 left in your account. Your car will have 25,000 miles of prior use. You can resell the car for $6,000. 3. Buy a new car for $20,000 but finance more of it on loan from the company. For example, if the entire amount is borrowed from the company, then the interest cost due will be $2,000; however, you will have more savings through the year. 4. Buy a used car for $9,000 but finance more of it on loan from the company. For example, if the entire amount is borrowed from the company, then the interest cost due will be $900 but you will have more savings through the year. ASSUMPTIONS: 1. Each car will perform at a satisfactory operating condition (as it was originally intended) and will provide the same total mileage before being sold. 2. Each car will have similar operating efficiency in terms of gallons per mile (alternatively, assume fixed values of gallons per mile for each car). 3. Interest earned on money remaining in savings is negligible (or assume a given rate of interest). STEP 3—Estimate the Cash Flows for Each Alternative (Principle 2 should be adhered to in this step.) a. Your net outflow of cash over two years will be ($10,000 + $10,000 + $1,000 – $10,000) = $11,000. b. Your net cash outflow will be ($9,000 – $6000) = $3,000. c. The net cash flow at the end of the period will be ($20,000 + $2,000 – $10,000) = $12,000. d. Your net cash outflow will be ($9,000 + $900 – $6000) = $3,900.

1-16

continued 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 prospective owner of a car). The value of the car to the owner is its market value (i.e., $20,000 for the new car and $9,000 for the used 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 reliability worth?). STEP 5—Analyze and Compare the Alternatives: Make sure you consider all relevant criteria (Principle 5). If interest on savings is negligible, there is no benefit to holding savings. In this case, options 3 and 4 are eliminated because they involve higher cash flows and the same end result as options 1 and 2. Alternative 2 is a good alternative to consider, because it spends the least amount of cash. The choice between 1 and 2 depends on how highly you value having a new car – i.e. is it worth $8,000 over two years? 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. How reliable will the used car be? How different would projected maintenance and repair costs on both cars be? How does mileage on both cars vary – is the new car much better than the old car? Do you have a complete history of the used car – such as whether it’s been in an accident before? How reliable are the projected end-period values of both cars? Based on the information in all previous steps, you decide that though the new car may be very good to drive, the sharp depreciation in its value makes it a poor choice to own for only two years. Alternative 2 was actually chosen. STEP 7—Monitor the Performance of Your Choice: This step goes hand-in-hand with Principle 7 (revisit your decisions). The used car turned out to be great. Mileage was great, and no repairs were needed. The systematic process of identifying and analyzing alternative solutions to this problem really paid off!

1-17

Employees might be able to come to work earlier and stay later if they are not restricted by public transport timings or by the need to come and go by carpool. If they were previously parking their cars farther away, then they might have been reluctant to work late; especially if the area in which they were parking was unsafe or had poor security. Finally, poor security for employees and their cars could have invited lawsuits and high legal costs if the company was found negligent in providing for its employees.

1-18

(a)

Value of metal in collection = (5,400/135 lb)(0.95)($3.80/lb) + (5,400/135 lb)(0.05) ($1.20/lb) = $146.8 Each penny is worth about 2.7 cents for its metal content. The numismatic value of each coin is likely to be much greater. Note: It is illegal to melt down coins.

(b)

This answer is left to the individual student. In general, the cost of purchases would go up slightly. The inflation rate would be adversely affected if all purchases were rounded up to the nearest nickel. Additional note: The cost of producing a nickel is almost 10 cents. Maybe the U.S. government should get out of the business of minting coins and turn over the minting operation to privately-owned subcontractors.

1-19

Left to student.

1-20

Left to student.

1-21

One assumption is that water at 60 degrees and 40 degrees has the same efficiency in washing and removing stains. If clothes washed at 40 degrees require longer or more frequent washing, or replacement, then the savings would not be as large. Another assumption would be related to the time for which the wash is run – for example, was there a rinse cycle, or a pre-wash cycle, and what temperatures were these run at? Another factor is whether the same detergent would be effective at both temperatures or whether different temperatures require different detergents. Finally, the material of clothing affects the temperature at which it can be most effectively washed.

Solutions to Chapter 2 Problems A Note To Instructors: Because of volatile energy prices in today's world, the instructor is encouraged to vary energy prices in affected problems (e.g. the price of a gallon of gasoline) plus and minus 50 percent and ask students to determine whether this range of prices changes the recommendation in the problem. This should make for stimulating inclass discussion of the results.

2-1

The total mileage driven would have to be specified (assumed) in addition to the variable cost of fuel per unit (e.g. $ per gallon). Also, the fixed cost of both engine blocks would need to be assumed. The efficiency of the traditional engine and the composite engine would also need to be specified

2-2

(a) (b) (c) (d) (e) (f) (g) (h)

4 – sunk 5 – opportunity 3 – fixed 2 – variable 6 – incremental 1 – recurring 7 – direct 8 – nonrecurring

2-3

(a)

# cows =

1,500,000 miles/year = 273.97 or 274 cows (365 days/year)(15 miles/day)

Annual cost = (1,500,000 miles/year)($10/60 miles) = $250,000 per year (b)

Annual cost of gasoline =

1,500,000 miles/year  $4/gallon = $200,000 per year 30 miles/gallon

It would cost $50,000 less per year to fuel the fleet of cars with gasoline.

2-4

Cost Annual rental fee Hauling cost Total cost

Site A = $100,000 (3)(150,000)($2) = $900,000 $1,000,000

Site B = $60,000 (4)(150,000)($2) = $1,200,000 $1,260,000

Note that the revenue of $8.00/yd3 is independent of the site selected. Thus, we can maximize profit by minimizing total cost. The solid waste site should be located in Site A.

2-5

Present cost to company = $1,300,000 + (44 repairs × 20 hrs/repair × $2,500/hr) = $3,500,000 With Ajax, the cost to company = $2,180,000 + 44 repairs × R hrs/repair × $2,500/hr Ajax is preferred if 2,180,000 + 110,000R ≤ $3,500,000 or R ≤ 12 hours/breakdown.

2-6

This reasoning is incorrect. The time and money spent on the ten hours so far cannot be recovered and it is a sunk cost, which should be ignored. Any further decision on whether to attend the classes or not should not be affected by the size of the sunk cost.

2-7

If the probability that you are able to find a tenant for the furnished flat is quite high then you should furnish the apartment. Your investment of $2,400 will be recovered in 8 months (the rent differential of $300 per month over 8 months). Even if you cannot find a tenant who will pay $800, but who is willing to pay, say $700, then the investment will still be recovered, though now in 12 months.

2-8

The opportunity cost to your friend of quitting her current job is losing a certain income of $80,000 per year. She expects to earn an additional (25 – 5) × 12 × $500 = $120,000 per year baking cookies, which yields a net increase in income of $120,000 – $80,000 – $10,000 = $30,000. Provided her projections about future events per month (25) are relatively certain, quitting her job and switching to catering fulltime would be a good decision. If she is able to cater at least 20 events a month, she will be able to turn a profit from switching careers.

2-9

(a) If you invest in risky property, you could end up losing all your money or not making any return at all. Your opportunity cost would be a guaranteed return of 5% per annum by investing funds in a bank deposit. (b) If you keep your money in a safe, you will not earn any financial return. Your opportunity cost will be at least 5% per annum, earned on a bank deposit. For a risk-loving person, the opportunity cost will be even higher because they might have considered the property market’s potential 25% over the next year as an alternative return.

2-10

Let X equal dollars per item delivered, and set total revenue equal to total cost: $X (15 items/hr) = $42.00/hr + ($0.50/item)(15 items/hr) X = ($49.50/hr) / (15 items/hr) X = $3.30 per item At least $3.30 per item delivered on Sunday will be needed to break even.

2-11

(a)

Students can use Equation (2-10) to determine D*. D* = Then

= 225 units/month

p = 75 – 0.1(225) = $52.50 per unit TR = 225 units × $52.50/unit = $11,812.50 and

CT = $1,000 + $30/unit × 225 units = $7,750 Maximum profit = TR – CT = $11,812.50 − $7,750 = $4,062.50 per month

(b)

Using the quadratic equation to solve for the breakeven points: 𝐷

4 0.1 1,000 0.2

𝐷

24 (rounded up from 23.5)

𝐷

426 (rounded down from 426.5)

40.3 0.2

Thus, the profitable range of demand is from 24 to 426 units per month.

2-12

Re-write the price-demand equation as follows: p = 3,200 – 0.1D. Then, TR = p D = 3,200D – 0.1D2 The first derivative of TR with respect to D is d(TR)/dD = 3,200 – 0.2D This, set equal to zero, yields the D̂ that maximizes TR. Thus, 3,200  0.2 Dˆ  0 D̂ = 16,000 units per month

What is needed to determine maximum monthly profit is the fixed cost per month and the variable cost per lash adjuster.

2-13

P = 500 − 0.2D; CF = $100,000; Cv = $150 Revenue = PD = (500 – 0.2D)D = 500D – 0.2D2 d(Revenue) / dD = 500 – 0.4D = 0 D = 500 / 0.4 = 1,250 units per year. d2R / dD = −0.4 which yields a maximum for revenue. Profit = PD − CF − Cv = (500 – 0.2D)D – 100000 – 150D = 500D – 0.2D2 – 100000 – 150D = 350D – 0.2D2 − 100000 d(Profit) / dD = 350 − 0.4D = 0 D = 875 units per year d2Profit / dD = −0.4 which yields a maximum for profit. Unit price at this level of demand is P = 500 – 0.2 (875) = $325. Profits at this price = $(325 – 150) × 875 – 100000 = $53,125

2-14

(a)

P = 450 – 0.1D; Cv = $25/racquet; CF = $25,000 + lease value Optimal D* = (a – Cv)/2b = (450 – 25)/2(0.1) = 2,125 units (Eqn 2-10).

(b)

Profit = [450 – 0.1(2,125)] × 2,125 – 25,000 – Lease value – 25(2,125) For profits to be at least 0, Lease value can be no greater than L = $426,562,50

2-15

(a)

Profit = PD − 2,000 − 35D = (5 + 4800/D – 3000/D2)D – 2,000 – 35D = −30D + 4800 – 3000/D − 2000 dProfit/dD = −30 + 3000/D2 = 0 D* = 10

(b)

d2Profit/dD = −6000/D3 < 0 at D* which implies that D* maximizes profit.

2-16

Profit = Total revenue – Total cost = (20X – 0.5X2) – (20 + 0.8X + 0.25X2) = 19.2X – 0.75X2 – 20 dProfit = 0 = 19.2  1.5X dX

X = 12.8 megawatts Note:

d 2 Profit =  1.5; thus, X = 12.8 megawatts maximizes profit dX 2

2-17

Annual sales (units) = $480,000/$40 = 12,000 units/yr. D’ = CF/(P – CV) = 150000/(40 – 25) = 10,000 units/yr. (from Eqn 2-13)

2-18

20,000 tons/year (2,000 pounds/ton) = 40,000,000 pounds of zinc per year are produced. The variable cost per pound is $30,000,000/40,000,000 pounds = $0.75 per pound. (a) Profit/year = (40,000,000 pounds/year)($1.00 – $0.75) – $10,000,000 = $10,000,000 – $10,000,000 = $0 per year Because Profit = 0, 20,000 tons per year is the breakeven point production level for this mine. A loss would occur for production levels < 20,000 tons/year and a profit for levels > 20,000 tons per year. (b) The mine will be profitable. If 21,000 tons (= 42,000,000 pounds) are produced, then Profit/yr = (42,000,000 pounds/year)($1.00 – $0.75) – $10,000,000 = $50,000

2-19

(a)

BE = $500,000 / ($50 − $10) = 12,500visitors per month.

(b)

New BE point = $500,000 / ($60 − $15) = 11,112visitors per month.

(c)

For 25,000 visitors per month, profit equals 25,000 ($60 − $15) − $500,000 = $625,000 per month, so the park makes a profit.

2-20

(a)

BE point = $70,000/($10 – $3) =10,000 tubs or 66.67% of capacity. Profits = 15,000 ($10 – $3) – $70,000 = $35,000.

(b)

Profits = 15,000 ($7 – $3) – $70,000 = –$10,000.

2-21

Annual savings are at least equal to ($120/lb)(500 lb) = $60,000. So the company can spend no more than $60,000 (conservative) and still be economical. Other factors include ease of maintenance and cleaning, passenger comfort, and aesthetic appeal of the improvements. Yes, this proposal appears to have merit, so it should be supported.

2-22

Afua is willing to pay $400 to eliminate the risk of data loss, in case she is unable to back up all her data to the portable hard drive, suggesting she is relatively risk-averse. Perhaps, she expects there could be further complications with an unstable hard drive that could make the computer unusable even with a portable hard drive. In such an eventuality, investing in a new laptop would eliminate the costs of any future repairs to the old computer.

2-23

Over 81,000 miles, the gasoline-only car will consume 2,700 gallons of fuel. The flex-fueled car will use 3,000 gallons of E85. So we have (3,000 gallons)(X) + $1,000 = (2,700 gallons)($3.89/gal) and X = $3.17 per gallon This is 18.5% less expensive than gasoline. Can our farmers pull it off – maybe with government subsidies?

2-24

(a)

𝐶 =𝑐 𝑋

Eqn. (2-8)

$2,000,000 = 𝑐 2000 𝑡𝑜𝑛𝑠 𝑐 = $1,000/ton = $0.50/ lb. Profit = [𝑠 - 𝑐 ]X - 𝐶 = [ $0.80/lb - $0.50/ lb] 4,000,000 lb - $700,000 = $500,000 (b)

X’

=𝐶 / 𝑠 -𝑐 )

Eqn. (2-13)

= $700,000 / ($0.80/lb - $0.50/ lb) = 2,333,333 lbs 𝑐 = 𝐶 / X’ = $700,000 / 2,333,333 lbs = $0.30 / lb OR X’ 𝑠 - 𝑐 ) = 𝑐 X’ at breakeven point 𝑐 (c)

𝑠 - 𝑐 ) = ($0.80/lb - $0.50/ lb) = $0.30/ lb

𝑐 = (X𝑐 + 𝐶 ) / X = [ 4,000,000 lb ($0.50/lb) + $700,000] / 4,000,000 lb = $0.675/ lb

2-25

Annual profits = Annual revenues (R) – Total fixed costs – Total variable costs $15,000 = R – $10,000 – .5R .5R = $25,000 R = $50,000 The correct choice is (b).

2-26 CT = Co + Cc = knv2 +

$1,500n v

dCT 1,500 = 0 = 2kv- 2 = kv3 - 750 dv v 750 k

v=3

To find k, we know that Co = $100/mile at v = 12 miles/hr n

Co = kv 2 = k(12) 2 = 100 n and k = 100 / 144 = 0.6944

so,

v = 3

750 = 10.25 miles / hr . 0.6944

The ship should be operated at an average velocity of 10.25 mph to minimize the total cost of operation and perishable cargo. Note: The second derivative of the cost model with respect to velocity is: d 2CT n = 1.388n + 3,000 3 2 dv v

The value of the second derivative will be greater than 0 for n > 0 and v > 0. Thus we have found a minimum cost velocity.

2-27

Solve for k: CG / n = kv

v in miles/hr

1/18 mils/gal = k (70 miles/hr) ; k = 1 hr – gal / 1,260 mi2 = 0.000794 hr-gal/mi2 CG = (0.000794 hr-gal/mi2)(3000)(v)($3.60/gal) = 8.5752 hr-gal/mi2 CFSS = ($15,000/hr)(1/v)

Find CT CT = ($8.5752 hr/mi2)(v mi/hr) + ($15,000/hr)(v-1hr/mi)($/mi)

d CT / dv = 8.5752 −15,000/v2 = 0 v2 = 15,000/8.5752 v* = 41.82 mi/hr

Check d2 CT / dv2 = 30,000v-3 which is positive for v > 0, therefore we have minimized the total cost

2-28 (293 kWh/106 Btu)($0.15/kWh) = $43.95/106 Btu

R11 R19 R30 R38 A. Investment cost $2,400 $3,600 $5,200 $6,400 6 B. Annual Heating Load (10 Btu/yr) 74 69.8 67.2 66.2 C. Cost of heat loss/yr $3,252 $3,068 $2,953 $2,909 D. Cost of heat loss over 25 years $81,308 $76,693 $73,836 $72,737 E. Total Life Cycle Cost = A + D $83,708 $80,293 $79,036 $79,137 Select R30 to minimize total life cycle cost.

2-29

(a)

C dC   I  CR t = 0 d 2

or, 2 = CI/CRt and, * = (CI/CRt)1/2; we are only interested in the positive root. (b)

d 2C d 2



2C I

3

 0 for  > 0

Therefore, * results in a minimum life-cycle cost value. (c)

Investment cost versus total repair cost

C CRꞏꞏt $



2-30

180 rpm

1 cycle = (12 jacks/refurb.)(1 +4 refurb.) / (7 jacks/hr) = 60/7 = 8.57 hrs Cost per cycle = 1 brush ---- $90 4 refurb. --- 4($30) Oper. Cost – 8.57 hr. ($70/hr) = $810/cycle Cost per jack = $810/60 = $13.50/jack

240 rpm

1 cycle = (8 x 3) / 10 = 2.4 hrs Cost per cycle = $90 + 2($30) + 2.4hr ($70/hr) = $318/cycle Cost per jack = $318/24 = $13.25/jack

300 rpm

1 cycle = (6 x 2) / 12 = 1hr Cost per cycle = $90 + 1($30) + 1hr ($70/hr) = $190/cycle Cost per jack = $190/12 = $15.83/jack

Select 240 rpm

2-31

(a)

With Dynolube you will average (20 mpg)(1.01) = 20.2 miles per gallon (a 1% improvement). Over 50,000 miles of driving, you will save 50,000 miles 50,000 miles   24.75 gallons of gasoline. 20 mpg 20.2 mpg

This will save (24.75 gallons)($4.00 per gallon) = $99. (b)

Yes, the Dynolube is an economically sound choice.

2-32

The cost of tires containing compressed air is ($200 / 50,000 miles) = $0.004 per mile. Similarly, the cost of tires filled with 100% nitrogen is ($220 / 62,500 miles) = $0.00352 per mile. On the face of it, this appears to be a good deal if the claims are all true (a big assumption). But recall that air is 78% nitrogen, so this whole thing may be a gimmick to take advantage of a gullible public. At 200,000 miles of driving, one original set of tires and three replacements would be needed for compressed-air tires. One original set and two replacements (close enough) would be required for the 100% nitrogen-filled tires. What other assumptions are being made?

2-33

(a)

Speed A B C

Drilling Rate (ft/min) 2 3 4

Bit life at this Speed (min) 10 6 3

Given: rock drill with 3 operating speeds Cost= bit cost + operator cost (+ blasting penalty) 1 cycle = 96 ft. of drilling Operator cost = $30/hr = $0.50/min Bit cost = $10.00/each Speed A – Cycle = 96ft / 2fpm = 48 min Bit cost- 48 min/ 10 min/bit = 4.8 bits x $10 each = $48.00 Oper. Cost- 48 min x $0.50/min

$24.00 $72.00/cycle

Cost/ft. = $72.00/cycle / 96 ft/cycle = $0.75/ft Speed B – Cycle = 96ft / 3fpm = 32 min Bit cost- 32 min/ 6 min/bit = 5.33 bits x $10 each = $53.33 Oper. Cost- 32 min x $0.50/min

$16.00 $69.33/cycle

Cost/ft. = $69.33/cycle / 96 ft/cycle = $0.72/ft Speed C – Cycle = 96ft / 4fpm = 24 min Bit cost- 24 min/ 3 min/bit = 8 bits x $10 each =

$80.00

Oper. Cost- 24 min x $0.50/min

$12.00 $92.00/cycle

Cost/ft. = $92.00/cycle / 96 ft/cycle = $0.96/ft

Choose Speed B

2-33

(b)

Penalty of $60.00/hour for cycle time greater than 30 minutes $60/hr = $1.00/min Speed A- Cycle Time = 48 min Total cost/cycle = $72.00 + (48-30)$1.00 = $90.00 (includes penalty) ($0.9375/ft) Speed B- Cycle Time = 32 min Total cost/cycle = $69.33 + (32-30)$1.00 = $71.33 (includes penalty) ($0.743/ft) Speed C- Cycle Time = 24 min (no penalty Total cost/cycle = $92.00 ($0.958/ft) Speed B still has the lowest cost per cycle, now the cost per foot is $0.743/ft

2-34

(a)

Manufacturing Option A Labor = (40 x 5)($10) = $2000/wk. Rental = $20,000 Σ = $22,000 Material = $15/unit Purchase Option B = $20/unit $22,000 + $15(x) = $20(x) $22,000/5 = 𝑥 𝑥 = 4,400 units/wk = Breakeven amount

(b)

$22,000 + $15(3500) [< or > ] $20 (3500) $74,500 > $70,000 Purchase this item

2-35

Strategy: Select the design which minimizes total cost for 125,000 units/year (Rule 2). Ignore the sunk costs because they do not affect the analysis of future costs. (a)

Design A Total cost/125,000 units = (12 hrs/1,000 units)($18.60/hr)(125,000) + (5 hrs/1,000 units)($16.90/hr)(125,000) = $38,463, or $0.3077/unit Design B Total cost/125,000 units = (7 hrs/1,000 units)($18.60/hr)(125,000) + (7 hrs/1,000 units)($16.90/hr)(125,000) = $33,175, or $0.2654/unit Select Design B

(b)

Savings of Design B over Design A are: Annual savings (125,000 units) = $38,463 − $33,175 = $5,288 Or, savings/unit = $0.3077 − $0.2654 = $0.0423/unit.

2-36

Profit per day = Revenue per day – Cost per day = (Production rate)(Production time)($30/part)[1-(% rejected+% tested)/100] – (Production rate)(Production time)($4/part) – (Production time)($40/hr) Process 1: Profit per day = (35 parts/hr)(4 hrs/day)($30/part)(1-0.2) – (35 parts/hr)(4 hrs/day)($4/part) – (4 hrs/day)($40/hr) = $2640/day Process 2: Profit per day = (15 parts/hr)(7 hrs/day)($30/part) (1-0.09) – (15 parts/hr)(7 hrs/day)($4/part) – (7 hrs/day)($40/hr) = $2155.60/day Process 1 should be chosen to maximize profit per day.

2-37

At 70 mph your car gets 0.8 (30 mpg) = 24 mpg and at 80 mph it gets 0.6(30 mpg) = 18 mpg. The extra cost of fuel at 80 mph is: (400 miles/18mpg – 400 miles/24 mpg)($4.00 per gallon) = $22.22 The reduced time to make the trip at 80 mph is about 45 minutes. Is this a good tradeoff in your opinion? What other factors are involved?

2-38

(a)

Operation 1

cycle time = 1 hr + 0.333 hr = 1.333 hr/cycle Cycles/day = (8hr/day)(1 cycle/ 1.333 hr) = 6 cycle/day Value added = (2000 parts/cycle)(6 cycles/day)($0.40/part) = $4,800/day

Cost1 = 8 hr/day ($20/hr) = $160/day Value – cost = $4,800 - $160 = $4,640/day Operation 2

cycle time = 2 hr + 0.5 hr = 2.5 hr/cycle

Cycles/day = (8hr/day)(1 cycle/ 2.5 hr) = 3.2 cycle/day Value added = (3500 parts/cycle)(3.2 cycles/day)($0.40/part) = $4,480/day Cost2 = 8 hr/day ($11/hr) = $88/day Value – cost = $4,480 - $88 = $4,392/day Select Operation 1 to maximize profit (b)

Output/day for Operation 1 = 12,000 parts and output/.day for Operation 2 = 11,200 parts. Downtime for Operation 1 = 6 x 20 min = 120 minutes/day and downtime for Operation 2 = 3.2 x 30 = 96 minutes/day. So increased production for Operation 1 is being traded off for increased tool changing time (downtime), and the balance is favorable for Operation 1 compared to Operation 2.

2-39

Apache: (24 hr/day)(7 days/wk) – 4 = 164 hrs/wk uptime (90 hits/min) (60 min/hour) = 5,400 hits/hr (5,400 hits/hr) (164 hrs/wk) = 885,600 hits/wk @ $0.015/hit = $13,284/wk Profit/yr. = ($13,284/wk)(52 wk/yr) = $690,768 Windows IIS: (24 hr/day)(7 days/wk) – 0.75 = 167.25 hrs/wk uptime (5,400 hits/hr) (167.25 hrs/wk) = 903,150 hits/wk @ $0.015/hit = $13,547.25/wk Profit/yr. = ($13,547.25/wk)(52 wk/yr) - $5,000 = $699,457 Go with Windows software.

2-40

Option A (Purchase): CT = (10,000 items)($8.50/item) = $85,000 Option B (Manufacture): Direct Materials = $5.00/item Direct Labor = $1.50/item Overhead = $3.00/item $9.50/item CT = (10,000 items)($9.50/item) = $95,000 Choose Option A (Purchase Item).

2-41

This statement makes sense, up to a point. Airlines do have very high fixed costs relative to variable costs, particularly no-frills airlines, which do not offer meals or other services that would vary on a perpassenger basis. However, the fuel carried by a plane does depend on the weight of the plane: with fewer passengers, planes can carry less fuel. As long as this variable cost is covered, any additional revenue from discounted ticket sales will help offset the fixed costs of operating the flight.

2-42

Assumptions: You can sell all the metal that is recovered Method 1:

Recovered ore = (0.62)(100,000 tons) = 62,000 tons Removal cost = (62,000 tons)($23/ton) = $1,426,000 Processing cost = (62,000 tons)($40/ton) = $2,480,000 Recovered metal = (300 lbs/ton)(62,000 tons) = 18,600,000 lbs Revenues = (18,600,000 lbs)($0.8 / lb) = $14,880,000 Profit = Revenues - Cost = $14,880,000 - ($1,426,000 + $2,480,000) = $10,974,000

Method 2:

Recovered ore = (0.5)(100,000 tons) = 50,000 tons Removal cost = (50,000 tons)($15/ton) = $750,000 Processing cost = (50,000 tons)($40/ton) = $2,000,000 Recovered metal = (300 lbs/ton)(50,000 tons) = 15,000,000 lbs Revenues = (15,000,000 lbs)($0.8 / lb) = $12,000,000

Profit = Revenues - Cost = $12,000,000 - ($750,000 + $2,000,000) = $9,250,000 Select Method 1 (62% recovered) to maximize total profit from the mine.

2-43

Profit per ounce (Method A) = $1,750 - $550 / [(0.90 oz. per ton)(0.90)] = $1,750 - $679 = $1,071 per ounce Profit per ounce (Method B) = $1,750 - $400 / [(0.9 oz. per ton)(0.60) =$1,750 - $741 = $1,009 per ounce Therefore, by a slim margin we should recommend Method A.

2-44

(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) (r) (s)

True False False False True True True True False True False True False True True True False True False

1,750,000 Btu  2-45

(a) Loss 

lb coal    12,000 Btu   486 lbs of coal 0.30

(b) 486 pounds of coal produces (486)(1.83) = 889 pounds of CO2 in a year.

2-46

(a) Let X = breakeven point in miles Fuel cost (car dealer option) = ($2.00/gal)(1 gal/20 miles) = $0.10/mile Motor Pool Cost = Car Dealer Cost ($0.36/mi)X = (6 days) ($30/day) + $0.20/mi + $0.10/miX $0.36X = 180 + $0.30X and X = 3,000 miles

(b)

6 days (100 miles/day) = 600 free miles If the total driving distance is less than 600 miles, then the breakeven point equation is given by: ($0.36/mi)X = (6 days)($30 /day) + ($0.10/mi)X X = 692.3 miles > 600 miles This is outside of the range [0, 600], thus renting from State Tech Motor Pool is best for distances less than 600 miles. If driving more than 600 miles, then the breakeven point can be determined using the following equation: ($0.36/mi)X = (6 days)($30 /day) + ($0.20/mi)(X - 600 mi) + ($0.10/mi)X X = 1,000 miles

(c)

The true breakeven point is 1000 miles.

The car dealer was correct in stating that there is a breakeven point at 750 miles. If driving less than 900 miles, the breakeven point is: ($0.34/mi)X = (6 days)($30 /day) + ($0.10/mi)X X = 750 miles < 900 miles However, if driving more than 900 miles, there is another breakeven point. ($0.34/mi)X = (6 days)($30/day) + ($0.28/mi)(X-900 mi) + ($0.10/mi)X X = 1800 miles > 900 miles The car dealer is correct, but only if the group travels in the range between 750 miles and 1,800 miles. Since the group is traveling more than 1,800 miles, it is better for them to rent from State Tech Motor Pool. This problem is unique in that there are two breakeven points. The following graph shows the two points.

2-46 continued

Total Cost

Car Dealer v. State Tech Motor Pool $1,000 $900 $800 $700 $600 $500 $400 $300 $200 $100 $0

X2' = 1,800 miles

X1' = 750 miles Car Dealer State Tech

500

1000

1500

2000

2500

Trip Mileage

2-47

This problem is location specific. We’ll assume the problem setting is in Tennessee. The eight years ($2,400 / $300) to recover the initial investment in the stove is expensive (i.e. excessive) by traditional measures. But the annual cost savings could increase due to inflation. Taking pride in being “green” is one factor that may affect the homeowner’s decision to purchase a corn-burning stove.

2-48 F

Net Income

250

$20,000 $1

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

$(20,000) $(40,000) $(60,000) $(80,000) $(100,000) Volume (Demand)

$600,000

$500,000

$400,000 Total Revenue

$300,000

Total Expense

$200,000

$100,000

10 00 15 00 20 00 25 00 30 00 35 00 40 00 45 00 50 00

$50 0

Monthly Price per Total Demand Unit Revenue Total Expense Net income 6 7 0 $ 180 $ $ 73,000 $ (73,000) 8 250 $ 175 $ 43,750 $ 93,750 $ (50,000) 9 500 $ 170 $ 85,000 $ 114,500 $ (29,500) 10 750 $ 165 $ 123,750 $ 135,250 $ (11,500) 11 1000 $ 160 $ 160,000 $ 156,000 $ 4,000 12 1250 $ 155 $ 193,750 $ 176,750 $ 17,000 13 1500 $ 150 $ 225,000 $ 197,500 $ 27,500 14 1750 $ 145 $ 253,750 $ 218,250 $ 35,500 15 2000 $ 140 $ 280,000 $ 239,000 $ 41,000 16 2250 $ 135 $ 303,750 $ 259,750 $ 44,000 17 2500 $ 130 $ 325,000 $ 280,500 $ 44,500 18 2750 $ 125 $ 343,750 $ 301,250 $ 42,500 19 3000 $ 120 $ 360,000 $ 322,000 $ 38,000 20 3250 $ 115 $ 373,750 $ 342,750 $ 31,000 21 3500 $ 110 $ 385,000 $ 363,500 $ 21,500 3750 $ 105 $ 393,750 $ 384,250 $ 9,500 22 4000 $ 100 $ 400,000 $ 405,000 $ (5,000) 23 4250 $ 95 $ 403,750 $ 425,750 $ (22,000) 24 4500 $ 90 $ 405,000 $ 446,500 $ (41,500) 25 4750 $ 85 $ 403,750 $ 467,250 $ (63,500) 26 5000 $ 80 $ 400,000 $ 488,000 $ (88,000) 27 5250 $ 75 $ 393,750 $ 508,750 $ (115,000) 28 5500 $ 70 $ 385,000 $ 529,500 $ (144,500) 29 30 31 32 Summary of impact of changes in cost components on optimum 33 demand and profitable range of demand. 34 35 Percent Change D 1' cv D2' CF D* 36 -10% -10% 2,633 724 4541 37 0% -10% 2,633 824 4443 38 10% -10% 2,633 928 4339 39 -10% 0% 2,425 816 4036 40 0% 0% 2,425 932 3918 41 10% 0% 2,425 1060 3790 42 -10% 10% 2,218 940 3495 43 0% 10% 2,218 1092 3343 44 10% 10% 2,218 1268 3167 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

$40,000

Volume (Demand)

$600,000 $500,000 $400,000 $300,000 Total Revenue $200,000

Total Expense Net income

$100,000 $$(100,000) $(200,000) 0

83 180 0.02

50 0 10 00 15 00 20 00 25 00 30 00 35 00 40 00 45 00 50 00 55 00

73,000

$ $ $

D Demand Start point (D) = Demand Increment =

Net Income

Cash Flow

2 3 4 5

Cash Flow

A Fixed cost/ mo. = Variable cost/unit = a= b=

Volume (Demand)

Reducing fixed costs has no impact on the optimum demand value, but does broaden the profitable range of demand. Reducing variable costs increase the optimum demand value as well as the range of profitable demand.

2-49

New annual heating load = (230 days)(72 °F  46 °F) = 5,980 degree days. Now, 136.7  106 Btu are lost with no insulation. The following U-factors were used in determining the new heating load for the various insulation thicknesses. R11 R19 R30 R38

Energy Cost

Investment Cost Annual Heating Load (106 Btu) Cost of Heat Loss/yr Cost of Heat Loss over 25 years Total Life Cycle Cost

U-factor 0.2940 0.2773 0.2670 0.2630

Heating Load 101.3  106 Btu 95.5  106 Btu 92  106 Btu 90.6  106 Btu

$/kWhr $0.086

$/106 Btu $25.20

R11 900 $

R19 1,350 $

R30 1,950 $

R38 2,400

101.3 $2,553 $63,814 $64,714

95.5 $2,406 $60,160 $61,510

92 $2,318 $57,955 $59,905

90.6 $2,283 $57,073 $59,473

2-50

In this problem we observe that "an ounce of prevention is worth a pound of cure." The ounce of prevention is the total annual cost of daylight use of headlights, and the pound of cure is postponement of an auto accident because of continuous use of headlights. Clearly, we desire to postpone an accident forever for a very small cost. The key factors in the case study are the cost of an auto accident and the frequency of an auto accident. By avoiding an accident, a driver "saves" its cost. In postponing an accident for as long as possible, the "annual cost" of an accident is reduced, which is a good thing. So as the cost of an accident increases, for example, a driver can afford to spend more money each year to prevent it from happening through continuous use of headlights. Similarly, as the acceptable frequency of an accident is lowered, the total annual cost of prevention (daytime use of headlights) can also decrease, perhaps by purchasing less expensive headlights or driving less mileage each year. Based on the assumptions given in the case study, the cost of fuel has a modest impact on the cost of continuous use of headlights. The same can be said for fuel efficiency. If a vehicle gets only 15 miles to the gallon of fuel, the total annual cost would increase by about 65%. This would then reduce the acceptable value of an accident to "at least one accident being avoided during the next 16 years." To increase this value to a more acceptable level, we would need to reduce the cost of fuel, for instance. Many other scenarios can be developed.