Solution Manual For Engineering Economy, 2025 18th Edition by William G. Sullivan Elin M Wicks Josep by welldoneassistant

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)

62 million tons per year × (0.05) = 3.1 million tons per year of reduced greenhouse gas $1.2 billion ÷ 3.1 million tons per year = $387.10 per ton

(b)

3 billion tons per year × (0.03) = 90 million tons per year of reduced greenhouse gas $1.2 billion $X billion = 3.1 million tons 90 million tons X = $34.84 billion

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 = $0.75/1.5 Watt-hours = $0.50 per Watt-hour At a cost of $0.50 per Watt-hour, it would cost (1,000)($0.50 per Watt-hour) = $500 per kilo Watt-hour for power from a single AAA battery. This is 5,000 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)

15,000 miles per year / 25 mpg = 600 gallons per year of E20 Savings = 600 gallons per year ($4.00 − $3.37) = $378 per year

(b)

Gasoline saved = 0.20 (600 gal/yr)(1,000,000 people) = 120 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 2,000 ft2 × $120/ ft2 = $240,000. The ICF home will cost $240,000 (1.10) = $264,000. So the cost difference to construct an ICF home is $24,000. The energy savings per month is 45% × $200 = $90 per month. Therefore, it will take $24,000/($90/month) = 267 months to pay back the extra ICF construction cost through monthly energy savings.

1-8

Increased lifetime earnings of a college graduate = $1,200,000(0.75) = $900,000

1-9

Strategy 1: Change oil every 3,000 miles. Cost = (15,000/3,000)($50) = $250 / year Strategy 2: Change oil every 5,000 miles. Cost = (15,000/5,000)($50) = $150 / year Savings = $100 per year

1-10

In six months you will spend approximately (180)(2)($1.15) = $414 on bottled water. The cost of the filter is $60, so you will save $354 every six months. This amounts to $708 over a year, and you don't need to bother recycling all those plastic bottles! An up-front expenditure of $60 can indeed save a lot of money each year.

1-11

110 gallons x $4.00 per gallon = $440 saved over 55,000 miles of driving. This comes down to $440 / 55,000 = $0.008 per mile driven. So Brand A saves 8/10 of 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 time, grade point average and energy that Pat is willing/able to exert, Problem A might be "How can Pat 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, Pat's Problem B could be "How can I use my brother's entry-level job as a springboard 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 their brother and take fewer courses per term, possibly graduating in the summer.

(2)

Quit partying and devote their extra time and limited funds to the task of graduating in the spring term (maybe Pat could get a scholarship to help with tuition, room and board).

Problem B - Some feasible solutions for Problem B would include: (1)

Work for their brother and take over the company to enable him to start another entrepreneurial venture.

(2)

Work part-time for their brother and continue to take courses over the next couple of years in order to graduate.

(3)

Work for their brother for one or two semesters to build up funds for their senior year. While interviewing, bring up the real life working experience and request a higher starting salary.

1-14

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 “Pick-Up 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 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 cost-based 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!

1-14

continued (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 difficult-to-predict 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 “Pick-Up-Sticks” 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

Therefore, order the pizza from “Fred’s” to minimize total cost per cubic inch. (d)

Typical other criteria you and your friends could consider are: (i) cost per square inch of pizza (select “Pick-Up-Sticks”), (ii) minimize total cost regardless of area or volume (select “Pick-UpSticks”), and (iii) “Fred’s” can deliver in 30 minutes but “Pick-Up-Sticks” cannot deliver for one hour because one of their ovens is not working properly (select “Fred’s”).

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. 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 part-time 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 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 one extra 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.

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 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.) 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 out-of-savings 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 hand-in-hand 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!

1-17

Imprudent use of electronic mail, for example, can involve legal issues, confidential financial data, trade secrets, regulatory issues, public relations goofs, etc. These matters are difficult to “dollarize” but add to the $30,000 annual savings cited in the problem. Surfing the web inappropriately can lead to legal prosecution for pornography violations.

1-18

(a)

Value of metal in collection = (5,000/130 lb)(0.95)($3.50/lb) + (5,000/130 lb)(0.05)($1.00/lb) = $129.81 Each penny is worth about 2.6 cents for its metal content. The numismatic value of each coin is most likely 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

A one degree difference in thermostat setting will reduce energy use by up to five percent when the heat loss in the winter and the heat gain in the summer are roughly equal. This assumes the efficiency of the heating system is about the same as that of the cooling system.

1-22

Assume a 50 week working year: (8 hours/day)(5 days/week)(50 weeks per year) = 2000 hours/year Savings per week = ($25/hour)(15 hours/week) = $375/week Saving per year = ($375/week)(50 weeks/year) = $18,750/year

1-23

Annual revenue = ($0.10/kWh)(150,000 kWh/year) = $15,000/year

1-24

$10,000/($200 per month savings) = 50 months

1-25

Compare material cost per year. A: $10/(5 years) = $2 per year B: $15/(8 years) = $1.875 per year On a $/year basis, Material B is more cost effective. However, if we assume that both materials have to be purchased twice to cover the 10 year period, A: $10 X 2 = $20 for 10 years = $2/year B: $15 X 2 = $30 for 10 years = $3/year If there is no further use for the material after 10 years, Material A would be more economical.

1-26

Annual savings = ($500,000/year)(0.03) = $15,000/year

1-27

Cost to outsource = ($15/hour)(1000 hours) = $15,000 Cost to automate = $10,000 It is more cost effective to automate the process.

1-28

$800/($150/year) = 5.33 years

1-29

Compute the total cost for 1,000 non-defective units. Cost to obtain 1 non-defective unit = ($/unit)/(% chance unit is non-defective) Supplier A = (1000 non-defective units)($20/unit)/0.98 = $20,408.16 Supplier B = (1000 non-defective units)($22/unit)/0.99 = $22,222.22 It is more cost effective to obtain the materials from Supplier A.

1-30

Money saved each day = 10 gallons/day X $0.005/gallon = $0.05 per day Time to recover $250 investment = $250/(0.05/day) = 5000 days = 13.7 years

1-31

$1,500 / ($5/hour) = 300 hours

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 =

1,000,000 miles/year = 182.6 or 183 cows (365 days/year)(15 miles/day)

Annual cost = (1,000,000 miles/year)($10 / 60 miles) = $166,667 per year (b)

miles/year Annual cost of gasoline = 1,000,000 × $4/gallon = $133,333 per year 30 miles/gallon It would cost $33,334 more per year to fuel the fleet of cars with gasoline.

2-4

Cost Rent Hauling Total

Site A = $5,000 (4)(200,000)($1.50) = $1,200,000 $1,205,000

Site B = $100,000 (3)(200,000)($1.50) = $900,000 $1,000,000

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 B.

2-5

Present cost to company = $1,000,000 + 40 repairs × 24 hrs/repair × $2,500/hr = $3,400,000 With Ajax, the cost to your company = $2,000,000 + 40 repairs × R hrs/repair × $2,500/hr Ajax is preferred if 2,000,000 + 100,000R ≤ $3,400,000 or R ≤ 14 hours/breakdown.

2-6

The $97 you spent on a passport is a sunk cost because you cannot get your money back. If you decide to take a trip out of the U.S. at a later date, the passport’s cost becomes part of the fixed cost of making the trip (just as the cost of new luggage would be).

2-7

If the value of the re-machining option ($60,000) is reasonably certain, this option should be chosen. Even if the re-machined parts can be sold for only $45,001, this option is attractive. If management is highly risk adverse (they can tolerate little or no risk), the second-hand market is the way to proceed to guarantee $15,000 on the transaction.

2-8

The bakery makes a per unit profit of $2.50 - $1.00 = $1.50 per pastry sold. To cover the cost of the new equipment, $20,000/($1.50 per pastry sold) = 13,333.33 units.

2-9

By automating the process, variable cost is reduced, saving $10 - $7 = $3 for every unit produced. This cost savings will breakeven with the cost to automate after $200,000/($3 per unit) = 66,667 units are produced.

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 CT = $1,000 + $30/unit × 225 units = $7,750 and 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 =

45 ± (−45) − 4(0.1)(1,000) 0.2 𝐷 =

45 ± 40.3 0.2

𝐷 = 24 (rounded up from 23.5) 𝐷 = 426 (rounded down from 426.5) 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 = 2,000 - 0.1D. Then, TR = p  D = 2,000D - 0.1D2. The first derivative of TR with respect to D is d(TR) / dD = 2,000 - 0.2D

 This, set equal to zero, yields the D that maximizes TR. Thus,  2,000 - 0.2 D = 0

 D = 10,000 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 = 150  0.01D

CF = $50,000

cv = $40/unit

Profit = 150D  0.01D2  50,000  40D = 110D  0.01D2  50,000 d(Profit)/dD = 110  0.02D = 0

 D = 5,500 units per year, which is less than maximum anticipated demand At D = 5,500 units per year, Profit = $252,500 and p = $150  0.01(5,500) = $95/unit.

2-14

(a)

D* = (a – Cv) / 2b = (700 – 131.50) / 0.10 = 568.5 / 0.10 = 5,685 103 board-feet Price = $700 – (0.05)(5,685) = $415.75 per 103 board feet Profit (per month) = pD – CF - CvD = $415.75(5,685) - $1,000,000 - $131.50(5,685) = 2,363,539 – 1,000,000 – 747,578 = $615,961/ month

(b)

–bD2 + (a – Cv)D – CF = 0 b= 0.05 (a – Cv) = 568.5 CF = 1,000,000 −0.05D2 + 568.5D – 1,000,000 = 0 D’1,2

= -568.5 ± [ (568.5)2 – 4(-0.05)(-1,000,000)]1/2 2 (-0.05) = -568.5 ± [ 323,192.25 – 200,000]1/2 -0.10

= (-568.5 + 351) / (-0.10) = 2,175

= (-568.5 - 351) / (-0.10) = 9,195

2-15

(a)

(b)

2700 5000   Profit = 38   2  D - 1000 - 40D D D   5000 = 38D + 2700 -1000 - 40D D 5000 Profit = -2D + 1700 D 5000 d (Profit) = -2 + =0 dD D2 5000 or, D2 = = 2500 and D* = 50 units per month 2 d 2 (Profit) dD

10,000 D3

< 0 for D > 1

Therefore, D* = 50 is a point of maximum profit.

2-16

Profit = Total revenue - Total cost = (15X - 0.2X2) - (12 + 0.3X + 0.27X2) = 14.7X - 0.47X2 - 12 dProfit = 0 = 14.7 - 0.94X dX

X = 15.64 megawatts Note:

d 2 Profit = - 0.94 thus, X = 15.64 megawatts maximizes profit dX 2

2-17

(a)

Using Equation (2-10), 𝐷∗ =

(b)

180 − 40 = 14 units per week 2(5)

Total profit = −5(142) + (180 – 40)(14) = −980 + 140(14) = $980/wk

2-18

20,000 tons/yr. (2,000 pounds / ton) = 40,000,000 pounds per year of zinc are produced. The variable cost per pound is $20,000,000 / 40,000,000 pounds = $0.50 per pound. (a) Profit/yr

= (40,000,000 pounds / year) ($1.00 - $0.50) - $17,000,000 = $20,000,000 - $17,000,000 = $3,000,000 per year

The mine is expected to be profitable. (b) If only 17,000 tons (= 34,000,000 pounds) are produced, then Profit/yr = (34,000,000 pounds/year)($1.00 - $0.50) - $17,000,000 = 0 Because Profit =0, 17,000 tons per year is the breakeven point production level for this mine. A loss would occur for production levels < 17,000 tons/year and a profit for levels > 17,000 tons per year.

2-19

(a)

BE = $1,500,000 / ($39.95 − $20.00) = 75,188 customers per month

(b)

New BE point = $1,500,000 / ($49.95 − $25.00) = 60,120 per month

(c)

For 63,000 subscribers per month, profit equals 63,000 ($49.95 − $25.00) − $1,500,000 = $71,850 per month This improves on the monthly loss experienced in part (a).

2-20

(a)

D =

CF $2,000,000 = = 40,000 units per year p - cv ($90 - $40) / unit

$10,000,000 $8,000,000

$9,000,000

Profit Breakeven Point

$6,000,000 $4,000,000

$6,000,000

Loss

$2,000,000

Fixed Cost D' = 40,000 units

$0 0

20,000

40,000

60,000

80,000

100,000

Number of Units

(b)

Profit (Loss) = Total Revenue - Total Cost (90% Capacity) = 90,000 ($90) - [$2,000,000 + 90,000 ($40)] = $2,500,000 per year (100% Capacity) = [90,000($90) + 10,000($70)] - [$2,000,000 + 100,000($40)] = $2,800,000 per year

2-21

Annual savings are at least equal to ($60/lb)(600 lb) = $36,000. So the company can spend no more than $36,000 (conservative) and still be economical. Other factors include ease of maintenance / cleaning, passenger comfort and aesthetic appeal of the improvements. Yes, this proposal appears to have merit so it should be supported.

2-22

Let X represent the number of TVs sold (and for every TV sold, 2 speakers are sold). For each TV sold, a profit of $150 is realized from the TV and a profit of 2 X $50 = $100 is realized from the sale of speakers. To breakeven with the fixed costs of $120,000, X = $120,000 / ($150 + $100) = 480 televisions and 960 speakers should be sold.

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)

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

2-25

(a)

Ownership cost = $120 + $0.60X where X = horsepower rating. Operating cost =

$0.055

hp-hour X

× 9000 hp-hour

Total annual cost = 120 + 0.60X +

dAC dX

( X

)

= 120 + 0.6X +

= 0.6 - 495X -2 = 0

X = 28.72 horsepower (b)

For X to be a minimum of the function in part (a), show that d2 AC dX 2

d2 AC dX 2

= 990X-3 which is greater than 0 for all positive values of X.

> 0.

2-26 CT = Co + Cc = knv2 +

$1,500n v

dCT 1,500 = 0 = 2kv- 2 = kv3 - 750 dv v v=3

750 k

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

Let X represent the number of new cars sold (and for every new car sold, 3 used cars are sold). For each new car sold, a profit of $2,000 is realized from the new car and a profit of 3 X $1,000 = $3,000 is realized from the sale of the used cars. To breakeven with the annual costs of $300,000, X = $300,000 / ($2,000 + $3,000) = 60 new cars and 180 used cars should be sold.

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 B. Annual Heating Load (106 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



2C I

3

 0 for  > 0

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

Investment cost versus total repair cost

C CR··t $



2-30

Total profit from residential = Total profit from commercial. If fixed costs are $60,000 annually, then each service type needs to cover $30,000. Residential = $30,000 / ($200 per job) = 150 residential jobs Commercial = $30,000 / ($500 per job) = 60 commercial jobs

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

Let X represent the number of cakes sold (and for every cake sold, 4 dozen cookies are sold). For each cake sold, a profit of $8 is realized from the cake and a profit of 4 X $3 = $12 is realized from the sale of cookies. To breakeven with the monthly fixed costs of $2,000, X = $2,000 / ($8 + $12) = 100 cakes and 400 dozens of cookies should be sold.

2-33

Let X represent the number of bicycles sold (and for every bicycle sold, 4 helmets are sold). For each bicycle sold, a profit of $100 is realized from the bicycle and a profit of 4 X $25 = $100 is realized from the sale of helmets. To breakeven with the fixed costs of $50,000, X = $50,000 / ($100 + $100) = 250 bicycles and 1,000 helmets should be sold.

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)/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) = $2166.5/day Process 1 should be chosen to maximize profit per day.