Solution Manual for Engineering Economy, 17th Edition

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 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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). © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-8 The certainty of making $200,000 – $120,000 = $80,000 net income is not particularly good. If your friend keeps her present job, she is turning away from a risky $80,000 gain. This “opportunity cost” of $80,000 balanced in favor of a sure $60,000 would indicate your friend is risk averse and does not want to work hard as an independent consultant to make an extra $20,000 next year. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-9 (a) If you purchase a new car, you are turning away from a risky 20% per year return. If you are a risk taker, your opportunity cost is 20%, otherwise; it is 6% per year. (b) When you invest in the high tech company’s common stock, the next best return you’ve given up is 6% per year. This is your opportunity cost in situation (b). © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-11 (a) Students can use Equation (2-10) to determine D*. D* = 75−30 0.2 = 225 units/month Then 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: ?′ = 45 ± √(−45)2 − 4(0.1)(1,000) 0.2 ?′ = 45 ± 40.3 0.2 ?1′ = 24 (rounded up from 23.5) ?2′ = 426 (rounded down from 426.5) Thus, the profitable range of demand is from 24 to 426 units per month. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 D1 = (-568.5 + 351) / (-0.10) = 2,175 D2 = (-568.5 – 351) / (-0.10) = 9,195 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-15 (a) (b) 2700 5000   Profit = 38   2  D – 1000 – 40D D D   5000 = 38D + 2700 -1000 – 40D D 5000 + 1700 Profit = -2D D 5000 d (Profit) = -2 + =0 dD D2 5000 or, D2 = = 2500 and D* = 50 units per month 2 d 2 (Profit) dD 2 = 10,000 D3 1 Therefore, D* = 50 is a point of maximum profit. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-17 (a) Using Equation (2-10), ?∗ = (b) 180 − 40 = 14 units per week 2(5) Total profit = −5(142) + (180 – 40)(14) = −980 + 140(14) = $980/wk © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 per year. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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). © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-20 (a) D = CF $2,000,000 = = 40,000 units per year p – cv ($90 – $40) / unit $10,000,000 $9,000,000 $8,000,000 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 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-22 Jerry’s logic is correct if the AC system does not degrade in the next ten years (very unlikely). Because the leak will probably get worse, two or more refrigerant re-charges per year may soon become necessary. Jerry’s strategy could be to continue re-charging his AC system until two re-charges are required in a single year. Then he should consider repairing the evaporator (and possibly other faulty parts of his system). © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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? © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-24 (a) ?? = ?? ? Eqn. (2-8) $2,000,000 = ?? (2000 ????) ?? = $1,000/ton = $0.50/ lb. Profit = [?? – ?? ]X – ?? = [ $0.80/lb – $0.50/ lb] 4,000,000 lb – $700,000 = $500,000 (b) X’ = ?? / (?? – ?? ) Eqn. (2-13) = $700,000 / ($0.80/lb – $0.50/ lb) = 2,333,333 lbs ?? = ?? / X’ = $700,000 / 2,333,333 lbs = $0.30 / lb OR X’(?? – ?? ) = ?? X’ at breakeven point ?? = (?? – ?? ) = ($0.80/lb – $0.50/ lb) = $0.30/ lb (c) ?? = (X?? + ?? ) / X = [ 4,000,000 lb ($0.50/lb) + $700,000] / 4,000,000 lb = $0.675/ lb © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-25 (a) Ownership cost = $120 + $0.60X where X = horsepower rating. $0.055 1 Operating cost = hp-hour (X) × 9000 hp-hour Total annual cost = 120 + 0.60X + 0.055(9000) dAC dX X 495 = 120 + 0.6X + X = 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 > 0. = 990X-3 which is greater than 0 for all positive values of X. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-26 C T = C o + C c = knv 2 + $1,500n v dC T 1,500 = 0 = 2kv – 2 = kv 3 – 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 CR··t $ CI  © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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? © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 = ?̂ ?̂ = 4,400 units/wk = Breakeven amount (b) $22,000 + $15(3500) [ ] $20 (3500) $74,500 > $70,000 Purchase this item © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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? © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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). © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-41 The two alternatives are “no jig with a skilled machinist” and “use a jig with a lesser skilled machinist.” No jig: (1.5 min/housing)/(60 min/hr)($25/hr)(4000 housings) = $2,500 With jig: (2 min/housing)/(60 min/hr)($15/hr)(4000 housings) + $500 = $2,500 Alternatives are equally attractive. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-44 (a) False; (d) False; (g) False; (j) False; (m) True; (p) False; (b) False; (e) True; (h) True; (k) True; (n) True; (q) True; (c) True; (f) True; (i) True; (l) False; (o) True; (r) True; (s) False © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 0 500 1000 1500 2000 2500 Trip Mileage © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-48 F G H I J K 0 L M N O P Net Income 250 $20,000 $1 2 3 4 5 6 7 8 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 0 $50 0 Volume (Demand) $600,000 $500,000 $400,000 $300,000 Total Revenue $200,000 Total Expense Net income $100,000 $$(100,000) $(200,000) 10 00 15 00 20 00 25 00 30 00 35 00 40 00 45 00 50 00 55 00 Monthly Price per Total Total Expense Net income Demand Unit Revenue 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 22 3750 $ 105 $ 393,750 $ 384,250 $ 9,500 23 4000 $ 100 $ 400,000 $ 405,000 $ (5,000) 24 4250 $ 95 $ 403,750 $ 425,750 $ (22,000) 25 4500 $ 90 $ 405,000 $ 446,500 $ (41,500) 26 4750 $ 85 $ 403,750 $ 467,250 $ (63,500) 27 5000 $ 80 $ 400,000 $ 488,000 $ (88,000) 28 5250 $ 75 $ 393,750 $ 508,750 $ (115,000) 29 5500 $ 70 $ 385,000 $ 529,500 $ (144,500) 30 31 32 Summary of impact of changes in cost components on optimum 33 demand and profitable range of demand. 34 35 Percent Change CF cv D1 ‘ D2 ‘ 36 D* 37 -10% -10% 2,633 724 4541 0% -10% 2,633 824 4443 38 10% -10% 2,633 928 4339 39 40 -10% 0% 2,425 816 4036 41 0% 0% 2,425 932 3918 10% 0% 2,425 1060 3790 42 -10% 10% 2,218 940 3495 43 44 0% 10% 2,218 1092 3343 45 10% 10% 2,218 1268 3167 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 0 83 180 0.02 E 50 0 73,000 $ $ $ D Demand Start point (D) = Demand Increment = Net Income $ C Cash Flow 2 3 4 5 B Cash Flow 1 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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 $ 6 Annual Heating Load (10 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 $/10 Btu $25.20 6 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 © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 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. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-51 Suppose my local car dealer tells me that it costs no more than $0.03 per gallon of fuel to drive with my headlights on all the time. For the case study, this amounts to (500 gallons of fuel per year) x $0.03 per gallon = $15 per year. So the cost effectiveness of continuous use of headlights is roughly six times better than for the situation in the case study. © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-52 p = 400 – D2 TR = p  D = (400 – D2) D = 400D – D3 TC = $1125 + $100  D Total Profit / month = TR – TC = 400D – D3 – $1125 – $100D = – D3 + 300D – 1125 dTP = -3D2 + 300 = 0  D2 = 100  D* = 10 units dD d 2 TP = -6D; at D = D*, dD 2 d 2 TP = – 60 dD 2 Negative, therefore maximizes profit. Select (a) © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-53 – D3 + 300D – 1125 = 0 for breakeven At D = 15 units; -153 + 300(15) – 1125 = 0 Select (b) © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-54 CF = $100,000 + $20,000 = $120,000 per year CV = $15 + $10 = $25 per unit p = $40 per unit D = CF $120,000 = = 8,000 units/yr p – c v ($40 – $25) Select (c) © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-55 Profit = pD – (CF + CVD) At D = 10,000 units/yr, Profit/yr = (40)(10,000) – [120,000 + (25)(10,000)] = $30,000 Select (e) © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-56 Profit = pD – (CF + CVD) 60,000 = 35D – (120,000 + 25D) 180,000 = 10D; D = 18,000 units/yr Select (d) © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-57 Annual profit/loss = Revenue – (Fixed costs + Variable costs) = $300,000  [$200,000 + (0.60)($300,000)] = $300,000  $380,000 = $80,000 Select (d) © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-58 Savings in first year = (7,900,000 chips) (0.01 min/chip) (1 hr/60 min) ($8/hr + 5.50/hr) = $17,775 Select (d) © 2019 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.