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Engineering Economy, 8th edition
Leland Blank and Anthony Tarquin
Chapter 13
Breakeven and Payback Analysis
Breakeven Analysis for a Project
13.1 (a) 89x = 5,000,000 + 45x
(b) At 100,000 units: Profit = revenue - cost
At 200,000 units: Profit = 89(200,000) - [5,000,000 + 45(200,000)]
13.2 For breakeven, 0 = revenue – cost
13.3 Let x = days per year to breakeven
13.4 (a) Let x = miles per year to breakeven
(b) No. days = 109,469/600
13.5 Let x = selling price
13.6 QBE = 500,000/(250-200)
13.7 0 = -150,000,000(A/P,10%,10) + 12,500(250)X0.5
13.8 (a) Let x = selling price per unit
13.9 Savings = 0.25(20) = $5.00
13.10 Let x = cost/ton
13.11 (a) Let x = hours per month billed to breakeven
(b) Billable hours per professional = 1246/10 = 124.6 hours
13.12 (a) First express all variable costs In terms of cost per unit and then find breakeven
production rate.
(b) Profit = 40,000(70) - [(240,000 + 40,000(42.50) + 40,000(2.40)]
(c) 1,000,000 = 70x - [(240,000 + x(42.50) + x(2.40)]
13.13 (a) By hand: Set PW of revenue equal to PW of expenses
4000(43,000 - 25,000)(P/A,12%,3) + G(43,000 - 25,000)(P/G,12%,3) = 250,000,000
(b) By spreadsheet:
Breakeven Analysis between Alternatives
13.14 x = number of units per year
13.15 (a) -2,300,000(A/P,8%,20) - 486 = -x(A/P,8%,10) – 774
(b) Develop the AW relations using the PMT function while allowing the n value for
13.16 (a) Calculate BTU/dollar for gasoline and set equal to cost for ethanol
(b) Energy in E85 = 0.85(75,670) + 0.15(115,600)
13.17 90,000(A/P,8%,5) + 450x = 800x
13.18 Set annual costs equal to each other and solve for FC, the fixed cost
13.19 By hand
(a) Set annual costs equal to each other and solve for A, annual maintenance cost
(b) Annual gravel road maintenance = 355,000(1.30) = $461,500; find n for the gravel
road
n is between 4 and 5 years. By interpolation, n = 4.3 years
By spreadsheet
13.20 (a) Let x = days per year to just breakeven
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The need must be 66 or more days per year to justify purchase
(b) Using Goal Seek is easier. Set up the annual cost using PMT with the breakeven days
13.21 Let x = square yards per year to breakeven
-109,000 – 2.75x = -225,000(A/P,8%,15) – 13x
13.22 Let x = gallons per day to breakeven
-465 - (485/720,000)x = -328 - (1280/950,000)x
13.23 (a) Fixed cost, C: FCC = -500,000(A/P,10%,5) + 0.25(500,000)(A/F,10%,5)
= -500,000(0.2638) + 0.25(500,000)(0.1638)
(b) Plot annual cost curves. Breakeven is only between B and C at slightly less than 4000
parts/year.
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(c) Make the fixed cost for method A the variable (cell A14). Use Goal Seek to force the
13.24 Let x = ads per year
13.25 (a) Set AWA = AWB, with PB = first cost of B. The final term in AWB removes the
(b) Set PB = -700,000 and solve for PA
13.26 Let x = days per year
13.28 Let T = number of tons/year. Solve relation AW1 = AW2 for T
1500 tons, select machine 1.
13.29 (a) Solve relation Revenue - Cost = 0 for Q = number of filters per year
50Q - [200,000(A/P,6%,5) + 25,000 + 20Q] = 0
(b) Solve the relation AWbuy = AWmake for Q = number of filters per year.
(c) Make: 5000 at $20 each.
(d) The spreadsheet below verifies the answers above
Payback Analysis
13.30 Payback analysis should be used only as a supplemental analysis tool because it
13.31 0 = -70,000 + (14,000 – 1850)(P/A,10%,np)
13.32 (a) Develop the PW relation and solve for np
13.33 0 = -39,000 + (13,500 – 6000)(P/A,10%,np)
13.34 (a) Develop the W relation and solve for np
13.35 42,000 + 1000np = 25,000(F/P,10%,np)
(a) By trial and error:
(b) By spreadsheet: Use Goal Seek to force the difference to be zero by changing cell B3.
13.36 (a) i = 0%: np = 200,000/(90,000 – 50,000)
(b) Develop the relation Revenue = Cost and solve for XBE = gallons per year
13.37 (a) i = 0%: Semi-automatic:
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Select only the semi-automatic machine
(b) i = 10%: Semi-automatic:
13.38 Let np = number of years until payback
13.39 Let np = number of months. Sample relation for equivalent worth per month for
13.40 –250,000 – 500np + 250,000(1 + 0.0075)np = 100,000
13.41 (a) Payback:
B: 0 = -300,000 + 10,000(P/A,8%,np) + 15,000(P/G,8%,np)
(b) Present worth:
A: PW = -300,000 + 60,000(P/A,8%,10)
(c) A sample spreadsheet solution follows. Answers are:
Spreadsheet Exercises
13.42 (a) Plot shows maximum quantity at approximately 1000 units. Profit estimate is $16,790
per month.
(b) Profit = R – TC = (-.008-.005) Q2 + (32-2.2)Q - 10
13.43 Let R = revenue for years 2 through 8. Set up PW = 0 relation.
Spreadsheet solution uses Goal Seek to find R = $73,195 with the remaining
revenue cells set equal to this value.
13.44 (a) Current: QBE = 400,000/(14-10) = 100,000 units
13.45 Current: Profit = 14Q -400,000 - 10Q = 4Q -400,000
13.46 Solve the relation AWI = AWO for N = number of tests per year
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83.25N = 209,340
N = 2514 tests per year
13.47 Spreadsheet used to calculate AW values for each N value. Functions are written in cell
13.48 By hand: It will raise the breakeven point. Outsourcing will cost $75, increasing to
By spreadsheet: Simply change the entries in the cost cells for outsourced. New
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13.49 It will decrease the breakeven point.
Additional Problems and FE Exam Review Questions
13.50 Production < breakeven point; select alternative with higher slope
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-400,000(A/P,10%,15) + (0.10)(400,000) (A/F,10%,15) – 300x = 720x
-400,000(0.13147) + (0.10)(400,000) (0.03147) – 300x = -720x
420x = 51,329
x = 122 cars per year
Answer is (b)
13.52 Set equations equal to each other and solve for QBE
13.53 Both the fixed cost and variable cost of Method X are higher than those of Y. Therefore,
13.54 Amount of gasoline to drive there and back = 54/18
13.55 Answer is (b)
13.58 Let VCIoT = variable cost of IoT-based process
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13.59 -320,000 + (98,000-40,000)(P/A,20%,np) = 0
(P/A,20%,np) = 5.5172
There is no P/A factor this large for any n value, that is, n → ∞. From observation, the
net income of $58,000 per year < $64,000 interest per year; the investment will never pay
off.
Answer is (d)
13.60 Breakeven = 500,000/(250 – 200)
13.61 Let x = number of ft2
13.62 VC = 40(4)/8
Solution to Case Study, Chapter 13
Sometimes, there is not a definitive answer to a case study exercise. Here are example responses.
WATER TREATMENT PLANT PROCESS COSTS
2. A decrease in the efficiency of the aerator motor renders the selected alternative of “sludge
3. If the cost of lime increased by 50%, the lime costs for “sludge recirculation only” and
4. If the efficiency of the sludge recirculation pump decreased from 90% to 70%, the net savings
5. If hardness removal were discontinued, the extra cost for its removal (column 4 in Table 13-
6. If the cost of electricity decreased to 8¢/kwh, the aeration only and sludge recirculation only
7. (a) For alternatives 1 and 2 to breakeven, the total savings would have to be equal to
