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Solutions to end-of-chapter problems
Engineering Economy, 8 th edition Leland Blank and Anthony Tarquin
Chapter 1
Foundations of Engineering Economy
Basic Concepts
1.1 Financial units for economically best.
1.2 Morale, goodwill, dependability, acceptance, friendship, convenience, aesthetics, etc.
1.3 Measure of worth is a criterion used to select the economically best alternative. Some
measures are present worth, rate of return, payback period, benefit/cost ratio.
1.4 The color I like, best fuel rating, roomiest, safest, most stylish, fastest, etc.
1.5 Sustainability: Intangible; installation cost: tangible; transportation cost: tangible ;
simplicity: intangible; taxes: tangible; resale value: tangible; morale: intangible; rate of return: tangible; dependability: intangible; inflation: tangible; acceptance by others: intangible; ethics: intangible.
1.6 Examples are: house purchase; car purchase, credit card (which ones to use); personal loans (and their rate of interest and repayment schedule); investment decisions of all types; when to sell a house or car.
Ethics
1.7 This problem can be used as a discussion topic for a team-based exercise in class.
(a) Most obvious are the violations of Canons number 4 and 5. Unfaithfulness to the client and deceptive acts are clearly present. (b) The Code for Engineer’s is only partially useful to the owners in determining sound bases since the contractor is not an engineer. Much of the language of the Code is oriented toward representation, qualifications, etc., not specific acts of deceit and fraudulent behavior. Code sections may be somewhat difficult to interpret in construction of a house. (c) Probably a better source would be a Code for Contractor’s or consulting with a real estate attorney.
1.8 Many sections could be identified. Some are: I.b; II.2.a and b; III.9.a and b.
1.9 Example actions are:
- Try to talk them out of doing it now, explaining it is stealing
- Try to get them to pay for their drinks
- Pay for all the drinks himself
- Walk away and not associate with them again
1.10 This is structured to be a discussion question; many responses are acceptable. Responses can vary from the ethical (stating the truth and accepting the consequences) to unethical (continuing to deceive himself and the instructor and devise some on-the-spot excuse).
Lessons can be learned from the experience. A few of them are:
- Think before he cheats again.
- Think about the longer-term consequences of unethical decisions.
- Face ethical-dilemma situations honestly and make better decisions in real time.
Alternatively, Claude may learn nothing from the experience and continue his unethical practices.
Interest Rate and Rate of Return
1.11 Extra amount received = 2865 - 25.80*100 = $
Rate of return = 285/ = 0.110 (11%)
Total invested + fee 2865 + 50 = $ Amount required for 11% return = 2915*1. = $3235.
1.12 (a) Payment = 1,600,000(1.10)(1.10) = $1,936,
(b) Interest = total amount paid – principal = 1,936,000- 1,600, = $336,
1.13 i = [(5,184,000 – 4,800,000)/4,800,000]*100% = 8% per year
1.14 Interest rate = interest paid/principal
= (312,000/2,600,000) = 0.12 (12%)
1.15 i = (1125/12,500)*100 = 9%
i = (6160/56,000)100 = 11% i = (7600/95,000)100 = 8% The $56,000 investment has the highest rate of return
1.16 Interest on loan = 45,800(0.10) = $4,
Default insurance = $ Set-up fee = 45,800(0.01) = 458 Total amount paid = 4,580 + 900 + 458 = $
Net cash flow = $3,170 ($3,170,000)
1.25 End-of-period amount for March: 50 + 70 = $120; Interest = 120*0.03 = $3.
End-of-period amount for June: 120 + 120 + 20 = $260; Interest = 260*0.03 = $7. End-of-period amount for September: 260 + 150 + 90 = $500; Interest = $15. End-of-period amount for Dec: 500 + 40 + 110 = $650; Interest = $19.
Equivalence
1.28 (a) i = (5000-4275)/4275 = 0.17 (17%)
(c) Price one year later = 28,000 * 1.04 = $29,
(d) Price one year earlier = 28,000/1.04 = $26,
(e) Jackson: Interest rate = (2750/20,000)* = 13.75%
Henri: Interest rate = (2295/15,000)* = 15.30%
(f) 81,000 = 75,000 + 75,000(i) i = 6,000/75, = 0.08 (8%)
1.29 (a) Profit = 8,000,000*0.
= $2,240,
(b) Investment = 2,240,000/0. = $14,933,
1.30 P + P(0.10) = 1,600,
1.1P = 1,600,
P = $1,454,
1.31 Equivalent present amount = 1,000,000/(1 + 0.10)
= $909, Discount = 790,000 – 909, = $119,
1.32 Total bonus next year = (this year’s bonus + interest) + next year’s bonus
= [4,000 + 4,000(0.10)] + 4, = $8,
1.33 (a) Early-bird: 20,000 – 20,000(0.10) = $18,
(b) Equivalent future amount = 18,000(1 + 0.06) = $19,
Savings = 20,000 - $19, = $
Simple and Compound Interest
1.34 (a) F = P + Pni
1,000,000 = P + P(3)(0.20) 1.60P = 1,000, P = $625,
3log (1 + i) = 0. log(1 + i) = 0. 1 + i = 1. i = 9.1% per year
Spreadsheet function: = RATE(3,,-10000,13000) displays 9.14%
1.41 Follow plan 4, Example 1.16 as a model
A is 9900 – 2000 = $ B is 7900(0.10) = $ C is 7900 + 790 = $ D is 8690 – 2000 = $
1.42 (a) Simple: F = P + Pni
2,800,000 = 2,000,000 + 2,000,000(4)(i) i = 10% per year
(b) Compound: F = P(1 + i) (1 + i) (1 + i) (1 + i) 2,800,000 = 2,000,000(1 + i) 4
(1 + i) 4 = 1. log(1 + i) 4 = log1. 4log(1 + i) = 0. log(1 + i) = 0. (1 + i) = 10
(1 + i) = 1. i = 8.77%
(c) Spreadsheet function: = RATE(4,,-2000000,2800000)
MARR and Opportunity Cost
1.43 Bonds - debt; stock sales – equity; retained earnings – equity; venture capital – debt; short
term loan – debt; capital advance from friend – debt; cash on hand – equity; credit card – debt; home equity loan - debt.
1.44 WACC = 0.40(10%) + 0.60(16%) = 13.60%
1.45 WACC = 0.05(10%) + 0.95(19%) = 18.55%
The company should undertake the inventory, technology, warehouse, and maintenance projects.
1.46 Let x = percentage of debt financing; Then, 1- x = percentage of equity financing
0.13 = x(0.28) + (1-x)(0.06) 0.22x = 0. x = 31.8%
Recommendation: debt-equity mix should be 31.8% debt and 68.2% equity financing
1.47 Money: The opportunity cost is the loss of the use of the $5000 plus the $100 interest.
Percentage: The 30% estimated return on the IT stock is the opportunity cost in percentage.
Exercises for Spreadsheets
1.48 (a) PV is P; (b) PMT is A; (c) NPER is n; (d) IRR is i; (e) FV is F; (f) RATE is i
1.49 (a) PV(i%,n,A,F) finds the present value P (b) FV(i%,n,A,P) finds the future value F (c) RATE(n,A,P,F) finds the compound interest rate i (d) IRR(first_cell:last_cell) finds the compound interest rate i (e) PMT(i%,n,P,F) finds the equal periodic payment A (f) NPER(i%,A,P,F) finds the number of periods n
1.50 (a) (1) F = ?; i = 8%; n = 10; A = $3000; P = $ (2) A = ?; i = 12%; n = 20; P = $-16,000; F = 0 (3) P = ?; i = 9%; n = 15; A = $1000; F = $ (4) n = ?; i = 10%; A = $-290; P = 0; F = $12, (5) F = ?; i = 5%; n = 5; A = $500; P = $-
(b) (1) negative (2) positive (3) negative (4) positive (years) (5) can’t determine if 5% per year will cover the 5 withdrawals of $
1.51 Spreadsheet shows relations only in cell reference format. Cell E10 will indicate $64 more
than cell C10.
i = 22% Answer is (c)
1.60 WACC = 0.70(16%) + 0.30(12%)
Answer is (c)
1.61 Amount available = total principal in year 0 + interest for 2 years + principal added year 1
- interest for 1 year = 850,000(1+0.15) 2
- 200,000 (1+0.15) = 1,124,125 + 230, = $1,354, Answer is (a)
Solution to Case Study, Chapter 1
There is no definitive answer to case study exercises. The following is only an example.
Renewable Energy Sources for Electricity Generation
- LCOE approximation uses 1/(1.05) 11 = 0.5847 and LCOE last year = 0.1022. Let X 11 = I 11 + M 11 + F 11
With the limited data, to estimate the value of X 11 set the LCOE for year 11 equal to the consumer cost for year 10.
(0.5847)X 11
(0.5847)(5.052 billion)
0.5847X 11 = (0.0005)(2.9539 billion)
X 11 = $2.526 million
If the sum of investments (I 11 ), M&O (M 11 ) and fuel (F 11 ) is significantly different than $2. million, the breakeven value for year 11 may change. Next step is to find the values of I, M and F for year 11.
Solutions to end-of-chapter problems
Engineering Economy, 8 th edition Leland Blank and Anthony Tarquin
Chapter 2
Factors: How Time and Interest Affect Money
Determination of F, P and A
2.1 (1) (F/P, 10%, 7) = 1.
(2) (A/P, 12%,10) = 0.
(3) (P/G,15%,20) = 33.
(4) (F/A,2%,50) = 84.
(5) (A/G,35%,15) = 2.
2.2 F = 1,200,000(F/P,7%,4)
2.3 F = 200,000(F/P,10%,3)
2.4 P = 7(120,000)(P/F,10%,2)
2.5 F = 100,000,000/30(F/A,10%,30)
2.6 P = 25,000(P/F,10%,8)
2.17 F = 125,000(F/A,10%,4)
2.18 F = 600,000(0.04)(F/A,10%,3)
2.19 P = 90,000(P/A,20%,3)
2.20 A = 250,000(A/F,9%,5)
2.21 A = 1,150,000(A/P,5%,20)
2.22 P = (110,000* 0.3)(P/A,12%,4)
2.23 A = 3,000,000(10)(A/P,8%,10)
2.24 A = 50,000(A/F,20%,3)
Factor Values
2.25 (a) 1. Interpolate between i = 8% and i = 9% at n = 15:
0.4/1 = x/(0.3152 – 0.2745) x = 0. (P/F,8.4%,15) = 0.3152 – 0. = 0.
- Interpolate between i = 16% and i = 18% at n = 10: 1/2 = x/(0.04690 - 0.04251) x = 0. (A/F,17%,10) = 0.04690 – 0. = 0.
(b) 1. (P/F,8.4%,15) = 1/(1 + 0.084) 15
= 0.
- (A/F,17%,10) = 0.17/[(1 + 0.17) 10 - 1] = 0.
(c) 1. = -PV(8.4%,15,,1) displays 0.
- = -PMT(17%,10,,1) displays 0.
2.26 (a) 1. Interpolate between i = 18% and i = 20% at n = 20:
1/2 = x/40. x = 20. (F/A,19%,20) = 146.6280 + 20. =166.
- Interpolate between i = 25% and i = 30% at n = 15: 1/5 = x/0. x = 0. (P/A,26%,15) = 3.8593 – 0. = 3.
(b) 1. (F/A,19%,20) = [(1+ 0.19) 20
- (P/A,26%,15) = [(1 + 0.26) 15 –1]/[0.26(1 + 0.26) 15 ] = 3.
2.30 P 0 = 500(P/A,10%,9) + 100(P/G,10%,9)
2.31 (a) Revenue = 390,000 + 2(15,000)
= $420,
(b) A = 390,000 + 15,000(A/G,10%,5) = 390,000 + 15,000(1.8101) = $417,151.
2.32 A = 9000 – 560(A/G,10%,5)
2.33 500 = 200 + G(A/G,10%.7)
500 = 200 + G(2.6216)
G = $114.
2.34 A = 100,000 + 10,000(A/G,10%,5)
F = 118,101(F/A,10%,5)
2.35 3500 = A + 40(A/G,10%,9)
3500 = A + 40(3.3724)
A = $3365.
2.36 In $ billion units,
P = 2.1(P/F,18%,5) = 2.1(0.4371) = 0.91791 = $917,910,
917,910,000 = 100,000,000(P/A,18%,5) + G(P/G,18%,5)
917,910,000 = 100,000,000(3.1272) + G(5.2312)
G = $115,688,
2.37 95,000 = 55,000 + G(A/G,10%,5)
95,000 = 55,000 + G(1.8101)
G = $22,
2.38 P in year 0 = 500,000(P/F,10%,10)
= 500,000(0.3855) = $192,
192,750 = A + 3000(P/G,10%,10)
192,750 = A + 3000(22.8913)
A = $124,
Geometric Gradient
2.39 Find (P/A,g,i,n) using Equation [2.32] and A 1 = 1
For n = 1: P (^) g = 1*{1 – [(1 + 0.05)/(1 + 0.10)] 1 }/(0.10 – 0.05) = 0.
For n = 2: P (^) g = 1*{1 – [(1 + 0.05)/(1 + 0.10)] 2 }/(0.10 – 0.05) = 1.
2.40 Decrease deposit in year 4 by 7% per year for three years to get back to year 1.
First deposit = 5550/(1 + 0.07) 3
= $4530.
2.41 P (^) g = 35,000{1 – [(1 + 0.05)/(1 + 0.10)] 6 }/(0.10 – 0.05) = $170,
2.42 P (^) g = 200,000{1 – [(1 + 0.03)/(1 + 0.10)] 5 }/(0.10 – 0.03) = $800,
Interest Rate and Rate of Return
2.47 1,000,000 = 290,000(P/A,i,5)
(P/A,i,5) = 3. Interpolate between 12% and 14% interest tables or use Excel’s RATE function By RATE, i = 13.8%
2.48 50,000 = 10,000(F/P,i,17)
5.0000 = (F/P,i,17) 5.0000 = (1 + i) 17
i = 9.93%
2.49 F = A(F/A,i%,5)
451,000 = 40,000(F/A,i%,5) (F/A,i%,5) = 11. Interpolate between 40% and 50% interest tables or use Excel’s RATE function By RATE, i = 41.6%
2.50 Bonus/year = 6(3000)/0.05 = $360,
1,200,000 = 360,000(P/A,i,10) (P/A,i,10) = 3. i = 27.3%
2.51 Set future values equal to each other
Simple: F = P + Pni = P(1 + 5*0.15) = 1.75P
Compound: F = P(1 + i) n
= P(1 + i) 5
1.75P = P(1 + i) 5
i = 11.84%
2.52 100,000 = 190,325(P/F,i,30)
(P/F,i,30) = 0. Find i by interpolation between 2% and 3%, or by solving P/F equation, or by Excel By RATE function, i = 2.17%
2.53 400,000 = 320,000 + 50,000(A/G,i,5)
(A/G,i,5) = 1. Interpolate between i = 22% and i = 24% i = 22.6%
Number of Years
2.54 160,000 = 30,000(P/A,15%,n)
(P/A,15%,n) = 5. From 15% table, n is between 11 and 12 years; therefore, n = 12 years By NPER, n = 11.5 years
2.55 (a) 2,000,000 = 100,000(P/A,5%,n)
(P/A,5%,n) = 20.
From 5% table, n is > 100 years. In fact, at 5% per year, her account earns $100,000 per year. Therefore, she will be able to withdraw $100,000 forever; actually, n is ∞.
(b) 2,000,000 = 150,000(P/A,5%,n) (P/A,5%,n) = 13. By NPER, n = 22.5 years
(c) The reduction is impressive from forever (n is infinity) to n = 22.5 years for a 50% increase in annual withdrawal. It is important to know how much can be withdrawn annually when a fixed amount and a specific rate of return are involved.
2.56 10A = A(F/A,10%,n)
(F/A,10%,n) = 10.
From 10% factor table, n is between 7 and 8 years; therefore, n = 8 years
2.57 (a) 500,000 = 85,000(P/A,10%,n)
(P/A,10%,n) = 5.
From 10% table, n is between 9 and 10 years.
(b) Using the function = NPER(10%,-85000,500000), the displayed n = 9.3 years.
