Solution Manual For Basics Of Engineering Economy, 2nd Edition

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1
Solutions to end-of-chapter problems
Basics of Engineering Economy, 2nd edition
Leland Blank and Anthony Tarquin
Chapter 1
Foundations of Engineering Economy
1.1 If the alternative that is actually the best one is not even recognized as an
alternative, it obviously will not be able to be selected by using any economic
analysis tools.
1.2 Non-economic factors are (b), (c), (e) and (f): morale, goodwill, public
acceptance and aesthetics, respectively.
1.3 Revenues and costs are examples of cash flows.
1.4 The analysis techniques that are used in engineering economic analysis are only
as good as the accuracy of the cash flow estimates.
1.5 The do-nothing alternative represents the as-is or status quo condition.
1.6 Evaluation criterion refers to the measurement standard or procedure that is used
to determine which alternative is “best”.
1.7 In engineering economy, the evaluation criterion is financial units (dollars, pesos,
etc).
1.8 Intangible factors are non-economic considerations that may have to be taken into
account in identifying the best alternative.
1.9 Examples of intangible factors are goodwill, morale, convenience, friendship,
implementation know-how, aesthetics, public acceptance
1.10 Time value of money means that there is a certain worth in having money and that
worth changes as a function of time.
1.11 Interest is a manifestation of the time value of money.
1.12 The most important fundamental dimension in economic analysis is time because
of time value of money.
1.13 The term that describes compensation for “renting” of money is time value of
money, which manifests itself as interest.
1.14 When an interest rate does not include the time period, the time period is assumed
to be one year.

2
1.15 The original amount of money in a loan transaction is known as the principal.
1.16 Minimum attractive rate of return is the lowest rate of return (interest rate) on a
project that companies or individuals consider to be high enough to induce them
to invest their money.
1.17 When the yield on a U.S. Government Bond is 3% per year, investors are
expecting the inflation rate to be near zero.
1.18 (a) Payment = 1,600,000(1.10)(1.10) = $1,936,000
(b) Interest = total amt paid – principal
= 1,936,000- 1,600,000
= $336,000
1.19 (a) Equivalent cost in 1 year = 38,000 + 38,000(0.10)
= $41,800
(b) Since $41,600 is less than $41,800, the firm should remodel 1ater (i.e. 1 year
from now).
1.20 P = 50,000/1.08
= $46,296.30
1.21 Rate of return = (1,450,000/10,000,000)(100%)
= 14.5% per year
1.22 Rate of return = (45/966)(100%)
= 4.7%
1.23 Rate of increase = [(29 – 22)/22](100%)
= 31.8%
1.24 Interest rate = (275,000/2,000,000)(100%)
= 13.75%
1.25 Rate of return = (2.3/6)(100%)
= 38.3%
1.26 F = P(1 + i)
2,360,000 = 2,000,000(1 + i)
i = 0.18 (18%)
1.27 F = P(1 + i)
53,000,000 = P(1 + 0.10)
P = $48,181,818

3
1.28 Amount of earnings in year one = 400,000,000(0.25)
= $100,000,000
1.29 Amt at end of year 1 = 280,000(1.15) = $322,000 < $425,000
Amt at end of year 2 = 322,000(1.15) = $370,300 < $425,000
Amt at end of year 3 = 370,300(1.15) = $425,845 > $425,000
Time = 3 years
1.30 MARR is set relative to the cost of capital.
1.31 The engineer is wrong, unless the MARR is exactly equal to the cost of capital.
Usually, the inequality ROR ≥ MARR > cost of capital is used, and the MARR is
established higher than the cost of capital so that profit, risk and other factors are
considered.
1.32 (a) Amt owed; compound interest = 150,000(1.05)(1.05)(1.05)
= $173,644
Amt owed; simple interest = P + Pni
= 150,000 + 150,000(3)(0.055)
= 150,000 + 24,750
= $174,750
Company should accept 5% compound interest rate
(b) Difference = 174,750 – 173,644
= $1106
1.33 F = P + Pni
850,000 = P + P(4)(0.10)
1.4P = 850,000
P = $607,143
1.34 F = P + Pni
100,000 = 1000 + 1000(n)(0.1)
99,000 = 1000(n)(0.10)
n = 990 years
1.35 Follow plan 4, Figure 1.3 as a model.
A is 9900 – 2000 = $7900
B is 7900(0.10) = $790
C is 7900 + 790 = $8690
D is 8690 – 2000 = $6690

4
1.36 F = P + Pni
35,000 = P + P(3)(0.07)
1.21P = 35,000
P = $28,925.60
1.37 P(1.20)(1.20) = 20,000
P = $13,888.89
1.38 (a) Interest rate = 900,000/6,000,000 = 0.15 = 15% per year
(b) Total due after 1 year = 6,000,000 + 900,000 = $6,900,000
Interest charged for 2nd year = 6,900,000(0.15) = $1,035,000
1.39 (a) F = P + Pni
2,800,000 = 2,000,000 + 2,000,000(4)(i)
Interest = 10% per year
(b) F = P(1 + i) (1 + i) (1 + i) (1 + i)
2,800,000 = 2,000,000(1 + i)4
(1 + i)4 = 1.4000
log(1 + i)4 = log1.400
4log(1 + i) = 0.146
log(1 + i) = 0.0365
(1 + i) = 100.0365
(1 + i) = 1.0877
i = 8.77%
1.40 F = P + P(n)(i)
3P = P + P(n)(0.20)
n = 10 years
1.41 (a) F = P + P(n)(i)
2P = P + P(4)(i)
i = 25%
(b) F = P(1 + i) (1 + i) (1 + i) (1 + i)
2P = P(1 + i)4
2 = (1 + i)4
log 2 = log(1 + i)4
0.301 = 4 log(1 + i)
log(1 + i) = 0.0753
(1 + i) = 100.0753
i = 18.93%

5
1.42 (a) Loan A: interest/year = 500,000(0.10) = $50,000
Loan B: interest/year = 500,000(0.10) = $50,000
The same amount of interest was be paid on each loan
(b) There was no difference paid between the two loans
1.43 All engineering economy problems will involve i and n
1.44 P = $13,000; F = $26,000; i = 6.8% per year; n = ?
1.45 P = $1000; F = $1270; n = 3; i = ?
1.46 P = $50,000; F = ?; i = 15%; n = 3
1.47 F= $200,000,000; n = 5; i = 12% per year; A = ?
1.48 A = $225,000; n = 3; i = 15% per year; F = ?
1.49 F = $400,000; n = 2; i = 20% per year; P = ?
1.50 P = $225,000; n = 4; i = 15% per year; A = ?
1.51 P = $1,800,000; n = 6; i = 12% per year; A = ?
1.52 P = $16,000,000; A = $3,800,000; i = 18% per year; n = ?
1.53 End-of-period-convention means end-of-interest period. It does not mean end of
the calendar year.
1.54 Period refers to interest period.
1.55 The difference between cash inflows and cash outflows is known as net cash
flow.
1.56 Office supplies = outflow; GPS surveying equipment = outflow;
Auctioning of used earth-moving equipment = inflow;
Staff salaries = outflow; Fees for services rendered = inflow;
Interest from bank deposits = inflow
1.57 The following items are inflows: salvage value, sales revenues, cost reductions
by subcontractors.
The following items are outflows: income taxes, loan interest, rebates to dealers,
accounting services.

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6
1.58 Assuming down is negative: down arrow of $40,000 in year 5; up arrow in year 0
identified as P =?; i = 15% per year.
1.59 Assuming down is negative: down arrow of $10,000 in year 0; up arrows in the
amount of $3000 in years 1 through 5; i = 10% per year; arrow in year 5
identified as F = ?.
1.60 Assuming down is negative: down arrow in year 0 identified as P=?; i = 15%
per year; up arrows in years 1 thru 5 in the amount of $75,000 - $30,000 =
$45,000
1.61 (a) FV is F (b) PMT is A (c) NPER is n (d) IRR is i (e) PV is P
1.62 (a) F = ?; i = 8%; n = 10; A = $2000; P = $-10,000
(b) A = ?; i = 12%; n = 30; P = $16,000; F = 0
(c) P = ?; i = 9%; n = 15; A = $1000; F = $700
(d) n = ?; i = 8.5%; A = $5000; P = $-50000; F = $20000 (calculator function)
1.63 (a) FV(i%,n,A,P) finds the future value, F
(b) IRR(first_cell:last_cell) finds the compound interest rate, i
(c) PMT(i%,n,P,F) finds the equal periodic payment, A
(d) PV(i%,n,A,F) finds the present value, P
1.64 For built-in spreadsheet functions, a parameter that does not apply can be left
blank when it is not an interior one. For example, if no F is involved when using
the PMT function, it can be left blank because it is an end parameter. When the
parameter involved is an interior one (like P in the PMT function), a comma must
be put in its position.
1.65 A common moral is a fundamental belief held by virtually all people. A personal
moral is the translation of a common moral into that which an individual believes
and uses as guidance for their own decisions and actions.
1.66 A code of ethics can be used as an evaluation measure for the decision and actions
of an individual who works in the discipline that honors the code.
1.67 (a) Assuming that Carol’s supervisor is a trustworthy and ethical person himself,
going to her supervisor and informing him of her suspicion is probably the
best of these options. This puts Carol on record (verbally) as questioning
something she heard at an informal gathering.
(b) Another good option is to go to Joe one-on-one and inform him of her
concern about what she heard him say at lunch. Joe may not be aware he is on
the bid evaluation team and the potential ethical consequences if he accepts
the free tickets from Dryer.

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7
1.68 Explain your own situation and what you think an ethical action might be.
1.69 Amount now = 10,000 + 10,000(0.10)
= $11,000
Answer is (c)
1.70 Answer is (a)
1.71 Move both cash flows to year 0 and solve for i
1000(1 + i) = 1345.60/(1 + i)
(1 + i)2 = 1345.60/1000
(1 + i) = 1.16
i = 16%
Answer is (d)
1.72 F in year 2 at 20% compound interest = P(1.20)(1.20) = 1.44P
For simple interest, F = P + Pni = P(1 + ni)
P(1 + 2i) = 1.44P
(1 + 2i) = 1.44P
i = 22%
Answer is (c)
1.73 F = P(1+i)n
16,000 = 8000(1 + i)9
21/9 = 1 + i
1.08 = 1 + i
i = 0.08 (8%)
Answer is (b)
1.74 Answer is (c)
1.75 Let i = compound rate of increase:
268 = 200(1 + i)5
(1 + i)5 = 268/200
(1 + i) = (1.34)0.2
(1 + i) = 1.0602
i = 6.02%
Answer is (b)

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1
Solutions to end-of-chapter problems
Basics of Engineering Economy, 2nd edition
Leland Blank and Anthony Tarquin
Chapter 2
Factors: How Time and Interest Affect Money
2.1 (a) (F/P,10%,20) = 6.7275
(b) (A/F,4%,8) = 0.10853
(c) (P/A,8%,20) = 9.8181
(d) (A/P,20%,28) = 0.20122
(e) (F/A,30%,15) = 167.2863
2.2 P = 30,000(P/F,10%,8)
= 30,000(0.4665)
= $13,995
2.3 F = 15,000(F/P,6%,25)
= 15,000(4.2919)
= 64,378.50
2.4 (a) F = 885,000 + 100,000(F/P,10%,3)
= 885,000 + 100,000(1.3310)
= $1,018,000
(b) Spreadsheet function is = -FV(10%,3,,100000) + 885000
Display is $1,018,000
2.5 (a) P = 19,000(P/F,10%,7)
= 19,000(0.5132)
= $9750.80
(b) If the calculator function is PV(10,7,0,19000), display is P = $-9750.00
(c) If the spreadsheet function is = -PV(10%,7,,19000), display is $9750.00
2.6 (a) Total for 7 lots is 7(120,000) = $840,000
P = 840,000(P/F,10%,2)
= 840,000(0.8264)
= $694,176
(b) If the calculator function is PV(10,2,0,840000), display is P = $-694,214.88
(c) If the spreadsheet function is = -PV(10%,2,,840000), display is $694,214.88

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2
2.7 (a) F = 3000(F/P,10%,12) + 5000(F/P,10%,8)
= 3000(3.1384) + 5000(2.1436)
= $20,133.20
(b) Sum two calculator functions
FV(10,12,,-3000) + FV(10,8,-5000)
9,415.29 + 10,717.94 = $20,133.23
(c) If the spreadsheet function is = – FV(10%,12,,3000) – FV(10%,8,,5000), the
display is $20,133.23
2.8 (a) P = 30,000,000(P/F,10%,5) – 15,000,000
= 30,000,000(0.6209) – 15,000,000
= $3,627,000
(b) If the spreadsheet function is = -PV(10%,5,,30000000) – 15,000000, the
display is $3,627,640
The increased decimal accuracy of a spreadsheet function indicates an increased
the required amount of $640.
2.9 F = 280,000(F/P,12%,2)
= 280,000(1.2544)
= $351,232
2.10 A = 12,700,000(A/P,20%,8)
= 12,700,000(0.26061)
= $3,309,747
2.11 P = 6000(P/A,10%,10)
= 6000(6.1446)
= $36,867.60
2.12 (a) A = 60,000(A/P,8%,5)
= 60,000(0.25406)
= $15,027.60
(b) If calculator function is PMT(8,5,-60000,0), the answer is $15,027.39
(c) A spreadsheet function of = -PMT(8%,5,60000) displays $15,027.39
2.13 A = 20,000,000(A/P,10%,6)
= 20,000,000(0.22961)
= $4,592,200

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3
2.14 A = 50,000(A/F,20%,3)
= 50,000(0.27473)
= $13,736.50
2.15 (a) 17,000,000(A/P,i,8) = 2,737,680
(A/P,i,8) = 0.16104
From interest tables at n = 8, i = 6% per year
(b) Calculator function is i(8,-2737680,17000000,0) to obtain i = 6.00%
(c) If the spreadsheet function is = RATE(8,-2737680,17000000), display is
6.00%
2.16 (a) A = 3,000,000(10)(A/P,8%,10)
= 30,000,000(0.14903)
= $4,470,900
(b) If calculator function is PMT(8,10,-30000000,0), the answer is $4,470,884.66
(c) If the spreadsheet function is = -PMT(8%,10,30000000), display is
A = $4,470,884.66
2.17 P = 1,400,000(F/P,7%,4)
=1,400,000(1.3108)
= $1,835,120
2.18 P = 600,000(P/F,12%,4)
= 600,000(0.6355)
= $381,300
2.19 (a) A = 225,000(A/P,15%,4)
= 225,000(0.35027)
= $78,811
(b) Recall amount = 78,811/0.10
= $788,110 per year
2.20 P = 100,000((P/F,12%,2)
= 100,000(0.7972)
= $79,720
2.21 F = 65,000(F/P,4%,5)
= 65,000(1.2167)
= $79,086

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4
2.22 P = (280,000-90,000)(P/A,10%,5)
= 190,000(3.7908)
= $720,252
2.23 F = 649(F/P,8%,2)
= 649(1.1664)
= $757
2.24 The value of the system is the interest saved on $20 million for 2 years.
F = 20,000,000(F/P,8%,2)
= 20,000,000(1.1664)
= $23,328,000
Interest = 23,328,000 - 20,000,000
= $3,328,000
2.25 P = 2,100,000(P/F,10%,2)
= 2,100,000(0.8264)
= $1,735,440
2.26 P = 40,000(P/F,12%,4)
= 40,000(0.6355)
= $25,420
2.27 (a) A = 850,000(A/F,18%,5)
= 850,000(0.13978)
= $118,813
(b) Spreadsheet function = PMT(18%,5,,850000) results in a minus sign.
2.28 P = 95,000,000(P/F,12%,3)
= 95,000,000(0.7118)
= $67,621,000
2.29 F = 375,000(F/P,10%,6)
= 375,000(1.7716)
= $664,350
2.30 F = 150,000(F/P,8%,8)
= 150,000(1.8509)
= $277,635
2.31 (a) P = 7000(P/F,10%,2) + 9000(P/F,10%,4) + 15,000(P/F,10%,5)
= 7000(0.8264) + 9000(0.6830) + 15,000(0.6209)
= $21,245.30

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5
(b) Three calculator functions are added.
-PV(10,2,0,7000) – PV(10,4,0,9000) – PV(10,5,0,15000)
Total is 5785.12 + 6147.12 + 9313,82 = $21,246.06
2.32 P = 600,000(0.10)(P/F,10%,2) + 1,350,000(0.10)(P/F,10%,5)
= 60,000(0.8264) + 135,000(0.6209)
= $133,406
2.33 P = 8,000,000(P/A,10%,5)
= 8,000,000(3.7908)
= $30,326,400
2.34 A = 10,000,000(A/P,10%,10)
= 10,000,000(0.16275)
= $1,627,500
2.35 A = 140,000(4000)(A/P,10%,4)
= 560,000,000(0.31547)
= $176,663,200
2.36 P = 1,500,000(P/A,8%,4)
= 1,500,000(3.3121)
= $4,968,150
2.37 A = 3,250,000(A/P,15%,6)
= 3,250,000(0.26424)
= $858,780
2.38 P = 280,000(P/A,18%,8)
= 280,000(4.0776)
= $1,141,728
2.39 A = 3,500,000(A/P,25%,5)
= 3,500,000(0.37185)
= $1,301,475
2.40 A = 5000(7)(A/P,10%,10)
= 35,000(0.16275)
= $5696.25
2.41 (a) F = 70,000(F/P,12%,6) + 90,000(F/P,12%,4)
= 70,000(1.9738) + 90,000(1.5735)
= $279,781

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6
(b) Spreadsheet function is = - FV(12%,6,0,70000) – FV(12%,4,0,90000) to
obtain $279,784.33
2.42 F = (458-360)(0.90)(20,000)(P/A,10%,5)
= 1,764,000(3.7908)
= $6,686,971
2.43 (a) Let CF4 be the amount in year 4
100,000(F/P,9%,3) + 75,000(F/P,9%,2) + CF4(F/P,9%,1) = 290,000
100,000(1.2950) + 75,000(1.1881) + CF4(1.0900) = 290,000
(1.09)CF4 = 71.392.50
CF4 = $65,497.71
(b) F in year 5 for 2 known amounts
= -FV(9%,3,0,100000) - FV(9%,2,0,75000)
P in year 4 of $290,000 minus amount above (assume it’s in cell H9)
= -PV(9%,1,0,290000-H9)
Answer is $65,495.05
2.44 P = 225,000(P/A,15%,3)
= 225,000(2.2832)
= $513,720
2.45 400,000 = 50,000(F/A,12%,n)
(F/A,12%,n) = 8.0000
From 12% interest table, n is between 5 and 6 years. Therefore, n = 6
2.46 F = P(F/P,10%,n)
3P = P(F/P,10%,n)
(F/P,10%,n) = 3.000
From 10% interest tables, n is between 11 and 12 years. Therefore, n = 12 years
2.47 (a) 1,200,000 = 400,000(F/P,10%,n) + 50,000(F/A,10%,n)
Solve for n by trial and error:
Try n = 5: 400,000(F/P,10%,5) + 50,000(F/A,10%,5)
400,000(1.6105) + 50,000(6.1051)
949,455 < 1,200,000 n too low
Try n = 8: 400,000(2.1436) + 50,000(11.4359)
1,429,235 > 1,200,000 n too high

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7
By continued interpolation, n is between 6 and 7. Therefore, n = 7 years
(b) Spreadsheet function = NPER(10%,-50000,-400000,1200000) displays 6.67
2.48 2,000,000(F/P,7%,n) = 158,000(F/A,7%,n)
Solve for n by trial and error (in $ thousands):
Try n = 30: 2,000,000(F/P,7%,30) = 158,000(F/A,7%,30)
2,000,000(7.6123) = 158,000(94.4608)
15,224,600 > 14,924,806 n too low
Try n = 32: 2,000,000(8.7153) = 158,000(110.2182)
17,430,600 > 17,414,476 n too low
Try n = 33: 2,000,000(9.3253) = 158,000(118.9334)
18,650,600 < 18,791,447 n too high
By interpolation, n is between 32 and 33, and close to 32 years.
Spreadsheet function is = NPER(7%,-158000,2000000) to display 32.1 years
2.49 (a) P = 26,000(P/A,10%,5) + 2000(P/G,10%,5)
= 26,000(3.7908) + 2000(6.8618)
= $112,284
(b) Spreadsheet: enter each annual cost in adjacent cells and use the NPV
function to display P = $112,284
Calculators have no function for gradients; use the PV function on each cash
flow and add the five P values to get $112,284.55
2.50 A = 72,000 + 1000(A/G,8%,5)
= 72,000 + 1000(1.8465)
= $73,846
2.51 (a) 84,000 = 15,000 + G(A/G,10%,5)
84,000 = 15,000 + G(1.8101)
G = $38,119
(b) The annual increase of over $38,000 is substantially larger than the first-year
cost of $15,000
2.52 A = 9000 – 560(A/G,10%,5)
= 9000 – 560(1.8101)
= $7986

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