Chapter 1
Foundations of Engineering Economy
Solutions to Problems
Nearest, tastiest, quickest, classiest, most scenic, etc
In simple interest, the interest rate applies only to the principal, while compound
interest generates interest on the principal and all accumulated interest.
Rate of return = (45/966)(100)
=4.65%
~~
Rate of return = (2.3/6)( 100)
= 38.3%
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(~ P + P(0.10) 1,600,000
1.1P = 1,600,000
P = $1,454,545
~?l9\
l./
80,000 + 80,000(i) = 100,000
25%
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( 1.29 \ (a) F ?; i = 7%; n = 10;
$2000; P = $9000
~ (b) A ?;i=11%;n=20;P $14,000;F=0
(c) P =?; i 8%; n 15; A=: $1000; F $800
(,1~
C/
rlJ~
Q
Highest to lowest interest rate is as follows: rate of return on risky investment,
minimum attractive rate of refilm, cost of capital, rate of return on safe
investment, interest on savings account, interest on checking account.
WACC = (0.25)(0.18) + (0.75)(0.10} = 12%
Therefore, MARR = 12%
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F=?
7"'-.
~ The cash flow diagram is:
10%
i
i
o
T
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Chapter 2
Factors: How Time and Interest Affect Money
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Solutions to Problems
2.7
P = 75(P/F,18%,2)
= 75(0.7182)
. $53.865 million
2.11 P
125,000(P!F,14%,5)
125,000(0.5149)
$ 64,925
2.15 P
7~
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",.
), OOO(P":."1..,-'­
/0,.))
75,000(2.1 065)
$157,988
2.18 P
(280,000-90,000)(P/A,10%,5)
190,000(3.7908)
$720,252
.­
2.21 P
200,000((P/A,10%,5)
200,000(3.7908)
= $758,160
2.27 (a) G
$200 (b) CFs = $1600
(c) n == 10
2.37 50 = 6(PlA,12%,6) + G(P/G,12%,6)
50 = 6(4.1114) + G(8.9302)
G = $2,836,622
2.41 P = 25,000{1
= $60,247
[(1+0.06)3/(1+0.15)3}]}/(0.15 ­ 0.06)
2.52 1,000,000 = 600,000(FIP,i,5)
(FIP,i,5) = 1.6667
1
10.8% (Excel)
2.60
100,000
1,600,000(PlF,18%,n)
(PlF,18%,n) = 0.0625
From 18% table, n is between 16 and 17 years; therefore, n
.­
17 years
Chapter 3 Combining Factors Solutions to Problems
3.2 P = 5o.,o.o.o.(56)(P/A,8%,4)(P1F,8%,I)
= 2,80.0.,0.0.0.(3.3121)(0..9259) $8.587 million 3.5 P = 150.,0.0.0. + 15o.,GGG(P/A,IG%,5)
= 150.,0.0.0. + 150.,0.0.0.(3.790.8)
= $718,620.
3.8 A = 40.0.0. + IGo.o.(FIA,IG%,4)(AlF,lo.%,7)
= 40.0.0. + 100.0.(4.6410.)(0..10541)
= $4489.21
3.19 10.0.,0.0.0. = A(FIA,7%,5)(FIP,7%,1O)
10.0.,0.0.0. A(5.75o.7)(1.9672) A = $8839.56 3.22 Amt, year 5 = lOo.o.(FIA,12%,4)(FIP,12%,2) + 2o.o.o.(P/A,12%,7)(p1F,12%,1)
= 10.0.0.(4.7793)(1.2544) + 20.0.0.(4.5638)(0..8929)
= $14,145
3.32 A 5o.Go.(AIP,12%,7) + 350.0. + 15o.G(F/A,12%,4)(AIF,12%,7)
= 50.0.0.(0..21912) + 350.0. + 150.0.(4.7793)(0..0.9912)
= $530.6.19
3.34 P = [4,1o.o.,o.o.G(P/A,6%,22) - 5o.,o.GG(P/G,6%,22)](PIF,6%,3)
+ 4,lOo.,o.GG(P/A,6%,3) = [4,100.,0.0.0.(12.0.416) - 50.,0.0.0.(98.9412](0..8396) + 4,10.0.,0.0.0.(2.6730.) $48,257,271 3.39 Find P in year 0. and then convert to A.
P = 40.0.0. + 4Go.G(P/A,15%,3) lOGG(P/G,15%,3) + [(6GGo.(P/A,15%,4)
+2GGo.(P/G,15%,4)](PIF,15%,3)
40.0.0. + 40.0.0.(2.2832) - 10.0.0.(2.0.712) + [(60.0.0.(2.8550.)
+20.0.0.(3.7864)](0..6575) = $27,30.3.69 A 27,3o.3.69(AIP,15%,7) 27,30.3.69(0..240.36) $6563 3.43 Find P in year-I; then find A in years 0~5.
Pg in yr 2 = (5)(4000){[1 - (1 + 0.08) 18/(1 + 0.10)18]/(0.10 - 0.08)} = 20,000(14.0640) = $281,280 Pin yr -1 = 281,280(PIF,10%,3) + 20,000(P/A,10%,3) 281,280(0.7513) + 20,000(2.4869) $261,064 A
261,064(AIP,10%,6)
261,064(0.22961 ) $59,943 3.49 P
3.52 P
5000 + 1000(P/A,12%,4) + [1000(P/A,12%,7) - 100(p/G,12%,7)](PIF,12%,4)
5000 + 1000(3.0373) + [1000(4.5638) 100(11.6443)](0.6355)
$10,198
2000+ 1800(P/A,15%,5)-200(P/G,15%,5)
= 2000 + 1800(3.3522) - 200(5.7751)
= $6878.94
Chapter 4
Nominal and Effective Interest Rates
Solutions to Problems
4.2 (a) quarterly (b) monthly (c) weekly
4.7 (a) 5% (b) 20%
4.16 (a) ilweek 0.068/26
(b) effective
0.262%
4.25 P = 1.3(P/A,I%,28)(PIF,1%,2)
= 1.3(24.3164)(0.9803)
= $30,988,577
4.28 F = 50(20,000,000)(FIP,1.5%,9)
= 1,000,000,000(1.1434)
= $1.1434 billion
(in $million)
-
4.35 First find P, then convert to A
p:::: 150,000{1 [(1-,-0.20)lo/(h-0.07io} ]}/(0.07 - 0.20) = 150,000(16.5197) $2,477,955 A == 2,477,955(AIP,7%,10) 2,477,955(0.14238) == $352,811 4.42 Move withdrawals to beginning ofperiods and deposits to end; then tlnd 1:.
F 1600(FIP,4%,5) +1400(FIP,4%,4) - 2600(FIP,4%,3) -+- 1000(FIP,4%,2)
-1000(FIP,4%,I)
= 1600(1.2167) + 1400(1.1699) -2600(1.1249) + 1000(1.0816) 000(1.04)
$701.44
4.50 i ;:;: eO,06 - 1 = 6.18% per year
P = 85,000(p1F,6.l8%,4)
85,000[1/(1 + 0.0618)4
= 85,000(0.78674) = $66,873 1
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