EGG 307: ENGINEERING ECONOMICS
2
nd
Midterm Exam
 Sample Solution
Problem 1
Solution
(a) Nominal interest rate/quarter, r = 12%/4 = 3%; Compounding periods, m = 3 (months);
N = 3x4 = 12 quarters
Effective interest rate per quarter = (1+r/m)
m
= (1 + 0.03/3)
3
– 1 = 3.03%
Therefore, A = $50,000(A/P, 3.03%, 12) = 50,000
⎥
⎦
⎤
⎢
⎣
⎡
−
+
+
1
0303
0
1
0303
0
1
0303
0
12
12
)
.
(
)
.
(
.
=
$5,032
(a) Nominal interest rate per month, r = 12%/12 = 1%; Compounding is continuous; N = 36 months
Therefore, A = P
1
1
−
−
rN
r
rN
e
e
e
)
(
= 50,000
1
1
36
01
0
01
0
36
01
0
−
−
x
.
.
x
.
e
)
e
(
e
=
$1,662
Problem 2 Solution
(a) To use present worth analysis, assume repeatability
of the investments and use a common analysis
period of N = 12 years
PW(A) = $300,000 –300,000(P/F, 15%, 4) – 300,000(P/F, 15%, 8) + 110,000(P/A, 15%, 12)
= $300,000 –300,000x0.5718 – 300,000x0.3269 + 110,000x5.421 = +$26,700
PW(B) = $650,000 – 650,000(P/F,15%, 6) + 200,000(P/F,15%, 6) + 200,000(P/F,15%, 12)
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 Spring '09
 Dr.Kaseko
 $2,500, Machine B, incremental irr, $1,662, $84.86, $551.31

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