CIVL 2812—Project Appraisal (Semester 2, 2019)
CIVL 2812—Project Appraisal (Semester 2, 2021)
Page 1
Tutorial 2 – Answers
ESSENTIAL QUESTIONS
Problem 4-60
The present equivalent amount at time zero (now) is:
P = −$30,000 (P/A, 7%, 10) − $80,000 (P/F, 7%, 15) − $80,000 (P/F, 7%, 20)
P = −$30,000 (7.0236) − $80,000 (0.3624) − $80,000 (0.2584) P = −$210,708 − $28,992 − $20,672 = −$260,372.
Then the equivalent annual amount over 30 years is: A = −$260,372 (A/P, 7%,30) = −$260,372 (0.0806)
= −$20,986.
Problem 4-79
Using time 1 as the reference point, set P1(LHS) = P1(RHS)
K(P/A,12%,2) (P/F,12%,2) = $100(P/A,12%,6) + $110(P/G,12%,6)
K(1.6901) (0.7972) = $100(4.1114) + $110(8.93) 1.3473K=$1,393.44
K = $1,034.25
Problem 4-81
A = [$2,000 (P/A,8%,4) + $400 (P/G,8%,4)] (P/F,8%,2) (A/F,8%,11)
= [$2,000 (3.3121) + $400 (4.650)] (0.8573) (0.0601) = $437.14
Problem 4-87
𝑷 =
𝟓𝟎𝟎[𝟏−(
𝑷
𝑭
,𝟏𝟐%,𝟏𝟓) (
𝒇
𝑷
,𝟔%,𝟏𝟓)]
𝟎.𝟏𝟐−𝟎.𝟎𝟔
= $4,684.51
Problem 4-102
a) (1 + 0.2 / M)M = 1.219, or M = 12 (i.e. monthly compounding)
b) (1 + 0.2 / M)M = 1.221, or M = 365 (i.e. daily compounding)
The F-equivalent amount of this series of deposits can be determined as follows:
F = $1,000 (F/A, 4%, 25) = $1,000 (41.6459) = $41,645.90.
CIVL 2812—Project Appraisal (Semester 2, 2021)
Page 2
EXTENSION QUESTIONS
Problem 4-73
Left side:
PL = −10 (P/F, 15%, 1) + H (P/A, 15%, 16−4) (P/F, 15%, 4) + 0.7H (P/A, 15%, 6) (P/F, 15%, 7)
= −10 (0.8696) + H (5.4206) (0.5718) + 0.7H (3.7845) (0.3759)
= 4.0953H – 8.696
Right side:
PR = −P0 + 2P0 (P/F, 15%, 10) = −P0 + 2P0 (0.2472) = −0.5056 P0
Set PR = PL: −0.5056 P0 = 4.0953H – 8.696
P0 = 17.2 – 8.1H
Problem 4-77
(a) F = $600 (P/G, i%, 6)(F/P i%, 6) = $10,000
If i = 7%, F = $9,884.81 < $10,000, i > 7%
If i = 8%, F= $10,019.37 > $10,000, i < 8%
Thus, 7% < i < 8%. Using linear interpolation,
i = 7.86%
(b) F = $600 (P/G, 5%, N) (F/P, 5%, N) = $10,000
If N = 6, F = $9,622.99 < $10,000, N > 6
If N = 7, F= $13,704.03 > $10,000, N < 7
Thus, 6 < N < 7. Using linear interpolation,
N = 6.1 periods. If an integer value of N is desired, choose N = 7 periods.
(c) F = $1,000(P/G,10%,12)(F/P,10%,12)
= $1,000(29.901)(3.1384) = $93,841.30
Problem 4-85
Note, N = 6 years,
𝑷 =
𝟐𝟎𝟎,𝟎𝟎𝟎[𝟏−(
𝑷
𝑭
,𝟐𝟎%,𝟔) (
𝒇
𝑷
,𝟗%,𝟔)]
𝟎.𝟐−𝟎.𝟎𝟗
P = 796,980
You can afford to spend as much as $796,980 for a higher quality heat exchanger.
Problem 4-88
𝑷(𝐺𝑎𝑠) =
𝟏𝟏,𝟐𝟎𝟎[𝟏−(
𝑷
𝑭
,𝟐𝟎%,𝟏𝟎) (
𝒇
𝑷
,𝟏𝟐%,𝟏𝟎)]
𝟎.𝟐−𝟎.𝟏𝟐 = $69,778
𝑷(𝑀𝑎𝑖𝑛. ) =
𝟓𝟔𝟓(𝟏.𝟏𝟖)[𝟏−(
𝑷
𝑭
,𝟐𝟎%,𝟏𝟎) (
𝒇
𝑷
,𝟏𝟖%,𝟏𝟎)]
𝟎.𝟐−.𝟎𝟏𝟖 = $5,141
A = ($69,778+ $5,141) (A/P, 20%, 10) = $17,868
CIVL 2812—Project Appraisal (Semester 2, 2021)
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EXTENSION QUESTIONS
Problem 4-97
P = $1,000 (P/F, 8%, 1) + $2,000 (P/F, 10%, 1)(P/F, 8%, 1) + $1,000 (P/F, 6%, 1)
(P/F, 8%, 1)(P/F, 10%, 1)(P/F, 8%, 1) + $2,000 (P/F, 10%, 1)(P/F, 6%, 2)(P/F, 8%, 1)(P/F, 10%, 1)(P/F, 8%, 1)
= $1,000 (0.9259) + $2,000 (0.9091)(0.9259) + $1,000 (0.9434)(0.9259)(0.9091)(0.9259)
+ $2,000 (0.9091)(0.8900)(0.9259)(0.9091)(0.9259)
= $4,606