CIVL 2812—Project Appraisal (Semester 2, 2019)
CIVL 2812—Project Appraisal (Semester 2, 2021)
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Tutorial 1 – Answers
ESSENTIAL QUESTIONS
Problem 4-4
Interest from the first investment = (3%)(P)(2) = 0.06P
Interest from second investment = (1.02)(1.02)P – P = 0.0404P
Therefore, the return on the first investment (simple interest at 3%) will be higher.
Problem 4-9
Here we are given, A=$624 per year, i=7%per year, N=10years. We want to determine F at the end of 10
years:
F=$624 (F/A, 7%,10) = $624 (13.81) = $8,621.4
Problem 4-11
The compounded future amount, given P = $6,000, i = 8% per year and N = 23 years, will be:
F = $6,000 (F/P, 8%, 23) = $35,229.
Problem 4-17
This is a “given P, find F” type of problem. Specifically, we use the following relationship to find “F” in 2024:
F = $74.5 (F/P, 7%, 7) = $74.5 (1.6058) = $119.63 per month.
Problem 4-25
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)
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EXTENSION QUESTIONS
Problem 4-5
Problem 4-6
Problem 4-12
CIVL 2812—Project Appraisal (Semester 2, 2021)
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EXTENSION QUESTIONS
Problem 4-21
(a) N = 2012 – 1985 = 27 years
$100,000 = $10,000 (1 + i)27
Barney and Lynne had a stock return at just below the standard for a good mutual fund.
(b) $100,000 = $800 (F/A, i′, 27)
(F/A, i′, 27) = 125.
i′ = 10.2% per year.
In this situation they did marginally better on their mutual fund investment.
Problem 4-49
(a) P = A (P/A, i%, N); $1,000 = $200(P/A, 12%, N); (P/A, 12%, N) = 5.0000
By looking at the 12% interest table in Appendix C under the P/A column,
(b) P = A (P/A, i%, N); $1,000 = $200 (P/A, i%,10); (P/A, i%, 10) = 5.0000
(P/A,15%, 10) = 5.0188 and (P/A, 18%, 10) = 4.4941, thus 15% < i < 18% By using linear interpolation, 18%−15% (5.0188−4.4941) = 𝑖%−15% 5.0188−5.0 ; i% = 15.11% (c) P = A (P/A, i%, N) = $200 (P/A, 12%, 5) = 200 (3.6048) = $720.96 (d) A = P (A/P, i%, N) = $1,000 (A/P, 12%, 5) = $1,000 (0.2774) = $277.40