CS计算机代考程序代写 Microsoft Word – CIVL2812-Tut3 Answers

Microsoft Word – CIVL2812-Tut3 Answers

CIVL 2812—Project Appraisal (Semester 2, 2021)

Page 1

Tutorial 3 – Answers

ESSENTIAL QUESTIONS
Solution 4-103

The number of instalments is 5 × 12 = 60 and the interest rate per month is 24/12 =2%.
A = P (A/P, 2%, 60) = $20,000(0.0288) = $576 per month for 60 months.

Solution 4-106
P = A(P/A, 3%, 6) = $200(5.4172) = $1,083.44.
The value of the washing machine is only $1,083, rather than $1,200.

Solution 4-113

10% = ei – 1, ln (1.1 ) = i
i = 9.53%.

Solution 5-1

All five projects are more profitable than the 10% being earned on municipal bonds and are therefore feasible
investments. We need to consider rank-ordered projects (based on annual rate of profit) versus cumulative
capital investment:

Project ranking C E A B D
Profit rate per year (%) 28 25 18 15 12

Cumulative investment ($millions) 40 65 95 105 120

Because of limited funds only projects C,E and A can be undertaken. The best rejected project is B, So TTW’s
MARR is 15% per year. Only whole projects are possible, so B and D cannot be funded.

Solution 5-2

PW(10%) = −$30 – 20(P/F, 10%,1) – 10(P/F, 10%,2) + 10(P/G, 10%,7) (P/F, 10%,2)
= $48.77 Millions
Since PW > 0, this is a profitable investment.

Solution 5-32

Let X = units produced per year. Then the breakeven equation becomes:
AW(12%) = −$650,000(A/P, 12%, 6) − $54,000 + X ($85 − $12.80) = 0
Solving yields X = 2,937 units per year

CIVL 2812—Project Appraisal (Semester 2, 2021)

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EXTENSION QUESTIONS
Solution 4-107

a) A = P(A/P,0.5%, 48) = $5,000(0.0235) = $117.50.
b) A = P(A/P,0.75%, 60) = $5,000(0.0208) = $104.

Solution 4-115

a)

b) P = $1,000(P/A,8%,12) = $1,000(7.4094) = $7,409.40
c) r = 8%/2% = 4%

F = 243(F/A, 4%, 12)

F = 243 x

F = 243 x
. ×

.

F = $3,668.30
d) F = $1,000(F/P, 8%, 9)

F = $2,054.40

Solution 4-123

(a) False; (b) False; (c) False; (d) True; (e) False;
(f) True; (g) False; (h) False; (i) False.

Solution 5-16

a) 𝐶𝑊(15%) =
$ ,

.
+

$ , , %,

.
, 15%, 1 = $63,224

b) Find the value for N for which (A/P,15%,N) = 0.15
From Table C-15, N=60 years

Solution 5-19

Let A = $2,900, G = −$100 (delayed 1 year)

F6 = −$2,000

P0 = $2,900 (P/A,6$,10) – 100(P/G,6%,9)(P/F,6%,1) −$2,000(P/F,6%,6) = $2,900 (7.3601) − $100
(24.5768) (0.9434) − $2,000 (0.7050) = $17,615.71

FW10 = $17,615.71(F/P, 6%, 10) = $17,615.71(1.7908) = $31,546.21

CIVL 2812—Project Appraisal (Semester 2, 2021)

Page 3

EXTENSION QUESTIONS
Solution 5-23

Opportunity Cost = Investment at BOY x i = Investment at BOY (0.15)

Capital Recovery Amount = Opportunity Cost + Loss in Value During Year

Year

Investment at
Beginning of Year

Opportunity
Cost of Interest
(i = 15%)

Loss in
Value
During Year

Capital
Recovery
Amount

1 $10,000 $10,000 (0.15) = 1,500 $3,000 1,500 + 3,000 = 4,500
2 10,000−3000 = 7,000 7,000 (0.15) = 1,050 2,000 1,050 + 2,000 = 3,050
3 7,000 − 2,000 = 5,000 5,000 (0.15) = 750 2,000 750 + 2,000 = 2,750
4 5,000 − 2,000 = 3,000 3,000 (0.15) = 450 1,000 450 + 1,000 = 1,450

P0 = $4,500 (P/F,15%,1) + $3,050 (P/F,15%,2) + $2,750 (P/F,15%,3) +$1,450 (P/F, 15%, 4)

= $4,500 (0.8696) + $3,050 (0.7561) + $2,750 (0.6575) + $1,450 (0.5718) = $ 8,856.54

A = $8,856.54 (A/P,15%,4) = $8,856.49 (0.3503) = $3,102.45

This same value can be obtained and confirmed with Equation (5-5):

CR(i%) = I (A/P, i%, N) − S (A/F, i%, N)

= $10,000 (A/P,15%,4) − $2,000 (A/F,15%,4)
= $10,000 (0.3503) − $2,000 (0.2003)

= $3,102.12

Note: The Annual Worth from the table and the CR amount from Equation (5-5) are the same.