Solution Manual
DmaH C. Newnan
J ~ e m P.
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7-18
Chapter 7
Rate Of Return Analysis
Chapter 7
7-19
Rate Of Return Analysis
x r n = q t g ( ? / ~ , ~ , s ) TV~L=B%
?
3500 =. 91% (3,993)
= 36bb
Tnq i =10%
9'.
Skh AmR? MANZ,
= 9153(3,791) = 3480
A m &WL~M.
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7-22 Chapter 7
Rate Of Return Analysis
Chapter 7
- $30 000(5 516015)=$l65 480 46
$15 000=$165,480 J6(1 T I ) "
$15 000(1 + 1)'' $165,400 46
1 + I = [(I65480 46i15000)]1"
I = 1071 - I or 7 1002"0> 5""
Unless $15,000 can be ~nvestedw ~ t ha return hlgher than 7 I%, ~tIS better to Walt for 35
years for the retirement fund $15 000 now IS onl) equ~valentto $165,480 46 35 years from
non ~fthe Interest rate now is 7 1% ~nsteadof the quoted 5"a
-
Rate Of Return Analysis
7-54
2000 - 9 1 05 (P/A 1*,30)
( P 4,1*,30)= 2000 91 05 = 2 l 966
from tables
%1
(P/A,r,30)
2
22 396
20 930
2 $2
I,, = 2% + 1 29.0((22 396-2 1 966) (22 396-20 930)
2 15'0 mo
Nom~nalROR r e ~ e i ~ eb)d tinance compan) - 12(2 15"0) 25 8"0
-
7-52
Using incremental analysis, compute the internal rate of return for the difference between the
two alternatives.
Alternative B - Alternative A
Year (n)
0
+ $12000
1
- 3000
2
- 3000
- 3000
3
4
- 3000
- 3000
5
6
- 3000
7
- 3000
8
- 4200
Note: Internal Rate of Return (IROR) equals the interest rate that makes the PW of costs
minus the PW of benefits equal to zero
12,000 - 3000(P/A. I', 7) - 4200(P/F, I., 8) = 0
TQ 1' = 18%. 12,000 - 3000(3.812) - 4200(0.2660) = -553
Try I' = 1746 12,000 - 3000(3 922) - 4200(0.2848) = 962
By str line interp.. I' = 17?/0+(1~')(962)1(962 +553)= 17.6%
THE CONTRACTOR SkiOU1,D CHOOSE 4L.T. B A N D L.EASE; 17 6 % , 15% MARR
7-53
-
$24,000
Total Annual Rekenues $00'12 months * 4 apt
Annual Revenues - Expenses = $24,000 - $8,000 = $16.000
To find Internal Rate ol Return the Net Pre5ent Worth must be 0
N P h - $ 16 000(P 4 I * 5)+$160,00O(P/F I* 5 )-$ I40 000
Try 1=12~/0NPW $8464
Tr) I = 15'0 h P W - - % 6 8 1 6
IKOR = 12'0 (3'0)[8 164 (8464+6816]] = 13 7'0
.
Part b. At 13 7'0 the apartment building is more attractive than the other options
7-23
3000 = 1 18.90 (P/A,r*,36)
(P/A.i*.36) = 300011 12 06 = 26.771
from tables:
1%
(P/A,i,36)
l %
27.661
1 314 26.543
I,,,,,
= I 1/2 96 + (1!4%)(27 661-26 771)1(27 661-26 543) = 1.699%
Nominal annual ROR = 12(1.699) = 20.4%
NPW
=
-300,000 + 20,000 (P/F, I * , 10)
t (67.000 - 3000) (P/A, I*, 10)
- 600 (P;G,
I*, 10)
looh
-300,000 + 20,000 (0.3855) + 64,000 (6.145) - 600 (22 891)
= 87,255
NPW greater than zero. The interest rate used is too low.
Try r
NPW
=
=
Try I = l8"0
NPW = -300.000
= -17,173
+
20,000 (0 191 1 )
+
64,000 (4 494) - 600 (14 352)
NPW negative Reduce
Try r = 15%
NPW = -300.000
t
I
20,000 (0 2472) + 64.000 (5 019) - 600 (16 979)
=
So the rate of return is between 15% and 18%. By linear interpolation:
9130
7-24
Chapter 7
Rate Of Return Analysis
Alternotrve
B
A
%300,000
'4-8
$615,000
$3 15,000
Maintenance & operating costs
25,000
10,000
- 1 5,000
Annual benefit
92,000
158,000
66,000
Salvage value
-5,000
65.000
70,000
F~rstcost
NPW = 0
N P W = -3 15,000 + (66,000 - (-15,000)] ( P / A . I * ,10)
+ 70,000 (P'F, I * , 10) = 0
Try
I
=
15%
3
-3 15,000 + 8 1.000 (5.0 19) + 70,000 (0.2472) '0
108,840 :, 0
:.A ROR .- MARR (= 15%). The higher cost alternative ,4 is the more desirable alternative
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