SOLUTIONS TO CHAPTER 14 PROBLEMS
Analytical and Simulation Approaches to Risk Analysis
14-1. j E[Aj]
σ
[ A j] V[Aj]
0 –$10,000 0 0
1-10 $2,000 200 40,000
(a) Annual receipts are statistically independent
E[PW] = -$10,000 + $2,000(P/A, 10%,10) = $2,289
10
14-2.(a) Pr(x < $5,000) = Pr
()
50.0)0SPr(
000,4$
000,5$000,5$
S=<=
−
<
000,5$0
−
174
σ
14-3. (a) By inspection, (i) E[Q] > E[X]
(ii)
σ
2 [X] >
σ
2 [Q]
(b) Alt Q : Pr(PW – C > $18M) = Pr )5.1SPr(
M2$
$15M – $18M
S >=
>
= 1 – Pr(S < 1.5) = 1- 0.93319 ≅ 0.07
$12M – $18M
>
14-4. PW = -$9,000 + $2,500(P/A,15%,N)
N PW Pr(N)
4 -$1,862.5 0.10
175
σ
14-5. CR = ($50,000-MV)(A/P,15%,N) + MV(0.15)
It is assumed that asset life, though uniformly distributed, is rounded to
integer values. Thus,
If random then
number is life is:
0-1 2
2-3 4
Trial
No.
Random
Number
Life
(yrs)
Random
Number
Market
Value
Capital Recovery
CR) Cost
1 2 4 9 $25,000
$12,507.50
2 7 8 8 15,000 10,051.50
If Life
Is:
And Random
Number Is:
Then Market
Value Is:
2 0-1 $15
,
000
2-6 20,000
7-9 25,000
176
14-6. AW = –I(A/P,10%,N) + A
Trial
No.
Random Normal
Deviate
(RND)
Investment (I)
[$1,000,000 + Random
RND* 000,000,16 Number
0-1: 5
yrs
Life 2-8:7yrs
9: 9 yrs
1 -1.724 $ 993,104 8 7yrs
Trial
No.
Random Number
As Decimal
(RN)
Net Annual Cash Flow (A)
[$120,000 +RN•($340,000-
$120,000)]
Annual
Worth
1 0.7955 $295,010 $ 91,026.44
2 0.0118 122,596 – 142,024.95
Estimate of AW mean: E[AW] =-$66,517.96/5 = -$13,303.59
Estimate of AW standard deviation:
177