ENGM 661 Engineering Economics for Managers. Risk Analysis. Learning Objectives. 1)Given a simple cash flow stream with uncertainty, be able to estimate the worth of a project using a worst case, best case, or expected value evaluation.
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Risk Analysis
1)Given a simple cash flow stream with uncertainty, be able to estimate the worth of a project using a worst case, best case, or expected value evaluation.
2) Given a simple probabilistic (discrete case) cash flow stream, be able to compute the distribution of the Net Present Worth (or Equivalent Uniform Annual Worth).
3) Given an analytic cumulative distribution function and a uniform random variable, U(0,1), be able to use the inverse method to compute a cash flow At.
4) Given a set of random variables (0,1), a cash flow stream, and an inverse formula, be able compute 1 realization of the Net Present Worth (or EUAW) using simulation techniques.
A1 A2 A3
3
,
000
p
1
/
4
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
BoundingLower Bound
A1 A2 A3
3
,
000
p
1
/
4
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
BoundingUpper Bound
A1 A2 A3
3
,
000
p
1
/
4
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
BoundingUpper & Lower Bounds
A1 A2 A3 A4 A5
MARR = 15%
1 2 3 4 5
10,000
Risk AnalysisNow Suppose the return in each year is a random
variable governed by the some probability distribution.
NPW = 10,000 + A1(1+i)1 + A2(1+i)2 + . . . + A5(1+i)5
A1 A2 A3 A4 A5
MARR = 15%
1 2 3 4 5
10,000
5
10
,
000
a
A
t
t
t
1
where
a
(
1
i
)
t
t
Risk AnalysisNPW = 10,000 + A1(1+i)1 + A2(1+i)2 + . . . + A5(1+i)5
Now suppose At iid N(3,000, 250)
NPW is a linear combination of normals
NPW Normal
Central Limit
Now suppose At iid N(3,000, 250)
NPW is a linear combination of normals
NPW Normal
NPW N(NPW, NPW)
E
[
NPW
]
E
[
10
,
000
a
A
NPW
t
t
t
1
MeanRecall: E[Z] = E[X1] + E[X2]
E[aX+b] = aE[X] + b
E
[
NPW
]
E
[
10
,
000
a
A
NPW
t
t
t
1
5
10
,
000
a
E
[
A
]
NPW
t
t
t
1
MeanRecall: E[Z] = E[X1] + E[X2]
E[aX+b] = aE[X] + b
10
,
000
a
E
[
A
]
NPW
t
t
t
1
5
10
,
000
a
[
3,000
]
NPW
t
t
1
Meanbut, E[At] = 3,000
10
,
000
a
[
3,000
]
NPW
t
t
1
5
a
= 10,000 + 3,000
t
t
1
5
(1+i)t
= 10,000 + 3,000
t
1
Mean= 10,000 + 3,000(P/A, i, 5)
2
2
(
10
,
000
a
A
)
NPW
t
t
2
2
a
2
NPW
A
t
t
VarianceRecall: 2(z) = 2(x) + 2(y)
2(ax+b)= a22
2
a
2
(At)
NPW
t
2
(
A
)
250
2
t
2
2
(
250
)
2
a
NPW
t
Variancebut,
= (250)2 [(1+i)2 + (1+i)4 + . . . + (1+i)10]
2
(
250
)
2
a
A=(250)2
NPW
t
1 2 3 4 5 6 7 8 9 10
P=s2NPW
Variance= (250)2 [(1+i)2 + (1+i)4 + . . . + (1+i)10]
Note that [(1+i)2 + (1+i)4 + . . . + (1+i)10 ] is just a
5 period annuity factor where the period is 2 years.
2
(
250
)
2
a
A=(250)2
NPW
t
1 2 3 4 5 6 7 8 9 10
P=s2NPW
Variance= (250)2 [(1+i)2 + (1+i)4 + . . . + (1+i)10]
= (250)2(P/A, ieff, 5) , ieff = (1+i)2 1
A1 A2 A3 A4 A5
1 2 3 4 5
At iid N(3,000, 250)
10,000
Risk AnalysisMARR = 15%
mNPW = 10,000 + 3,000(P/A, 15, 5)
= 10,000 + 3,000(3.3522)
= $56
A1 A2 A3 A4 A5
1 2 3 4 5
At iid N(3,000, 250)
10,000
Risk AnalysisMARR = 15%
ieff = (1.15)2  1 = 32.25%
NPW = 10,000 + 3,000(P/A, 15, 5)
= 10,000 + 3,000(3.3522)
= $56
A1 A2 A3 A4 A5
1 2 3 4 5
At iid N(3,000, 250)
10,000
Risk AnalysisMARR = 15%
ieff = (1.15)2  1 = 32.25%
NPW = 10,000 + 3,000(P/A, 15, 5)
= 10,000 + 3,000(3.3522)
= $56
s2NPW = (250)2(P/A, 32.25, 5)
= 62,500(2.3343) = 145,894
A1 A2 A3 A4 A5
1 2 3 4 5
At iid N(3,000, 250)
10,000
Risk AnalysisMARR = 15%
NPW = $56
s2NPW = 145,894 s = 382
NPW N(56, 382)
A1 A2 A3 A4 A5
N(56, 382)
1 2 3 4 5
At iid N(3,000, 250)
10,000
1,090
56
1,202
Risk AnalysisMARR = 15%
NPW N(56, 382)
A1 A2 A3 A4 A5
N(56, 382)
1 2 3 4 5
10,000
1,090
56
1,202

m

NPW
0
56
<
=
<
NPW
P
(
NPW
0
)
P
s
382
NPW
Risk AnalysisNPW N(56, 382)
A1 A2 A3 A4 A5
N(56, 382)
1 2 3 4 5
10,000
1,090
56
1,202

m

NPW
0
56
<
=
<
NPW
P
(
NPW
0
)
P
s
382
NPW
Risk AnalysisNPW N(56, 382)
= P( Z < 0.15 ) = 0.44
A1 A2 A3 A4 A5
35,000
If Ai iid
N(10,000, 300)
Class ProblemYou are given the following cash flow diagram:
If the MARR = 15%, what is the probability this
investment alternative is no good?
2
2
4
10
(
NPW
)
300
[(
1
.
15
)
(
1
.
15
)
.
.
.
(
1
.
15
)
]
Class ProblemE[NPW] = 35,000 + 10,000(P/A, 15, 5)
= 35,000 + 10,000(3.3522)
=  1,478
= 3002 (2.3343)
= 210,087
0
(
1
,
478
)
P
458
1
.
0
Class ProblemNPW N(1,478 , 458)
P{NPW < 0} =
= P{Z < 3.27}
A1 A2 A3 A4 A5
35,000
A Critical ThunkIf Max Ai 10,900
NPW = 35,000 + 10,900(P/A, 15, 5)
= 35,000 + 10,900(3.3522)
= 1,539
A1 A2
10,000
A TwistSuppose we have the following cash flow diagram.
If Ai iid U(5000, 7000)
Now how can we compute the distribution
of the NPW? MARR = 15%.
A1 A2
10,000
SimulationLet us arbitrarily pick a value for A1 and A2 in the uniform range (5000, 7000). Say A1 = 5,740 and
A2 = 6,500.
A1 A2
5,740 6,500
10,000
10,000
SimulationWe now have one realization of NPW for a given
realization of A1 and A2.
A1 A2
5,740 6,500
10,000
10,000
SimulationWe now have one realization of NPW for a given
realization of A1 and A2. Choose 2 new values for
A1, A2.
A1 A2
6,820 6,218
10,000
10,000
A1 = 6,820 A2 = 6,218NPW = 10,000 + 6,820(1.15)1 + 6,218(1.15)2
= 632.14
A1 A2
5,273 6,422
10,000
10,000
A1 = 5,273 A2 = 6,422NPW = 10,000 + 5,273(1.15)1 + 6,422(1.15)2
= (558.83)
A1 A2 NPW
5,740 6,500 (93.96)
6,820 6,218 632.14
5,273 6,422 (558.83)
.
.
.
6,855 5,947 457.66
NPW
1,871 0 1,380
SimulationWith enough realizations of A1, A2, and NPW, we can begin to get a sense of the distribution of the NPW.
What we now need is a formal method of selecting random values for A1 and A2 to avoid selection bias.
5,000 7,000
SimulationWhat we now need is a formal method of selecting random values for A1 and A2 to avoid selection bias.
Recall the uniform
f(x)
0
,
x
5
,
000
1
x
5
,
000
F
(
x
)
2
,
000
1
,
x
7
,
000
5,000 7,000
SimulationThe uniform has cumulative distribution given by:
Recall that 0 < F(x) < 1. Using modulo arithmetic, it is fairly easy to generate random numbers between 0 and 1 (Rand Function in Excel). Let P be a random variable uniformly from 0 to 1.
P U(0,1)
Recall that 0 < F(x) < 1. Using modulo arithmetic, it is fairly easy to generate random numbers between 0 and 1. Let P be a random variable uniformly from 0 to 1.
P U(0,1)
Algorithm:
1. Randomly generate P
2. Let P = F(x)
3. Solve for x = F1(p)
1
x
5
,
000
F
(
x
)
2
,
000
5,000 7,000
Simulation1. Randomly generate P U(0,1). P = .7
5,000 < x < 7,000
1
x
5
,
000
F
(
x
)
2
,
000
5,000 7,000
Simulation1. Randomly generate P U(0,1). P = .7
2. Let P = F(x).
.7
5,000 < x < 7,000
1
x
5
,
000
F
(
x
)
2
,
000
5,000 7,000
Simulation1. Randomly generate P U(0,1). P = .7
2. Let P = F(x).
3. x = F1(p).
.7
5,000 < x < 7,000
6,400
5
,
000
F
(
x
)
2
,
000
x
5
,
000
P
F(x)
7
,
000
5
,
000
1
x
5
,
000
2
,
000
5,000 7,000
Formal DerivationRecall, for
5,000 < x < 7,000. Then
P
x
5
,
000
2
,
000
P
1
5,000 7,000
Formal DerivationSolving for x = F1(p),
P
Note:
1. P = 0 x = 5,000
2. P = 1 x = 7,000
x
A1 A2 A3
7,000
(
x
1
,
000
)
/
3
,
000
F
(
x
)
1
e
Class ProblemYou are given the following cash flow diagram. The Ai are iid shifted exponentials with location parameter a = 1,000 and scale parameter = 3,000. The cumulative is then given by
, x > 1,000
A1 A2 A3
7,000
(
x
1
,
000
)
/
3
,
000
F
(
x
)
1
e
Class ProblemYou are given the first 3 random numbers U(0,1) as follows:
P1 = 0.8
P2 = 0.3
P3 = 0.5
You are to compute one realization for the NPW.
MARR = 15%.
1000
P
1
e
3000
x
1000
e
3000
1
P
x
1000
ln(
1
P
)
3000
x
1
,
000
3
,
000
ln(
1
P
)
Class Problem
1
,
000
3
,
000
ln(
1
P
)
Class ProblemA1 = 1,000  3000 ln(1  .8)
= 5,828
A2 = 1,000  3000 ln(1  .3)
= 2,070
A3 = 1,000  3000 ln(1  .5)
= 3,079
3,079
2,070
7,000
Class ProblemNPW = 7,000 + 5,828(1.15)1 + 2,070(1.15)2 + 3,079(1.15)3
= 1,657
A1 A2 A3
3
,
000
p
1
/
4
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
Class ProblemSuppose we have the following cash flow diagram (MARR = 15%).
Determine if the project is worthwhile.
A1 A2 A3
3
,
000
p
1
/
4
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
Central Limit TheoremPreliminary
A1 A2 A3
3
,
000
p
1
/
4
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
Central Limit TheoremDistribution of NPW
A1 A2 A3
3
,
000
p
1
/
4

=
867
E
[
NPW
]
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
Central Limit TheoremDistribution of NPW
A1 A2 A3
3
,
000
p
1
/
4
N(867, 938)

=
867
E
[
NPW
]
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
3,681
867
1,947
Central Limit TheoremDistribution of NPW
A1 A2 A3
3
,
000
p
1
/
4
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
SimulationA1 A2 A3
3
,
000
p
1
/
4
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
SimulationP{NPW > 0} = 5/20 = 0.25
Analytic
P{NPW > 0} = 0.188
C.L.T.
P{NPW > 0} = 0.178
Simulation
P{NPW > 0} = 0.25
@Risk
P{NPW > 0} = 0.20 Next WEEK