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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ENGM 661Engineering Economics for Managers
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
Lower 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
Upper 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
Upper & Lower Bounds
A1 A2 A3 A4 A5
MARR = 15%
1 2 3 4 5
10,000
Now 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
NPW = 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
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)
5
E
[
NPW
]
E
[
10
,
000
a
A
NPW
t
t
t
1
Recall: E[Z] = E[X1] + E[X2]
E[aX+b] = aE[X] + b
5
E
[
NPW
]
E
[
10
,
000
a
A
NPW
t
t
t
1
5
10
,
000
a
E
[
A
]
NPW
t
t
t
1
Recall: E[Z] = E[X1] + E[X2]
E[aX+b] = aE[X] + b
5
10
,
000
a
E
[
A
]
NPW
t
t
t
1
5
10
,
000
a
[
3,000
]
NPW
t
t
1
but, E[At] = 3,000
5
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
= 10,000 + 3,000(P/A, i, 5)
2
2
(
10
,
000
a
A
)
NPW
t
t
Recall: 2(z) = 2(x) + 2(y)
2(ax+b) = a22
2
2
(
10
,
000
a
A
)
NPW
t
t
2
2
a
2
NPW
A
t
t
Recall: 2(z) = 2(x) + 2(y)
2(ax+b)= a22
2
2
a
2
(At)
NPW
t
2
(
A
)
250
2
t
2
2
(
250
)
2
a
NPW
t
but,
= (250)2 [(1+i)2 + (1+i)4 + . . . + (1+i)10]
2
2
(
250
)
2
a
A=(250)2
NPW
t
1 2 3 4 5 6 7 8 9 10
P=s2NPW
= (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
2
(
250
)
2
a
A=(250)2
NPW
t
1 2 3 4 5 6 7 8 9 10
P=s2NPW
= (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
MARR = 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
MARR = 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
MARR = 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
MARR = 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
MARR = 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
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
NPW N(56, 382)
= P( Z < 0.15 ) = 0.44
A1 A2 A3 A4 A5
35,000
If Ai iid
N(10,000, 300)
You are given the following cash flow diagram:
If the MARR = 15%, what is the probability this
investment alternative is no good?
2
2
2
4
10
(
NPW
)
300
[(
1
.
15
)
(
1
.
15
)
.
.
.
(
1
.
15
)
]
E[NPW] = 35,000 + 10,000(P/A, 15, 5)
= 35,000 + 10,000(3.3522)
=  1,478
= 3002 (2.3343)
= 210,087
NPW
0
(
1
,
478
)
P
458
1
.
0
NPW N(1,478 , 458)
P{NPW < 0} =
= P{Z < 3.27}
A1 A2 A3 A4 A5
35,000
If Ai iid
N(10,000, 300)
Max Ai 10,900
A1 A2 A3 A4 A5
35,000
If 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
Suppose 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
Let 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
NPW = 10,000 + 5,740(1.15)1 + 6,500(1.15)2
= (93.96)
A1 A2
5,740 6,500
10,000
10,000
We now have one realization of NPW for a given
realization of A1 and A2.
A1 A2
5,740 6,500
10,000
10,000
We 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
NPW = 10,000 + 6,820(1.15)1 + 6,218(1.15)2
= 632.14
A1 A2 NPW
5,7406,500(93.96)
6,8206,218632.14
Choose 2 new values.
A1 A2
5,273 6,422
10,000
10,000
NPW = 10,000 + 5,273(1.15)1 + 6,422(1.15)2
= (558.83)
A1 A2 NPW
5,7406,500(93.96)
6,8206,218632.14
5,2736,422(558.83)
Choose 2 new values.
A1 A2 NPW
5,7406,500(93.96)
6,8206,218632.14
5,2736,422(558.83)
.
.
.
6,8555,947457.66
Freq.
NPW
1,871 0 1,380
With 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.
1/2,000
5,000 7,000
What we now need is a formal method of selecting random values for A1 and A2 to avoid selection bias.
Recall the uniform
f(x)
F(x)
0
,
x
5
,
000
1
x
5
,
000
F
(
x
)
2
,
000
1
,
x
7
,
000
5,000 7,000
The 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)
F(x)
1
x
5
,
000
F
(
x
)
2
,
000
5,000 7,000
1. Randomly generate P U(0,1). P = .7
5,000 < x < 7,000
F(x)
1
x
5
,
000
F
(
x
)
2
,
000
5,000 7,000
1. Randomly generate P U(0,1). P = .7
2. Let P = F(x).
.7
5,000 < x < 7,000
F(x)
1
x
5
,
000
F
(
x
)
2
,
000
5,000 7,000
1. 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
x
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
Recall, for
5,000 < x < 7,000. Then
P
F(x)
x
5
,
000
2
,
000
P
1
5,000 7,000
Solving for x = F1(p),
P
x
F(x)
x
5
,
000
2
,
000
P
1
5,000 7,000
Solving 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
You 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
You 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%.
x
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
)
x
1
,
000
3
,
000
ln(
1
P
)
A1 = 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
5,828
3,079
2,070
7,000
NPW = 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
Suppose 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
Preliminary
A1 A2 A3
3
,
000
p
1
/
4
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
Distribution 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
Distribution 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
Distribution of NPW
A1 A2 A3
N(867, 938)
1 2 3
10,000
3,681
867
1,947
Distribution of NPW
P{NPW >0} = .031
.063
.031
.016
.031
.016
.188
Analytic
P{NPW > 0} = 0.188
C.L.T.
P{NPW > 0} = 0.178
A1 A2 A3
3
,
000
p
1
/
4
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
A1 A2 A3
3
,
000
p
1
/
4
Ai
4
,
000
p
1
/
2
1 2 3
5
,
000
p
1
/
4
10,000
P{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
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