TM 661 Engineering Economics for Managers

1
TM 661Engineering Economics for Managers

• Risk Analysis

2
Motivation

• Suppose we have the following cash flow diagram.

NPW -10,000 A(P/A, 15, 5)
3
Motivation

• Now suppose that the annual return A is a random
  variable governed by the discrete distribution

4
Motivation
For A 2,000, we have
NPW -10,000 2,000(P/A, 15, 5) -3,296
5
Motivation
For A 3,000, we have
NPW -10,000 3,000(P/A, 15, 5) 56
6
Motivation
For A 4,000, we have
NPW -10,000 4,000(P/A, 15, 5) 3,409
7
Motivation

• There is a one-for-one mapping for each value of
  A, a random variable, to each value of NPW, also
  a random variable.

A 2,000 3,000 4,000 p(A)
1/6 2/3 1/6 NPW
-3,296 56 3,409 p(NPW) 1/6
2/3 1/6
8
Spreadsheet
9
Risk Analysis
Now Suppose the return in each year is a
random variable governed by the some probability
distribution.
NPW -10,000 A1(1i)-1 A2(1i)-2 . . .
A5(1i)-5
10
Risk Analysis
NPW -10,000 A1(1i)-1 A2(1i)-2 . . .
A5(1i)-5
11
Risk Analysis

• Now suppose At iid N(3,000, 250)
• NPW is a linear combination of normal random
• variables.

?
?
12
Risk Analysis

• Now suppose At iid N(3,000, 250)
• NPW is a linear combination of normal random
  variables.
• NPW Normal

?
?
?
Central Limit
13
Risk Analysis

• Now suppose At iid N(3,000, 250)
• NPW is a linear combination of normal random
  variables.
• NPW Normal
• NPW N(?NPW, ?NPW)

?
?
?
?
14
Mean
Recall EZ EX1 EX2 EaXb
aEX b
15
Mean
Recall EZ EX1 EX2 EaXb
aEX b
16
Mean
but, EAt 3,000
17
Mean
-10,000 3,000(P/A, i, 5)
18
Variance
Recall ?2(z) ?2(x) ?2(y) ?2(axb)
a2?2
19
Variance
Recall ?2(z) ?2(x) ?2(y) ?2(axb)
a2?2
20
Variance
but,
(250)2 (1i)-2 (1i)-4 . . . (1i)-10
21
Variance
(250)2 (1i)-2 (1i)-4 . . . (1i)-10
Note that (1i)-2 (1i)-4 . . . (1i)-10
is just a 5 period annuity factor where the
period is 2 years.
22
Variance
(250)2 (1i)-2 (1i)-4 . . . (1i)-10
(250)2(P/A, ieff, 5) , ieff (1i)2 -1
23
Risk Analysis
MARR 15
mNPW -10,000 3,000(P/A, 15, 5) -10,000
3,000(3.3522) 56
24
Risk Analysis
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
25
Risk Analysis
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
26
Risk Analysis
MARR 15
NPW 56 s2NPW 145,894 s 382
NPW ? N(56, 382)
27
Risk Analysis
MARR 15
NPW ? N(56, 382)
28
Risk Analysis
NPW ? N(56, 382)
29
Risk Analysis
NPW ? N(56, 382)
P( Z lt -0.15 ) 0.44
30
For Normal Distribution

• The standard normal table gives you the
  probability of P(Xlta).
• P(Xgta) 1- P(Xlta)
• P(Xlt -a) P(Xgta) 1- P(Xlta)
• P(Xgt-a) P(Xlta)

31
Class Problem

• You are given the following cash flow diagram

Compute the distribution for the Net Present
Worth if the MARR 15.
32
Class Problem
33
Class Problem

• You are given the following cash flow diagram

If the MARR 15, what is the probability this
investment alternative is no good?
34
Class Problem

• ENPW -35,000 10,000(P/A, 15, 5)
• -35,000 10,000(3.3522)
• - 1,478

3002 (2.3343) 210,087
???????
35
Class Problem

• NPW N(-1,478 , 458)
• PNPW lt 0

?
PZ lt 3.27
36
A Critical Thinking
Max Ai ? 10,900
37
Critical Thinking
If Max Ai ? 10,900
NPW -35,000 10,900(P/A, 15, 5) -35,000
10,900(3.3522) 1,539
38
A Twist

• 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.
39
Solution Alternatives

• Assume normality
• Upper/Lower Bounds
• Laplace Transforms
• Transformation/Convolution
• Simulation

40
Simulation
Let us arbitrarily pick a value for A1 and A2 in
the uniform range (5000, 7000). Say A1 5,740
and A2 6,500.
41
Simulation
NPW -10,000 5,740(1.15)-1
6,500(1.15)-2 (93.96)
42
Simulation
We now have one realization of NPW for a given
realization of A1 and A2.
43
Simulation
We now have one realization of NPW for a given
realization of A1 and A2. Choose 2 new values
for A1, A2.
44
A1 6,820 A2 6,218
NPW -10,000 6,820(1.15)-1
6,218(1.15)-2 632.14
45
Summary

• A1 A2 NPW
• 5,740 6,500 (93.96)
• 6,820 6,218 632.14

Choose 2 new values.
46
A1 5,273 A2 6,422
NPW -10,000 5,273(1.15)-1
6,422(1.15)-2 (558.83)
47
Summary

• A1 A2 NPW
• 5,740 6,500 (93.96)
• 6,820 6,218 632.14
• 5,273 6,422 (558.83)

Choose 2 new values.
48
Summary

• 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

49
Simulation

• With enough realizations of A1, A2, and NPW, we
  can begin to get a sense of the distribution of
  the NPW.

50
Simulation

• What we now need is a formal method of selecting
  random values for A1 and A2 to avoid selection
  bias.

51
Simulation

• 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)
52
Simulation

• The uniform has cumulative distribution given by
  

53
Simulation

• Recall that 0 lt F(x) lt 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)

?
54
Simulation

• Recall that 0 lt F(x) lt 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 F-1(p)

?
55
Simulation

• 1. Randomly generate P U(0,1). P .7

?
5,000 lt x lt 7,000
56
Simulation

• 1. Randomly generate P U(0,1). P .7
• 2. Let P F(x).

?
.7
5,000 lt x lt 7,000
57
Simulation

• 1. Randomly generate P U(0,1). P .7
• 2. Let P F(x).
• 3. x F-1(p).

?
.7
5,000 lt x lt 7,000
6,400
58
Formal Derivation

• Recall, for

5,000 lt x lt 7,000. Then
P
59
Formal Derivation

• Solving for x F-1(p),

P
x
60
Formal Derivation

• Solving for x F-1(p),

P
Note 1. P 0 x 5,000 2. P
1 x 7,000
x
61
Risk Analysis (Excel)
62
Class Problem

• 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 gt 1,000
63
Class Problem

• 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.

64
Class Problem
65
Class Problem
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
66
Class Problem
NPW -7,000 5,828(1.15)-1 2,070(1.15)-2
3,079(1.15)-3 1,657
67
Multivariate

• If there are more than one random variable in the
  equation (e.g. NPW, AW, etc.) you need to find
  the mean and variance for the new random
  variable.
• For example, if costs and revenues are both
  random variables, you need to deal with two
  random variables. In this case, the NPW is
  defined in terms of two random variables.

68
Example

• Consider the following information for a 10
  year investment proposal (The first cost, annual
  worth income, and the salvage value are all
  random variables).
• P Pr(P) A Pr(A)
  S Pr(S)
• 100,000 0.6 15,000 0.4
  20,000 0.7
• 125,000 0.4 20,000 0.6
  30,000 0.3
• If the minimum attractive rate of return is 10,
  determine the
• probability that the Net Present Worth in
  greater than 0.

69
Solution

• NPW - P A(P/A, i, n) S (P/F, i, n)
• We need to find the distribution for NPW mean
  and
• variance
• ENPW-E P E(A)(A/P,10,10) E(S)(P/F,10,
  10)
• E(P) 100,000(.6) (125,000)(0.4) 110,000
• E(A) 15,000(0.4) (20,000)(0.6) 18,000
• E(S) 20,000(0.7) (30,000)(0.3) 23,000

70
Solution

• We need to find the variance for P, A, and S.
• VarP E(P2) E(P)2
• E(P2) (0.6)(100,000)2 (0.4)(125,000)2
• VarP 150,000,000
• VarA6,000,000
• VarS21,000,000

71
Solution

• ENPW -110,000 18,000(6.1446) 23,000(
  0.3855)
• 9469.3
• Var NPW 6,000,000(P/A, ieff, 10)
• 21,000,000(P/F, 10, 10)2
• Ieff (1i)2 1
• (10.1)2 -1
• 21
• VarNPW6,000,000(4.054) 21,000,000(2.5937)
• 78791700
• s 8876.47

72
Solution

• NPW N9469.3, 8876.472
• P(NPWgt0) PZgt (9469.3 0)/8876.47)
• P (Zgt1.07)
• 1- P(Zlt1.07)
• 1 0.85769
• 0.14231