ENGM 661Engineering Economics

for Managers

Risk AnalysisRisk Analysis

Motivation

Suppose we have the following cash flow diagram.

NPW = -10,000 + A(P/A, 15, 5)

1 2 3 4 5

A A A A A

10,000

MARR = 15%

Motivation

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

A

p

p

p

2 000 1 6

3 000 2 3

4 000 1 6

, /

, /

, /

Motivation

A

p

p

p

2 000 1 6

3 000 2 3

4 000 1 6

, /

, /

, /

For A = 2,000, we have

NPW = -10,000 + 2,000(P/A, 15, 5)

= -3,296

Motivation

A

p

p

p

2 000 1 6

3 000 2 3

4 000 1 6

, /

, /

, /

For A = 3,000, we have

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

= 56

Motivation

A

p

p

p

2 000 1 6

3 000 2 3

4 000 1 6

, /

, /

, /

For A = 4,000, we have

NPW = -10,000 + 4,000(P/A, 15, 5)

= 3,409

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

Spreadsheet

MARR = 15%P(A ) = 1/6 2/3 1/6

A 2,000 3,000 4,000t CF CF CF0 (10,000) (10,000) (10,000)1 2,000 3,000 4,0002 2,000 3,000 4,0003 2,000 3,000 4,0004 2,000 3,000 4,0005 2,000 3,000 4,000

NPV = (3,296) 56 3,409P(NPV) = 1/6 2/3 1/6

Risk Analysis

1 2 3 4 5

A1 A2 A3 A4 A5

10,000

MARR = 15%

Now Suppose the return in each year is a randomvariable governed by the some probability distribution.

NPW = -10,000 + A1(1+i)-1 + A2(1+i)-2 + . . . + A5(1+i)-5

Risk Analysis

10 000

1

5

, a Att

t

1( )where a it

t

1 2 3 4 5

A1 A2 A3 A4 A5

10,000

MARR = 15%

NPW = -10,000 + A1(1+i)-1 + A2(1+i)-2 + . . . + A5(1+i)-5

Risk Analysis

Now suppose At iid N(3,000, 250)

NPW is a linear combination of normals

Risk Analysis

Now suppose At iid N(3,000, 250)

NPW is a linear combination of normals

NPW Normal

Introduction toProbability & Statistics

The Central Limit TheoremThe Central Limit Theorem

Introduction toProbability & Statistics

The Central Limit TheoremThe Central Limit Theorem

Central Limit

Risk Analysis

Now suppose At iid N(3,000, 250)

NPW is a linear combination of normals

NPW Normal

NPW N(NPW, NPW)

Mean

NPW tt

tE NPW E a A [ ] [ ,10 000

1

5

Recall: E[Z] = E[X1] + E[X2]

E[aX+b] = aE[X] + b

Mean

NPW tt

tE NPW E a A [ ] [ ,10 000

1

5

Recall: E[Z] = E[X1] + E[X2]

E[aX+b] = aE[X] + b

10 000

1

5

, [ ]a E Att

t NPW

Mean

10 000

1

5

, [ ]a E Att

t NPW

but, E[At] = 3,000

10 000

1

5

, [ ]a 3,000tt

NPW

Mean

10 000

1

5

, [ ]a 3,000tt

NPW

= -10,000 + 3,000 a tt

1

5

= -10,000 + 3,000 (1+i)-t

t

1

5

= -10,000 + 3,000(P/A, i, 5)

Variance

2 2 10 000NPW t ta A ( , )

Recall: 2(z) = 2(x) + 2(y)

2(ax+b) = a22

Variance

2 2 10 000NPW t ta A ( , )

Recall: 2(z) = 2(x) + 2(y)

2(ax+b) = a22

2 2 2NPW t Aa t

Variance

but, 2 2250( )At

2 2 2

NPW ta (At)

2 2 2250NPW ta ( )

= (250)2 [(1+i)-2 + (1+i)-4 + . . . + (1+i)-10]

Variance

2 2 2250NPW ta ( )

= (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.

A=(250)2

1 2 3 4 5 6 7 8 9 10

P=2NPW

Variance

2 2 2250NPW ta ( )

= (250)2 [(1+i)-2 + (1+i)-4 + . . . + (1+i)-10]

A=(250)2

1 2 3 4 5 6 7 8 9 10

P=2NPW

= (250)2(P/A, ieff, 5) , ieff = (1+i)2 -1

Risk Analysis

1 2 3 4 5

A1 A2 A3 A4 A5

10,000

MARR = 15%

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

= -10,000 + 3,000(3.3522)

= $56

At iid N(3,000, 250)

Risk Analysis

1 2 3 4 5

A1 A2 A3 A4 A5

10,000

MARR = 15%

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

= -10,000 + 3,000(3.3522)

= $56

At iid N(3,000, 250)

ieff = (1.15)2 - 1 = 32.25%

Risk Analysis

1 2 3 4 5

A1 A2 A3 A4 A5

10,000

MARR = 15%

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

= -10,000 + 3,000(3.3522)

= $56

2NPW = (250)2(P/A, 32.25, 5)

= 62,500(2.3343) = 145,894

At iid N(3,000, 250)

ieff = (1.15)2 - 1 = 32.25%

Risk Analysis

1 2 3 4 5

A1 A2 A3 A4 A5

10,000

MARR = 15%

NPW = $56

2NPW = 145,894 = 382

At iid N(3,000, 250)

NPW 382)

Risk Analysis

1 2 3 4 5

A1 A2 A3 A4 A5

10,000

MARR = 15%

At iid N(3,000, 250)

NPW 382)

N(56, 382)

Risk Analysis

1 2 3 4 5

A1 A2 A3 A4 A5

10,000

NPW 382)

N(56, 382)

P NPW PNPW NPW

NPW

( )

00 56

382

Risk Analysis

1 2 3 4 5

A1 A2 A3 A4 A5

10,000

NPW 382)

N(56, 382)

P NPW PNPW NPW

NPW

( )

00 56

382

= P( Z < -0.15 ) = 0.44

Review

You are given the following cash flow diagram:

Class Problem

5, 000 , 0.6

7, 000 , 0.4

If Ai iid

A1 A2

10,000

Compute the distribution for the Net Present Worth if the MARR = 15%.

Class Problem

MARR = 0.15

t R1 R2 R3 R4

0 (10,000) (10,000) (10,000) (10,000)

1 5,000 5,000 7,000 7,000

2 5,000 7,000 5,000 7,000

NPW (1,871) (359) (132) 1,380

P(NPW) 0.36 0.24 0.24 0.16

You are given the following cash flow diagram:

Class Problem

If the MARR = 15%, what is the probability this investment alternative is no good?

A1 A2 A3 A4 A5

35,000 If Ai iid N(10,000, 300)

Class Problem

E[NPW] = -35,000 + 10,000(P/A, 15, 5)

= -35,000 + 10,000(3.3522)= - 1,478

2(NPW)3002 [(1.15) 2 (1.15) 4 . ..(1.15) 10 ]

= 3002 (2.3343)

= 210,087

Class Problem

NPW N(-1,478 , 458)

P{NPW < 0} =

PNPW

0 ( 1,478)

458

= P{Z < 3.27}1.0

A Critical ThunkA1 A2 A3 A4 A5

35,000

If Ai iid N(10,000, 300)

Max Ai 10,900

A Critical ThunkA1 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

If Ai iid U(5000, 7000)

A Twist

Suppose we have the following cash flow diagram.

A1 A2

10,000

Now how can we compute the distributionof the NPW? MARR = 15%.

Solution Alternatives

Assume normality

Solution Alternatives

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

Simulation

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.

Simulation

A1 A2

10,000

5,740 6,500

10,000

NPW = -10,000 + 5,740(1.15)-1 + 6,500(1.15)-2

= (93.96)

Simulation

A1 A2

10,000

5,740 6,500

10,000

We now have one realization of NPW for a given realization of A1 and A2.

Simulation

A1 A2

10,000

5,740 6,500

10,000

We now have one realization of NPW for a given realization of A1 and A2. Choose 2 new values forA1, A2.

A1 = 6,820 A2 = 6,218

A1 A2

10,000

6,820 6,218

10,000

NPW = -10,000 + 6,820(1.15)-1 + 6,218(1.15)-2

= 632.14

Summary

A1 A2 NPW

5,740 6,500 (93.96)6,820 6,218 632.14

Choose 2 new values.

A1 = 5,273 A2 = 6,422

A1 A2

10,000

5,273 6,422

10,000

NPW = -10,000 + 5,273(1.15)-1 + 6,422(1.15)-2

= (558.83)

Summary

A1 A2 NPW

5,740 6,500 (93.96)6,820 6,218 632.145,273 6,422 (558.83)

Choose 2 new values.

Summary

A1 A2 NPW

5,740 6,500 (93.96)6,820 6,218 632.145,273 6,422 (558.83) . . .6,855 5,947 457.66

Simulation

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

-1,871 0 1,380 NPW

Freq.

Simulation

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

Simulation

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

Recall the uniform

5,000 7,000

1/2,000

f(x)

Simulation

The uniform has cumulative distribution given by:

5,000 7,000

1

F(x)

F( x)

0 , x5,000

1 , x7,000

x 5,0002,000

Simulation

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)

Simulation

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 = F-1(p)

Simulation

1. Randomly generate P U(0,1). P = .7

5,000 7,000

1

F(x)

F( x)x 5,000

2,000

5,000 < x < 7,000

Simulation

1. Randomly generate P U(0,1). P = .72. Let P = F(x).

5,000 7,000

1

F(x)

F( x)x 5,000

2,000

5,000 < x < 7,000

.7

Simulation

1. Randomly generate P U(0,1). P = .72. Let P = F(x).3. x = F-1(p).

5,000 7,000

1

F(x)

F( x)x 5,000

2,000

5,000 < x < 7,000

.7

6,400

Formal Derivation

Recall, for

5,000 7,000

1

F(x)

F( x)x 5,000

2,0005,000 < x < 7,000. Then

P

Px

5 000

7 000 5 000

,

, ,

x 5 000

2 000

,

,

Formal Derivation

Solving for x = F-1(p),

5,000 7,000

1

F(x)

P

x

x P 5 000 2 000, ,

Formal Derivation

Solving for x = F-1(p),

5,000 7,000

1

F(x)

P

x

x P 5 000 2 000, ,

Note: 1. P = 0 x = 5,000

2. P = 1 x = 7,000

Risk Analysis (Excel)

A t = a + (b-a) P NPW = 282

t P = U(0,1) At n NPWn

0 (20,000) 1 (672)

1 0.6979 6,396 2 726

2 0.1273 5,255 3 (903)

3 0.5949 6,190 4 (789)

4 0.8161 6,632 5 (1,006)

5 0.4013 5,803 6 904

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

7,000

A1 A2 A3

F x e x( ) ( , )/ , 1 1 000 3 000 , x > 1,000

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%.

7,000

A1 A2 A3

F x e x( ) ( , )/ , 1 1 000 3 000

Class Problem

P1 e

x 1000

3000

e

x 1000

3000

1 P

x 1000

3000

ln(1 P)

x1,000 3,000 ln(1 P)

Class Problemx1,000 3,000ln(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

Class Problem

7,000

5,8282,0703,079

NPW = -7,000 + 5,828(1.15)-1 + 2,070(1.15)-2 + 3,079(1.15)-3

= 1,657

Class Problem

You are given the following cash flow diagram. The Ai are iid gammas with shape parameter = 4 and scale parameter = 3,000. The density function is given by 7,000

A1 A2 A3

f x x e x( )( )

/

1 , x > 0

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%.

7,000

A1 A2 A3

Class ProblemFor = integer, the cumulative distribution function is given by

Set P = F(x), solve for x

0,!

)/(1)(

1

0

/

Xj

xexF

j

jx

Class Problem

For general (not integer),

F(x) = not analytic

Class Problem

For general (not integer),

F(x) = not analytic

No Inverse

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