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# ENGM 661 Engineering Economics for Managers PowerPoint PPT Presentation

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.

ENGM 661 Engineering Economics for Managers

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## ENGM 661Engineering 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.

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.

### Solution Methodologies

• Bounding

• Analytic

• C.L.T. (Assume Normality)

• Simulation

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

### Bounding

Upper & Lower Bounds

A1 A2 A3 A4 A5

MARR = 15%

1 2 3 4 5

10,000

### Risk Analysis

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

### Risk Analysis

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

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)

5

E

[

NPW

]

E

[

10

,

000

a

A

NPW

t

t

t

1

### Mean

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

### Mean

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

### Mean

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

### Mean

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

2

2

(

10

,

000

a

A

)

NPW

t

t

### Variance

Recall: 2(z) = 2(x) + 2(y)

2(ax+b) = a22

2

2

(

10

,

000

a

A

)

NPW

t

t

2

2

a

2

NPW

A

t

t

### Variance

Recall: 2(z) = 2(x) + 2(y)

2(ax+b)= a22

2

2

a

2

(At)

NPW

t

2

(

A

)

250

2

t

2

2

(

250

)

2

a

NPW

t

### Variance

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

### 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

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 Analysis

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

### 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

A1 A2 A3 A4 A5

1 2 3 4 5

At iid N(3,000, 250)

10,000

### 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

A1 A2 A3 A4 A5

1 2 3 4 5

At iid N(3,000, 250)

10,000

### Risk Analysis

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

### Risk Analysis

NPW  N(56, 382)

= P( Z < -0.15 ) = 0.44

A1 A2 A3 A4 A5

35,000

If Ai iid

N(10,000, 300)

### Class Problem

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

)

]

### Class Problem

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

### Class Problem

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

### A Critical Thunk

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

### 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%.

A1 A2

10,000

### 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.

A1 A2

5,740 6,500

10,000

10,000

### Simulation

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

### Simulation

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

### Simulation

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

### A1 = 6,820 A2 = 6,218

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

= 632.14

### Summary

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

### A1 = 5,273 A2 = 6,422

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

= (558.83)

### Summary

A1 A2 NPW

5,7406,500(93.96)

6,8206,218632.14

5,2736,422(558.83)

Choose 2 new values.

### Summary

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

### Simulation

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

### Simulation

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

### 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)

F(x)

0

,

x

5

,

000



1



x

5

,

000

F

(

x

)



2

,

000



1

,

x

7

,

000



5,000 7,000

### Simulation

The uniform has cumulative distribution given by:

### 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)

F(x)

1

x

5

,

000

F

(

x

)

2

,

000

5,000 7,000

### Simulation

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

### Simulation

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

### Simulation

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

2. Let P = F(x).

3. x = F-1(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

### Formal Derivation

Recall, for

5,000 < x < 7,000. Then

P

F(x)

x

5

,

000

2

,

000

P

1

5,000 7,000

### Formal Derivation

Solving for x = F-1(p),

P

x

F(x)

x

5

,

000

2

,

000

P

1

5,000 7,000

### Formal Derivation

Solving for x = F-1(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 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 > 1,000

A1 A2 A3

7,000

(

x

1

,

000

)

/

3

,

000

F

(

x

)

1

e

### 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%.

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

)

### 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

5,828

3,079

2,070

7,000

### Class Problem

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

### Class Problem

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

### Central Limit Theorem

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

### Central Limit Theorem

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

### Central Limit Theorem

Distribution of NPW

A1 A2 A3

N(-867, 938)

1 2 3

10,000

-3,681

-867

1,947

### Central Limit Theorem

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

### Simulation

P{NPW > 0} = 5/20 = 0.25

### Simulation

Analytic

P{NPW > 0} = 0.188

C.L.T.

P{NPW > 0} = 0.178

Simulation

P{NPW > 0} = 0.25

### @Risk

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