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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 661 engineering economics for managers

ENGM 661Engineering Economics for Managers

Risk Analysis


Learning objectives

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

Solution Methodologies

  • Bounding

  • Analytic

  • C.L.T. (Assume Normality)

  • Simulation


Bounding

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

Lower Bound


Bounding1

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 Bound


Bounding2

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


Risk analysis

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


Risk analysis1

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 analysis2

Risk Analysis

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

NPW is a linear combination of normals


Risk analysis3

Risk Analysis

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

NPW is a linear combination of normals

NPW Normal

Central Limit


Risk analysis4

Risk Analysis

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

NPW is a linear combination of normals

NPW Normal

NPW N(NPW, NPW)


Engm 661 engineering economics for managers

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


Engm 661 engineering economics for managers

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


Engm 661 engineering economics for managers

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


Engm 661 engineering economics for managers

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)


Variance

2

2

(

10

,

000

a

A

)

NPW

t

t

Variance

Recall: 2(z) = 2(x) + 2(y)

2(ax+b) = a22


Variance1

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


Variance2

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]


Variance3

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.


Variance4

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


Risk analysis5

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


Risk analysis6

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


Risk analysis7

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


Risk analysis8

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)


Risk analysis9

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 Analysis

MARR = 15%

NPW  N(56, 382)


Risk analysis10

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)


Risk analysis11

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


Class problem

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?


Class problem1

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




Class problem2

NPW

0

(

1

,

478

)





P







458



1

.

0

Class Problem

NPW N(-1,478 , 458)

P{NPW < 0} =

= P{Z < 3.27}


A critical thunk

A1 A2 A3 A4 A5

35,000

If Ai iid

N(10,000, 300)

A Critical Thunk

Max Ai 10,900


A critical thunk1

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


A twist

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%.


Simulation

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.


Simulation1

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)


Simulation2

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.


Simulation3

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.


A 1 6 820 a 2 6 218

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

Summary

A1 A2 NPW

5,7406,500(93.96)

6,8206,218632.14

Choose 2 new values.


A 1 5 273 a 2 6 422

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)


Summary1

Summary

A1 A2 NPW

5,7406,500(93.96)

6,8206,218632.14

5,2736,422(558.83)

Choose 2 new values.


Summary2

Summary

A1 A2 NPW

5,7406,500(93.96)

6,8206,218632.14

5,2736,422(558.83)

.

.

.

6,8555,947457.66


Simulation4

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.


Simulation5

Simulation

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


Simulation6

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)


Simulation7

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:


Simulation8

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)


Simulation9

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)


Simulation10

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


Simulation11

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


Simulation12

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


Formal derivation

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


Formal derivation1

F(x)

x

5

,

000

2

,

000

P

1

5,000 7,000

Formal Derivation

Solving for x = F-1(p),

P

x


Formal derivation2

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


Risk analysis excel

Risk Analysis (Excel)


Class problem3

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


Class problem4

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%.


Class problem5

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

)

Class Problem


Class problem6

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


Class problem7

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


Class problem8

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.


Central limit theorem

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

Preliminary


Central limit theorem1

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


Central limit theorem2

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


Central limit theorem3

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


Central limit theorem4

A1 A2 A3

N(-867, 938)

1 2 3

10,000

-3,681

-867

1,947

Central Limit Theorem

Distribution of NPW


Engm 661 engineering economics for managers

P{NPW >0} = .031

.063

.031

.016

.031

.016

.188


Engm 661 engineering economics for managers

Analytic

P{NPW > 0} = 0.188

C.L.T.

P{NPW > 0} = 0.178


Simulation13

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


Simulation14

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


Simulation15

Simulation


Simulation16

Simulation


Simulation17

Simulation

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


Simulation18

Simulation

Analytic

P{NPW > 0} = 0.188

C.L.T.

P{NPW > 0} = 0.178

Simulation

P{NPW > 0} = 0.25


@risk

@Risk


@risk1

@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


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