engm 661 engr. economics for managers decisions under uncertainty

ENGM 661Engr. Economics for

ManagersDecisions Under Decisions Under

UncertaintyUncertainty

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%

MotivationNow suppose that the annual return A is a

random variable governed by the discrete distribution:

Appp

2 000 1 63 000 2 34 000 1 6

, /, /, /

MotivationThere 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 4,000p(A) 1/6 2/3 1/6NPW -3,296 56 3,409p(NPW) 1/6 2/3 1/6

Motivation

While we note that we now know the distribution of NPW, we have not made a decision.

1 2 3 4 5

A A A A A

10,000

MARR = 15%

NPWppp

-3 296 1 62 31 6

, / 56 /

3 409, /

Motivation

There are lots of decision rules:1. Maximax/Minimin2. Maximin/Minimax3. Weighted Max4. Minimax Regret5. Dominance6. Expectation/Variance7. Most Probable8. Aspiration-Level

1 2 3 4 5

A A A A A

10,000

MARR = 15%

NPWppp

-3 296 1 62 31 6

, / 56 /

3 409, /

Motivation

For now, let us use the expectation rule which states to use that decision which maximizes (minimizes) expected NPW (EUAW).

1 2 3 4 5

A A A A A

10,000

MARR = 15%

NPWppp

-3 296 1 62 31 6

, / 56 /

3 409, /

Motivation

E[NPW] = -3,296(1/6) + 56(2/3) + 3,409(1/6)

= $ 56.2

E[NPW]> 0 INVEST

1 2 3 4 5

A A A A A

10,000

MARR = 15%NPWppp

-3 296 1 62 31 6

, / 56 /

3 409, /

Decision TreeWe can model this same problem with as a decision tree. Assume our decision alternatives are to invest or not to invest. Then

Invest

Do not invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = -3,296

NPW = 0

Decision TreeWe can model this same problem with as a decision tree. Assume our decision alternatives are to invest or not to invest. Then

Invest

Do not invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = -3,296

NPW = 0

Note: Sequential decisions can be modeled with Decision Trees

Notation

A1

A2

( 11; p 11

)

(12; p12)

(21; p21)

A11

A12

D1

D2

(111; p111)

(112; p112)

(121; p121)

(122; p122)

V111

V112

V121

V122

V12

V21

Notation

A1

A2

( 11; p 11

)

(12; p12)

(21; p21)

A11

A12

D1

D2

(111; p111)

(112; p112)

(121; p121)

(122; p122)

V111

V112

V121

V122

V12

V21

= a decision point

= fork in tree where chance events can influence outcomes

Di = ith decision Aik= kth alternative available given decision 1 was alternative i(; p) = outcome , having associated value v(; p), which occurs with probability pVikj = value associated with branch ikj

Maximize Epected Gain

A1

A2

( 11; p 11

)

(12; p12)

(21; p21)

A11

A12

D1

D2

(111; p111)

(112; p112)

(121; p121)

(122; p122)

V111

V112

V121

V122

V12

V21

1. At decision point D2, calculate expected gains for A11 and A12

E[A11] = p111V111 + p112V112

E[A12] = p121V121 + p122V122

If E[A11] > E[A12] Choose A11


A1

A2

( 11; p 11

)

(12; p12)

(21; p21)

D1 V12

V21

2. Replace D2 with E[A11]

E[A11]


A1

A2

( 11; p 11

)

(12; p12)

(21; p21)

D1 V12

V21

3. Calculate expected gains for alternatives A1 and A2

E[A1] = p11E[A11] + p12V12

E[A2] = p21 V21 = V21

If E[A1] > E[A2] Choose A1

E[A11]


A1

A2

( 11; p 11

)

(12; p12)

(21; p21)

A11

A12

D1

D2

(111; p111)

(112; p112)

(121; p121)

(122; p122)

V111

V112

V121

V122

V12

V21

Summary:Choose Alternative A1 at D1.


A1

A2

( 11; p 11

)

(12; p12)

(21; p21)

A11

A12

D1

D2

(111; p111)

(112; p112)

(121; p121)

(122; p122)

V111

V112

V121

V122

V12

V21

Summary:Choose Alternative A1 at D1.Given A1, either outcome 11 or 12 will occur.


A1

A2

( 11; p 11

)

(12; p12)

(21; p21)

A11

A12

D1

D2

(111; p111)

(112; p112)

(121; p121)

(122; p122)

V111

V112

V121

V122

V12

V21

Summary:Choose Alternative A1 at D1.Given A1, either outcome 11 or 12 will occur.If 11 occurs, choose alternative A11 at D2.


A1

A2

( 11; p 11

)

(12; p12)

(21; p21)

A11

A12

D1

D2

(111; p111)

(112; p112)

(121; p121)

(122; p122)

V111

V112

V121

V122

V12

V21

Summary:Choose Alternative A1 at D1.Given A1, either outcome 11 or 12 will occur.If 11 occurs, choose alternative A11 at D2.Given A11, either outcome 111 or 112 will occurresulting in values gained of V111 or V112.

Problem Revisited

1 2 3 4 5

A A A A A

10,000

Appp

2 000 1 63 000 2 34 000 1 6

, /, /, /

Problem Revisited

1 2 3 4 5

A A A A A

10,000

Appp

2 000 1 63 000 2 34 000 1 6

, /, /, /

Invest

Do not invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = -3,296

NPW = 0

Problem Revisited

Invest

No invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = -3,296

NPW = 0

Problem Revisited

Invest

No invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = -3,296

NPW = 0

E[Invest] = -3,296(1/6) + 56(2/3) + 3,409(1/6)= $ 56.2

56.2

Problem Revisited

E[Invest] = -3,296(1/6) + 56(2/3) + 3,409(1/6)= $ 56.2

E[No Invest] = 0

Invest

No invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = -3,296

NPW = 0

56.2

0

Problem Revisited

Invest

No invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = -3,296

NPW = 0

E[Invest] > E[No Invest]

Choose to Invest

56.2

0

K-Corp

Re-engineer

New Design

do nothing

No Competition (.8)

Competition (.2)

Competition (.5)

No Comp. (.5)

New

do noNo Comp. (.3)

Comp. (.7)

125

-90

40

60

80

20-50

K-Corp

Re-engineer

New Design

do nothing

No Competition (.8)

Competition (.2)

Competition (.5)

No Comp. (.5)

New

do noNo Comp. (.3)

Comp. (.7)

125

-90

40

60

80

20-50

(38)

(60)

K-Corp

Re-engineer

New Design

do nothing

No Competition (.8)

Competition (.2)

Competition (.5)

No Comp. (.5)

New

do noNo Comp. (.3)

Comp. (.7)

125

-90

40

60

80

20-50

(38)

(60)[60]

K-Corp

Re-engineer

New Design

do nothing

No Competition (.8)

Competition (.2)

Competition (.5)

No Comp. (.5)

New

do noNo Comp. (.3)

Comp. (.7)

125

-90

40

60

80

20-50

(38)

(60)[60]

(82)

(50)

(-50)

K-Corp

Re-engineer

New Design

do nothing

No Competition (.8)

Competition (.2)

Competition (.5)

No Comp. (.5)

New

do noNo Comp. (.3)

Comp. (.7)

125

-90

40

60

80

20-50

(38)

(60)[60]

(82)

(50)

(-50)

[82]

Class ProblemPerico is considering a capacity expansion project based on a 3-point sales estimate:

Annual Demand Probability 200,000 0.3 150,000 0.5 75,000 0.2

Perico has 3 alternatives for expansion. The net return on each is dependent on annual sales demand.

Class ProblemTotal Return after investment (in millions) over 5 years:

Annual DemandAlternative 200,000 150,000 75,000Build new plant $ 3.75 $ 2.50 ($ 2.00)Expand existing 1.50 1.00 0.75Do nothing 0.00 0.00 0.00

Class ProblemEach alternative also has an associated capital investment requirement:

Alternative Capital InvestBuild New $1.25 mil.Expand Existing $0.75 mil.Do Nothing $0.00

Class ProblemSet up the decision tree for Perico Inc. Use the resulting tree to determine if Perico should build a new plant, expand the existing facilities, or do nothing.

Class Problem

Decision Tree

A

B

C

Build

New

Expand

Do Nothing

200,000 (.3)

150,000 (.5)

75,000 (.2)200,000 (.3)

150,000 (.5)

75,000 (.2)

0

Total Revenue $ 3.75

$ 2.50($2.00)$ 1.50

$ 1.00$ 0.75

$ 0.00

(1.25

)

(0.75)

(0.0)

Decision Tree

A

B

C

E[R] = 1.125

E[R] = 1.250

E[R] = (0.40)E[R] = 0.450

E[R] = 0.500

E[R] = 0.150

E[R] = 0.000


$ 2.50($2.00)$ 1.50

$ 1.00$ 0.75

$ 0.00

[1.975]

[1.10]

[0.0]

(1.25

)

(0.75)

(0.0)

Decision Tree

A

B

C

E[R] = 1.125

E[R] = 1.250

E[R] = (0.40)E[R] = 0.450

E[R] = 0.500

E[R] = 0.150

E[R] = 0.000


$ 2.50($2.00)$ 1.50

$ 1.00$ 0.75

$ 0.00

[1.975]

[1.10]

[0.0]

Net = .7

25

Net = 0.35

Net = 0.0

Decision TreeE[R] = 1.125

Net = .7

25

A

B

C

E[R] = 1.250

E[R] = (0.40)E[R] = 0.450

E[R] = 0.500

E[R] = 0.150

E[R] = 0.000


$ 2.50($2.00)$ 1.50

$ 1.00$ 0.75

$ 0.00

[1.975]

[1.10]

[0.0]

Net = 0.35

Net = 0.0

Value of Perfect Information

Invest

No invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = -3,296

NPW = 0

56.2

0

Recall our simple cash flow example:

1 2 3 4 5

A A A A A

10,000


Although we expect to make money (at 15%), it is entirely possible we will lose $3,296 if sales are low.

Invest

No invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = -3,296

NPW = 0

56.2

0


Although we expect to make money (at 15%), it is entirely possible we will lose $3,296 if sales are low.

Invest

No invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = -3,296

NPW = 0

56.2

0

Idea: If I knew exactly whether sales will be high,medium, or low, I can know precisely whether or not it is worthwhile for me to invest.

[56.2]


Under certainty, we would not invest if sales are low. We can then modify our decision treeas follows:

Invest

No invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = 0

NPW = 0

605.5

0


That is, under certainty, we will know whether to invest or not. Under these conditions, the expected NPW is now 605.5

Invest

No invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = 0

NPW = 0

605.5

0


Then the expected value of perfect information is:

EVPI = $605.5 - $ 56.2 = $549.3

Invest

No invest

High (1/6)

Medium (2/3)

Low (1/6)

NPW = 3,409

NPW = 56

NPW = 0

NPW = 0

605.5

0

Bayes’ Decision Rule

E[A1] > E[A2]

choose A1

PayoffAlternative Market No Market ExpectationDevelop 400 -100 100Sell Land 90 90 90Chance 0.40 0.60

Decision Tree

Develop

Sell

Market (0.4)

No Market (.6)

400

-100

90

100

90

Posterior Probabilities

Suppose we can do a market survey which would cost us $50,000. Would it be worth it?


Suppose we can do a market survey which would cost us $50,000. Would it be worth it?

Develop

Sell

Market (0.4)

No Market (.6)

400

-100

90

100

90


Develop

Sell

Market (0.4)

No Market (.6)

400

-100

90

100

90

E[payoff perfect info] = .4(400) + .6(90) = 214

E[payoff w/o survey] = 100


Develop

Sell

Market (0.4)

No Market (.6)

400

-100

90

100

90

E[payoff perfect info] = .4(400) + .6(90) = 214

E[payoff w/o survey] = 100

EVPI = 214 - 100 = 114

Expected Value Survey

If we spend $50,000 to conduct survey, we have one of 4 possibilities Survey says good when there is a market Survey says bad when there is no market Survey says good when there is no market Survey says bad when there is a market


Survey says good when there is a market Survey says bad when there is no market Survey says good when there is no market Survey says bad when there is a market

P(good survey | market) = 0.8P(bad survey | no market) = 0.7


Survey says good when there is a market Survey says bad when there is no market Survey says good when there is no market Survey says bad when there is a market

P(good survey | market) = 0.8P(bad survey | no market) = 0.7

P(bad survey | market) = 0.2P(good survey | no market) = 0.3


Want

P(market | good survey) = ?P(no market | bad survey) = ?


No Market Market 0.6 0.4

Surveysays 0.3market

Survey says no 0.7market

Survey0.8 says market

Survey 0.2 says no market


No Market Market 0.6 0.4

Surveysays 0.3market

Survey says no 0.7market

Survey0.8 says market

Survey 0.2 says no market

Want P(Market | Survey says Good) = ?



P(NM) = 0.6 P(M) = 0.4

P GS P GS M P GS NMP GS M P M P GS NM P NM

( ) ( ) ( )( | ) ( ) ( | ) ( ). ( . ) . ( . ).

0 8 0 4 0 3 0 60 5



P(NM) = 0.6 P(M) = 0.4 P(GS) = 0.5

P BS P BS M P BS NMP BS M P M P BS NM P NM

( ) ( ) ( )( | ) ( ) ( | ) ( ). ( . ) . ( . ).

0 2 0 4 0 7 0 605



P(NM) = 0.6 P(M) = 0.4 P(GS) = 0.5 P(BS) = 0.5

P M GS P M GSP GS

P GS M P MP GS

( | ) ( )( )

( | ) ( )( )

. ( . ).

.

0 8 0 405

0 64

0.64



P(NM) = 0.6 P(M) = 0.4 P(GS) = 0.5 P(BS) = 0.5

P NM GS P NM GSP GS

P GS NM P NMP GS

( | ) ( )( )

( | ) ( )( )

. ( . ).

.

0 3 0 605

0 36

0.640.36



P(NM) = 0.6 P(M) = 0.4 P(GS) = 0.5 P(BS) = 0.5

P M BS P M BSP BS

P BS M P MP BS

( | ) ( )( )

( | ) ( )( )

. ( . ).

.

0 2 0 405

016

0.16



P(NM) = 0.6 P(M) = 0.4 P(GS) = 0.5 P(BS) = 0.5

P NM BS P NM BSP BS

P BS NM P NMP BS

( | ) ( )( )

( | ) ( )( )

. ( . ).

.

0 7 0 605

084

0.16

0.84

Decision Tree

Do Survey

No Survey

Favorable

Unfavorable develop

develop

develop

market

none

market

none

sell

market

none

sell

sell

Decision Tree

Do Survey

No Survey

0.5

0.5 develop

develop

develop

0.64

0.36

0.16

0.84

sell

0.4

0.6

sell

sell

350

-150

90

350

-150

90

400

-100

90


E[A1] > E[A3] > E[A2]

choose A1

PayoffAlternative Low Gr Med. Gr. High Gr. Expectation

Crane 1 43,000 60,000 68,000 59,000Crane 2 37,000 52,000 75,000 55,900Crane 3 30,000 57,000 80,000 58,500Prob 0.2 0.5 0.3

Expectation

E[A1] > E[A3] > E[A2]

choose A1



Expectation

E[A1] > E[A2] > E[A3]

choose A1



Expectation

E[A1] > E[A3] > E[A2]

choose A1



Laplace PrincipleIf one can not assign probabilities, assume each state equally probable.

Max E[PAi] choose A1



Expectation-Variance

If E[A1] = E[A2] = E[A3]

choose Aj with min. variance

PayoffAlternative Low Gr Med. Gr. High Gr. Expectation Variance

Crane 1 43,000 60,000 68,000 59,000 76,000,000Crane 2 37,000 52,000 75,000 55,900 188,490,000Crane 3 30,000 57,000 80,000 58,500 302,250,000Prob 0.2 0.5 0.3


E[A1] > E[A2]

choose A1


Sensitivity Payoff

Alternative Market No MarketDevelop 400 -100Sell Land 90 90Chance p 1-p

Suppose probability of market (p) is somewherebetween 30 and 50 percent.

Sensitivity



Sensitivityp Develop Sell0.3 50 900.5 150 90

Sensitivityp Develop Sell0.3 50 900.5 150 90

Sensitivity Plot

0

50

100

150

200

0 0.1 0.3 0.4 0.5

Prob. of Market

Expe

cted

Val

ue

Develop

Sell

Aspiration-Level

Aspiration: max probability that payoff > 60,000

PayoffAlternative Low Gr Med. Gr. High Gr.

Crane 1 43,000 60,000 68,000Crane 2 37,000 52,000 75,000Crane 3 30,000 57,000 80,000Prob 0.2 0.5 0.3

P{PA1 > 60,000} = 0.8P{PA2 > 60,000} = 0.3P{PA3 > 60,000} = 0.3

Choose A2 or A3

Maximin

Select Aj: maxjminkV(jk)e.g., Find the min payoff for each alternative.

Find the maximum of minimums Sell LandChoose best alternative when comparing worst possible outcomes for each alternative.

Alternative Oil DryDrill fer Oil 700 -100Sell Land 90 90Chance 0.25 0.75

engm 661 engr. economics for managers decisions under uncertainty

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