engm 661 engr. economics for managers decisions under uncertainty
DESCRIPTION
Motivation Now suppose that the annual return A is a random variable governed by the discrete distribution: A p p p ,/,/,/TRANSCRIPT
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
Maximize Epected Gain
A1
A2
( 11; p 11
)
(12; p12)
(21; p21)
D1 V12
V21
2. Replace D2 with E[A11]
E[A11]
Maximize Epected Gain
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]
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
Summary:Choose Alternative A1 at D1.
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
Summary:Choose Alternative A1 at D1.Given A1, either outcome 11 or 12 will occur.
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
Summary:Choose Alternative A1 at D1.Given A1, either outcome 11 or 12 will occur.If 11 occurs, choose alternative A11 at D2.
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
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]
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]
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
Total Revenue $ 3.75
$ 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
Total Revenue $ 3.75
$ 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
Total Revenue $ 3.75
$ 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
Value of Perfect Information
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
Value of Perfect Information
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]
Value of Perfect Information
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
Value of Perfect Information
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
Value of Perfect Information
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?
Posterior Probabilities
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
Posterior Probabilities
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
Posterior Probabilities
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
Expected Value Survey
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
Expected Value Survey
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
Expected Value Survey
Want
P(market | good survey) = ?P(no market | bad survey) = ?
Expected Value 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
Expected Value 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
Want P(Market | Survey says Good) = ?
Expected Value Survey
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
Expected Value Survey
Want P(Market | Survey says Good) = ?
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
Expected Value Survey
Want P(Market | Survey says Good) = ?
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
Expected Value Survey
Want P(Market | Survey says Good) = ?
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
Expected Value Survey
Want P(Market | Survey says Good) = ?
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
Expected Value Survey
Want P(Market | Survey says Good) = ?
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
Bayes’ Decision Rule
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
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[A2] > E[A3]
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
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
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
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
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
Laplace PrincipleIf one can not assign probabilities, assume each state equally probable.
Max E[PAi] choose A1
PayoffAlternative Low Gr Med. Gr. High Gr. Expectation
Crane 1 43,000 60,000 68,000 56,943Crane 2 37,000 52,000 75,000 54,612Crane 3 30,000 57,000 80,000 55,611Prob 0.333 0.333 0.333
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
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
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
Suppose probability of market (p) is somewherebetween 30 and 50 percent.
PayoffAlternative Market No Market ExpectationDevelop 400 -100 50Sell Land 90 90 90Chance 0.30 0.70
Sensitivity
Suppose probability of market (p) is somewherebetween 30 and 50 percent.
PayoffAlternative Market No Market ExpectationDevelop 400 -100 150Sell Land 90 90 90Chance 0.50 0.50
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
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