EMIS 7300

Decision Trees

Updated 2 December 2005

2

Example 1

There is a 0.65 probability of no growth in the investment climate and 0.35 probability of rapid growth. The payoffs are $500 for a bond investment in a no-growth state, $100 for a bond investment in a rapid-growth state, -$200 for a stock investment in a no-growth state, and a $1100 payoff for a stock investment in a rapid-growth state.

STATE OF NATURE

No Growth (0.65) Rapid Growth (0.35)

DECISIONALTERNATIVE

BondsStocks

$500-$200

$100$1100

3

EMV Criterion

The expected monetary value for the bonds decision alternative is

EMV(bonds) = $500(0.65) + $100(0.35) = $360

The expected monetary value for the stocks decision alternative is

EMV(stocks) = -$200(0.65) + $1100(0.35) = $255

Select bonds under the EMV criterion.

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Example 1 as a Decision Tree

Decision Node

Bonds

Stocks

State-of-Nature Node

1

2

No Growth (0.65)

Rapid Growth (0.35)

No Growth (0.65)

Rapid Growth (0.35)

Payoffs$500

$100

$-200

$1100

5

Calculate EMV for Each Node

Bonds

Stocks

1

2

No Growth (0.65)

Rapid Growth (0.35)

No Growth (0.65)

Rapid Growth (0.35)

Payoffs$500

$100

$-200

$1100

EMV for Node 1 = $360

EMV for Node 2 = $255

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Completed Decision Tree

Bonds

Stocks

1

2

No Growth (0.65)

Rapid Growth (0.35)

No Growth (0.65)

Rapid Growth (0.35)

Payoffs$500

$100

$-200

$1100

$360

$255

$360

Select Bonds

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Expected Value with Perfect Information

• Suppose that before we invest, we can consult an oracle who knows with certainty which state of nature will occur. Our investment policy will be:

1. If the oracle predicts no growth, then invest in bonds and receive a payoff of $500.

2. If the oracle predicts rapid growth, then invest in stocks and receive a payoff of $1,100.

• EVwPI = (0.65)($500) + (0.35)($1,100) = $710.

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Example 2Banana Computer Company manufactures memory chips in batches of ten chips. From past experience, Banana knows that 80% of all batches contain 10% (1 out of 10) defective chips, and 20% of all batches contain 50% (5 out of 10) defective chips. If a good (that is, 10% defective) batch of chips is sent on to the next stage of production, processing costs of $1000 are incurred, and if a bad batch (50% defective) is sent on to the next stage of production, processing costs of $4000 are incurred. Banana also has the option of reworking a batch at a cost of $1000 before sending it to the next stage of production. A reworked batch is sure to be a good batch.

9

Example 2 Continued

• Develop a decision tree for this problem and use it to determine a policy for minimizing Banana’s expected total cost per batch.

10

Decision Tree

Decision Node

State of Nature Node

2Good Batch (0.8)

Bad Batch (0.2)

Send Directly to next stage

-$1,000

-$4,000

Rework every batch -$2,0001

EMV(2)=(0.8)(-$1,000)-(0.2)($4,000) = -$1,600

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Optimal Policy

• Send every batch directly to the next stage

• EMV = -$1,600

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Example 3

• For a cost of $100, Banana can test one chip from each batch in an attempt to determine whether the batch is defective.

1. Develop a decision tree for this problem and use it to determine a policy for minimizing Banana’s expected total cost per batch.

2. What is the expected value of the sample information (EVSI) obtained from testing chips?

3. What is the expected value of perfect information (EVPI) for this problem?

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Policies to Consider

1. Rework every batch before sending it Stage 2. This policy will cost $2000 for every batch: $1000 to rework it and $1000 to process the reworked batch at stage 2.

2. Send every batch to Stage 2 without reworking it. E[cost] = (0.8)($1000) + (0.2)($4000) = $1600

3. Test a chip from each batch and decide what to do based on the result.

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Expected Cost of Policy 3

• Case 1: The chip tested is good (not defective). – The expected cost of sending the batch to stage

2 is ($1000)P(GB|GC) + ($4000)P(BB|GC).

• Case 2: The chip is defective. – The expected cost of sending the batch to stage

2 is ($1000)P(GB|DC) + ($4000)P(BB|DC).

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0.2P(BB)

8.0P(GB)

:iesProbabilitPrior

Banana Computer Solution

5.0BB)|P(DC0.1GB)|P(DC

5.0BB)|P(GC9.0GB)|P(GC

:iesProbabilit lConditionaGiven

.82.0

)2.0)(5.0()8.0)(9.0(

BB)P(BB)|P(GC GB)P(GB)|P(GC

BB)&P(GCGB)&P(GCP(GC)

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Decision Tree

Decision Node

State of Nature Node

Test a Chip

3Defective Chip

(0.18)

Good Chip (0.82)

2Good Batch (0.8)

Bad Batch (0.2)

Send Directly to next stage

-$1,000

-$4,000

Rework

4Good Batch (?)

Bad Batch (1-?)Direct to Stage 2 -$1,100

-$4,100

-$2,100

Rework every batch -$2,0001

Rework

5Good Batch (??)

Bad Batch (1-??)

-$1,100

-$4,100

-$2,100

Direct to Stage 2

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.12.0GC)|P(BB

.88.0

82.0

)8.0)(9.0(

82.0

GB)P(GB)|P(GC

P(GC)

GB)&P(GC

P(GC)

GC)&P(GBGC)|P(GB

Banana Computer Solution: Posterior Probabilities

18

.44.0DC)|P(GB

.56.018.0

)2.0)(5.0(

18.0

BB)P(BB)|P(DCP(DC)

BB)&P(DC

P(DC)

DC)&P(BBDC)|P(BB

Banana Computer Solution: Posterior Probabilities

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Decision Tree

Decision Node

State of Nature Node

Test a Chip

3Defective Chip

(0.18)

Good Chip (0.82)

2Good Batch (0.8)

Bad Batch (0.2)

Send Directly to next stage

-$1,000

-$4,000

Rework

4Good Batch (0.88)

Bad Batch (0.12)Direct to Stage 2 -$1,100

-$4,100

-$2,100

Rework every batch -$2,0001

Rework

5Good Batch (0.44)

Bad Batch (0.56)

-$1,100

-$4,100

-$2,100

Direct to Stage 2

EMV(2)=(0.8)(-$1,000)-(0.2)($4,000) = -$1,600

EMV(4)=(0.88)(-$1,100)-(0.12)($4,100) = -$1,460

EMV(5)=(0.44)(-$1,100)-(0.56)($4,100) = -$2,780

EMV(3)=(0.82)(-$1,460)-(0.18)($2,100) = -$1575.2

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Optimal Policy

• Test one chip– If the chip is good, then send the batch directly

to the next stage– If the chip is defective, then rework the batch

before sending it to the next stage

• EMV = -$1,575.25

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Expected Value of Sample Information (EVSI)

1. The optimal decision is to test a chip.• The expected payoff is -$1,575.25.• If the test were free, the expected payoff would be

-$1,475.25 2. The optimal decision without the test is to send

all batches directly to the next stage • The expected payoff is -$1,600.

3. The expected value of the test (EVSI) is given by (-$1,475.25) – (-$1,600) = $124.75

• If the test cost more than $124.75 we wouldn’t use it.

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EVPI

1. The optimal decision is to pay to test the chip.The expected payoff is -$1,575.25

2. The optimal decision with perfect information is

1. If the batch is good, then send it to the next stage.

2. If the batch is bad, then rework it before sending it to the next stage

3. EMVwPI = (-$1,000)(0.8)+(-2,000)(0.2) = -$1,200.

3. The expected value of perfect information is -$1,200 – (- $1,575.25) = $375.25