DECISION

THEORY

It’s deals with a very scientific and quantitative way of coming to decision.

It has 4 phases.1.Action or acts.2.State of nature or events or outcome.3.Pay off and pay off table or pay off matrix.Decision A decision problem may be represented by tree

diagram

Decision making problems deals with the selection of single act from a set of acts.

There can be 2 or more acts denoted by A1,A2,A3….An.

action space A = {A1 A2 A3 ……….An} Decision tree of acts

Tabular form of reprehensive acts

Action or acts

action acts

A1 A2 ………An

A1 A2 . An

Each act is associated with one or more events or state of natures.

There events are the outcome of consequence of an act.

Events are denoted by E1 E2 ….En E = {E1 E2 ………En} is a set of events. Tree diagram of events.

Tabular form of events.

Events or state of natures

E E

E1 E2 ………En

E1 E2 . En

In decision problems it is required to measure the degree to which the decision maker’s objectives is achieved.

Monetary value is used or a measure to represent achievement or lack of achievement.

This monetary gain or loss is called a pay off. Pay off is expressed as profit, loss cost satisfaction

etc.

Pay off & pay off table

E A

E1 E2 …… En

A1 P11 P12 P1nA2 P21 P22 P2n. . . . .Am

Am1

Am2

. AmnPAY OFF TABLE TREE DIAGRAM OF PAY OFF

Once a pay off table is read no its turn to some decision.

There are 3 decisions making situations.1. Decision under uncertainty.(without

problem)2. Decision under risk.(with problem)3. Decision under certainty.

Decision making situations

The probabilities of the states of nature is not known.

Decision is taken on the basis of 4 criteria.

1. Maxi min or mini max2. Maxi max or mini min3. Mini max reg.4. Laplace.

Decision under uncertainty.(without problem)

Maxi min => maximize the minimum Minimax => minimize the maximum Maximin : find the pay off using maximin

Minimum profit/pay off for

Miximum pay off of minimum profit.

A2 act is chosen.

Pacimistic approach

A1 8A2 40A3 -25

E A

E1 E2 E3

A1 8 70 50A2 50 45 40A3 -25 -10 0

A2 40

Minimax :- find the pay off using minimax

Maximum cost

minimax of maximum

cost = 100A3 act is chosen.

A1 700

A2 900

A3 100

E A

E1 E2 E3

A1 50 700

500

A2 10 500

900

A3 100

60 80

Maximax => maximum of maximum profit (optimistic approach)

Maximum pay off =

Maximum of maximum pay off = A1 = act is chosen according to

maximaxMinimum criteria.Minimine pay off

minimum of minimum cost =

A3 act is chosen according to minimum

A1 7A2 4A3 6

E A

E1

E2

E3

E4

A1 -5 0 7 0A2 -4 -3 3 4A3 -6 -7 6 2

A1 = 7

A3 = -7

A1 -5A2 -4A3 -7

2)

1)Cal the maximum of E (regret pay off) 3)take max of each rowmax reg. minimum of4)take minimum of this

(max of reg. pay off)A3 act is chosen

Minimax regret or minimax opportunity loss

E A

E1

E2

E3

E4

A1 18

12

14

9

A2 15

14

11

11

A3 13

16

17

16

E A

E1

E2

E3

E4

A1 18

12

14

9

A2 15

14

11

11

A3 13

16

17

16

E1 18E2 16E3 17E4 16

A1 7A2 6A3 5

200

Find the average pay off for each act. Find the maximum av from step(1)

A1 is chosen.

Laplase(equally likely criteria)

E A

E1 E2 E3 AV

A1 200

200

200

200

A2 175

205

195

195.6

A3 150

180

210

180

In such problems uncertainty is there but probability is given may be from past experience.

In such problems 2 methods are used:1. Using EMV(expected monetary value)2. Using EOL(expected opportunity loss)

Decision making under risk(probability given)

A baker buys veg cutlet at rs.2 & sell it for rs.5. at the end of the day unsold veg cutlets are given to the poor for free of cost.

The following table shows the sales of veg cutlets during the past 100 days.

total = 100days

Decision under risk by(EMV) method consider

Daily sale 10 11

12

13

No. of days 15 20

40

25

Now the question is how many veg cutlets the baker has to stock every day in order to maximise his profit?

The 4 events are:E1 = demand for 10 cutletsE2 = .. .. 11 ..E3 = .. .. 12 ..E4 = .. .. 13 ..

The 4 acts are:A1 = stock of 10 cutlets profit on 1 cutlet

= rs.3A2 =.. .. 11 ..A3 =.. .. 12 ..A4 =.. .. 13 ..

net profit is called conditional pay off

Conditional pay off for each act event combinationPay off for A1.E1 = 10×3 = 30Pay off for A1.E2 = 10×3 = 30 as 50 onPay off for A2.E1 = 10×3 -2 = 28Pay off for A2.E2 = 11×3 =33 as soon.P(selling 10 cutlets) = 15/100 = 0.15P(selling 11 cutlets) = 20/100 = 0.20P(selling 12 cutlets) = 40/100 = 0.40P(selling 13 cutlets) = 25/100 = 0.25

E A

E1 10

E2 11

E3 12

E4 13

10 A1

30 30 30 30

11 A2

28 33 33 33

12 A3

26 31 36 36

13 A4

24 29 34 39

Expected conditional pay off is given by the multiplying each conditional pay off by the corresponding probability,

expected conditional pay off for A1.E1 = 30(.15) = 4.5 expected conditional pay off for A1.E2 = 30(.20) = 6 expected conditional pay off for A1.E3 = 30(.40) = 12 expected conditional pay off for A1.E4 = 30(.25) = 7.5And so on…Table for expected conditional pay off

E A

E1 E2 E3 E4

A1 4.5 6 12 7.5A2 4.2 6.6 13.

28.25

A3 3.9 6.2 14.4

9

A4 3.6 5.8 13.6

9.75

EMV (expected monetary value) for A1 = 4.5+ 6 + 12 + 7.5 = 30EMV (expected monetary value) for A2 = 32.5EMV (expected monetary value) for A3 = 33.5EMV (expected monetary value) for A4 = 32.5

Since, EMV is maximum for act 3 i.e. A3 = 33.5act A3 is chosen.

i.e. 12 veg cutlets are to be stocked every day for maximum profit

It is same as (EMV) method only the difference is; After finding the conditional pay off regret pay off

has to found. This new table is called conditional opportunity loss table(COL).

The product of col and the corresponding probability given expected COL.

The sum of all expected COL is act wise given EOL. The minimum of EOL is selected as or act.

Decision under risk by(EOL) method

A newspaper boy purchases magazines at rs.3 each & sales them at rs.5 each. He cannot return the unsold magazines. The probability distribution of the demand for the magazine is given below.

Determine how many copies of magazines should he purchases daily by EOL method

Demand 16

17 18 19 20

probability 0.1

0.15

0.2

0.25

0.3

demand

stock conditional pay off table

For conditional pay off;cp = 3 & sp = 5

Profit is rs.2 on each magazine.conditional pay off forA1E1 = 16 × 2 = 32A2E1 = 32 – 3 = 29A3E1 = 32 – 6 = 26

E A

E1 16

E2 17

E3 18

E4 19

E5 20

A1 16

32 32 32 32 32

A2 17

29 34 34 34 34

A3 18

26 31 36 36 36

A4 19

23 28 33 38 38

A5 20

20 25 30 35 40

0.1 0.15 0.2 0.25 0.3

CONDITIONAL OPPORTUNITY LOSS TABLE

For E1 = (32~x) , x Є E1 For E2 = (34~x) , x Є E2 and so on…

E A

E1 E2 E3 E4 E5

A1 0 2 4 6 8A2 3 0 2 4 6A3 6 3 0 2 4A4 9 6 3 0 2A5 12 9 6 3 0

Expected COL = p(E) × COL

minimum EOLA3 is chosen18 magazine should be

purchased

Both COL & EMV are same result

E A

E1 E2 E3 E4 E5 EOL

A1 0 0.3 0.8

1.5 2.4 5

A2 0.3

0 0.4

1 1.8 3.5

A3 0.6

0.45

0 0.5 1.2 2.75

A4 0.9

0.9 0.6

0 0.6 3

A5 1.2

1.35

1.2

0.75

0 4.5

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