
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
 acts
A1
A2
.
An
Tabular form of reprehensive acts



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.
E
E
E1
E2
.
En
E1 E2 ………En
Tabular form of events.


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.
A
E E1 E2
A1 P11 P12
…… En
P1n
.
A2 P21 P22
.
.
.
.
P2n
Am Am1 Am2 .
Amn
PAY 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.


The probabilities of the states of nature is not known.

Decision is taken on the basis of 4 criteria.
1.
Maxi min or mini max
2.
Maxi max or mini min
3.
Mini max reg.
4.
Laplace.


Maxi min => maximize the minimum

Minimax => minimize the maximum

Maximin : find the pay off using maximin
E E1 E2 E3
A
A1 8 70 50
A2 50 45 40
A3 -25 -10 0

Minimum profit/pay off for
A1 8
A2 40
A3 -25

Miximum pay off of minimum profit.
A2 40

A2 act is chosen.


Minimax :- find the pay off using minimax
A
E E1 E2 E3
A1 50 700 500
A2 10 500 900
A3 100 60 80

Maximum cost
A1 700
A2 900
A3 100 minimax of maximum cost = 100
A3 act is chosen.


Maximax => maximum of maximum profit (optimistic approach)
E E1 E2 E3 E4
A
A1 -5 0 7 0
A2 -4 -3 3 4
A3 -6 -7 6 2

Maximum pay off =
A1 7
A2 4
A3 6

Maximum of maximum pay off =
A1 = 7
A1 = act is chosen according to maximax
Minimum criteria.
Minimine pay off
A1 -5
A2 -4
A3 -7 minimum of minimum cost =
A3 = -7
A3 act is chosen according to minimum

E1 E2 E3 E4
E E1 E2 E3 E4
A
E
A1 18 12 14 9
A
A1 18 12 14 9
A2 15 14 11 11
2)
A2 15 14 11 11
A3 13 16 17 16 A3 13 16 17 16
1)Cal the maximum of E
E1 18
E2 16
E3 17
E4 16
(regret pay off)
3)take max of each row
A1 7
A2 6
A3 5 max reg. minimum of
4) take minimum of this
(max of reg. pay off)
A3 act is chosen

E E1 E2 E3 AV
A

A1 200 200 200 200
A2 175 205 195 195.6
A3 150 180 210 180
 200

Find the average pay off for each act.

Find the maximum av from step(1)
A1 is chosen.


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)


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.
Daily sale
No. of days
10 11 12 13
15  20 40 25  total = 100days


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 cutlets
E2 = ..
.. 11 ..
E3 = ..
.. 12 ..
E4 = ..
.. 13 ..

The 4 acts are:
A1 = stock of 10 cutlets  profit on 1 cutlet = rs.3
A2 = .. .. 11 ..
A3 = .. .. 12 ..
A4 = .. .. 13 ..

A
E 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 net profit is called conditional pay off
Conditional pay off for each act event combination
Pay off for A1.E1 = 10×3 = 30
Pay off for A1.E2 = 10×3 = 30 as 50 on
Pay off for A2.E1 = 10×3 -2 = 28
Pay off for A2.E2 = 11×3 =33 as soon.
P(selling 10 cutlets) = 15/100 = 0.15
P(selling 11 cutlets) = 20/100
P(selling 12 cutlets) = 40/100
P(selling 13 cutlets) = 25/100
= 0.20
= 0.40
= 0.25


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) = 7.5
And so on…
Table for expected conditional pay off
A
A1
A2
A3
E
A4
E1 E2 E3 E4
4.5
3.6
6
4.2
6.6
13.2
8.25
3.9
6.2
14.4
9
5.8
12
13.6
7.5
9.75

EMV (expected monetary value) for A1 = 4.5+ 6 + 12 +
7.5 = 30
EMV (expected monetary value) for A2 = 32.5
EMV (expected monetary value) for A3 = 33.5
EMV (expected monetary value) for A4 = 32.5
Since, EMV is maximum for act 3 i.e. A3 = 33.5
act 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.


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.
Demand probability
16 17 18 19 20
0.1 0.15 0.2
0.25 0.3

Determine how many copies of magazines should he purchases daily by EOL method

stock
A
E demand
0.1 0.15 0.2 0.25 0.3
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
 conditional pay off table
A4 19 23 28 33 38 38
A5 20 20 25 30 35 40
For conditional pay off; cp = 3 & sp = 5
Profit is rs.2 on each magazine.
conditional pay off for
A1E1 = 16 × 2 = 32
A2E1 = 32 – 3 = 29
A3E1 = 32 – 6 = 26

CONDITIONAL OPPORTUNITY LOSS TABLE
A
A1
A2
A3
E E1 E2 E3 E4 E5
0
3
6
2
0
3
4
2
0
6
4
2
8
6
4
A4
A5
9 6
12 9
3
6
0
3
2
0

For E1 = (32~x) , x Є E1

For E2 = (34~x) , x Є E2 and so on…


Expected COL = p(E) × COL
A
E E1 E2 E3 E4 E5
A1 0 0.3
0.8
1.5
2.4
A2 0.3
0 0.4
1 1.8
A3 0.6 0.45
0 0.5
1.2
A4 0.9
0.9
0.6
0 0.6
A5 1.2
1.35
1.2
0.75
0
EOL
5
3.5
2.75
3
4.5
Both COL & EMV are same result
 minimum EOL
A3 is chosen
18 magazine should be purchased