1. DECISION THEORY

2.  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

3.  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

4.  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 E1 E2 ………En E1 E2. En

5.  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 E1E2……En A1P11P12P1n A2P21P22P2n..... AmAm1Am2.Amn PAY OFF TABLE TREE DIAGRAM OF PAY OFF

6.  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

7.  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. Decision under uncertainty.(without problem)

8.  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 A18 A240 A3-25 E A E1E2E3 A187050 A2504540 A3-25-100 A240

9.  Minimax :- find the pay off using minimax  Maximum cost minimax of maximum cost = 100 A3 act is chosen. A1700 A2900 A3100 E A E1E2E3 A150700500 A210500900 A31006080

10.  Maximax => maximum of maximum profit (optimistic approach)  Maximum pay off =  Maximum of maximum pay off = A1 = act is chosen according to maximax Minimum criteria. Minimine pay off minimum of minimum cost = A3 act is chosen according to minimum A17 A24 A36 E A E1E2E3E4 A1-5070 A2-4-334 A3-6-762 A1 = 7 A3 = -7 A1-5 A2-4 A3-7

11. 2) 1)Cal the maximum of E (regret pay off) 3)take max of each row max reg. minimum of 4) take minimum of this (max of reg. pay off) A3 act is chosen Minimax regret or minimax opportunity loss E A E1E2E3E4 A11812149 A2151411 A313161716 E A E1E2E3E4 A11812149 A2151411 A313161716 E118 E216 E317 E416 A17 A26 A35

12.   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 E1E2E3AV A1200 A2175205195195.6 A3150180210180

13.  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)

14.  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 sale10111213 No. of days15204025

15.  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..

16. 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= 0.20 P(selling 12 cutlets) = 40/100= 0.40 P(selling 13 cutlets) = 25/100= 0.25 E A E1 10 E2 11 E3 12 E4 13 10 A130 11 A22833 12 A3263136 13 A424293439

17.  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 E A E1E2E3E4 A14.56127.5 A24.26.613.28.25 A33.96.214.49 A43.65.813.69.75

18. 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

19. 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

20.  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 Demand1617181920 probability0.10.150.20.250.3

21. demand stock  conditional pay off table 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 E A E1 16 E2 17 E3 18 E4 19 E5 20 A1 1632 A2 172934 A3 18263136 A4 1923283338 A5 202025303540 0.1 0.15 0.2 0.25 0.3

22. CONDITIONAL OPPORTUNITY LOSS TABLE  For E1 = (32~x), x Є E1  For E2 = (34~x), x Є E2 and so on… E A E1E2E3E4E5 A102468 A230246 A363024 A496302 A5129630

23.  Expected COL = p(E) × COL  minimum EOL A3 is chosen 18 magazine should be purchased Both COL & EMV are same result E A E1E2E3E4E5EOL A100.30.81.52.45 A20.300.411.83.5 A30.60.4500.51.22.75 A40.9 0.60 3 A51.21.351.20.7504.5

24. THANK YOU