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Decision Theory and Decision Trees salesmen. On the other hand, if the new line is handled by the existing sales force, using the existing facilities, the initial investment would only be Rs 30,000, principally for training his present salesmen. The new product sells for Rs 250. The representative normally receives 20 per cent of the sales price on each unit sold, of which 10 per cent is paid as commission to handle the new product. The manufacturer offers to pay 60 per cent of the sale price of each unit sold to the representative, if the representative sets up a separate sales organization. Otherwise the normal 20 per cent will be paid. In either case the salesman gets a 10 per cent commission. Based on the size of the

territory and their experience with other products, the representative estimates the following probabilities for annual sales of the new product: Sales (in units) : Probability :

1,000 0.10

2,000 0.15

3,000 0.40

4,000 0.30

5,000 0.05

(a) Set up a regret table. (b) Find the expected regret of each course of action. (c) Which course of action would have been best under the maximin criterion?

HINTS AND ANSWERS 1. 2. 3. 4.

(i) S2, (ii) S2 or S3, (iii) S3, (iv) S1, (v) S2 (a) Full, (b) Minimal, (c) Full or partial, (d) Partial (a) S3; Rs 150 (b) S3; Rs 80 (c) S1; Rs 40 (d) S3; Rs 55 Choose A: Rs 120, Choose B: Rs 300, Choose C: Rs 176.6, Choose C: Rs 50 5. Let S1 = install new sales facilities Let S2 = continue with existing sales facilities.

11.5

Therefore, payoff function corresponding to S1 and S2 would be S1 = –1,00,000 + 250 × {(30 – 10)/100} × α = –1,00,000 + 50α S2 = –30,000 + 250 × {(20 – 10)/100} × α = – 30,000 + 25α Equating the two, we get – 1,00,000 + 50α = – 30,000 + 25α or α = 2,800.

DECISION-MAKING UNDER RISK

In this decision-making environment, decision-maker has sufficient information to assign probability to the likely occurrence of each outcome (state of nature). Knowing the probability distribution of outcomes (states of nature), the decision-maker needs to select a course of action resulting a largest expected (average) payoff value. The expected payoff is the sum of all possible weighted payoffs resulting from choosing a decision alternative. The widely used criterion for evaluating decision alternatives (courses of action) under risk is the Expected Monetary Value (EMV) or Expected Utility.

11.5.1

Expected Monetary Value (EMV)

The expected monetary value (EMV) for a given course of action is obtained by adding payoff values multiplied by the probabilities associated with each state of nature. Mathematically, EMV is stated as follows: m

EMV (Course of action, Sj) = Σ pij pi i =1

where

m = number of possible states of nature p i = probability of occurrence of state of nature, Ni pij = payoff associated with state of nature Ni and course of action, Sj

Expected monetary value is obtained by adding payoffs for each course of action, multiplied by the probabilities associated with each state of nature.

The Procedure 1. 2. 3.

Construct a payoff matrix listing all possible courses of action and states of nature. Enter the conditional payoff values associated with each possible combination of course of action and state of nature along with the probabilities of the occurrence of each state of nature. Calculate the EMV for each course of action by multiplying the conditional payoffs by the associated probabilities and adding these weighted values for each course of action. Select the course of action that yields the optimal EMV.

Example 11.5 Mr X flies quite often from town A to town B. He can use the airport bus which costs Rs 25 but if he takes it, there is a 0.08 chance that he will miss the flight. The stay in a hotel costs Rs 270 with a 0.96 chance of being on time for the flight. For Rs 350 he can use a taxi which will make 99 per cent chance of being on time for the flight. If Mr X catches the plane on time, he will conclude a business transaction that will produce a profit of Rs 10,000, otherwise he will lose it. Which mode of transport should Mr X use? Answer on the basis of the EMV criterion.

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348 Operations Research: Theory and Applications Solution

Computation of EMV associated with various courses of action is shown in Table 11.3. Courses of Action

States of Nature

Catches the flight

Bus Cost

10,000 – 25 0.92 = 9,975

Miss the flight

Table 11.3

– 25

Stay in Hotel

Prob. Expected Value

0.08

Expected monetary value (EMV)

Cost

Prob.

9,177

10,000 – 270 = 9,730

0.96

– 2.0

– 270

0.04

Taxi

Expected Value

Cost

Prob.

Expected Value

9,340.80

10,000 – 350 = 9,650

0.99

9,553.50

– 10.80

– 350

0.01

– 3.50

9,175

9,330

9,550

Since EMV associated with course of action ‘Taxi’ is largest (= Rs 9,550), it is the logical alternative. Example 11.6 The manager of a flower shop promises its customers delivery within four hours on all flower orders. All flowers are purchased on the previous day and delivered to Parker by 8.00 am the next morning. The daily demand for roses is as follows. Dozens of roses : Probability :

70 0.1

80 0.2

90 0.4

100 0.3

The manager purchases roses for Rs 10 per dozen and sells them for Rs 30. All unsold roses are donated to a local hospital. How many dozens of roses should Parker order each evening to maximize its profits? What is the optimum expected profit? [Delhi Univ., MBA, Dec. 2004] Solution The quantity of roses to be purchased per day is considered as ‘course of action’ and the daily demand of the roses is considered as a ‘state of nature’ because demand is uncertain with known probability. From the data, it is clear that the flower shop must not purchase less than 7 or more than 10 dozen roses, per day. Also each dozen roses sold within a day yields a profit of Rs (30 – 10) = Rs 20 and otherwise it is a loss of Rs 10. Thus Marginal profit (MP) = Selling price – Cost = 30 – 10 = Rs 20 Marginal loss (ML) = Loss on unsold roses = Rs 10 Using the information given in the problem, the various conditional profit (payoff) values for each combination of decision alternatives and state of nature are given by Conditional profit = MP × Roses sold – ML × Roses not sold

=

RS 20D, T 20D − 10(S − D) = 30D − 10S ,

if D ≥ S if D < S

where D = number of roses sold within a day and S = number of roses stocked. The resulting conditional profit values and corresponding expected payoffs are computed in Table 11.4. States of Nature (Demand per Day)

Table 11.4 Conditional Profit Value (Payoffs)

Probability

Conditional Profit (Rs) due to Courses of Action (Purchase per Day)

Expected Payoff (Rs) due to Courses of Action (Purchase per Day)

(1)

70 (2)

80 (3)

90 (4)

100 (5)

70 (1)×(2)

80 (1)×(3)

90 (1)×(4)

100 (1)×(5)

70

0.1

140

130

120

110

14

13

12

11

80

0.2

140

160

150

140

28

32

30

28

90

0.4

140

160

180

170

56

64

72

68

100

0.3

140

160

180

200

42

48

54

60

140

157

168

167

Expected monetary value (EMV)

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Decision Theory and Decision Trees

349

Since the highest EMV of Rs 168 corresponds to the course of action 90, the flower shop should purchase nine dozen roses everyday. Example 11.7 A retailer purchases cherries every morning at Rs 50 a case and sells them for Rs 80 a case. Any case that remains unsold at the end of the day can be disposed of the next day at a salvage value of Rs 20 per case (thereafter they have no value). Past sales have ranged from 15 to 18 cases per day. The following is the record of sales for the past 120 days. Cases sold : Number of days :

15 12

16 24

17 48

18 36

Find out how many cases should the retailer purchase per day in order to maximize his profit. [Delhi Univ., MCom, 2000; Ajmer Univ., MBA, 2003]

Solution Let Ni (i = 1, 2, 3, 4) be the possible states of nature (daily likely demand) and Sj ( j = 1, 2, 3, 4) be all possible courses of action (number of cases of cherries to be purchased). Marginal profit (MP) = Selling price – Cost = Rs (80 – 50) = Rs 30 Marginal loss (ML) = Loss on unsold cases = Rs (50 – 20) = Rs 30 The conditional profit (payoff) values for each combination of decision alternatives and state of nature are given by Conditional profit = MP × Cases sold – ML × Cases unsold = (80 – 50) (Cases sold) – (50 –20) (Cases unsold)  30S =  (80 – 50) S – 30(N – S) = 60S – 30N 

if S ≥ N if S < N

The resulting conditional profit values and corresponding expected payoffs are computed in Table 11.5. States of Nature (Demand per Week)

Probability

(1)

Conditional Profit (Rs) due to Courses of Action (Purchase per Day) 15 (2)

16 (3)

17 (4)

Expected Payoff (Rs) due to Courses of Action (Purchase per Day)

18 (5)

15 (1)×(2)

16 (1)×(3)

17 (1)×(4)

18 (1)×(5)

15

0.1

450

420

390

360

45

42

39

36

16

0.2

450

480

450

420

90

96

90

84

17

0.4

450

480

510

480

180

192

204

192

18

0.3

450

480

510

540

135

144

153

162

450

474

486

474

Expected monetary value (EMV)

Table 11.5 Conditional Profit Value (Payoffs)

Since the highest EMV of Rs 486 corresponds to the course of action 17, the retailer must purchase 17 cases of cherries every morning. Example 11.8 The probability of demand for hiring cars on any day in a given city is as follows: No. of cars demanded Probability

: :

0 0.1

1 0.2

2 0.3

3 0.2

4 0.2

Cars have a fixed cost of Rs 90 each day to keep the daily hire charges (variable costs of running) Rs 200. If the car-hire company owns 4 cars, what is its daily expectation? If the company is about to go into business and currently has no car, how many cars should it buy? Solution Given that Rs 90 is the fixed cost and Rs 200 is variable cost. The payoff values with 4 cars at the disposal of decision-maker are calculated as under: No. of cars demanded : 0 1 2 3 4 Payoff : 0 – 90 × 4 200 – 90 × 4 400 – 90 × 4 600 – 90 × 4 800 – 90 × 4 (with 4 cars) = – 360 = – 160 = 40 = 240 = 440

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350 Operations Research: Theory and Applications Thus, the daily expectation is obtained by multiplying the payoff values with the given corresponding probabilities of demand: Daily Expectation = (– 360)(0.1) + (– 160)(0.2) + (40)(0.3) + (240)(0.2) + (440)(0.2) = Rs 80 The conditional payoffs and expected payoffs for each course of action are shown in Tables 11.6 and 11.7. Demand of Cars

0 1 2 3 4

Table 11.6 Conditional Payoff Values

0.1 0.2 0.3 0.2 0.2

Demand of Cars

0 1 2 3 4

Table 11.7 Expected Payoffs and EMV

Probability

0.1 0.2 0.3 0.2 0.2

EMV

Conditional Payoff (Rs) due to Decision to Purchase Cars (Course of Action)

Probability 0

1

2

3

0 0 0 0 0

– 90 110 110 110 110

– 180 20 220 220 220

– 270 – 70 130 330 330

4 – 360 – 160 40 240 440

Conditional Payoff (Rs) due to Decision to Purchase Cars (Course of Action) 0

1

2

3

4

0 0 0 0 0

–9 22 33 22 22

– 18 4 66 44 44

– 27 – 14 39 66 66

– 36 – 32 12 48 88

0

90

140

130

80

Since the EMV of Rs 140 for the course of action 2 is the highest, the company should buy 2 cars.

11.5.2

Expected Opportunity Loss (EOL)

Expected opportunity loss (EOL), also called expected value of regret, is an alternative decision criterion for decision making under risk. The EOL is defined as the difference between the highest profit (or payoff ) and the actual profit due to choosing a particular course of action in a particular state of nature. Hence, EOL is the amount of payoff that is lost by not choosing a course of action resulting to the minimum payoff in a particular state of nature. A course of action resulting to the minimum EOL is preferred. Mathematically, EOL is stated as follows. m

EOL (State of nature, Ni ) = Σ lij pi i =1

where

lij = opportunity loss due to state of nature, Ni and course of action, Sj p i = probability of occurrence of state of nature, Ni

The Procedure 1. Prepare a conditional payoff values matrix for each combination of course of action and state of nature along with the associated probabilities. 2. For each state of nature calculate the conditional opportunity loss (COL) values by subtracting each payoff from the maximum payoff. 3. Calculate the EOL for each course of action by multiplying the probability of each state of nature with the COL value and then adding the values. 4. Select a course of action for which the EOL is minimum. Example 11.9 A company manufactures goods for a market in which the technology of the product is changing rapidly. The research and development department has produced a new product that appears to have potential for commercial exploitation. A further Rs 60,000 is required for development testing. The company has 100 customers and each customer might purchase, at the most, one unit of the product. Market research suggests that a selling price of Rs 6,000 for each unit, with total variable costs of manufacturing and selling estimate as Rs 2,000 for each unit.

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Decision Theory and Decision Trees

351

From previous experience, it has been possible to derive a probability distribution relating to the proportion of customers who will buy the product as follows: Proportion of customers : 0.04 0.08 0.12 0.16 0.20 Probability : 0.10 0.10 0.20 0.40 0.20 Determine the expected opportunity losses, given no other information than that stated above, and state whether or not the company should develop the product. Solution If p is the proportion of customers who purchase the new product, the company’s conditional profit is: (6,000 – 2,000) × 100 p – 60,000 = Rs (4,00,000 p – 60,000). Let Ni (i = 1, 2, . . ., 5) be the possible states of nature, i.e. proportion of the customers who will buy the new product and S1 (develop the product) and S2 (do not develop the product) be the two courses of action. The conditional profit values (payoffs) for each pair of Nis and Sjs are shown in Table 11.8. Proportion of Customers (State of Nature)

0.04 0.08 0.12 0.16 0.20

Conditional Profit = Rs (4,00,000 p – 60,000) Course of Action S1 (Develop)

S2 (Do not Develop)

– 44,000 – 28,000 – 12,000 4,000 20,000

0 0 0 0 0

Table 11.8 Conditional Profit Values (Payoffs)

Opportunity loss values are shown in Table 11.9. Proportion of Customers (State of Nature)

Probability

0.04 0.08 0.12 0.16 0.20

0.1 0.1 0.2 0.4 0.2

Conditional Profit (Rs)

Opportunity Loss (Rs)

S1

S2

S1

S2

– 44,000 – 28,000 – 12,000 4,000 20,000

0 0 0 0 0

44,000 28,000 12,000 0 0

0 0 0 4,000 20,000

Using the given estimates of probabilities associated with each state of nature, the expected opportunity loss (EOL) for each course of action is given below: EOL (S1) = 0.1 (44,000) + 0.1 (28,000) + 0.2 (12,000) + 0.4 (0) + 0.2 (0) = Rs 9,600 EOL (S2) = 0.1 (0) + 0.1 (0) + 0.2 (0) + 0.4 (4,000) + 0.2 (20,000) = Rs 5,600 Since the company seeks to minimize the expected opportunity loss, the company should select course of action S2 (do not develop the product) with minimum EOL.

11.5.3

Expected Value of Perfect Information (EVPI)

If decision-makers can get perfect (complete and accurate) information about the occurrence of various states of nature, then choosing a course of action that yields the desired payoff in the presence of any state of nature is easy. The EMV or EOL criterion helps the decision-maker to select a particular course of action that optimizes the expected payoff, without any additional information. Expected value of perfect information (EVPI) represents the maximum amount of money required to pay for getting additional information about the occurrence of various states of nature before arriving to a decision. Mathematically, it is stated as: EVPI = (Expected profit with perfect information) – (Expected profit without perfect information) =

Table 11.9 Opportunity Loss Values

Expected value of perfect information is an average (or expected) value of an additional information if it were of any worth.

m

pij ) – EMV* ∑ pi max( j

i =1

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352 Operations Research: Theory and Applications where

EVPI provides an instant way to check whether getting any additional information might be worthwhile.

pij = best payoff when action, Sj is taken in the presence of state of nature, Ni p i = probability of state of nature, Ni EMV* = maximum expected monetary value

Example 11.10 A company needs to increase its production beyond its existing capacity. It has narrowed down on two alternatives in order to increase the production capacity: (a) expansion, at a cost of Rs 8 million, or (b) modernization at a cost of Rs 5 million. Both approaches would require the same amount of time for implementation. Management believes that over the required payback period, demand will either be high or moderate. Since high demand is considered to be somewhat less likely than moderate demand, the probability of high demand has been set at 0.35. If the demand is high, expansion would gross an estimated additional Rs 12 million but modernization would only gross an additional Rs 6 million, due to lower maximum production capability. On the other hand, if the demand is moderate, the comparable figures would be Rs 7 million for expansion and Rs 5 million for modernization. (a) Calculate the conditional profit in relation to various action-and-outcome combinations and states of nature. (b) If the company wishes to maximize its expected monetary value (EMV), should it modernize or expand? (c) Calculate the EVPI. (d) Construct the conditional opportunity loss table and also calculate EOL. [Delhi Univ, MBA, 2004] Solution (a) States of nature: High demand and Moderate demand, and Courses of action: Expand and Modernize. Since probability of high demand is estimated at 0.35, the probability of moderate demand must be (1 – 0.35) = 0.65. The calculations for conditional profit values are shown in Table 11.10. State of Nature (Demand)

Conditional Profit (million Rs) due to Course of Action Expand (S1)

Table 11.10 Conditional Profit Table

High demand (N1)

12 – 8 =

Moderate demand (N2)

Modernize (S2)

4

6 – 5 = 1

7 – 8 = –1

5 – 5 = 0

(b) The payoff table (Table 11.10) can be rewritten as follows along with the given probabilities of states of nature. State of Nature (Demand)

Probability

Conditional Profit (million Rs) Due to Course of Action Expand

Table 11.11 Payoff Table

Modernize

High demand

0.35

4

1

Moderate demand

0.65

–1

0

The calculation of EMV for each course of action S1 and S2 is given below: EMV(S1) = 0.35(4) + 0.65( – 1) = Rs 0.75 million EMV(S2) = 0.35(1) + 0.65(0) = Rs 0.35 million Since EMV (S1) = 0.75 million is maximum, the company must choose course of action S1(expand). (c) To calculate EVPI, first calculate EPPI by choosing optimal course of action for each state of nature. Multiply conditional profit associated with each course of action by the given probability to get weighted profit, and then add these weights as shown in Table 11.12. State of Nature (Demand)

High demand Moderate demand Table 11.12

Probability

0.35 0.65

Optimal Course of Action

S1 S2

Profit from Optimal Course of Action (Rs million) Conditional Profit

Weighted Profit

4 0

4 × 0.35 = 1.40 0 × 0.65 = 0 EPPI = 1.40

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Decision Theory and Decision Trees

If optimal EMV* = Rs 0.75 million corresponding to the course of action S1, then EVPI = EPPI – EMV(S1) = 1.40 – 0.75 = Rs 0.65 million. In other words, to get perfect information on demand pattern (high or moderate), company should consider paying up to Rs 0.65 million. (d) The opportunity loss values are shown in Table 11.13. State of Nature (Demand)

Probability

High demand (N1) Moderate demand (N2)

0.35 0.65

Conditional Profit (Rs million) Due to Course of Action

Conditional Opportunity Loss (Rs million) Due to Course of Action

S1

S2

S1

S2

4 –1

1 0

0 1

3 0

Table 11.13

Since probabilities associated with each state of nature, P (N1) = 0.35, and P(N2) = 0.65, the expected opportunity losses for the two courses of action are: EOL(S1) = 0.35(0) + 0.65(1) = Re 0.65 million EOL(S2) = 0.35(3) + 0.65(0) = Rs 1.05 million Since the expected opportunity loss, EOL (S1) = Re 0.65 million minimum, decision-maker must select course of action S1, so as to have smallest expected opportunity loss. Example 11.11 A certain piece of equipment has to be purchased for a construction project at a remote location. This equipment contains an expensive part that is subject to random failure. Spares of this part can be purchased at the same time the equipment is purchased. Their unit cost is Rs 1,500 and they have no scrap value. If the part fails on the job and no spare is available, the part will have to be manufactured on a special order basis. If this is required, the total cost including down time of the equipment, is estimated at Rs 9,000 for each such occurrence. Based on previous experience with similar parts, the following probability estimates of the number of failures expected over the duration of the project are provided below: Failure : 0 1 2 Probability : 0.80 0.15 0.05 (a) Determine optimal EMV* and optimal number of spares to purchase initially. (b) Based on opportunity losses, determine the optimal course of action and optimal value of EOL. (c) Determine the expected profit with perfect information and expected value of perfect information. Solution (a) Let N1 (no failure), N2 (one failure) and N3 (two failures) be the possible states of nature (i.e. number of parts failures or number of spares required). Similarly, let S1 (no spare purchased), S2 (one spare purchased) and S3 (two spares purchased) be the possible courses of action. The conditional costs for each pair of course of action and state of nature is shown in Table 11.14. State of Nature (Spare Required)

Course of Action (Number of Spare Purchased)

Purchase Cost (Rs)

Emergency Cost (Rs)

Total Conditional Cost (Rs)

 0  N1  0   0

0

0

0

0

1

1,500

0

1,500

2

3,000

0

3,000

 1  N2  1   1

0

0

9,000

9,000

1

1,500

0

1,500

2

3,000

0

3,000

 2  N3  2   2

0

0

18,000

18,000

1

1,500

9,000

10,500

2

3,000

0

3,000

Table 11.14

Using the conditional costs and the probabilities of states of nature, the expected monetary value can be calculated for each of three states of nature as shown in Table 11.15.

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354 Operations Research: Theory and Applications State of Nature (Space Required)

Probability

Conditional Cost Due to Course of Action S1

N1

0.80

0

Weighted Cost Due to Course of Action

S2

S3

S1

S2

S3

1,500

3,000

0.80 (0)

1,200

2,400

225

450

= 900

525

150

EMV = 2,250

1,950

3,000

= 0

Table 11.15 Expected Monetary Value

N2

0.15

9,000

1,500

3,000

0.15 (9,000) = 1,350

N3

0.05

18,000

10,500

3,000

0.05 (18,000)

Since weighted cost = Rs 1,950 is lowest due to course of action, S2, it should be chosen. If the EMV is expressed in terms of profit, then EMV* = EMV(S2) = – Rs 1,950. Hence, the optimal number of spares to be purchased initially should be one. (b) The calculations for conditional opportunity loss (COL) to determine EOL are shown in Table 11.16. State of Nature (Space Required) Table 11.16 Conditional Opportunity Loss (COL)

Conditional Cost Due to Course of Action

N1 N2 N3

Conditional Opportunity Loss Due to Course of Action

S1

S2

S3

S1

0 9,000 18,000

1,500 1,500 10,500

3,000 3,000 3,000

0 7,500 15,000

S2

S3

1,500 0 7,500

3,000 1,500 0

Since we are dealing with conditional costs rather than conditional profits, the lower value for each state of nature shall be considered for calculating opportunity losses. The calculations for expected opportunity loss are shown in Table 11.17. State of Nature (Space Required)

Probability

Conditional Opportunity Loss (Cost) Due to Course of Action S1

Table 11.17 Expected Opportunity Loss (EOL)

Weighted Opportunity Loss (Cost) Due to Course of Action

S2

S3

S1

N1

0.80

0

1,500

3,000

N2

0.15

7,500

0

1,500

N3

0.05

15,000

7,500

0

S2

S3

0.80(0) = 0 0.15 (7,500) = 1,125 0.05 (15,000) = 750

1,200

2,400

0

225

375

0

EMV = 1,875

1,575

2,625

Since minimum, EOL* = EOL(S2) = Rs 1,575, therefore adopt course of action S2 and purchase one spare. (c) The expected profit with perfect information (EPPI) can be determined by selecting the optimal course of action for each state of nature, multiplying its conditional values by the corresponding probability and then adding these products. The EPPI calculations are shown in Table 11.18. States of Nature

Probability

Optimal Course of Action

(Space Required) N1 N2 N2 Table 11.18

0.80 0.15 0.05

S1 S2 S3

Cost of Optimal Course of Action (Rs) Conditional Cost (Minimum Value)

Weighted Opportunity Loss

0 1,500 3,000

0.80(0) = 0 0.15 (1,500) = 225 0.05 (3,000) = 150 Total = 375

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Decision Theory and Decision Trees

355

Thus expected profit with perfect information is, EPPI = Rs 375. Expected value of perfect information then is: EVPI = EPPI – EMV* = – 375 – (– 1,950) = Rs 1,575. It may be observed that, EVPI = EOL* = Rs 1,575 Example 11.12 XYZ Company manufactures parts for passenger cars and sells them in lots of 10,000 parts each. The company has a policy of inspecting each lot before it is actually shipped to the retailer. Five inspection categories, established for quality control, represent the percentage of defective items contained in each lot. These are given in the following table. The daily inspection chart for past 100 inspections shows the following rating or breakdown inspection: Due to this the management is considering two possible courses of action: (i) S1: Shut down the entire plant operations and thoroughly inspect each machine. Rating Excellent (A) Good (B) Acceptable (C) Fair (D) Poor (E)

Proportion of Defective Items

Frequency

0.02 0.05 0.10 0.15 0.20

25 30 20 20 5 Total = 100

(ii) S2 : Continue production as it now exists but offer the customer a refund for defective items that S2 : are discovered and subsequently returned. The first alternative will cost Rs 600 while the second alternative will cost the company Re 1 for each defective item that is returned. What is the optimum decision for the company? Find the EVPI. Solution Rating

A B C D E

Calculations of inspection and refund cost are shown in Table 11.19. Defective Rate

Probability

0.02 0.05 0.10 0.15 0.20

Cost

Opportunity Loss

Inspect

Refund

Inspect

Refund

0.25 0.30 0.20 0.20 0.05

600 600 600 600 600

200 500 1,000 1,500 2,000

400 100 0 0 0

0 0 400 900 1,400

1.00

600*

670

EOL = 170*

240

Table 11.19 Inspection and Refund Cost

The cost of refund is calculated as follows: For lot A : 10,000 × 0.02 × 1.00 = Rs 200 Similarly, the cost of refund for other lots is calculated. Expected cost of refund is: 200 × 0.25 + 500 × 0.30 + . . . + 2,000 × 0.05 = Rs 670 Expected cost of inspection is: 600 × 0.25 + 600 × 0.30 + . . . + 600 × 0.05 = Rs 600 Since the cost of refund is more than the cost of inspection, the plant should be shut down for inspection. Also, EVPI = EOL of inspection = Rs 170. Example 11.13 A toy manufacturer is considering a project of manufacturing a dancing doll with three different movement designs. The doll will be sold at an average of Rs 10. The first movement design using ‘gears and levels’ will provide the lowest tooling and set up cost of Rs 1,00,000 and Rs 5 per unit of variable cost. A second design with spring action will have a fixed cost of Rs 1,60,000 and variable cost of Rs 4 per unit. Yet another design with weights and pulleys will have a fixed cost of Rs 3,00,000 and variable cost Rs 3 per unit. The demand events that can occur for the doll and the probability of their occurrence is given below:

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356 Operations Research: Theory and Applications Demand (units)

Probability

25,000 1,00,000 1,50,000

0.10 0.70 0.20

Light demand Moderate demand Heavy demand

(a) Construct a payoff table for the above project. (b) Which is the optimum design? (c) How much can the decision-maker afford to pay in order to obtain perfect information about the demand? Solution

The calculations for EMV are shown in Table 11.20. Payoff = (Demand × Selling price) – (Fixed cost + Demand × Variable cost) = Revenue – Total variable cost – Fixed cost

States of Nature (Demand)

Table 11.20 EMV and Payoff Values

Light Moderate Heavy

Probability

0.10 0.70 0.20

Conditional Payoff (Rs) Due to Courses of Action (Choice of Movements)

Expected Payoff (Rs) Due to Courses of Action

Gears and Levels

Spring Action

Weights and Pulleys

Gears and Levels

Spring Action

Weights and Pulleys

25,000 4,00,000 6,50,000

– 10,000 4,40,000 7,40,000

– 1,25,000 4,00,000 7,50,000

2,500 2,80,000 1,30,000

– 1,000 3,08,000 1,48,000

– 12,500 2,80,000 1,50,000

4,12,500

4,55,000

4,17,500

EMV

Since EMV is largest for spring action, it is the one that must be selected.

Table 11.21 Expected Payoff with Perfect Information

States of Nature (Demand)

Probability

Light Moderate Heavy

0.10 0.70 0.20

Courses of Action Gears and Spring Weights Levels Action and Pulleys 25,000 4,00,000 6,50,000

– 10,000 4,40,000 7,40,000

– 1,25,000 4,00,000 7,50,000

Maximum Payoff

Maximum Payoff × Probability

25,000 4,40,000 7,50,000

2,500 3,08,000 1,50,000 Total = 4,60,500

The maximum amount of money that the decision-maker would be willing to pay in order to obtain perfect information regarding demand for the doll will be EVPI = Expected payoff with perfect information – Expected payoff under uncertainty (EMV) = 4,60,500 – 4,55,000 = Rs 5,500 Example 11.14 A TV dealer finds that the cost of holding a TV in stock for a week is Rs. 50. Customers who cannot obtain new TV sets immediately tend to go to other dealers and he estimates that for every customer who cannot get immediate delivery he loses an average of Rs. 200. For one particular model of TV the probabilities of demand of 0, 1, 2, 3, 4 and 5 TV sets in a week are 0.05, 0.10, 0.20, 0.30, 0.20 and 0.15, respectively. (a) How many televisions per week should the dealer order? Assume that there is no time lag between ordering and delivery. (b) Compute EVPI. (c) The dealer is thinking of spending on a small market survey to obtain additional information regarding the demand levels. How much should he be willing to spend on such a survey. [Delhi Univ., MBA, 2000] Solution If D denotes the demand and S the number of televisions stored (ordered), then the conditional cost values are computed and are shown in Table 15.22. 50S + 200 ( D − S ) , when D ≥ S

Cost function =  50 D + 50 ( S − D ) , when D < S

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Decision Theory and Decision Trees State of Nature (Demand)

Probability

Conditional Cost (Rs.) Course of Action (Stock) 0

1

2

3

4

5

0

0.05

0

50

100

150

200

250

1

0.10

200

50

100

150

200

250

2

0.20

400

250

100

150

200

250

3

0.30

600

450

300

150

200

250

4

0.20

800

650

500

350

200

250

5

0.15

1000

850

700

550

400

250

Expected Cost = 590

450

330

250

230

250

State of Nature (Demand)

357

Probability (1)

Minimum Cost for Perfect Information (2)

Table 11.22 Expected Cost

Expected Cost for Perfect Information (3) = (1) × (2)

0

0.05

0

1

0.10

50

0 5

2

0.20

100

20

3

0.30

150

45

4

0.20

200

40

5

0.15

250

37.5

Table 11.23 Expected Payoff with perfect Information

ECPI = 147.5

EVPI = Conditional Cost – ECPI = 230 – 147.5 = Rs. 82.5. The pay-off for EMV is shown in Table 15.24 State of Nature (Demand)

0

Probability

Conditioinal Payoff Course of Action (Stock)

0.05

0

1

2

3

4

5

0

– 50

– 100

– 150

– 200

– 250

1

0.10

0

150

100

50

0

– 50

2

0.20

0

150

300

250

200

150

3

0.30

0

150

300

450

400

350

4

0.20

0

150

300

450

600

550

5

0.15

0

150

300

450

600

750

EMV = 0

140

260

340

360

340

State of Nature (Demand) 0 1 2 3 4 5

Probability (1)

Payoff for Perfect Information (2)

Expected Payoff for Perfect Information (3) = (1) × (2)

0.05 0.10 0.20 0.30 0.20 0.15

0 150 300 450 600 750

0 15 60 135 120 112.5 442.5

Table 11.24 Computation of EMV

Table 11.25 Computation of EPPI

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358 Operations Research: Theory and Applications The expected value of perfect information is given by EVPI = EPPI – EMV* = 442.5 – 360 = Rs. 82.5. (c)

On the basis of the given data, the dealer should not be willing to spend more than Rs. 82.5 for the market survey.

Example 11.15 XYZ company is considering issuing 1,00,000 shares to raise capital needed for expansion. It is estimated that if the issues were made now, it would be fully taken up at a price of Rs. 30 per share. However, the company is facing two crucial situations, both of which may influence the share price in the near future, namely: (i) A wage dispute with tool room operators which could lead to a strike and have an adverse effect on the share price. (ii) The possibility of a substantial contract with a large company overseas which would increase the share price. The four possible outcomes and their expected effect on the company’s share prices are: E1 : No strike and contract obtained–share price rises to Rs. 34. E2 : Strike and contract obtained–share price stays at Rs. 30. E3 : No strike and contract lost–share price raises to Rs. 32. E4 : Strike and contract lost–share price falls to Rs. 16. Management has identified three possible strategies that the company could adopt, namely A1 : Issue 1,00,000 shares now A2 : Issue 1,00,000 shares only after the outcomes of (i) and (ii) are known A3 : Issue 5,00,000 shares now and 50,000 shares after the outcomes of (i) and (ii) are known. (a) Determine the maximax solution. What alternative criterion might be used? (b) It has been estimated that the probability of a strike is 55 per cent and that there is a 65 per cent chance of getting the contract, these probabilities being independent. Determine the optimum policy for the company using the criterion of maximizing expected pay-off. (c) Determine the expected value of perfect information for the company. Solution States of Nature E1 E2 E3 E4 Table 11.26

The payoff values are shown in Table 11.26 Probability

(1 – 0.55) × 0.65 0.55 × 0.65 (1 – 0.55)(1 – 0.65) 0.55 × (1 – 0.65)

Conditional Payoff (Rs. lakh) Alternative Strategy A1 A2 A3 = 0.2925 = 0.3575 = 0.1575 = 0.1925

Expected Values

30 30 30 30

34 30 32 16

30

34

0.5 × 30 + 0.5 × 34 0.5 × 30 + 0.5 × 30 0.5 × 30 + 0.5 × 32 0.5 × 30 + 0.5 × 16

Conditional Loss (Rs. lakh) Alternative Strategy A1 A2 A3 = 31 = 30 = 30.5 = 23

4 0 2 0

0 0 0 14

3 0 1.5 7

31

A strategy with highest minimum (maximin) payoff (i.e. 30) is A1 and a strategy with highest (maximax) payoff (i.e. 34) is A2. Since highest pay-off of Rs. 30 lakh is obtained corresponding to strategy A1, the company should adopt strategy A1. (c) Calculations for expected value of the perfect information are shown in Table 15.27.

Table 11.27 EVPI

State of Nature

Maximum Pay-off

Probability

Expected Pay-off with Perfect Information

E1 E2 E3 E4

34 30 32 16

0.2925 0.3575 0.1575 0.1925

9.94 10.72 5.04 5.78 31.48

Hence, expected value of perfect information is : 31.48 – 30 = Rs. 1,48,000.

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Decision Theory and Decision Trees

Example 11.16 A vegetable seller buys tomatoes for Rs. 45 a box and sells them for Rs. 80 per box. If the box is not sold on the first selling day, it is worth Rs. 15 as salvage. The past records indicate that demand is normally distriubuted, with a mean of 30 boxes daily and a standard deviation of 9 boxes. How many boxes should he stock? Solution The probability of selling at least one additional unit (box) to justify the stocking of additional unit is given by IL (45 − 15) p= = = 30 = 0.462 IP + IL (80 − 45) + (45 − 15) 65 where Incremental loss (IL) = Cost price – Salvage price Incremental profit (IP) = Selling price – Cost price This implies that the vegetable seller must be 46.2 per cent sure of selling at least one additional unit before he should pay to stock an additional unit. The vegetable seller should stock additional boxes until point A is reached. If more units are stocked, then the probability will fall below 0.462. The point A is at 0.1 standard deviation to the right of the mean x = 30. Since the standard deviation of the distribution of past demand is 9 boxes, point A can be located as follows: Point A = Mean + Standard deviation = 30 + 0.1 × 9 = 30.1  31 boxes Hence, the fruit seller should stock 31 boxes.

Fig. 11.1

Example 11.17 A stall at a certain railway station sells for Rs. 1.50 paise a copy of daily newspaper for which it pays Rs. 1.20. Unsold papers are returned for a refund of Re. 0.95 a copy. Daily sales and corresponding probabilities are as follows : Daily sales : 500 600 700 Probability : 0.5 0.3 0.2 (a) How many copies should it order each day to get maximum expected profit? (b) If unsold copies cannot be returned and are useless, what should be the optimal order each day? Use increment analysis. Solution Given that, Incremental profit (IP) = Rs. (1.50 – 1.20) = Re. 0.30 Incremental loss (IL) = Rs. (1.20 – 0.95) = Re. 0.25 The probability (p) of selling at least one additional copy of the newspaper to justify keeping that additional copy of newspaper is: IL 0.25 = = 0.45. p= IP + IL 0.30 + 0.25 Thus to justify the ordering of an additional copy, there must be at least 0.45 cumulative probability of selling that copy. Cumulative probabilities are computed below: Daily sales : 500 600 700 Probability : 0.50 0.30 0.20 Comulative probability : 1.00 0.50 0.20 Hence, the optimal order size is 600 copies. (b) If unsold copies are non-refundable, then IL 1.20 1.20 = = = 0.80 p= IP + IL 0.30 + 1.20 1.50 Hence, optimal order size is, 500 copies. Example 11.18 The demand pattern of the cakes made in a bakery is as follows: No. of cakes demanded : 0 1 2 3 4 5 Probability : 0.05 0.10 0.25 0.30 0.20 0.10 If the preparation cost is Rs 30 per unit and selling price is Rs 40 per unit, how many cakes should the baker bake for maximizing his profit?

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360 Operations Research: Theory and Applications Solution Given that incremental cost (IC) to prepare a cake is Rs 30 per unit and incremental price (IP) to sell a cake is Rs 40 per unit. The cumulative probability ( p) of selling at least an additional unit of cake to justify the stocking of that additional unit of cake is given by p=

30 IC = = 0.428 IC + IP 30 + 40

The cumulative probabilities of greater than type are computed as shown in Table 11.28. Demand ( No. of Cakes)

Probability P (Demand = k)

0 1 2 3 4 5

0.05 0.10 0.25 0.30 0.20 0.10

Table 11.28

Cumulative Probability P (Demand ≥ k) 1.00 0.95 0.85 0.60 ← 0.30 0.10

Since P(demand ≥ k) that exceeds the critical ratio, p = 0.428 is k = 3 units of cake, the optimal decision is to prepare only 3 cakes.

11.6

Posterior probabilities are the revised probabilities of the states of nature obtained after conducting a test to improve the prior probabilities of respective nature.

POSTERIOR PROBABILITIES AND BAYESIAN ANALYSIS

An initial probability statement to evaluate expected payoff is called a prior probability distribution, but if the probability statement has been revised due to additional information, then such a probability statement is called a posterior probability distribution. In this section we will discuss the method of computing posterior probabilities, given prior probabilities using Bayes’ theorem. The analysis of problems using posterior probabilities with new expected payoffs and additional information, is called prior-posterior analysis.

Baye’s Theorem Statement Let A1, A2, . . ., An be mutually exclusive and collectively exhaustive outcomes. Their probabilities P(A1), P(A2), . . ., P(An) are known. There is an experimental outcome B for which the conditional probabilities P(B | A1), P(B | A2), . . ., P(B | An) are also known. Given the information that outcome B has occurred, the revised conditional probabilities of outcomes Ai, i.e. P(Ai | B), i = 1, 2, . . ., n are determined by using the following relationship: P ( Ai and B ) P ( Ai ∩ B ) P ( Ai ) P ( B | Ai ) P ( Ai | B ) = = = P( B) P( B) P( B) where

P( B) =

n

∑ P( Ai ) P( B | Ai )

i =1

Since each joint probability can be expressed as the product of a known marginal (prior) and conditional probability, i.e., P ( Ai ∩ B ) = P ( Ai ) × P ( B | Ai )

Baye’s decision rule uses the prior probabilities to determine the expected payoff for each decision alternatives and then chooses the one with the largest expected payoff.

Example 11.19 A company is considering to introduce a new product to its existing product range. It has defined two levels of sales as ‘high’ and ‘low’ on which it wants to base its decision and has estimated the changes that each market level will occur, together with their costs and consequent profits or losses. This information is summarized below: States of Nature

Probability

Courses of Action Market the Product (Rs ’000)

Do not Market the Product (Rs ’000)

High sales

0.3

150

0

Low sales

0.7

– 40

0

The company’s marketing manager suggests that a market research survey may be undertaken to provide further information on which the company should base its decision. Based on the company’s past

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Decision Theory and Decision Trees

experience with a certain market research organization, the marketing manager assesses its ability to give good information in the light of subsequent actual sales achievements. This information is given below: Actual Sales

Market Research (Survey outcome)

Market ‘high’

Market ‘low’

‘High’ sales forecast

0.5

0.1

Indecisive survey report

0.3

0.4

‘Low’ sales forecast

0.2

0.1

The market research survey costs Rs 20,000, state whether or not there is a case for employing the market research organization. [Delhi Univ., MBA, 2000] Solution

The expected monetary value (EMV) for each course of action is shown in Table 11.29.

States of Nature High sales (N1) Low sales (N2)

Prior Probability

Courses of Action

Expected Profit (’000 Rs)

Do not Market

Market

Do not Market

150 – 40

0 0

45 – 28

0 0

EMV = 17

= 0

0.3 0.7

Market

Table 11.29

With no additional information, the company should choose course of action ‘market the product’. However, if the company had the perfect information about the ‘low sales’, then company would not go ahead with the decision because expected value would be (–) Rs 28,000. Thus, the value of perfect information is the expected value of low sales. Let outcomes of the research survey be: high sales (S1), indecisive report (S2) and low sales (S3) and states of nature be: high market (N1) and low market (N2). The calculations for prior probabilities of forecast are shown in Table 11.30. Outcome

Sales Prediction High Market ( N1 )

Low Market ( N2 )

High sales (S1) Indecisive report (S2)

P(S1 | N1) = 0.5 P(S2 | N1) = 0.3

P(S1 | N2) = 0.1 P(S2 | N2) = 0.4

Low sales (S3)

P(S3 | N1) = 0.2

P(S3 | N2) = 0.5

Table 11.30

With additional information, the company can now revise the prior probabilities of outcomes to get posterior probabilities. The calculations of the revised probabilities, given the sales forecast are shown in Table 11.31. States of Nature

Prior Probability P(Ni)

Conditional Probability P(Si | Ni)

High sales (N1)

0.3

P(S1 | N1) = 0.5

0.15

–

–

P(S2 | N1) = 0.3

–

0.09

–

P(S3 | N1) = 0.2

–

–

0.06

P(S1 | N2) = 0.1

0.07

–

–

P(S2 | N2) = 0.4

–

0.28

–

P(S3 | N2) = 0.5

–

–

0.35

0.22

0.37

0.41

Low sales (N2)

Marginal Probability

0.7

Joint Probability P(Si ∩ Ni) = P(Ni) P(Si | Ni)

Table 11.31 Revised Probabilities

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362 Operations Research: Theory and Applications The posterior probabilities of actual sales, given the sales forecast, are: Outcome (Si)

Probability P(Si)

States of Nature (Ni)

Posterior Probability P(Ni | Si) = P(Ni ∩ Si)/P(Si)

S1

0.22

N1 N2

0.15/0.22 = 0.681 0.07/0.22 = 0.318

S2

0.37

N1 N2

0.09/0.37 = 0.243 0.28/0.37 = 0.756

S3

0.41

N1 N2

0.06/0.41 = 0.146 0.35/0.41 = 0.853

Given the additional information, the revised probabilities to calculate net expected value with respect to each outcome are shown in Table 11.32. Sales Forecast States of Nature

Table 11.32

Revised Conditional Profit (Rs)

High

Indecisive

Low

Prob.

EV (Rs)

Prob.

EV (Rs)

Prob.

EV (Rs)

High sales

130

0.681

88.53

0.243

31.59

0.146

18.98

Low sales

– 60

0.318

–19.08

0.756

– 45.36

0.853

– 51.18

Expected value of sales forecast

69.45

– 13.77

– 32.20

Probability of occurrence Net expected value (Expected value × Probability)

0.22

0.37

0.41

15.279

– 5.095

13.202

CONCEPTUAL QUESTIONS B 1. Given the complete set of outcomes in a certain situation, how is the EMV determined for a specific course of action? Explain in your own words. 2. Explain the difference between expected opportunity loss and expected value of perfect information. 3. Indicate the difference between decision-making under risk, and uncertainty, in statistical decision theory.

4. Briefly explain ‘expected value of perfect information’ with examples. 5. Describe a business situation where a decision-maker faces a decision under uncertainty and where a decision based on maximizing the expected monetary value cannot be made. How do you think the decision-maker should make the required decision?

SELF PRACTICE PROBLEMS B 1. You are given the following payoffs of three acts A1, A2 and A3 and the events E1, E2, E3. States of Nature

Three Acts A1

A3

A3

E1

25

– 10

– 125

E2

400

440

400

E3

650

740

750

The probabilities of the states of nature are 0.1, 0.7 and 0.2, respectively. Calculate and tabulate the EMV and conclude which would prove to be the best course of action. 2. The management of a company is faced with the problem of choosing one of three products that it wants to manufacture. The potential demand for each product may turn out to be good, moderate or poor. The probabilities for each of the states of nature were estimated as follows.

Product X Y Z

Nature of Demand Good

Moderate

Poor

0.70 0.50 0.40

0.20 0.30 0.50

0.10 0.20 0.10

The estimated profit or loss in rupees under the three states may be taken as: Product

Good

Moderate

X Y Z

30,000 60,000 40,000

20,000 30,000 10,000

Poor 10,000 20,000 – 15,000

Prepare the expected value table, and advise the management about the choice of product.

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