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# Decision theory

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### Decision theory

1. 1. 1 Decision Theory Definition of Decision Analysis: Decision analysis is defined as the use of a rational process/technique for selecting the best of several alternatives. The “goodness” of a selected alternative depends on the quality of the data used in describing decision situation. Example: Mr X is a bright student, has received full academic scholarships from three institutions: U of A, U of B and U of C. To select a university Mr. X lists two three main criteria: location, academic reputation and political violence. Being the excellent student he is, he judges academic reputation to be five times as important as location and less political violence, which gives a weight of approximation 80% to reputation, 15% to location and 5% to political violence. He then use a systematic analysis to rank the three universities from the standard point of location, reputation and political violence. The analysis produces the following estimates Location: U of A (13%) U of B (28%) U of C (59%) Reputation U of A (54%) U of B (27%) U of C (19%) Political Violence U of A (25%) U of B 45% U of C (30%) The structure of the decision problem is summarized below in Figure. Select a University Location (0.15) Reputation (0.80) Political Violence 0.05 U of A 0.13 U of B 0.28 U of C 0.59 U of A 0.54 U of B 0.27 U of C 0.19 U of A 0.25 U Of B 0.45 U of C 0.30 Sazzad Hossain 128
2. 2. 2 To select the university here we use the technique of decision analysis. Classification of Decision Situation Decision making process can be classified into the following three broad categories (1) Decision making under certainty in which the data are known deterministically (2) Decision making under risk in which the data can be described by probability distributions. 3. Decision making under uncertainty in which the data cannot be assigned relative weights that represent their degree of relevance in the decision process. Decision Making Under Certainty: Linear programming and transportation models provide an example of decision making under certainty. Here only one state of nature exists, there is complete certainty about the future. The decision maker simply finds the best payoff in that one column and chooses the associated alternative. Few complex managerial decision making problems, however ever enjoy the luxury of having complete information about the future and hence decision making under uncertainty is of little consequential interest. This section presents a different approaches for the situation in which ideas, feelings and emotions are quantified to provide a numeric scale for prioritizing decision alternatives. The approach is known as the Decision Making Under Risk Here, more than one states of nature exist and decision maker has sufficient information to assign probabilities to each of these states. For this reason decision making under risk is usually based on the expected value criteria in which decision alternatives are compared based on the maximization of expected profits or the minimization of expected cost. This approach has limitations in the sense that it may not be applicable to certain situations Expected Value Criteria: It is also known as the expected monetary value (EMV) criteria, it consists of the following steps. (i) Construct a payoff table listing the alternative decisions and the various states of nature. Enter the conditional profit for each decision-event combination along with the associated probabilities. In general, a decision problem may include n states of nature and m alternatives. If jp is the probability of occurrence for state of nature j and let ija is the payoff of alternative i given state of nature j (i= 1, 2,……,m; j = 1, 2,……,n)
3. 3. 3 State j Probability( jp ) Alternatives 1, 2, ……………. i………. ….. m 1 2 . . . j . . n 1p 2p . . . jp . . np 11a 21a ………………. i1a ……………. m1a 12a 22a ………………… i2a …………… m2a . ……………………………………….. . ……………………………………….. . ……………………………………….. 1ja 2ja ………………….. ija ……………… mja . ……………………………………………. . ……………………………………….. 1na 2na …………………… ina ……………… mna (ii) Calculate the EMV for each decision alternative by multiplying the conditional profits by the assigned probabilities and adding the resulting conditional values. The expected monetary value or expected payoff for alternative I is computed as i ij j 1 EMV = a p ; i = 1, 2,.....,m n j  (iii) The best alternative is the one associated with * i i iEMV =max [EMV ] or * i i iEMV =min [EMV ] depending, respectively, on whether the payoff of the problem represents profit (income) or loss (expense) Problem: A newspaper boy has the following probabilities of selling a sports magazine No. of Copies Sold Probabilities 10 0.01 11 0.15 12 0.20 13 0.25 14 0.30 Cost of a copy is TK 30 and sale price is TK 50. He cannot return unsold copies. How many copies should ne order? Solution: The numbers of copies for purchases and for sales are 10, 11, 12, 13 and 14. These are his sales magnitudes. There is no reason for him to buy less than 10 or more than 14 copies. Stocking 10 copies each day will always result in profit TK200 irrespective of demand. For instance, even if the demand on some day is more than 10 copies, he can sell only 10 and hence his conditional profit is TK200. When he stocks 11 copies, his conditional profits is TK220 on days buyers request 11, 12, 13 or 14 copies. But on days when he has 11 copies on stock and buyers buy only 10 copies, his profit will be TK 170. The profit of TK 200 on the 10 copies sold
4. 4. 4 must be reduced by TK 30, the cost of one copy left unsold. The same will be true when he stocks 12, 13, or 14 copies. The conditional profit is given below; Payoff = (20 copies sold-30 copies unsold ) TK  The payoff table is constructed below; Table 1: Payoff Table: Conditional profit in TK Possible Demand No. of Copies Prob. Possible stock action 10 Copies 11 Copies 12 Copies 13 Copies 14 Copies 10 11 12 13 14 0.10 0.15 0.20 0.25 0.30 200 200 200 200 200 170 220 220 220 220 140 190 240 240 240 110 160 210 260 260 80 130 180 230 280 Now the expected value of each decision alternative is obtained by adding the multiplying its conditional profit and the associated probability. The is shown in Table below; Table: Expected profit Possible Demand No. of Copies Prob. Expected profit from stocking in TK 10 Copies 11 Copies 12 Copies 13 Copies 14 Copies 10 11 12 13 14 0.10 0.15 0.20 0.25 0.30 20 30 40 50 60 17 33 44 55 66 14 28.5 48 60 72 11 24 42 65 78 8 19.5 36 57.5 84 Total Expected Profit 200 215 222.5 220 205 Thus the newsboy must order 12 copies to earn highest possible average daily profit. This stocking will maximize the total profits over a period of time. Of course there is no guarantee that he will make a profit of TK222.5 tomorrow. However, if he stocks 12 copies each day under the condition given, he will have average profit of TK 222.5 per day. This is the best he can do because choice of any of the other four possible stock actions result in a lower daily profit. Expected Opportunity Loss (EOL) Criterion This approach is an alternative to EMV. EOL or expected value of regrets is the amount by which maximum possible profit will be reduced under various possible actions. The course of action that minimize these losses is the optimal decision alternative. It consists of the following steps: Step 1: Construct the conditional profit table for each decision-event combination and write the associated probabilities
5. 5. 5 Step 2: For each event, calculate the conditional opportunity loss (COL) by subtracting the payoff from the maximum payoff for the event. Step 3: Calculate the expected opportunity loss (EOL) for each decision alternative by multiplying the COL’s by the associated probabilities and then adding the values. Step 4. Select the alternative that yields the lowest EOL Problem: A newspaper boy has the following probabilities of selling a sports magazine No. of Copies Sold Probabilities 10 0.01 11 0.15 12 0.20 13 0.25 14 0.30 Cost of a copy is TK 30 and sale price is TK 50. He cannot return unsold copies. How many copies should ne order? Solution: The best alternative for demand 10 copies is to order 10 copies, resulting in optimal profit of TK 200. The conditional opportunity loss for each stock action is obtained just by subtraction the respective conditional profit from TK 200. Likewise, for demand of 11, 12, 13 and 14 copies we subtract the conditional payoff values for each of these rows from the maximum of that row. The resulting conditional opportunity loss table is shown below; Table 1: Conditional Loss Table in TK Possible Demand No. of Copies Prob. Possible stock action 10 Copies 11 Copies 12 Copies 13 Copies 14 Copies 10 11 12 13 14 0.10 0.15 0.20 0.25 0.30 0 20 40 60 80 30 0 20 40 60 60 30 0 20 40 90 60 30 0 20 120 90 60 30 0 Now the EOL value of each decision alternative is obtained by adding the multiplying its conditional opportunity loss and the associated probability. The is shown in Table below; Table: Expected profit Possible Demand No. of Copies Prob. Expected profit from stocking in TK 10 Copies 11 Copies 12 Copies 13 Copies 14 Copies 10 11 12 13 14 0.10 0.15 0.20 0.25 0.30 0 3 8 15 24 3 0 4 10 8 6 4.5 0 5 12 9 9 6 0 6 12 13.5 12 7.5 0 Total Expected Profit 50 35 27.5 30 45
6. 6. 6 The optimal stock action is the one which will minimize expected opportunity losses. If the newsboy stock 12 copies each day, the expected loss will be minimum. Note : It may be pointed out that EMV and EOL decision criteria are completely consistent and yield the same optimal decision alternative. However, while the decisions alternative will be always based on finding the minimum EOL value under EOL criterion irrespective of whether the problem is of maximization expected profit or minimizing expected loss. Expected Value of Perfect Information (EVPI): Complete and accurate information about the future demand referred to as perfect information, would remove all uncertainty from the problem. With this perfect information, the decision maker would know in advance exactly about the future demand. EVPI represents the maximum amount he would pay to get this additional information on which may be based the decision alternative. Problem: A dairy firm wants to determine the quantity of butter it should produce to meet the demand. The past records have shown the following demand patterns; Quantity Demanded (in kg) No. of Days Demand Occurred 15 20 25 30 35 40 50 6 14 20 80 40 30 10 The stock levels are restricted to the range of 15 kg to 50 kg and the butter left unsold at the end of the day must be disposed of due to inadequate storing facilities. Butter costs TK 80 and sold TK 110 per kg. (i) Construct a conditional profit table (ii) Determine the action alternative associated with the maximization of expected profit (iii) Determine the EVPI Solution: The dairy firm would not produce butter less than 15 kg and more than 50 kg. Form the given data we can calculate the conditional profit for each stock action and event (demand combination. If CP is the conditional profit, S is the quantity in stock and D is the quantity demanded, then CP is given by; 30S; when D S CP= 110D-80S; when D<S    The probability of demand of 15 kg is = 6/200 = 0.03. The probabilities associated with other demand levels are given below win payoff table below:
7. 7. 7 Possible Demand (Event in kg) (D) Prob. Possible stock action (Alternative ) (in kg) (S) 15 20 25 30 35 40 50 15 20 25 30 35 40 50 0.03 0.07 0.10 0.40 0.20 0.15 0.05 450 450 450 450 450 450 450 50 600 600 600 600 600 600 -350 200 750 750 750 750 750 -750 -200 350 900 900 900 900 -1150 -600 -50 500 1050 1050 1050 -1550 -1000 -450 100 650 1200 1200 -2350 -1800 -1250 -700 -150 400 1500 The expected profit for each alternative is given below Table: Expected profit Possible Demand (Event in kg) (D) Prob. Possible stock action (Alternative ) (in kg) (S) 15 20 25 30 35 40 50 15 20 25 30 35 40 50 0.03 0.07 0.10 0.40 0.20 0.15 0.05 13.5 31.5 45.0 180 90 67.5 22.5 1.5 42 60 240 120 90 30 -10.5 14 75 300 150 112.5 37.5 -22.5 -14 35 360 180 135 45 -34.5 -42 -5 200 210 157.5 52.5 -46.5 -70 -45 40 130 180 60 -70.5 -126 -125 -280 -30 60 75 EMV 450.0 583.5 678.5 718.5 538.5 248.5 -496.5 Since the maximum EMV is TK 718.5 for stock of 30 kg of butter, thus the dairy firm may produce 30 kg of butter and can expect average daily profit of TK 718.5 Now the expected profit with perfect information is given below in Table () Table : Expected profit with perfect information Market Demand (in kg) Probabilities Conditional Profit under certainty Expected profit with perfect information 15 20 25 30 35 40 50 0.03 0.07 0.10 0.40 0.20 0.15 0.05 450 600 750 900 1050 1200 1500 13.5 42 75 360 210 180 75 EPPI 955.5
8. 8. 8 The expected value of perfect information is given by EVPI = EPPI- max(EMV) = 955.5-718.5 = 237 TK Use of Incremental (Marginal ) Analysis. Any additional unit purchased will be either sold or remain unsold. If p denotes the probability of selling one additional unit, then (1-p) must be the probability of not selling it. If the additional unit sold, the conditional profit will increase as a result of the profit earned from this additional unit. This is termed as the incremental profit or marginal profit. Let the define the marginal profit is MP. If the additional unit is not sold, the conditional profit is reduced and the amount of reduction is called the incremental loss or marginal loss. Let us define the marginal loss is ML. Thus the expected incremental profit will be pMP while the expected incremental loss will be (1-p) ML. Thus the units should be stocked up to the point such that pMp (1-p)ML or at least p(MP+ML) = ML or at least p=ML/(MP+ML)  Thus additional unit should be stocked so long as the probability of selling at least one additional unit is greater than p. Problem: A milkman buys milk at Tk 30 and sells for TK 40 per litre. Unsold milk has to be thrown away. The daily demand has the following probability distribution Demand (Liter) Probability 46 48 50 52 54 56 58 60 62 64 0.01 0.03 0.06 0.10 0.20 0.25 0.15 0.10 0.05 0.05 If each day’s demand is independent of previous days’ demand, how many liters should he ordered every day. Solution: The marginal profit is given as MP = 10 Tk And the marginal loss is given as
9. 9. 9 ML = TK 30 The milkman should stock additional liters of milk so long as the probability of selling at least an additional liter of milk is greater than p where ML p= MP+ML 30 0.75 10 30    The value of 0.75 for p implies that in order to justify the stocking of an additional unit, there must be at least 0.75 cumulative probability of selling that unit. The cumulative probability of sales are computed below; Demand (Liter) Probability Cumulative probability that sales will be at this level or higher 46 48 50 52 54 56 58 60 62 64 0.01 0.03 0.06 0.10 0.20 0.25 0.15 0.10 0.05 0.05 1.00 0.99 0.96 0.90 0.80 0.60 0.35 0.20 0.10 0.05 The optimum number of liters of milk to be stocked is 54 liters. If the number increased to 56 liters, the cumulative probability will become 0.60 which is less than the required probability 0.75. The expected incremental profit is given by; EIP = pMP=0.75 10=7.5 TK The expected incremental loss is given by; EIL = (1-p)ML = 0.25 0.30=7.5 If the milkman stocks 56 liters of milk, the expected incremental loss will be more than expected incremental gain. The approach give you the same decision as provided by the EMV and EOL approaches. However, the computational effort required in this approach is much less. Problem: A vegetable seller buys a box of tomatoes for TK 300 and sells them for TK 450 a box. If the box is not sold on the first selling day, it is worth of TK 200 as salvage. The past records indicate that the demand is normally distributed with a mean of 30 boxes daily and standard
10. 10. 10 deviation is 9. How many boxes should be stock? Decision Trees Analysis: A decision tree is a graphical representation of the decision process indicating decision alternatives, states of nature, probabilities attached to the state of nature and conditional benefits and losses. It consists of a network of nodes and branches. Two types of nodes are used: (i) decision node represented by a square and state of nature (chance or event) node represented by a circle. Alternative courses of action (strategies) originate from the decision nodes as main branches (decision branches). At the end of each decision branch, there is a state of nature node from which emanate chance events in the form of sub-branches (chance branches). The respective payoff and the probabilities associated with alternative courses and the chance events are shown alongside these branches. At the terminal of the chance branches are shown the expected value of the outcome. The general approach used in decision tree analysis is to work backward through the tree from right to left, computing the expected value of each chance node. We then choose the particular branch leaving a decision node which leads to the chance node with the highest expected value. This approach is known as roll back or fold back process. Example: A client asks an estate agent to sell three properties A, B, and C for him and agrees to pay him 5% commission on each sale. He specifies certain conditions. The estate agent must sell property A first, and this he must do within 60 days. If and when A is sold the agent receives his 5% commission on that sale. He can then either back out at this stage or nominate and try to sell one of the remaining two properties within 60 days. If he does not succeed in selling the nominated property in that period, he is not given the opportunity to sell the third property on the same conditions. The prices, selling costs( incurred by the estate agent whenever a sale is made) and the estate agent’s estimated probabilities of making a sale are given below; Property Price of Property Selling Cost Probability of Sale A 125000 5000 0.70 B 175000 4000 0.60 C 225000 6000 0.50 (i) Draw up an appropriate decision tree for the state agent (ii) What is the estate agent’s best strategy under EMV approach? Solution: The state agent’s gets 5% commission if he sells the properties and satisfies the specified conditions. The amount he receives as commission on sale of properties A, B and C will be TK 6250, TK 8750, and TK 11250 respectively. Since the selling costs incurred by his are TK 5000, TK 4000 and TK 6000 respectively and his conditional profits are TK 1250, TK 4750 and TK
11. 11. 11 5250 respectively. The decision tree for the problem is shown below; EMV of Node D = TK [(5250 0.50)+(0 0.50)]=TK2625  EMV of Node E = TK [(4750 0.60)+(0 0.40)]=TK2850  EMV of Node 3 = Max of TK [ 2625, 0] = TK2625 EMV of Node 4 = Max of TK [ 2850, 0] = TK 2850 EMV of Node B = TK [(4750+2625) 0.60)+(0 0.40)]=TK4425  EMV of Node C = TK [(5250+2850) 0.50)+(0 0.50)]=TK4050  EMV of Node 2 = Max of [ 4425, 4050] = TK 4425 EMV of Node A = TK [(1250+4425) 0.70)+(0 0.30)]=TK3972.5  EMV of Node 1 = TK 3972.5 Thus the optimal strategy for the estate agent is to sell A, if he sells A, then try to sell B and if he sells B and then try to sell C to get an optimum amount of TK 3972.5 Sells C, 5250. 0.5 Takes c Stops, TK 0 Sells Does Not C, TK 0 4750, 0.60 Stop, TK0 4750, 0.6 Sells A, Takes B Does Not Sell B Sells B TK 0, 0.40 Takes B Does Accepts TK1250, 0.70 Take C Sells, 5250, 0.50 Not Sell Does not Sell, TK 0 Does Not Sell C Stops, TK0 TK0 0.30 0.50, TK 0 1 A 2 B C 3 4 D E