CHAPTER 5 DECISION ANALYSIS

Decision theory provides a framework and methodology for rational decision making. It treats decisions against nature, where the result (return) from a decision depends on action of another player (nature). For example, if the decision is to carry an umbrella or not, the return (get wet or not) depends on what action nature takes. It is important to note that, in this model, the returns accrue only to the decision maker. Nature does not care what the outcome is. This condition distinguishes decision theory from game theory. In game theory, both players have an interest in the outcome.

In many cases, solving your problem involves choosing among alternatives. Your objective is to choose the alternative that is best, where “best” depends on what your goals are. Indeed, the first rule of decision making is to know what your goals are. For example, if your decision problem is which movie to see, then “best” means “most entertaining” (assuming being entertained is your goal).

Having identified your goals, you next have to identify your alternatives. For some decision-making problems, your alternatives are obvious. For instance, if you are deciding which movie to see, then your alternatives are the movies playing plus, possibly, not seeing any movie at all. For other problems, however, identifying your alternatives is more difficult. For instance, if you are deciding which personal computer to buy, then it can be quite difficult to identify all your alternatives (e.g., you may not know all the companies that make computers or all the optional configurations available).

Your choice of alternative will lead to some consequence. Depending on the decisionmaking problem you face, the consequence of choosing a given alternative will be either Operations Research

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known or uncertain. If you are driving in your neighborhood, then you know where you will end up if you turn left at a given intersection. If you are investing in the stock market, then you are uncertain about what returns you will earn. Typically, we will suppose that even if you are uncertain about which particular consequence will occur, you know the set of possible consequences. For instance, although you don’t know what your stock price will be a year from now; you do know that it will be some non-negative number. Moreover, you likely know something about which stock prices are more or less likely. For example, you may believe that it is more likely that your stock’s price will change by 20% or less than it will change by 21% or more.

5.1 Prototype Example The X Company owns a land that may contain oil. A geologist has reported to management that he/she believes there is a chance of ¼ of oil. Because of this an oil company has offered to purchase the land for LE90,000. However, X is considering holding the land in order to drill for oil itself. If oil is found, the company’s expected profit will be approximately LE700,000. A loss of LE100,000 will be incurred if the land is dry (no oil).

In this example, the decision maker must choose an action d from a set of possible actions. The set contains all the feasible alternatives under consideration for how to proceed. Table 5.1 summarizes the example data.

Table 5.1 Prospective profit for the X Company State of nature

decision Oil

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Drill for oil Sell the land

LE700,000 LE90,000

Chance of state

¼

129

Dry -LE100,000 LE90,000 ¾

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This choice of an action must be made in the face of uncertainty, because the outcome will be affected by random factors that are outside the control of the decision maker. Each of these possible situations is referred to as a state of nature. For each combination of a decision di and a state of nature j, the decision maker knows what the resulting payoff would be. The payoff is a quantitative measure of the value to the decision maker of the consequences of the outcome. Frequently, payoff is represented by the monetary gain (profit)

5.2 General Representation In general, the decision analysis problem can be represented as given in Table 5.2. This table shows that the fundamental piece of data for decision theory problems is the payoff. This table is also called the payoff table. Table 5.2 provides the payoff for each combination of a decision d and state of nature j.

Table 5.2 Payoff table for the decision analysis 1

State of nature 2 …. m

r11 r21 ..... rn1

r12 r22 ..... rn2

decision d1 d2 …. dn

…. …. .... ….

r1m r2m .... rnm

The entries rij are the payoffs for each possible combination of decision and state of nature. The decision process can be summarized as follow: - The decision maker selects one of the possible actions d1, d2, …, dn. Say di. - After this decision is made, a state of nature occurs. Say state j. - The return received by the decision maker is rij. - The payoff table then should be used to find an optimal decision according to the appropriate criterion.

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The question faced by the decision maker is: which decision to select? The decision will depend on the decision maker's belief concerning what nature will do, that is, which state of nature will occur. If we believe state j will occur, we select the decision di associated with the largest number rij in column j of the payoff table. Different assumptions about nature's behavior lead to different procedures for selecting “the best" decision.

If we know which state of nature will occur, we simply select the decision that yields the largest return for the known state of nature. In practice, there may be infinitely many possible decisions. If these possible decisions are represented by a vector d and the return by the real-valued function r(d), the decision problem can then be formulated as: max r(d) subject to feasibility constraints on d As indicated in Table 5.1, the X Company has two possible actions (decisions; d1 and d2): drill for oil or sell the land. With each decision, there is two possible states of natures: the land contains oil or not. The prior probabilities of the two states of nature are 0.25 and 0.75 respectively.

5.3 Decisions Under Uncertainty Choosing a specific decision depends upon the decision maker either if he/she is a risk averse or a risk taker. Thus, the following three methods will be considered: Maximin payoff criterion; the maximum likelihood criterion, and the expected value.

5.3.1

The Max-min Payoff Criterion

In this method, for each possible decision, the minimum payoff over all states of nature is determined. Then, select the maximum of these minimum payoffs. Finally, choose the decision whose minimum payoff gives this maximum. Going back to the prototype example of section 5.1, the minimum payoff for the first decision "Drill for oil" is LE100,000 and that for the second decision "Sell the land" is LE90,000. Then the selected decision is "Sell the land" as it has the maximum of the minimum payoffs.

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Such method provides the best guarantee of the payoff that will be obtained. This criterion is not often used in decision analysis as it is extremely conservative criterion. This criterion normally is of interest only to a very cautions decision maker.

5.3.2

The Maximum Likelihood Criterion

This method focuses on the most likely state of nature (i.e., the state of nature with highest probability of occurrence). Using this method, one should identify the most likely state of nature and then select the decision with the highest payoff in this state of nature. In the prototype example of section 5.1, the "Dry" state of nature has the maximum probability of occurrence "0.75". At this state of nature, the highest payoff is LE90,000 corresponding to second decision. Thus, the selected decision is "Sell the land".

This method assumes that the most important state of nature is the most likely one, and the chosen decision is the best one for the most important state of nature. However, this method does not rely on the probabilities of the other states of nature. This represents the major drawback of this method as it ignores all other information.

5.3.3

The Expected Return

This method is the most commonly one. It uses the best available estimates of the probabilities of the respective states of nature. Then, it calculates the expected value of the payoff for each of the possible decisions, and then chooses the action with maximum expected payoff. Expected value is a criterion for making a decision that takes into account both the possible outcomes for each decision alternative and the probability that each outcome will occur.

Illustrative Example Consider that two persons (a and b) are playing with rolling a die. Player "a" will pay to the other LE9, and then a fair die will be rolled. If the die comes up a 3, 4, 5,

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or 6, then player "b" will pay "a" LE15. If the die comes up 1 or 2, player "a" looses the LE9. Furthermore, player "b" agrees to repeat this game as many times as player "a" wishes to play. Should player "a" agree to play this game?

If a six-sided die is fair, there is a 1/6 probability that any specified side will come up on a roll. Therefore there is a 4/6 (2/3) probability that a 3, 4, 5, or 6 will come up and you will win. At first glance, this may not look like a good bet since "a" may lose LE9, while he/she can only win LE6. However, the probability of winning the LE6 is 2/3, while the probability of losing the LE9 is only 1/3. Perhaps this isn't such a bad bet after all since the probability of winning is greater than the probability of losing. The payoff table is shown below.

Table 5.2: Payoff Table of the Illustrative Example decision

State of nature Win Loose

Play Do not play

LE6 LE0

-LE9 LE0

Chance of state

2/3

1/3

The key to analyze this decision is that "b" allows "a" plays this game as many times as he/she wants. For example, how often would you expect to win if you play the game 1,500 times? Based on probability theory, you know that the proportion of games in which you will win over the long run is approximately equal to the probability of winning a single game. Thus, out of the 1,500 games, you would expect to win approximately (2/3) x 1500 = 1000 times. Therefore, over the 1,500 games, you would expect to win a total of approximately 1000 x LE6 + 500 x (LE9) = LE1500.

Based on this logic, what is each play of the game worth? If 1500 plays of the game are worth LE1500, then one play of the game should be worth LE1500 / 1500 = LE1. Accordingly, you will make an average of LE1 each time you play the game. Operations Research

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A little thought about the logic of these calculations shows that you can directly determine the average payoff from one play of the game by multiplying each possible payoff from the game by the probability of that payoff, and then adding up the results. For the die tossing game, this calculation is (2/3) x LE6 + (1/3) x LE(-9) = LE1.

The quantity calculated is called the expected value for an alternative; this value is a good measure of the value of an alternative since over the long run this is the average amount that you expect to make from selecting the alternative. Expected Value for an uncertain alternative is calculated by multiplying each possible outcome of the uncertain alternative by its probability, and summing the results. The expected value decision criterion selects the alternative that has the best expected value. In situations involving profits where "more is better," the alternative with the highest expected value is best, and in situations involving costs, where "less is better," the alternative with the lowest expected value is best.

In calculating the expected return, we make the assumption that there is more than one state of nature and that the decision maker knows the probability with which each state of nature will occur. Let pj be the probability that state j will occur. If the decision maker makes decision di, then the expected return ERi is: ERi = ri1p1 + ri2p2 + ….. + rimpm. The decision di* that maximizes ERi will be chosen namely ERi* = maximum over all i of ERi. Example 5.1 Let us consider the example of the newsboy problem: a newsboy buys papers from the delivery truck at the beginning of the day. During the day, he sells papers. Leftover papers at the end of the day are worthless. Assume that each paper costs 15 cents and sells for 50 cents and that the following probability distribution is known. Operations Research

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p0 = Prob { demand = 0 } = 2/10 p1 = Prob { demand = 1 } = 4/10 p2 = Prob { demand = 2 } = 3/10 p3 = Prob { demand = 3 } = 1/10 How many papers should the newsboy buy from the delivery truck?

To solve this exercise, we first construct the payoff table. Here rij is the reward achieved when i papers are bought and a demand j occurs.

Decision 0 1 2 3

0 0 -15 -30 -45

State of nature 1 2 0 0 35 35 20 70 5 50

3 0 35 70 105

Next, we compute the expected returns for each possible decision.

ER0 = 0(2/10)

+ 0(4/10)

+ 0(3/10)

+ 0(1/10)

=0

ER1 = -15(2/10)

+ 35(4/10)

+ 35(3/10)

+ 35(1/10)

= 25

ER2 = -40(2/10)

+ 10(4/10)

+ 60(3/10)

+ 60(1/10)

= 30

ER3 = -45(2/10)

+ 5(4/10)

+ 50(3/10)

+ 105(1/10)

= 18.50

The maximum occurs when the newsboy buys 2 papers from the delivery truck. His expected return is then 30 cents. The fact that the newsboy must make his buying decision before demand is realized has a considerable impact on his revenues. If he could first see the demand being realized each day and then buy the corresponding number of newspapers for that day, his expected return would increase by an amount known as the expected value of perfect information. Millions of dollars are spent every year on market research projects; geological tests etc, to determine what state of nature will occur in a wide variety of applications. Operations Research

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It should be pointed out that the criterion of maximizing expected return can sometimes produce unacceptable results. This is because it ignores downside risk. Most people are risk averse, which means they would feel that the loss of x dollars is more painful than the benefit obtained from the gain of the same amount. Decision theory deals with this problem by introducing a function that measures the “attractiveness" of money. This function is called the utility function. Instead of working with a payoff table containing the dollar amounts rij, one would instead work with a payoff table containing the utilities, say uij. The optimal decision di* is that which maximizes the expected utility: EUi = ui1p1 + ui2p2 + ……+ uimpm

over all i

5.4 Decision Trees Decision trees provide a useful way of visually displaying the problem and then organizing the computational work. These trees are especially helpful when a sequence of decisions must be made.

Example 5.2 Company ABC has developed a new line of products. Top management is attempting to decide on the appropriate marketing and production strategy. Three strategies are being considered, which we will simply refer to as A (aggressive), B (basic) and C (cautious). The market conditions under study are denoted by S (strong) or W (weak). Management's best estimate of the net profits (in millions of dollars) in each case is given in the following payoff table. Management's best estimates of the probabilities of a strong or a weak market are 0.45 and 0.55 respectively. Which strategy should be chosen?

State of nature

decision A B C

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S

W

30 20 5

-8 7 15

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Using the approach introduced earlier, we can compute the expected return for each decision and select the best one, just as we did for the newsboy problem.

ERA = 30(0.45) - 8(0.55)

= 9.15

ERB = 20(0.45) + 7(0.55)

= 12.85

ERC = 5(0.45) + 15(0.55)

= 10.50

The optimal decision is to select B.

A convenient way to represent this problem is through the use of decision trees, as in Figure 5.1. A square node represents a point at which a decision must be made, and each line leading from a square will represent a possible decision. A circular node represents situations where the outcome is uncertain, and each line leading from a circle will represent a possible outcome.

Figure 5.1: Decision tree for Example 2

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Using a decision tree to find the optimal decision is called solving the tree. To solve a decision tree, one works backwards. This is called folding back the tree. First, the terminal branches are folded back by calculating an expected value for each terminal node. See Figure 5.2. Management now faces the simple problem of choosing the alternative that yields the highest expected terminal value. So, a decision tree provides another, more graphic, way of viewing the same problem. Exactly the same information is utilized, and the same calculations are made.

Figure 5.2: Reduced decision tree for Example 2

Solving trees for expected-value

1. For each of the rightmost nodes proceed as follows: (a) If the node is a decision node, determine the best alternative to take. The payoff from this alternative is the value of this decision node. Arrow the best alternative. (b) If the node is a chance node, calculate the expected value. This expected value is the value of this chance node. 2. For the nodes one to the left proceed as follows:

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(a) If the node is a decision node, determine the best alternative to take, using as needed, and the values of future nodes (nodes to the right) as payoffs. The payoff from this alternative is the value of this decision node. Arrow the best alternative. (b) If the node is a chance node, calculate the expected value, using, as needed, the values of future nodes (nodes to the right) as payoffs. The expected value is the value of this chance node. 3. Repeat Step 2 as needed, until the leftmost node is reached. Following the arrows from left to right gives the sequence of appropriate decisions to take.

5.5 Sensitivity Analysis The expected return of strategy A is: ERA = 30 P(S) – 8 P(W) or, equivalently, ERA = 30 P(S) - 8(1 - P(S)) = - 8 + 38 P(S) Thus, this expected return is a linear function of the probability that market conditions will be strong. Similarly ERB = 20 P(S) + 7(1 – P(S)) = 7 + 13 P(S) ERC = 5 P(S) + 15(1 - P(S)) = 15 – 10 P(S) We can plot these three linear functions on the same set of axes (see Figure 5.3). This diagram shows that Company ABC should select the basic strategy (strategy B) as long as the probability of a strong market demand is between P(S) = 0.348 and P(S) = 0.6. This is reassuring, since the optimal decision in this case is not very sensitive to an accurate estimation of P(S). However, if P(S) falls below 0.348, it becomes optimal to

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choose the cautious strategy C, whereas if P(S) is above 0.6, the aggressive strategy A becomes optimal.

Figure 5.3: Expected return as a function of P(S)

5.6 Sequential Decisions 5.6.1

Conditional Probability

Example 5.3 An experiment consists of rolling a die once. Let X be the outcome. Let F be the event that X = 6, and let E be the event that X > 4. We assign the distribution function m(ω) = 1/6 for ω = 1, 2, ……., 6. Thus, P(F) = 1/6. Now, suppose that the die is rolled and we are told that the event E has occurred. This leaves only two possible outcomes: 5 and 6. In the absence of any other information, we would still regard these outcomes to be equally likely, so the probability of F becomes 1/2, making P(F/E) = 1/2.

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Example 5.4 In the Life Table, one finds that in a population of 100,000 females, 89.835% can expect to live to age 60, while 57.062% can expect to live to age 80. Given that a woman is 60, what is the probability that she lives to age 80?

This is an example of a conditional probability. In this case, the original sample space can be thought of as a set of 100,000 females. The events E and F are the subsets of the sample space consisting of all women who live at least 60 years, and at least 80 years, respectively. We consider E to be the new sample space, and note that F is a subset of E. Thus, the size of E is 89,835, and the size of F is 57,062. So, the probability in question equals 57,062/89,835 = 0.6352. Thus, a woman who is 60 has a 63.52% chance of living to age 80.

5.6.2

Decisions Under Conditional Probability

Frequently, additional testing can be done to improve the preliminary estimates of the probabilities of the respective state of nature provided by the prior probabilities.

Going back to the prototype example of section 5.1, an available option before making a decision of drill for oil or sell the land is to conduct a seismic survey of the land to obtain a better estimate of the probability of oil. The cost of this seismic study is LE30,000. This seismic survey indicates whether the geological structure is favorable to the presence of oil.

The seismic survey has two states of nature: S = 1 (oil is fairly likely); S = 0 (oil is fairly unlikely). From past experience, if there is oil such seismic study were encouraging (favorable) 60% of the time and they were discouraging (unfavorable) 40% of the time. If there is no oil such seismic study were encouraging (favorable) 20% of the time and they were discouraging (unfavorable) 80% of the time. This can be represented using the decision tree as follows (Figure 5.4)

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670 Oil Drill

f Dry

-130

c Sell

60

Unfavorable S=0

670 Oil b

g Favorable S=1

Do seismic survey

Dry

Drill d

-130

Sell

60

700 Oil

a h

No seismic survey

Dry

-100

Drill e

90

Sell

Figure 5.4: Decision tree of the prototype example

Example 5.5 As stated in Example 2, although the basic strategy B is appealing for Company ABC, ABC's management has the option of asking the marketing research group to perform a market research study. Within a month, this group can report on whether the study was encouraging (E) or discouraging (D). In the past, such studies have tended to be in the right direction: When market ended up being strong, such studies were encouraging 60% of the time and they were discouraging 40% of the time. Whereas, when market ended up Operations Research

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being weak, these studies were discouraging 70% of the time and encouraging 30% of the time. Such a study would cost $500,000. Should management request the market research study or not?

Figure 5.5: Test versus no-test decision tree

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Let us construct the decision tree for this sequential decision problem. See Figure 5.5. It is important to note that the tree is created in the chronological order in which information becomes available. Here, the sequence of events is: - Test decision - Test result (if any) - Make decision - Market condition The leftmost node corresponds to the decision to test or not to test. Moving along the "Test" branch, the next node to the right is circular, since it corresponds to an uncertain event. There are two possible results. Either the test is encouraging (E), or it is discouraging (D). The probabilities of these two outcomes are P(E) and P(D) respectively. How does one compute these probabilities?

The information we are given is conditional. Given S, the probability of E is 60% and the probability of D is 40%. Similarly, we are told that, given W, the probability of E is 30% and the probability of D is 70%. We denote these conditional probabilities as follows: P(E|S) = 0.6

P(E|W) = 0.3

P(D|S) = 0.4

P(D|W) = 0.7

In addition, we know P(S) = 0.45 and P(W) = 0.55. This is all the information we need to compute P(E) and P(D). Indeed, for events S1; S2; ….., Sn that partition the space of possible outcomes and an event T, one has: P(T) = P(T|S1) P(S1) + P(T|S2) P(S2) + ….. + P(T|Sn) P(Sn) Here, this gives: P(E)

= P(E|S) P(S) + P(E|W) P(W) = (0.6) (0.45) + (0.3) (0.55) = 0.435

and

P(D)

= P(D|S) P(S) + P(D|W) P(W) = (0.4) (0.45) + (0.7) (0.55) = 0.565

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As we continue to move to the right of the decision tree, the next nodes are square, corresponding to the three marketing and production strategies. Still further to the right are circular nodes corresponding to the uncertain market conditions: either weak or strong. The probability of these two events is now conditional on the outcome of earlier uncertain events, namely the result of the market research study, when such a study was performed. This means that we need to compute the following conditional probabilities: P(S|E); P(W|E); P(S|D) and P(W|D). These quantities are computed using the formula: P(R|T) = P(T|R) P(R) / P(T) This is valid for any two events R and T. Here, we get P(S|E) = P(E|S) P(S) / P(E) = (0.6) (0.45) / 0.435 = 0.621 Similarly,

P(W|E) = 0.379 P(S|D) = 0.318

P(W|D) = 0.682

Now, we are ready to solve the decision tree. As earlier, this is done by folding back. See Figures 5.6, 5.7 and 5.8. You fold back a circular node by calculating the expected returns. You fold back a square node by selecting the decision that yields the highest expected return. The expected return when the market research study is performed is 12.96 million dollars, which is greater than the expected return when no study is performed. So the study should be undertaken.

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Figure 5.6: Solving the tree

Figure 5.7: Solving the tree Operations Research

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Figure 5.8: Solving the tree 5.7 Solved Examples 5.7.1

Example 1

An art dealer has a client who will buy the masterpiece Rain Delay for $50,000. The dealer can buy the painting now for $40,000 (making a profit of $10,000). Alternatively, he can wait one day, when the price will go down to $30,000. The dealer can also wait another day when the price will be $25,000. If the dealer does not buy by that day, then the painting will no longer be available. On each day, there is a 2/3 chance that the painting will be sold elsewhere and will no longer be available. (a) Draw a decision tree representing the dealer’s decision making process. (b) Solve the tree. What is the expected profit? When should he buy the painting?

Solution a.

Figure 5.9: Decision tree for Example 5.6 Operations Research

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b. Folding back the tree, the value is 10, for an expected profit of $10,000. He should buy the painting immediately.

5.7.2

Example 2

The Scrub Professional Cleaning Service receives preliminary sales contracts from two sources: its own agent and building managers. Historically, 3/8 of the contracts have come from the Scrub agent and 5/8 from building managers. Unfortunately, not all preliminary contracts result in actual sales contracts. Actually, only 1/2 of those preliminary contracts received from building managers result in a sale, whereas 3/4 of those received from the Scrub agent result in a sale. The net return to Scrub from a sale is $6400. The cost of processing and following up on a preliminary contract that does not result in a sale is $320. What is the expected return associated with a preliminary sales contract?

Solution The decision tree for scrub professional cleaning service:

Reduced decision tree:

Therefore, the expected return is = 3/8 * 4720 + 5/8 * 3440 = 3920 Operations Research

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5.7.3

Example 3

Walter's Dog and Pony show is scheduled to appear in Cedar Rapids on July 4. The profits obtained are heavily dependent on the weather. In particular, if the weather is rainy, the show loses $28,000 and if sunny, the show makes a profit of $12,000. (We assume that all days are either rainy or sunny.) Walter can decide to cancel the show, but if he does, he forfeits a $1,000 deposit he put down when he accepted the date. The historical record shows that on July 4, it has rained 1/4 of the time for the last 100 years.

(a) What decision should Walter make to maximize his expected net dollar return?

Walter has the option to purchase a forecast from Victor's Weather Wonder. Victor's accuracy varies. On those occasions when it has rained, Victor has been correct (i.e. predicted rain) 90% of the time. On the other hand, when it has been sunny, he has been right (i.e. predicted sun) only 80% of the time.

(b) If Walter had the forecast, what strategy should he follow to maximize his expected net dollar return?

Solution a) According to the decision tree given below Walter should show.

Decision tree for Walter:

=

Expected net dollar return = 2000

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b) If Walter had the forecast. Prob (predicted sun) = Prob (predicted sun|sun) * Prob (sun) + Prob (predicted sun|rain) * Prob (rain)

Prob (sun|predicted sun) = [Prob (predicted sun|sun) * Prob (sun)] / Prob (predicted sun) = (0.8 * ¾) / (0.8 * ¾ + 0.1 * ¼) =

= 0.96

Prob (rain|predicted sun) = 1 – 0.96 = 0.04 Prob (rain|predicted rain) = (0.9 * 1/4) / (0.9 * 1/4 + 0.2 * 3/4) = 0.6 Prob (sun|predicted rain) = 1 – 0.6 = 0.4

If predicted sun: show: expected value = 0.96 * 12000 – 0.04 * 28000 = 10400 cancel: expected value = - 1000 If predicted rain: show: expected value = 0.4 * 12000 – 0.6 * 28000 = - 12000 cancel: expected value = - 1000

The strategy should be followed is: If predicted sun, then show. If predicted rain, then cancel. Operations Research

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5.7.4

Example 4

A firm can delay its launch of a new product or not delay. The success of the product depends on the growth rate of the economy. If the economy enjoys a high rate of growth, then the firm stands to earn $20 million. If the rate of growth is moderate, then the firm makes $5 million. Finally if there is low or no growth, then firm will lose $5 million. If the firm delayed its launch of the new product, this will allow the firm to learn the state of the economy before making its launch decision. Delay is not without cost. IF the firm delayed its launch of the new product, this will cost the firm $D million. The probabilities that the economy rate growth being high, moderate, or low are 02, 0.4, and 0.4, respectively. a. Draw the decision tree for this example showing the payoffs and the probabilities of each state on the tree. b. What is the maximum value of D if the company decided to delay the launch of its product? Solution

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b. From the above decision tree, we see the firm should delay if $6 million – 3/5D ≥ $4 million. Or, solving for D, if D ≤ $10/3 million.

5.8 Exercises 1.

As winter approaches, you are concerned that your dewatering system will not be able to keep the site dry and will delay progress. You know that any significant amount of rain will flood the site. Additional pumps will cut into a slim profit margin, if not generate losses. Also, delays combined with liquated damages will reduce profit. Which of these two options (do nothing or install additional pumps) is the desired option? Considering that, the probability that the rain will be less than 6 inch in 12-hour period is 0.5 (S1), the probability that the rain will reach 6 inch one time in 12-hour period is 0.3 (S2), and the probability that the rain will reach 6 inch many times in 12-hour period is 0.2 (S3). The pound values representing costs are shown in the following table.

Alternative

S1

S2

S3

Install pumps

15,000

15,000

65,000

Do noting

0

20,000

100,000

a. Identify the actions, state of nature, and payoff table and draw the decision tree. b. Determine the better decision.

2.

A university must decide between two plans for starting a graduate program in a new academic year. The goal is to maximize the increase in student population, but it is unclear whether the interest in this new area will be high, medium, or low. Projected increases in student populations and their probabilities are shown below for each plan.

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Increase Interest

Probability Plan 1

Plan 2

High

0.6

220

200

Medium

0.3

170

180

Low

0.1

110

150

a. Develop a decision analysis formulation for this problem by identifying the actions, state of nature, and payoff table and draw the decision tree. b. What is the optimal action? 3.

You, as a primary contractor, are suing a subcontractor over failure to complete a portion of a construction project. The subcontractor’s bonding company has offered to settle out of court for LE410,000. You are considering making a counteroffer of LE500,000. The bonding company may accept the offer, refuse the offer ad go to trial, or make a counteroffer. Your lawyer has dealt with the bonding company on similar suits. He/she says there is a 20% chance the company will counter your counteroffer with LE450,000 and 40% chance it will go to the court. If the bonding company counters your counteroffer, you can accept or decline and go to court. If the case goes to court, there is a 10% chance you will be awarded LE1,000,000 a 35% chance of a LE500,000 award, and a 55% chance of the suit’s being dismissed as trivial. What course of action should you pursue?

4.

Two pumping systems A and B are suggested for supplying water. The construction cost for systems A and B is LE250,000 and LE750,000, respectively. If partial failure occurs, it is expected that damage cost for systems A and B is LE800,000 and LE150,000, respectively. If complete failure occurs, it is expected that damage cost for systems A and B is LE1,500,000 and LE600,000, respectively. The probabilities of partial and complete failures are 5% and 1 %, respectively.

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a. Identify the actions, state of nature, and payoff table and draw the decision tree. b. Determine the better system to minimize losses.

5.

Consider a project that needs 20-Mgal/day of water. Two alternatives are under study. Alternative A is to use a primary pump of 20-Mgal/day and a backup ump with the same capacity. Alternative B is to use two primary pups of 10-Mgal/day each and a backup pump f 10-Mgal/day. The probabilities of failure for both systems are shown in the table below. Probability Alternative Q=0

Q = 10

Q = 20

A

0.0025

0.0

0.9975

B

0.001

0.006

0.993

The capital cost for construction of a pumping station, in millions of dollars, s a function of flow q in million gallons/day (Mgal/day) is C = 0.035q1.25. Failure to deliver 20 Mgal/day of water is assumed to cause inconvenience and economic loss. The annual losses are estimated to be as given in the table below. q (Mgal/day)

0

10

20

Annual loss

$250M

$25M

0

Draw a decision tree and determine the better pump system.

6.

Wildcat Oil is considering spending $100,000 to drill at a particular spot. The result of such a drilling is either a “DryWell" (of no value), a “Wet Well" (providing $150,000 in revenues) or a “Gusher" (providing $250,000 in revenues). The probabilities for these three possibilities are 0.5, 0.3 and 0.2 respectively.

(a) Draw a decision tree for the problem of deciding whether to drill or not.

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Dr. Emad Elbeltagi

(b) Solve the decision tree assuming the goal is to maximize the expected net revenue. Should the company drill? (c) Close-enough Consultants offer to use their specialized seismic hammer. This hammer returns either encouraging or discouraging results. In the past, when applied to a Gusher, the hammer always returned encouraging results. When applied to a Wet Well, it was encouraging 75% of the time and discouraging 25% of the time. When applied to a Dry Well, it was encouraging one-third of the time and discouraging two-thirds of the time. What strategy should be followed to maximize the expected net return?

7.

Omar has been employed at Gold Real Estate Company at a salary of LE 2000 per month during the last year. Because Omar is considered to be a top salesman, the manager of the company is offering him one of three salary plans for the next year: 1) a 25% increase to LE 2500 per month; 2) a base salary of LE 1000 per month plus LE 600 per house sold; or, 3) a straight commission of LE 1000 per sold house. Suppose that during the past year the following is Omar’s distribution of home sales. Number of months

1

2

1

2

1

3

2

Home sales

0

1

2

3

4

5

6

(a) Calculate the monthly salary payoff table for Omar; showing the possible decisions, their state of natures, and probability of occurrence. (b) Find Omar’s optimal decision using: the conservative approach; and the optimistic approach. (c) Assuming that the above table is a typical distribution for Omar’s monthly sales, using the expected value, what salary plan should Omar select?

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REFERENCES Anderson, D.R., Sweeney, D.J., and Williams, T.A. (2005). An Introduction to Management Science: Quantitative Approaches to Decision Making, 11th Edition, Thompson South-Western College Publishing

Cornuejols, G., and Trick, M. (2002). Quantitative Methods for the Management Sciences, Course Notes, Graduate School of Industrial Administration, Carnegie Mellon University, Pittsburgh, PA, USA Griffis, F.H., Farr, J.V., and Morris, M.D. (2000). Construction Planning for Engineers, McGraw-Hill, Inc., New York. Hillier, F.S., and Liberman, G.J. (1990). Introduction to Mathematical Programming, McGraw-Hill, Inc., New York.

Hillier, F.S., and Liberman, G.J. (1995). Introduction to Operations Research, McGrawHill, Inc., New York.

Revelle. C.S., Whitlatch. E.E., and Wright, J.R. (2004). Civil and Environmental Systems Engineering, Prentice Hall, Inc., upper Saddle River, New Jersey.

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