SUPPLEMENT TO

CHAPTER 5 Decision Analysis Supplement Outline

Introduction and Decision Tables, 2 Decision Trees and Influence Diagrams, 4 Value of Information, 7 Sensitivity Analysis, 9 OM in Action: Oglethorpe Power, 11 Summary, 12

Key Terms, 12 Solved Problems, 12 Discussion and Review Questions, 14 Internet Exercises, 15 Problems, 15 Mini-Case: Southern Company, 21 Mini-Case: ICI Canada, 22

Learning Objectives After completing this supplement, you should be able to:

Model a single-stage decision problem as a decision table and use the expected value approach to solve it.   LO 2 Construct a decision tree and use it to analyze a multistage problem; describe an influence diagram.   LO 3 Calculate the expected value of perfect and imperfect (sample) information.   LO 4 Conduct sensitivity analysis on a simple decision problem.   LO 1

2

PART THREE   System Design

Introduction and Decision Tables

LO 1

Decision analysis is a scientific and practical method for making important decisions. It was introduced in the 1960s (see http://decision.stanford.edu/library/ronald-a.-howard/­ Decision%20Analysis-%20Applied%20Decision%20Theory.pdf). Decision analysis involves identification, clear representation, and formal assessment of important aspects of a decision and then determination of the best decision by applying the maximum expected value criterion. Decision analysis is suitable for a wide range of operations management decisions where uncertainty is present. Among them are capacity planning, product design, equipment selection, and location planning. Decisions that lend themselves to a decision analysis approach tend to be characterized by the following elements: 1. There is at least one random variable with a set of possible values (or future conditions or states of nature) that will have a bearing on the result of the decision. 2. There is a decision with a list of alternatives for the manager/decision maker to choose from. 3. There is a known payoff (or value) for each alternative under each possible future condition. To use this approach, a decision maker would employ the following process: 1. Determine the goal or objective, e.g., maximize expected profit or net present value, or minimize expect cost. 2. Develop a list of possible alternatives for the decision in order to achieve the goal. 3. Identify possible future conditions or states of nature for each random variable (e.g., demand will be low, medium, or high; the equipment will or will not fail; the competitor will or will not introduce a new product) that will affect the goal. 4. Determine or estimate the payoff (or value) associated with each alternative for every possible future condition. 5. Estimate the likelihood of each possible future condition for each random variable. 6. Evaluate the alternatives according to the goal or decision criterion, and select the best alternative. payoff table  Table showing the payoffs (values) for each alternative in every possible state of nature of a onerandom-variable problem.

The data for a one random-variable problem can be summarized in a payoff table, which shows the payoffs (value) for each alternative under the various possible states of nature of the variable. These tables are helpful in choosing among alternatives because they facilitate comparison of alternatives. Consider the following payoff table, which illustrates a capacity planning problem. Possible Future Demand Alternatives

Low

Moderate

High

Small facility

$10*

$10

$10

Medium facility

7

12

12

Large facility

(4)

2

16

*Net present value in $ millions.

The payoffs are shown in the body of the table. In this instance, the payoffs are in terms of net present values, which represent equivalent current dollar values of estimated future income less costs. This is a convenient measure because it places all alternatives on a comparable basis. If a small facility is built, the payoff will be $10 million for all three possible states of nature. For a medium facility, low demand will have net present value of $7 million, whereas both moderate and high demand will have net present values of

SUPPLEMENT TO CHAPTER 5   Decision Analysis

3

$12 million. A large facility will have loss of $4 million if demand is low, net present value of $2 million if demand is moderate, and net present value of $16 million if demand is high. The problem for the decision maker is to select one of the alternatives, taking the net present values into account. In order to do so, one needs to determine the probability of occurrence for each state of nature (unless one of the alternatives dominates the rest, i.e., would have the highest NPV under each state of nature. Because the states of nature for a random variable should be mutually exclusive and collectively exhaustive, these probabilities must add upto 1.00. These probabilities are usually subjective opinions of experts. The decision analyst interviewing an expert should be careful not to ask leading questions and should be aware of psychological biases such as overconfidence or intentional underestimation by marketing/sales managers (to make meeting these goals easier). Given the probabilities for each state of nature, a widely used approach is the expected value method. The expected value is calculated for each alternative, and the one with the highest expected value is selected (when more is better such as for NPV). The expected value for an alternative is the sum over the states of nature of the payoff multiplied by probability of each state of nature.

Using the expected value method, identify the best alternative for the following payoff table given these probabilities for demand: low = .30, moderate = .50, and high = .20.

Example S-1

Possible Future Demand Alternatives

Low

Moderate

High

Small facility

$10*

$10

$10

Medium facility

7

12

12

Large facility

(4)

2

16

*Net present value in $ millions.

Find the expected value of each alternative by multiplying the probability of occurrence for each state of nature by the payoff of the alternative for that state of nature and summing them: EVsmall = .30($10) + .50($10) + .20($10) = $10 EVmedium = .30($7) + .50($12) + .20($12) = $10.5 EVlarge = .30(−$4) + .50($2) + .20($16) = $3 Hence, choose the medium facility because it has the highest expected net present value.

The expected value method is most appropriate when a decision maker is neither risk averse nor risk seeking, but is risk neutral. Typically, well-established organizations facing decisions of this nature tend to use the expected-value method because it provides the long-run average payoff. That is, the expected-value amount (e.g., $10.5 million in Example S-1) is not an actual payoff but an expected or average amount that would be achieved if a large number of identical decisions were to be made. Hence, if a decision maker applies this method to a large number of similar decisions, the expected payoff for the total will approximate the sum of the individual expected payoffs. If the decision maker is not risk neural, then he tends to consider the spread or risk of payoffs, and even individual payoffs. For a risk-averse decision maker, the minimum payoff, if the medium facility is built, $7 million, may still be acceptable. If it is not, he may look for a new alternative (e.g., seek

Solution

PART THREE   System Design

4

FIGURE 5S-1

of State

Format of a decision tree 1 eA

State

s

oo

Ch

� e A1

of na

ture 2

Ch

oo

se

Choos

A

2

State

Decision point

ure 1

f nat

o State

of na

ture 2

Payoff 2

2 Possible second decisions

Initial decision

1

Payoff 1

e1

natur

Choos

e A2�

Payoff 3

� e A3

Payoff 4

Choos

2

Choos

e A4�

Payoff 5

Payoff 6

Chance event

more information) or choose to build a small facility that has a higher minimum payoff ($10 million). An approach to deal with risk-averse decision makers is called the utility theory. More complex decision problems will have more than one decision variable (i.e., a multistage decision problem) and/or more than one random variable. In this case, the payoffs cannot be represented by a table. Instead, decision trees are used to graphically model the decision problem.

Decision Trees and Influence Diagrams

LO 2

decision tree  A graphical representation of the decision variables, random variables, probabilities, and payoffs.

Example S-2

A decision tree is a graphical representation of the decision variables, random variables and their probabilities, and the payoffs. The term gets its name from the treelike appearance of the diagram (see Figure 5S-1). Decision trees are particularly useful for analyzing situations that involve sequential or multistage decisions. For instance, a manager may initially decide to build a small facility but she has to allow for the possibility that demand may be higher than anticipated. In this case, the manager may plan to make a subsequent decision on whether to expand or build an additional facility. A decision tree is composed of a number of nodes that have branches emanating from them (see Figure 5S-1). Square nodes denote decision points, and circular nodes denote chance events. Read the tree from left to right. Branches leaving square nodes represent alternatives; branches leaving circular nodes represent the states of nature. After a decision tree has been drawn and necessary data are determined, it is analyzed from right to left; that is, starting with the last decision that might be made it is “rolled” back. For each decision, choose the alternative that will yield the greatest return (or the lowest cost). For each chance node, calculate the expected value of the payoffs of its states of nature. If chance events follow a decision, choose the alternative that has the highest expected value (or lowest expected cost).

A manager must decide on the size of a video arcade to construct. The manager has narrowed the choices to two: large or small. Information has been collected on payoffs, and the following decision tree has been constructed. Analyze the decision tree and determine

SUPPLEMENT TO CHAPTER 5   Decision Analysis

5

which initial alternative (build small or build large) should be chosen in order to maximize expected value.

Low

all

High

m ds

uil

B

nd

a dem

(.4)

dem

and

$40

D

(.6)

$40

g

hin

ot on

Exp

and ing

oth

n Do

Bu

ild

lar

ge

and

em wd

Lo

High

dem

and

(.4)

Red

uce

(.6)

pric

es

$55 ($10)

$50

$70

The dollar amounts at the branch ends indicate the estimated payoffs if the sequence of decisions and chance events occurs. For example, if the initial decision is to build a small facility and it turns out that demand is low, the payoff will be $40 (thousand). Similarly, if a small facility is built, and demand turns out high, and a later decision is made to expand, the payoff will be $55. The figures in parentheses on branches leaving the chance nodes indicate the probabilities of those states of nature. Hence, the probability of low demand is .4, and the probability of high demand is .6. Payoffs in parentheses indicate losses. Analyze the decisions from right to left: 1. Determine which alternative would be selected for each possible second decision. For a small facility with high demand, there are two choices: do nothing, or expand. Because expand has higher payoff, you would choose it. Indicate this by placing a double slash through do nothing alternative. Similarly, for a large facility with low demand, there are two choices: do nothing or reduce prices. You would choose reduce prices because it has the higher expected value, so a double slash is placed on the other branch. 2. Determine the product of the chance probabilities and their respective payoffs for the remaining branches: Build small Low demand .4($40) = $16 High demand .6($55) = 33 Build large Low demand .4($50) = 20 High demand .6($70) = 42 3. Determine the expected value of each initial alternative: Build small $16 + $33 = $49 Build large $20 + $42 = $62 Hence, the choice should be to build a large facility because it has higher expected value than the small facility.

Solution

PART THREE   System Design

6

The above problem was entered in the decision analysis software DPL, from Syncopation Software (www.syncopation.com). The solution is shown below:

Small

Demand [49.000]

Low

[40.000]

40%

40.000

High

React to high Demand [55.000]

60%

Build [62.000]

Large

Demand [62.000]

React to low demand [50.000]

Low

Do nothing

[40.000]

Expand

[55.000]

40.000

Do nothing Reduce prices

40%

55.000 [–10.000] –10.000 [50.000] 50.000

High

[70.000]

60%

70.000

How to use DPL software  In the DPL’s window, the top pane is for the influence diagram (described next), and the bottom pane is for the decision tree (see the snapshot below). The decision tree was drawn by first adding decision nodes (by clicking on the square icon at the top and moving it into the top pane) and discrete chance nodes (by clicking on the green circle icon at the top and moving it into the top pane), and then moving them to the bottom pane in order to draw the decision tree. Each node is then double-clicked and its properties defined (name, alternatives/outcomes, probabilities, nature of node’s branches: symmetric vs. asymmetric). Next, each payoff is specified by double-clicking each branch. Finally, the expected value is computed by clicking on Run (at the top) and then Decision Analysis.

React to Demand Build

Demand React to low demand

Demand

Low 40 Build

React to Demand Do nothing

Small

40

High Expand 55

Large Demand Do nothing –10

Low

Reduce prices 50 High 70

SUPPLEMENT TO CHAPTER 5   Decision Analysis

Influence Diagrams Influence diagrams (see example below) can graphically represent complex decision problems that have many random variables (chance events) and one or more decision variables. Influence diagrams are more concise than decision trees because they do not show the alternatives branches coming out of the decision nodes and the states of nature branches coming out of the chance nodes. Constructing and validating an influence diagram improves communication and consensus building at the beginning of the decision modelling process. The following is an example of the influence diagram representing the decision of whether or not to introduce a new product. The green circles show the random variables (chance events) and the rounded yellow squares show the payoff or part of it. This influence diagram for a new product decision also involves a pricing decision. The uncertainties (i.e., random variables) are units sold, which are affected by the pricing decision, fixed cost, and variable cost. Profit is the ultimate payoff, which is influenced by the total cost and revenue. Decision analysis software such as DPL are able to measure the effect of variability of each random variable on the payoff (e.g., profit) and identify those crucial random variables that have the greatest influence on profit. The result is plotted in a vertical bar chart called a Tornado diagram, named as such because the more important random variables, those that affect the payoff most and hence have a longer bar, are drawn on the top, making the chart look like a tornado (see e.g., http://en.wikipedia.org/wiki/Tornado_diagram). Fixed Cost

Units Sold

Price?

Variable Cost Total Cost

Revenue Introduce Product? Profit

Source:  K. Chelst, “Can’t See the Forest Because of the Decision Trees: A Critique of Decision Analysis in Survey Texts,” Interfaces (28)2, March–April 1998, pp. 80–98.

Value of Information

LO 3

In certain situations, it is possible to know with more certainty which state of nature of a random variable will actually occur in the future or be able to prepare better for various states of nature. For instance, the choice of location for a restaurant may weigh heavily on whether a new highway will be constructed nearby. A decision maker may have probabilities for this random variable; however, it may be possible to delay a decision until it is more clear which state of nature will occur. This might involve taking an option to buy the land. If the state of nature is favourable, the option can be exercised; if it is unfavourable, the option can be allowed to expire. The question to consider is whether the cost of the option will be less than the expected gain due to delaying the decision. Possible ways of obtaining information about a random variable depend on the nature of the random variable. Information about consumer preferences might come from market research, and information about potential of a product could come from product testing. If the information is perfect (i.e., it is 100 percent accurate and will pinpoint the state of nature of the random variable), then the expected gain in payoff is called the expected value of perfect information, or EVPI.

7

8

PART THREE   System Design

expected value of perfect information (EVPI)  The difference between the expected payoff with perfect information and the expected payoff without the information.

Expected value of perfect information (EVPI) is the difference between the expected payoff with perfect information (under certainty) and the expected payoff without the information (under risk).

Example S-3

Solution

Expected value of Expected payoff Expected payoff = − perfect information under certainty under risk

(5S-1)

Using the information from Example S-1, determine the expected value of perfect information using Formula 5S-1. First, calculate the expected payoff under certainty. To do this, identify the best payoff under each state of nature. Then combine these by weighting each payoff by the probability of that state of nature and adding the amounts. Thus, the best payoff under low demand is $10, the best payoff under moderate demand is $12, and the best payoff under high demand is $16. The expected payoff under certainty is, then: .30($10) + .50($12) + .20($16) = $12.2 Possible Future Demand Alternatives

Low

Moderate

High

Small facility

$10*

$10

$10

Medium facility

7

12

12

Large facility

(4)

2

16

*Net present value in $ millions.

The expected payoff under risk, as calculated in Example S-1, is $10.5. The EVPI is the difference between these: EVPI = $12.2 − $10.5 = $1.7

expected value of sample information (EVSI)  The difference between the expected payoff with sample (imperfect) information and the expected payoff without sample information.

EVPI indicates the upper limit on the amount the decision maker should be willing to spend to obtain information. Thus, if the cost exceeds EVPI, the decision maker would be better off not spending additional money and simply going with the alternative that has the highest expected payoff. In most cases, the information that can be obtained about a random variable is useful but not perfect. In these cases, a statistical result called Bayes Rule can be used to update the (prior) probability distribution of the states of nature of the random variable. The difference between the expected payoff with imperfect (sample) information and the expected payoff without sample information is called the expected value of sample information (EVSI). Bayes Rule:  Let A be an “information” event, and let B1 and B2 be mutually exclusive and

exhaustive states of nature. Let P(B1) and P(B2) be the prior probabilities of the states of nature, and P(A | B1) and P(A | B2) be the likelihood of observing A given B1 and B2, respectively (i.e., the conditional probabilities). Then, the posterior probabilities P(B1 | A) and P(B2 | A) are:

(

) (P (A | B ) P (B ) + P (A | B ) P (B ))

P (B 1 | A ) = P (A | B1 ) P (B1 )

P (B 2 | A ) = 1 − P (B 1 | A )

1

1

Note: P(A) = P(A | B1) P(B1) + P(A | B2) P(B2  ).

2

2

SUPPLEMENT TO CHAPTER 5   Decision Analysis

A manager is considering random drug testing of his employees. However, the test is not 100 percent accurate. The conditional probabilities are as follows: If a person uses drugs, then the test will be positive 93 percent of the time, whereas if he is not, the test will be negative 97 percent of the time. Suppose that 5 percent of the employees are drug users.

9

Example S-4

a. What are the posterior probabilities? b. If the cost of not identifying (and dismissing) a drug user is $1,000, cost of falsely accusing a non-user is $200, and other costs are zero, what is the maximum the manager should be willing to pay for a drug test, i.e., EVSI? a. Consider a random employee. Let B1 = the event that he is a drug user, B2 = the event that he is not a drug user. We have P(B1) = .05 and hence P(B2) = 1 – .05 = .95. Let A + = the event that test result is positive. We have P(A + | B1) = .93 and P(A + | B2) = 1 – .97 = .03. Then,

(

) ( .93 ( .05 ) + .03 ( .95 )) = .62

P ( B 1 | A + ) = .93 ( .05 )

Solution

and

P ( B2 | A + ) = 1 − .62 = .38.

Also, P(A +) = P(A + | B1) P(B1) + P(A + | B2) P(B2) = .93(.05) + .03(.95) = .075, And P(A−) = 1 – P(A+) = 1 – .075 = .925  ote the relatively low (62 percent) probability that a drug user will be identified and a N relatively high (38 percent) probability that a non-drug user will be implicated. Similarly (A – | B1) = 1 – .93 = .07 and P(A – | B2) = .97, and

(

) ( .07 ( .05 ) + .97 ( .95 )) = .0038

P ( B 1 | A − ) = .07 ( .05 )

and

P ( B2 | A − ) = 1 − .0038 = .9962.

b. EVSI can be determined from the following decision tree. Note that negative payoffs are costs. Positive Yes

Test result [–9.215]

Test [–9.215]

Employee 1 [–76.000]

7.50%

User 62.00% Non-user 38.00%

Negative

Employee 2 [–3.800]

User 0.38% Non-user

92.50%

99.62% No

Employee 3 [–50.000]

User 5.00% Non-user 95.00%

[–1000.000] –1000.000 [0.000] 0.000

The maximum that the manager should pay for a drug test (i.e., EVSI) is ­$50 - $9.215 = $40.785.

Sensitivity Analysis

LO 4

Generally speaking, both the payoffs and the probabilities in a decision problem are estimated values. Consequently, it can be useful for the decision maker to have some indication

[0.000] 0.000 [–200.000] –200.000 [–1000.000] –1000.000 [0.000] 0.000

10

sensitivity analysis  Determining the range of probability for which an alternative has the best expected payoff.

Example S-5

PART THREE   System Design

of how sensitive the choice of an alternative is to changes in one or more of these values. Unfortunately, it is impossible to consider all possible combinations of every change in a typical problem. Nevertheless, the sensitivity of the chosen alternative to probability estimates can be determined for a simple problem. Sensitivity analysis provides the range of probability over which an alternative has the best expected payoff. The approach illustrated here is useful when there are two states of nature. It involves constructing a graph and then using algebra to determine the range of probabilities for which a given alternative is best. In effect, the graph provides a visual indication of the range of probability over which various alternatives are optimal, and the algebra provides exact values of the endpoints of the ranges. Example S-5 illustrates the procedure. Given the following (revenue) payoff table, determine the range of probability for state of nature #2, that is, P2, for which each alternative is optimal using the expected value method. State of Nature

Alternative

Solution

#1

#2

A

4

12

B

16

2

C

12

8

First, plot the expected payoff of each alternative relative to P2. To do this, plot the #1 payoff on the left side of the graph and the #2 payoff on the right side. For instance, for alternative A, plot 4 on the left side of the graph and 12 on the right side. Then, connect these two points with a straight line. The three alternatives are plotted on the graph as shown below. The graph shows the range of values of P2 over which each alternative is optimal. Thus, for low values of P2 (and thus high values of P1, since P1 + P2 = 1.0), alternative B will have the highest expected value; for intermediate values of P2 alternative C is best; and for higher values of P2 alternative A is best. To find exact values of the ranges, determine where the upper parts of the lines intersect. Note that at the intersections, the two alternatives represented by the lines would be equivalent in terms of expected value; hence, the decision maker would be indifferent between the two at that point. To determine the intersections, you must obtain the equation of each line. Because these are straight lines, they have the form y = a + bx, where a is the y-intercept value at the left axis, b is the slope of the line, and x is P2. Slope is defined as the change in y for a one-unit change in x. In this case, the distance between the two vertical axes is 1.0. Consequently, the slope of each line is equal to the right-hand value minus the left-hand value. The slopes and equations are: 16

16 B

14

14

12 10 #1 Payoff

A

C

12 10

8

#2 8 Payoff

6

6

4

4

2

B best 0

.2

C best .4

.6 P (2)

2

A best .8

1.0

SUPPLEMENT TO CHAPTER 5   Decision Analysis #1

#2

Slope

A

4

12

B

16

2

2 − 16 = −14

16 − 14 P 2

C

12

8

8 − 12 = −4

12 − 4 P 2

12 − 4 = +8

11

Equation 4 + 8 P 2

From the graph, we can see that alternative B is best from the point P2 = 0 to the point where the alternative B line intersects the alternative C line. To find that point, solve for the value of P2 at their intersection. This requires setting the two equations equal to each other and solving for P2. Thus, 16 − 14 P2 = 12 − 4 P2 Rearranging terms yields 4 = 10 P2 Solving yields P2 = .40. Thus, alternative B is best from P2 = 0 up to P2 = .40. Alternatives B and C are equivalent at P2 = .40. Alternative C is best from that point until its line intersects alternative A’s line. To find that intersection, set those two equations equal and solve for P2. Thus, 4 + 8 P2 = 12 − 4 P2 Rearranging terms results in 12 P2 = 8 Solving yields P2 = .67. Thus, alternative C is best from P2 > .40 up to P2 = .67, where alternatives A and C are equivalent. For values of P2 greater than .67 up to P2 = 1.0, alternative A is best. Note: If a problem calls for ranges with respect to P1, you could find the P2 ranges as above, and then subtract each P2 from 1.00 (e.g., .40 becomes .60, and .67 becomes .33). OM in Action 

Oglethorpe Power

O

glethorpe Power is an electric generation and distribution cooperative supplying 20 percent of the electricity used in Georgia. Most of the remaining demand for electricity in Georgia is supplied by Georgia Power Co. Oglethorpe and Georgia Power jointly own most of the power transmission lines in Georgia, and supply their excess electricity to Florida. Some years ago, Florida Power and Light (FPL) indicated to Oglethorpe that they were interested in the construction of another major transmission line between Florida and Georgia. Oglethorpe invited the consulting company Applied Decision Analysis Inc. to assist it in making a decision. A team was formed to investigate the decision variables, random variables, and their payoffs. Brainstorming resulted in an influence diagram and identification of the three decision variables (each with alternatives): line (joint with Georgia Power, alone, or no line), nature of control (Oglethorpe or FPL), and whether to upgrade the associated facilities (joint with Georgia Power, alone, or no upgrade). Five random variables (each with states of nature) were identified: construction cost (low, medium, or high), competitive situation in Florida (good, fair,

or bad), Florida demand (low, medium, or high), Oglethorpe’s share of demand (very low, low, medium, high, or very high), and future spot price for electricity (low, medium, or high). The payoffs were measured in terms of net present value. The team estimated the probability of the states of nature for each random variable and the payoff for each combination of decision alternatives and states of nature. Using the DPL software, the alternative (in brackets) for each decision variable with highest expected value was: line (alone), control (Oglethorpe), and upgrade (no). Then, the risk profile (the distribution of NPV) of this solution was determined by DPL. Because this showed considerable possible negative NPV, the team identified the random variable that most affected the downside risk: the competitive situation in Florida. The next step was to collect further information on this random variable in order to reduce the range of values for the payoff. The team reported all its findings to top management, who started negotiating with FPL. Applied Decision Analysis, Inc. is now part of PricewaterhouseCoopers. Source:  A. Borison, “Oglethorpe Power Corporation Decides about Investing in a Major Transmission System,” Interfaces (25)2, March– April 1995, pp. 25–36.

PART THREE   System Design

12

Summary

Key Terms

Solved Problems

Decision analysis is a formal approach to decision making that is useful in operations management. Decision analysis provides a framework for the analysis of decisions. It involves identifying the decision alternatives, chance events (random variables), and payoffs. A simple decision problem can be represented as a decision table. The common alternative selection method is the expected value method. Two visual tools useful for representing decision problems are decision trees and influence diagrams. Information can be used to reduce the variability of the important random variables. Two tools that can be used in deciding whether or not to buy information are expected value of perfect information and expected value of sample information. Graphical sensitivity analysis can be used to determine the effect of variability of probabilities assigned to states of nature on the best alternative.

payoff table, 2 sensitivity analysis, 10

decision tree, 4 expected value of perfect information (EVPI), 8 expected value of sample information (EVSI), 8

«Xtags error: Style name ambiguous: tag @...:»Solved

The following solved problems 1–4 refer to this payoff (profits) table:

Alternative capacity for new store

New Bridge Built

No New Bridge

A

1

14

B

2

10

C

4

 6

where A = small, B = medium, and C = large.

Problem 1

Using graphical sensitivity analysis, determine the probability for “No New Bridge” state of nature for which each alternative would be optimal.

Solution

Plot a straight line for each alternative. Do this by plotting the payoff for “new bridge built” on the left axis and the payoff for “no new bridge” on the right axis, and then connecting the two points with a straight line. Each line represents the expected profit for an alternative for the entire range of probability of “no new bridge.” Because the lines represent expected profit, the line that is highest for a given value of P(no new bridge) is optimal. Thus, from the graph, you can see that for low values of this probability, alternative C is best, and for higher values, alternative A is best (alternative B is never the highest line, so it is never optimal, i.e., it is dominated). Payoff if new bridge

14 A

B

Payoff if no new bridge 10

C

6

4 2 1

C best 0

A best 1.0

.27 P (no new bridge)

0

SUPPLEMENT TO CHAPTER 5   Decision Analysis

13

The dividing line between the ranges where alternatives C and A are optimal occurs where the two lines intersect. To find that probability, first formulate the equation for each line. To do this, let the intersection with the left axis be the y intercept; the slope equals the right-side payoff minus the left-side payoff. Thus, for C you have 4 + (6 − 4)P, which is 4 + 2P. For A, 1 + (14 − 1) P, which is 1 + 13P. Setting these two equal to each other, you can solve for P: 4 + 2P = 1 + 13P Solving, P = 3/11 = .27. Therefore, the ranges for P(no new bridge) for each alternative to be best are: A: .27 ≤ P ≤ 1.00 B: never optimal C: 0 ≤ P ≤ .27 Using the probabilities of .60 for a new bridge and .40 for no new bridge, calculate the expected value of each alternative, and identify the alternative that would be selected using the expected value method. A: .60(1) + .40(14) = 6.20 [best]

Problem 2

Solution

B: .60(2) + .40(10) = 5.20 C: .60(4) + .40(6) = 4.80 Calculate the EVPI using the data from the previous problem.

Problem 3

Using Formula 5S-1, the EVPI is the expected payoff under certainty minus the expected payoff under risk (i.e., the maximum ­expected value). The expected payoff under certainty involves multiplying the best payoff in each column by the column probability and then summing those amounts. The best payoff in the first column is 4, and the best in the second column is 14. Thus,

Solution

Expected payoff under certainty = .60(4) + .40(14) = 8.00 The maximum expected value is 6.20 from the previous problem. Therefore, EVPI = 8.00 − 6.20 = 1.80 Suppose that the values in the payoff table represent costs instead of profits.

Problem 4

a. Using sensitivity analysis, determine the range of P(no new bridge) for which each alternative would be optimal. b. If P(new bridge) = .60 and P(no new bridge) = .40, find the alternative to minimize expected cost. a. The graph is identical to that shown in Solved Problem 1. However, the lines now represent expected costs, so the best alternative for a given value of P(no new bridge) is the lowest line. Hence, for very low values of P(no new bridge), A is best; for intermediate values, B is best; and for high values, C is best. You can set the equations of A and B, and B and C, equal to each other in order to determine the values of P(no new bridge) at their intersections. Thus, A = B: 1 + 13P = 2 + 8P; solving, P = 1/5 = .20 B = C: 2 + 8P = 4 + 2P; solving, P = 2/6 = .33 Hence, the ranges are: A best: 0 ≤ P