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Steps to Good Decisions

¨ Define problem and influencing factors¨ Establish decision criteria¨ Select decision-making tool (model)¨ Identify and evaluate alternatives using

decision-making tool (model)¨ Select best alternative¨ Implement decision¨ Evaluate the outcome

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The Decision-Making Process

Problem Decision

Quantitative Analysis

LogicHistorical DataMarketing ResearchScientific AnalysisModeling

Qualitative Analysis

EmotionsIntuitionPersonal Experience and MotivationRumors

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Decision Problem

Alternatives

States of Nature

Out-comes

¨ Decision trees¨ Decision tables

Ways of Displaying a Decision Problem

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Preference Matrix¨ Used in evaluating an alternative ¨ Preference matrix is a table that allows the

manager to rate an alternative according to several performance creteria.

Example :The Following table shows the

performance criteria, weights and scores for a new product:

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A thermal storage air Conditioner¨ If Mgt. wants to introduce just one new

product & the highest total score of any of other product ideas is 800

¨ Should firm pursue making the air conditioner.

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Performance Criteria

Weight Score Weighted score

Market potential 30 8 240

Unit profit margin 20 10 200

Op. Compatibility 20 6 120

Competitive adv. 15 10 150

Investment Req. 10 2 20

Project risk 5 4 20

Totals 100 From 1-10 750

So Mgt. would not pursue the thermal Storage air Cond. 750 < 800.

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Decision Theory

¨ Is a general approach to decision making when outcomes associated with alternatives are often in doubt.

¨ It helps op. manager with decisions on process, capacity, locations & inventory because such decisions are about an uncertain future.

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The procedure is as follows:

¨ List the feasible alternatives ( one of them do nothing )

¨ List the events ( chance events )

¨ Calculate the payoff for each alternative in each event. ( payoff is the total cost or total profit )

¨ Esitimate the likelihood of each event

¨ Select a decision rule to evaluate the alternatives

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Fundamentals of Decision Theory

The three types of decision models:

¨ Decision making under uncertainty¨ Decision making under risk ¨ Decision making under certainty

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Fundamentals of Decision Theory - continued

Terms:¨ Alternative: course of action or choice¨ State of nature: an occurrence over which the

decision maker has no control

Symbols used in decision tree: A decision node from which one of several

alternatives may be selected A state of nature node out of which one state of

nature will occur

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Decision Table

States of Nature

Alternatives State 1 State 2

Alternative 1 Outcome 1 Outcome 2

Alternative 2 Outcome 3 Outcome 4

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Decision Making Under Certainty

¨ Decision rule here is pick the alternative with the best payoff for the known event.

¨ Costs the lowest ¨ Profits the highest

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Example¨ A manager is deciding whether to build a

small or a large facility.¨ The manager knows with certainty the

payoffs that will result under each alternative shown in the following payoff table

¨ The payoffs in $ are the present value of the future revenues – costs for each alternative in each event.

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¨ What is the best choice if the future demand be low?

¨ Solution :the highest value in the low future demand

is 200,000 which is the small facility.

Alternatives Possible future demand

Low High

Small 200 270

Large 160 800

Do nothing 0 0

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Decision Making Under Uncertainty

¨ Maximax - Choose the alternative that maximizes the maximum outcome for every alternative (Optimistic criterion)

¨ Maximin - Choose the alternative that maximizes the minimum outcome for every alternative (Pessimistic criterion)

¨ Equally likely ( laplace) - chose the alternative with the highest average outcome.

¨ Minimax regret – the best “worst regret” ¨ Calculate table of regrets ( opportunity losses) in

which the rows represent the alternative & the columns represent the events

¨ Regret is the difference between a given payoff & the best payoff in the same column.

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Example¨ Reconsider the payoff table of the previous

example¨ what is the best alternative for each decision

rule?

( maximin, maximax, laplace, minimax regret)

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Maximin¨ The best of the worst ¨ The worst payoffs are

The best of the worst is $200,000 small

Alternative Worst payoffs

Small 200large 160

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Maximax¨ Best of the best

¨ The best is 800,000 large

Alternative best payoffs

Small 270

large 800

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Laplace ( Equally Likely )¨ We have here two events so we assign each a

probability of ½¨ But if we have 3 events we assign each a probability

of 1/3

¨ The best is 480,000 , Large

Alternative Weighted payoffSmall ½ X 200 + ½ X 270 = 235Large ½ X 160 + ½ X 800 = 480

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Minimax regret

Alternative Regret Maximum regretLow High

Small 200-200=0 800-270=530 530

large 200-160=40 800-800=0 40

The worst regret appears in the maximum regret column

To minimize the maximum regret we choose 40 ( large )

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Example - Decision Making Under Uncertainty

States of NatureAlternatives Favorable

MarketUnfavorable

MarketMaximum

in RowMinimumin Row

RowAverage

Constructlarge plant

$200,000 -$180,000 $200,000 -$180,000 $10,000

Constructsmall plant

$100,000 -$20,000 $100,000 -$20,000 $40,000

$0 $0 $0 $0 $0

Maximax Maximin Equally likely

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¨ Probabilistic decision situation¨ States of nature have probabilities of

occurrence¨ Select alternative with largest expected

monetary value (EMV)¨ EMV = Average return for alternative if decision

were repeated many times

Decision Making Under Risk

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Expected Monetary Value Equation

Probability of payoffEMV A V P V

V P V V P V V P V

i ii

i

N N

( ( )

( ) ( ) ( )

) =N

= *

= * + * + + *

1

1 1 2 2

Number of states of nature

Value of Payoff

Alternative i

...

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Example¨ Reconsider the previous example ¨ If the probability of the low demand is estimated to

be 0.4 & the probability of the high demand is estimated to be 0.6 .

¨ Choose the Large Facility ( the highest EMV )

Alternatives Expected Monetary Value ( EMV )

Small Emv= 0.4 X 200 + 0.6 X 270 = 242Large Emv= 0.4 X 160 + 0.6 X 800 = 544

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Example - Decision Making Under Risk

States of NatureAlternatives Favorable

MarketP(0.5)

UnfavorableMarket P(0.5)

Expectedvalue

Constructlarge plant

$200,000 -$180,000 $10,000

Constructsmall plant

$100,000 -$20,000 $40,000

Do nothing $0 $0 $0

Best choice

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¨ Graphical display of decision process¨ Used for solving problems

¨ With 1 set of alternatives and states of nature, decision tables can be used also

¨ With several sets of alternatives and states of nature (sequential decisions), decision tables cannot be used

¨ EMV is criterion most often used

Decision Trees

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Analyzing Problems with Decision Trees

¨ Define the problem¨ Structure or draw the decision tree¨ Assign probabilities to the states of nature¨ Estimate payoffs for each possible

combination of alternatives and states of nature

¨ Solve the problem by computing expected monetary values for each state-of-nature node

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Decision trees

Introduction

In many problems chance (or probability) plays an important role. Decision analysis is the general name that is given to techniques for analyze problems containing

risk/uncertainty/probabilities.

Decision trees are one specific decision analysis technique and we will illustrate the technique by use of an example.

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Decision Tree

1

2

State 1 E1( P(E1))

State 2 E2(P(E2))

State 1

State 2

Alternative 1

Alternative 2

Decision Node

Outcome 1Outcome 1

Outcome 2Outcome 2

Outcome 3Outcome 3

Outcome 4Outcome 4

State of Nature Node

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¨ After drawing the decision tree, we solve it by working from right to left, calculating the expected payoff for each node as follows:

1. For event node , we multiply the payoff of each event branch by event’s probability. We add these products to get the expected payoff.

2. For a decision node , we pick the alternative that has the best expected payoff. If an alternative leads to an event node , its payoff is equal to that node’s expected payoff( already calculated).

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Example

¨ Retailer must decide whether to build a small or large facility at a new location.

¨ Demand can be either low or high, with probability estimated to be 0.4 and 0.6 respe.

¨ If a small facility is built and demands prooves to be high the manager may choose not to expand payoff 223 or to expand payoff 270

¨ If a small facility is built and demand is low, there is no reason to expand and the payoff is $200,000.

¨ If large facility is built and demand proves to be low, the choice is do nothing $40,000 or to stimulate demand through local advertising.

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Example continue

¨ The response to advertising may be either modest or sizable, with their probability estimated to be 0.3 and 0.7,rspect.

¨ If it is modest, the payoff is estimated to be only $20,000; the payoff grows to $220,000 if the response is sizable.

¨ Finally , if a large facility is built and demand turns out to be high, the payoff is $800,000.

Draw and analyze the decision tree.

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Format of a Decision Tree

Low demand [0.4]

B

$200

High demand[0.6]

$ 223

$ 270

2

Don’t Expand

Expand

$ 800High Demand [0.6]

3

$ 40Do nothing

AdvertiseLow Demand [0.4]

Smal

l fac

ility

Large facility

1

Decision Point

Chance Event

$ 270

Modest [0.3]

Sizable [0.7]

$ 20

$ 220$ 160

$ 160

$ 544

$ 242

$ 544

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Analyzing Solution¨ For the event node dealing with advertising

the expected payoff is 160 or the sum of each event’s payoff weighted by its probability [ 0.3 ( 20 ) + 0.7 ( 220) ]

¨ The expected payoff for decision node 3 is 160 because 160 is better than do nothing 40, Prune the do nothing Alternative.

¨ And so on

The best alternative is to built a large facility.

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ExampleA company faces a decision with respect to a product (codenamed M997) developed by one of its research laboratories. It has to decide whether to proceed to test market M997 or whether to drop it completely. It is estimated that test marketing will cost $100. Past experience indicates that only 30% of products are successful in test market. If M997 is successful at the test market stage then the company faces a further decision relating to the size of plant to set up to produce M997. A small plant will cost $150 to build and produce 2000 units a year whilst a large plant will cost $250 to build but produce 4000 units a year. The marketing department have estimated that there is a 40% chance that the competition will respond with a similar product and that the price per unit sold (in $) will be as follows (assuming all production sold): Large plant Small plant Competition respond 20 35 Competition do not respond 50 65 Assuming that the life of the market for M997 is estimated to be 7 years and that the yearly plant running costs are $50 (both sizes of plant - to make the numbers easier!) should the company go ahead and test market M997?

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In that figure we have three types of node represented:

• Decision nodes; • Chance nodes; and • Terminal nodes.

Decision nodes represent points at which the company has to make a choice of one alternative from a number of possible alternatives e.g. at the first decision node the company has to choose one of the two alternatives "drop M997" or "test market M997".

Chance nodes represent points at which chance, or probability, plays a dominant role and reflect alternatives over which the company has (effectively) no control. Terminal nodes represent the ends of paths from left to right through the decision tree.

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Step 1

In this step we, for each path through the decision tree from the initial node to a terminal node of a branch, work out the profit (in $) involved in that path. Essentially in this step we work from the left-hand side of the diagram to the right-hand side.

• path to terminal node 2 - we drop M997

Total revenue = 0 Total cost = 0Total profit = 0

Note that we ignore here (and below) any money already spent on developing M997 (that being a sunk cost, i.e. a cost that cannot be altered no matter what our future decisions are, so logically has no part to play in deciding future decisions).

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• path to terminal node 4 - we test market M997 (cost $100) but then find it is not successful so we drop it

Total revenue = 0 Total cost = 100 Total profit = -100 (all figures in $)

• path to terminal node 7 - we test market M997 (cost $100), find it is successful, build a small plant (cost $150) and find we are without competition (revenue for 7 years at 2000 units a year at

$65 per unit = $910) Total revenue = 910 Total cost = 250 + 7x50 (running cost) Total profit = 310

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• path to terminal node 8 - we test market M997 (cost $100), find it is successful, build a small plant (cost $150) and find we have competition (revenue for 7 years at 2000 units a year at $35 per unit = $490)

Total revenue = 490 Total cost = 250 + 7x50 Total profit = -110

• path to terminal node 10 - we test market M997 (cost $100), find it is successful, build a large plant (cost $250) and find we are without competition (revenue for 7 years at 4000 units a year at $50 per unit = $1400)

Total revenue = 1400 Total cost = 350 + 7x50 Total profit = 700

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• path to terminal node 11 - we test market M997 (cost $100), find it is successful, build a large plant (cost $250) and find we have competition (revenue for 7 years at 4000 units a year at $20 per unit = $560)

Total revenue = 560 Total cost = 350 + 7x50 Total profit = -140

• path to terminal node 12 - we test market M997 (cost $100), find it is successful, but decide not to build a plant

Total revenue = 0Total cost = 100Total profit = -100

Note that, as mentioned previously, we include this option because, even if the product is successful in test market, we may not be able to make sufficient revenue from it to cover any plant construction and running costs.

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Hence we can form the table below indicating, for each branch, the total profit involved in that branch from the initial node to the terminal node. Terminal node Total profit (£K) 2 0 4 -100 7 310 8 -110 10 700 11 -14012 -100

So far we have not made use of the probabilities in the problem - this we do in the second step where we work from the right-hand side of the diagram back to the left-hand side.

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Step 2

Consider chance node 6 with branches to terminal nodes 7 and 8 emanating from it. The branch to terminal node 7 occurs with probability 0.6 and total profit $310 whilst the branch to terminal node 8 occurs with probability 0.4 and total profit $-110. Hence the expected monetary value (EMV) of this chance node is given by

0.6 x (310) + 0.4 x (-110) = 142 ($) Essentially this figure represents the expected (or average) profit from this chance node (60% of the time we get $310 and 40% of the time we get $ -110 so on average we get

(0.6 x (310) + 0.4 x (-110)) = 142 ($)).

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The EMV for any chance node is defined by

"sum over all branches, the probability of the branch multiplied by the monetary ($) value of the branch".

Hence the EMV for chance node 9 with branches to terminal nodes 10 and 11 emanating from it is given by

0.6 x (700) + 0.4 x (-140) = 364 ($) node 10 node 11

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We can now picture the decision node relating to the size of plant to build as below where the chance nodes have been replaced by their corresponding EMV's.

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Hence at the plant decision node we have the three alternatives:

Alternative 3: build small plant EMV = 142K

Alternative 4: build large plant EMV = 364K Alternative 5: build no plant EMV = -100K

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It is clear that, in $ terms, alternative number 4 is the most attractive alternative and so we can discard the other two alternatives, giving the revised decision tree shown below.

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We can now repeat the process we carried out above. The EMV for chance node 3 representing whether M997 is a success in test market or not is given by

0.3 x (364) + 0.7 x (-100) = 39.2 ($) plant decision node node 4

Hence at the decision node representing whether to test market M997 or not we have the two alternatives:

Alternative 1: drop M997 EMV = 0 Alternative 2: test market M997 EMV = 39.2$ It is clear that, in $ terms, alternative number 2 is preferable and so we should decide to test market M997.

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Summary

Let us be clear then about what we have decided as a result of the above process:

• We should test market M997 and this decision has an expected monetary value (EMV) of $39.2

• If M997 is successful in test market then we anticipate, at this

stage, building a large plant (recall the alternative we chose at the decision node relating to the size of plant to build). However it is plain that in real life we will review this once test marketing has been completed

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Note here that the EMV of our decision (39.2 in this case) DOES NOT reflect what will actually happen - it is merely an average or expected value if we were to have the tree many times - but if fact we have the tree once only.

If we follow the path suggested above of test marketing M997 then the actual monetary outcome will be one of

[-100, 310, -110, 700, -140, -100] corresponding to terminal

nodes 4,7,8,10,11 and 12 depending upon future decisions and chance events.

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