Module 2

Advanced Decision Analysis

2.1

Introduction

2/1

2.2

Continuous Probability Distributions

2/2

2.3

Assessment of Subjective Probabilities

2/7

2.4 2.4.1

Revising Probabilities Technical Note

2/13

2.5

Concluding Remarks

2/19

2/18

Review Questions

2/20

Case Study 2.1

2/22

Case Study 2.2

2/23

Case Study 2.3

2/24

This is an advanced module which extends the scope of decision analysis. It may be omitted on a first time through the course. It looks in some depth at ways of estimating and manipulating the probabilities which are frequently the most sensitive parts of an application of the technique. Methods of using continuous distributions, of estimating subjective probabilities and of revising probabilities when new data become available are described. By the end of the module readers should be better equipped to handle what is perhaps the most important element of decision analysis, the probabilities of chance events. Prerequisite

Modules

Module 1: Decision Analysis

The most difficult aspect of decision analysis (and also the aspect which in practice seems to give managers the greatest sense of unease) is the derivation of the probabilities of chance events. There is likely to be considerable doubt whether some outcome has a probability of, say, 10% or 15%. Yet the decision maker may be aware instinctively, and sensitivity analysis can verify systematically, that the optimal decision path changes if the probability is 15% instead of 10%. Are there methods by which probabilities can be estimated or calculated sufficiently accurately to point to 10% rather than 15%? In one sense the fact that the optimal decision is so highly dependent upon one probability is valuable information since it suggests that the decision problem

has no clear-cut answer. It also indicates the area to which attention should be given and further work done. This significant event is the one to which resources in the form of, say, market research, additional manpower or additional finances should be directed in order to reduce the uncertainty. Or, on the other hand, such sensitivity may point to the need for delay. In spite of this, it would still be preferable to have greater confidence in the probability estimate. This module attempts to demonstrate how this may be achieved by dealing with three aspects of probability assessment. The first topic is that of continuous probability distributions. So far in decision analysis the outcomes of chance events have been described in terms of discrete probabilities. Chance events with outcomes which have continuous probabilities can also be incorporated into the technique. Second, systematic methods of making a subjective estimate of an event's probability are described. Here, the emphasis is on demonstrating that subjective probabilities are a matter of scientificmethodology, not merely wild guesses. Finally, the module will deal with Bayesian revision. When additional information about an event becomes available, the probabilities associated with it are likely to change. For instance, market research information will change the probabilities of success or failure for a new consumer product. The method by which these changes can be calculated is called Bayes' theorem (hence the alteration of probabilities is usually referred to as Bayesian revision).

The decision trees encountered up to now have had chance event nodes with a limited number of outcomes, usually no more than three. For example, the product launch case analysed in the previous module showed the demand for the product as being 'strong', 'weak' or 'non-existent'. In business, however, variables such as product demand are likely to have a continuous distribution. The decision analyst might be presented with a distribution represented by a curve, rather than a table giving a few outcomes each with its associated probability. Figure 2.1 is an example of a continuous probability distribution of the estimated weekly demand for a new product. Such a distribution may have been assessed subjectively: the product manager believed that demand would be in the range 200-700 with a peak at 350. Or it may have been the result of some very accurate data measurement: the weekly demands for similar products over many weeks were collected and formed into a distribution. The next section (on probability assessment) looks in greater detail at the origins of distributions like this. How can probabilities be derived from distributions like Figure 2.1? With difficulty! It is done by measuring the areas under parts of the curve. For example, to find the probability that demand will be in the range 250-350 first measure the area under that part of the curve. This can be done by counting squares on the graph paper or, if by some means that mathematical equation of the curve were known, by a mathematical technique called integration. The probability is this area as a proportion of the whole area. If the area between 250 and 350 were 38% of the whole then the probability of the demand being between 250 and 350 is:

200

250

300

350

L700

400 Demand

Measuring areas under curves is a messy business and so a demand distribution is often represented in a different way, that of a cumulative distribution. This shows the probability of a variable taking a value less than or equal to or more than or equal to a specified value. In the 'less than or equal to' form the demand distribution of Figure 2.1 would appear as in Figure 2.2. Since demand is in the range 200-700, there is zero probability that demand will be 'less than or equal to' 199 and 100% probability that demand will be 'less than or equal to' 700. Between 200 and 700 the probability increases at varying rates, depending upon the nature of the distribution. The next section shows how such a distribution can be assessed.

~

:0 rn

46

.0

c..~ 8

o 200 250 300 350 400

500 Demand

A 'more than or equal to' curve gives the probability of the variable being more than or equal to a specified value. The curve starts at 100% and decreases through to 0%. The advantage of the cumulative form is that one can read probabilities from the vertical scale rather than having to measure areas. For example, to find the probability that demand will be between 250 and 350, Figure 2.2 shows that the probability of a demand less than or equal to 250 is 8% and for 350, 46%: P(x ::::;350) = 46%

and

P(x ::::;250) = 8% Therefore

P(250 ::::;x ::::;350) = P(x ::::;350) - P(x ::::;250)

= 46%

- 8%

=38%

When the outcome of a chance event node is a continuous distribution, whether or not it is cumulative, it is represented in a decision tree by a fan, as in Figure 2.3.

The method of handling continuous outcomes is to replace them by a number of representative values. Once the fans have been replaced by the cumulative values then the EMV of the node is calculated simply as if the outcomes were discrete. The customary way of selecting the representative values is to use bracket medians. The procedure is as follows. (a) Decide how many representative values are required. This will depend upon: (i) The degree of accuracy required: how close need the approximation be to the continuous distribution in view of the decision being taken? (ii) The accuracy of the distribution: if the assessment has been very approximate, is there any point in being over-precise in choosing representative values? (iii) The intricacy of the distribution: are there many changes in slope and shape which should be reflected in the representative values chosen?

(iv) The computational load entailed: will the representation result in unnecessarily heavy computations, especially if sensitivity analysis will be required? As a rule of thumb, at least five but not more than ten representative values are normally chosen. However, if the calculations are computerised, a large number of representative values can be used for particularly intricate or sensitive distributions. (b) Divide the whole range of values into a number of equiprobable ranges, the number being the same as the number of representative values chosen. Equiprobable means that there should be an equal chance of a demand falling in any of the ranges. For example, suppose there are to be five representative values for the demand data of Figures 2.1 and 2.2. From the cumulative form of the distribution, read off the demands relating to 0%, 20%, 40%, 60%, 80% and 100%. There is then an equal chance (20%) that the demand will fall in any of the ranges marked off by these demands. (c) Select one representative value from each of the ranges by taking the median of the range. This value is the bracket median. For example, for the 0% to 20% range the bracket median is the demand corresponding to the probability 10%. The five bracket medians are thus simply the demands relating to the probabilities 10%, 30%, 50%, 70% and 90%. (d) Let the bracket median stand for all the demands in the range and assign to it the probability of all the values in the range. In the example there would then be five branches, each with a probability of 20% and for which the payoffs are the bracket medians. The calculations can now proceed as if it were a normal decision tree with discrete outcomes.

What

is the EMV for the chance event node shown in Figure 2.47 The payoffs

are given by the cumulative distribution represent the distribution.

of Figure 2.5. Use ten bracket medians to

100 80 ;,R

~

VI

60

~

1i co .0

2 35 c..

40 20

The ten bracket medians will be associated with ten equiprobable ranges, relating to probabilities 0 to 0.1 (i.e. 0% to 10%), 0.1 to 0.2, 0.2 to 0.3 and so on. The bracket medians will therefore correspond to the probabilities 0.05, 0.15, 0.25, etc. The bracket medians themselves must be read off the cumulative distribution. For instance, the bracket median relating to the probability 0.35 is -10 (see Figure 2.5). Table 2.1 gives all the bracket medians. Bracket medians for Figure 2.5 Range

Mid-point

0-0.1 0.1-0.2 0.2-0.3 0.3-0.4 0.4-0.5

0.05 0.15 0.25 0.35 0.45

0.5-0.6 0.6-0.7 0.7-0.8 0.8-0.9 0.9-1.0

0.55 0.65 0.75 0.85 0.95

Bracket median

-130 -80 -45 -10 15 40 60 100 180 360

Using the bracket medians the chance event node in Figure 2.4 can be considered as if it were an ordinary chance node with ten branches each of which has a probability of 0.1 and a payoff given by the relevant bracket median. EMV

=

(0.1 x - 130) + (0.1 x - 80) + (0.1 x - 45) + (0.1 x -

10)

+ (0.1 x 15) + (0.1 x 40) + (0.1 x 60) + (0.1 x 100) + (0.1 x 180) + (0.1 x 360) -

= 49

13 - 8 - 4.5 - 1 + 1.5 + 4 + 6 + 10 + 18 + 36

The example illustrates the difference between equal ranges and equiprobable ranges. The ranges are not equal although they have equal probabilities. Each range has the same probability, 0.1, but covers a different span of payoffs. For example (from Figure 2.5), the 0-0.1 range covers approximately 100 in terms of payoff (-200 to -100) whereas the 0.1-0.2 range covers 40 (-100 to -60).

Once the reasoning behind the use of bracket medians is clear it is, of course, unnecessary to bother about the equiprobable ranges themselves. It is sufficient to go straight to the bracket medians. If the distribution is not in cumulative form, the same process can be applied but not quite so easily. The equiprobable ranges have to be found by dividing the whole area under the distribution into equal areas. These areas have in turn to be divided into two equal parts (this means two equal areas, not equal ranges) to find the bracket medians. Since this is a much more difficult procedure than reading these values directly from a graph, it is better to insist on having distributions in cumulative form in the first place. The requirement for bracket medians to relate to equiprobable ranges is usual but not strictly necessary. A bracket median can be determined for any range, whether or not it is equiprobable with other ranges: the first bracket median may be associated with the probability range 0% to 20%, the next with 20% to 30% and so on. Circumstances may dictate that bracket medians have to relate to ranges with unequal probabilities. An example of such a situation is given in a case study at the end of the module. The continuous distributions discussed so far have been observed distributions: they were formed from past data or by subjective assessment. It is possible that the outcome of a chance event node might be in the form of a standard distribution, such as the normal distribution. In this case the EMV is simply the mean of the distribution. For example, if the distribution were normal then the EMV is the mean of that normal distribution. It may nevertheless still be useful to represent the distribution by bracket medians in order to draw up a sensible risk profile. The distribution table can be used in performing this task. For example, for a normal distribution ten bracket medians can be found from the normal curve table by determining the points that leave an area of 0.05 in the tail, 0.15 in the tail, 0.25 in the tail and so on.

In the decision analysis problems encountered so far it has been assumed that probabilities for the branches leaving chance event nodes have been easily available. It is this assumption that gives some managers cause to doubt the technique. To the cursory view it can appear that these probabilities are the product of pure guesswork. It does not seem worthwhile carrying out a complicated analysis when the main underlying data are so flimsy. While it is true that the provision of these probabilities is not automatic, their estimation can be a precise and painless activity. The procedure whereby an analyst assists a manager in estimating a cumulative probability distribution has several stages. The analyst has to:

(a) Set the context for the estimation by discussing the range of values the variable might take. This allows the analyst to introduce the basics of the procedure gradually and allows the manager to focus his or her ideas on what is required. This stage is rather like a doctor exchanging pleasantries with a patient before proceeding to conduct an intimate examination. (b) Ask the manager to give a number such that it is equally likely that the variable takes a value that is lower or higher. In effect the manager is being asked to say what the median (in this context called the 50% fractile) is. (c) Direct attention in turn to the ranges on either side of the median and divide each into two equiprobable halves. In other words persuade the manager to specify the 25% and 75% fractiles. (d) Make some preliminary checks for consistency. The manager should feel that there is an equal chance that the variable falls between the 25% and 75% fractiles as outside these points (since both events have an equal probability, 50%). If the manager feels that this is not so then there is some inconsistency. The analyst must go over the estimation of 25%, 50% and 75% fractiles with the manager once more and try to achieve consistency. (e) Divide each of the four equiprobable ranges (0% to 25%, 25% to 50%, 50% to 75% and 75% to 100%) into two equiprobable halves by estimating the 12.5%,37.5%, 62.5% and 87.5% fractiles. (f) Check consistency by asking questions such as whether it is equally likely that the variable would fall in the range marked by the 25% and 37.5% fractiles or that marked by the 75% and 87.5%fractiles. (g) Continue the process of subdivision until the manager feels that he or she no longer has sufficient knowledge to make the process a sensible one. In practice a further subdivision into 16 equiprobable ranges is usually the most that can be attempted. (h) Mark all fractiles on a graph and draw a smooth freehand curve between them. This is the cumulative probability distribution for the variable, just like the demand distribution of Figure 2.2. It is used in just the way described in the previous section. Example Suppose the distribution of likely demand for a new product in its first year is being assessed. The conversation between decision analyst and product manager might be as follows: Analyst: As you know, our purpose is to transform what you know of the product into a distribution of first year demand. But first, I'd like to know what your views of this product are, in general terms. Manager: I have been involved in marketing some similar products in the past. Lessons have been learned and this product is superior in many ways. We also have a better appreciation of how to market it. On the other hand costs have risen significantly and our advertising budget has been restricted by the Finance Director. Overall, we should do better than previously, but as to forming a distribution I really don't think it's going to be possible. My ideas just aren't sufficiently detailed.

~

Analyst: Manager:

I think you will be surprised by how much you do know. I'm not even sure I could make a best guess at the demand.

Analyst: If I were to ask you to do that I'd have to tell you exactly what I meant by a 'best guess'. What would 'best' mean? I'd like to try a slightly different approach. But first I'd like to try to get at the broad picture. Do you think it is more likely that the demand is above or below 50000? Manager: Above, certainly. We have plenty of advance orders which are already above that level. I would say the mimimum is 100000. Analyst: Do you think it is more likely that the demand is above or below 800000, which, according to the background information, is your maximum production capacity for year 1? Manager: Another easy one. Demand for this type of product has never reached such a level and our customer research shows that the basic demand just isn't there. 800000 is the absolute maximum. Analyst: A harder question now. Can you give me a number such that it would be difficult for you to decide whether demand would be above or below it? In other words for what value is it equally likely that demand falls above or below it? Let's talk in units of thousands. Manager: That is a harder question. I would think between 400 and 500, say 450. But what exactly does 'equally likely' mean? Analyst: 'Equally likely' means that you are indifferent between forecasting a demand above or below 450. If all your worldly goods were at stake on the outcome, you would still not be able to say whether demand was going to be above or below 450. Manager: Even if my life were at stake it would be difficult but I think that would concentrate my mind wonderfully. I might even be able to be a bit more precise. I have a gut feeling that 450 is fractionally too high. On the basis that my life is at stake, I will have to go for 440. Analyst: So far you have told me that 440 divides the demand range into two parts that are, in your judgement, equally likely. Let's repeat the division process, concentrating on the lower range. Do you think it is more likely that demand is between 200 and 440 or less than 200? Manager: Analyst: Manager: Analyst:

Between 200 and 440. Between 400 and 440 or less than 400? Lessthan 400. Can you go further

and give me a number x such that you think

it

equally likely that demand is below x or between x and 440? Manager:

Er. ..

Analyst: I can ask the question in a different way. Suppose I could tell you that demand will definitely be less than 440. Can you give me a number x

such that you would be indifferent between receiving a prize of a Mercedes convertible conditional upon demand being below x or receiving the identical prize conditional upon demand being between x and 440? Manager:

I would say x should be 360.

Analyst: You're telling me that it is equally likely that demand is less than 360 or between 360 and 440. Manager:

I think that's right.

Analyst: Let's now look at higher levels of demand - when it exceeds 440. Can you divide this range into two equally likely parts? In other words can you give me a number y such that it is equally likely that demand is between 440 and y or above y? Manager: I think I see what you're trying to do although I'm worried I'm giving you some highly doubtful numbers. I think y should be fairly close to 440, say 500. Analyst: I'm aware that making such judgements as these gives rise to concern about over-precision. But we're making progress. Let's summarise what we've done so far. You have said that in your judgement it is just as likely that demand is in anyone of the four ranges (less than 360), (360-440), (440-500) and (above 500). Manager:

Agreed.

Analyst: Now for some checking. This is not to catch you out, but to make sure we are being consistent in our judgements. Virtually everybody is to some extent inconsistent the first time round and has to make adjustments. Let's look from a different perspective. Would you rather bet that demand would be inside or outside the interval 360 to 500? Manager: I'd have to say outside. The future for such a new product is by no means certain and the demand could cover a fairly wide range. My answer to your question means that I'm being inconsistent, doesn't it? Analyst: Yes, but this is not unexpected. I would have been surprised if you had not been. Think again about your previous estimates. Where do you think the inconsistency crept in? Manager: Well, I don't want to change the 440. I thought about it as if my life and a Mercedes were on the line. I think the 360 estimate is the most dubious number. On second thoughts I would change this to 340. I think it's a 50-50 bet that demand lies inside or outside the range 340 to 500. Analyst: You think it's equally likely that demand is less than 340 or between 340 and 440? Manager: That's right, but I am concerned that if we were starting from scratch again, I would get different numbers. I might easily have come up with 325, 430 and 480 instead of 340, 440 and 500. Analyst: These numbers are close to the original ones so I'm not too worried. Nevertheless, they could make a difference to the optimal decision and we

'--./

should bear in mind your feelings of uncertainty when we come to apply decision analysis and do some sensitivity analysis on the results. I would, however, be extremely worried if you told me that on your second run-through you could have come up with 200, 350 and 400. Manager:

I am not as uncertain

as that.

Analyst: The next step is to carry out some more subdivisions. three numbers you've given me are called the 25% fractile fractile (440) and the 75% fractile (500).

Incidentally, the (340), the 50%

The next subdivisions will have the objective of providing the 12.5%, 37.5%, 62.5% and 87.5% fractiles. First, the range of less than 340. What's the number that would divide this range into two equally likely parts? In other words, what is z so that a demand less than z is as likely as a demand between z and 340? Manager:

I suppose about 260.

Analyst: Then the 12.5% fractile is 260. What's the number that range 340 to 440 into two equally likely parts? Manager:

divides the

I would say 400.

Analyst: This is the 37.5% fractile. And what about the value which in the same way divides the 440 to 500 range into two parts? Manager: Analyst:

Roughly 465. This is the 62.5% fractile.

Finally what about the 87.5% fractile?

Manager: I think 600. I think this means that demand between 500 and 600 as to be more than 600? Analyst:

Exactly right.

I've just been drawing Fractile %

Now we can summarise your judgements

in this table

up: Fractile value

12.5

260

25

340

37.5

400

50

440

62.5

465

75

500

87.5

600

Manager:

is just as likely to be

What do we do with these results?

Analyst: The results are used to draw up a cumulative probability distribution on graph paper. In fact, we should now carry out some further consistency checks, but I'll draw up the graph just to show what we're aiming at. The graph will have cumulative probability as the vertical axis and demand as the horizontal axis. Each line of the table above is a point on the graph. We plot the points and then join them in as smooth a curve as we can manage (see Figure 2.6).

100 <;?

>< v

75

"0

c

'"E e. Q)

50

g :0

'"e

.0

25

0-

0 100

200

300

400

Demand

500

600

700

800

(OOOs)

Manager: This is the distribution from which we obtain the bracket medians that we use for the chance event nodes? Analyst: That's right. You will notice that I've taken 100 and 800, the values we discussed at the outset, to be the limits of the distribution, the 0% and 100% fractiles. We might have to take a closer look at this before we come to construct the final version of the distribution. However, when I look at even this first version it seems that you knew more about the distribution of demand than you originally thought. Now, let's get back to the question of consistency for the last fractiles you gave me ...

What has been described is not the whole story. The process can be varied to suit the circumstances. For example, at each stage of the subdivision it is not necessary always to halve the ranges. The ranges could be split into three (or more) equiprobable parts. It depends upon what the manager feels to be the easier and more sensible. Also, when the distribution appears to have lengthy tails more attention might be given to the tails than to the body of the distribution. For instance, the 37.5% and 62.5% fractiles might not be discussed while the 6.25%and 93.75%fractiles might be estimated in addition to the 12.5% and 87.5% fractiles. If it is important to have more detailed knowledge of the tails (because large profits or losses are involved at the extremes) and if the manager feels able to do it, then this extra attention should be given. Nor is the above the only process for assessing probabilities. There are other approaches although the one described is perhaps the most common. An alternative estimation process is one that is based on associating values of the variable with familiar probabilities such as the result of throwing a die or selecting from a pack of cards. Although it is subjective probabilities which cause managers to doubt or reject the technique, in practice the assessment process is usually welcomed. Managers appreciate the opportunity to give their views in precise form. Instead of having to provide point estimates and having to think how to adjust them for selfprotection and for contingencies, they are allowed to express their position fully by giving a range of values and associated probabilities. In other words, the

barrier to the use of subjective probabilities is an initial one and in most cases resistance turns to grateful co-operation once the procedure gets under way.

However scientific the process for subjectively assessing probabilities may be, it is still reassuring to know that not all probabilities have to be estimated in this way. Where sample information is involved (or market research or any other form of additional information) some probabilities can be calculated. Recall the product launch example of the previous module. Probabilities of strong, weak or non-existent demand were given. When a test market was undertaken, these probabilities, as expected, changed in the light of this new information. The source of the revised probabilities was not discussed. Had they been obtained subjectively,there would surely have been a danger that they were not consistent with one another and with the earlier probabilities. In fact they were calculated using Bayes' theorem. The revision process is known as Bayesian revision. The probabilities given in the product launch example were: 1

2

P(S) = P(Strong demand) = 0.4 P(W) = P(Weak demand) = 0.4 P(N) = P(Non-existent demand) P(S I F) = P(W I F) = P(N I F) = P(S I U) = P(W I U) = P(N I U) =

=

0.2

P(Strong, given a favourable test) = 0.72 P(Weak, given a favourable test) = 0.24 P(Non-existent, given a favourable test) = 0.04 P(Strong, given an unfavourable test) = 0.08 P(Weak, given an unfavourable test) = 0.56 P(Non-existent, given an unfavourable test) = 0.36

1

These are called prior probabilities. They are 'prior' in the sense that they are the probabilities estimated before any sample information is obtained from the test market.

2

These are called posterior probabilities. They are 'posterior' in the sense that they are the probabilities of the different demand levels after sample information (i.e. the test market) is made available. They are conditional probabilities, the condition being the nature of the test market information. The notation is the usual notation for conditional probabilities, written with a vertical line separating the condition:

There is nothing to prevent prior and posterior both being estimated subjectively. Prior probabilities are often estimated in this way, although they could of course be derived from historical data if available. Posterior probabilities on the other hand are better estimated using Bayes' theorem along with:

(a) prior probabilities; (b) information about the known accuracy of the particular sample information that is available. The prior and posterior probabilities will then be consistent with one another and with the sample information; they may not have been had they been assessed independently. Information about the known accuracy of, for instance, a test market is usually in the form: if there really is a strong market demand then the probability that the test will give a favourable result is x, and the probability that the test will give an unfavourable result is y. Probabilities similar to x and y will relate to other levels of market demand. (Note that the sum of x and y must be 1.). Putting this in mathematical notation, information about the accuracy of the test market might be: p(FIS) = 0.9(x above) p(Uls)= O.l(y above)

p(FIW) = 0.3 p(UIW)= 0.7

p(FIN) = 0.1

P(UIN)= 0.9

These data have not been calculated. They were given by the devisers of the test market who used their knowledge of the structure of the test and their experience of using it to establish the probabilities. The accuracy of sample or tests (whether in marketing or medicine or wherever) is in this form because this is how the tests are checked, i.e. their performances in known situations were measured. One can now show mathematically and diagrammatically what Bayes' theorem does - see Figure 2.7.

Prior probabilities P(S), P(W), P(N)~ Posterior probabilities P(S/F), P(W/F), etc. Accuracy information ~ P(F/S), P(U/S), etc.

Bayes' theorem is a series of formulae for calculating posterior probabilities, such as P(S IF). Before considering the algebraic approach to the theorem, it is probably easier to start with a diagrammatic method. A Venn diagram is a square which represents the probabilities. There are three steps in using a Venn diagram. (a) Divide the square vertically to represent the prior probabilities. Thus rectangles are created whose areas are proportional to the prior probabilities. (b) Divide these rectangles horizontally to represent the 'sample accuracy' probabilities. Thus further rectangles are created whose areas are proportional to the conditional probabilities which are the measures of the accuracy of the information provided by the sample. (c) Combine and divide areas to obtain the conditional probabilities which are the posterior probabilities. The procedure is best understood by carrying it out for the product launch example for which the relevant probabilities are given above. The first step is to divide the square vertically according to the prior probabilities. This is shown in Figure 2.8.

Strong

Weak

Nonexistent

The second step is to divide the diagram horizontally according to the 'sample accuracy' data. This is shown in Figure 2.9. The rectangle relating to a strong market is divided into two parts: (a) a part relating to an unfavourable test (area D), covering 10%of the rectangle because P(Unfavourable IStrong) = 0.1; (b) a part relating to a favourable test (area A), covering 90% of the rectangle because P(Favourable IStrong) = 0.9.

P(Strong demand given a favourable test market)

=

P(SIF)

= Proportion of area representing 'favourable' that relates to 'strong'

= Area = =

A/ (Area A + Area B + Area C) (0.9 x 0.4)/0.5 0.72

P(Weak demand given a favourable test market) Area B/ (Area A + Area B + Area C) = (0.3 x 0.4)/0.5 = 0.24

=

P(WIF)

=

P(Non-existent demand given a favourable test market) = Area C/(Area A + Area B + Area C) = (0.1 x 0.2)/0.5 = 0.04

=

P(Strong demand given an unfavourable test market) = Area D/ (Area D + Area E + Area F) = (0.1 x 0.4)/0.5 = 0.08

p(SIU)

P(Weak demand given an unfavourable test market) = Area E/(Area D + Area E + Area F) = (0.7 x 0.4)/0.5 = 0.56

=

=

P(NIF)

p(WIU)

P(Non-existent demand given an unfavourable test market) = Area F/ (Area D + Area E + Area F) = (0.9 x 0.2)/0.5

=

p(NIU)

= 0.36

All these calculations are summarised in Table 2.2. This is constructed in such a way that it shows how the calculations are built up. It is the conventional form for displaying the results of Bayesian revision. The posterior probabilities shown in column 6 are the ones obtained when the problem was analysed using the Venn diagram. As always an intuitive check on the probabilities can be made by working in frequencies. The posterior probabilities in column 6 are also the ones used in the previous module in the product launch problem although it was not explained at the time that their derivation was by Bayesian revision. In general, Bayes' theorem is used whenever probabilities are revised in the light of some new information. In essence Bayes' theorem is a method for combining experience with numbers. The posterior probabilities are based on the prior probabilities (experience) combined with sample information (numbers).

Area 0 (=P(UlS)) 0.1 Unfavourable

Area E (=P(UIW)) 0.7 Unfavourable

Area F (=P(U/N)) 0.9 Unfavourable

Area B (=P(FIW)) 0.3 Favourable

AreaC (=P(F/N)) 0.1 Fa-

Area A (=P(F/S)) 0.9 Favourable

vourable

Similarly the rectangle referring to a weak market is divided into area E (70% of it, because P(U I W) = 0.7) and area B (30% of it, because P(F I W) = 0.3). The rectangle referring to a non-existent market is divided into areas F (90% of it, because P(U I N) = 0.9) and C (10% of it, because P(F I N) = 0.1). In Figures 2.8 and 2.9 probabilities are associated with areas, which can be measured in the usual way by multiplying length by width. Probabilities, including posterior probabilities, can be calculated by considering the area they are represented by: P(Favourable test market)

Total area associated with favourable test Area A + Area B + Area C = (0.9 X 0.4) + (0.3 x 0.4) + (0.1 X 0.2) = 0.5 =

=

P(Unfavourable test market) = Total area associated with unfavourable test = Area D + Area E + Area F =~.lxO.~+~.7xO.~+~.9xO.~ = 0.5

For conditional probabilities the range of possible outcomes is limited by the condition. In other words there is a preliminary restriction to a subsection of the whole square. A conditional probability is given, therefore, not by the area representing the event, but by this area as a proportion of the area that represents the condition. For P(strong demand given a favourable test market) the preliminary restriction is to the parts of the square relating to a favourable test market (areas A, B and C). P(S I F) is then calculated as the proportion of A + B + C that is associated with a strong demand:

Table 2.2

Bayesian revision of product launch problem

Test

Market

outcome

demand

For U (1) F

Prior probabilities

Test

Posterior

accuracy

probabilities

x

P(SIF) etc.

peS) etc.

P(FIS) etc.

(2)

(3)

(4)

5

0.4

0.36

(= Area A)

0.72

W

0.4

0.9 0.3

0.12

(= Area B)

0.24

N

0.2

0.1

0.02

(= Area C)

0.04

S, War

N

(3)

(4)

(5)

--

(6)

P(F) = 0.50 U

5

0.4

0.1

0.04

(= Area D)

0.08

W

0.4

0.7

0.28

(= Area E)

0.56

N

0.2

0.9

0.18

(= Area F)

0.36

--

P(U) = 0.50

Most people find the Venn diagram approach to Bayes' theorem easier to understand than the algebraic approach. However, the latter is not especially difficult although the formulae and notation can make it seem tedious. Bayes' theorem stems from the multiplication law of probability. Keeping to the product launch example:

P(F) . P(SIF) = P(S) . P(FIS) P(SIF) = P(S) . P(FIS)jP(F)

Equation (2.1) is nearly in the form needed for Bayes' theorem. The posterior probability (P(S I F)) is on the left-hand side of the equation. The prior probability (P(S))and the test accuracy measure (P(F I S)) are on the right-hand side. It would be possible to calculate the posterior probabilities if P(F) were known. This can be found from the addition law of probabilities. A favourable outcome can occur in anyone of three ways: in conjunction with the true demand being strong, weak or non-existent. These three ways are mutually exclusive, therefore the addition law applies:

The right-hand side is a combination of prior probabilities and test accuracy measurements, all of which are assumed to be known. Therefore it is possible

to calculate P(F). The formula for the posterior probability thus becomes, from equations (2.1) and (2.2): P(SIF)

=

P(S) . p(FIS) -P(-S)-.-P-(F-IS-) +-P-(-W-) -. P-(F-!W-) +-P(-N-) -.P-(F-!N-)

Equation (2.3) can be related to the Venn diagram given in Figure 2.9 since this formula is no more than: P(SIF)

= --------

Area A

Area A + Area B + Area C

Bayes' theorem merely amounts to a series of formulae corresponding to equation (2.3) for calculating all the posterior probabilities.

In Module 1 the basic technique of decision analysis was seen as a method for obtaining the optimal solution to certain types of decision problem. This optimal solution was to be used in practice as an aid to taking the decision. Even without the optimal solution the technique performed a valuable function in helping the decision maker to structure and understand the problem. A potential weakness, it was suggested, was the derivation of probabilities. This module has strengthened this area by looking at ways of measuring and expressing the probabilities of chance events even more accurately. This extension of the technique could have gone further. For example, not only can continuously distributed chance outcomes be handled, they can also in some circumstances be revised in the light of additional information. Bayes' theorem can therefore apply to the revision of continuous as well as discrete probabilities. The refinement of probabilities is not the only direction in which decision analysis can be extended. The payoffs can also be handled differently to deal with another of the perceived weaknesses of decision analysis which is that the EMV criterion can mask large payoffs by averaging them out. A utility function represents monetary payoffs by utilities and allows EMVs to be used universally. A utility is a measurement which reflects the true value of a monetary amount to the decision maker. For example, in strict numerical terms a £2 million loss is double a £1 million loss, whereas the true consequence of a £2 million loss (bankruptcy?) may be more than double that of a £1 million loss (a difficult year ahead?). The utility value associated with a £2 million loss would in these circumstances be more than double that associated with a £1 million loss. Similarly a profit of £2 million might have a utility less than double that of a £1 million profit because currently the company does not have enough profitable projects in which it could invest the extra £1 million. To use the concept of utilities the decision maker's utility function is drawn in a similar manner to the assessment of subjective probabilities. The utility function is in effect a graph relating utility and money. Payoffs can then be converted from money into utilities. The roll-back deals with expected utility instead of expected monetary value. The end result is an optimal path which

does not hide any 'unallowed-for' disastrous outcomes since they have already been incorporated into the utilities. A utility function is unique to the decision maker and is indicative of the real worth of financial amounts to the decision maker. Details of the application of utility functions go beyond the scope of this module. As with most business techniques new research is continually being carried out into decision analysis and new experience of applications gained. Where useful, these findings are gradually incorporated into the accepted practice of the technique, which therefore improves as time goes by. The manager should always be ready to put into practice new developments in the analytical techniques he or she uses. This is especially true of the development of software for computerising decision analysis, Bayes' theorem, and probability assessment. This should not, however, obscure the more important task of making proper and full use of what is already known and available. This is especially true for decision analysis which, because of its subjective probabilities and expected values, is probably viewed as one of the more mysterious management techniques.

P(Demand:::; 100) = 33%. P(Demand :::; 200) = 74%. P(Demand :::; 300) = 92%. P(Demand :::;400) = 100%. What is P(Demand between A

100 and 300)?

18%.

B

59%.

C

74%.

D

93%.

'--,/

Questions 2.2 to 2.4 refer to the following distribution: Probability Variable

o

10

20

30

40

50

60

70

80

90

100

100

160

200

240

270

320

380

450

550

750

1000

2.2 The first of these ranges covers A

100-160.

B

100-200.

C

100-240.

D

160-240.

'less than or equal to' cumulative

x

values:

2.3 The first of the five bracket medians is:

A

10.

B

20.

C

100.

D

160.

2.4 The expected value of x is: A

320.

B

384.

C

550.

D

some other number.

2.5 The choice of the number of equiprobable ranges with which to represent an assessed distribution is influenced by the following factors: I

The complexity

II

The range of values taken by the variable.

of the distribution.

III

The accuracy with which the distribution

Which one of the following A

I only.

B

I and III only.

C

II and III only.

D

I, II and III.

has been assessed.

is correct?

2.6 In the assessment of the distribution of a variable the 50% fractile is 40, the 25% fractile is 30 and the 75% fractile is 65. The range of the variable for probabilities 25-50% is 10 and that for 50-75% probabilities is 25. Which of the following is correct? The estimates: A

are inconsistent and should be revalued.

B

are inconsistent but should be left alone.

C

are consistent.

D

in themselves say nothing

2.7 Is the following Probabilities

about consistency.

statement true or false?

indicating

the accuracy of sample information

are in the form:

P{Sample result given state of the world) because they are calculated from Bayes' theorem. Questions 2.8 to 2.10 refer to the following

situation:

A newly developed technical process will either work in practice or fail. The prior probabilities are: P{Success)= 0.7; P{Failure) = 0.3 A test has been designed to predict the eventual success or failure of the process. Its accuracy is given by: P{Predicts success given process really is successful) = 0.9 P{Predicts failure given process really is a failure) = 0.8

2.8 What is the probability that the test predicts success? A

31%.

B

50%. 63%. 69%.

e D

2.9 What is the probability A 0.09. B 0.63. e 0.77. D

of success given that success has been predicted?

0.91.

2.10 What is the probability A 0.06. B 0.09. e 0.23. D 0.77.

of failure given that success has been predicted?

Case Study 2. 1 A manufacturer of household goods is preparing to launch a newly developed, microchip-based dishwasher. In the first year after the launch only one standard model will be offered. The price, which is largely dictated by the competitive environment, will be ÂŁ650. The decision to offer a fuller range of models will be taken towards the end of the first year and will depend upon results up to that time. If a fuller range is marketed, the number of variants on the standard model will be decided at the same time. Table 2.3

End-of-first year decisions

Demand

Subsequent 0-5000 5000-10000 10000-15000 15000-20000

20000+

decision

Re-model standard version or abandon Introduce

2 variants

Introduce

1 or 2, 3 Introduce 4 or Introduce 1 or

or 4 variants

5 variants 2 variants

Market research has indicated what the demand for the standard model in the first year might be. This has been expressed as the cumulative probability distribution shown in Figure 2.10. This gives the probability that the demand will be greater than that shown on the horizontal axis. The different decisions that could be taken at the end of the first year will relate to the demand during the first year according to Table 2.3. 1 Express the problem as a decision tree, including probabilities but excluding payoffs (which have not of course been given in the case). 2 What is the expected demand for the standard dishwasher in the first year?

~ 80 ~ .0 60 ro

.0

e 40 c..

10 Demand

15

20

(OOOs)

An overseas government is trying to determine the best policy for controlling a pest which is damaging the tropical fruits which are one of its major exports. There are three ways of attacking the pest: (a) Spray with a DDT-type substance. (b) Spray with a hormone which prevents development to adulthood. (c) Use a two-stage programme to upset breeding patterns. First, trap males with the help of a scent. Second, sterilise these males, then release them. Approach (a) is tried and tested and the outcome is known with near certainty. The DDT-type substances can·control but not eradicate so, although the cost of this approach is not high, neither are the benefits. This approach is the standard against which improvements brought about by the other approaches can be compared. In other words the net financial benefit of approach (a) is taken to be zero and the benefits of approaches (b) and (c) measured as increments to the benefits that approach (a) could bring. The cost of approach (b) is £4 million. There is no more than a 0.2 chance this will work, but if it does the benefit in terms of increased exports is as high as £60 million because it would eradicate the pest. If it does not work, there will be a loss of £5 million, in addition to the cost of the approach, because by the time the failure is known it will be too late to spray with DDT and exports will be lost. If approach (c) is followed, the first question is the success or failure of the scent. There is a 0.5 chance that a high number of males will be trapped, a 0.5 chance that a low number will be trapped. Once this is known a decision to spray with DDT or to sterilise and then release the trapped males must be made. The cost of the scent programme is £6 million and the cost of sterilisation is an additional £8 million. If the two-stage programme is successful, the benefit

in terms of the value of fruit exports, but excluding costs of the approach, is ÂŁ40 million. If the sterilisation is unsuccessful, this result will not be known in time to spray with DDT this year and the net loss will be ÂŁ5 million as for approach (b). If the first stage traps a lot of the males, there is an 85% chance of success by releasing the sterile males; if the first stage traps few males, there is only a 15% chance of success. 1 Use decision analysis to indicate the optimal solution. Show the costs of approaches (b) and (c) as gates on the tree. 2 What is the maximum that should be paid for a series of laboratory trials to research the uncertainties inherent in all three approaches? 3 A department of a local university offers to carry out some research work on the effectiveness of the hormone. The researchers believe that the accuracy of their work will be 95/80. This means that if the hormone really will work there is a 95% chance that the research will predict it; if it really will not work there is an 80% chance that the research will predict it. What is the maximum that should be .paid for this research? 4 What assumptions and non-quantitative information should be taken into account in coming to a decision?

A team of advisers has just completed an analysis of the prospects next year for five well-known stocks. The prospects depend upon the state of the economy. The team has supposed that the economy will be one of the four states: (i) further decline (sl); (ii) holding steady (s2); (iii) slight improvement (s3); (iv) major expansion (s4). The per pound sterling growth for the stocks corresponding to each state of the economy is given by: Stock A B C D E

51 -0.35 -0.25 -0.15 -0.05 0

State of economy 52 53 -0.15 +0.15 -0.05 0 0 +0.05 +0.05 +0.10 +0.05 +0.05

54 +0.60 +0.35 +0.20 +0.20 +0.10

1 Which stock is preferable, stock C or stock D? Why? 2 The team estimates the probabilities of different states of the economy as: P(sl)

= 0.1; P(s2) = 0.3; P(s3) = 0.4; P(s4) = 0.2

Using an expected monetary value approach, which stock is preferable?

'---../

3 The team considers commIssIOning a group of economic consultants to provide an opinion on the likely state of the economy next year. Preliminary discussions reveal that the consultants will be able to give an opinion whether, in the context of stock performance, the economy will be (i) good (01) (ii) average (02) or (iii) poor (03). The reliability of their opinions is given in the table below, derived from their previous performances. The table shows the probability of the consultants' opinions given the true state of the economy. Opinion Good (01) Average (02) Poor (03)

51 0.1 0.2 0.7

State of economy 52 53 0.4 0.2 0.4 0.6 0.2 0.2

54 0.6 0.3 0.1

What is the EVSI (expected value of sample information) of the consultants' advice?

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