Appendix 1
Solutions to Review Questions and Case Studies
Suggested Answers to Review Questions Suggested Solutions to Case Studies
A1/1 A 1/12
1.1 Only A is a characteristic. A is correct since at a decision node the choice is between different branches having a different quantity (usually an EMV) associated with each. Probabilities are attached to the branches stemming from chance nodes, not decision nodes, thus B is not correct. C is not correct because the payoffs are not necessarily in financial units. There is no restriction on the number of branches coming from each node, therefore D is not a characteristic. 1.2 C: Both representations are equally correct. The logic of A and B is the same and either version can be used. It is not necessary for the sequence of subdecisions and chance events to be strictly decision/ chancel decision/ chance, etc. If it is clearer, two (or more) nodes of the same type can be adjacent.
Expected cost = (0.25 x 4) + (0.25 x 3) + (0.5 x 2) =1+0.75+1 = 2.75
1.4 D is correct. EMV at point B is 80 (obtained by eliminating the less favourable branch). EMV at point A = (0.1 x 80) + (0.5 x 60) + (0.4 x50) = 8 + 30  20 = 18
1.5 B: Contract out. The lowest expected cost should be chosen. Option 3 is the lowest but when the cost of the report is included, the expected cost becomes ÂŁ195000. Option 2 is the lowest.
EVSI = Expected cost without information  Expected cost with information (but excluding its cost) = 190000  175000 = 15000
1.7 C is correct. The payoffs at X and Yare 300 and 100, in respect of successful and nonsuccessful development of the device. Since costs are ignored, the payoff for not developing the device, point Z, is O.
EMV at B = (0.75 x 300) + (0.25 x  100) = 225  25 = 200
1.9 A is correct. Perfect information will say whether or not the development, if attempted, would be successful. If the development were successful, then the correct action would be to develop the advice and make a profit of 300. There is a 75% probability that perfect information would say this. There is a 25% chance that the perfect information would say that the development would not be successful. In this case the correct action would be not to attempt the development and the payoff would be zero. EVPI = EMV with perfect information  EMV without perfect information = (0.75 x 300) + (0.25 x 0)  200 = 225  200 = 25
P(Demand 100  300) = P(Demand ~ 300)  P(Demand ~ 100) = 92%  33% = 59%
2.2 B is correct. The first of five equiprobable ranges would cover the first 20% of the distribution, i.e. up to the 20% fractile.
2.3 D is correct. The bracket median for the 020% range is the 10% fractile. This is x = 160. 2.4 B is correct. The five bracket medians are 160,240,320, 450 and 750. These, along with their probabilities of 0.2 (because there are five equiprobable ranges), are used to represent the distribution. The expected value of x: x 160) + (0.2 x 240) + (0.2 x 320) + (0.2 x 450) + (0.2 x 750) + 48 + 64 + 90 + 150
= (0.2 = 32
= 384
2.5 B is correct. If the distribution is complex, for example having many changes of slope, then more bracket medians are needed to represent it. Thus I is a factor. If it has not been possible to assess the distribution to a great degree of accuracy then there is little point in using many bracket medians to get a close approximation to it. Thus III is a factor. Factor II has no bearing since this is merely a question of the units of measurement. 2.6 D is correct. To check for consistency the assessor would have to be asked whether he thought the two ranges were equally likely, as they should be if the fractiles have been correctly assessed. The difference in the two ranges merely expresses the lack of symmetry of the distribution. 2.7 False. The probabilities are in the form given since that is how sample accuracy is measured, by finding out the likelihood of particular sample results in given situations (states of the world). They are used in Bayes' theorem calculations but that is' not where they come from. 2.8 D is correct. The Venn diagram is given in Figure ALL P(Success predicted) = Area relating to predicted success and real success + Area for predicted success and real failure = (0.9 = 0.63 = 0.69
x 0.7) + (0.2 x 0.3) + 0.06
Success
0.7 Success predicted
Success predicted
0.2
0.9
0.8
0.1
2.9 D is correct. The probability required is the proportion of the area relating to a prediction of success that refers to success in reality: = =
0.63/0.69 0.91
2.10 B is correct. Since there are only two possible outcomes when success is predicted: P(Failure given that success has been predicted) = 1  P(Success given that success has been predicted) =
=
1  0.91 0.09
3.1 A
True. By definition, decisions, whether quantitative or not, have decision variables of some form and an objective which is a criterion distinguishing a good decision from a bad one.
B
False. LP problems have the three elements but, in addition, it must be possible to represent the objective function and the constraints as linear functions of the decision variables.
C
False. It is possible to solve twovariable problems graphically (and even threevariable problems if a threedimensional drawing is made) but in practice it is easier to solve all LP problems on a computer.
D
True. There is no reason why a LP problem of any size cannot be solved, but if the computer is small in relation to the problem, the solution, if not impossible, might take so much time as to be impracticable.
E
True. Some companies run planning LP models covering their whole business and these models do have thousands of variables and constraints.
3.2 B is correct. There are two decision variables, namely, the numbers of each type of chair to produce. Answers A and 0 are not possible since they do not allow a distinction to be drawn between the different costs, prices and production times of the two types of chair. Answer C would make it difficult to devise algebraic expressions for the objective and the constraints. 3.3 B is correct. The objective function would be to maximise contribution. A is incorrect because it ignores costs. The maximum revenue does not necessarily provide the maximum profit. C is incorrect because it would lead to a solution for which only the minimum sales requirements were produced, ignoring the fact that higher production levels will be more profitable. 0 is incorrect because minimising time on the machines is unlikely to lead to greater financial returns. In any case there is no information that extra machine time can be used for anything else. 3.4 C is correct. The constraints are: Times available for each operation: Minimum sales requirements for each type of chair Maximum sales requirements for each type of chair
Constraints 4 2 2
8 Note that nonnegativity constraints are not required since they are covered by the minimum sales requirements. 3.5 C is correct. This objective is the one that maximises contribution. A is incorrect because it is the maximisation of revenue. B is incorrect because it is the maximisation of revenue but with the coefficients interchanged. 0 is incorrect because it is the minimisation of costs. 3.6 0 is correct. For the lathe L23 requires six hours and L31, two hours. Thus the total hours used must be 6x + 2y. This must be no more than the total availability, 120 (= 3 lathes x 40 hours each). The constraint is therefore 6x + 2y ::; 120. Note that B is incorrect since constraints are always, ::;, = or 2, never strict inequalities.
3.7 D is correct. Each L23 requires three hours of polisher time; each L31 requires five hours of polisher time. The total machine time taken is thus 3x + 5y. This must be no more than the total available, 120 hours (3 machines x 40 hours each). The polisher constraint is 3x + 5y :::::120. In addition, the two variables must be restricted to be nonnegative, i.e. x ~ 0, y ~ O. 3.8 C is correct. The optimal point is x = 0, y = 24, i.e. no L23s should be made, but 24 L31s should be made. The graph of the feasible region and objective function shows that as the objective increases, the line will move upwards and the last point at which it leaves the region will be point A. 3.9 C is correct. Substituting x of 2880.
= 0, y = 24 in the objective function gives an optimal objective
3.10 D is correct. The lathe constraint is 6x + 2y ~ 120. Substituting x = 0, y = 24 shows that 48 of the available 120 hours are used and therefore the slack is 120  48 = 72. 3.11 B is correct. The lefthand side of a constraint expresses the amount of a resource used up (in terms of the decision variables), while the righthand side is the total amount of that resource available. Both LHS and RHS must be in the same units. A is wrong. An objective function can be to maximise profit (in ÂŁs) while constraints could be the availability of time (hours), the level of demand (in whatever the unit of production is) and many other things. Cis wrong. For example, a decision variable could be in tonnes of production, with an objective of maximising profit in ÂŁs. D is wrong because it includes both I and III, which have been explained above. 3.12 C is correct. x + 3y :::::12 means that x must be less than 12 (x = 12 when y = 0); 2x + 3y :::::18 means that x must be less than 9 (x = 9 when y = 0). Consequently x ::::: 15 is a weaker constraint than the other two and is redundant.
4.1 False. Each dual value refers to a constraint. 4.2 False. A dual value can be greater or less than the price paid for the resource. It is not always greater since extra resources are not always profitable.
4.3 C is correct. The dual value for the polisher is 24. This means that if an additional polisher hour were available, the objective function would increase by 24, i.e. become 2904. 4.4 A is correct. Since 72 hours (the slack value) of lathe time are already unused the value of an additional hour (and this is the dual value) must be o. 4.5 C is correct. The righthand side range for the polisher constraint shows that the dual value (24 for each extra hour) applies as the available hours are increased from 120 up to 300. Only the first 180 of the 200 subcontractor hours therefore carry the dual value of 24. The total extra profit is 180 x 24 = 4320. 4.6 B is correct. The ranges for the objective function coefficients show that the coefficient for L31 can be decreased to 67 without altering the optimal values of the decision variables. If the new value of this coefficient is 100, the solution to the LP is still x = 0 and y = 24. The new optimal value of the objective function must then be 40 x 0 + 100 x 24 = 2400. 4.7 True. A transportation problem is characterised by the fact that, in the constraints, the coefficients of the decision variables are all either 0 or 1. 4.8 D is correct. This is the definition of an integer programme. A and B are false because the 'integer' in integecprogramming does not refer to the objective function value or the righthand sides. C is false because not all the variables have to take whole number values only. 4.9 B is correct. I and III could describe a nonlinear programming problem because they refer to the appearance of squares or logarithms of decision variables in the formulation. II does not describe a nonlinear problem since it is the logarithm of a constant that appears and this is merely another constant. 4.10 A is correct. Many integer, nonlinear and stochastic problems do occur in practice and many are potentially solvable. But solutions may take so much computer time that it is not worthwhile to do so even though computers are powerful enough to solve them.
5.1 C and D are true. A is untrue since physical models are used  the example of store layout has been given. B is untrue since deterministic simulations which do not involve probabilities have many applications. 5.2 False. Risk analysis is the term given to the application of the Monte Carlo technique to financial problems. 5.3 A is correct. I is not an advantage: the amounts of data depend upon the situation rather than the technique. III is an advantage of optimisation over simulation. 5.4 D is correct. The numbers should be allocated as follows: Demand Numbers Probability 0.20 000199 3 200449 0.25 4 450699 0.25 5 700849 0.15 6 7 850999 0.15 1.00 Total 5.5 B is correct. The random number 75 is in the range 7589 which refers to a demand of five. 5.6 A is correct. Several simulation runs using different random number streams help to average out bias that could result from a particular set of random numbers. 5.7 False. The time period covered by the simulation should reflect what happens in practice. For example, if the business situation being simulated has an annual cycle, then the simulation should cover a year. 5.8 D is correct. Project Y seems to a narrower spread. the decision maker not possible to say
be better on average but project X is less risky, having Deciding which is preferable depends therefore on how trades off risk against return. Without knowing this it is which project is superior.
5.9 False. A robust 'best' decision is one that stays best when assumptions vary to some substantial extent. However, the exact definition of substantial is not specified and it is certainly unlikely that any decision would always be the best, however much assumptions were varied.
5.10 True. Simulation depends on the decision maker to put forward different policies or decisions to tryout. In many situations the decision maker, at the outset, may have very little idea what type of policy may be best. To avoid a lengthy search trying out an almost limitless number of options, a structured approach should be adopted so that the best decision is arrived at as quickly as possible.
6.1 False. The first step is to list the activities. This is a crucial step involving decisions as to the appropriate level to break down or aggregate activities. If the network were drawn without giving separate, prior attention to the definition of activities the wrong decisions might be made. 6.2 A and C are correct, Band D are incorrect. A is correct since it describes how activities are depicted on a network. However, they are not allowed to have exactly the same starting and finishing nodes therefore B is incorrect. Some activities can take place at the same time thus C is correct. D is incorrect since an activity may have several immediate predecessors. 6.3 A and D are not valid reasons, Band Care. There are two situations in which dummy activities are used and they are described by Band C. 6.4 A, Figure 6.17, is the correct diagram. B, Figure 6.18, is wrong because activities D and E have exactly the same starting and finishing nodes. C, Figure 6.19, is wrong because it implies that D and E, as well as B, are predecessors of C. 6.5 C is correct. A forward pass through the network gives the results shown in Figure A1.2. The earliest finish time for the final activity, F, is 68 days and this is also the shortest time in which the project can be completed.
6.6 B is correct. A backward pass through the network gives the results shown in Figure A1.3. The earliest and latest start times for activity C are days 30 and 38 respectively.
(48,20,48) (48, 68) F
A (0,20,20) (0, 20)
Float
=
Latest start time  Earliest start time
= 38  30
= 8 days 6.8 B is correct. Activities with zero float lie on the critical path. These are A E and F. 6.9 B is correct. The average or expected duration is calculated as: Expected
= = =
Variance
=
= = =
(Optimistic + 4 x Most likely + Pessimistic)/6 (2 + 4 x 4 + 12)/6 5 days
Pessimistic 
6 Optimistic)
2
(
C 2) 2
;
2
25/9 2.77
6.11 D is correct. All three assumptions have to be made in calculating confidence limits by the method described in the text.
6.12 True. An activity's crash time is the shortest time in which it can be completed and its crash cost is the cost of achieving this.
7.1 C, a laser printer. An accounting package, an electronic diary and a spreadsheet are all software. 7.2 A, I and II are correct. II is part of the definition of a smart card and I is one of the current applications. III is not correct since a smart card is not a piece of workstation equipment. 7.3 B, electric flex is not used as network cabling. Depending upon the purpose and size of the network  and cost factors  any of coaxial cable, optic fibres or telephone wire might be used for a network. 7.4 True. The computerisation on mainframes of payroll, inventory systems, etc. was generally successful but the use of these systems to produce management information often met with failure. 7.5 C, II and III are correct. The point about user participation is that although users did usually participate, they were often not allowed to influence events. Therefore I is not correct. II is correct since little effort was made to format the data as useful information. III is correct' since users often had unrealistic expectations of the capabilities of computer systems. 7.6 B is correct. Payroll systems are usually large and companywide and are typical of mainframe applications. Electronic mail, analysis of downloaded data and wordprocessing are all applications needing local, i.e. decentralised, facilities. 7.7 D, strategic. The system enhances customer service, giving competitive advantage, and is therefore a strategic benefit. 7.8 False. An IT strategy may include a list of approved suppliers but this is a very small part of a true IT strategy, which is a plan showing how IT is to be used to meet corporate objectives.
7.9 B, business needs. 'Topdown' means asking what business needs could IT help to meet rather than asking where IT could be applied. 7.10 B, an expert system. The consultant's system has stored knowledge and mimics the role of an expert. Therefore it is an expert system.
1 Follow the five stages of decision analysis. The process is illustrated in Figure Al.4.
Payoff
400
80 300
Stage 1: Draw the tree. Drawing the tree is straightforward. The managing director has three options: to buy 250, 200 or 150 cars. This decision is shown at the first node of the tree. Each of these options leads to a chance event node, the outcome of which is one of the three possible states of the economy.
./
Stage 2: Insert the probabilities. The three probabilities given in the case relate to the three states of the economy. They are the same for each chance event node. Each of the three branches issuing from the chance event nodes should have the appropriate probability attached. Stage 3: Insert the payoffs. There are nine payoffs in all, one for each state of the economy for each decision option. These payoffs have been added to the end of each branch of the tree in Figure AlA. The units are £OOOs. Stage 4: The rollback. The rollback starts at chance event node A, proceeds to nodes Band C and ends at the initial decision node  see Figure AlA. EMV = (0.3 x 400) + (004 x 230) + (0.3 x  80) 120 + 92  24 188 EMV =
EMV =
(0.3 x 300) + (004 x 250) + (0.3 x 60) 90 + 100 + 18 208 (0.3 x 250) + (004 x 200) + (0.3 x 120) 75 + 80 + 36
191 At the initial decision node, the choice is therefore between EMVs of 188, 208 and 191. The correct decision is to buy 200 cars. The branches referring to purchases of 250 and 150 cars should be eliminated. Stage 5: Summarise the optimal path and draw up the risk profile. The optimal decision is to buy 200 cars. The risk profile showing all outcomes that could result from this strategy along with their probabilities is: Outcome Upswing Level Recession
Payoff £300000 £250000 £60000
Probability 0.3 004 0.3
1.0 This risk profile is rather simple and contains no surprises concerning the chances of making profits or losses. 2 The second part of the question asks what is the maximum amount the managing director should pay for some additional information. The most that should be paid for any extra information is the expected value of perfect information (EVPI). Were perfect information available, it would tell the managing director what the economic climate was going to be. He could then take whatever action was the most profitable in the light of that information. His responses are shown below:
Information says Upswing Level Recession
= (0.3 = 120
x
400) + (0.4
Payoff 400 250 120
Action Buy 250 Buy 200 Buy 150
x
250) + (0.3
x
120)
+ 100 + 36
= 256
The maximum that the managing director should pay for extra information is the EVPI, i.e. ÂŁ48000. Marking scheme Out of 20: Marks Decision tree Probabilities in correct places Payoffs in correct places Rollback: EMVs at chance nodes Elimination of initial decision options
EVPI:
4 1
2 4 1 3
EMV under perfect information Subtraction of EMV without information
1 What recommendations should be made to the board? Follow the five stages of a decision analysis problem. The stages are illustrated in Figure A1.5. Stage 1: Draw the tree. The initial decision (node A) is to go ahead with project A or not. If the company is unsuccessful (at node B), everything comes to an end. If successful, there is a decision to be made at node C. Careful consideration of the problem reveals what might not have been obvious from a less analytical approach: that either project B or project C can be attempted after project A. The company also has the option to stop at this point. If project B is attempted after project A, then chance node Dl represents the outcome. Again, failure brings everything to an end. Success at Dl
26
0.7
F1
Success
with C
Fail with C 6
A 26 9.268
18
0 1
19
brings the option (at node E1) to stop or to proceed with project C. Node F1 represents success or failure with project C. lf project C is attempted after A, the tree has the same structure as above but with projects Band C interchanged. The nodes are at D2, E2 and F2. Stage 2: Insert probabilities. The probabilities are taken directly from the information given in this case. Stage 3: Insert payoffs. The basic data from which the payoffs are calculated are shown below. The commission is calculated as 10% of the value of the project: Commission if successful (ÂŁ million) 2.5 8.0
20.0 The payoffs for the branch ends are calculated as below. Initial part of tree: Refuse deal: 0 Attempt A, but fail:  2 Succeed with A, then stop: 2.5  2
=
0.5
Upper section of tree: Succeed with A but fail with B: 2.5  2  1 = 0.5 Succeed with A and B, then stop: 2.5  2  1 + 8 = 7.5
Succeed with A and B, fail with C: 2.5  2  1 + 8  1.5 = 6 Succeed with all three: 2.5  2  1 + 8  1.5 + 20 = 26 Succeed Succeed Succeed Succeed
with with with with
A but fail with C: 2.5  2  1.5 = 1.0 A and C, then stop: 2.5  2  1.5 + 20 = 19 A and C, fail with B: 2.5  2  1.5 + 20  1 all three: 26 as before.
=
18
Stage 4: The rollback. The rollback starts with the chance node F1 and works back to decision node C. At FI:
EMV = (0.7 x 26) + (0.3 x 6) 18.2 + 1.8
20 AtE1: Eliminate the 'stop' option. At 01:
EMV
= (0.8 x 20)  (0.2 xO. 5) 16  0.1 15.9
No decision can be taken at C unless the rollback goes forward again to F2 and works back to C through the lower section. At F2:
EMV = (0.8 x 26) + (0.2 x 18) 20.8 + 3.6 24.4
At E2: Eliminate the 'stop' option. At 02:
EMV = (0.7 x 24.4)  (0.3 x 1) =
17.08  0.3
16.78 Now the decision at C can be taken and the rollback can continue to the initial decision node A. At C: Eliminate the 'stop' and 'try B' options. At B:
EMV
= (0.6 x 16.78)  (0.4 x 2) 10.068  0.8 9.268
At A: Eliminate the 'refuse' option. Stage 5: Summarise path is:
the optimal path and draw up the risk profile. The optimal
Attempt project A. If successful, attempt project C. If successful, attempt project B. Failure at any stage brings matters to a halt. The expected profit is ÂŁ9268000. The risk profile, in descending order of payoff, is:
Outcome Success with Success with Success with Failure with
Payoff all three projects A, C; failure with B A; failure with C A
26
18 1
2
0.336 0.084 0.180 00400 1.000
Probability (= 0.6 x 0.7 x 0.8) (= 0.6 x 0.7 x 0.2) (= 0.6 x 0.3)
The probabilities are calculated using the multiplication law. The probability of success or failure of a project is not affected by the success or failure of projects attempted earlier. For example: P(Success with A and C; Failure with B) = P(Success A) x P(Success C1Success A) x P(Failure B[Success A and C) = P(Success A) x P(Success C) x P(Failure B) = 0.6 x 0.7 x 0.2 = 0.084
From these probabilities the chances of a profit or loss can be calculated. There is a 42% chance of making a profit, a 58% chance of making a loss. But the size of the losses are very small compared to the profits. ~ 2 What is the expected value of perfect information? Perfect information would say which projects would be successfully coordinated and which not. There are eight possibilities: (Success/fail A) combined with (Success/fail B) combined with (Success/fail C) i.e. 2 x 2 x 2 = 8 possibilities. Note that perfect information does not recognise the limitations imposed by the tree. For example, perfect information might say: failure with A, success with Band C, whereas if project A fails it is not possible to proceed to attempt Band C. The best response of the decision maker to each of these possibilities is shown in the table below. Perfect information says: Success A, success B, success C Success A, fail B, success C Success A, success B, fail C Success A, fail B, fail C Fail A, success B, success C Fail A, success B, fail C Fail A, fail B, success C Fail A, fail B, fail C
Action Try all Try A, Try A, Try A,
three try C then stop try B, then stop then stop
Payoff 26 19 7.5 0.5
Probability 0.336 0.084 0.144
0.036
The probabilities of the outcomes shown in the table are calculated, as before, from the multiplication law: EM~ with perfect information = (0.336 = 8.736
x 26) + (0.084 x 19) + (0.144 x 7.5) + (0.036 x 0.5) + (0.400 x 0) + 1.596 + 1.08 + 0.Q18 + 0
= 11.43
EVPI
=
EMV with perfect information 
EMV without perfect information
= 11.43  9.268 = 2.162
The maximum that should be paid for any additional information is ÂŁ2162000. Marking scheme Out of 40: Recommendation to the board: Drawing the tree Spotting option of trying project C after A Inserting probabilities Calculating and inserting payoffs Rollback Summary of optimal path Risk profile EVPI List of outcomes Actions in response to perfect information Probabilities EMVjEVPI calculations
1 Should the company go ahead with a series? Follow the five stages of a decision analysis problem. The stages are illustrated in Figure A1.6. Stage 1: Draw the tree. The initial decision (node A) is to go ahead with a series or not. If the series is unsuccessful, the project comes to an end. If successful, a second series would certainly be made, therefore the chance node (C) for the second series follows with no intervening decision node. An unsuccessful second series again brings the project to an end. A successful second series leads to a decision node (D) giving options of proceeding with further series or making a feature film. If the film is made there is a chance node (E) relating to the three possible outcomes.
Stage 2: Insert probabilities. The probabilities are taken directly from the information given in the case. Stage 3: Insert payoffs. Note that the payoffs given in the case are cumulative rather than being given for individual series, e.g. the net profit accruing from one successful and one unsuccessful series is given as a total and not for each series separately. Note also that the payoff for the option not to . make the series is assumed to be zero, any costs incurred being sunk at the time the decision is being made. Stage 4: The rollback. The 'rollback starts with the chance node E and works back to the initial decision node A. EMV = (0.2 x 25) + (0.5 x 2.5) + (0.3 x  7.5) 5 + 1.25  2.25 4
At 0: Choose the option to make the film. This option has an EMV of 4. Eliminate the option to make further series (EMV = 3). EMV =
(0.5 x 4) + (0.5 x 0.2) 2 + 0.1 2.1
EMV = (0.25 x 2.1) + (0.75 x 0.5) 0.525  0.375 0.15
At A: Choose the option to make a series (EMV = 0.15). Strike out the option to do nothing (EMV = payoff = 0).
Stage 5: Summarise the optimal path and draw up the risk profile. The optimal path is to make a series of these comedy programmes. If it is successful a second series will always be made. If two series have been successfut the option to make a feature film should be chosen. Under this optimal route there are five possible outcomes. The risk profile shows them in descending order: Outcome Major film success Minor film success Series success/ then fail First series fails Film flop
Payoff
25 2.5 0.2
0.5 7.5
Probability 0.025 0.0625 0.125 0.75
(= 0.25 x 0.5 x 0.2) (= 0.25 x 0.5 x 0.5) (= 0.25 x 0.5)
0.0375 1.0000
The calculation of many of the probabilities in this risk profile are based on the multiplication law of probability. For example, the probability of a major film success depends on a successful first series being followed by a second series in turn being followed by a major film success. It is thus calculated by multiplying the individual probabilities together: P(Major film success)
=
= =
P(lst series success) x P(2nd series success/1st series success) x P(Major film success/2 successful series) 0.25 x 0.5 x 0.2 0.025
2 What is the probability that the company makes a loss if it chooses the optimal decision path? The risk profile will help to answer the question. A loss can occur either through a film flop or through an unsuccessful first series, therefore: P(Loss)
P(Film flop) + P(lst series failure) 0.0375 + 0.75 = 0.7875 =
=
3 Should a pilot programme be made? The possibility of a pilot programme in effect puts a third branch on the initial decision at node A. Following the stages of decision analysis for this branch alone gives the tree shown in Figure A1.7. Stage 1: Draw the tree. The option to make a pilot leads to a chance event node (AI) which is the success or failure of the pilot. Failure brings the project to an end; success leads to a decision node (A2) giving the options to sell the idea and pilot or proceed with a series. The latter branch continues as in Figure A1.6 and there is no need to draw in all of the later branches.
'
0.5 1st series success
2.1 from node C in Fig. A1.6
Stage 2: Insert probabilities. The probability of a successful pilot is 0.4. The probability of a successful series following a successful pilot is 0.5. Otherwise the probabilities are the same as they were without the pilot. Stage 3: Insert payoffs. There is a 'gate' over the branch representing the pilot option. This has the effect of subtracting £125000, the cost of the pilot. Failure of the pilot results in a zero payoff, the net costs having already been included in the £125000. If, after a successful pilot, the idea is sold, the payoff is £750000. Success of the pilot leading to the making of a series gives payoffs just as .previously  there is no need to insert these payoffs. The previously calculated EMV,£2.1 million, corning from the branches that have not been drawn is inserted as if it were a payoff.
At Bl:
EMV = (0.5 x 2.1) + (0.5 x  0.5) 1.05  0.25 0.8
AtA2: At AI:
Choose to make the series. Eliminate the 'sell' option. EMV = (0.4 x 0.8) + (0.6 x 0) = 0.32 Subtract 0.125 from the EMV because of the gate. EMV = 0.32  0.125 = 0.195
At A:
Choose to make a pilot programme. This carries an EMV of 0.195, higher than the EMV of 0.15 which could be obtained without a pilot.
Stage 5: Summarise optimal path and draw up the risk profile. The optimal path is to make a pilot; if successfut a series should be made; if two series are successful, the option of a feature film should be taken up. The resulting risk profile is: Payoff (ÂŁ million)
Pilot succeeds and: Major film success Minor film success Series success/ then fail 1st series fails Film flop Pilot fails
24.875 2.375 0.075 0.625 7.625 0.125
0.02 0.05 0.10 0.20 0.03 0.60 1.0000
(= (= (= (= (=
0.4 x 0.4 x 0.4 x 0.4 x 0.4 x
0.5 x 0.5 x 0.5 x 0.5) 0.5 x
0.5 x 0.2) 0.5 x 0.5) 0.5) 0.5 x 0.3)
The payoffs shown in the risk profile have been reduced by the cost of the pilot. The option of making a pilot programme carries an 83% chance of making a loss. However, a significant amount of this risk is associated with a small loss, that of having a failed pilot. 4 What is the expected value of sample information? In this situation the sample information is the indication of success or failure that comes from the pilot programme. Without subtracting the cost of the pilot programme, the EMV resulting from this branch is 0.32, as shown in Figure A1.7. The EVSIis the difference between this amount and the EMV if the sample information is not used:
5 How do nonoptimal paths compare with the optimal in terms of EMV and risk? EMV sometimes ceases to be a valid decision criterion when high payoffs are involved. The optimal paths, both with and without the pilot programme, carry small chances of a very high loss resulting from a 'flop' film. It could well be the case that a riskaverse company would prefer another option which, although having lower EMV,did not carry any chance of a high loss. One such alternative would be to make the pilot programme but then sell the idea and pilot to another television company. Diagrammatically this option means following Figure A1.7 until node A2 and then choosing to sell rather than proceed. First, calculate the EMV of this option.
At A2: The sell option has been chosen, therefore the EMV at this node is 0.75, not 0.8. At AI:
EMV = (0.4 x 0.75) + (0.6 x 0) 0.3
= =
0.3  0.125 0.175
This EMV is close to the EMV (0.195) of the optimal path. It is more than the EMV (0.15) of the optimal path without a pilot. The risk profile is: Outcome Sell pilot Pilot fails
Payoff 0.625 0.125
Probability 0.4 0.6 1.0
Decision path Optimal without pilot Optimal with pilot Pilot/ sell idea
EMV (ÂŁ million) 0.15 0.195 0.175
Loss 79 83 60
Probability (%) of: High loss High profit 2.5 4 2 3 0 0
The table suggests that the option of selling a successful pilot might well be preferred in spite of its lower EMY.For a slightly lower EMV (0.175 against 0.195), the option carries a much lower probability of a loss (60% against 83%) and, perhaps more,significantly, no chance at all of a high loss. On the other hand the sell option carries no chance of making a really high profit. Other companies might prefer an option just because there is a chance of a high profit. They might even prefer it in spite of a lower EMY. The choice between alternative decisions depends upon the attitude of the decision maker to risk. Decisions should not be based solely on EMY. Especially when high payoffs are involved, the risk profiles of alternative decisions should be brought into consideration.
Marking scheme Out of 40: 1
2 3
4 5
Decision without a pilot: Tree Probabilities Payoffs Rollback Summary of optimal path Risk profile Chance of making a loss Decision with a pilot: Tree Probabilities Payoffs (including gate) Rollback Summary of optimal path Risk profile (including calculation of probabilities) W8 Comparison of alternatives
4 1 1 4 1 3 2
Total
2 1 2 3 1 5 4 6 40
1 Express the problem as a decision tree, including probabilities. The decision tree is shown in Figure A1.8. There are five demand ranges, each of which leads to two or three decision options as specified in Table 2.3. The probabilities for each demand are taken from Figure 2.10. P(Demand P(Demand P(Demand P(Demand P(Demand
2: 0) = 100 2: 5 000) = 85 2: 10 000) = 45 2: 15 000) = 22 2: 20 000) = 12
P(Demand P(Demand P(Demand P(Demand P(Demand
a  5 000) = 100  85 = 15% 5 000  10 000) = 85  45 = 40% 10 000  15 000) = 45  22 = 23% 15 000  20 000) = 22  12 = 10% 20 000 + ) = 12%
2 What is the expected demand for the standard dishwasher in the first year? The expected demand should be calculated by summing over all demand ranges:
The probabilities of the ranges have already been calculated, but which demands should be chosen to represent each demand range? The bracket medians should be used. This means finding the demand that is at the centre (in terms of probability) of each range. The midpoint in terms of demand (for example 7500 as the midpoint of the range 500010000) is an alternative but this does not reflect the changing slope of the cumulative probability distribution. For example, although 2500 is the midpoint of the 05000 range, it is (according to the distribution) far more likely that the demand will be between 2500 and 5000 than between 0 and 2500. In this sense the midpoint of the range is unrepresentative of demand. The bracket medians are found by reading off the graph the demand associated with the probability midpoint of each range. For example, the range 05000 relates to cumulative probabilities of 100% (for demand = 0) and 85% (for demand = 5000). The bracket median is therefore the demand for probability (100 + 85)/2 = 92.5. From the graph this demand is (very approximately) 3500. All the bracket medians are shown in the following table:
Demand range 05000 500010000 1000015000 1500020000 20000+ Expected demand
Bracket median 3500 7400 12000 17500 23000
= Sum of (Bracket median x Probability)
=
(3500 x 0.15) + (7400 x 0.40) + (12000 xiJ.23) + (17500 x 0.10) + (23000 x 0.12) 525 + 2960 + 2760 + 1750 + 2760 10755
Marking scheme Out of 20 marks: Expressing problem as a decision tree: Tree structure Taking probabilities from demand distribution Calculating expected demand: Bracket medians Rest of calculation
1 What is the optimal policy? Follow the five stages of decision analysis: The resulting decision tree including payoffs, probabilities and EMVs is shown in Figure A1.9. Stage 1: The tree is relatively straightforward. The only complexity is in approach (c) where there is a subdecision to spray or sterilise once the outcome of the trapping process is known. Stage 2: The probabilities are just as given in the case. Stage 3: The payoffs are as given in the case and are measured against the norm, approach (a). The costs of approaches (b) and (c) are shown as 'gates' on the tree. Stage 4: Results of the rollback are also shown in Figure A1.9. Stage 5: The optimal decision is to adopt approach (c). If the trapping process is successful, go on to sterilisation; if not spray with the DDTtype substance. The EMV of this optimal path is ÂŁ6625000. The risk profile is:
EMV
= 6.625
Outcome Two stages successful Trap successful, then fail Trap unsuccessful! spray
Payoff 26
(= 40  6  8)
19 (=568) 6 (= 0  6)
Probability 0.425 (= 0.5 x 0.85) 0.075 (= 0.5 x 0.15) 0.5
A net loss is more likelythan a net benefit but there are no surprises hidden away in the risk profile. 2 What is the maximum
that should be paid for a series of laboratory trials to research the uncertainties inherent in all three approaches? That is, what is the expected value of perfect information? Perfect information would say whether the trapping, sterilising and hormone processes would be successful. The responses to this information, and the corresponding payoffs and probabilities, are shown in the following table.
The payoffs in the table are taken from Figure A1.9, including the pest control costs which are shown as 'gates' in the decision tree. The probabilities are calculated by multiplying the probabilities of the events shown in the lefthand side of the table.
Perfect information says success (S) or failure (F) of Trapping Sterilise Hormone S S S F S S S F S F S F S F S F S F S F F F F F
Response Use hormone Trap/Sterilise Use hormone DDT spray Use hormone Trap/Sterilise Use hormone DDT spray
PaYoff
56 26
56
o 56 26
56
o
Probability 0.085 0.340 0.015 0.060 0.015 0.060 0.085 0.340 1.000
From the table: EMV with perfect information (56 x 0.085) + ('26 x 0.34) + (56 x 0.015) + (0 x 0.06) + (56 x 0.015) + (26 x 0.06) + (56 x 0.085) + (0 x 0.34) = 4.76 + 8.84 + 0.84 + 0 + 0.84 + 1.56 + 4.76 + 0 21.60
EVPI
= 21.60 =
 6.625 14.975
The maximum to be paid for any additional sample information is ÂŁ14975000. 3 What is the maximum that should be paid for the university research? The initial decision in the tree of Figure A1.9 should now have a fourth option added. It should be constructed by following the first four stages of decision analysis. The fifth stage, defining the optimal path, cannot be carried out until the cost of the research is known. Stage 1: The first chance event is the result of the research, whether it indicates that the hormone will work or not. Once this result is known, it must be decided either to use the hormone or revert to the previous optimal route of the twostage approach. Using the hormone leads to a chance node showing its success or failure  see Figure A1.10. Stage 2: The payoffs are the same as originally. The cost of using the hormone is shown as a gate. The decision branch representing reversion to the twostage approach is truncated and has a payoff equal to the EMV (6.625)derived above. Had this branch been drawn out to its full extent the effect would have been exactly the same.
EMV=
30.295 4 Use hormone
0.35 Research says hormone successful
Stage 3: The probabilities (e.g. P(Success of hormone given the research predicts success) have to be calculated with Bayes' theorem. The Venn diagram (Figure A1.11)is drawn: (i) Insert a vertical line to split the square according to the prior probabilities (0.2 chance of success, 0.8 chance of failure). (ii) Insert horizontal lines according to the research accuracy data. The 'prior success' rectangle is split 955%; the 'prior failure' rectangle is split 2080%. (iii) Calculate the posterior probabilities from the Venn diagram. From Figure A1.11: P(Research says 'success')
=
Area C + Area D x 0.95) + (0.8 x 0.2)
= (0.2
= 0.35
P(Research says 'failure')
=
Area A + Area B x 0.05) + (0.8 x 0.8)
= (0.2
= 0.65
Prior probabilities Success
Failure
0.2
0.8
Research results
A
B
0.8 Research says 'failure'
C
0
P(Success/Research says 'success')
=
0.2 Research says 'success'
C/(C + D)
= 0.19/0.35 = 0.543 P(Failure/Research says 'success')
=
D/(C + D)
= 0.16/0.35 = 0.457 P(Success /Research says 'failure')
=
P(Failure/Research says 'failure')
=
+ B) 0.01/0.65 = 0.015
= A/(A
B/(A + B)
= 0.64/0.65 = 0.985 Stage 4: The rollback is shown in Figure A1.IO. It gives an EMV of ÂŁ13509000.
EVSI
= =
=
EMV with research  EMV without research 13.509  6.625 6.884, i.e. ÂŁ6884000
Presuming that the research costs less than this amount, the research option is now the optimal path. 4 What assumptions
and nonquantitative information should be taken into account in coming to a decision? The assumptions are: (i) The payoffs are discounted cash flows and therefore must be sensitive to the choice of time horizon and the discount rate, neither of which is mentioned in the case. (ii) The payoffs are net benefits/losses based on the incremental effect of the different programmes. These figures must be open to doubt since they are bound to be hard to measure. (iii) The research does not cause delays which affect the payoffs. Nonquantitative considerations: (i) Approach (b), and to a lesser extent approach (c), carry qualitative benefits in that they provide information that may be valuable for future work in this area. (ii) DDT and hormone spraying presumably have environmental disbenefits.
Marking scheme Out of 40: Marks 1
2
3
4
Optimal decision without research information: Tree Payoffs / probabilitjes Rollback Risk profile
2
3 3
4
EVPI:
List of outcomes Payoffs / probabilities EMV calculations Research branch: Tree Payoffs Bayesian revision Rollback Assumptions/nonquantitative
4 3 3 2 1 6
3 information
6
Total
40
1 Which stock is preferable, Cor D? Why? Stock D is preferable. It is at least as good as stock C whatever the state of the economy. The row in the payoff table relating to stock D is either equal to or more than that for stock C for all states of the economy. Stock C can therefore be eliminated from the analysis immediately since it could never be superior to D. 2 What stock has the highest EMV? The EMV of each stock can be calculated by summing (Probability x Return) for all four states of the economy. This is equivalent to drawing a decision tree which has, for each stock, a chance node with four branches, one for each state of the economy. EMV of A
= = =
(0.1 x  0.35) + (0.3 x  0.15) + (0.4 x 0.15) + (0.2 x 0.60) 0.035  0.045 + 0.060 + 0.120 0.10
EMV of B = (0.1 x  0.25) + (0.3 x  0.05) + (0.4 x 0) + (0.2 x 0.35) = 0.025  0.Q15 + 0 + 0.070 = 0.03 EMV of D
= = =
EMV of E
= = =
(0.1 x  0.05) + (0.3 x 0.05) + (0.4 x 0.10) + (0.2 x 0.20) 0.005 + 0.015 + 0.040 + 0.040 0.09 (0.1 x 0) + (0.3 x 0.05) + (0.4 x 0.05) + (0.2 x 0.10) 0 + 0.Q15+ 0.020 + 0.020 0.055
3 What is the EVSI for the consultants' economic opinions? The probabilities of different states of the economy have to be revised, using Bayes' theorem, in the light of what the opinions might be. The Venn diagram is shown in Figure A1.12. First divide the square vertically in respect of the prior probabilities, the initial probabilities of different states of the economy. Second, subdivide each of the rectangles relating to prior probabilities according to the reliability factors given in the table in the question. Probabilities of different opinions being given can be obtained from this Venn diagram. For example: P(ol)
= = =
(0.1 x 0.1) + (0.2 x 0.3) + (0.4 x 0.4) + (Q.6 x 0.2) 0.01 + 0.06 + 0.16 + 0.12 0.35
State of the economy
s2
s3
0.3
0.4
01 0.1

0.2 0.4
02 0.2 0.6 c
0
'c
0.6
'0. 0
0.4 03
0.7 0.3 0.2
0.2 0.1
Revised probabilities of the states of economy can also be calculated using Bayes' theorem. For example: P(sl/ol)
=
(0.1 x 0.1)/0.35
= 0.029
All the revised probabilities are shown in Table A1.1. Once the revised probabilities have been obtained, the situation can be represented by the decision tree shown in Figure Al.13. The payoffs are the same as previously. The rollback procedure suggests the following optimal path: If the opinion is 'good': choose stock A. If the opinion is 'average' or 'poor': choose stock D. The EMV if this optimal route is pursued is 0.131. EVSI = EMV with opinion  EMV without opinion. = 0.131  0.100 = 0.031
This figure is per ÂŁ invested. The actual monetary EVSI will therefore vary with the level of the investment.
Appendix
1 I Solutions to Review Questions
Table A1.1 Opinion
and Case Studies
Bayesian revision of stock selection problem State of economy
Prior
(3)
Reliability
x
pro lls1) ete.
probabilities P(s1) ete.
(4)
Posterior probabilities Col. (5)1P(o 1) ete. P(sll0 1) ete.
(1)
(2)
(3)
(4)
(5)
(6)
01
51 52
0.1 0.3 0.4
0.1 0.2 0.4
0.029 0.171 0.457
0.2
0.6
0.01 0.06 0.16 0.12
53 54
P(01) 02
51 52 53 54
0.1 0.3 0.4
0.2 0.6 0.4
0.2
0.3
0.1
0.7
0.3 0.4
0.2 0.2 0.1
P(02) 03
51 52 53 54
0.2
P(03)
=
=
=
0.35 0.02 0.18 0.16 0.06 0.42 0.07 0.06 0.08 0.02
0.343 0.048 0.429 0.381 0.142 0.304 0.261 0.348 0.087
0.23
4 What is the EVPI? Perfect information would say what the state of the economy was to be. The investor would then choose the stock with the highest return for that state of the economy. Perfect information says: sl s2 s3 s4
Choose stock E DorE A A
Payoff
0 0.05 0.15 0.60
Probability
0.1 0.3 0.4 0.2
EMV with perfect information = 0 + 0.015 + 0.06 + 0.12 = 0.195 EVPI = EMV with perfect information  EMV without perfect information = 0.195  0.100 = 0.095
This when multiplied by the amount invested gives the maximum that should be paid for any information.
0.05 +0.05 +0.10 +0.20
0.048 0.429 0.381 0.142
0.35
0.25
0.15
0.05
+0.15
0 +0.35
51
0.05
52
+0.05
53
+0.10
54
+0.20
+0.10
0.35 0.15
0.05 +0.05 +0.10 +0.20
Marking scheme
Out of 50: Superiority of Dover C for all states of economy EMV using only prior probabilities EVSI: Bayesian revision Application of posterior probabilities to tree Rollback, selection of optimal path Calculation of EVSI EVPI
x y
= =
Marks 5 5
number of acres planted with flowers number of acres planted with fruit
Objective Function
The objective is to maximise the contribution, i.e. maximise 320x + 400y.
There are constraints on labour time in each of the four seasons, and on the total acreage available. In addition there are nonnegativity constraints: lOx + lOy 40x
+ 20y
lOx
+ lOy x +y
50y
x,y
100 (winter) 200 (spring) :::; 200 (summer) :::; 100 (autumn) :::; 6 (acreage) ~ 0 :::;
:::;
The constraints can be reduced and tidied up. First, since the winter and autumn constraints are identical, one is redundant and can be eliminated. Second, the remaining winter/autumn constraint can be divided through by 10 to give:
This is always less stringent than the acreage constraint, i.e. if x + y :::;6 then x + y must also be :::;10. The remaining autumn/winter constraint can therefore also be eliminated. The spring constraint can be divided through by 20, the summer constraint by 50, leaving:
2x + Y :::;
10
Y:::;
4
x +Y:::; X,Y 2:
6 0
Which of A, B, C or 0 is the optimal point can be judged either by accurate drawing or by considering the slopes of the constraints relative to the slope of the objective function. 10 4 Y= 6 X+Y= 320x + 400y = Obj. 2x+y=
(Slope = 2) (Slope = 0) (Slope = 1) (Slope = 4/5)
The slopes are calculated either from the graph or by putting the equations in the form y = mx + c where m is the slope. The objective function is less steep than x + y = 6 but steeper than y = 4. If we were to draw the objective function of Figure A1.14, its slope would
be somewhere between the slopes of the lines AB and BC As the objective increases, B will therefore be the last point of contact between objective function and feasible region. B is the optimal point. At B: y =
x+y=
4
6
The coordinates are therefore
y
= 4, x = 2.
The objective at this point is (320 x 2) + (400 x 4) = 2240. The farmer should therefore plant two acres with flowers and four with fruit to obtain a return of ÂŁ2240. Marking scheme Out of 20: Marks
~.
Formulation: Decision variables Objective function Time constraints Acreage constraint Nonnegativity constraints Solution: Identifying redundant constraints Graphical representation including feasible region Identification of optimal point Optimal objective function value
2 2 4 1 1
Total
x
=
number of type A assembled per day
y = number of type B assembled per day
2 4 3 1 20
x +y x y x y O.3x + O.Sy
x y
300 shells
::;
::; 250 timer A ::; 280 timer B ::; 190 wiring A ::; 300 wiring B
140 labour 2: 100 min. production A 2: 80 min. production B ::;
The constraints for timer A and wiring B are redundant, i.e. they are already covered by other constraints (wiring A and timer B). The constraints reduce to:
x
300 ::; 280 ::; 190 ::; 1400 2: 100
y
2: 80
x +y y
x 3x + Sy
::;
y 300
280
\ \
200 \ Obj.
= 30 000
" \ \ \ \ \ \ \ \ \ \ \
\ \
100
\120
Figure A1.15 shows a graphical representation of the problem. The equation 3x + 5y :s; 1400 is well outside the feasible region and is not shown. Drawing the objective function when its value is 30000 demonstrates that as the objective is increased, the optimal point is point A. A is the intersection of the lines x must be
= 190 and
x +y
= 300. The optimal point
Substituting in the objective function gives: Optimal objective value = 64000 Marking scheme Out of 20: Formulation: Decision variables Objective function 6 less than constraints 2 more than constraints Solution: Identifying redundant constraints Graphical representation including feasible region Identification of optimal point Optimal objective function
1 The optimal values of the variables are given as 14.23 and 12.69. Therefore 14230 boardfeet of lumber should be produced and 12690 square feet of plywood. 2 The optimal objective value is ÂŁ46340 which is the profit contribution at the production levels given in (1) above. This can be checked by substituting L = 14.23 and P = 12.69 in 1250L + 2250P. 3 The question marks are in the positions of the slacks for the two minimum order level constraints. For L :::::6, the slack is 14.23 6 = 8.23 For P :::::12, the slack is 12.69 12 = 0.69. Note that for::::: constraints it is more usual to refer to these amounts as 'surplus' not 'slack'.
Marking scheme Out of 10: Marks Correct values for decision variables Correct objective function value Correct slacks
4 2
4
10
Let x Let y
= number of chairs manufactured.
=
number of tables manufactured.
Objective Function
Maximise contribution:
80 assembly time 120 finishing time
4x + 2y :::; 2x + 5y :::;
o
x, y 2':
The graphical solution is shown in Figure A1.16. The optimal point is A at the intersection of the assembly and finishing constraints. The x and y values can be read off from the graph or found from the simultaneous equations: 4x + 2y = 80 2y + 5y = 120
Multiply the second equation by 2 and subtract the first: 4x + 2y =
240 80
8y =
160
y =
20
4x+10y
=
4x+40=80
x
=
10
.. ,
,, , '"
= 3,200 ,Profit ,
,, ,, ,
Feasible region
10
30',
,,4{),,, , , " , 'Profit
= 2 200
. Profit = 1 800
x y
= =
10 chairs 20 tables
Objective value
= =
60 x 10 + 80 x 20 ÂŁ2200
If an extra hour of finishing time were available, the optimal point would be Al instead of A (see Figure A1.17) and its coordinates would be given by the
two equations: 4x + 2y = 80 2x + 5y = 121
4x + lOy =
242
80 162 20.25
4x +2y = 8y =
y=
4x + 2(20.25) = 80 4x = 39.5
x
=
9.875
The new optimum is 20.25 tables, 9.875 chairs, so the extra time allows output of the more profitable product (tables) to be produced. It is assumed that '0.25 table' and '0.875 chair' refer to partly finished products, otherwise integer programming would have to be used. The new objective function value is: 60 x 9.875 + 80 x 20.25 = 592.5 + 1620 = 2212.5
New objective value 
Old objective value
= =
2212.5  2200 12.5
An extra hour of finishing time increases the contribution by £12.5. The normal cost of finishing time is already included in the objective which measures contribution not revenue. The dual value (£12.5) should be compared with the extra overtime cost of finishing time (£8). Since it is greater, overtime should be used. Marking scheme Out of 20: Marks Formulation Graphical solution Knowing how to find dual values Calculation of new optimal solution Calculation of new optimal objective Calculation of dual values Interpretation of dual values
4 4
3 4 2 1
2
20
Let x = number of bags of JDJ. Let y = number of bags of PRP.
There are constraints for the requirements of each of nitrogen, phosphorus and potassium. If x bags of JDJ are used, then 40x kg of JDJ are used (40 kg per bag). Since JDJ contains 20% nitrogen compounds, the quantity of nitrogen applied will be 20% of 40x, i.e. 8x kg.
Similarly, the nitrogen in y bags of PRP is 10% of 60y, i.e. 6y kg. The total amount of nitrogen applied will therefore be:
This must be greater than the amount required which is 60 kg. The constraint must be:
2x + 6y ~ 24 (phosphorus) 8x + 3y ~ 40 (potassium)
Nonnegativity constraints are also required. The complete formulation is: Min. Subject to
6x + 5y 8x + 6y
~
2x + 6y
~
8x + 3y
~
x,y
60 Constraint 1 24 Constraint 2 40 Constraint 3
~ 0
The graphical solution is shown in Figure A1.1S. Note that when a constraint is 'greater than', the feasible area defined by it is on the opposite side of the line than that for a 'less than' constraint. Note also that the objective line is moved downwards in the minimising process. A carefully drawn graph shows that the optimal point is A, at the intersection of constraints 1 and 2. Its coordinates are given by the simultaneous equations: 8x + 6y = 60 2x + 6y = 24
The answer to the problem is to use six bags of JDJ and two bags of PRP per acre at a cost of ÂŁ46/ acre.
10 9 8 7
4
6
Constraint 2
5
If 9 kg less of phosphorus is required the associated constraint (2) will be 2x + 6y ~ 15 and the graphical solution will be as in Figure A1.19. The optimal point is B, still at the intersection of constraints 1 and 2, but also on the x axis. It can be seen from Figure A1.19 that at B:
The new objective function value is (6 x 7.5) + (5 x 0) = £45. The farmer will therefore save £1/ acre if the second analyst's advice is thought to be correct. The dual value for phosphorus is the change in the objective function if the requirement for phosphorus is changed by one. It has not been changed by one but by nine (from 24 to 15), in which case the objective function changed by £1. The dual value of phosphorus must therefore be:
Constraint 2
2'/,
Marking scheme Out of 20: Marks Formulation Graphical solution Recalculating new optimal solution Calculation of new optimal objective Interpretation of new solution Calculation of dual value from new solution
4
4 6 2 2 2
20
1 The optimal values of the variables are given as 14.23 and 12.69. Therefore 14230 boardfeet of lumber should be produced and 12690 square feet of plywood. 2 The dual value for constraint 1 (referring to spruce) is £961.50. This means that an extra 1000 boardfeet of spruce would produce a further £961.50 in profit contribution. An extra 500 boardfeet could therefore be expected to produce an extra £480.75. It should be checked that an extra 500 boardfeet is within the allowable righthand side range. From Table 4.5 this range is 44600 to 48100 board feet. Therefore the dual value applies for an increase of 500 boardfeet. 3 The dual values apply for up to 48100 boardfeet of spruce and for up to 80700 boardfeet of fir. Consequently the mill can accept an additional 3100 boardfeet of spruce and still make an extra profit contribution (equal to the dual value times 3.1). Or it can accept an extra 700 boardfeet of fir and make an extra profit of 0.7 times the dual value for fir. Note that the extra must be either spruce or fir since dual values apply to increases in one constraint at a time. 4 The allowable ranges for the coefficients in the objective function mean that the profit contribution for lumber can be between £1230 and £1610 without the optimal production mix changing. Likewise for plywood the allowable range is 1750 to 2290. Therefore the optimal production mix would be unchanged if the profit contribution of plywood were £2290. Marking scheme Out of 20: Mark Basic solution Dual values RHS ranges Coefficient ranges
4 4
6 6
20
First associate twodigit numbers with the different demand levels in proportion to the probabilities. Second, let the midpoint of each category represent the whole category, e.g. the 1012 class is represented by 11. This is an approximation, but since the original distribution was subjectively estimated, the approximation is reasonable in the context (Table A1.2). Numbers in the range 0004 are associated with a demand of 11000 units, 0524 with 13000 units and so on. In particular 79/ 18/ and 52 will be associated with 17000 units, 13000 units and 15000 units respectively.
Table A1.2 Midpoint
Units
Probabilities
(000)
(000)
(%)
11
1012 1214 1416 1618 1820 2025
5 20 35 25 10
13 15 17
19 22.5
5
0004 0524 2559 6084 8594 9599
Marking scheme
Out of 10: Marks
Using midpoints Associating numbers with demand levels Correct demands for the three numbers
2 5 3
10
The simulation will proceed minute by minute between the hours of 9.30 and 12.00. The flowchart for the simulation is given in Figure A1.20. Marking scheme
Out of 10: Marks
Correct Correct Correct Overall
decision nodes action nodes information recordipg structure
5 5 5
5 20
~
~~~~~~~" ~
Cashier file One file for each desk manned. It records for each minute: Calculate (i) average queue (ii) % cashier busy
Busy/idle Queue length Time next customer served Queue length distribution to date
Customer file One file for each customer records: 
Time of arrival Queue joined Time start service Time finished
1 If the distributions of present values are normal, the attributes curves can be used to calculate the probabilities (Table Al.3).
of normal
Profit Loss Product A Product B
Less than 1 % 16%
Since 68% of a normal distribution the mean, from Figure A1.21:
+
ÂŁ300 million Virtually 0
16%
lies between
Âąl standard
deviation
of
I
150
300
+ +1
z=
2 Product B has a higher average present value than A. If the company has many such investment opportunities then B would be preferable because on average it will produce the better return. Product B is also more risky. It has a larger chance of a loss and of a high profit. If the company has few investment opportunities, and if the potential losses are large in comparison to its total capital, the company might prefer not to risk making a loss and select A. On the other hand, the chance of a high profit might in itself be attractive in spite of the potential losses, and therefore B might be preferable.
In short the choice between high average PV fhigh scatter and low average PV flow scatter depends upon the finances and risk attitudes of the company.
Marking scheme Out of 10: Marks Use of normal distribution Probability (profit or loss) for each product Argument about launch decision
2 4 4
10
1 A Monte Carlo simulation should be carried out with arrival rate and service time as the stochastic variables. The distributions would be sampled with random numbers and the values so obtained used to make cost calculations. 2 The flowchart is given in Figure A1.22. Note that there is no 'right answer' for flowcharts. Two different charts may still capture reality adequately enough for the simulation to be constructed. So, for example, the situations described in Cases 5.2 and 5.4 are similar (people v. buses) but the flowcharts are slightly different. Although different, both are correct since they are both good approximations of reality. 3 Economically the argument involves the tradeoff between the time buses/crews are kept waiting for inspection against the time the inspection units may be idle waiting for buses to arrive. Safety might be affected if the queues of waiting buses were long so that the inspection mechanics hurried their work and made mistakes. Overall the relevant factors would seem to be: Operational: Average time/week units are used Average time a bus is queuing for inspection Average number of buses in a queue Maximum queue length Financial: Cost of crews having to wait Cost of buses having to wait Cost of units being idle Saving in running costs from having two units only
(
TIme period 1 )
Marking scheme Out of 20: Marks Brief description of Monte Carlo method Flowchart Factors in decision: operational financial
5 5 5 5
20
1 The stages in a network analysis are: 1
List all activities in the project.
2
Identify the predecessors o( each activity.
3
Draw the network diagram.
4
Estimate activity durations as average times or, if uncertainty is to be dealt with explicitly, as optimistic/most likely/pessimistic times.
5
Produce an activity schedule by making a forward then a backward pass through the network.
6
Determine the critical path by calculating activity floats.
7
Measure uncertainty by calculating activity variances, the variance of the whole project and finally confidence limits for the duration of the project.
8
Draw costtime curves for each activity.
9
Crash the network and construct a costtime graph for the project.
Steps 8 and 9 are not relevant to this case since crashing the network is not required. Following steps 17: 1
A list of the activities is already available.
2
The immediate predecessors can be taken straight from the case description and are as given overleaf:
'../
Appendix 1 / Solutions to Review Questions and Case Studies
Duration (days) A B
C D E F G H I
J ML
Activity Receive kit Order engine Manufacture trim Assemble shell Paint shell Test engine Add trim Insert engine Test Ship = Most Likely Opt.
A D B
=
C,E G, F H I Optimistic
Opt. 15 5 2 1 1 1 1 1
ML
4
1 Pess.
=
20 15 3 2 2 2 2 2 5 3 Pessimistic
Pess. 25 25 4
3 3 9
3 6 12 8
3
The network diagram is shown in Figure A1.23.
4
Estimates of the durations of the activities are given in the case but an expected duration has to be calculated from the optimistic, most likely and pessimistic estimates. This is shown below: Duration (days) ML Pess. Expected 20 25 20 15 25 15 3 4 3
A B
C D E F G H I
J ML DecisionMaking
Predecessors
Techniques
=
Activity Predecessors Opt. Receive kit 15 Order engine 5 Manufacture 2 trim Assemble shell A 1 Paint shell D 1 Test engine B 1 C,E Add trim 1 Insert engine G, F 1 Test H 4 Ship I 1 Pess. Most Likely Opt. = Optimistic
Edinburgh Business School
2 2 2 2 2 5 3
=
3 3 9
3 6
12 8 Pessimistic
2 2 3 2 2.5 6
3.5
IA1/Ssl
5
A forward followed by a backward pass through the network gives the result shown in Figure A1.24. Earliest start ~
I
Expected
Earliest
dur~tion
finish
Latest (20, 2, 22 Latest start (20, 0, 22r finish
o
(26, 2.5, 285) (28.5, 6, 34.5)(34.5, 3.5, 38) (26, 0,28.5) (28.5,0,34.5)(34.5, 0,38)
(0, 3, 3) (21,21,24)
C
H
I
J
F ~ (15,3,18) (23,8,26)
The results can be used to produce the activity schedule shown below, giving the earliest and latest start and finish of each activity: Activity Receive kit Order engine Manufacture trim Assemble shell Paint shell Test engine Add trim Insert engine Test Ship 6
ES
Earliestllatest Start/Finish LS EF
o
o o o
8 21 20 22 23 24 26 28.5 34.5
20 22
15 24
26 28.5 34.5
20
15 3 22 24 18 26 28.5 34.5 38
LF 20 23 24
22 24 26
26 28.5 34.5 38
The float for each activity (Earliest startLatest start) is shown in Figure A1.24. The critical path is those activities with zero float and is therefore: ADEGHIJ.
. Vanance
=
(pessimistic  Optimistic) 6
2
Appendix 1 / Solutions to Review Questions and Case Studies
The results are shown below:
A B C D
E F G
H I
J
~,
ML
=
Duration (days) Variance Activity Pess. (dayi) Opt. ML Expected 20 25 2.8 Receive kit 15 20 11.1 Order engine 5 15 25 15 Manufacture 0.1 2 3 4 3 trim Assemble shell 1 2 3 2 0.1 Paint shell 1 2 3 2 0.1 1.8 Test engine 1 2 9 3 Add trim 1 2 3 2 0.1 2.5 Insert engine 0.7 1 2 6 1.8 12 Test 4 5 6 1.4 3.5 Ship 1 3 8 Pess. = Pessimistic Most Likely Opt. = Optimistic
2 The variance of the whole project is calculated as the sum of the variances
of the activities on the critical path: Variance of project
= 2.8 + 0.1 + 0.1 + 0.1 + 0.7 + 1.8 + 1.4 = 7.0
The standard deviation is therefore )7, i.e. 2.6. The 95% confidence limits for the duration of the project are the expected completion time plus or minus two standard deviations: = 38 =
Âą2x
2.6
32.8 to 43.2 days
Marking scheme Out of 25: Listing predecessors Drawing network diagram Calculating expected durations Passes through network for ES/LS/EF/LF Calculating float, defining critical path Calculating activity variances Calculating project variance Calculating confidence limits
Marks 2 3 2
8 3 3 2
2
25
1
List all activities in the project.
2
Identify the predecessors of each activity.
3
Draw the network diagram.
4
Estimate activity durations as average times or, if uncertainty is to be dealt with explicitly, as optimistic/most likely/pessimistic times.
5
Produce an activity schedule by making a forward then a backward pass through the network.
6
Determine the critical path by calculating activity floats.
7
Measure uncertainty by calcul~ting activity variances, the variance of the whole project and finally confidence limits for the duration of the project.
8
Draw costtime curves for each activity.
9
Crash the network and construct a costtime graph for the project.
Step 7 is not relevant in this case since the durations of activities are given as single estimates. Optimistic, most likely and pessimistic times would be needed to measure uncertainties in the project. Otherwise, step by step: 1
The activities have already been defined and are listed in the case.
2
The predecessors of each activity are described in the case and are shown below:
A
B C D E F G H I
J K
1A 1/581
Activity Implications Order materials Personnel hire Plan equip. modifications Deliver materials Modify Quality control training Prod. control Production Test Final stage
Predecessors
Duration (weeks)
2 3
A A A
14 12
B D C,E,F
20 18 2
C,E,F G,H I
18
3 5
3
J
3
The network diagram is shown in Figure A1.25.
4
Estimates of the durations of the activities are given in the case.
Edinburgh Business School
DecisionMaking
Techniques
5
A forward followed by a backward pass through the network gives the results shown in Figure A1.26. The project will take 61 weeks to complete.
(32,2,34) (33,1,35) G
6
The float for each activity (Earliest stateLatest start) is shown in Figure A1.26. The critical path is those activities with zero float and is therefore: ADFHIJK.
.
7
Step omitted for this case.
8
A planner can usually reduce the project completion time by spending more money, i.e. by crashing the network. Crashing begins by looking at the costtime curve for each activity. Information given in the table in the case shows the crash cost and the crash time. From this information, the ratio: Cost increase Time saved
is calculated. This is in effect the 'crashing cost per unit of time' for the activity. It shows which activities can bring about time reductions at the lowest cost and which should therefore be considered first. The table below shows normal and crash costs and times, as well as the maximum possible time reduction and the crashing ratio, for the activities that are capable of being crashed.
Activity
C 0 F I 9
Hiring personnel Plan equip. mods Modifications Production
Normal Time Cost (weeks) (ÂŁ000) 14 12 18 18
100 450 2600 1800
Crash Time Cost (weeks) (ÂŁ000) 10 9 12 14
200 630 2900 2400
Max. time redn 4 3 6 4
Crash ratio 25 60 50 150
Crashing involves the original analysed network (Figure A1.26) and the crashing ratios given above. Only critical activities are crashed since they are the only ones that can reduce the overall project completion time. The activities on the critical path are examined to find the one with the least crashing cost (= lowest crashing ratio). This activity is crashed by reducing its duration either until the maximum possible reduction has been made or until another parallel path is critical.
The critical activities are A, 0, F, H, I, J and K. Of these only 0, F and I are capable of being crashed. F has the lowest crashing ratio, 50. It can be reduced by the full amount possible, six weeks, without altering the critical path. The new network is shown in Figure A1.27. It has a total duration of 55 weeks (the original 61 minus the reduction of six).
(26,2,28) (27,1,29)
G
The activities are reexamined to find the activity on the critical path with the next lowest ratio. This is 0, with a crashing ratio of 60. If it were crashed to its maximum of three weeks a parallel path, through activities B and E, would become critical since these activities have a float of only one week. If activity 0 is reduced by more than one week the Path OF will no longer be part of the critical path, which would then take the route ABEHIJK. Further reductions in activity 0 alone are therefore pointless since the overall time for the project will not change. To continue crashing the network requires that an activity on the branch BE be reduced simultaneously with activity O. However, neither B nor E is capable of being crashed. Therefore 0 is reduced
by just one week, giving the network shown in Figure A1.28. The duration the project has been reduced to 54 weeks.
of
Once the network of Figure A1.28 has been achieved the project is fully crashed. The only activity not yet crashed is I but crashing this would incur a cost of £150 000 per week. This is more than the penalty cost of £100 000 per week and is therefore not worth doing. It is better to leave the project completion time at 54 weeks and pay four weeks' penalty charges.
(25,2,27) (26,1,28) G
It is therefore in the best financial interests of the company to crash the network to 54 weeks at a cost £360 000 (6 x 50 + 1 x 60). This will save £700 000 (7 weeks x 100) in penalty costs. Whether the government, which wanted completion in 50 weeks, will be happy with this financially oriented view is another matter. The company may also need to consider whether its future business with the government is being put at risk. Marking scheme Out of 25: Listing predecessors Drawing network diagram Passes through network to calculate completion time Calculating float, defining critical path Calculating crash ratios Crashing activity F Crashing activity D Stopping without crashing I
Marks 2 3 8 2
1 3 3
3
25
Driver should take the following points into account (relating to the three main job objectives defined in Table 7.1). Only major issues are discussed.
(a) To introduce Internet technology into offices where it may benefit the company. (i) The introduction of Internet technology is clearly his prime brief but any individual projects should relate to business objectives rather than be 'the Internet for the sake of the Internet'. All systems should be evaluated and justified against sound investment criteria. Driver will have to develop these appraisal methods, relating them to BE's standard procedures for capital expenditures. (ii) The first task is to stimulate interest and involvement, perhaps by giving demonstrations at a few major sites and allowing people to find out for themselves what the Internet could do for them. This could be reinforced if there were live applications to act as examples of what could be done. Looking for projects already in the pipeline is probably the quickest way to find live applications. This approach should promote user participation from the start  and of course their involvement should be contmued. (iii) Live applications would also establish the existence of local 'Internet champions' to set up local user groups and to act as focal points for stimulating further applications. (iv) Houghton takes a positive, optimistic approach towards this introduction and wants to give people at all levels in the organisation a chance to become familiar with new technology in the expectation that this will contribute to innovative management thinking as well as more effective business practices. This attitude means that Driver has excellent top management support not just for installing IT but for using IT for broad management purposes rather than just cost or manpower savings. However, it is important for this support to be transmitted through to senior management in the companies as well as at group level. (ii), (iii) and (iv) together should be aimed at developing attitudes of commitment and motivation amongst users at all levels of management. (b) To ensure a maximum degree of cohesion within and between different departments in order to reduce costs and maximise utilisation of resources. (i) Providing a 'maximum degree of cohesion' within a company that allows a great deal of local management autonomy will require an approach of patient persuasion rather than distant diktat. A Group Internet strategy could be seen by some as an unwanted extension of Group control over local affairs. There is also likely to be a belief that each company, and perhaps each site, has problems peculiar to their circumstances that cannot be resolved by Group policy. Group management has no authority to impose an Internet strategy on companies  it will have to be 'sold' to managing directors of each company in the Group and to the managers and staff who would use the systems. Any strategy will have to show tangible benefits to local business needs and be sufficiently flexible to meet different needs. Where there is a
requirement for some form of standardisation, the benefits will have to be clearly stated in business rather than technical terms, for example in terms of reduced costs rather than software compatibility. (ii) Since he is to coordinate the Internet implementation throughout the Group, Driver will have to use rigorous planning methods, such as CPM or PERT. (iii) There is a technical aspect to 'cohesion'. Driver will have to work out where computer communications within and between departments would be beneficial and plan appropriate networks. (iv) The introduction of new technology is always liable to cause organisational problems. This will especially be the case at a time of industrial and economic uncertainty. Driver will have to ensure that he has the right 'people policies'. In particular the following are likely to be problems: Fear of computing. There may be reluctance for anyone other than secretaries and typists to use the Internet, particularly among older people. This may in turn indicate that there are deeprooted anxieties about the impact of new technology on managerial and office work. Lack of understanding of capabilities and business applications of the Internet. A great deal of groundwork will have to be done to explain what the Internet can do, what is available, at what prices. This will have to be done before people can identify how they can best apply systems to their particular requirements. Perceptions of software costs and development times. Traditionally, mainframe software has been expensive and has taken a long time to develop. The existing expertise among local ITMs and in BEC would tend to back this view. If such expected costs were added to the low cost of Internet hardware, the costjustification argument for; say, an Internet solution versus a terminal linked to a mainframe would fail to take account of the advantage of much lower cost (and possibly free) Internet software. Expectations of long lead times for software development could also turn opinion against the belief that a quick introduction of the Internet was feasible or desirable. Age and culture clash. The age profile of the Electronics Group was middle aged and many people, like Jim Hamilton, had worked in their companies for decades. Driver, however, was in his twenties. Hamilton's description of him as 'a typical whizzkid' implied that he could be seen by some as young, brash and perhaps with full theoretical enthusiasm for business but lacking the practical understanding that comes from real working experience. (c) To evolve a Group Internet Strategy to include specific recommendations. A strategy will have to include many items but the following should be given particular attention.
(i) Driver's business appraisal methods will also have to insist that alternative nonInternet solutions are considered in order to avoid his recommendations being seen as 'the Internet for the sake of the Internet'. (ii) There will have to be a technical part to the strategy, especially dealing with the need to create some form of technical standardisation. This also follows from Driver's goal of trying to achieve cohesion within the Group. It will have to cover questions about the nature of that standardisation, such as: What items of hardware and software should be covered? How can all companies in the Group be encouraged to accept it? How can the standards be made to support a variety of applications with a flexibility that satisfies different users, without undermining the advantages of standardisation? There are other technical issues to cover such as communications between and within sites, and.how to ensure systems have the flexibility to change and grow. (iii) There are 'distinct cultural streams' in the different companies in the group and Driver will have to manage transitions sensitively and thoroughly. (iv) The demarcation between Driver's area of responsibility and that of BEe is likely to be an area of potential organisational and technical conflict. There are relatively clear distinctions between the kinds of applications different systems are used for. Mainframes are typically applied to corporate and administrative tasks involving large volumes of data such as accounting, stock control, manufacturing, resource planning, payroll and corporate databases; microcomputers are used for more personal tasks, such as wordprocessing, job costing, spreadsheets and relatively smallscale local accounting, databases and technical calculation activities. The Internet will have to find a place between these massive areas. However, as the capabilities of the Internet grow, the distinction between what it can and cannot do is increasingly blurred. There is likely to be a 'clash of computing cultures' as the potential of the Internet grows: the difference in cultural perspective between the traditional IT person and the Internet enthusiast may be far more important than technical differences between technologies. (v) Driver's responsibilities, however, will also encroach on the domains of local ITMs, making a potential source of conflict that will not be resolved through formal company statements. (vi) A support strategy must be devised so that users have access to help but do not make demands upon Driver's time that will detract from his main objectives. A support strategy will maintain momentum in the longer term. (vii)Driver will have to incorporate longterm monitoring into his plans to prevent interest dying and further developments being held back.
Monitoring should ensure the Internet is being used effectively and prevent its being used inefficiently or not being developed beyond initial applications. Marking scheme The following issues should have been discussed but not necessarily under the same headings as shown. Out of 30: Marks (a)
Introducing IT: Businessrelated projects Stimulation of interest Champions Senior management support (b) Ensuring cohesion: Persuasive not dictatorial approach Planning methods Technical cohesion People policy (c) Strategy: Need to consider alternatives Standardisa tion Managing transition Dealing with BEC and the ITMs Support strategy Monitoring
3 2 2 3 3 2 2 3
1 2 2 2 2
1 Total
30
Published on Dec 13, 2010
Expectedcost=(0.25x4)+(0.25x3)+(0.5x2) =1+0.75+1 =2.75 1.4 Discorrect. EMVatpointBis80(obtainedbyeliminatingthelessfavourablebranch). 1.2 C:...