1 2019 EXAMINATIONS PART II (Second, Third and Final Year) MANAGEMENT SCIENCE MSCI 224 Techniques for Management Decision Making CLOSED BOOK EXAMINATION (3 HOURS) FORMULA SHEET PROVIDED Calculators and standard dictionaries are permitted. In Part A, answer the one question that has been set. In Part B, answer three out of the four questions. Answer each question in a separate booklet. If you write answers to all four questions in Part B, please cross out the answer that should not be marked. Otherwise, the marker will mark only the first three answers encountered. Show all your workings. DO NOT TURN THIS PAPER OVER UNTIL YOU ARE TOLD TO DO SO 2 Formula Sheet Networks PERT: Mean Duration of an Activity = 6 4 bma ++ Variance of the Duration of an Activity = 2 6 ab where a = optimistic estimate b = pessimistic estimate m = most likely estimate Forecasting (Exponential Smoothing) Single (Simple) Exponential Smoothing tmt Sy =+ (m = number of periods ahead to forecast) 1)1( += ttt SyS αα ( 10 << α ) Smoothed level Holt’s Method ttmt mbSy +=+ (m = number of periods ahead to forecast) ))(1( 11 + += tttt bSyS αα ( 10 << α ) Smoothed level 11 )1()( + = tttt bSSb ββ ( 10 << β ) Smoothed trend Winters’ Method Lmtttmt FmbSy ++ += )( (m as before; L = length of the seasonal cycle) ))(1()/( 11 + += ttLttt bSAFyAS ( 10 << A ) Smoothed level Ltttt FBSyBF += )1()/( ( 10 << B ) Smoothed seasonality 11 )1()( + = tttt bCSSCb ( 10 << C ) Smoothed trend Simulation (Random Sampling) Negative exponential distribution: )1ln( Rmx = Uniform distribution, U[a,b]: Rabax )( += where m = mean, a and b are lower and upper limits, and R is a uniform [0.1] rand. var. (paper continues over) 3 Area in one tail for the standard normal distribution Z = (X – m) / s where m is the mean and s is the standard deviation Second decimal place for Z Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.50000 0.49601 0.49202 0.48803 0.48405 0.48006 0.47608 0.47210 0.46812 0.46414 0.1 0.46017 0.45620 0.45224 0.44828 0.44433 0.44038 0.43644 0.43251 0.42858 0.42465 0.2 0.42074 0.41683 0.41294 0.40905 0.40517 0.40129 0.39743 0.39358 0.38974 0.38591 0.3 0.38209 0.37828 0.37448 0.37070 0.36693 0.36317 0.35942 0.35569 0.35197 0.34827 0.4 0.34458 0.34090 0.33724 0.33360 0.32997 0.32636 0.32276 0.31918 0.31561 0.31207 0.5 0.30854 0.30503 0.30153 0.29806 0.29460 0.29116 0.28774 0.28434 0.28096 0.27760 0.6 0.27425 0.27093 0.26763 0.26435 0.26109 0.25785 0.25463 0.25143 0.24825 0.24510 0.7 0.24196 0.23885 0.23576 0.23270 0.22965 0.22663 0.22363 0.22065 0.21770 0.21476 0.8 0.21186 0.20897 0.20611 0.20327 0.20045 0.19766 0.19489 0.19215 0.18943 0.18673 0.9 0.18406 0.18141 0.17879 0.17619 0.17361 0.17106 0.16853 0.16602 0.16354 0.16109 1.0 0.15866 0.15625 0.15386 0.15151 0.14917 0.14686 0.14457 0.14231 0.14007 0.13786 1.1 0.13567 0.13350 0.13136 0.12924 0.12714 0.12507 0.12302 0.12100 0.11900 0.11702 1.2 0.11507 0.11314 0.11123 0.10935 0.10749 0.10565 0.10383 0.10204 0.10027 0.09853 1.3 0.09680 0.09510 0.09342 0.09176 0.09012 0.08851 0.08691 0.08534 0.08379 0.08226 1.4 0.08076 0.07927 0.07780 0.07636 0.07493 0.07353 0.07215 0.07078 0.06944 0.06811 1.5 0.06681 0.06552 0.06426 0.06301 0.06178 0.06057 0.05938 0.05821 0.05705 0.05592 1.6 0.05480 0.05370 0.05262 0.05155 0.05050 0.04947 0.04846 0.04746 0.04648 0.04551 1.7 0.04457 0.04363 0.04272 0.04182 0.04093 0.04006 0.03920 0.03836 0.03754 0.03673 1.8 0.03593 0.03515 0.03438 0.03362 0.03288 0.03216 0.03144 0.03074 0.03005 0.02938 1.9 0.02872 0.02807 0.02743 0.02680 0.02619 0.02559 0.02500 0.02442 0.02385 0.02330 2.0 0.02275 0.02222 0.02169 0.02118 0.02068 0.02018 0.01970 0.01923 0.01876 0.01831 2.1 0.01786 0.01743 0.01700 0.01659 0.01618 0.01578 0.01539 0.01500 0.01463 0.01426 2.2 0.01390 0.01355 0.01321 0.01287 0.01255 0.01222 0.01191 0.01160 0.01130 0.01101 2.3 0.01072 0.01044 0.01017 0.00990 0.00964 0.00939 0.00914 0.00889 0.00866 0.00842 2.4 0.00820 0.00798 0.00776 0.00755 0.00734 0.00714 0.00695 0.00676 0.00657 0.00639 2.5 0.00621 0.00604 0.00587 0.00570 0.00554 0.00539 0.00523 0.00508 0.00494 0.00480 2.6 0.00466 0.00453 0.00440 0.00427 0.00415 0.00402 0.00391 0.00379 0.00368 0.00357 2.7 0.00347 0.00336 0.00326 0.00317 0.00307 0.00298 0.00289 0.00280 0.00272 0.00264 2.8 0.00256 0.00248 0.00240 0.00233 0.00226 0.00219 0.00212 0.00205 0.00199 0.00193 2.9 0.00187 0.00181 0.00175 0.00169 0.00164 0.00159 0.00154 0.00149 0.00144 0.00139 3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 3.1 0.00097 0.00094 0.00090 0.00087 0.00084 0.00082 0.00079 0.00076 0.00074 0.00071 3.2 0.00069 0.00066 0.00064 0.00062 0.00060 0.00058 0.00056 0.00054 0.00052 0.00050 3.3 0.00048 0.00047 0.00045 0.00043 0.00042 0.00040 0.00039 0.00038 0.00036 0.00035 3.4 0.00034 0.00032 0.00031 0.00030 0.00029 0.00028 0.00027 0.00026 0.00025 0.00024 3.5 0.00023 0.00022 0.00022 0.00021 0.00020 0.00019 0.00019 0.00018 0.00017 0.00017 3.6 0.00016 0.00015 0.00015 0.00014 0.00014 0.00013 0.00013 0.00012 0.00012 0.00011 3.7 0.00011 0.00010 0.00010 0.00010 0.00009 0.00009 0.00008 0.00008 0.00008 0.00008 3.8 0.00007 0.00007 0.00007 0.00006 0.00006 0.00006 0.00006 0.00005 0.00005 0.00005 3.9 0.00005 0.00005 0.00004 0.00004 0.00004 0.00004 0.00004 0.00004 0.00003 0.00003 4.0 0.00003 0.00003 0.00003 0.00003 0.00003 0.00003 0.00002 0.00002 0.00002 0.00002 (paper continues over) 0 Z 4 Random Number Table (paper continues over) 5 PART A: Answer Question A1 in this section of the paper. Question A1 (25 marks) A dental hospital has two dental nurses and one dentist. The dentist has specialist knowledge and skills for dealing with patients who require specialist treatments (e.g. fillings, root canals). If the dentist is busy and a patient requires a specialist treatment they must wait until the dentist becomes free. Patients with general dental problems can see any of the staff, but the dentist will only attend to a general patient if both dental nurses are busy. Customers wait in the waiting room if they cannot be seen immediately. The hospital opens at 9am and closes at 5pm each day. The time between general patients arriving at the dental hospital follows a negative exponential distribution with mean of 0.2 hours, and the time between special treatment patients arriving at the dental hospital follows a negative exponential distribution with mean of 0.5 hours. The service time of a general patient is described by a uniform distribution between 0.5 and 1 hours, and the service time of a specialist treatment patient can be described by a uniform distribution between 1 and 2 hours. a) How can random numbers can be used to model the variability in this situation Specify the formulae you would use. (25% of marks) b) Using discrete event simulation, simulate the system up to the time of departure of the 1st patient requiring general treatment. How long does each customer wait to be seen by one of the dental staff Do all your calculations to 3 decimal places. Use 4-digit random numbers from the formulae sheet. For each process, use the following number stream: General patient inter-arrival times (5th row) : 6524 3011 7654 4608 8595 5921 2692 8923 2024 2108 … Special Treatment patient inter-arrival times (7th row): 7493 5070 3768 5243 5010 3662 3924 6180 0823 9804 General patient service times (1st row): 1869 8478 8578 4023 9576 2595 5664 6042 0073 3181 Special Treatment patient service times (3rd row): 1139 9393 4279 6670 6454 5597 4593 5746 5139 0827 (40% of marks) c) The staff at the dental hospital are interested in improving patient experience. On the outcome of a survey, long waiting times appear to be the largest cause of poor customer experiences. Firstly, explain how simulation can be used to investigate the waiting time of customers in the current system. Secondly, suggest a simple change that could be made to the system to improve waiting times, and how simulation could be used to investigate the impact of this change. (25% of marks) d) Would it be appropriate to use a warm up period for this system Justify your answer. (10% of marks) - please turn over for Part B 6 PART B: Answer three of the four questions in this section of the paper. Question B1 (25 marks) A sports goods manufacturer makes snowboards. The following table shows the actual sales (in ￡000s) each quarter in years 2017 and 2018. Do all your calculations to 3 decimal places. Year Period Sales 2017 1 493 2 752 3 1194 4 1348 2018 1 521 2 802 3 1501 4 1719 a) Forecast the sales for all quarters of year 2019 by using 3-points moving average method after de-seasonalising the data with the proper moving average method. Use multiplicative seasonal method. (30% of marks) The following table is calculated using the Winter’s exponential smoothing method using smoothing constants A, B and C all equal to 0.15. The table shows the actual sales y (in ￡000s), smoothed values S, seasonal factors F, and trend factors b for each quarter of 2016, 2017 and 2018. Year Period Y (sales) S F b 2016 1 421 766.503 0.562 25.083 2 669 798.568 0.826 27.178 3 1022 823.499 1.247 26.504 4 1142 846.359 1.359 25.411 2017 1 493 873.406 0.563 25.902 2 752 902.639 0.828 26.901 3 1194 937.927 1.255 29.417 4 1348 974.713 1.366 31.628 2018 1 521 982.059 0.553 24.343 2 802 995.061 0.821 20.941 3 1501 1070.010 1.299 37.142 4 1719 1152.530 1.404 50.756 b) Calculate the sales for all 4 quarters in year 2019 in ￡000s using Winter’s method. Use the appropriate values given in the table above in your calculations. (20% of marks) c) The actual sales for 2019 quarter 1 were 647 (in ￡000s). Calculate the S, F and b values for 2019 period 2. (20% of marks) - question continues over 7 Q B1, continued d) If the actual sales in year 2019 are 647, 955, 1803 and 2011 (in ￡000s), what is the mean error, mean absolute error and mean square error of moving average and Winter’s methods in year 2019. What can you say about the accuracy of these two methods (20% of marks) e) One of the difficulties in using Winter’s method is that initial values are required for the smoothed level, trend and seasonal factors. One way of starting Winter’s method is to use initial values of the smoothing level, trend and seasonal factors that are based on the first few values of the series. Explain a suitable procedure to obtain such initial values. (10% of marks) Question B2 (25 marks) A construction project consists of the following activities, with their predecessors, durations (in weeks) and resource requirements (in workers): Activity Duration Predecessors Resource Requirement A 5 2 B 4 2 C 3 1 D 5 A,B 1 E 4 B,C 2 F 2 D 2 G 2 D,E 1 H 6 F 1 I 5 G 1 a) Draw a network diagram for the project. Calculate the earliest and latest start and finish times for each activity, and the total, free and interfering floats for each activity. State the project duration and the critical path. (30% of marks) b) Draw the Gantt Chart and resource allocation graph for the project. Calculate the weekly resource requirement throughout the project if each activity starts at their earliest start time. Report the maximum resource level requirement throughout the project. (20% of marks) It is recognised that the previous analysis does not take into account the uncertainties in activity durations. Estimates have been made of the expected means and variances of each of the activity times, for the same activities and network as before, as shown in the table below. Assume that the lengths of the project activities are mutually independent. - question continues over 8 Q B2 continued Activity Predecessors Mean Completion Times (Weeks) Variance of Completion Times (Weeks) A 5 1 B 4 10 C 3 2 D A,B 5 3 E B,C 4 4 F D 2 1 G D,E 2 5 H F 6 2 I G 5 5 c) What is the variance of the critical path duration Do all your calculations to 3 decimal places. (5% of marks) d) Calculate the probability that the project duration will be less than 19 weeks. (10% of marks) e) Calculate the duration x for which there is a 98% chance that the project duration will be less than x weeks. (15% of marks) f) There are a few activities with quite high variance of completion times. Have you considered these high variances in your calculations What would be the weakness of your calculations regarding these activities (20% of marks) Question B3 (25 marks) Lonsdale Drinks Company produces three smoothie drinks made by blending different combinations of three fruit juice liquids. The three fruit juice liquids are called Banana, Strawberry, and Raspberry (named by their main ingredient). The three smoothie drinks are called Strawnana, Raspnana and Berry and the proportions of the liquids in each smoothie are as follows: Banana Strawberry Raspberry Strawnana 40% 50% 10% Raspnana 50% 10% 40% Berry 5% 55% 40% The profits on each smoothie are Strawnana ￡2 per litre, Raspnana ￡1.50 per litre, Berry ￡1 per litre. In the coming planning period, there are 5000 litres of Banana liquid, 4000 litres of Strawberry liquid and 7000 litres of Raspberry liquid available. The maximum production capacity of the plant is 14000 litres of smoothies. The company will be able to sell all that it produces. a) Formulate Lonsdale Drinks Company’s problem as a linear programme to maximise profit. (30% of marks) - question continues over 9 Q B3 continued Parts of the solution and sensitivity analysis obtained using Excel Solver are: Answer report: Objective Cell (Max) Cell Name Original Value Final Value $E$2 profit 20714.286 20714.286 Variable Cells Cell Name Original Value Final Value Integer $B$9 Strawnana 7142.857 7142.857 Contin $C$9 Raspbana 4285.714 4285.714 Contin $D$9 Berry 0 0 Contin Constraints Cell Name Cell Value Formula Status Slack $E$4 banana used 5000 $E$4<=$F$4 Binding 0 $E$5 strawberry used 4000 $E$5<=$F$5 Binding 0 $E$6 raspberry used 2428.571 $E$6<=$F$6 Not Binding 4571.429 $E$7 production capacity 11428.571 $E$7<=$F$7 Not Binding 2571.429 Sensitivity report Variable Cells Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$9 Strawnana 7142.857 0 2 5.5 0.139 $C$9 Raspbana 4285.714 0 1.5 0.192 1.1 $D$9 Berry 0 -0.179 1 0.179 1E+30 Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $E$4 banana used 5000 2.619 5000 1350 1800 $E$5 strawberry used 4000 1.905 4000 2250 3000 $E$6 raspberry used 2428.571 0 7000 1E+30 4571.429 $E$7 production capacity 11428.571 0 14000 1E+30 2571.429 b) From this output, state the optimal production plan of how much of each smoothie Lonsdale Drinks Company should make and how much profit they will obtain. (10% of marks) c) In the optimal production plan, one of the smoothies is not produced. State how much the profit per litre for this smoothie would need to increase in order for it to be included in the optimal production plan. (10% of marks) 10 - question continues over Q B3 continued d) Lonsdale Drinks Company may be able to purchase up to 1500 litres of additional Strawberry liquid and up to 1000 litres of additional Raspberry liquid from an alternative supplier, although both prices would be ￡0.50 per litre higher than from the current supplier. Assuming these quantities are available, explain with reasons how much of the additional Strawberry liquid and how much of the additional Raspberry liquid the company should purchase and calculate the effect on the profit. (20% of marks) e) Assume now that the additional Strawberry and Raspberry liquids in part d) are not available. The company is considering a change in manufacturing procedures which would have the effect of increasing the profit per litre of Strawnana by ￡0.50 per litre and decreasing the profit per litre of Raspnana by ￡0.70. Explain with reasons whether it is possible to calculate the effect of this change from the Excel Solver output, and, if so, calculate the change in profit from making this change. (20% of marks) f) Lonsdale Drinks Company is considering relaxing slightly the percentages of banana and strawberry while keeping Raspberry liquid percentage at 10% in Strawnana. They will keep both Banana and Strawberry liquids percent at least 30%. What changes should they implement in the existing formulation in order to model this new problem (Assume that the profit from Strawnana is unchanged) (10% of marks) Question B4 (25 marks) A clothing store has to place its order to a clothing manufacturer in autumn for the spring/summer season. Due to the large lead times, only one order can be placed in advance, and no further orders will be possible. The clothing store must decide whether to make a large or small order. The profit it earns depends heavily on the level of demand for their products. It is unknown whether demand will be high or low, but the probability of high demand is estimated to be 0.45. If demand is low, the store knows it will make ￡30,000 in profit for a small order and a loss of ￡10,000 for a large order. If demand is high, the store knows it will make ￡40,000 in profit for a small order and ￡70,000 for a large order. a) Formulate the information above as a payoff table (in ￡000s) (5% of marks) b) Consider the following decision criteria: Expected Value Maximin Maximax Minimax regret Which order size maximises the payoff under each of these criteria (30% of marks) - question continues over 11 Q B4 continued c) The clothing store management is unsure that the probability of high demand is exactly 0.45. Using sensitivity analysis, at what value of the probability of high demand does the decision under the expected value criterion change (15% of marks) d) Assuming the clothing store management is now sure the probability of high demand is 0.45, it has the option of carrying out some market research into the demand for its products. Using a decision tree calculate the perfect gain in information assuming the market research company has a perfect record. (15% of marks) e) In reality the market research company does have a perfect record. The past record of the market research company is as follows: Market research report True Market Outcome Negative Positive Low Demand 0.7 0.3 High Demand 0.2 0.8 Note that it will cost ￡5,000 to carry out market research. Using a decision tree, what is the best course of action using the expected value criterion (35% of marks) END OF PAPER
