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90-760: Decision & Risk Modeling Midterm Solution, Spring 2017Question #1: (12 points)What is the total waiting time if the following four jobs are serviced in the indicated service disciplines?Job #1: 5 minutesJob #2: 12 minutesJob #3: 1 minutesJob #4: 8 minutesa) FCFS (i.e., FIFO)b) SPTc) LCFS (i.e., LIFO)Solution: a) 3*5 + 2*12 + 1*1 = 40b) 3*1 + 2*5 + 1*8 = 21c) 3*8 + 2*1 + 1*12 = 38Question #2: (12 points)Customers have been arriving to an M/M/3 queueing system at an average rate of 10 per hour since dinosaursruled the world. Service times average 15 minutes. a) Fill in the blanked out cells E3:E4.b) If you arrived at a random time, how many customers would you see waiting for service?c) On average, how long do customers spend in the system before they complete service?d) Over an eight hour day, how long would a server expect to be idle?Solution:a) 4 and 3, respectively.b) Lq = 3.51c) 0.6 hours or 36 minutesd) 8 / 6 = 4/3 of an hour or 80 minutesQuestion #3: (16 points)The Borough of Upper Scranton operates 100 buses that each break down completely at random an average ofonce a month. When a mechanic, who is paid $200 per day, works on a bus, the repair time is exponentiallydistributed with a mean of one day. When a bus is not in service, the Borough pays $500 a day to rent areplacement. I evaluated queueing performance characteristics with the Q.xls workbook for various numbers ofmechanics, obtaining the results below. a) If the Borough employs 3 mechanics, on average how long is a bus out of service when it break down?b) What is the expected total daily cost to the Borough if 4 mechanics are employed?c) What number of mechanics minimizes the Borough’s average daily cost?d) If the Q.xls spreadsheet didn’t give you the utilization, explain how you could compute it from othernumbers that it does give. (You’ll need to think this through from first principles, not apply any specificformula we used in class.)Solution:a) 4.41 daysb) $200 * 4 + $500 * 4.93 = $3,265.c) 5d) Utilization = (L – Lq) / s, where s = # of serversInformation for questions #4 - #6. Suppose that three management science students (unwisely) divvy up theeight problems on a homework as indicated blow, work on their problems independently in order, and whenthey are all done staple the results together and submit the combined work. They also (unwisely) wait until sixhours before the homework is due to begin, and they are interested in whether they will finish the assignment ontime. Since each individual homework problem involves a very large number of separate tasks, each taking a variableamount of time, completion time for each problem can reasonably be modeled as being a normally distributedrandom variable with the following parameters.Question #4: (15 points)Suppose these unwise students forget all about uncertainty and just assume that every homework problem’scompletion time will be exactly equal to its expected value. E.g., the first problem will take exactly 2.5 hours,and so on.a) Draw the CPM project network for this problem augmented with dummy nodes as appropriate.b) Execute the forward and backward passes to find the earliest and latest starting times for each homeworkproblem.c) What is the critical path?Solution:c) The critical path is the dummy origin then nodes #4, #5, #6 and the dummy final node.Question #5: (15 points)If the students suddenly remember there is uncertainty but can’t remember how to run a simulation and so takethe PERT approach to problem management:a) What do they think is the mean, variance, and standard deviation of the project completion time?b) What type of random variable do they think describes the project completion time and why?c) What do they judge to be the probability that they will finish the homework on time and how do youknow? (The explanation can be very brief but must include the key insight.)Solution:a) Bertha’s path is critical by the CPM with activity times set equal to their expected values, and thevariance of a sum is equal to the sum of the variances when – as PERT assumes – completion timesare independent, so mean = 2 + 2 + 2 = 6, variance = 0.25 + 0.16 + 0.09 = 0.5 and standard deviation= SQRT(0.5) = 0.7071.b) Normal because the sum of normally distributed random variables is always normal.c) 50% because the normal distribution is symmetric around its mean. Since they think the projectcompletion time is normally distributed with a mean of 6, then think there is a 50% chance they willfinish the homework on time.Question #6: (20 points)Suppose, much to the relief of their professor, the students suddenly remember the limitations of PERT and howto run a simulation. They build and run the simulation with 10,000 trials producing the output that is pastedbelow (with a few cells blanked out). Note: the formula in cell K5 is “=COUNTIF(J$24:J$10023,J5)/10000”,and similarly for cells K6 and K7.a) Now approximately what do they think is the probability that they will complete the homework on time?b) What is a balanced 90% confidence interval for how long it will take them to finish the homework?c) If they want to be 90% sure they will submit the homework on time, how long in advance of the duedate should they start?d) What is the Excel formula in cell I23?e) What is the Excel formula in cell I7?Solution:a) Somewhere between 20% and 30% and not that close to either extreme. (Although not shown, the=PERCENTRANK() function reveals that 25.1% of these trials had a completion time of 6 hours orless.b) 5.46 – 7.56 hoursc) 7.27 hours beforehand.d) =MAX(D14:D16)e) =I$3-1.65*I$4/SQRT(10000) or =I$3-1.65*I$4/100. (Dollar signs not needed) Question #7: (24 points)Sketch the typical form of the following graphs, labeling both axes and adding a smiley face in the ideal corner.a) Tradeoff curve when considering crashing a project in project management.b) A risk-return frontier for profit that shows the width of each alternative’s 90% confidence interval.c) Queue length as a function of utilization.d) Cumulative risk profile for two ways of saving lives with Alternative B exhibiting 1st order stochasticdominance over Alternative A.Solution:a) Cost on vertical axis vs. completion time on horizontal axis. Decreasing and convex. Smiley is in lowerleft, by the origin.b) Width of 90% CI on vertical axis and EV on horizontal axis (or v.v. is OK). Smiley face is in lower right(or upper left if axes are reversed). Graph

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