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IOE 419 Mark S. Daskin Service Operations Management IOE Department Winter, 2017 University of Michigan Problem set 6 DUE: WEDNESDAY - March 29, 2017 Points:

IOE 419 Mark S. Daskin Service Operations Management IOE Department Winter, 2017 University of Michigan Problem set 6 DUE: WEDNESDAY - March 29, 2017 Points: 100 points total Problem 1: The capacitated fixed charge location model tries to find the balance between the fixed facility costs and the transportation costs. In its basic form, there is a single capacity associated with each candidate site. In fact, companies often can determine the capacity of a facility at the same time that they determine where to locate the facilities and how many to locate. We will try to formulate such a model in this problem. In particular, we will define: Inputs: I J Lj Set of demand nodes Set of candidate sites Set of possible capacities for candidate facility j J k jl The l th possible capacity value for a candidate facility at j J f The fixed cost of building a facility at j J with the l th possible capacity jl hi cij at node j J demand at node i I cost per unit shipped from facility j J to demand node i I Decision Variables: X jl Y ij 1 if we build at facility at j J at capacity level l L j 0 if not fraction of demand at node i I that is assigned to a facility at j J 1 IOE 419 Mark S. Daskin Service Operations Management IOE Department Winter, 2017 University of Michigan With this notation the problem can be formulated as follows: Min f jl X jl + cij hiY ij jJ lL j s.t. iI jJ X jl 1 j J lL j Y ij = 1 i I Y ij X jl i I; j J hiY ij k jl X jl j J jJ lL j lL j iI X jl {0,1} l L j; j J Y ij 0 i I; j J The objective function minimizes the sum of the fixed cost and the transport cost. The first constraint says that we select at most one capacity level for each candidate site. The next constraint ensures that all demand is satisfied. The next constraint links the assignment and location decisions and states that demand at node i I cannot be assigned to a facility at j J unless we locate at j J . The next constraint is the capacity constraint. Note that the right hand side of this constraint is the total installed capacity at node j J . Finally, we have the binary and non-negativity constraints. Note that the final constraints can be replaced by single sourcing constraints of the form i I; j J . Y ij {0,1} Formulate this problem in EXCEL for the data given in ANN ARBOR Fixed Charge Var Cap INPUTS.xlsx. The decision variables are already set up for you and they are shown with a light green background. Complete the following table where the second term in the objective function is actually written as the cost per item mile TIMES the total item-miles (the p-median objective function value). Cost per Item Mile 0.001 Assignments Integer Facility Transport Total Total Capacity % Utilization 0.001 Non-Neg 0.01 Integer 0.01 Non-Neg 0.05 Integer MED: 4035, 4036, 4052; BIG: 4005; $9,815 $5,345 $15,160 116,296 98.97% 0.05 Non-Neg SML: 4036; MED: 4035; BIG: 4002, 4052; $9,899 $5,195 $15,094 117,977 97.56% 0.1 Integer 0.1 Non-Neg SIZE: Where 2 IOE 419 Mark S. Daskin Service Operations Management IOE Department Winter, 2017 University of Michigan Problem 2: The M/Ek/s queue has Poisson arrivals, Erlang-k service times and s servers. As far as I know, there is no closed form expression for the steady-state service characteristics of this queue. In this problem, you will set up a simulation of 5,000 customers. Recall that the Erlang-k distribution is the distribution of the sum of k independent identically distributed exponential random variables. Set up the following columns: B Customer Number C Arrival Time D Service Time (must be computed from columns AA-AY) E-N Next free time for each of servers 1-10 O First available time (the minimum time of the first s servers from columns E-N) P Assign customer to server (tells you which server has the minimum available time Q Start time of the customer R End time of the customer S Wait time of the customer AA-AY 25 exponentially distributed times with mean 1/mu. Used to compute the service time in column D. Be careful how you do this, since there is more to computing the service time than simply adding up the first k values in these columns. Complete the following table: Lambda 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4 4 4 Mu 5 5 5 1 1 1 0.5 0.5 0.5 1 1 1 Kval 1 9 25 1 9 25 1 9 25 1 9 25 Simulated SD Avg Theoretical Avg Wait Wait Avg Wait Theory Servers Time Time Time From? 1 This is the 1 sample This tells us 1 This is the standard This is the where you 5 average deviation theoretical got the 5 of 10 of the 10 value which formula 5 simulated simulated you should from (e.g., average average find using the 10 M/M/1, wait wait times best formula 10 M/G/1, times for divided by that you can 10 M/M/s, each row the use 5 GI/G/s square 5 root of 10 5 Significant This tells us whether or not your sample average value is significantly different from the theoretical value Note that for the first 9 rows, the utilization ratio is constant at 0.9, while it decreases to 0.8 for the last three rows. In each group of three rows, the value of k increases, meaning that the variability of the service time distribution decreases. Recall that the mean service time is given by 1 and the variance of the service time distribution is given by 1 . Therefore, the first three rows correspond k 2 to fast service but a single server, the next three correspond to slower service, but 5 servers, while 3 IOE 419 Mark S. Daskin Service Operations Management IOE Department Winter, 2017 University of Michigan rows 7-9 represent even slower service but 10 servers. The last three rows are similar to rows 4-6, except that the arrival rate has been decreased from 4.5 to 4. In the column labeled \"Significant\" indicate whether the average waiting time is significantly different from the theoretical value at the 95% level. For each row of the table above, you should have another 10 columns that represent 10 realizations of the simulation and have 10 values of the average waiting time. a) For given values of lambda, mu, and servers, what is true of the waiting time as the value of k increases? b) What value of the t-distribution do you use in the last column? c) For given values of lambda and k, what happens to the average waiting time as you increase the number of servers, holding mu*servers fixed? In other words, compare the first, fourth and seventh rows of the table, the second, fifth and eighth rows, and the third, sixth and ninth rows. d) What happens to the variability of the average service time as the utilization ratio decreases (compare rows 4-6 with rows 10-12). Problem 3: Consider a small private parking facility in Ann Arbor. The facility only has 5 spots. Customers arrive at the rate of 2 per hour. The time in the system is exponentially distributed. Complete the following table: Avg Number in System Effective Arrival Rate Avg Time in System Avg Time Waiting Avg Number Waiting Expected Revenue per hour 1 Avg Time in System 2 3 4 Compute the expected revenue per hour assuming that customers pay $2.5 just to park at the facility and $1 per hour for the time they spend in the parking lot. Think about this in terms of how much revenue an attendant would get per hour if she asks everyone who actually parks in the lot how much time they will spend in the lot (and everyone is perfectly honest and accurate) and then charges everyone accordingly. In doing this, you should use the \"non-normalized probability\" method of computing the state probabilities. Problem 4: Ms. Edith Granny Smith is a computer repairperson for Prof. Seth Guy Keema. Prof. Keema has 10 highly specialized computers that Ms. Smith maintains. When working, each computer fails at a rate of 1/30th per day and failures occur according to a Poisson process. Repair times are exponentially distributed with a mean of 1.5 days per failed computer. Find: a) The average number of broken computers b) The expected arrival rate of broken computers to Ms. Smith's shop c) The average time a broken computer spends in the repair shop d) The average time a broken computer spends waiting to be repaired e) The average number of broken computers waiting to be repaired. You should again use the \"non-normalized probability\" method to compute the state probabilities. 4

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