Question: Table: Lq Values for the Multi-server Queue Values of Lq for s servers, with mean utilization rate r, assuming Poisson arrivals and exponential service times.

Table: Lq Values for the Multi-server Queue Values of Lq for s servers, with mean utilization rate r, assuming Poisson arrivals and exponential service times. Utilization 0.10 0.20 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.62 0.64 0.66 0.68 0.70 0.72 0.74 0.76 0.78 0.80 0.82 0.84 0.86 0.88 0.90 0.92 0.94 0.96 0.98 0.99 1 0.011 0.050 0.129 0.189 0.267 0.368 0.500 0.672 0.900 1.012 1.138 1.281 1.445 1.633 1.851 2.106 2.407 2.766 3.200 3.736 4.410 5.283 6.453 8.100 10.580 14.727 23.040 48.020 98.010 2 0.002 0.017 0.059 0.098 0.152 0.229 0.333 0.477 0.675 0.774 0.888 1.019 1.170 1.345 1.550 1.791 2.079 2.424 2.844 3.366 4.027 4.885 6.041 7.674 10.139 14.271 22.570 47.535 97.518 Number of Servers (s) 3 0.000 0.006 0.030 0.055 0.094 0.152 0.237 0.358 0.532 0.621 0.725 0.845 0.985 1.149 1.342 1.572 1.847 2.180 2.589 3.098 3.746 4.591 5.735 7.354 9.806 13.924 22.209 47.160 97.136 4 0.000 0.002 0.016 0.033 0.061 0.105 0.174 0.277 0.431 0.511 0.605 0.716 0.846 1.000 1.183 1.403 1.667 1.989 2.386 2.883 3.519 4.353 5.383 7.090 9.529 13.634 21.906 46.844 96.813 5 0.000 0.001 0.009 0.020 0.040 0.074 0.130 0.219 0.354 0.427 0.513 0.615 0.737 0.882 1.055 1.265 1.519 1.830 2.217 2.703 3.327 4.149 5.268 6.862 9.289 13.382 21.641 46.566 96.427 mms.xls M/M/s Queueing Formula Spreadsheet Inputs: l m Outputs: s 4 2 Definitions of terms: l = arrival rate m = service rate s = number of servers Lq = average number queue L = average number in the system Wq = average time spent in the queue (avg. wait in queue) W = average time spent in the system (avg. wait in system) P(0) = probability of zero customers in the system P(delay) = probability that an arriving customer has to wait r = Utilization ( fraction of time servers are busy) Lq L Wq W P(0) P(Delay) 0 1 infinity infinity infinity infinity 0.000000 1.000000 2 infinity infinity infinity infinity 0.000000 1.000000 3 0.888889 2.888889 0.222222 0.722222 0.111111 0.444444 4 0.173913 2.173913 0.043478 0.543478 0.130435 0.173913 5 0.039801 2.039801 0.009950 0.509950 0.134328 0.059701 6 0.009009 2.009009 0.002252 0.502252 0.135135 0.018018 7 0.001924 2.001924 0.000481 0.500481 0.135298 0.004811 8 0.000382 2.000382 0.000095 0.500095 0.135329 0.001146 9 0.000070 2.000070 0.000018 0.500018 0.135334 0.000246 10 0.000012 2.000012 0.000003 0.500003 0.135335 0.000048 11 0.000002 2.000002 0.000000 0.500000 0.135335 0.000008 12 0.000000 2.000000 0.000000 0.500000 0.135335 0.000001 13 0.000000 2.000000 0.000000 0.500000 0.135335 0.000000 14 0.000000 2.000000 0.000000 0.500000 0.135335 0.000000 15 0.000000 2.000000 0.000000 0.500000 0.135335 0.000000 16 0.000000 2.000000 0.000000 0.500000 0.135335 0.000000 17 0.000000 2.000000 0.000000 0.500000 0.135335 0.000000 18 0.000000 2.000000 0.000000 0.500000 0.135335 0.000000 r 1.000000 1.000000 0.666667 0.500000 0.400000 0.333333 0.285714 0.250000 0.222222 0.200000 0.181818 0.166667 0.153846 0.142857 0.133333 0.125000 0.117647 0.111111 19 20 0.000000 0.000000 2.000000 2.000000 0.000000 0.000000 0.500000 0.500000 0.135335 0.135335 0.000000 0.000000 0.105263 0.100000 Problem 3. Suppose you observe the following sequence of interarrival and service times for 6 arrivals. Assume that the first arrival finds the system empty. The queue has 1 server. Customer # IAT Time of Arrival 1 2 3 4 5 6 12 40 34 14 72 3 (IAT+AT-1) 12 12+40 = 52 34+52 = 86 14+86 = 100 72+100 = 172 3+172 = 175 Service time Time service starts Time service ends 65 15 45 85 15 65 12 77 92 137 222 237 +65 = 77 +15 = 92 +45 = 137 +85 = 222 +15 = 237 302 1. (1 point) What delay was experienced by arrival #3? Delay = 92-86 = 6 minutes 2. (1 point) Find the average wait time experienced by these 6 arrivals. Problem 4. A manager of a medium-sized call center needs to determine the optimal number of operators that she should hire. The call center serves two types of clients: Business (B) and Leisure (L) customers. She has determined that B calls arrive according to a Poisson process at a rate of 18 calls per hour. On the other hand, L calls arrive according to a Poisson process at a rate of 50 calls per hour. She decides to hire highly skilled operators who can serve both type of customers and the average amount of time that they need is 9 minutes per call independent of the type (service times are exponentially distributed). The cost of hiring an H operator is $35 per hour. The manager has also estimated that the cost of keeping a B customer waiting in queue is $80 per customer per hour while the cost of keeping an L customer waiting in queue is $20 per customer per hour. The manager is considering having a single queue for both types of calls and hire only H operators that will serve both types of calls. Calls will be served on a First-ComeFirst-Serve basis independent of their type. Find: 1. (1 points) Minimum feasible number of operators 2. (2 points) Average number of B and L customers in the queue if the call center is staffed by 12 operators 3. (1 point) Total cost incurred by the call center per hour of operating if it is staffed by 12 operators 4. (2 points) Nopt: the optimal number of operators that minimizes the cost incurred by the call center, and the corresponding cost, Wq, and utilization. Please show all your work. Be sure to include the formula for the wait cost computation

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related General Management Questions!