Question: answer the following under the assumption that the facility is staffed with a single (one) server: What is the average server utilization ? What is



answer the following under the assumption that the facility is staffed with a single (one) server:
What is the average server utilization ?
What is the average number of customers waiting in line?
What is the average number of customers in the system? (Waiting in line + being served)
What is the average time (in minutes) spent waiting in line?
What is the average time (in minutes) spent in the system (Waiting in line + being served)
Continuing from the previous problem, suppose later in the day the customer arrival rate (customers per hour) will be five times as many per hour vs. the original problem. We respond by using five servers instead of just one. Answer the following for this busier time period:
What is the average server utilization ?
What is the average number of customers waiting in line?
What is the average number of customers in the system? (Waiting in line + being served)
What is the average time (in minutes) spent waiting in line?
What is the average time (in minutes) spent in the system (Waiting in line + being served)
1 Queuing Problem: 2 3 Customers arrive to a service facility on average 27 minutes apart 4 (exponentially distributed time between arrivals) 5 6 The average service time is 25 minutes per customer 7 (exponentially distributed service times) 8 9 10 11 12 13 14 15 A B C D E F G H 1 J L M N P Q R S T 10 0.1000 12 0.0833 Average number of arrivals per period Average number of periods between arrivals Average number of customers served per period, when the server is busy Average service time ( number of periods ) per customer 1 Single Server Waiting Line Model (M/M/1) 2 Exponential time between arrivals, exponential service times 3 4 Arrival rate 2 = 5 Average time between arrivals 1/2 = 6 Service rate HE 7 Average service time 1/u = 8 9 10 System Utilization p= 11 Probability system is empty Po = 12 Average number of customers in queue 13 Average number of customers in system Ls = 14 Average time in queue per customer 15 Average time in system per customer Ws = 16 0.8333 0.1667 4.1667 5.0000 0.4167 0.5000 W, = 17 n = 18 P( 4 units in system) 19 PlStep by Step Solution
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