1 Ardavan Asef-Vaziri Jan-2011Operations Management: Waiting Lines 2 Example: The average arrival rate to a GAP store is 6 customers per hour. The average.

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1 Ardavan Asef-Vaziri Jan-2011Operations Management: Waiting Lines 2 Example: The average arrival rate to a GAP store is 6 customers per hour. The average service time is 5 min per customer. Problem 1 R = 6 customers per hour Rp =1/5 customer per minute, or 60(1/5) = 12/hour U= R/Rp = 6/12 = 0.5 a) How long does a customer stay in the processor (with the server)? Tp = 5 minutes b) On average how many customers are there with the server? RTp = Ip = 6(5/60) = 0.5 Alternatively; Ip = cU =1(0.5) = 0.5

2 Ardavan Asef-Vaziri Jan-2011Operations Management: Waiting Lines 2 Problem 2 What if the arrival rate is 11 per hour? Processing rate is still Rp=12 U= R/Rp U=11/12 a) How long does a customer stay in the processor (with the server)? Tp = 5 minutes b) On average how many customers are there with the server? RTp = Ip = (11/60)(5) = 11/12 Alternatively Ip = cU = 1(11/12) = 11/12

3 Ardavan Asef-Vaziri Jan-2011Operations Management: Waiting Lines 2 A local GAP store on average has 10 customers per hour for the checkout line. The store has two cashiers. The average service time for checkout is 5 minutes. Problem 3 Arrival rate: R = 10 per hour Average inter-arrival time: Ta = 1/ R = 1/10 hr = 6 min Average service time: Tp = 5 min Number of servers: c =2 Rp = c/Tp = 2/5 per min or 24 per hour U= R/Rp = 10/24 = 0.42

4 Ardavan Asef-Vaziri Jan-2011Operations Management: Waiting Lines 2 Problem 3 a) How long does a customer stay in the processors (with the servers)? Average service time: Tp = 5 min b) On average how many customers are there with the servers? Ip =? RTp = Ip = (1/12)(10) = 0.84 Alternatively Ip = cU = 2(0.42) = 0.84

5 Ardavan Asef-Vaziri Jan-2011Operations Management: Waiting Lines 2 A call center has 11 operators. The average arrival rate of calls is 200 calls per hour. Each of the operators on average can serve 20 customers per hour. U = 200/220 =0.91. Problem 4 a) How long does a customer stay in the processors (with the servers)? Average service time: Tp = 60/20 = 3 min b) On average how many customers are there with the servers? Ip =? RTp = Ip = (200/60)(3) = 10 Alternatively Ip = cU = 11(200/220) = 10