Chap8 - Chapter 8: Managing Flow Variability or the Taming...

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Chapter 8: Managing Flow Variability or the Taming of the Queue Some days we don’t let the line move at all. We call those weekdays” – the Dept. of Motor Vehicles 1

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Which Queue Configuration Do You Prefer? 2
3 Agenda What causes waiting lines to form? How can we estimate the impact of variability on process performance measures? Simple approximation formula Exponential probability distribution Queuing software Performance.xls Why do we NOT want to target high capacity utilization especially in service operations? What does Goldratt say about statistical fluctuations? The psychology of waiting Simulating processes

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Recall: Processes with Variability Can be Stable Stable if process capacity > R i Stable if process capacity utilization < 100% Safety capacity = process capacity - R i If there is variability in a stable process, queues will build up and decrease periodically For stable processes, we will use R = throughput = avg. outflow rate = avg. inflow rate T = average flow time I = average inventory in the process 4
Single Phase Stable Process T i + T p = T Average flow time I i + I p = I Average inventory Little’s Law: I = R × T I i = R T i I p = R T p Let R p = average capacity of the resource pool Capacity Utilization = R i / R p = ρ η (0< ρ η <1) Input Buffer or Queue resource pool R i =R 5 R o = R

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Queuing Jargon Flow unit = customer Resource unit = server (also called a channel) Processing time = service time Customers who are being “served” or processed are not considered to be in the queue Waiting line = queue (text also calls it the input buffer) 6
7 All Times Constant Suppose a customer arrives every 6 min and service takes exactly 5 min. There is 1 server. R i = -- R s = R p = ρ = Avg time customer spends in queue, T i ? Avg flow time, T? Avg number of customers in queue, I i ? Hourly throughput? Avg # of customers in process, I?

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8 Number of Customers in Process Minutes 6 12 18 # in process 1 2 0
Average Number of Customers in Process Minutes 6 12 18 Average # in process 1 2 0 0.833 9

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A Problem The Bank of Banks in Banks County, Georgia, has been getting complaints from customers about the waiting times in line at the tellers. They know that customer service times are not all the same and that customers arrivals are not evenly spaced. They want a way to analyze the performance of this process so they can find ways to improve it. They start by collecting data on arrival times and service times. 10
Number Arrival Time Inter- arrival Time (min) 1 8:00 - 2 8:10 10 3 8:20 10 4 8:22 2 5 8:42 10 6 8:43 1 7 8:46 3 8 8:53 7 9 9:02 9 10 9:04 2   Arrival   Times   Customer interarrival times are the times between consecutive customer arrivals Average interarrival time = μ i = 54min/9 = 6 min The standard deviation of the interarrival times? σ i = 3.937 min Coefficient of variation =

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This note was uploaded on 05/12/2008 for the course MIS 560 taught by Professor Slaton during the Spring '08 term at University of Arizona- Tucson.

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Chap8 - Chapter 8: Managing Flow Variability or the Taming...

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