Revision Chapter 5 on Queue management with example.xls

Revision Chapter 5 on Queue management with example.xls -...

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Data a = 2 (average interarrival time) u = Cva = 1 (interarrival time coefficient of variation) Tq = p = 3 (average service time) T = CVp = 1.5 (service time coefficient of variation) Ip = m = 3 (number of sAnswer b Iq = I = TWT = 5 (target waiting time) SL =
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Inter-arrival, a, is the period of tim p refers to processing time or activity time Hint to use waitng or queue
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Results Meaning 0.50 (utilization) a = average interarrival time 0.92 (time in queue) CV'a = std deviation/ a 3.92 (time in system) p = average service time 1.500 (inventory in service) CV'p= std deviation of activi 0.458 (inventory in queue) Std deviation of activity 1.958 (inventory in system) m = number of servers or re 0.93954347 (service level given TWT) TWT = target waiting time or u = Utilisation Tq = time in queue, excludin T = total flowtime = tq + p Ip = inventory in service or b Iq = inventory in queue I = overall inventory in the s
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me a flow unit comes e management is interrarrival time, processing time or service time and st
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e (e.g. one flow unit come in every XX min) ity time/ p y time = service time variability (e.g. Given standard deviation of processing esources providing the service r service level ng processing time or average service time being processed or being service system = Ip + Iq
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td deviation
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time is 4.5min. Then, the service time variability has changed from 4.5min to
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o 1.5min, this means standard deviation is now 1.5min)
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Template for Predicting Waiting Time and Service Level for Multiple Parallel Resources Data Results a = 2 (average interarrival time) u = 0.67 (utilization) Cva = 1 (interarrival time coefficient of variation) Tq = 1.19 (time in queue) answer a p = 4 (average service time) T = 5.19 (time in system) CVp = 0.5 (service time coefficient of variation) Ip = 2.000 (inventory in service) m = 3 (number of servers) Iq = 0.596 (inventory in queue) I = 2.596 (inventory in system) answer b TWT = 0 (target waiting time) SL = 0.5235361392 (service level given TWT) 0.4764638608 Comments: average interarrival time, a = 2min because an email or flow unit arrive on an average of every 2min CV'a = std deviation/ mean = 2min/ 2min = 1 average service time, p = 4min as it takes on average 4min to write a response email CV'p = Std deviation of procesing or activity time/ Mean of processing or activity time = 2min/4min = 0.5 Number of servers, m = 3 because there are three employees TWT is not given therefore 0 A B C D E F G H I J K L M N O P Q 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
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Q2a Comment CV'a = 1 (because std dev is not given, hence is assumed 1) p=3min bec it takes an average 3min to respond CV'p = 4.5/3 m=2 Template for Predicting Waiting Time and Service Level for Multiple Parallel Resources Data Results Comment a = 2 (average interarrival time) u = 0.75 (utilization) Cva = 1 (interarrival time coefficient of variation) Tq = 6.43 (time in queue) Answer a 6.43min is the average waiting time for a customer before reaching a teller (i.e. Tq is the time the flow unit has to spend in the queue waiting for the service to begin. Tq does not include the processing time) p = 3 (average service time) T = 9.43 (time in system) CVp = 1.5 (service time coefficient of variation) Ip = 1.500 (inventory in service) m = 2 (number of servers) Iq = 3.213 (inventory in queue) I = 4.713
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