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# 09 - Tw = Ta= Ra=.75 Tp= 305 Rp=120 Service Time Factor...

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FMIS 3301 QUEUING EXERCISE 1 A photo printing service has a single machine that processes pictures as they arrive. Arrivals occur an average of every 45 seconds (a rate of 80/hour), with a standard deviation of 27 seconds . The machine can print 1 picture in a constant 30 seconds = Tp . Service time factor Tp=30 s=1 Tp/S = 30 = 30 seconds Utilization Factor U = 80/120 = .67 U/ (U-1) .67/.33 = 2 Variation Factor Cva = 27/45 SD/ average arrival = .6 CVp = 0 (.6squared + 0)/2 = .18 ALL: Tw = 30*2*.18 = 10.8 sec T=10.8 + 30 = 40.8 What is the mean time in hours that a picture spends in the system? How many pictures would be waiting to be printed? # waiting = Ib = RaTw = (80/3600) 10.8 = .24 or 24 sec Customers arrive at QuickNacho at an average rate of 0.75 per minute, and are served by a (S) four servers at an average rate of Tp 0.25 per minute. Assume Poisson arrival and service rates. What is the average waiting time in the queue?

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Unformatted text preview: Tw = Ta= Ra= .75 Tp= 305 Rp=120 Service Time Factor Tp=4m, S=4, Tp/s = 4/4 = 1 Utilization Factor U= 45/60 = .75 (Sqroot 2* (S+1))-1 (square root 2 *5) -1 = 2.162 (.75)>2.162 / (.25) = .537/.25 = 2.15 Variation factor 1 Tw = 1*2.15 *1= 2.15 min (129 seconds) What is the average number of customers at QN? T= 2.15 + 4 = 6.15 minutes ; I=RaT I= .75* 6.15 = 4.6 customers How long would a customer spend at QuickNacho if one more server were added? S=5 U=.75/5*.25=.6 Service Time Factor 4/5 = .8m Ut Factor ( An office server can be accessed by remote or direct connections. During peak time, queries arrive through remote every 10 seconds and through direct every 5 seconds both with a SD of 2s; the server can handle an average of 25 queries per minute with a SD of 3s About how much time on the average does a query wait for response? Approximately how many queries will be waiting to be processed, on average?...
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09 - Tw = Ta= Ra=.75 Tp= 305 Rp=120 Service Time Factor...

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