Ch 13 - Solutions

Ch 13 - Solutions - Chapter 13 Queuing Theory 6 a b c There...

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Chapter 13 Queuing Theory 6. a. There are 30/60 = 0.5 arrivals per minute, on average. b. In a ten minute interval we would expect 10*0.5 = 5 arrivals to occur. c. P(x=0) = (5 0 e -5 )/(0!) = 0.00674 (NOTE: 0!=1) P(x=1) = (5 1 e -5 )/(1!) = 0.03369 P(x=2) = (5 2 e -5 )/(2!) = 0.08422 P(x=3) = (5 3 e -5 )/(3!) = 0.14037 d. P(x>3)=1-P(x 3) = 1-0.00674-0.03369-0.08422-0.14037 = 0.73497 7. a. Expected service time = 1/40 = 0.025 hours (or 1.5 minutes) b. P(t 1/60) = 1- e -40/60 = 0.4865 c. P(2/60 t 5/60) = e -40*2/60 - e -40*5/60 = 0.2279 P(t < 4/60) = 1- e -40*4/60 = 0.931 P(t 3/60) = 1- P(t 3/60) = 1-.8646 = 0.1354 8. Use the M/M/s template in Q.xls with: Arrival Rate = 30, Service Rate = 40, Number of Servers = 1. a. 0.25 b. 0.75 c. 2.25 d. 0.1 hours or 6 minutes e. 0.075 hours or 4.5 minutes f. 60 per hour 9. Use the M/M/s template in Q.xls with: Arrival Rate = 7 per hour, Service Rate = 3 per hour, and vary the number of servers. With 3 servers the average waiting time is 0.3057 hours or 18.34 minutes.
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This note was uploaded on 11/01/2009 for the course ACCT 3203 taught by Professor Shelton during the Spring '09 term at Montana.

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Ch 13 - Solutions - Chapter 13 Queuing Theory 6 a b c There...

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