HW9 solution

HW9 solution - a. 0.25 b. 0.75 c. 2.25 d. 0.1 hours or 6...

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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 t 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 =

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to occur. 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

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8. Use the M/M/s template in Q.xlsx with: Arrival Rate = 30, Service Rate = 40, Number of Servers = 1.

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Unformatted text preview: 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 12. Use the M/M/s template in Q.xlsx with: Arrival Rate = 8, Service Rate = 8, Number of Servers = 2. a. 40 minutes b. 0.333 c. 0.333 d. No balking or reneging. 16. Use M/M/s with finite queue length template in Q.xlsx with: Arrival Rate = 60 (per hour), Service Rate = 4 (per hour), Numbe Queue Length = 10. a. 3.538 b. 0.063 hours c. 0.0643 d. With 19 modems there is only a 0.49% chance of a caller rec modem connections would need to be added. er of Servers = 15, and Maximum eiving a busy signal. Thus, 4...
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This note was uploaded on 04/19/2011 for the course CBA 0060 taught by Professor Barrymitnick during the Spring '09 term at Pittsburgh.

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HW9 solution - a. 0.25 b. 0.75 c. 2.25 d. 0.1 hours or 6...

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