The Govt. of Denver investigators are now at a branch on Wanker St., which has u
ID: 349145 • Letter: T
Question
The Govt. of Denver investigators are now at a branch on Wanker St., which has unbalanced demand for the services. The peak visitor arrivals occur in 15 minute intervals between 1pm – 1:15pm. According to the records, they found that the arrival rate is 6 visitors per 5-minute interval while the service rate is 8 visitors per 5-minute interval. One server (i.e., a staff) is paid $21/hour and the loss of goodwill was estimated at 25 cents per minute of wait time. The investigators attempt to find the best configuration of servers for the “peak 15-minute” solution.
Requirement:
For k = 1, 2, and 3 servers, determine the following values, filling in the table with your answers. Calculate all values manually. Show your work in steps using proper formula.
k = 1
k = 2
k = 3
Utilization Factor
Probability of no units in system
Average Queue Length
Average Number in the System
Average Wait Time
Average Time in the System
Total Cost (Wait + Server)
k = 1
k = 2
k = 3
Utilization Factor
Probability of no units in system
Average Queue Length
Average Number in the System
Average Wait Time
Average Time in the System
Total Cost (Wait + Server)
Explanation / Answer
So we have
So one server is best
Option 1 Kendall's notation M/M/1 Comment Arrival rate A 72 (6*12) per hour Service rate S 96 (8*12) U Utilization ratio U=A/S 0.75000 <1 Length of queue diminishing L Expected number of people in system Ls A/(S-A) 3.00 W Avg waiting time in the system Ws 1/(S-A) 0.04167 hours 2.5000 mins Wq Avg waiting time in the Queue Wq A/S(S-A) 0.03125 hous 1.8750 mins Lq Avg no of customer waiting to be served Lq A^2/S(S-A) 2.2500 Po Prob of 0 units in system Po 1-(A/S) 0.25000 25% Option 2 Kendall's notation M/M/2 Comment Arrival rate A 72 Service rate S 96 U Utilization ratio U=A/2*S 0.37500 <1 Length of queue diminishing L Expected number of people in system 0.873 W Avg waiting time in the system Ws 0.01200 hours 0.7200 mins Wq Avg waiting time in the Queue Wq 0.00200 hous 0.1200 mins Lq Avg no of customer waiting to be served Lq 0.1230 Po Prob of 0 units in system Po 1-(A/S) 0.45455 45% Option 3 Kendall's notation M/M/3 Comment Arrival rate A 72 Service rate S 96 U Utilization ratio U=A/3*S 0.25000 <1 Length of queue diminishing L Expected number of people in system 0.765 W Avg waiting time in the system Ws 0.01100 hours 0.6600 mins Wq Avg waiting time in the Queue Wq 0.00000 hous 0.0000 mins Lq Avg no of customer waiting to be served Lq 0.0150 Po Prob of 0 units in system Po 1-(A/S) 0.47059 47%Related Questions
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