Single channel queuing model with Poisson arrivals and exponential service times
ID: 451984 • Letter: S
Question
Single channel queuing model with Poisson arrivals and exponential service times (M/M/1)
=arrival rate=#/per hour
4
µ=service rate=#/hours
7
hours per work day
8
Cw=
$35
m, number of service channels
1
Cs=service cost per hour per channel
$25.00
the average number of customers in the system is 1.33
the average time a customer spends waiting in the queue is less than 0.19
the percent idle time is 0.52
total daily service cost is (in hours) 200.00
the average time a customer spends in the system is $0.33
the average number of customers in the queue is 0.76
total costs of the queuing system is more than total waiting cost
total waiting cost is $213.33
the utilization factor for the system is not 0.47
=arrival rate=#/per hour
4
µ=service rate=#/hours
7
hours per work day
8
Cw=
$35
m, number of service channels
1
Cs=service cost per hour per channel
$25.00
Explanation / Answer
A.
The average number of customers in the system = /( µ-) = 4/(7-4) = 4/3 = 1.33
B.
The average time a customer spends waiting in the queue = /(µ*(µ-)) = 4/(7*(7-4) = .19
D.
Total Daily Service cost = Service cost per hour* No. of hours = 25*8 = $200
E.
Average time a customer spends in the queuing system (W) = 1/( µ-) = 1/(7-4) = .33
F.
Average number of customers in the queue = ^2/(µ*(µ-)) = 4^2/(7*(7-4) = .76