Homework Answers
This system of queuing is M/M/1 model.
Given that, Arrival rate, λ = 1.5 calls per minute
Service rate, μ = 60/10 = 6 calls per minute
a) Probability that the operator is busy = λ/μ = 1.5/6 = 25% = 0.25
b) Average waiting time = λ/μ/(μ-λ) = 1.5/6/(6-1.5) = 0.06 minute
c) Average number of calls waiting to be answer = λ2/μ/(μ-λ) = (2.25 / 6)/(6-1.5) = 0.833
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Table: Average Number of Customer in Waiting
Line
Poisson Arrivals, Exponential Service Times
Number of Service Channels, M
rhoρ
1
2
3
4
5
0.1
0.0111
0.15
0.0264
0.0008
0.2
0.05
0.002
0.25
0.0833
0.0039
0.3
0.1285
0.0069
0.35
0.1884
0.011
0.4
0.2666
0.0166
0.45
0.3681
0.0239
0.0019
0.5
0.5
0.0333
0.003
0.55
0.6722
0.0449
0.0043
0.6
0.9
0.0593
0.0061
0.65
1.2071...
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same $12 per hour. The total cost ( service cost + waiting cost) =
$___per hour (round your response to two
decimal places)
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Table: Average Number of Customer in Waiting
Line
Poisson Arrivals, Exponential Service Times
Number of Service Channels, M
rhoρ
1
2
3
4
5
0.1
0.0111
0.15
0.0264
0.0008
0.2
0.05
0.002
0.25
0.0833
0.0039
0.3
0.1285
0.0069
0.35
0.1884
0.011
0.4
0.2666
0.0166
0.45
0.3681
0.0239
0.0019
0.5
0.5
0.0333
0.003
0.55
0.6722
0.0449
0.0043
0.6
0.9
0.0593
0.0061
0.65
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