In: Statistics and Probability
The admin office has one centralized copy machine located in a small room in the office. Jobs arrive at this copy center according to a Poisson process at a mean rate of 2 per day. The processing time to perform each job has an exponential distribution with a mean of 1/4 day. Because the copy jobs are bulky, those not being worked on are currently being stored in a room some distance from the copy machine. However, to save time in fetching the jobs, the office manager is proposing to add enough in-process storage space next to the copy machine to accommodate 3 jobs in addition to the one being processed. (Excess jobs will continue to be stored temporarily in the distant room.)
Answer:
(a) What is the probability that no jobs are in the machine?
We write X~Poisson (λ). e= 2.7183
P(X)= prob. of X events
E(X)= λ=np
Here X=0
λ = 1/2
Therefore P(X=0) = 0.6065
b) With this new proposal, what is the probability that queue length is full?
λ=2
μ=4
λ/μ= 2/4
=0.5
P0=1-λ/μ=1-0.5=0.5
Probability of having 4( for example 3 employments in pausing and one preparing) =0.5*(0.5)^4=0.03125
So the probability that queue length is full =0.03125
c) Under this proposal, what is the probability of time will this storage space next to the copy machine be adequate to accommodate not more than 4 waiting jobs?
Probability that the space will be satisfactory ( not exactly or equivalent to 4 jobs) = 0.96875
Probability that the space will be satisfactory for ;under five jobs ( 4 waiting and one in process)
= P(0)+P(1)+P(2)+P(3)+P(4)+P(5)
= 0.96875+0.015625
= 0.98437
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