Question

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.)

a,What is the probability that no jobs are in the machine?

b,With this new proposal, what is the probability that queue length is full?

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?

Answer 1

Solution :

a)

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) = .6065

b)

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)

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

Let me know in the comment section if anything is not clear. I will reply ASAP!
If you liked the answer, please give an upvote. This will be quite encouraging for me.Thank-you!