On average 2 customers arrive per hour at a Foto-Mat to process film. There is o

ID: 2747008 • Letter: O

Question

On average 2 customers arrive per hour at a Foto-Mat to process

film. There is one clerk in attendance that on average spends 15
minutes per customer.

1.What is the average queue length and average number of customers in the system?

2.What is the average waiting time in queue and average time spent in the system?

3.What is the probability of having 2 or more customers waiting in queue?

4.If the clerk is paid $4 per hour and a customer’s waiting cost in queue is considered $6 per hour. What is the total system cost per hour?

5. What would be the total system cost per hour, if a second clerk were added and a single queue were used?

L=Arrival rate= 2 customers per hour

m=Service rate= 15 mins per customer= 4 customers/ hr

L/m=2/4=0.5

1- Avg queue length =Lq=L^2/(m(m-L))=2^2/(4*(4-2))=0.5

avg number of units in the system=Lq+L/m=0.5+0.5=1

2-Avg waiting time in queue=Wq=Lq/L=0.5/2=0.25 hr

avg time in system=Wq+1/m=0.25+1/4=0.5 hr

3- Probability of 2 of more customers in system =1-Prob of 1 customer in system=1-(L/m)^1 *(1-L/m)

=1-0.5*0.5=0.75

4-total cost= Cw*avg. units in system+Cs*number of channels=6*1+4*1=10

5- if 2 channels are there then at L/m=0.5 Po(prob. of no units in system)=0.6

Lq= (L/m)^kLm/((k-1)!*(km-L)^2) *Po= 0.5^2 *4*2/(1*(2*4-2) *0.6=0.2

L=Lq+L/m=0.2+0.5=0.7

hence total cost= Cw*avg. units in system+Cs*number of channels= 6*0.7+4*2=12.2

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