Question

1. Keuka Park Savings and Loan currently has one drive-in teller window. Cars arrive at a mean rate of 10 per hour. The mean service rate is 12 cars per hour. a. What is the probability that the service facility will be idle? b. If you were to drive up to the facility, how many cars would you expect to see waiting and being serviced? c. What is the average time waiting for service? d. What is the probability an arriving car will have to wait?

2. To improve its customer service, Keuka Park Savings and Loan (Problem 1) wants to investigate the effect of a second drive-in teller window. How would the answers to a, b, c and d in Problem 1 change?

3. If the maintenance cost for each drive-in teller is $50 per hour and the customer waiting time is valued at $25 per hour, should Keuka Park Savings and Loan expand to the two drive-in teller system? Explain.

**** I cant answer questions 2 and 3

Answer 1

1.
a)In a single server queuing model, if arrival rate is λ and service time is μ, the probability of system being idle will be 1-(λ/μ)
= 1- (10/12) = 0.1666667
b)cars expect to see waiting and being serviced = total cars in the system =
the average number of units in the system:
= Lq+λ/μ
Lq = the average number of units waiting for service
= λ^2/(μ(μ-λ))
= 10^2/(12*2)
= 100/24
= 4.16667
the average number of units in the system:
= 4.16667 + 10/12 = 4.16667 + 0.83333 = 5
c. What is the average time waiting for service?
the average time a unit spends waiting for service:
= Wq = Lq/λ = 4.16667/10 = 0.41667
d. What is the probability an arriving car will have to wait?
λ/μ = 10/12 = 0.83333
2.M/M/2 process

a)the probability of system being idle will be

= 1/(((λ/μ)^0/0!+(λ/μ)^1/1!)+((λ/μ)^2*μ/(1!*2μ-λ)))
= 1/((10/12)^0/0!+(10/12)^1/1!)+(((10/12)^2*12/(1!*(2*12-10))))
= 1/(1+10/12+(10/12)^2*12/(14))
= 0.4117647
b)
the average number of units in the system:
L = Lq + λ/μ
Lq = ((λ/μ)^k)*λ*μ*P0/(((k-1)!(k*μ-λ)^2))
= ((10/12)^2)*10*12*0.4117647/(((2-1)!(2*12-10)^2))
= 0.17507
L = 0.17507 + 10/12 = 0.17507 + 0.83333 = 1.0084
c)the average time a unit spends waiting for service:
Wq = Lq/λ = 0.17507/10 = 0.017507
d)
What is the probability an arriving car will have to wait?
Pw = 1/2!*((λ/μ)^2)*(k*μ*P0/(k*μ-λ)
= 1/2*((10/12)^2)*(2*12*0.4117647/(14))
= 0.245098