Question

1) A fast-food franchise is considering opening a drive-up window food service operation. Assume that customer arrivals follow a Poisson distribution ( interarrival times follow an exponential distribution), with a mean arrival rate of 24 cars per hour, and that service times follow an exponential probability distribution. Arriving customers place orders at an intercom station at the back of the parking lot and then drive up to the service window to pay for and receive their order. The following four service alteratives are being considered: (A) A single channel operation where one employee fills the order and takes the money from the customer. The average service time for this alternative is 2 minutes. (B) A single channel operation where two employees work as a team to serve each customer. The average service time for this alternative is 1.25 minutes. (C) A two channel operation with two service windows and two employees. Newly arriving customers join a queue which feeds both service windows. The employee stationed at each window fills the order and takes the money for customers arriving at that window. The average service time for this alternative is 2 minutes. (D) There are two service windows and two employees. There is a separate queue for each of the service windows. Newly arriving customers randomly choose one of two queues. The employee stationed at each window fills the order and takes the money for customers arriving at that window. The average service time for this alternative is 2 minutes. Compute the operating characteristics (Tal. T.D) for each of the service alternatives and fill in the following table: Service alternative ( AB Average waiting time (in minutes). Average number of cars waiting for service. Z Average time in the system (in minutes), I Average number of cars in the system. I

Answer 1

(A)

Average arrival rate, λ = 24 per hr.
Average service rate, μ = 1 in 2 minutes = 30 per hr.
No. of servers = 1

Tq = λ / {μ.(μ - λ)} = 24 / (30*(30 - 24)) = 0.1333 hrs. = 8 minutes

Iq = Tq * λ = 0.1333*24 = 3.2 cars

T = Tq + 1/μ = 0.1333 + 1/30 = 0.1667 hrs. = 10 minutes

l = T * λ = 0.1667*24 = 4 cars.

(B)

Average arrival rate, λ = 24 per hr.
Average service rate, μ = 1 in 1.25 minutes = 48 per hr.
No. of servers = 1

Tq = λ / {μ.(μ - λ)} = 24 / (48*(48 - 24)) = 0.0208 hrs. = 1.25 minutes

Iq = Tq * λ = 0.0208*24 = 0.50 cars

T = Tq + 1/μ = 0.0208 + 1/48 = 0.04167 hrs. = 2.5 minutes

l = T * λ = 0.04167*24 = 1 car

(C)

Average arrival rate, λ = 24 per hr.
Average service rate per server, μ = 1 in 2 minutes = 30 per hr.
No. of servers, s = 2

Use the following two formulae to compute the idle server probability (P0) and the average waiting time in the queue (Tq).

— for s.ua ua/u)* (s-1)!(s.и – а)2*0 **

So,

P0 = 1 / ((1 + (24/30)^1) + (1/2)*(24/30)^2*(2*30)/(2*30-24)) = 0.4286

Tq = 30*(24/30)^2*0.4286/(1*(2*30-24)^2) = 0.006349 hrs. = 0.38 minutes

Iq = Tq * λ = 0.006349*24 = 0.152 cars

T = Tq + 1/μ = 0.006349 + 1/30 = 0.0397 hrs. = 2.38 minutes

l = T * λ = 0.0397*24 = 0.95 cars.

(D)

For each single queue:

λ = 24/2 = 12 per hr.
μ = 1 in 2 minutes = 30 per hr.

Tq = λ / {μ.(μ - λ)} = 12 / (30*(30 - 12)) = 0.0222 hrs. = 1.333 minutes

Iq = Tq * λ = 0.0222*12 = 0.2667 cars

T = Tq + 1/μ = 0.0222 + 1/30 = 0.0556 hrs. = 3.333 minutes

l = T * λ = 0.0556*12 = 0.667 cars.

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So, for two queues combined:

Tq = 1.333 minutes

Iq = 0.2667*2 = 0.5333 cars

T = 3.333 minutes

I = 0.667*2 = 1.333 cars