Consider a flow shop with two workstations in series. Each workstation can be de

Question

Consider a flow shop with two workstations in series. Each workstation can be defined as a single-server queueing system with a queue of infinite capacity. All service times are independent and follow an exponential distribution. The mean service times are 4 minutes at workstation 1 and 5 minutes at workstation 2. Raw parts arrive to workstation 1 following a Poisson process with a mean of 10 parts per hour.

(a) (5 points) What is the probability that no raw parts arrive to workstation 1 in 15 min.?

(b) (10 points) Find the joint steady-state distribution of the number of parts at workstation 1 and the number of parts at workstation 2.

(c) (5 point) What is the probability that both servers are busy?

(d) (5 points) What is the probability that both queues are empty?

(e) (5 points) Find the expected total number of parts in the flow shop.

Explanation / Answer

Solution:

The problem given is a tandem queue where the arrival rate follows poisson process with rate = 10/hours. The service time at both workstations follows exponential distribution with µ1 = 15 and µ2 = 12.

a. Probability(No raw parts arrive to worksattion 1in 15 minutes) = e^-10x0.025 (10 x 0.025)^0/0! = 0.08208

b. Let there are n customers at workstation 1, and m customers at workstation 2, Therefore, the joint steady state distribution is

P (n, m) = ( / µ1)^n (1 - / µ1) ( / µ1)^m (1 - / µ2) ----------------- (A)

c. P (both servers are busy) = 1 - P (both servers are free)

P (both servers are busy) = 1 - P00

P (both servers are busy) = 1 - (1 - 10/15) (1 - 10/12)

P (both servers are busy) = 1 - 1/18

P (both servers are busy) = 0.9444

d. P (both queue are empty) = P00 + P01 + P10 + P11

P (both queue are empty) = 1/18 + 5/108 + 1/27 + 5/162

P (both queue are empty) = 0.16975 ~ 0.17

e. Expected number = ( / µ1)^n (1 - / µ1) ( / µ1)^m (1 - / µ2)

Substituting the values we get

Expected number = 10/(15-10) + 10/(12-10) = 7 customers.

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