In a waiting line situation, arrivals occur at a rate of 2 per minute, and the service times average 18 seconds. Assume the Poisson and exponential distributions.

a. What is l?

b. What is m?

c. Find probability of no units in the system.

d. Find average number of units in the system.

e. Find average time in the waiting line.

f. Find average time in the system.

g. Find probability that there is one person waiting.

h. Find probability an arrival will have to wait.

 

As the arrivals follow Poisson distribution and service time follows exponential distribution and there is a single server so this is a M/M/1 queueing system.

a) Arrival rate (l) =  2 per minute

b) Service rate (m) = 60/18 per minute = 10/3 per minute

Using the M/M/1 queueing system formulas the followings are calculated.

c. P(no units in the system) = 1-l/m = 1-2/(10/3) = 0.4

d. Average number of units in the system = l/(m-l) = 2/(10/3-2) = 1.5

e. Average time in the waiting line = l/(m*(m-l)) = 2/(10/3*(10/3-2)) = 0.45 minutes or 27 seconds

f. Average time in the system = 1/(m-l) = 1/(10/3-2) = 0.75 minutes or 45 seconds

g. P(there is one person waiting) = P(there are 2 persons in the system) = (1-l/m)*(l/m)² = (1-2/(10/3))*(2/(10/3))² = 0.144

h. P(an arrival will have to wait) = P(at least one unit in the system) = 1-P(no units in the system) = 1-0.4 = 0.6; From part c