Question

A system has a common checkout line for customers to wait to be served by one of four servers. Every customer arrives based on an exponential distribution with mean .5 minutes and it takes Tria(1, 2, 3) minutes for checkout processing. Ignoring the time to walk from the queue to the server, what is the steady state utilization? Is the system stable?

Answer 1

If a system behavior is independent of initial conditions and the elapsed time, then the behavior of Queue is said to be steady state behaviour.

The state of queuing system is represented by a single number n, the number of customers currently in the system.

 It utilizes memory-less property of exponential distribution. As per this property the time since the last arrival and the time the current customer has been in the service process are irrelevant to the future behavior of the system.

 Consider the system to be in steady state, which means that the system has been running for so long that the current state doesn’t depend on the starting condition.

 By computing the long run probabilities of being in each state, we will determine the performance measures of queuing models as long term steady state performance measures

 Hence, the customers arrive only one customer at a time. The system state can change only by one unit at a time.