Feasible solution found at step:        0

 

            Variable           Value

                  AR        25.00000

                 STM        6.000000

                 STH       0.1000000

                  FB       0.5000000E-01

                  NS        5.475485

Solution: EZQUEUE

Because we cannot have fractional servers, we need at least six servers to meet our requirement that no more than 5% of customers find all servers busy.

Suppose that you install a sufficient number of incoming lines so customers finding all servers busy can wait. Further, you will still use six servers. You want to find the following:

1.the fraction of customers finding all servers busy,
2.the average waiting time for customers who wait,
3.the average overall waiting time, and
4.the average number waiting.

The following variation on the previous model computes these four statistics:

! Arrival rate of customers/ hour;

  AR = 25;

! Service time per customer in minutes;

  STM = 6;

! Service time per customer in hours;

  STH = STM/ 60;

! The number of servers;

  NS = 6;

! The PEL function finds number of servers

 needed, NS;

  FB = @PEB( AR * STH, NS);

! The conditional wait time for those who wait;

  WAITC = 1 / ( NS / STH - AR);

! The unconditional wait time;

  WAITU = FB * WAITC;

! The average number waiting;

  NWAIT = AR * WAITU;

Note how we now use the @PEB function, rather than @PEL, to account for the presence of a queue to hold callers finding all lines busy. The solution to the modified model is:

          Variable           Value

                AR        25.00000

               STM        6.000000

               STH       0.1000000

                NS        6.000000

                FB       0.4744481E-01

             WAITC       0.2857143E-01

             WAITU       0.1355566E-02

             NWAIT       0.3388915E-01

Remember the unit of time is an hour, so the expected waiting time for those who wait is .2857 * 60 = 1.7 minutes.