Question

Consider the following set of processes, the length of the CPU burst time given in milliseconds P1->6, P2->8, P3->7,P4->3. Assuming the above process being scheduled with the RR (time slice=3ms) scheduling algorithm. The average turnaround time and waiting time is

a.

18.5ms, 12.5ms

b.

18.4ms, 12.4ms

c.

16.25ms, 10.25ms

d.

12.25ms, 6.25ms​

Answer 1

Answer:

->6, P2->8, P3->7,P4->3. Assuming ans c.

Answer 2

(A) The Average turnaround time = 18.5 ms and the Average waiting time = 12.5 ms

Explanation:

• Here, we need to calculate the mean turnaround time and also waiting time of the given processes getting executed in the CPU and scheduled with the given Round Robin scheduling algorithm.

The process schedule is as follows:

        PROCESS        A.T.         B.T.        T.A.T.         W.T.

              P1                 0             6             15              9

               P2                0             8             23             15

               P3               0             7             24              17

               P4               0             3              12              9

        --------------------------------------------------------------------------

                                                                 74             50

The process Gantt chart is as follows:

P1 (0-3)  

P2 (3-6)

P3 (6-9)

P4 (9-12)

P1 (12-15)

P2 (15-18)

P3 (18-21)

P2 (21-24)

P3 (23-24)

Here,

Average turnaround time = Total Turn around time / no of processes

                                           = 74/4

                                           = 18.5 ms

Mean waiting time = Summation of all waiting time / no of processes

                                     = 50/4

                                      = 12.5 ms

Thus, The mean turnaround time = is 18.5 ms, and the mean waiting time = is 12.5 ms.