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find average waiting time using Round Robin (time slice=4ms)? ProcessArrival TimeBurst TimeP108P214P329P435 Select one:a. 12.25b. 11.75c. 11d. 10

Question

find average waiting time using Round Robin (time slice=4ms)?

Process Arrival Time Burst Time
P1 0 8
P2 1 4
P3 2 9
P4 3 5

Select one:
a. 12.25
b. 11.75
c. 11
d. 10

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Solution

To find the average waiting time using Round Robin scheduling, we need to calculate the waiting time for each process and then find the average. The time slice given is 4ms.

Here's how we can calculate it:

  1. For process P1, it arrives at 0 and has a burst time of 8. Since the time slice is 4ms, it will run for 4ms, then it will wait for the remaining processes to complete their time slice. The remaining burst time for P1 is 4ms. The waiting time for P1 is the time it waits after the first time slice until it gets the CPU again, which is the sum of the time slices of P2, P3, and P4 (4ms each), so 12ms.

  2. For process P2, it arrives at 1 and has a burst time of 4. It will run for 4ms after P1's first time slice. So, it waits for 4ms (P1's first time slice). The waiting time for P2 is 4ms.

  3. For process P3, it arrives at 2 and has a burst time of 9. It will run for 4ms after P1 and P2 have had their time slices. So, it waits for 8ms (P1's first time slice + P2's time slice). After its first time slice, it will wait for P1's and P4's time slices (4ms each), so another 8ms. The total waiting time for P3 is 16ms.

  4. For process P4, it arrives at 3 and has a burst time of 5. It will run for 4ms after P1, P2, and P3 have had their time slices. So, it waits for 12ms (P1's first time slice + P2's time slice + P3's first time slice). The waiting time for P4 is 12ms.

Now, we can calculate the average waiting time by adding up the waiting times and dividing by the number of processes:

(12ms for P1 + 4ms for P2 + 16ms for P3 + 12ms for P4) / 4 = 11ms

So, the average waiting time using Round Robin scheduling with a time slice of 4ms is 11ms. The correct answer is c. 11.

This problem has been solved

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