Question

Consider the following set of processes A, B, C, D with the following CPU burst time and I/O. Find the average waiting time with RR of quantum 10ms and context switch time=2ms. Draw the Gantt chart clearly and give necessary explanations. Process Arrival Time CPU Burst Time VO CPU Burst Time A 0 24 30 15 B 15 23 25 21 C 0 32 D 25 12

Answer 1

Note ----> I/O is not included in waiting Time . Waiting time is also counted when process is ready and waiting in ready Queue to get the CPU

IN RR scheduling every process gets chance once time quantum of current process ends.

In gantt Chart we have 2 ms blocks that depicts Context Switch now Context Switch means when one Processes is leaving the CPU and context of other processes in loading onto the CPU

See table for each processes completion and waiting time and Gantt Chart Images

Process AT BT Ilo BI WT WT CT 24 2 3 30 25 A B c D 0 15 O 25 15 130 MB 21 | 155 52 키 32 113 12 104 167 Queue ACABCPABCDBC Ready Gantt Chart A B Tolle D 46 48 5860 70 72 o Jo 12 2224 3436

A B CO B BIC 72 76 78 88 to 100 102 104 106 109 111 113 As A has completed Flo so it resumes Π1 Α 1/10 B 13 116 134 155 Waiting Time of A 14+38+0 = 52 Waiting Time of B = 21+39+18 to = 71 Waiting Time of c = 12 +26+32+ 11 = 81 waiting time go 67

Explanation-->

1. First we have A and C arrived so schedule A for T = 0 to 10 msec

2. C is also ready so now C runs from T=12 to T=22 and now time T=10 to T=12 is for context switch

3. Again A comes back and runs

4.Now as B has Arrived At T = 15 so now it will run from T = 24 to 34

See Gantt chart for further execution sequence

5. D Arrives a T = 25 but it gets CPU at T= 60

6. Now A gets competed At T = 76 and started it's I/O time of 30 msec

7.Similarly B also finishes at T =113 and started it's I/O of 25 msec

8 All processes  completed their first CPU Burst Time till 113 and Now CPU is idle till T =116 as A I/O completes At T=116 So 3 msec CPU will be idle

8 From T = 116 to T=131 A completes it's second CPU Burst Time and gets Completed

8. Now CPU is again free as B I/O completes at T = 134 So Again for 3 sec CPU is idle

9. At last B completes it's second CPU time and gets completed at T = 155 msec

So the main concept here was how to handle the I/O time along with CPU Time .And as only A and B has I/O so this time is not included in their Waiting Time

Hence waiting time of process is

1 Waiting time of A = 14 ( i.e T=10 - T = 24 ) + 38 ( T = 34 - T= 72 ) = 52

2. Waiting Time of B = 21 + 32+ 18 = 71

3. Waiting Time of C = 81

4. Waiting Time of D = 67

Average Waiting Time = 52 + 71 + 81 + 67 / 4 = 271/4 = 67.75

This is how using RR scheduling you can calculate the waiting time of each process

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