GATE 2020
Q50. Consider the following set of processes, assumed to have arrived at time 0. Consider the CPU scheduling algorithms Shortest Job First (SJF) and Round Robin (RR). For RR, assume that the processes are scheduled in the order P!, P2, P3, P4.
|
Processes |
P1 |
P2 |
P3 |
P4 |
|
Burst time (in ms) |
8 |
7 |
2 |
4 |
If the time quantum for RR is 4 ms, then the absolute value of the difference between the average turn around times (in ms) of SJF and RR (round off to 2 decimal places) is _____________________.
Shortest Job First (SJF)
|
Process |
Burst Time |
Completion Time |
Turn Around Time |
|
P1 |
08 |
21 |
21 |
|
P2 |
07 |
13 |
13 |
|
P3 |
02 |
02 |
02 |
|
P4 |
04 |
06 |
06 |
Average Turn Around TIme (21+13+2+6)/4 = 10.5
Round Robin (RR)
|
Process |
Burst Time |
Completion Time |
Turn Around Time |
|
P1 |
08 |
18 |
12 |
|
P2 |
07 |
21 |
21 |
|
P3 |
02 |
10 |
10 |
|
P4 |
04 |
14 |
14 |
Average Turn Around Time (18+21+10+14)/4 = 15.75
Absolute value of the difference = | 10.75 – 15.75 | =5.25