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# GATE CS 2020 CPU Scheduling PYQ

### 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 _____________________.
Solution.

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