GATE CSE 2020 Q60: Difference of average turnaround times for SJF and RR (q = 4)

Operating System
2 marks
NAT
No negative marking

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 $P_1, P_2, P_3, P_4$ .

Processes	$P_1$	$P_2$	$P_3$	$P_4$
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 turnaround times (in ms) of SJF and RR (round off to 2 decimal places) is _______.

Try it: numerical answer: type the value

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Numerical answer. No negative marking.

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Answer: 5.25 (official GATE 2020 final answer key)

Concept used: With all arrivals at 0, turnaround time equals completion time; draw a Gantt chart for each policy.

First step: SJF runs $P_3, P_4, P_2, P_1$ finishing at 2, 6, 13, 21. For RR with quantum 4, the slices are $P_1, P_2, P_3, P_4, P_1, P_2$ .

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Source: official GATE 2020 CS question paper and final answer key, organised by IIT Delhi. Question paper · Answer key

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GATE CSE 2020 Q60: Difference of average turnaround times for SJF and RR (q = 4)

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