Process-Scheduling

Question 21

For the processes listed in the following table, which of the following scheduling schemes will give the lowest average turnaround time?

A	First Come First Serve
B	Non-preemptive Shortest Job First
C	Shortest Remaining Time
D	Round Robin with Quantum value two

Question 21 Explanation:

FCFS:

TAT for A = completion time(A) – AT(A) = 3 – 0 = 3
TAT of B = 9 – 1 = 8
TAT of C = 13 – 4 = 9
TAT of D = 15 – 6 = 9
∴ Avg. TAT = 29/4
SJF:

TAT of A = 3 – 0 = 3
TAT of B = 9 – 1 = 8
TAT of C = 15 – 4 = 11
TAT of D = 11 – 6 = 5
∴ Avg. TAT = 27/4
SRTF:

TAT of A = 3 – 0 = 3
TAT of B = 15 – 1 = 14
TAT of C = 8 – 4 = 4
TAT of D = 10 – 6 = 4
∴ Avg. TAT = 25/4
RR:

TAT of A = 5 – 0 = 5
TAT of B = 15 – 1 = 14
TAT of C = 13 – 4 = 9
TAT of D = 11 – 6 = 5
∴ Avg. TAT = 33/4
∴ SRTF has lowest Avg. TAT.

Correct Answer: C

Question 21 Explanation:

FCFS:

TAT for A = completion time(A) – AT(A) = 3 – 0 = 3
TAT of B = 9 – 1 = 8
TAT of C = 13 – 4 = 9
TAT of D = 15 – 6 = 9
∴ Avg. TAT = 29/4
SJF:

TAT of A = 3 – 0 = 3
TAT of B = 9 – 1 = 8
TAT of C = 15 – 4 = 11
TAT of D = 11 – 6 = 5
∴ Avg. TAT = 27/4
SRTF:

TAT of A = 3 – 0 = 3
TAT of B = 15 – 1 = 14
TAT of C = 8 – 4 = 4
TAT of D = 10 – 6 = 4
∴ Avg. TAT = 25/4
RR:

TAT of A = 5 – 0 = 5
TAT of B = 15 – 1 = 14
TAT of C = 13 – 4 = 9
TAT of D = 11 – 6 = 5
∴ Avg. TAT = 33/4
∴ SRTF has lowest Avg. TAT.