GATE Papers >> CSE >> 2015 >> Question No 257

Question No. 257 CSE | GATE 2015

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

Process Arrival
Time
Processing
Time
A 0 3
B 1 6
C 4 4
D 6 2

Answer : (C) Shortest Remaining Time


Solution of Question No 257 of GATE 2015 CSE Paper
Process Arrival Time B.T.
A 0 3
B 1 6
C 4 4
D 6 2

(i)   FCFS:

Avg TAT $ =\frac{3+8+9+9}4=\frac{29}4=7.25 $

(ii)   Non-peemptive SJF:

Avg TAT $ =\frac{3+8+5+11}4=6.75 $

(iii)  SRTF:

Avg TAT $ =\frac{3+14+4+4}4=\frac{25}4=6.75 $

(iv)  RR:

Avg TAT $ =\frac{5+14+9+5}4=\frac{33}4=8.25 $

$\therefore$  SRTF has lowest turn around time.

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