Question

	Burst Time	Arrival Time
P1	54	2
P2	12	3
P3	26	4
P4	16	5
P5	8	6
P6	92	7

use SRTF

(1) Gant chart

(2) Waiting time and Turn around time for every process

(3) Average WT and Average TAT

Answer 1

Please find the answer below,

First, let's see some information about SRTF (Shortest Remaining Time First)

1. In this algorithm, processes are scheduled based on the burst time

2. It's preemptive by nature means the process will be stopped and another process with less burst time would be scheduled.

Gantt Chart:

P4 | PIPS | PT Pਪ | P 3P1 | P6 3 6 01 233965 128 2 z 0.

Please find the process table below:

Process	Burst Time	Arrival Time	Completion Time	Turn Around Time(TAT)	Waiting Time(WT)
P1	54	2	118	116	62
P2	12	3	23	20	8
P3	26	4	65	61	35
P4	16	5	39	34	18
P5	8	6	14	8	0
P6	92	7	210	203	111

Average Waiting time = 234 / 6 = 39

Average Turn Around Time = 442 / 6 = 73.67

Explanation:

--> At time 0 no process was there for execution.

--> At time 2, first process P1 arrives and starts execution till time 3.

--> At time 3, Process P2 arrives with burst time 12, so CPU would be assigned to P2. P1 has 53 units remaining for execution.

--> At time 4, Process P3 arrives with burst time of 26 so it would be put in the queue and P2 continue execution.

--> At time 5, Process P4 arrives with burst time of 16 so it would be put in the queue and P2 continue execution.

--> At time 6, Process P5 arrives with a burst time of 8 which is less than P2 so P2 will be stopped and P5 is assigned with CPU. P2 has 9 units of burst time remaining.

--> At time 7, Process P6 arrives with burst time of 92 so it would be put in the queue and P5 continue execution.

--> At time 14, Process P5 completes execution and P2 which has the lowest burst time among the queued process will be brought into CPU.

--> At the time 23, Process P2 completes execution and P4 which has the lowest burst time among the queued process will be brought into CPU.

--> At the time 39, Process P4 completes execution and P3 which has the lowest burst time among the queued process will be brought into CPU.

--> At the time 65, Process P3 completes execution and P1 which has the lowest burst time among the queued process will be brought into CPU.

--> At time 118, Process P1 completes execution and the last process P6 will be brought into CPU.

--> At time 210 P6 completes the execution.

Now find out the completion time of each process.

--> Using completion time calculate Turn Around Time which is (Completion Time - Arrival Time)

--> Using Turn Around Time calculate waiting time which is (Turn Around Time - Burst Time)

Please let us know in the comments if you have any doubt.