Question

4. The burst times and start times of four processes A,B,C,D are as follows:

Process Start Time Burst Time

A 0   15

B   10 10

C   11 3

D   15 8

Show scheduling of the 4 processes using a Gantt chart for STCF. Compute the wait times for each process, and the average wait time, under STCF.

Answer 1

Solution:-

SCTF - Shortest Time to Completion First.

SCTF-P = Pre-emptive version of SCTF.

STCF: run whatever job has the least amount of work to do before it finishes (or blocks for an I/O) .

Completion time : Time taken to complete a process.

Turnaround time : time from job submission to returned result.

Waiting time : total time spent waiting in the ready queue.

Given Information:

Procedure :

At starting time 0 Process A is in Ready Queue.

Here Starting time = 15.

At starting time 15 Process B, C, D are in Ready Queue.

According to rule of SCTF a Shortest Burst time Process should complete first.

In above 3 Processes C has shortest Burst time of 3 units.Therefore C should complete first.

Here Starting time = 18.

At starting time 18 Process B and D are in Ready Queue.

According to rule of SCTF a Shortest Burst time Process should complete first.

In above 3 Processes D has shortest Burst time of 8 units.Therefore D should complete first.

Starting time = 26.

Remaining Process is B having burst time 10 units.

Resulting Gantt Chart:

Calculating Completion Time of the Processes.Here CT(P_i) = Completion Time of the Process

CT(A) = 15

CT(B) = 36

CT(C) = 18

CT(D) = 26

Calculating TurnAround Time of the Processes.Here TAT(P_i) = CT(P_i) - ST(P_i).

TAT(A) = CT(A) - ST(A) = > 15 - 0 = 15

TAT(B) = CT(B) - ST(B) = > 36 - 10 = 26

TAT(C) = CT(C) - ST(C) = > 18 - 11 = 7

TAT(D) = CT(D) - ST(D) = > 26 - 15 = 11

Calculating Waiting Time of the Processes.Here WT(P_i) = TAT(P_i) - BT(P_i).

WT(A) = TAT(A) - BT(A) = 15 - 15 = 0

WT(B) = TAT(B) - BT(B) = 26 - 10 = 16

WT(C) = TAT(C) - BT(C) = 7 - 3 = 4

WT(D) = TAT(D) - BT(D) = 11 - 8 = 3

Calculating Average Waiting Time of the Processes.

Average Waiting Time = (WT(A)+WT(B)+WT(C)+WT(D)) / no.of processes.

                                => (0 + 16 + 4 + 3) / 4

                                => 23/4 = 5.75

Resulting Table:-