Question

[8 What is the waiting time of each process for each of the scheduling algorithms? Marks] Job Arrival Time Size (msec) P1 0 9 Priority 3 (Silver) 1 (Diamond) 3 (Silver) P2 3 P3 3 3 P4 6 5 4 (Bronze) 2 (Gold) P5 7 4 (a) First-Come-First-Served (FCFS) scheduling (b) Preemptive PRIORITY scheduling (c) Non-preemptive PRIORITY scheduling (d) Round Robin (RR) (quantum = 3) scheduling

The question is this. there is nothing given

Answer 1

1. FIRST COME FIRST SERVE (FCFS)::

THE PROCESS WHICH COME FIRST WILL BE SERVED FIRST IT IS NON-PREEMPTIVE ALGORITHM.

GANTT CHART:: (arrivaltime- finish time) of each processor

P1 P2 P3 P4 P5

0-09 9-10 10-13 13-18 18-22

waiting time:: finish time-arrival time-execution time

waiting time of each processor

p1 --> 9-0-9 =0

p2-->10-3--1=6

p3-->13-3-3=7

p4-->18-6-5=7

p5-->22-7-4=11

2.PREEMPTIVE PRIORITY::

process having highest priority will execure first in preemptive mode

GANT CHART:: (TIME SLOT,REMAINING TIME)

wating time=finish time-arrival time-execution time

P1 P2(finished) P1 P1 P5(finished) P1(finished) P3(finished) P4 (finished)

0-3,6 3-4,0 4-6,4 6-7,3 7-11,0 11-14,0 14-17,0 17-22,0

finish time arrival time execution time waiting time

P1 14 0 9 5

P2 4 3 1 0

P3 17 3 3 11

P4 22 6 5 11

P5 11 7 4 0

3.NON-PREEMPTIVE PRIORITY SCHEDULING::

process having highest priority will execure first in non-preemptive mode

Gantt chart:: (Time Slot)

P1 P2 P5 P3 P4

0-9 9-10 10-14 14-17 17-22

finish time arrival time execution time waiting time

P1 9 0 9 0

P2 10 3 1 6

P3 17 3 3 11

P4 22 6 5 11

P5 14 7 4 3

4.ROUND-ROBBIN:::

each process execute in given time interval given to that process.

Gantt Chart::(TIME SLOT,REMAING TIME)

P1 P2(fin) P3(fin) P1 P4 P5 P1(FIN) P4(fin) P5(FIN)

0-3,6 3-4,0 4-7,0 7-10,3 10-13,2 13-16,1 16-19,0 19-20,0 20-22,0

finish time arrival time execution time waiting time

P1 19 0 9 10

P2 4 3 1 0

P3 7 3 3 1

P4 20 6 5 9

P5 22 7 4 11