Question

Process	Burst Time
P1	6ms
P2	2ms
P3	7ms
P4	3ms
P5	8ms
P6	10ms

Using the table above calculate the average wait time using First Come First Served (FCFS) and Shortest Job First (SJF) CPU scheduling.

Answer 1

1. Calculate average time using FCFS First come First served

Consider the processes P1, P2, P3, P4, P5, P6 given table and given Burst Time, arrives for execution in the same order, with Arrival Time 0, let's find the average waiting time using the FCFS scheduling algorithm.

Average time is = (0 + 6 +8 + 15 +18 +26 )/6

= 73/6

=12.1666667 ms

Diagram:

ALGORITHM:

1-  Input the processes along with their burst time (bt).
2-  Find waiting time (wt) for all processes.
3-  As first process that comes need not to wait so 
    waiting time for process 1 will be 0 i.e. wt[0] = 0.
4-  Find waiting time for all other processes i.e. for
     process i -> 
       wt[i] = bt[i-1] + wt[i-1] .
5-  Find turnaround time = waiting_time + burst_time 
    for all processes.
6-  Find average waiting time = 
                 total_waiting_time / no_of_processes.
7-  Similarly, find average turnaround time = 
                 total_turn_around_time / no_of_processes.

Output of the algorithm will come like this :

2. Calculate average time using SJF Shortest Job First

SJF is of 2 types :

• Non-Preemptive SJF
• Preemptive SJF

Let us consider the arrival time for the following processes are:

Process	Burst Time	Arrival time
p1	6	2
p2	2	0
p3	7	3
p4	3	1
p5	8	4
p6	10	5

Average time = (0+1+3+8+14+21)/6

= 47/6

= 7.833335 ms

Diagram:

Java Code to calulate the average time is for your reference:

import java.util.*;

public class demo{

public static void main(String args[])

{

Scanner sc = new Scanner(System.in);

System.out.println ("enter no of process:");

int n = sc.nextInt();

int pid[] = new int[n];

int at[] = new int[n];

int bt[] = new int[n];

int ct[] = new int[n];

int ta[] = new int[n];

int wt[] = new int[n];  

int f[] = new int[n];

int st=0, tot=0;

float avgwt=0, avgta=0;

for(int i=0;i<n;i++)

{

System.out.println ("enter process " + (i+1) + " arrival time:");

at[i] = sc.nextInt();

System.out.println ("enter process " + (i+1) + " brust time:");

bt[i] = sc.nextInt();

pid[i] = i+1;

f[i] = 0;

}

boolean a = true;

while(true)

{

int c=n, min=999;

if (tot == n)

break;

for (int i=0; i<n; i++)

{

if ((at[i] <= st) && (f[i] == 0) && (bt[i]<min))

{

min=bt[i];

c=i;

}

}

if (c==n)

st++;

else

{

ct[c]=st+bt[c];

st+=bt[c];

ta[c]=ct[c]-at[c];

wt[c]=ta[c]-bt[c];

f[c]=1;

tot++;

}

}

System.out.println("\npid  arrival brust  complete turn waiting");

for(int i=0;i<n;i++)

{

avgwt+= wt[i];

avgta+= ta[i];

System.out.println(pid[i]+"\t"+at[i]+"\t"+bt[i]+"\t"+ct[i]+"\t"+ta[i]+"\t"+wt[i]);

}

System.out.println ("\naverage tat is "+ (float)(avgta/n));

System.out.println ("average wt is "+ (float)(avgwt/n));

sc.close();

}

}

Output: