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Comprehension

Consider the following set of processes with the arrival time and length of CPU burst time given in milli secends (ms)
Process Arrival Time Burst Time
P1 0 5
P2 2 3
P3 2 2
P4 5 3
P5 6 1

What is the average waiting time for these processes with non-preemptive Shortest Job First (SJF) scheduling Algorithm?

  1. 3.0 ms
  2. 2.1 ms
  3. 4.6 ms
  4. 3.2 ms

Answer (Detailed Solution Below)

Option 4 : 3.2 ms

Detailed Solution

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Scheduling Algorithm: Non-preemptive Shortest Job First (SJF)

Gantt chart:

P₁

P₃

P₅

P₂

P₄

0       5       7       8       11       14

Process Table:

Process

Arrival Time (AT)

Burst Time (BT)

Completion Time (CT)

Turn Around Time (TAT)

Waiting Time (WT)

P₁

0

5

5

5

0

P₃

2

2

7

5

3

P₅

6

1

8

2

1

P₂

2

3

11

9

6

P₄

5

3

14

9

6

 

Average waiting time = \(\frac{{0 + 3 + 1 + 6 + 6}}{5} = 3.2\)

Important Points:

WT = TAT - BT

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More Process Scheduling Questions

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