Operating
System- CS604
Assignment
# 2
Solution
Fall 2010
Marks: 20
Due Date
Your assignment must be uploaded before or on 26th Nov 2010.
Objective
The objective of this assignment is to familiarize
with the Scheduling Algorithms
Instructions
Ø
Avoid Plagiarism. No marks will be given in case of cheating
or copying from the internet or from other students.
Ø Submit
the assignment through your account on VULMS. No assignment will be accepted
through email after the due date.
Ø If
you have any problem related to assignment, feel free to discuss it by email at
cs604@vu.edu.pk
Question # 1:
Part a: [10]
Using
Non-preemptive SJF algorithm answer
the following questions using the table below:
|
Process
|
Arrival Time
|
Burst Time
|
|
P2
|
0.0
|
12
|
|
P3
|
3.0
|
8
|
|
P4
|
5.0
|
4
|
|
P1
|
10.0
|
10
|
|
P5
|
12.0
|
6
|
i-
Develop Gantt
showing the execution of the processes in the order mentioned.
ii-
Waiting times for
each process.
iii-
Average waiting
time.
i. Gantt
Chart
|
P2
|
P4
|
P5
|
P3
|
P1
|
0 12 16 22 30 40
ii.
Waiting Time
for P1 = 30-10 = 20
Waiting Time
for P2 = 0-0 = 0
Waiting Time
for P3 = 22-3 = 19
Waiting Time
for P4 = 12-5 = 7
Waiting Time
for P5 = 16-12 = 4
iii.
Average
Waiting Time = (20 + 0 + 19 + 7 + 4) / 5
= 50 / 5
= 10
Part b: 10
Using
Preemptive SJF algorithm answer the
following questions using the same table:
i-
Develop Gantt
showing the execution of the processes in the order mentioned.
ii-
Waiting times for
each process.
iii-
Average waiting
time.
i. Gantt
Chart
|
P2
|
P3
|
P4
|
P3
|
P3
|
P3
|
P5
|
P2
|
P1
|
0 3 5 9 10
12 15 21 30 40
ii.
Waiting Time
for P1 = 40-10-10 = 20
Waiting Time
for P2 = 30-0-12 = 18
Waiting Time
for P3 = 15-3-8 = 4
Waiting Time
for P4 = 9-5-4 = 0
Waiting Time
for P5 = 21-12-6 = 3
iii.
Average
Waiting Time = (20 + 18 + 4 + 0 + 3) / 5
= 45 / 5
= 9
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