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CS604 –Operating System Dear Students, Assignment No.3 has been uploaded on LMS interface. Due date is 21-01-2016. Wish you best of luck!

Assignment No. 03
Semester: Fall 2015

CS604 –Operating System

Total Marks: 20

 

Due date:21-01-2016

Assignment Objective:

  • Implementation of Banker’s algorithm
  • To know algorithm working to check the safe and unsafe state
  • Allocation, available and maximum utilization of resources checking
  • Finding the safe sequence of process execution
  • Uploading instructions:

 

  • Your assignment must be in .doc format (Any other formats like scan images, PDF, bmp, etc will not be accepted).
  • Save your assignment with your ID (e.g. bc020200786.doc).
  • No assignment will be accepted through email.

 

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  • The assignment is submitted after due date.
  • The submitted assignment does not open or file is corrupted.
  • Your assignment is copied from internet, handouts or from any other student

      (Strict disciplinary action will be taken in this case).

 

Assignment

 

 

Question

You have to show the working of Banker’s algorithm in this assignment. Consider four processes P0, P1, P2, P3 and three resource types A, B, C. Resource types A has 9 instances, Resource type B has 6 instances and resource type C has 4 instances. Maximum and allocated resources are given according to the following table.

 

Process

Allocation

Max

Available

A

B

C

A

B

C

A

B

C

P0

1

1

1

8

4

2

 

 

 

P1

3

2

0

4

3

1

 

 

 

P2

2

0

0

5

0

2

 

 

 

P3

2

1

1

7

4

3

 

 

 

 

 

 

You have to give the following:

 

 

  • Give available resources in the table
  • Show execution of the Banker’s algorithm by giving the table step wise
  • Finally find the safe sequence of execution of processes

 

 

 

NOTE: Do not put any query on MDB about this assignment, if you have any query then email at cs604@vu.edu.pk.

 

No need to write theory. Just give tables.

 

 

Deadline: Your assignment must be uploaded/submitted on or before 21-01-2016

 

 

 

 

 

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Please Discuss here about this assignment.Thanks

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

This assignment is not hard to solve but the hard thing is to get the first three values of A, B, C if anyone find these first values so share it because I got A = 1, B = 3 and C = 2 but I still have a doubt please share it don't share whole solution

A=1

B=2

C=2

kindly share the solution plz

plzzzz any one give solution......

ShariqMayo

kya hum just available ki value put krya gya bs ? work wala table to ni bnana

Bro ap ye jo link he es ko dekhen  foran samjh a jae ge jo Shariq bro ne diya hua he.

kindly tell me sol

A=1

B=2

C=2

Question
You have to show the working of Banker’s algorithm in this assignment. Consider four processes P0, P1, P2, P3 and three resource types A, B, C. Resource types A has 9 instances, Resource type B has 6 instances and resource type C has 4 instances. Maximum and allocated resources are given according to the following table.

Process Allocation Max Available
A B C A B C A B C
P0 1 1 1 8 4 2
P1 3 2 0 4 3 1
P2 2 0 0 5 0 2
P3 2 1 1 7 4 3

You have to give the following:

• Give available resources in the table
• Show execution of the Banker’s algorithm by giving the table step wise
• Finally find the safe sequence of execution of processes

Solution:

Process Allocation Max Available
A B C A B C A B C
P0 1 1 1 8 4 2
P1 3 2 0 4 3 1
P2 2 0 0 5 0 2
P3 2 1 1 7 4 3

Need = Max – Allocation

Process Need
A B C
P0 7 3 1
P1 1 1 1
P2 3 0 2
P3 5 3 2

Available = instance - Max

Available (A) = instance (A) – Max (A) => 9 – 8 = 1
Available (B) = instance (B) – Max (B) => 6 – 4 = 2
Available (C) = instance (C) – Max (C) => 4 – 2 = 2

Process Allocation Need Available
A B C A B C A B C
P0 1 1 1 7 3 1 1 2 2
P1 3 2 0 1 1 1
P2 2 0 0 3 0 2
P3 2 1 1 5 3 2

Process Allocation Need Available
A B C A B C A B C
P0 1 1 1 7 3 1 1 2 2
P1 3 2 0 1 1 1 4 4 2
P2 2 0 0 3 0 2
P3 2 1 1 5 3 2

Safe Sequence:

Process Allocation Need Available
A B C A B C A B C
P0 1 1 1 7 3 1 1 2 2
P1 3 2 0 1 1 1 4 4 2
P2 2 0 0 3 0 2 6 4 2
P3 2 1 1 5 3 2

Safe Sequence:

Process Allocation Need Available
A B C A B C A B C
P0 1 1 1 7 3 1 1 2 2
P1 3 2 0 1 1 1 4 4 2
P2 2 0 0 3 0 2 6 4 2
P3 2 1 1 5 3 2 8 5 3

Safe Sequence:

The Safety Algorithm concludes that the system is in safe with < P1, P2, P3, P0 > being the safe sequence.

Final Value of Available:

Available
A B C
9 6 4

is it correct correct and complete any one tell us plzzzzz

instense ki vale kaisy ati hai it means k 9,6,4 ki value kaisy ae hai plz reply me

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