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Assignment No. 03 CS604 –Operating System 
Total Marks: 20
Due date:21012016 

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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.
You have to give the following:
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 21012016 

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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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