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Banker’s Algorithm:


Assumptions

1) Multiple instances of resources

2) Each process must claim the maximum use a priori

3) When a process requests a resource it may have to wait

4) When a process gets all its resources it must return them in a finite amount of time

Consists of safety algorithm and resource request algorithm.


Data Structures for the Banker's algorithm


Let n = number of processes and m = number ofresourcetypes
1)Available: vector of length m. 
If available [j] k, k instances of resource type are available.
2)Max: n x m matrix
If Max[i, j] = k, 
process  may request at most k in instances of resources types  
3)Allocation: n x m matrix
If Need[i, j] = k,
then Pi may need k more instances of  to complete its task
Need[i, j] =Max[i, j] - Allocation[i, j].



Banker's algorithm- safety procedure


1) Let Work and Finish be vectors of length m and n, respectively. Initialize, 

                 Work = Available

                  Finish [i] = false for i = 0, 1,..., n-1

2) Find process i such that:

       a) Finish [i] == false; and

       b)  ≤ Work

       If no such i exists, go to step 4 

3) Work = work +   ≤ Work

      Finish [i] = true

      go to step 2

4) If Finish [i] = = true for all i, then the system is in a safe state. Otherwise, it is in an unsafe state.


Banker's Algorithm- Resource Request algorithm for process  


Request = request vector for process. If [j] = k then process Pi wants k instances of resource type 

1) If  <= go to step 2. Otherwise, raise error condition, since process has exceeded its maximum claim.

2) If <= Available, go to step 3. Otherwise  must wait, since resources are not available.

3) Pretend to allocate requested resources to  by modifying the state as follows: 

Available = Avalable – ;

 = + ;

 = – ;


Call safety algorithm

  1. If sae => the resources are allocated to .
  2. If unsafe =>  must wait, and the old resource- allocation state is restored


Example of Banker's Algorithm


5 process P0 through P4; 3 resource type A(10 instances), B(5 instances), and C (7 instances) and Snapshot at time T0:


Allocation
Max
Available

A
B
C
A
B
C
A
B
C
P0
0
1
0
7
5
3
3
3
2
P1
2
0
0
3
2
2



P2
3
0
2
9
0
2



P3
2
1
1
2
2
2



P4
0
0
2
4
3
3




The content of the matrix. Need is defined to be Max- Allocation.

Need


A
B
C
P0
7
4
3
P1
1
2
2
P2
6
0
0
P3
0
1
1
P4
4
3
1


The system is in a safe state since the sequence < P1, P3, P4, P2, P0 satisfies safety criteria.


P1 request (1, 0, 2) 

- Check that request <= Available (that is, (1, 0, 2) <= (3, 3, 2)) => true


Allocation
Need
Available

A
B
C
A
B
C
A
B
C
P0
0
1
0
7
4
3
2
3
0
P1
3
0
2
0
2
0



P2
3
0
1
6
0
0



P3
2
1
1
0
1
1



P4
0
0
2
4
3
1




- Executive safety algorithm shows that sequence < P1, P3, P4, P0, P2> satisfies safety requirement.



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