Question

Part B (10) Deadlock Avoidance Consider the following maximum-claim reusable resource system with four processes and three resource types. The maximum claim matrix is given by C [4 3 5 11 1 41 1 4 6 13 1 6] where Cij denote maximum claim of process i for resourcej. The total units of each resource type are given by the vector (5, 8, 15). The current allocation of resources is given by the matrix To 1 4] 2 0 1 1 2 1 (1 03] where Aij denotes the units of resources of type j currently allocated to process į. The current state of the system is safe. The processes and resources are both indexed from 0. Determine if a request by process 1 for 2 units of resource 2 can be safely granted. Note: If you are using Microsoft Word to edit this file, you can copy paste one of the above matrices and modify it, rather than creating it from scratch.

Answer 1

Consider processes as P0, P1, P2, P3, and resources as R0, R1, R2.

Given 2 Matrices Maximum Matrix C which shows the maximum number of resources of each type required by each process during its execution and the Allocation matrix A shows the current number of resources of each type allocated to each process

Resources available are [R0 R1 R2]=[5 8 15]

Need Matrix=Maximum Matrix-Allocation Matrix

Now we have,

	Maximum Matrix ( C)			Allocation Matrix (A)			Need Matrix
	R0	R1	R2	R0	R1	R2	R0	R1	R2
P0	4	1	4	0	1	4	4	0	0
P1	3	1	4	2	0	1	1	1	3
P2	5	6	13	1	2	1	4	4	12
P3	1	1	6	1	0	3	0	1	3

and by the mentioned in the question, the system is in a safe state for now.

The processes and resources are indexed from 0.

Checking whether a request by process p1 for 2 units of resource 2 can be safely granted or not:

P1 requests [R0 R1 R2]=[0 0 2] i.e., process 1 requests 2 units of resource 2.

Available Resources [R0 R1 R2]=[5 8 15]

Clearly,

The instances available currently, the requirements or requests of process1 P1 can be satisfied.

So Bankers algorithm assumes that the request from process 1 is entertained and then it modifies the matrix as

Note: Modified data is represented in Bold

	Maximum Matrix ( C)			Allocation Matrix (A)			Need Matrix
	R0	R1	R2	R0	R1	R2	R0	R1	R2
P0	4	1	4	0	1	4	4	0	0
P1	3	1	4	2	0	3	1	1	1
P2	5	6	13	1	2	1	4	4	12
P3	1	1	6	1	0	3	0	1	3

Now Available resources are

[5 8 15] - [0 0 2] = [5 8 13]

Now, it follows the safety algorithm to check whether this resulting state is a safe state or not.

if it is a safe state, then the request can be safely permitted otherwise not.

Step 1:

• With the instances available currently, the requirement of the process P0 can be satisfied.
• So, process P0 is allocated to the requested resources.
• It completes its execution and then frees up the instances of resources held by it.

Then available instances are

= [5 8 13] + [0 1 4]

= [5 9 17]

Step 2:

• With the instances available currently, the requirement of the process P1 can be satisfied.
• So, process P1 is allocated to the requested resources.
• It completes its execution and then frees up the instances of resources held by it.

Then available instances are

= [5 9 17] + [2 0 3]

= [7 9 20]

Step 3:

• With the instances available currently, the requirement of the process P2 can be satisfied.
• So, process P2 is allocated to the requested resources.
• It completes its execution and then frees up the instances of resources held by it.

Then available instances are

= [7 9 20] + [1 2 1]

= [8 11 21]

Step 4:

• With the instances available currently, the requirement of the process P3 can be satisfied.
• So, process P3 is allocated to the requested resources.
• It completes its execution and then frees up the instances of resources held by it.

Then available instances are

= [8 11 21] + [1 0 3]

= [9 11 24]

Thus ,

There exists a safe sequence P0-->P1-->p2-->P3 in which all processes are executed.

So,The system is in a safe state.

Thus, The request by process 1 can be safely granted.