Question

Consider the following snapshot of a system:

Allocation                   Max                 Available

A B C D              A B C D                    A B C D

P0                    0 0 1 2                0 0 1 2                      1 5 2 0

P1                    1 0 0 0                1 7 5 0

P2                    1 3 5 4                2 3 5 6

P3                    0 6 3 2                0 6 5 2

P4                   0 0 1 4               0 6 5 6

Answer the following questions using the banker’s algorithm:

a. What is the content of the matrix Need?

b. Is the system in a safe state? Demonstrate the reason for your answer.

c. If a request from process P1 arrives for (0,4,2,0), can the request be granted immediately? Demonstrate the reason for your answer.

IN TEXT ONLY!!!

Answer 1

Allocated Max

	A	B	C	D		A	B	C	D
P0	0	0	1	2		0	0	1	2
P1	1	0	0	0		1	7	5	0
P2	1	3	5	4		2	3	5	6
P3	0	6	3	2		0	6	5	2
P4	0	0	1	4		0	6	5	6

Available = ( 1 5 2 0 )

a) Need = Max - Allocated

Need

	A	B	C	D
P0	0	0	0	0
P1	0	7	5	0
P2	1	0	0	2
P3	0	0	2	0
P4	0	6	4	2

b) Available = 1 5 2 0

P0 allocated = 0 0 1 2 (P0 releases all the resources allocated to it)

Available = 1 5 3 2

P2 allocated = 1 3 5 4

Available = 2 8 8 6

P1 allocated = 1 0 0 0

Available = 3 8 8 6

P3 allocated = 0 6 3 2

Available = 3 14 11 8

P4 allocated = 0 0 1 4

Available =3 14 12 12

All processes are served successfully so System is in a safe state.

c) If P1 is allocated with ( 0 4 2 0 ) then Allocated and need matrices, Available vector will be changed.

Allocated Need

	A	B	C	D		A	B	C	D
P0	0	0	1	2		0	0	0	0
P1	1	4	2	0		0	3	3	0
P2	1	3	5	4		1	0	0	2
P3	0	6	3	2		0	0	2	0
P4	0	0	1	4		0	6	4	2

Available = ( 1 1 0 0 )

P0 allocated = 0 0 1 2

Available = 1 1 1 2

P2 allocated = 1 3 5 4

Avaiable = 2 4 6 6

P1 allocated = 1 4 2 0

Available = 3 8 8 6

P3 allocated = 0 6 3 2

Available = 3 14 11 8

P4 allocated = 0 0 1 4

Avaialable = 3 14 12 12

All processes are served successfully So system is safe if we serve P1. So request can be granted.