operating system ans

a)      Need matrix

 Need= maximum – allocation

                            0   0   0   0

0         7  5  0

1         0  0  1

1          1  2  1 

5   6   4   0

Currently the system is in safe state since available is ( 1   5   2   2)

Safe sequence

Need of each process is compared with the available resource, if need< available the resources are allocated.

If  need> available next process is compared

In the given problem

P1 is ( 0 0 0 0) and available is ( 1   5   2   2)

Since need  resourse

After allocation process is completed .

Now the available resource is available=available+allocation(which was released by process p0)

= (1,5,2,2)+(0,0,1,1)

=(1,5,3,3)currently available

Now for process p2  the need( 0,7,5,0)<(1,5,3,3)

=è process is allocated

After completion the resources are released

Available = available +allocation

=(1,5,3,3)+(1,0,0,1)==è (2,5,3,4)currently available

Now for process p3 the need( 1    0    0    1)<(2,5,3,4)

=è process is allocated

After completion the resources are released

Available = available +allocation

====(2,5,3,4)+(1,3,5,1)=(3,8,8,5)

Now for process p4 the need(1   1   2   1)<(3,8,8,5)

è process is allocated

After completion the resources are released

Available = available +allocation

=(3,8,8,5)+(0   5   3   1)

=(3,13,11,6)

Now for process p5 the need(5   6   4    0)<(3,13,11,6)

è process is allocated

After completion the resources are released

Available = available +allocation

== (3,  13,11,6)+( 0  0  1   1)

=(3,13,12,7)

Therefore the safe sequence is [p1,p2,p3,p4,p5]

REQUEST FROM P1 CAN BE GRANTED IMMEDIATELY SINCE THE REQUEST IS( 0  3   2   0)

AND THE AVAILABLE IS (1   5  2   2)