Determine resources, need matrix using banker's algorithm, Computer Engineering

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Consider the following system snapshot using data structures in the Banker's algorithm, with resources A, B, C, and D, and process P0 to P4:

                          Max                             Allocation                        Need                             Available

 

P0

A

6

B

0

C

1

D

2

A

4

B

0

C

0

D

1

A    B   C   D

A

B

C   D

P1

1

7

5

0

1

1

0

0

 

 

 

 

P2

2

3

5

6

1

2

5

4

 

 

 

 

P3

1

6

5

3

0

6

3

3

 

 

 

 

P4

1

6

5

6

0

2

1

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

2

1    1

Using Banker's algorithm, answer the following questions.

(a)   How many resources of type A, B, C, and D are there?

(b)   What are the contents of the Need matrix?

(c)   Is the system in a safe state? Why

(d)  If a request from process P4 arrives for additional resources of (1,2,0,0,), can  the  Banker's  algorithm  grant  the  request  immediately? Show the new system state and other criteria.

(a) Resources of type A, B, C, and D are as: A-9; B-13;C-10;D-11

(b) Need[i, j]=Max[i,j]-Allocation[i,j] therefore content of Need matrix is

            A          B          C          D

P0        2          0          1          1

P1        0          6          5          0

P2        1          1          0          2

P3        1          0          2          0

P4        1          4          4          4

(c)  The  system  is  in  a  safe  state  like  the  processes  can  be  complete  in  the sequence P0, P2, P4, P1 and P3.

(d) If a request from process P4 attains for additional resources of (1,2,0,0,), and if such request is granted so the new system state would be in tabulated form asfollows.

                  Max                    Allocation              Need                         Available

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

P0        6    0    1    2         4    0    0    1          2    0    1    1

P1        1    7    5    0         1    1    0    0          0    6    5    0

P2        2    3    5    6         1    2    5    4          1    1    0    2

P3        1    6    5    3         0    6    3    3          1    0    2    0

P4        1    6    5    6         1    4    1    2          0    2    4    4

2    0    1    1

After PO finishes P3 can be assigned. 1020 from released 6012 and also available 2011(Total 80 23) and is a safe sequence.


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