Test by Bikram | Operating Systems | Test 2 | Question: 22

Question

A uniprocessor system has two resources of type A and B shared by four processes. There are $4$ units of each resource type available after allocation. Let “current” indicate the resources allocated to each process and “Need” indicate the remaining number of resources required by each process for completion. If there exist a safe sequence, then which process will complete last$$\overset{\textbf{Current}}{\begin{array}{|c|c|c|} \hline  & A & B \\ \hline P0 & 2 & 1 \\ \hline P1 & 1 & 2 \\ \hline P2 & 0 & 2 \\ \hline P3 & 1 & 0 \\ \hline  \end{array}} \qquad\overset{\textbf{Need}}{\begin{array}{|c|c|c|} \hline  & A & B \\ \hline P0 & 6 & 2 \\ \hline P1 & 4 & 5 \\ \hline P2 & 3 & 2 \\ \hline P3 & 5 & 6 \\ \hline  \end{array}}$$

• A. $P1$
• B. $P0$
• C. $P3$
• D. No safe sequence exists

Answer 1

There are 3 rules for Safety Algorithm:

1.  we calculate Available resources ( in a Matrix structure )

2. Need <=(less than and equal to) Available 

3. Available =  available + Allocation

so in this question our Allocation matrix and Need matrix are respectively:

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

question says, 

there are 4 units of each resource type is available , so our Available matrix becomes

A  B

4   4  

so start from P0 it have need 6,2 it not satisfy 4,4 which is available. Only P2 satisfy it has 3,2 so our new Available matrix become (4,4 + 0,2) = 4,6 so now P3 can not satisfy it. P0 also can not, P1 have 4,5  it can satisfy.  so now our safe sequence is <P2 , P1> . 

next Available become (4,6 + 1,2)= 5,8 so now P3 comes next , it can satisfy. new Available become 5,8 + 1,0= 6,8 now P0 can satisfy it . so final safe sequence become < P2 , P1. P3, P0 > .

Safe Sequence:  < P2 ,P1 ,P3 ,P0 >  so  P0 will execute last.

Comments

HOW ?

IT MAY B ANY IF WE HAVE 8 RESOURCE OF EACH AVAILABLE.

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@wanted

will describe you tomorrow how it calculates, for today plz see

https://www.cs.jhu.edu/~yairamir/cs418/os4/sld027.htm

http://www.cs.cornell.edu/courses/cs4410/2013su/slides/lecture10.pdf

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yes @wanted, i am solving this problem now.

This question is based on Safety Algorithm. 

To become a safe sequence  the system must be in a safe state.

There are 3 rules for Safety Algorithm:

1.  we calculate Available resources ( in a Matrix structure )

2. Need <=(less than and equal to) Available 

3. Available =  available + Allocation

so in this question our Allocation matrix and Need matrix is respectively:

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

question says, 

there are 4 units of each resource type is available , so our Available matrix become

A  B

4   4  

so start from P₀ it have need 6,2 it not satisfy 4,4 which is available. Only P₂ satisfy it have 3,2 so our new Available matrix become (4,4 + 0,2) = 4,6 so now P₃ can not satisfy it. P₀ also can not, P₁ have 4,5  it can satisfy.  so now our safe sequence is <P2 , P₁> . 

next Available become (4,6 + 1,2)= 5,8 so now P₃ comes next , it can satisfy. new Available become (5,8 + 1,0) = 6,8 now P₀ can satisfy it . so final safe sequence become < P₂ , P₁. P₃, P₀ > HENCE P₀ will complete in last.

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sir ,

please mention in the question that 4 resources of each type are available after allocating it to the respective resources.