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OS LAB EVEN SEM 24-25

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A single processor system has 3

resource types X,Y,Z.

 there are 5 units of each resource type.

 Consider the scenario where column allocation

denotes the number of units of each resource types allocated to each process.

And the column request denote the number of units of each resource type

requested by a process in order to complete execution .  state which process finish last?

 

Allocation

Request

 

X

Y

Z

X

Y

Z

P1

1

2

1

1

0

3

P2

2

0

1

0

1

2

P3

2

2

1

1

2

0

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 let there be 5 processes and 3 resources type A,B,C

Resource type A has 10 instances.

Resource type B has 5 instances.

Resource type C has 7 instances.

Consider the snapshot at time T0

 

Allocation

Max

Available

P1

0

1

0

7

5

3

3

3

2

P2

2

0

0

3

2

2

 

 

 

P3

3

0

2

9

0

2

 

 

 

P4

2

1

1

2

8

2

 

 

 

P5

0

0

2

4

3

3

 

 

 

 

Which of the following statement is true

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Semaphore S=1, there are 5 processes po,p1,p2,p3,p4 execute the following code

p0 to p3 execute 

while (1)

   P(S)

   <CS.>

   V(S)

}

p4 execute 

   {P(S)

   <CS.>

   V(S)}

How many maximum number of processes  can be present in critical section at any point of time?

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Binary semaphore S1=1  and S2=1

int x=10

p0()

{

          p(s1)

          p(s2)

          x=x+1

         v(s2)

         v(s1)

}

                 ------------                  p1()

{

          p(s2)

          p(s1)

           x=x+1

       v(s2)

       v(s1)

}

what will be the value of x  after execution of p0 and p1?

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Processes p1,p2,p3,p4,p5  share a semaphore variable mutex, initialized to 1. 

Suppose a processes  executes in the following manner :

signal(mutex);

.....

critical section

.....

wait(mutex);

In this situation :

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Semaphore S=1, there are 5 processes po,p1,p2,p3,p4 execute the following code

p0 to p3 execute 

while (1)

   P(S)

   <CS.>

   V(S)

}

                             

p4 execute 

   {V(S)

   <CS.>

   V(S)}

How many maximum number of processes  can be present in critical section at any point of time?

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Consider the methods used by processes P1 and P2 for accessing their

critical sections whenever needed, as given below. The initial values of

shared boolean variables S1 and S2 are  0.

Method used by P1 :

while(S1==S2);

Critical section

S1 = S2;

 

Method used by P2 :

while(S1!=S2);

Critical section

S2 = not(S1);

which process will execute first? the above construct  fails to provide ---

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What is value of binary  semaphore "S" after executing 10

P (Wait) operations if the initial value of "S" is 1? how many process will be blocked?

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boolean blocked[2];

int turn;

void P(int id) {

while(true) {

blocked[id] = true;

while(turn != id) {

while(blocked[1-id])

/* do nothing */;

turn = id;

}

/* critical section */

blocked[id] = false;

/* remainder */

}

}

void main() {

blocked[0] = false;

blocked[1] = false;

turn = 0;

parbegin(P(0), P(1)); //RUN P0 and P1 parallel

The  above code provide satisfies progress   or not?

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Binary semaphore S1=1  and S2=0

int x=10

p0()

{

          p(s1)

          x=x+1

       v(s2)

}

        ---------------        p1()

{

          p(s2)

          x=x+1

       v(s1)

}

what will be the value of x  after execution of p0 and p1?

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