Question

The following program consists of 3 concurrent processes and 3 binary semaphores. The semaphores are initialized as S0=1, S1=0, S2=0.

Process P0	Process P1	Process P2
While (true) { Wait (S0); Print ‘0’ Release (S1); Release (S2); }	Wait (S1); Release (S0);	Wait (S2); Release (S0);

 

How many times will process P0 print ‘0’? 

1. At least twice
2. Exactly twice
3. Exactly thrice
4. Exactly once

Answer 1

Correct Answer - Option 1 : At least twice

The correct answer is option 1

Concept:

Binary semaphore which can take only two values 0 and 1 and ensure mutual exclusion.

A process is entered into the critical section only when the value of semaphore(s) = 1

Otherwise, it will wait until semaphore(s) >0

Explanation:

The semaphores are initialized as S0=1, S1=0, S2=0.

Because S0 =1 then P0 enter into the critical section and other processes will wait until either S1=1 or S2 =1

The minimum number of times 0 printed:

• S0 =1 then P0 enter into the critical section
• print '0'
• then release S1 and S2 means S1 =1 and s2 =1
• now either P1 or P2 can enter into the critical section
• if P1 enter into the critical section
• release S0
• then P2 enter into the critical section
• release S0
• P1 enter into the critical section
• print '0'

The minimum number of time 0 printed is twice when executing in this order (p0 -> p1 -> p2 -> p0)

The Maximum number of times 0 printed:

• S0 =1 then P0 enter into the critical section
• print '0'
• Then release S1 and S2 means S1 =1 and s2 =1
• Now either P1 or P2 can enter into the critical section
• If P1 enter into the critical section
• Release S0 means S0 =1
• S0 =1 then P0 enter into the critical section
• print '0'
• Then P2 enter into the critical section
• Release S0 means S0 =1
• S0 =1 then P0 enter into the critical section
• print '0'

Maximum no. of time 0 printed is thrice when execute in this order (p0 -> p1 -> p0 -> p2 -> p0)

So, At least twice will process P0 print ‘0’