Question

Fifteen processes share a critical section implemented by using a binary semaphore named x initialized to 1. Eleven of these processes use the correct code wait(x); critical section; signal(x). However, four processes erroneously use the code signal(x); critical section; signal(x). Assume that each process executes its critical section only once.) Answer the following questions. Calculate the maximum number of processes that can be in the critical section at the same time, and explain your answer. Does your answer change if x is a counting semaphore? Explain.

Answer 1

Check the explanation

Step-by-step explanation:

Given data:

INFORMATION THAT WE HAVE:

Number of Process:15

Binary Semaphore X=1

wait(X): It decrease X value by 1

wait operation can only be done when X>0

Signal(X): It increase X value by 1

Signal operation can be done when X<1

11 process use:(P1-P11)

wait(X)

CS

signal(X)

4 processes use:(P12-P15)

signal(X)

CS

signal(X)

We can conclude that:

As it is binary semaphore,it value can only be 1 or 0.

Process can enter into CS by making X=0 and after completion of CS task makes X=1.

In case of 11 processes,at a time only 1 process can enter into CS.

Let's say process P9 enter into CS by changing X=0.

A,first instruction of P12 is signal(X) then it can also enter into CS by changing X=1.As signal operation can be performed when X=0 but wait operation cannot be performed.

So,now in CS there is two process: P9 and P12

First instruction of P13 is also signal(X).If it is going to try make X++ it can not do because it is binary semaphore and already X=1

P9 cannot also do signal operation.So it's kind of deadlock.

a)

In critical section,at the same time at most 2 processes can be present.

And now we consider that this is Counting semaphore

So,We can conclude that:

Counting Semaphore X=1

wait(X): It decrease X value by 1

wait operation can only be done when X>0

Signal(X): It increase X value by 1

Signal operation can be done at any value of X

Let's say process P1 enter into CS by changing X=0.

A,first instruction of P12 is signal(X) then it can also enter into CS by changing X=1.As signal operation can be performed when X>=0 but wait operation cannot be performed.

P13 can also enter by changing X=2 [ First operation is signal(X) ]

P14 can also enter by changing X=3 [ First operation is signal(X) ]

P15 can also enter by changing X=4 [ First operation is signal(X) ]

In the question , every process can execute critical section only once is mentioned

P2 can enter by changing X=3 [First operation is wait(X) ]

P3 can enter by changing X=2 [First operation is wait(X) ]

P4 can enter by changing X=1 [First operation is wait(X) ]

P5 can enter by changing X=0 [First operation is wait(X) ]

So,now available process in CS :(P1,P2,P3,P4,P5,P12,P13,P14,P15)

B)

In serious section, at the same time when nine processes can be present at most, that is if every process can execute CS only once.

* in case there is no situation whereby every process can execute CS only once then at a time at most all the processes can be present.