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Concurrent processes and shared variables

explain

4Comments
Gopi Shankar @groovyg1
12 Nov 2014 05:54 am

Here in each of the option except option 2 we are able to see a cycle, for instance in option 1 we have X engaging resource a -> resource c & Z engaging resource c -> resource a. Similar thing happens in option 3 and 4 as well.

Gopi Shankar @groovyg1
12 Nov 2014 05:58 am

One can try enumerating each resource in some priority order as well, if sharing of resource involve more than two processes, to unfold the cycle.

Vivek Vikram Singh @vivek14
12 Nov 2014 01:08 pm

Answer would be B. 

In A, in 1st P() operation, X ,Y and Z all will be proceeding by decrementing a from 1 to 0, b from 1 to 0 and c from 1 to 0. Current values of a,b and c are 0  and d=1.

in 2nd P() operation in all processes, X and Y will be stuck as b and c are having 0. The only process ,which will be proceeding , is Z.values of a,b,c,d=0.

in 3rd and final operation for Z, Z is doing wait operation on a, which is having value 0. So Z will also be  waiting. So this case all processes are deadlocked. option A is eliminated. This way you can eliminate all other options C and D as well.

Lets look at the option B.

In 1st set of P() operations of all processes, X and Y doing signal operation on b.Lets assume each case separately.

1) Suppose X proceed first while Y get stuck in waiting. Z will be proceeding too as it can independently execute signal on a. So as of now we have 2 progressing process X and Z. values of a and b is 0 whereas c and d are having values 1.

in 2nd set of P() operation, X will be stuck as it is doing P() operation on a, which is having value 0,while Z can move further as c had value of 1 which will be decremented to 0. Current values of a,b,c is 0 and d=1.

in 3rd set of P() operations, Z will be decrement the value of d from 1 to 0 and goes to Critical Section. It is a deadlock free operation. 

Got the Answer? Yes??? Nooooo. You are still missing one more case on the same option with Y proceeding instead of X.

2) in 1st set of operations , suppose this time Y gets proceed instead of X. X is stuck and Y and Z will move to next set of P() operations. values of a,b is 0 and c,d is 1.

in 2nd set of P(), either process can go ahead by decrementing value of c from 1 to 0 and in next set of P() operations, it can decrement value of d from 1 to 0 and enter critical section.

So in both the cases, this option B lead to Deadlock free operation.

Parimal Andhalkar @parimal_andhalkar
13 Nov 2014 09:09 am

got it thanks ...........