Consider the program: var s: int:= 1, i: int co i:= 1 to 2 rightarrow do true ri

Question

Consider the program: var s: int:= 1, i: int co i:= 1 to 2 rightarrow do true rightarrow (await s > 0 rightarrow s: = s - 1) (s:= s + 1) Above, S_i is a statement list that is assumed not to modify shared variable s. Develop complete proof outlines for the two processes. Demonstrate that the proofs of the processes are interference-free. Then use the proof outlines and the method of Exclusion of Configurations (2.25) to show that S_1 and S_2 cannot execute at the same time and that the program is deadlock-free. What scheduling policy is required to ensure that a process delayed at its first await statement will eventually be able to proceed? Explain.

Explanation / Answer

int i=1

do(i=1 to 2)

{

wait(s>0){

s=s-1

Si

}

s=s+1

}

to prove that it should be deadlock free it should support mutual exclusion,progress,bounded waiting.

mutual exclusion:-

p1:

int i=1

do(i=1 to 2) //entry section

{

wait(s>0){ //critical section

s=s-1

Si

}

s=s+1

} //exit section

so while p1 is exexcuting p2 while s enters into critical section as we don't update any values so mutual exclusion is possible and can be formed.

progress:-

when p1 is in non-critical section p2can execute in critical section.program gurantees progress.

bounded-waiting :-

suitable for ony countable no of procedures and processes.

as it satisfies all properties ....so it's deadlock free.

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