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Is is true that...If there is deadlock then system can not progress?
in Operating System
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yes if a system in deadlock then no need to check bounded waiting as well as progress. progress said that if one process its not interested than other process definiltly goes to critical section but here its not going another process also .so its fail to progress condition.
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But there could be bounded waiting..
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@Hardesh Patel...deadlock implies no progression but deadlock has no relation with bounded waiting..
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But, Does deadlock implies no bounded waiting ?

I think No because bounded waiting is per process and deadlock is for the system.

Moreover, In a deadlock, bounded waiting conditon is not violated and bounded waiting is not with respect to time, but with respect to the number of times the processes enter the critical section before intended process enters the critical system .
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1 Answer

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Best answer

If u look at the definition of progress , quoted from Galvin , it says :

If no process is executing in its critical section and some
processes wish to enter their critical sections, then only those processes
that are not executing in their remainder sections can participate in
deciding which will enter its critical section next, and this selection cannot
be postponed indefinitely.
 

So this at least one of the process willing to go into critical section  should be allowed to do so provided the critical section is empty at that instant..

Also we know deadlock is the scenario where all processes will get blocked and hence unable to go into the critical section..So it violates the progress condition required fr proper synchronisation construct..

So 

Deadlock  ==> Progress Violation

But converse is not necessarily true..

By converse I mean some specific process is not allowed to enter into CS even though it is empty , thus violating the progress condition..But the other process may go into CS as in the case of Dekker's Algorithm..Thus it is not deadlock..

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4 Comments

Since no process can enter cs and if we consider this as a function, then f(n)=0 and so there is a bound, so bounded waiting may be satisfied, I think.
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https://gateoverflow.in/74002/doubt-in-some-basic-things-synchronisation

I have read this question, and the conclusion i am getting is that it is not necessary that every 2 process soln satisfy BW, as in the while loop process P0 is cont checking the cond when P1 is in CS. even that no of cond testing is finite but not bound. Hence it is not satisfying BW. Please check it whether my conclusion is true or not.

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Is this solution right?
 

 

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