UGC NET CSE | June 2012 | Part 3 | Question: 65

Question

Consider the methods used by processes $P_1$ and $P_2$ for accessing their critical sections.  The initial values of shared Boolean variables $S_1$ and $S_2$ are randomly assigned,

$P_1$ while ($S_1$==$S_2$); critical section $S_1$=$S_2$;

$P_2$ while ($S_1$==S2); critical section $S_1$=$S_2$;

Which one of the following statements describes the properties achieved?

• A. Mutual exclusion but not progress
• B. Progress but not mutual exclusion
• C. Neither mutual exclusion nor progress
• D. Both mutual exclusion and progress

Comments

Answer should be A

Here also not bound and waiting

Answer 1

OPTION B: PROGRESS BUT NOT MUTUAL EXECUTION

WHY NOT ME:

SUPPOSE S1=S2=5

1.NOW LET P1 EXECUTED THE WHILE LOOP WHICH IS TRUE  AND PREEMPTED

2.THEN P2 COMES AND EXECUTES THE WHILE LOOP WHICH IS TRUE SO ACCESSING THE CRITICAL SECTION 

3. THEN P1 COMES AND ENTERS CRITICAL SECTION

BOTH PROCESS IN CRITICAL SECTION  THEREFORE NOT ME

BUT PROGRESS IS THERE:

SUPPOSE  P1 IS NOT INTRESTED IN CRITICAL SECTION SO IT WILL NOT EXECUTE THE CODE..AND P2 WILL EXECUTE ITS CODE ...SINCE THERE IS NO DEPENDENCY ON P1(S1,S2 VALUES WILL NOT DEPENDS UP ON P1'S EXECUTION) SO P2 CAN EXECUTE ANY NUMBER OF TIMES IT WANT

Comments

is progress guaranteed  if S1=S2 the both P1,  P2  will execute infinitely preventing them to  enterinto  C.S and no progress is possible.

and off ans  was  also C

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I think Answer should be Option 3

Why Not ME

S1=0 ,S2=1

first p1 will check While condition and preempt and then s2 will check and enter into C.S. and preempt and p1 will load and enter into C.S.

Why Not Progress

S1=S2 then P1 and P2 both Are Deadlock 

and we Know if there is deadlock, then no progress.

 

Correct me If i am wrong.....