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Consider a schedule
S: r1(A); W1(B); W1(C); r2(A); W2(B); W2(C); abort2; r3(A); W3(B); W3(C); C1; C3 

How many conflict equivalent serial schedules are possible for the given schedule ?

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$T1\rightarrow t2\rightarrow t3$

Only one such conflict serial schedule is possible and conflict equal serial schedule is the topological order of the precedence graph.

3 Comments

actually this answer is not given.
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answer is given 3.
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But how 3 ??
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@resilient. T2 is performing abort in between. So, we can consider the non-conflicting operations only before 'abort' operatioon.

1st schedule: Given serial schedule

2nd schedule: W1(c) and W2(B) can be swapped to produce conflict equivalent schedule.

So, I am getting only 2 schedules.

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