ME question

Question

Consider the following transactions:

The number schedules of T1 and T2 recoverable ?

(ANS GIVEN IS 20 )

Comments

exactly!! i am also getting same.

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@Shubhanshu

can you explain how?

 

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Recoverable = Total - Irrecoverable.

Total = $\frac{7!}{3!2!} = 35$

and Irrecoverable schedule are 6.

So, Recoverable = $35-6=29.$

To, find irrecoverable remember two conditions those are:-

1. The schedule should have at least 1 dirty read.

2. The transaction which has dirty read must commit before the transaction due to which its read become dirty.

In above $T_2.r(B)$ is a dirty read because of $T_1.w(B)$ apply some combinations and then you will get the irrecoverable schedule as 6.

Hence, 29 is the correct answer.