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Total View Serializable schedules will be 6. (Refer: https://gateoverflow.in/13348/solve)

These 6 are:

T1->T2 T3->T4 T5->T6 1st order

T1->T2 T5->T6 T3->T4  2nd order 

T5->T6 T3->T4 T1->T2   3rd order 

T3->T4 T5->T6 T1->T2   4th order 

T5->T6 T1->T2 T3->T4   5th order 

T3->T4 T1->T2 T5->T6   6th order 

Out of these, except for the 1st order, none of the others are conflict equivalent to the given schedule. This is because the given schedule has conflicting Read-Write operations successively and hence it can't be transformed into another conflict equivalent schedule via swapping.

Therefore, answer should be 5.

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