Question

Consider the following schedule:
$S:W_1\left(A\right)W_2\left(B\right)W_3\left(B\right)W_4\left(B\right)$
The number of schedule which are view equal to S is _________

• A. 12
• B. 8
• C. 10
• D. 6

Answer 1

The correct option is B 8

For schedule to be view equal, three thing has to be considered:
(a) Initial Read
(b) Updated Read
(c) Final Write
For this schedule, initial read and updated read does not count because no read operation is
considered. So According to final write Transaction $T_4$ should perform write on B finally. W2(B) and W3(B) can be executed in two ways and W1(A) can be executed in four ways.
So total possibilities = 4 * 2 = 8.