Consider the following non serial schedule

$S : W_{4} (E), R_{1} (C), R_{3} (A), R_{1} (E), W_{2} (A), R_{4} (A), W_{3} (A), R_{3} (B), W_{4} (A), W_{2} (D), W_{1} (A), R_{4} (C), W_{1} (B), R_{4} (D)$

Which of the following is true statement?

Open in App

Solution

Precedence graph:

As the precedence graph has cycle and the schedule is not conflict serializable $\begin{matrix} T_{1} & T_{2} & T_{3} & T_{4} W(E) R(C) R(A) R(E) W(A) R(A) W(A) R(B) W(A) W(D) W(A) R(C) W(B) R(D) \end{matrix}$

The schedule is to be view serializable

$R_{3} (A)$ occur before $W_{2} (A)$

$W_{2} (A)$ occur before $R_{4} (A)$

$R_{4} (A)$ occur before $W_{3} (A)$

Which causes a cycle from $T_{3} - T_{2} - T_{4} - T_{3},$ and the schedule is not view serializable.

Suggest Corrections

Similar questions

S : R_{1} (A), R_{3} (B), W_{1} (A), W_{2} (A), W_{2} (B), W_{3} (B)

R_{x} (Z) denotes Trans X Reads item

Z, W_{x} (Z) denotes Trans X writes item Z.

S_{1} : R_{1} (A), W_{1} (A), R_{2} (A), W_{2} (A), Commit T_{2}, R_{1} (B), W_{1} (B

S_{2} : R_{1} (A), R_{2} (A), W_{2} (A), Commit T_{2}, R_{1} (A), W_{1} (A), Commit

S_{3} : R_{2} (A), R_{1} (A), W_{1} (A), W_{2} (A), W_{2} (B) Commit T_{2}, W_{1} (B)

S : R_{1} (A), R_{4} (B), W_{3} (A), W_{2} (B), W_{4} (C), c o m m i t_{4}, W_{3} (C), W_{1} (A), c o m m i t 1, R_{2} (D), W_{3} (D),

Join BYJU'S Learning Program