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For having two schedules as view equivalent (hence view serializable),  we have to check for three conditions to be true in both schedules :

1.Initial Read

2.Write – Read pair

3.Final Write 

Here in this question as T3 reads A first so in every view equivalent schedule R3(A) remains same . Similarly T1 makes final write on A  so W1(A) remains same.

Now considering W-R pair since T1 reads value of  A written by T2,hence every view equivalent schedule must have same order .Here we can swap W3(A) and W2(A) means we can perform W3(A) before W2(A) and R1(A).

So one more possible view serializable schedule of the schedule S is R3(A), W3(A), W2(A), R1(A), W1(A);

Answer is one 

 

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pranjalgennext asked Jan 26, 2017
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The number of view equal serial schedules possible for the given schedule are:S: w1(A) r2(A) w3(A) r4(A) w5(A) r6(A) w7(A) r8(A)