Question

Consider the three transactions T1, T2, and T3, and the schedules S5 and S6 given below. Show all conflicts and draw the serializability (precedence) graphs for S5 and S6, and state whether each schedule is serializable or not. If a schedule is serializable, write down the equivalent serial schedule(s).

T1: r1(x); r1(z); w1(x);
T2: r2(z); r2(y); w2(z); w2(y);

T3: r3(x); r3(y); w3(y);

S5: r1(X); r2(Z); r1(Z); r3(X); r3(Y); w1(X); c1; w3(Y); c3; r2(Y); w2(Z); w2(Y); c2;

S6: r1(X); r2(Z); r1(Z); r3(X); r3(Y); w1(X); w3(Y); r2(Y); w2(Z); w2(Y); c1; c2; c3;

Answer 1

SOLUTION-

Given,

T1: r1(x); r1(z); w1(x);

T2: r2(z); r2(y); w2(z); w2(y);

T3: r3(x); r3(y); w3(y);

S5: r1(X); r2(Z); r1(Z); r3(X); r3(Y); w1(X); c1; w3(Y); c3; r2(Y); w2(Z); w2(Y); c2;

S6: r1(X); r2(Z); r1(Z); r3(X); r3(Y); w1(X); w3(Y); r2(Y); w2(Z); w2(Y); c1; c2; c3;

For S5 -

For S6 -

• First we found all the conflicting instructions.
• Conflicting instructions are those instructions which work on different transactions and wok on the same data value and one of them is write operation.
• Then we found the serializability of the schedules. Both the schedules are serializable as there is no cycle in the graph.
• Then at last we draw the equivalent serial schedule.

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