Solution

The correct answer is Only S1 is conflict-serializable

Concept:

• Using precedence graph method, we can find out Schedule is conflict serializable or not.
• If precedence graph has any loop then that schedule is going to be not conflict-serializable.
• If precedence graph doesn’t have any loop then that schedule is going to be conflict-serializable.

Explanation:

Schedule S1:

T1	T2	T3
r(X)
		r(Y)
		r(X)
	r(Y)
	r(Z)
		w(Y)
		w(Z)
r(Z)
w(X)
w(Z)

 

Precedence graph of S1:

Since there is no loop in S1, it is is conflict serializable

Order: T2 → T1 → T2

Schedule S2: 

r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z) 

T1	T2	T3
r(X)
		r(Y)
	r(Y)
		r(X)
r(Z)
	r(Z)
		w(Y)
w(X)
	w(Z)
w(Z)

 

Precedence graph of S2:

• So there is loop in S2 so S2 is not conflict serializable.
• Schedule S1 is equivalent to only this serial schedule T2 : T3 : T1.

Hence option 1 is the correct answer.