Consider the following transaction schedules S1 and S2

Choose the correct option thatholds for the schedules S1 and S2.

S1 is non-serializable, S2 is serializable

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Both S1 and S2 are non-serializable

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S1 is serializable, S2 is non serializable

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Both S1 and S2 are serializable

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Solution

The correct option is C S1 is serializable, S2 is non serializable

schedule S1: r1(A) w1(A) r2(B) w2(B) r1(B)

swap r2(B) and w1(A): r1(A) r2(B) w1(A) w2(B) r1(B)

swap r2(B) and r1(A): r2(B) r1(A) w1(A) w2(B) r1(B)

swap w2(B) and w1(A): r2(B) r1(A) w2(B) w1(A) r1(B)

swap w2(B) and r1(A): r2(B) w2(B) r1(A) w1(A) r1(B)

schedule S2: r1(A) w1(A) r2(A) w2(A) r1(A)

swapping r1(A) and w2(A) is not possible because read and write in two different transactions occur on the same database (A).

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Similar questions

Q. Consider the following schedules involving three transaction :

S_{1} : r_{1} (x), r_{2} (z), r_{1} (z), r_{3} (x), r_{3} (y), w_{1} (x), w_{3} (y), r_{2} (y), w_{2} (z), w_{2} (y)

S_{2} : r_{1} (x), r_{2} (z), r_{3} (x), r_{1} (z), r_{2} (y), r_{3} (y), w_{1} (x), w_{2} (z), w_{3} (y), w_{2} (y)

Which of the following schedules is conflict serializable?

Q. In the following schedules,

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

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

Consider the following schedules:

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)

Which of the following statements are true?

Consider the methods used by processes P1 and P2 for accessing their critical sections whenever needed, as given below. The initial values of shared boolean variables S1 and S2 are randomly assigned.

Method used by P1	Method used by P2
while (S1 == S2); Critica1 Section S1 = S2;	while (S1 != S2); Critica1 Section S2 = not (S1);

Which one of the following statements describes the properties achieved?

Q. Consider the following C program:
#include <stido.h>
void fun 1 (char *s1, char *s2) {
char *tmp
tmp = s1;
s1 = s2;
s2 = tmp;
}
void fun2 (char **s1, char **s2) {
char *tmp;
tmp = * s1;
*s1 = *s2;
*s2 = tmp;
}
int main ( ) {
char * str1 = "Hi", *str2 = "Bye";
fun1 (str1, str2); printf("%s, %s", str1, str2);
fun2 (&str1, &str2);printf("%s %s", str1, str2);
return 0;
}
The output of the program above is

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schedule S1:	r1(A)	w1(A)	r2(B)	w2(B)	r1(B)
swap r2(B) and w1(A):	r1(A)	r2(B)	w1(A)	w2(B)	r1(B)
swap r2(B) and r1(A):	r2(B)	r1(A)	w1(A)	w2(B)	r1(B)
swap w2(B) and w1(A):	r2(B)	r1(A)	w2(B)	w1(A)	r1(B)
swap w2(B) and r1(A):	r2(B)	w2(B)	r1(A)	w1(A)	r1(B)
schedule S2:	r1(A)	w1(A)	r2(A)	w2(A)	r1(A)

Consider the following transaction schedules S1 and S2 Choose the correct option thatholds for the schedules S1 and S2.

Consider the following transaction schedules S1 and S2

Choose the correct option thatholds for the schedules S1 and S2.