dbms10

Question

Comments

it is a very basic question ( option D ), if you didn't read the concept, no one can help you !

Answer 1

Till time stamp t3 no worries

In t4 W1(X) and t12 R2(X) and t13 W2(X) gives T1-->T2

In t5 W3(Y) and t8 R1(Y) and t9 W1(Y) gives T3-->T1

In t6 W3(Z) and t7 R2(Z)  gives T3-->T2

In t9 W1(Y) and t10 R2(Y) and t11 W2(Y) gives T1-->T2

Therefore T3-->T1-->T2

so option d is  correct.

Answer 2

1. #f nodes =#f transaction

2. Find all the conflicts related to the schedule, and directed the edge i->j.

3. Make the precedence graph, If cycle not exists then you can achieve equivalent some serial schedule