0 votes 0 votes How to solve this?? Databases made-easy-test-series databases transaction-and-concurrency + – Lucky sunda asked Jan 9, 2017 • edited Mar 6, 2019 by adeebafatima1 Lucky sunda 531 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Kantikumar commented Jan 9, 2017 reply Follow Share Ans : 20? 0 votes 0 votes Lucky sunda commented Jan 9, 2017 reply Follow Share 8 is given. I don't if it is correct. No longer I trust Made Easy. 0 votes 0 votes Kantikumar commented Jan 9, 2017 reply Follow Share I think it should be 40. What I did is this : $\left ( \frac{6!}{3!*3!} \right )$ * 2 = 40 Total possibility= 6! (as there are 6 operations) W1(B) W2(B) W3(B) should appear in this exact order $\therefore$ divide by 3! W1(A) R2(A) R3(A) can appear in this exact order or W1(A) R3(A) R2(A) in this order. $\therefore$ divide by 3! then multiply with 2. 1 votes 1 votes sachin486 commented Nov 29, 2020 reply Follow Share https://gateoverflow.in/156383/doubt-question-on-transaction-schedules 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Answer is 1. Make precedence graph from the given schedule and find the no of topological sortings possible. No. of topological orderings= No. of conflict equivalent schedules. In this case T1->T2->T3 is the only conflict equivalent schedule Ashwani Kumar 2 answered Jan 9, 2017 Ashwani Kumar 2 comment Share Follow See 1 comment See all 1 1 comment reply Lucky sunda commented Jan 9, 2017 reply Follow Share No..u r answering serial schedules. I also did it like that. I don't this is correct solution for this question. 0 votes 0 votes Please log in or register to add a comment.
