0 votes 0 votes how to find total number of concurrent schedule possible when any schedule is given ? for example consider this schedule T1 : R1(A) , W1(A) , R1(B) , W1(B) ; T2 : R2(A) , W2(A) , R2(B) , W2(B) ; Databases databases transaction-and-concurrency + – Rahul_Rathod_ asked Dec 28, 2018 Rahul_Rathod_ 1.1k views answer comment Share Follow See all 11 Comments See all 11 11 Comments reply Lakshman Bhaiya commented Dec 28, 2018 i edited by Lakshman Bhaiya Dec 28, 2018 reply Follow Share 68? Concurrent schedule means non-serial schedule. what is the answer? 0 votes 0 votes Rahul_Rathod_ commented Dec 28, 2018 reply Follow Share no @Lakshman Patel RJIT 0 votes 0 votes balchandar reddy san commented Dec 28, 2018 reply Follow Share 12? 0 votes 0 votes Rahul_Rathod_ commented Dec 28, 2018 reply Follow Share yes how? please give approach to solve this type of question... @balchandar reddy san 0 votes 0 votes balchandar reddy san commented Dec 28, 2018 i edited by balchandar reddy san Dec 28, 2018 reply Follow Share There are in total 12 ways..i have divided into 4 sets of RW and in each case there are 3 possiblities in positioning 2&3. if we break R() W() of same data item then the graph will lose it's serializability. so W of a data item on any transaction should be executed right after the R of the same data item on the same transaction. concurrent 1: R1(A) , W1(A) , R2(A) , W2(A) , R1(B) , W1(B) , R2(B) , W2(B) ; concurrent 2: R2(A) , W2(A) ,R1(A) , W1(A) , R1(B) , W1(B) R2(B) , W2(B) ; concurrent 3: R1(A) , W1(A) , R2(A) , W2(A) , R2(B) , W2(B) R1(B) , W1(B) ; concurrent 4: R2(A) , W2(A) ,R1(A) , W1(A) , R2(B) , W2(B) R1(B) , W1(B) ; each one can be represented in 3 ways as shown in the image let me know if u need any clarification.. 0 votes 0 votes Magma commented Dec 28, 2018 reply Follow Share We need to maintain order of operations of an individual transaction total operations : 8 $\binom{8}{4} \times \binom{4}{4} = 70$ 70 - 2 {two are serial schedule} = 68 ans OR $\frac{(4+4)!}{4! \times 4!} = 70$ 70 - 2 {two are serial schedule} = 68 0 votes 0 votes Lakshman Bhaiya commented Dec 28, 2018 reply Follow Share serial schedule$=2!=2$ 0 votes 0 votes Magma commented Dec 28, 2018 reply Follow Share 2 ! = 2 x1 = 2 0 votes 0 votes Lakshman Bhaiya commented Dec 28, 2018 reply Follow Share oh sorry,it's my mistake. 0 votes 0 votes balchandar reddy san commented Dec 28, 2018 reply Follow Share i think they meant concurrent schedules which are serializable. 0 votes 0 votes KUSHAGRA गुप्ता commented Oct 18, 2019 reply Follow Share I think the question is not written properly. 70 concurrent 12 serializable-- 2 serial & 10 non-serial. 58 Non-serializable 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Number of concurrent schedules = All schedules possible – serial schedule =(4+4)!/4!4! – 2! = 8!/4!4! – 2 aaaakash001 answered Nov 21, 2022 aaaakash001 comment Share Follow See all 0 reply Please log in or register to add a comment.