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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) ;

in Databases 420 views
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68?
Concurrent schedule means non-serial schedule.
what is the answer?
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12?
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yes how?

please give approach to solve this type of question... @balchandar reddy san

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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..

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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

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serial schedule$=2!=2$
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2 ! = 2 x1 = 2
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oh sorry,it's my mistake.
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i think they meant concurrent schedules which are serializable.
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I think the question is not written properly.

70 concurrent

12 serializable-- 2 serial & 10 non-serial.

58 Non-serializable

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