number of recoverable schedule possible

Question

if transactions are T1:r(A),w(A),r(B),w(B),c

                               T2: r(A),r(B),c

then howmany recoverable schedule possible over T1,T2?

Answer 1

Number of concurrent schedule=$\binom{8}{3}=56$

Number of irrecoverable schedule=check here are 2 dirty read . So, anyone of them causing dirty read

$\text{Case 1:}$ when $R_{2}(A)$ placing at top,  $W_{1}(B)$ and $R_{2}(B)$ causing dirty read. So, only 1 schedule possible

$T_{1}$	$T_{2}$
	$R(A)$
$R(A)$
$W(A)$
$R(B)$
$\color{red}{W(B)}$
	$\color{red}{R(B)}$
	$Commit$
$Commit$

 

$\text{Case 2:}$ In case of $R_{2}(A)$ placing after $R_{1}(A)$

Here also 1 Irrecoverable schedule possible

$T_{1}$	$T_{2}$
$R(A)$
	$R(A)$
$W(A)$
$R(B)$
$\color{red}{W(B)}$
	$\color{red}{R(B)}$
	$Commit$
$Commit$

 

$\text{Case 3:}$ When $R_{2}(A)$ is placing after $W_{1}(A)$

Here we already getting dirty read for $A$, So, no need to concentrate more for dirty read

$T_{1}$	$T_{2}$
$R(A)$
$\color{red}{W(A)}$
	$\color{red}{R(A)}$
$R(B)$
$W(B)$
	$R(B)$
	$Commit$
$Commit$

 

Here number of irrecoverable schedule $3+2+1=6$

$\text{Case 4:}$ Here $R_{2}(A)$ is paced after $R_{1}(B)$

$T_{1}$	$T_{2}$
$R(A)$
$\color{red}{W(A)}$
$R(B)$
	$\color{red}{R(A)}$
$W(B)$
	$R(B)$
	$Commit$
$Commit$

 

Here 3 irrecoverable schedules,

$\text{Case 5:}$ Schedule as given i question

$T_{1}$	$T_{2}$
$R(A)$
$\color{red}{W(A)}$
$R(B)$
$\color{red}{W(B)}$
	$\color{red}{R(A)}$
	$\color{red}{R(B)}$
	$Commit$
$Commit$

 

Here only 1 schedule is possible

 

So, total numberof recoverable schedule=$56-12=44$

refer:https://gateoverflow.in/31867/how-many-recoverable-schedules-are-possible-from-t1-and-t2

Comments

updated answer will be 41

plz chk, if anything wrong

@Shaik I got 15 irrecoverable schedule

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@srestha mam,

HOW YOU ARE GETTING 9 , IN CASE 3?

it is $\frac{(2+2)!}{2! * 2!} = \frac{4!}{4} = 6$ only , check it.

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yes thank u:)

order cannot change, so will be 6 for case 3

I havenot checked it manually