GATE CS 2016 Set 2 Q.61 (CS Q.51): Transactions, Serializability and Concurrency Control

Consider the following database schedule with two transactions, $T_1$ and $T_2$ .

S = r_2(X); r_1(X); r_2(Y); w_1(X); r_1(Y); w_2(X); a_1; a_2

where $r_i(Z)$ denotes a read operation by transaction $T_i$ on a variable $Z$ , $w_i(Z)$ denotes a write operation by $T_i$ on a variable $Z$ and $a_i$ denotes an abort by transaction $T_i$ .

Which one of the following statements about the above schedule is TRUE?

$S$ is non-recoverable

$S$ is recoverable, but has a cascading abort

$S$ does not have a cascading abort

$S$ is strict

Official answer and solution

Official answer: C
From: Official GATE 2016 answer key

Solution

Inspect the schedule for uncommitted reads and write-write dependencies to evaluate recoverability and cascading aborts.

Analyze the sequence of operations in schedule $S$ :

$r_2(X)$
$r_1(X)$
$r_2(Y)$
$w_1(X)$
$r_1(Y)$
$w_2(X)$
$a_1$
$a_2$

Check for Dirty Reads (Reading Uncommitted Writes):

$T_1$ writes $X$ at step 4 ( $w_1(X)$ ). Does $T_2$ read $X$ after step 4? Step 6 is $w_2(X)$ , which is a write, not a read.
$T_2$ writes $X$ at step 6 ( $w_2(X)$ ). No read follows step 6.
Neither transaction reads any data value written by the other transaction. Thus, there are NO dirty reads in $S$ .

Evaluate Schedule Properties:

Recoverability: Since there are no dirty reads, no transaction reads uncommitted changes. Therefore, $S$ is recoverable.
Cascading Aborts: A cascading abort occurs when the abort of one transaction forces another transaction to abort because of dirty read dependencies. Because no dirty reads exist, the abort of $T_1$ does not cause $T_2$ to abort. The abort $a_2$ is independent. Thus, $S$ does not have a cascading abort.
Strictness: A schedule is strict if a transaction neither reads nor writes a data item until the last transaction that wrote it has committed or aborted. Here, $T_1$ writes $X$ at step 4, and $T_2$ writes $X$ at step 6 before $T_1$ aborts at step 7. Therefore, $S$ is not strict.

Therefore, statement C (' $S$ does not have a cascading abort') is TRUE.

Final answer

Common trap

Assuming that $w_2(X)$ after $w_1(X)$ is a read operation or causes cascading aborts, confusing write-write conflicts with dirty reads.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2016 Computer Science and Information Technology (CS) Session 2
Question: 61 of 65, printed as CS Q.51
Section: Computer Science and Information Technology

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