Official PYQ · 2016Set 2 · Q51Transactions and Concurrency ControlMCQ1 mark

GATE CS 2016 Set 2 Q51 — Transactions and Concurrency Control

Consider the following database schedule with two transactions $T_{1}$ and $T_{2}$ .

$S= r_{2}\left(X\right); r_{1}\left(X\right); r_{2} \left(Y\right); w_{1} \left(X\right); r_{1} \left(Y\right); w_{2} \left(X\right); a_{1}; a_{2}$

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?

View worked solution

Answer: C

To determine the correct statement about the schedule $S$ , let us analyze the dependencies and properties of the transactions:

$S = r_{2}(X); r_{1}(X); r_{2}(Y); w_{1}(X); r_{1}(Y); w_{2}(X); a_{1}; a_{2}$

1. Checking for Dirty Reads (Cascading Aborts)

A dirty read occurs when a transaction $T_i$ reads a data item that has been modified by another active transaction $T_j$ (which has not yet committed or aborted).

$T_1$ writes to $X$ ( $w_1(X)$ ). After this, no other transaction reads $X$ .
$T_2$ writes to $X$ ( $w_2(X)$ ). After this, no other transaction reads $X$ .
There are no other write operations in the schedule.

Since no transaction reads uncommitted data written by another transaction, there are no dirty reads in the schedule. Consequently, the abort of $T_1$ ( $a_1$ ) does not force $T_2$ to abort, and vice-versa. Thus, the schedule is cascadeless (i.e., it does not have cascading aborts).

2. Checking for Recoverability

Since the schedule is cascadeless, it is automatically recoverable. Therefore, Option A is false.

3. Checking for Strictness

A schedule is strict if a transaction can only read or write a data item $Z$ after all previous transactions that wrote $Z$ have committed or aborted.

$T_1$ writes to $X$ ( $w_1(X)$ ) at step 4.
$T_2$ writes to $X$ ( $w_2(X)$ ) at step 6, before $T_1$ aborts at step 7 ( $a_1$ ).

Since $T_2$ performs a write on $X$ while $T_1$ 's write on $X$ is still uncommitted/unaborted, the schedule is not strict. Therefore, Option D is false.