Test by Bikram | Databases | Test 1 | Question: 11

Question

Consider the following database schedule with $3$ transactions, $T1$ ,$T2$ and $T3$. Where $r$ stands for READ, w for WRITE.

$S : r2 (A); r1 (B); w2 (A); r2 (B); r3 (A); w1 (B); w3(A); w2 (B)$

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

• A. the schedule is  serializable
• B. the schedule is not  serializable
• C. schedule will not execute
• D. this schedule have deadlock

Answer 1

 The schedule is not conflict serializable as there is a cycle in the precedence graph 

cycle in precedence graph therefore not conflict serializable,

now we have to check for is blind write existing or not but there is no blind write therefore not conflict serializable and not view serializable therefore not serializable answer is B .

Comments

How to check , is clear from this diagram .

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what is view rule?

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The Definition to become view equivalent serial schedule 

1) Initial read must be same.

2)  Any write by Ti  followed by a read in Tj , the same condition should be followed

[ that means We need to preserve Write-Read dependency( W-R ) . ]

3) last operation on some item should be same in view equivalence schedule.

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why not deadlock?