Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Consider the following schedule S of transact...
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Consider the following schedule S of transactions T1,T2,T3,T4:
Which one of the following statement/s is correct?
- a)S is conflict-serializable but not recoverable.
- b)S is not conflict-serializable but is recoverable.
- c)S is both conflict-serializable and recoverable.
- d)S is neither conflict-serializable not is it recoverable.
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Consider the following schedule S of transactions T1,T2,T3,T4:Which on...
We know that:
Using precedence graph method we can find out Schedule is conflict serializable or not.
If precedence graph has any loop then that schedule is going to be not conflict-serializable.
If precedence graph doesn’t have any loop then that schedule is going to be conflict-serializable.
If one transaction (T1) is writing some value × and that value is read by some other transaction (T2) then T1 should commit first then only (T2) should read that value and commit in order to become recoverable schedule.
There is no loop in this graph so schedule is conflict serializable.
T2: T3: T1: T4 is the only serial schedule that is equivalent to given schedule. (Using Topological Sorting)
There are two write read dependency.
1. T1 write (x) and T4 read (x)
2. T2 write (y) and T4 read (y)
In first case T1 is committed first then read by T4 then T4 is committed so this is recoverable.
In second case T2 is committed first then read by T4 then T4 is committed so this is also recoverable so schedule S is recoverable.
Hence, the correct answer is (C).
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Community Answer
Consider the following schedule S of transactions T1,T2,T3,T4:Which on...
We know that:
Using precedence graph method we can find out Schedule is conflict serializable or not.
If precedence graph has any loop then that schedule is going to be not conflict-serializable.
If precedence graph doesn’t have any loop then that schedule is going to be conflict-serializable.
If one transaction (T1) is writing some value × and that value is read by some other transaction (T2) then T1 should commit first then only (T2) should read that value and commit in order to become recoverable schedule.
There is no loop in this graph so schedule is conflict serializable.
T2: T3: T1: T4 is the only serial schedule that is equivalent to given schedule. (Using Topological Sorting)
There are two write read dependency.
1. T1 write (x) and T4 read (x)
2. T2 write (y) and T4 read (y)
In first case T1 is committed first then read by T4 then T4 is committed so this is recoverable.
In second case T2 is committed first then read by T4 then T4 is committed so this is also recoverable so schedule S is recoverable.
Hence, the correct answer is (C).
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Consider the following schedule S of transactions T1,T2,T3,T4:Which one of the following statement/s is correct?a)S is conflict-serializable but not recoverable.b)S is not conflict-serializable but is recoverable.c)S is both conflict-serializable and recoverable.d)S is neither conflict-serializable not is it recoverable.Correct answer is option 'C'. Can you explain this answer?
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