MCQ on various aspects in DBMS including database disk access and querying, cost for reading disk block pages, count the number of disk pages required to store a table, serializability, serial schedules, RDBMS Multiple Choice Questions with Answers
Database management systems - MCQ exam questions
1. Which of the following is TRUE about serializable schedules?
a) Every serializable schedule is recoverable.
b) Every serializable schedule contains no conflicting actions.
c) Every 2PL schedule is serializable.
d) Every view serializable schedule is also a conflict serializable
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Answer: (c) Every 2PL schedule is serializable Two-phase locking protocol ensures conflict serializability.Consider any transaction. If the transaction has acquired its final lock, we say that it has reached the end of growing phase. The point in the schedule is referred as lock point. Now, transactions can be ordered according to their lock points—this ordering is, in fact, a serializability ordering for the transactions. Every serializable schedule is recoverable – FALSE Every serializable schedule contains no conflicting actions – FALSE. View serializable schedules may have conflicting actions. Every view serializable schedule is also a conflict serializable – FALSE. Alternate of this statement is true. |
2. Let us join two relations R(a, b) and S(a, c) on attribute a using a merge join. Suppose B(R)=10,000 and B(S)=5,000. Assume a memory size of 101 blocks. If both R and S are already sorted on a, how many disk IOs are needed for a merge join?
a) 10000
b) 15000
c) 5000
d) 150000000
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Answer: (b) 15000 Cost of merge join for sorted relations = Sum of number of disk blocks occupied by the relations. = B(R) + B(S) = 10000 + 5000 = 15000 |
3. Which of the following is TRUE about recoverable schedules?
a) In a recoverable schedule, if a transaction T commits, then any other transaction that T read from must also have committed.
b) In a recoverable schedule, no transaction will ever be aborted because a transaction that it read from has aborted.
c) Every recoverable schedule is serializable.
d) Every cascadeless schedule is also non-recoverable.
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Answer: (a) In a recoverable schedule, if a transaction T commits, then any other transaction that T read from must also have committed. Let us suppose that there are two transactions T1 and T2 in a schedule. Also, let us suppose that T2 reads a data item (say A) which was written by T1. That is, WRITE(A) of T1 appears before READ(A) of T2. Then, the COMMIT instruction of T1 must appear before the COMMIT of T2. |
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