Database Management System
Question 1 |
Consider the following partial Schedule S involving two transactions T1 and T2. Only the read and the write operations have been shown. The read operation on data item P is denoted by read(P) and the write operation on data item P is denoted by write(P).
T2 must be aborted and then both T1 and T2 must be re-started to ensure transaction atomicity | |
Schedule S is non-recoverable and cannot ensure transaction atomicity | |
Only T2 must be aborted and then re-started to ensure transaction atomicity | |
Schedule S is recoverable and can ensure atomicity and nothing else needs to be done |
Question 1 Explanation:
T2 is reading the value written by T1 and getting committed before T1 commits. So it is non-recoverable schedule.
Question 2 |
Consider a B+ tree in which the search key is 12 bytes long, block size is 1024 bytes, record pointer is 10 bytes long and block pointer is 8 bytes long. The maximum number of keys that can be accommodated in each non-leaf node of the tree is ____ .
50 | |
60 | |
70 | |
80 |
Question 2 Explanation:
Explanation:
Suppose that ‘k’ is order of the non-leaf node
k(8) + (k-1)12 ≤ 1024
20k ≤1036
k ≤ ⌊1036/20⌋ ⇒ k ≤ 51
As the order is 51, maximum we can store 50 keys.
k(8) + (k-1)12 ≤ 1024
20k ≤1036
k ≤ ⌊1036/20⌋ ⇒ k ≤ 51
As the order is 51, maximum we can store 50 keys.
There are 2 questions to complete.