-
In a B+ tree, if the search-key value is 8 bytes long, the block size is 512 bytes and the block pointer size is 2 bytes, then the maximum order of the B+ tree is ______.
-
- 25
- 52
- 13
- 31
Correct Option: B
As given that: Search key = 8 bytes Block size = 512 bytes. Block pointer (BP) = 2 bytes. Then maximum order of B+ tree is Let K is the order then, K * Bp + (K - 1)*Key ≤ Block size
K *2 + (K - 1)*8 ≤ 512
10K ≤ (512 + 8)
10K ≤ 520
| K = |  |  | = 52 | |
| 10 |
Hence, 52 is correct answer.
