Algorithms
October 3, 2023Algorithms
October 3, 2023Database-Management-System
| Question 3 |
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 _________.
| 52 | |
| 53 | |
| 54 | |
| 55 |
Question 3 Explanation:
Given,
Block size = 512 bytes
Block pointer = 2 bytes
Search key = 8 bytes
Let Maximum order of B+ tree = n
n(Pb) + (n – 1)(k) ≤ 512
n(2) + (n – 1)(8) ≤ 512
2n + 8n – 8 ≤ 512
10n ≤ 520
n ≤ 520/10
(n = 52)
Block size = 512 bytes
Block pointer = 2 bytes
Search key = 8 bytes
Let Maximum order of B+ tree = n
n(Pb) + (n – 1)(k) ≤ 512
n(2) + (n – 1)(8) ≤ 512
2n + 8n – 8 ≤ 512
10n ≤ 520
n ≤ 520/10
(n = 52)
Correct Answer: A
Question 3 Explanation:
Given,
Block size = 512 bytes
Block pointer = 2 bytes
Search key = 8 bytes
Let Maximum order of B+ tree = n
n(Pb) + (n – 1)(k) ≤ 512
n(2) + (n – 1)(8) ≤ 512
2n + 8n – 8 ≤ 512
10n ≤ 520
n ≤ 520/10
(n = 52)
Block size = 512 bytes
Block pointer = 2 bytes
Search key = 8 bytes
Let Maximum order of B+ tree = n
n(Pb) + (n – 1)(k) ≤ 512
n(2) + (n – 1)(8) ≤ 512
2n + 8n – 8 ≤ 512
10n ≤ 520
n ≤ 520/10
(n = 52)
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