Operating-Systems
October 4, 2023Logical-Reasoning
October 4, 2023Database-Management-System
|
Question 187
|
In a database file structure, the search key field is 9 bytes long, the block size is 512 bytes, a record pointer is 7 bytes and a block pointer is 6 bytes. The largest possible order of a non-leaf node in a B+ tree implementing this file structure is
|
23
|
|
|
24
|
|
|
34
|
|
|
44
|
Question 187 Explanation:
From the structure of B+ tree we can get this question:
n × p + (n – 1) × k ≤ B (for non-leaf node)
Here, n = order, p = tree/block/index pointer, B = size of block
So,
n × p + (n – 1) × k ≤ B
n × 6 + (n – 1) × 9 ≤ 512
n ≤ 34.77
∴ n = 34
n × p + (n – 1) × k ≤ B (for non-leaf node)
Here, n = order, p = tree/block/index pointer, B = size of block
So,
n × p + (n – 1) × k ≤ B
n × 6 + (n – 1) × 9 ≤ 512
n ≤ 34.77
∴ n = 34
Correct Answer: C
Question 187 Explanation:
From the structure of B+ tree we can get this question:
n × p + (n – 1) × k ≤ B (for non-leaf node)
Here, n = order, p = tree/block/index pointer, B = size of block
So,
n × p + (n – 1) × k ≤ B
n × 6 + (n – 1) × 9 ≤ 512
n ≤ 34.77
∴ n = 34
n × p + (n – 1) × k ≤ B (for non-leaf node)
Here, n = order, p = tree/block/index pointer, B = size of block
So,
n × p + (n – 1) × k ≤ B
n × 6 + (n – 1) × 9 ≤ 512
n ≤ 34.77
∴ n = 34
Subscribe
Login
0 Comments