Question

The order of an internal node in a b+ tree index is the maximum number of children it can have. Suppose that a child pointer takes 6 bytes, the search field value takes 14 bytes, and the block size is 512 bytes. What is the order of the internal node

Answer 1

Answer:

The order of the internal code is 26 bytes.

Explanation:

Size of the Key= 14 bytes (given)

Size of Child pointer = 6 bytes (given)

We assume the order of B+ tree to be ‘x’.

Block size = (x – 1) x key size + x x child pointer

512 = (x – 1) x 14 + x x 6

512 = 14 x x – 14 + 6 x x

x = (512 + 14) / 20

x = 526 / 20

x = 26.3

x = 26 bytes (approx)