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?
(A) 24
(B) 25
(C) 26
(D) 27
Answer: (C)
Explanation:
Key size = 14 bytes (given)
Child pointer = 6 bytes (given)
We assume the order of B+ tree to be ‘n’.
Block size = (n – 1) * key size + n * child pointer
512 >= (n – 1) * 14 + n * 6
512 >= 14 * n – 14 + 6 * n
n = (512 + 14) / 20
n = 526 / 20
n = 26.3
n = 26
Thus, option (C) is correct.
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