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UGC-NET | UGC NET CS 2017 Jan – II | Question 20

  • Last Updated : 26 Mar, 2018

The order of a leaf node in a B+ tree is the maximum number of children it can have. Suppose that block size is 1 kilobytes, the child pointer takes 7 bytes long and search field value takes 14 bytes long. The order of the leaf node is ________.

(A) 16
(B) 63
(C) 64
(D) 68


Answer: (A)

Explanation:

Key size = 14 bytes (given)

Child pointer = 7 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 * 7
512 >= 14 * n – 14 + 7 * n 
n <= (1024 + 14) / 20
n <= 1038 / 21
n <= 49.42

So, option (A) is correct.

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