UGC-NET | UGC NET CS 2017 Jan – II | Question 20
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.
Quiz of this Question
Last Updated :
26 Mar, 2018
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...