# GATE | GATE-CS-2004 | Question 52

• Last Updated : 28 Jun, 2021

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

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

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

Thus, option (C) is correct.

Please comment below if you find anything wrong in the above post.

Quiz of this Question

My Personal Notes arrow_drop_up