GATE | GATE-CS-2004 | Question 52
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?
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
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Thus, option (C) is correct.
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