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An internal node of a B+ tree must fit in a 1,024-byte block. A key takes 8 bytes and a tree pointer takes 8 bytes. What is the maximum num…

1538Medium DSSSB TGT Computer Science 0 replies 1 views

An internal node of a B+ tree must fit in a 1,024-byte block. A key takes 8 bytes and a tree pointer takes 8 bytes. What is the maximum number of keys that an internal node can hold?

A128
B64
C127
D63

Correct answer: D

Explanation

With n pointers and n − 1 keys, 8n + 8(n − 1) ≤ 1024 gives n ≤ 64.5, so n = 64 pointers and 63 keys.

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