GATE | GATE-CS-2017 (Set 2) | Question 60

In a B+ tree, if the search-key value is 8 bytes long, the block size is 512 bytes and the block pointer is 2 bytes, then the maximum order of the B+ tree is ____.

Note: This question appeared as Numerical Answer Type.
(A) 51
(B) 52
(C) 53
(D) 54


Answer: (B)

Explanation:

Order of a B+ tree node is maximum number of children 
in an internal node

Let the order be x. Number of keys in a node is equal to
number children minus 1.
So a full node has (x-1) keys and x children.

(x-1)*(search key) + x * block ptr <= block size
==> (x-1)*8 + x*2 <= 512 ==> 10x <= 520 ==> x <= 52


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