GATE | Sudo GATE 2020 Mock I (27 December 2019) | Question 23

In a B+ tree, if the search-key value is 10 bytes long, the block size is 2 Kb and the block pointer is 1 byte, then the minimum number of keys in any internal node is ________.

(A)

187

(B)

186

(C)

93

(D)

83

Answer: (C)
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)*10 + x*1 ≤ 2048
= 11 * x ≤ 2058
= x ≤ 187.09 
= 187 

Now, minimum number of keys in any internal node,

= Ceil(x / 2) - 1 
= 93

So, option (C) is correct.

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