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