The order of a leaf node in a B+ tree is the maximum number of children it can have. Suppose that block size is 1 kilobytes, the child pointer takes 7 bytes long and search field value takes 14 bytes long. The order of the leaf node is ________.
Key size = 14 bytes (given) Child pointer = 7 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 * 7 512 >= 14 * n – 14 + 7 * n n <= (1024 + 14) / 20 n <= 1038 / 21 n <= 49.42
So, option (A) is correct.
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