The order of an internal node in a $B+$ tree index is the maximum number of children it can have. Suppose that a child pointer takes $6$ bytes, the search field value takes $14$ bytes, and the block size is $512$ bytes. What is the order of the internal node?
lets assume order of b+ tree is n.
B+ tree internal node has n-1 keys and n node (child) pointer.
block size = (n-1) *( key size+data pointer) + n * child pointer is not considered
Because above formula is for B tree internal node structure
internal node = cp1 key1 cp2 key2 cp3 ...... cp(n-1)key(n-1)cpn
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