The order of an internal node in a tree index is the maximum number of children it can have. Suppose that a child pointer takes bytes, the search field value takes bytes, and the block size is bytes. What is the order of the internal node?
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Answer: C
In a tree, an internal node of order (where is the maximum number of child pointers) contains:
- At most child pointers
- At most search key values
Let:
- be the order of the internal node.
- (size of a child pointer).
- (size of a search key value).
- (block size).
To fit the internal node within a single disk block, the total size must satisfy the inequality:
Substituting the given values:
Since the order must be an integer, the maximum possible value is .
Therefore, the order of the internal node is 26 (Option C).
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