Question

Node size = 4096 B
Node pointer = 10 B
Search key = 11 B
Record pointer = 12 B
Find the maximum number of node pointers.

Hint:

In B+ Trees, if there are \( p \) pointers in an internal node, then there are always \( p-1 \) search keys.

Answer 1

Correct Answer: 195

Step 1: Understanding the Topic
This question is about calculating the order (or fan-out) of an internal node in a B+ Tree. The order is defined as the maximum number of child pointers the node can hold. This is determined by the fixed size of a node (or block) and the space required to store the pointers and search keys within that node.
Step 2: Key Approach - Formulating the Space Inequality
An internal node in a B+ Tree with $p$ node (child) pointers contains $p-1$ search keys. The total space occupied by these pointers and keys must not exceed the total available node size. We can express this as an inequality: \[ (p \times \text{Size of a node pointer}) + ((p-1) \times \text{Size of a search key}) \le \text{Total node size} \] The record pointer size is only relevant for leaf nodes, not internal nodes.
Step 3: Detailed Calculation
Let $p$ be the maximum number of node pointers. Substitute the given values into the inequality: \[ (p \times 10) + ((p-1) \times 11) \le 4096 \] Now, we solve this inequality for $p$: \[ 10p + 11p - 11 \le 4096 \] Combine the terms with $p$: \[ 21p - 11 \le 4096 \] Add 11 to both sides: \[ 21p \le 4096 + 11 \] \[ 21p \le 4107 \] Divide by 21: \[ p \le \frac{4107}{21} \] \[ p \le 195.571... \] Since the number of pointers, $p$, must be an integer, we take the floor of this value.
\[ p_{max} = \lfloor 195.571... \rfloor = 195 \] Step 4: Final Answer
The maximum number of node pointers that can fit in an internal node is 195. \[ \boxed{195} \]