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

This is a B+ Tree internal node size calculation problem.
In an internal node of B+ Tree:
If there are \( p \) node pointers, then there are \( (p-1) \) search keys.
Step 1: Write the space equation.
Total size used in a node: \[ p \times (\text{node pointer size}) + (p-1) \times (\text{search key size}) \] Substitute given values: \[ p(10) + (p-1)(11) \] \[ = 10p + 11p - 11 \] \[ = 21p - 11 \] This must be less than or equal to total node size: \[ 21p - 11 \le 4096 \] Step 2: Solve the inequality.
\[ 21p \le 4107 \] \[ p \le \frac{4107}{21} \] \[ p \le 195.57 \] Since \( p \) must be an integer, \[ p = 195 \] Final Answer: \[ \boxed{195} \]