A node size is 4096B. Node pointer size is 10B. Search key is 11B. Record pointer is 12B. What is the Max node pointers that can be stored in a non-leaf node of a B+ tree?

The correct answer is 195.

In a B+ tree, non-leaf nodes (also known as internal nodes) are used for routing searches. They contain only search keys and node pointers (which point to child nodes). Unlike leaf nodes, non-leaf nodes do not store record pointers or actual data. Therefore, the record pointer size of 12 B is not used in this calculation.

Let us define the variables as follows:
- Let $d$ be the maximum number of node pointers (also called child pointers).
- A node with $d$ pointers will contain $d-1$ search keys.
- Node size = 4096 B
- Node pointer size = 10 B
- Search key size = 11 B

The total space occupied by the node pointers and search keys must be less than or equal to the total node size. This gives us the following inequality:

$$(d��\text{Node Pointer Size})+\left(\right(d-1)\times \text{Search Key Size})\le \text{Node Size}$$

Substituting the given values into the inequality, we get:

$$(d\times 10)+\left(\right(d-1)\times 11)\le 4096$$

Now, we simplify the expression step-by-step:

$$10d+11d-11\le 4096$$

$$21d-11\le 4096$$

Add 11 to both sides of the inequality:

$$21d\le 4107$$

Divide by 21 to solve for $d$:

$$d\le \frac{4107}{21}$$

$$d\le 195.57$$

Since the number of node pointers must be an integer, we take the floor of the value (round down to the nearest whole number):

$$d=195$$

Therefore, the maximum number of node pointers that can be stored in a non-leaf node is 195.