Consider a database implemented using B+ tree for file indexing and installed on a disk drive with block size of 4 KB. The size of search key is 12 bytes and the size of tree/disk pointer is 8 bytes. Assume that the database has one million records. Also assume that no node of the B+ tree and no records are present initially in main memory. Consider that each record fits into one disk block. The minimum number of disk accesses required to retrieve any record in the database is ________.

To find the minimum number of disk accesses, the B+ tree must have the minimum height (i.e., maximum node capacity/fan-out at each level).

Given:

• Block size, $B = 4\text{ KB} = 4096\text{ bytes}$

• Search key size, $K = 12\text{ bytes}$

• Pointer size, $P = 8\text{ bytes}$

• Number of records, $N = 1,000,000$

1. Internal Node Order ($p$): An internal node containing $p$ block pointers and $p - 1$ search keys must satisfy:

2. Leaf Node Capacity ($n$): Each leaf node entry has a search key and a data/record pointer (and optionally 1 pointer to next leaf node):

3. Minimum Height of the B+ Tree:

• Number of leaf nodes needed for $10^6$ records:

  $$\left\lceil \frac{10^6}{204} \right\rceil = 4902 \text{ leaf nodes}$$

• Level 1 above leaves:

  $$\left\lceil \frac{4902}{205} \right\rceil = 24 \text{ nodes}$$

• Level 2 above leaves (Root):

  $$\left\lceil \frac{24}{205} \right\rceil = 1 \text{ node (Root)}$$

Thus, the B+ tree has $3$ levels:

• Level 1: Root

• Level 2: Intermediate level

• Level 3: Leaf level

4. Total Disk Accesses to Retrieve a Record:

• Number of disk accesses for index tree traversal $= 3$ (Root $\to$ Internal Node $\to$ Leaf Node)

• Number of disk accesses to fetch the actual data record from disk $= 1$