Consider a join (relation algebra) between relations $r\ (R)$ and $s\ (S)$ using the nested loop method. There are 3 buffers each of size equal to disk block size, out of which one buffer is reserved for intermediate results. Assuming $\text{size}(r(R)) < \text{size}(s(S))$, the join will have fewer number of disk block accesses if (A) relation $r\ (R)$ is in the outer loop. (B) relation $s\ (S)$ is in the outer loop. (C) join selection factor between $r\ (R)$ and $s\ (S)$ is more than 0.5. (D) join selection factor between $r\ (R)$ and $s\ (S)$ is less than 0.5.

Let $B(r)$ and $B(s)$ be the number of disk blocks occupied by relations $r$ and $s$, respectively.

We are given that $3$ buffer blocks are available in total, with $1$ buffer block reserved for the output/intermediate results. This leaves $2$ buffer blocks for the input relations (i.e., $1$ buffer block for the outer relation and $1$ buffer block for the inner relation).

For a block nested-loop join:

• If relation $r$ is the outer relation, the total number of block accesses is:

  $$\text{Cost} = B(r) + B(r) \times B(s)$$

• If relation $s$ is the outer relation, the total number of block accesses is:

  $$\text{Cost} = B(s) + B(s) \times B(r)$$

Comparing the two costs:

Since $\text{size}(r(R)) < \text{size}(s(S))$, we have $B(r) < B(s)$, which means $B(r) - B(s) < 0$.

Therefore, placing the smaller relation $r(R)$ in the outer loop yields fewer disk block accesses. Hence, option (A) is correct.