سوال ۹۶ کنکور ارشد مهندسی کامپیوتر ۱۴۰۲

The elevator / SCAN (LOOK) algorithm sweeps the head in one direction, servicing every request it passes, then reverses.

Requests sorted by cylinder: $2, 6, 10, 20, 22, 38, 40$.

The head starts at cylinder $20$ and is told to move upward first. So it services the requests at or above $20$ in increasing order, then turns around and services the lower requests in decreasing order.

Upward sweep (from $20$): $20 \to 22 \to 38 \to 40$. The arm only needs to reach the highest pending request ($40$), so the upward travel is $40 - 20 = 20 \text{ cylinders.}$

Downward sweep (reverse, from $40$): $40 \to 10 \to 6 \to 2$. The arm travels down to the lowest pending request ($2$): $40 - 2 = 38 \text{ cylinders.}$

Total head movement: $20 + 38 = 58 \text{ cylinders.}$

Each single-cylinder move costs $6$ time units, so the total seek time is $58 \times 6 = 348.$

This equals option (2). (The intended unit is microseconds: $348\,\mu s$; the printed word «milliseconds» is a known typo in this booklet, but the matching numeric value is unambiguously $348$.)

Exam trick: SCAN/LOOK distance is just (top reached − start) + (top reached − bottom reached). Total movement is independent of how many requests sit in between — only the extreme cylinders in each direction matter. Note that because the head reverses only at the last request ($40$, not a fixed disk end), this is technically the LOOK variant.