Question 40

Given requests: $45, 20, 90, 10, 50, 60, 80, 25, 70$.
Initial head position: track $50$.
Note that request $50$ is serviced immediately with distance $0$.

1. SSTF (Shortest Seek Time First):

• From $50$, remaining requests are $\{10, 20, 25, 45, 60, 70, 80, 90\}$.
• Closest to $50$ is $45$ (distance $|50 - 45| = 5$).
• From $45$, closest is $60$ (distance $|45 - 60| = 15$, whereas $|45 - 25| = 20$).
• From $60$, closest is $70$ (distance $|60 - 70| = 10$).
• From $70$, closest is $80$ (distance $|70 - 80| = 10$).
• From $80$, closest is $90$ (distance $|80 - 90| = 10$).
• From $90$, the head must reverse and service remaining left requests: $25 \to 20 \to 10$.
  Distance from $90$ to $10 = 90 - 10 = 80$.
• Total SSTF head movement:
  $$\text{Dist}_{\text{SSTF}} = 5 + 15 + 10 + 10 + 10 + 80 = 130\text{ tracks}.$$

2. SCAN (Elevator) moving towards track 100:

• Head starts at $50$ and moves upward to track $100$:
  Distance $= 100 - 50 = 50\text{ tracks}$, servicing $60, 70, 80, 90$.
• Head reverses at track $100$ and moves downward to the lowest request (track $10$):
  Distance $= 100 - 10 = 90\text{ tracks}$, servicing $45, 25, 20, 10$.
• Total SCAN head movement:
  $$\text{Dist}_{\text{SCAN}} = (100 - 50) + (100 - 10) = 50 + 90 = 140\text{ tracks}.$$

3. Difference:

The additional distance is $10$ tracks.