GATE CS 2009 Q31 — I/O Scheduling

In the Shortest Seek Time First (SSTF) disk scheduling algorithm, the request closest to the current head position is serviced next.

The set of pending requests is: {2, 4, 6, 7, 10, 15, 19, 20, 34, 73}

Starting from the initial head position at 50, the step-by-step head movement is as follows:

1. 50 → 34: Seek distance = $|50 - 34| = 16$
2. 34 → 20: Seek distance = $|34 - 20| = 14$
3. 20 → 19: Seek distance = $|20 - 19| = 1$
4. 19 → 15: Seek distance = $|19 - 15| = 4$
5. 15 → 10: Seek distance = $|15 - 10| = 5$
6. 10 → 7: Seek distance = $|10 - 7| = 3$
7. 7 → 6: Seek distance = $|7 - 6| = 1$
8. 6 → 4: Seek distance = $|6 - 4| = 2$
9. 4 → 2: Seek distance = $|4 - 2| = 2$
10. 2 → 73: Seek distance = $|73 - 2| = 71$

Now, we sum the individual seek distances to find the total head movement:

Given that it takes $1\text{ ms}$ to move across one cylinder, the total time taken is: