Disk scheduling is one of the most calculation-heavy corners of Operating Systems, and also one of the most predictable. Almost every question hands you a queue of cylinder requests and a starting head position, then asks for total head movement, service order, or how many times the head reverses. The scoring skill is mechanical: apply FCFS, SSTF or SCAN carefully and add the distances without arithmetic slips. GATE, UGC NET and state recruitment papers all draw from this same well.
Work each one on paper first. The algorithm fixes the service order; everything after that is arithmetic on the gaps between consecutive head positions, and that is where the marks actually go. The theory behind FCFS, SSTF, SCAN and the cyclic variants sits in the disk scheduling learn module.
Basics and FCFS versus SSTF
Q1. Moving the read/write head of a disk in or out to position it on the correct track is known as _____. (DSSSB 2021)
(a) waiting time
(b) access time
(c) seek time
(d) latency time
Answer: (c) seek time.
Seek time is the interval spent moving the head to the target track, and it is usually the largest component of disk access time. Rotational latency is the separate wait for the right sector to spin under the head. Because seek dominates, every scheduling algorithm is really trying to cut total head movement.
Q2. Which of the following disk strategies is likely to give the best throughput? (GATE 1999, see the full solution)
(a) Farthest cylinder next
(b) Nearest cylinder next
(c) First come first served
(d) Elevator algorithm
Answer: (b) Nearest cylinder next.
Nearest cylinder next is another name for Shortest Seek Time First, which always serves the closest pending request. By minimising head movement per step, it maximises the number of requests served per unit time. The trade-off is possible starvation of far-away requests, but for raw throughput SSTF wins here.
Q3. An operating system loads and executes a single sequential user process at a time using FCFS disk head scheduling. If FCFS is replaced by SSTF, claimed to give 50% better benchmarks, what improvement in I/O performance of user programs is expected? (GATE 2004, see the full solution)
(a) 50%
(b) 40%
(c) 25%
(d) 0%
Answer: (d) 0%.
Disk scheduling only helps when a queue of competing requests can be reordered. With a single sequential process there is never more than one outstanding request, so there is nothing to reorder. SSTF and FCFS behave identically, and the improvement is zero.
Q4. Five requests (P, 155), (Q, 85), (R, 110), (S, 30), (T, 115) sit in the queue with the head at cylinder 100 under SSTF. Which statement is FALSE? (GATE 2020, see the full solution)
(a) T is serviced before P
(b) Q is serviced after S, but before T
(c) The head reverses its direction of movement between servicing of Q and P
(d) R is serviced before P
Answer: (b) Q is serviced after S, but before T.
Working SSTF from 100 gives the service order R, T, Q, S, P, since the nearest pending request is picked each time. In that order Q actually comes before S and after T, so the claim that Q is serviced after S and before T is wrong. That makes (b) the false statement.
Q5. A disk system has 100 cylinders. Requests arrive for 4, 34, 10, 7, 19, 73, 2, 15, 6, 20 with the head at cylinder 50. If moving one cylinder takes 1 ms and SSTF is used, how long to satisfy all requests? (GATE 2009, see the full solution)
(a) 95 ms
(b) 119 ms
(c) 233 ms
(d) 276 ms
Answer: (b) 119 ms.
SSTF from 50 serves the requests in the order 50, 34, 20, 19, 15, 10, 7, 6, 4, 2, 73, always taking the nearest first. Adding the step distances 16, 14, 1, 4, 5, 3, 1, 2, 2 and the final jump of 71 gives 119 cylinders of movement. At 1 ms per cylinder that is 119 ms.
SCAN, C-SCAN and LOOK
Q6. On a disk with 1000 cylinders (0 to 999), the last serviced request was at track 345 and the head is moving toward track 0. Using SCAN, how many tracks must the arm move to serve the FIFO queue 123, 874, 692, 475, 105, 376? (UGC NET 2012, see the full solution)
(a) 2013
(b) 1219
(c) 1967
(d) 1507
Answer: (b) 1219.
Under SCAN the head continues in its current direction to the end of the disk, so it travels from 345 down to cylinder 0, covering 345 tracks. It then reverses and sweeps up to the farthest pending request at 874, covering another 874 tracks. The total is 345 plus 874, which is 1219 tracks.
SCAN, C-SCAN, LOOK and C-LOOK differ only in where the head turns, and on this queue two of the distractors are exactly what the other names produce. LOOK turns at the lowest pending request instead of running down to cylinder 0: 345 down to 105 is 240 tracks, then 105 up to 874 is 769, giving 1009. C-SCAN runs down to 0, jumps the full width of the disk to 999 and sweeps down again to the last pending request at 376: 345 plus 999 plus 623 is 1967, which is option (c). C-LOOK turns at 105, jumps straight to 874 and works down to 376: 240 plus 769 plus 498 is 1507, option (d). Read the algorithm name before you start adding distances.
Q7. A 200-track disk (0 to 199) holds the queue 98, 183, 37, 122, 14, 124, 65, 67 with the head at track 53 and moving toward higher tracks. Under SCAN, what is the total head movement? (MPPSC 2025, see the full solution)
(a) 236
(b) 299
(c) 331
(d) 208
Answer: (c) 331.
SCAN runs to the boundary before it turns, so the head covers 53 up to 199, which is 146 tracks, then reverses and sweeps down to the lowest pending request at 14, another 185 tracks. The total is 331. Both near misses are other algorithms on this same queue: turning at the highest request 183 instead of at the boundary is LOOK, giving 130 plus 169, which is 299, and serving the nearest request throughout is SSTF, which totals 236.
Head-movement and direction-change numericals
Q8. A disk has 200 tracks (0 to 199). The head is servicing track 120 and the previous request was track 90. Pending requests in arrival order are 30, 70, 115, 130, 110, 80, 20, 25. How many times does the head change direction under SSTF and FCFS? (GATE 2004, see the full solution)
(a) 2 and 3
(b) 3 and 3
(c) 3 and 4
(d) 4 and 4
Answer: (c) 3 and 4.
The head reaches 120 moving upward from 90, so a first move downward already counts as a change. SSTF serves 115, 110, 130, 80, 70, 30, 25, 20: down, down, up, then down all the way, which is three changes including the one on entry. FCFS follows the arrival order 30, 70, 115, 130, 110, 80, 20, 25: down, up, up, up, down, down, down, up, which is four. So the counts are 3 and 4.
Q9. A hard disk has 2048 tracks and its head serves requests by SSTF, with a tie in distance never turning the head. If the head may start from any track, what is the largest request set for which the head changes direction after servicing every request? (GATE 2007, see the full solution)
(a) 9
(b) 10
(c) 11
(d) 12
Answer: (c) 11.
Measure every request by its distance from the starting track, with successive requests on alternating sides. After serving one at distance x, the nearest request across is x plus its own distance away, while the next one on the same side is its distance minus x away, so the head turns only when that second gap is the bigger of the two. The smallest integers that satisfy the rule all the way down are 1, 2, 5, 10, 21, 42, 85, 170, 341, 682 and 1365: eleven requests. The outermost pair sits 1365 above the start and 682 below it, a span of 2047 tracks, which is exactly what 2048 tracks allow. Start at track 682 and the set is 683, 680, 687, 672, 703, 640, 767, 512, 1023, 0 and 2047. A twelfth request would have to sit 2730 tracks from the start, off the disk. Pin the head to track 180 instead of letting it start anywhere and the same ladder stops at nine, which is the trap in option (a).
Q10. A disk head follows SSTF starting at track 180. Which request set causes the head to change direction after servicing every request, given that a tie does not change direction and all requests arrive before servicing? (GATE 2007, see the full solution)
(a) 11, 139, 170, 178, 181, 184, 201, 265
(b) 10, 138, 170, 178, 181, 185, 201, 265
(c) 10, 139, 169, 178, 181, 184, 201, 265
(d) 10, 138, 170, 178, 181, 185, 200, 265
Answer: (b) 10, 138, 170, 178, 181, 185, 201, 265.
Set (b) is the first eight rungs of that same ladder, anchored at track 180: distances 1, 2, 5, 10, 21, 42, 85 and 170. In (a) and (c) the head stops at 181 and finds 178 and 184 both three tracks away; a tie does not turn the head, so it carries on upward. In (d) the head stops at 185 with 170 and 200 both fifteen tracks away, the same failure. Only (b) never ties.
Q11. A disk queue holds requests for cylinders 98, 183, 37, 122, 14, 124, 65, 67 and the head starts at 53. Under FCFS, what is the total number of head movements? (BPSC 2024)
(a) 640
(b) 620
(c) 630
(d) More than one of the above
Answer: (a) 640.
FCFS follows the queue as given, so the head moves 53 to 98, then 183, 37, 122, 14, 124, 65 and 67. The distances are 45, 85, 146, 85, 108, 110, 59 and 2. Their sum is 640.
Q12. For the queue 98, 183, 37, 122, 14, 124, 65, 67 with the head at 53 under SSTF, arrange the head positions A. 67, B. 37, C. 65, D. 98, E. 14 in order of traversal. (UGC NET 2025)
(a) A, B, C, D, E
(b) C, A, E, B, D
(c) C, A, B, E, D
(d) A, C, D, B, E
Answer: (c) C, A, B, E, D.
SSTF from 53 picks 65 first (distance 12), which is C, then 67 (distance 2), which is A. Next comes 37, which is B, then 14, which is E, and finally 98, which is D. So the ordering of these labelled positions is C, A, B, E, D.
Where these 12 fit in your preparation
Papers draw from three pockets. The definitions and the FCFS-versus-SSTF contrast carry the one-mark questions (Q1 to Q5). Where a sweep turns, and how the cyclic variants differ from it, is its own pocket (Q6 and Q7). The head-movement and direction-change numericals take the rest of the marks (Q8 to Q12). If your totals come out wrong, the culprit is almost always a distance added twice or a missed reversal, so lay out the service order in a single line before summing.
For the theory and more worked sweeps, use the disk scheduling learn module and place the topic in context with our Operating Systems for GATE breakdown. GATE aspirants get the full OS sequence inside GATE Guidance by Sanchit Sir; NET aspirants can start from the NET CS category page. Solve each queue by hand once, then re-solve from memory a week later.
