Phase 6Ch 6.2 Disks and I/O
ACompressed recap
Capacity = surfaces × tracks × sectors × bytes; access = seek + half-rotation latency + transfer; rotation = 60/RPM. Disk scheduling on the running queue (head 53): FCFS 640, SSTF 236, SCAN ↓ 236 / ↑ 331, LOOK ↑ 299 / ↓ 208, C-SCAN ↑ 382, C-LOOK ↑ 322. Programmed I/O busy-waits; interrupts cost per transfer; DMA one interrupt per block, cycle stealing; buffering, caching, spooling.
BMerged rule card
6.2.1 Disk structure and access time
- Track, sector, cylinder; one head per surface.
- Capacity = surfaces × tracks × sectors × bytes.
- Access = seek + latency + transfer; rotation = 60/RPM s; average latency = half a rotation.
- Transfer rate = bytes per track × rotations per second. Linear sector = (c × heads + h) × sectors + s.
- SSD: no seek, no rotation.
6.2.2 Disk scheduling: FCFS, SSTF, SCAN, C-SCAN, LOOK, C-LOOK
- FCFS: arrival order. SSTF: nearest next (starvation).
- SCAN: sweep to the disk's end, reverse. LOOK: reverse at the last request.
- C-SCAN: sweep to the end, jump to the other end, continue same direction. C-LOOK: jump from last request to the farthest one.
- Running queue (head 53): FCFS 640, SSTF 236, SCAN ↓ 236 / ↑ 331, LOOK ↑ 299 / ↓ 208, C-SCAN ↑ 382, C-LOOK ↑ 322.
6.2.3 I/O techniques: programmed I/O, interrupts, DMA, buffering and spooling
- Programmed I/O: CPU polls and moves data (busy waiting).
- Interrupt I/O: CPU fraction = transfers/s × overhead per transfer.
- DMA: one interrupt per block; cycle stealing fraction = words/s × memory cycle.
- Buffer (in transit), double buffering, cache (copy for reuse), spool (queue of jobs for a non-sharable device).
- Blocking, non-blocking, asynchronous I/O calls.
CInterleaved GATE-style questions
Q1 · NAT · 1 mark(s) — Latency
A disk spins at 6000 rpm. Average rotational latency in ms?
Numerical answer: ________
Q2 · NAT · 2 mark(s) — Capacity
4 platters with 2 surfaces each, 4096 tracks per surface, 64 sectors per track, 512 bytes per sector. Capacity in GB?
Numerical answer: ________
Q3 · NAT · 2 mark(s) — FCFS
Head at 100; requests 23, 89, 132, 42, 187 served FCFS. Total head movement?
Numerical answer: ________
Q4 · NAT · 2 mark(s) — SSTF
Head at 100; requests 23, 89, 132, 42, 187 (SSTF). Total head movement?
Numerical answer: ________
Q5 · NAT · 2 mark(s) — SCAN
Head at 100 moving towards 199 (cylinders 0–199); requests 23, 89, 132, 42, 187 (SCAN). Total head movement?
Numerical answer: ________
Q6 · MCQ · 1 mark(s) — C-SCAN
The main advantage of C-SCAN over SCAN is:
- (A) less total movement always
- (B) more uniform waiting time across cylinders
- (C) no need to know the head direction
- (D) it serves the nearest request first
Q7 · NAT · 1 mark(s) — Interrupts
A device sends 100,000 words per second, one interrupt per word, 2 µs per interrupt. Percentage of CPU time used?
Numerical answer: ________
Q8 · MCQ · 1 mark(s) — DMA
Cycle stealing refers to:
- (A) interrupts being lost
- (B) the disk skipping sectors
- (C) the DMA controller using memory cycles that delay the CPU
- (D) the CPU taking time slices from processes
DAnswer Vault
Try the questions first. Open when you are ready to check.
🔐 Open the Answer Vault
Q1 Answer: 5 · 1 mark(s)
10 ms per rotation / 2.
Re-read: Lesson 6.2.1 — Disk structure and access time
Q2 Answer: 1 · 2 mark(s)
8 × 2¹² × 2⁶ × 2⁹ = 2³⁰.
Re-read: Lesson 6.2.1 — Disk structure and access time
Q3 Answer: 421 · 2 mark(s)
77 + 66 + 43 + 90 + 145.
Re-read: Lesson 6.2.2 — Disk scheduling: FCFS, SSTF, SCAN, C-SCAN, LOOK, C-LOOK
Q4 Answer: 273 · 2 mark(s)
100 → 89 → 132 → 187 → 42 → 23.
Re-read: Lesson 6.2.2 — Disk scheduling: FCFS, SSTF, SCAN, C-SCAN, LOOK, C-LOOK
Q5 Answer: 275 · 2 mark(s)
99 up to 199, then 176 down to 23.
Re-read: Lesson 6.2.2 — Disk scheduling: FCFS, SSTF, SCAN, C-SCAN, LOOK, C-LOOK
Q6 Answer: (B) · 1 mark(s)
Every cylinder is reached at regular intervals.
- (A) is wrong: Not always less.
- (C) is wrong: It sweeps one way.
- (D) is wrong: That is SSTF.
Re-read: Lesson 6.2.2 — Disk scheduling: FCFS, SSTF, SCAN, C-SCAN, LOOK, C-LOOK
Q7 Answer: 20 · 1 mark(s)
0.2 s per second.
Re-read: Lesson 6.2.3 — I/O techniques: programmed I/O, interrupts, DMA, buffering and spooling
Q8 Answer: (C) · 1 mark(s)
Both share the memory bus.
- (A) is wrong: No.
- (B) is wrong: No.
- (D) is wrong: That is pre-emption.
Re-read: Lesson 6.2.3 — I/O techniques: programmed I/O, interrupts, DMA, buffering and spooling