Disk structure and access time

0 · Recall Warm-up

Answer from memory first. Answers are in the Answer Vault at the bottom.

1. Max file size with an inode? from Lesson 6.1.2
2. Rights given by octal 7? from Lesson 6.1.1
3. Why does a page fault take milliseconds? from Lesson 5.3.1

1 · The Hook

A hard disk is a stack of spinning plates with a moving arm, like an old record player. Reading data means moving the arm to the right ring, waiting for the plate to spin the data under the head, and then reading it. Each step has its own time, and GATE combines them in "average access time" and "capacity" numericals.

3 · Core Concept

Predict First — the answer

6000 rpm = 100 rotations per second → 10 ms per rotation. On average half a rotation: 5 ms.

In plain words

Each plate (platter) surface has rings (tracks) divided into sectors. The same ring on all surfaces forms a cylinder. Time to read = time to move the arm to the cylinder (seek) + time for the sector to arrive (rotational latency) + time to read the bytes (transfer).

1. Geometry

• Platters, each with 2 surfaces and one read/write head per surface; all heads move together on one arm.

• Track: one ring on a surface. Sector: a slice of a track (e.g. 512 bytes). Cylinder: the set of tracks at the same arm position on all surfaces.

• Capacity = surfaces × tracks per surface × sectors per track × bytes per sector.

  16 surfaces, 1024 tracks, 256 sectors, 512 bytes → 2⁴ × 2¹⁰ × 2⁸ × 2⁹ = 2³¹ B = 2 GB.

• Sector address (cylinder c, head h, sector s) as one number: (c × heads + h) × sectors per track + s.

2. Access time

Access time = seek time + rotational latency + transfer time (+ controller overhead if given).

• Seek time: moving the arm; questions give an average.

• Rotation time = 60 / RPM seconds. Average rotational latency = half a rotation.

  7200 rpm → 8.33 ms per rotation → 4.17 ms average latency.

• Transfer time = data size / transfer rate, or (sectors read / sectors per track) × rotation time.

• Transfer rate = bytes per track × rotations per second. 256 × 512 B = 128 KB per track at 120 rps → 15,360 KB/s.

Example: 7200 rpm, average seek 6 ms, 256 sectors of 512 B per track, read 4 KB (8 sectors): 6 + 4.17 + (8/256) × 8.33 = 6 + 4.17 + 0.26 ≈ 10.43 ms. Seek and latency dominate, so reading big contiguous chunks is far cheaper per byte.

3. Solid-state drives (SSDs)

No moving parts: no seek or rotational latency, so request order matters much less. Writes wear out cells; erasure happens in large blocks.

4 · Visual

The arm moves the head to the right track (seek); the platter spins the wanted sector under it (rotational latency); then the bytes are read (transfer). The same track on every surface forms a cylinder.

5 · Worked Examples

Example 1 Easy

A disk spins at 15,000 rpm. Average rotational latency?

Solution

60/15,000 s = 4 ms per rotation → 2 ms.

Example 2 Medium

A disk has 8 surfaces, 2048 tracks per surface, 128 sectors per track and 512-byte sectors. Capacity?

Solution

2³ × 2¹¹ × 2⁷ × 2⁹ = 2³⁰ B = 1 GB.

Example 3 GATE

16 heads, 256 sectors per track. What is the linear sector number of (cylinder 2, head 3, sector 10)?

Solution

(2 × 16 + 3) × 256 + 10 = 35 × 256 + 10 = 8970.

6 · GATE Traps

• Average latency is half a rotation, not a full one.
• RPM is per minute: rotation time = 60 / RPM seconds.
• Capacity uses surfaces (or heads), not platters, unless told each platter is one surface.
• Seek time is usually the largest term; transfer of a few sectors is tiny.

7 · Memory Hook

"Seek + half a turn + transfer. Capacity = surfaces × tracks × sectors × bytes."

8 · Connect the Dots

Seek time is the part we can reduce by choosing which request to serve next. The next lesson compares disk-scheduling algorithms on one request queue.

9 · Self-Test

GATE format. Try every question before opening the vault.

Q1 · NAT · 1 mark(s) — Latency

A disk spins at 7200 rpm. Average rotational latency in ms (2 decimals)?

Numerical answer: ________

Q2 · NAT · 1 mark(s) — Capacity

16 surfaces, 1024 tracks per surface, 256 sectors per track, 512 bytes per sector. Capacity in GB?

Numerical answer: ________

Q3 · NAT · 2 mark(s) — Access time

7200 rpm, average seek 6 ms, 256 sectors (512 B each) per track. Average time to read one 4 KB block (2 decimals)?

Numerical answer: ________

Q4 · NAT · 2 mark(s) — Transfer rate

A track holds 256 sectors of 512 bytes and the disk spins at 7200 rpm. Transfer rate in KB/s?

Numerical answer: ________

Q5 · NAT · 1 mark(s) — Addressing

8 heads and 64 sectors per track. Linear sector number of (cylinder 5, head 2, sector 7)?

Numerical answer: ________

10 · Rule Card

• 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.

11 · Concept Ledger

Unlocked in this lesson. From here on, these terms may be used.

plattersurfacetracksectorcylinderread/write headdisk capacityseek timerotational latencytransfer timetransfer rateaccess timelinear sector addressingsolid-state drive