Secondary storage and bandwidth: disk access time, transfer rate, interleaved memory

0 · Recall Warm-up

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

• 4K × 8 memory from 1K × 4 chips: how many chips? from Lesson 4.1.4
• DRAM refresh overhead formula? from Lesson 4.1.3
• Hierarchical vs simultaneous AMAT? from Lesson 4.1.2

1 · The Hook

🎣

Below main memory sits the disk, where a single access costs milliseconds, a million times a cache hit. And beside main memory sits a trick that makes block transfers faster: split memory into banks that work at the same time. Both turn up in GATE as short calculations.

2 · Predict First

🤔

A disk spins at 7,200 rpm. On average, how long do you wait for the right sector to rotate under the head?

Make a guess in your head. The answer opens Section 3.

3 · Core Concept

💡

Predict First — the answer

One rotation = 60 / 7200 s = 8.33 ms; on average you wait half a rotation: 4.17 ms.

Disk geometry

• Platters with two surfaces each; each surface has concentric tracks; each track is divided into sectors (e.g., 512 bytes).
• A cylinder = the tracks at the same position on all surfaces.
• Capacity = surfaces × tracks per surface × sectors per track × bytes per sector.

Access time

Access time = seek time + rotational latency + transfer time

• Seek: move the head to the right track (given, often an average).
• Rotational latency: wait for the sector; average = half a rotation = 0.5 × 60/rpm seconds.
• Transfer time = bytes to transfer ÷ transfer rate.
• Transfer rate = bytes per track × rotations per second.

Memory bandwidth

Bandwidth = bytes delivered per unit time. A memory delivering 8 bytes every 10 ns has bandwidth 0.8 GB/s (decimal).

Interleaved memory

Main memory is split into k modules (banks) that can work at the same time.

• Low-order interleaving: the low address bits choose the module, so consecutive words sit in consecutive modules. A block of k consecutive words can be fetched from all modules at once.
• High-order interleaving: the high bits choose the module; consecutive words sit in the same module (good for separate programs, no help for one block).

Block fetch time, with module access time t_a and bus transfer time t_b per word:

• No interleaving: k × (t_a + t_b).
• k-way low-order interleaving: t_a + k × t_b (all modules access together, then words go out one per t_b).

Example: t_a = 40 ns, t_b = 10 ns, k = 4: 4 × 50 = 200 ns vs 40 + 40 = 80 ns. Lesson 4.3.2 uses this for cache miss penalties.

4 · Visual

Disk: seek (move arm) -> rotational latency (wait) -> transfer (read sector)

        track         sector
     .-------.      /
    /  .---.  \    /             one rotation at 7200 rpm = 8.33 ms
   |  ( hub )  |  <---- head     average latency = 4.17 ms
    \  '---'  /
     '-------'

Low-order interleaving, 4 modules:   word address mod 4 picks the module
  M0: 0 4 8 ...   M1: 1 5 9 ...   M2: 2 6 10 ...   M3: 3 7 11 ...

5 · Worked Examples

Example 1 Easy

A disk has 8 surfaces, 1,000 tracks per surface, 200 sectors per track, 512 bytes per sector. Capacity in bytes?

Solution

8 × 1000 × 200 × 512 = 819,200,000 bytes (781.25 MB in binary units).

Example 2 Medium

The same disk spins at 7,200 rpm. What is its transfer rate in bytes per second?

Solution

A track holds 200 × 512 = 102,400 bytes; 7200 rpm = 120 rotations per second → 12,288,000 B/s.

Example 3 GATE

Average seek time 5 ms, 7,200 rpm, 200 sectors per track. Average time to read one sector (ms, 2 decimals)?

Solution

Latency 4.17 ms; transfer = one sector = 1/200 of a rotation = 8.33/200 = 0.04 ms. Total ≈ 5 + 4.17 + 0.04 = 9.21 ms.

6 · GATE Traps

⚠️

• Average rotational latency is half a rotation.
• rpm → rotations per second: divide by 60.
• Transfer of one sector = rotation time ÷ sectors per track.
• Low-order interleaving helps block transfers; high-order does not.

7 · Memory Hook

🧠

"Seek, spin, read." Access = seek + half a rotation + transfer.

8 · Connect the Dots

Chapter 4.1 is complete: hierarchy, AMAT, chips, interfacing, disks and interleaving. Chapter 4.2 opens the cache itself: where a memory block may be placed, how an address is split into tag, line and offset, and how hits are detected.

9 · Self-Test

GATE format. Try every question before opening the vault.

Q1NAT1 markRotational latency

A disk spins at 6,000 rpm. Average rotational latency in ms?

Numerical answer: ________

Q2NAT2 marksCapacity

A disk has 4 platters (both surfaces used), 2,048 tracks per surface, 128 sectors per track and 512 bytes per sector. Capacity in GB (binary, 1 GB = 2³⁰ bytes)?

Numerical answer: ________

Q3NAT2 marksAccess time

Seek 4 ms, 10,000 rpm, 400 sectors per track. Average time to read one sector (ms)?

Numerical answer: ________

Q4NAT2 marksInterleaving

Module access time 50 ns, bus transfer 5 ns per word. Time to fetch an 8-word block with 8-way low-order interleaving (ns)?

Numerical answer: ________

Q5MCQ1 markInterleaving kind

Which interleaving places consecutive word addresses in different modules?

• (A) Low-order
• (B) Both
• (C) High-order
• (D) Neither

10 · Rule Card

• Capacity = surfaces × tracks × sectors × bytes. Cylinder = same track on all surfaces.
• Access = seek + ½ rotation + transfer; rotation = 60/rpm s; transfer rate = bytes per track × rps.
• Low-order interleaving: block of k words in t_a + k·t_b vs k(t_a + t_b).

11 · Concept Ledger

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

platter, surface, track, sector, cylinderdisk capacityseek timerotational latencytransfer time and ratememory bandwidthmemory modules (banks)low-order vs high-order interleavinginterleaved block fetch time

12 · Answer Vault