DSE ICT | Computer Hardware and Peripherals Lesson-Prep Guide (Tutor Edition) | CPU Performance, Memory Hierarchy and I/O Bottlenecks

2026-11-02

Tutor lesson-prep guides · Secondary ICT

Students can say one machine is faster, but not which part is doing the work. This lesson splits speed into clock rate, cache hit rate and disk rotation with three sums.

Objectives — Students can: (1) compute instructions per second from clock rate and CPI, and explain why a higher clock does not always win; (2) compute average memory access time from hit rate and miss penalty (AMAT = 6 cycles) and explain what the cache buys; (3) compute a 4 MB disk read at about 46.5 ms from seek time, rotation and transfer rate, and explain why an SSD is roughly 465 times faster.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Clock rate	How many cycles per second is 3.2 GHz	3.2 billion
Unit conversion	How many MB is 8 GB	8,192 MB, or 2^33 bytes
Averages	Ninety-five per cent hit, five per cent miss, what is the average	Weight by probability, not the midpoint

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Five parts of a machine	CPU, memory, storage, input, output	Trace the data path
10-22 min	CPU performance	Clock, CPI, instructions per second	Compute 1.28 billion per second
22-34 min	Memory hierarchy	Registers, cache, RAM, disk	Compute AMAT of 6 cycles
34-44 min	The I/O bottleneck	Seek, rotational delay, transfer rate	Compute 46.5 ms
44-54 min	Choosing peripherals	Scanners, printers, touch, resolution	Justify each choice
54-60 min	Exit check	Three questions marked on the spot	All three right

03 Teaching script (read it out)

Scene 1: The clock is a heartbeat, CPI is steps per beat

A CPU runs at 3.2 GHz, or 3.2 billion cycles a second. If the average instruction needs 2.5 cycles (CPI), it executes 3.2 x 10^9 / 2.5 = 1.28 x 10^9 instructions per second, about 1.28 GOPS.
So a higher clock wins only sometimes: a 2.8 GHz chip at CPI 1.4 reaches 2.0 x 10^9, which is faster. The figure that matters is clock divided by CPI, plus core count.
Cores do not simply multiply: two cores help only if the program can run in parallel, otherwise one works while the other waits.
Peripherals sit on the same picture: keyboard and mouse are input, monitor and printer are output, touchscreens and network cards do both. A slow device makes a fast CPU wait.

Ask students: why a 3.2 GHz machine can be slower than a 2.8 GHz one (different CPI, whether cores can work in parallel).

Scene 2: The memory hierarchy trades time against hit rate

From fast to slow: registers (within a cycle), L1/L2 cache, RAM, SSD, hard disk.
A cache hit costs 1 cycle and a miss costs a 100-cycle penalty. At a 95% hit rate, AMAT = 1 + 0.05 x 100 = 6 cycles.
If the hit rate falls to 80%, AMAT = 1 + 0.2 x 100 = 21 cycles, three and a half times slower. How a program lays out its data (locality) therefore changes real speed.
Know the conversions: 8 GB = 8 x 1,024 MB = 8,192 MB; with 4 KB pages, 8 GB holds 8 x 1024^3 / 4096 = 2,097,152 pages.
RAM is volatile and loses data on power loss; SSD and hard disk are non-volatile. That distinction appears almost every year.

Ask students: why a fall from 95% to 80% hit rate raises the average from 6 to 21 cycles (the miss penalty is amplified).

Scene 3: Three delays add up to the real disk time

Reading a 4 MB file: seek time 9 ms; at 7,200 rpm one revolution takes 60 / 7,200 = 8.33 ms, so average rotational delay is half, 4.17 ms; at 120 MB/s the transfer takes 33.33 ms. The total is about 46.5 ms.
The same file on an SSD takes about 0.1 ms, roughly 465 times faster, because there are no moving parts and no waiting for rotation.
So an SSD transforms booting and loading, but barely affects pure computation, which is the distinction examiners look for.
Peripheral choices need a purpose: document scanning needs optical resolution and a feeder; photo printing needs droplet size and gamut; touchscreens need sampling rate, not just size.

Ask students: why an SSD speeds up booting but not video encoding (I/O bound against CPU bound).

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation	Instructions per second	Clock rate / CPI	1.28 x 10^9
Calculation	Average memory access time	Hit time + miss rate x penalty	6 cycles
Calculation	Disk access time	Seek + rotational delay + transfer	46.5 ms
Explanation	Choosing a peripheral	Resolution, speed and cost for the stated use	Every claim needs a reason

Sanity check (teach this): GHz means 10^9, not 10^6; a unit slip throws the answer out by a thousand.
The miss rate in AMAT is 1 minus the hit rate, not the hit rate.
Average rotational delay is half a revolution, not a full one.
1 GB is 1,024 MB in binary, not 1,000 MB, unless the question says decimal.
Core count and clock rate cannot be multiplied into a performance figure, because parallelism depends on the program.

05 Independent practice (3 questions)

Question	Answer	What to watch
Instructions per second at 3.2 GHz and CPI 2.5	1.28 x 10^9	3.2e9 / 2.5
AMAT at 95% hit rate with a 100-cycle penalty	6 cycles	1 + 0.05 x 100
Transfer time for 4 MB at 120 MB/s	33.33 ms	4 / 120 x 1000

06 Common mistakes and what to say

Mistake	Student writes	Say this
Multiplying by the hit rate	Answering 1 + 0.95 x 100	The miss rate is the multiplier: 0.05 x 100, and a hit costs only 1.
Forgetting to halve the rotational delay	Answering that 8.33 ms is added	On average you wait half a revolution, so 4.17 ms.
Treating 1 GB as 1,000 MB	Answering that 8 GB is 8,000 MB	Binary uses 1,024, so 8 GB is 8,192 MB.
Assuming an SSD speeds up everything	Answering that an SSD doubles video encoding speed	It removes I/O delay; CPU work is unchanged.

07 Homework (paper, section and question numbers)

Secondary ICT chapter index — All chapters and practice
Secondary lesson-prep overview — Compare other subjects

08 Exit check (last 5 minutes)

Question	Answer	Decision
Instructions per second at 3.2 GHz, CPI 2.5	1.28 x 10^9	All three right means move to I/O
AMAT at a 95% hit rate	6 cycles	One wrong means redo miss rates
Average rotational delay	4.17 ms	Two wrong means repeat act three

09 Tutor FAQ

Must students memorise model specifications?

No. They must compute from given clock, CPI and rpm, and say where the bottleneck is.

Is a bigger cache always better?

Hit rate rises, but cost, power and area rise too, and returns fall once the cache exceeds the working set.

Does an SSD have drawbacks?

Yes: higher cost per GB, limited write endurance and harder data recovery. Two of these are enough.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.