If a microcomputer operates at 5 MHz with an 8-bit bus and a newer version…
2011
If a microcomputer operates at 5 MHz with an 8-bit bus and a newer version operates at 20 MHz with a 32-bit bus, the maximum speed-up possible approximately will be
- A.
2
- B.
4
- C.
8
- D.
16
Attempted by 19 students.
Show answer & explanation
Correct answer: D
Concept
When a system is upgraded along two independent performance dimensions, the maximum theoretical speed-up is the PRODUCT of the individual improvement factors, not their sum. Each factor is an independent multiplier on throughput, so the gains compound.
For a bus-based microcomputer, data throughput is proportional to (clock frequency) × (bus width): the clock frequency sets how many bus cycles occur per second, and the bus width sets how many bits move per cycle. Improving either one multiplies throughput; improving both multiplies the two multipliers together.
Application
Clock-frequency factor = new clock ÷ old clock = 20 MHz ÷ 5 MHz = 4.
Bus-width factor = new width ÷ old width = 32-bit ÷ 8-bit = 4.
Maximum speed-up = (clock factor) × (bus factor) = 4 × 4 = 16.
Cross-check / contrast
Using only the clock factor gives 4 — this ignores the wider bus, so it understates the gain.
Adding the factors gives 4 + 4 = 8 — independent throughput multipliers compound, they are never added.
A value of 2 reflects neither factor fully and corresponds to no consistent calculation from the given figures.
Only the product 4 × 4 = 16 captures both independent improvements together, so it is the maximum theoretical speed-up.
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