Multi-Cycle CPU Performance Comparison: 25-Comp-A3 Computer Architecture, May 2014, Q5

Question 5: Multi-Cycle CPU Performance Comparison (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Cycle counts: original LI=4, ARITHMETIC=3, MEMORY READ=5, MEMORY STORE=4, BRANCH=3 (modified LI=5, all others unchanged). Instruction mix: LI 10%, ARITHMETIC 50%, MEMORY READ 20%, MEMORY STORE 10%, BRANCH 10%. The modified design's clock frequency is 3% HIGHER than the original's.

Find. Which implementation executes the given mix faster, and by what percentage.

Approach. Compute each design's average CPI from the weighted cycle counts, convert to actual execution time by scaling for the modified design's faster clock (shorter cycle time), then compare.

Average CPI, original design. $$CPI_{orig}=0.10(4)+0.50(3)+0.20(5)+0.10(4)+0.10(3)=0.4+1.5+1.0+0.4+0.3=\boxed{3.6}\ \text{cycles/instr}.$$
Average CPI, modified design (only LI's own cycle count rises, from 4 to 5). $$CPI_{mod}=0.10(5)+0.50(3)+0.20(5)+0.10(4)+0.10(3)=0.5+1.5+1.0+0.4+0.3=\boxed{3.7}\ \text{cycles/instr}.$$
Convert CPI to execution time per instruction. Let $T$ be the original clock's cycle time. The modified clock runs at $1.03\times$ the original frequency, so its cycle time is SHORTER, $T/1.03$: $$t_{orig}=CPI_{orig}\times T=3.6T,\qquad t_{mod}=CPI_{mod}\times\frac{T}{1.03}=\frac{3.7}{1.03}T=3.5922\,T.$$
Compare. Since $3.5922T\lt3.6T$, $$\boxed{t_{mod}\lt t_{orig}\ \Rightarrow\ \text{the MODIFIED implementation executes programs faster.}}$$ The original design takes $t_{orig}/t_{mod}=3.6/3.5922=1.00216$ times as long per instruction as the modified one — the modified design is about $\dfrac{t_{orig}-t_{mod}}{t_{orig}}\times100\%\approx0.22\%$ faster (equivalently, the original is about $\dfrac{t_{orig}-t_{mod}}{t_{mod}}\times100\%\approx0.22\%$ slower). The 3% clock-frequency gain (applying to EVERY instruction) modestly outweighs the CPI penalty from LI needing one extra cycle (weighted only 10% of the mix), unlike a case where the CPI penalty is heavier or the clock gain smaller.

Final results — Question 5
Quantity	Original	Modified
Average CPI	3.6	3.7
Relative cycle time	$T$	$T/1.03=0.9709T$
Avg. execution time/instr.	$3.6T$	$3.5922T$
Faster implementation	Modified, by ≈0.22% (original is ≈0.22% slower)

Topic: CPU performance equation — CPI times clock period, not CPI alone.

Key relations: $\text{ExecTime}=\text{InstrCount}\times CPI\times T_{cycle}$; average $CPI=\sum_i freq_i\times cycles_i$; if frequency rises by fraction $x$, cycle time scales by $1/(1+x)$.

Why this works: a HIGHER average CPI does not automatically mean SLOWER execution once the clock period also changes — a small, uniform clock-frequency gain applies to every single instruction executed, so it can outweigh a CPI penalty concentrated in one infrequent instruction class.

Common pitfall: comparing CPI alone (3.6 < 3.7) and concluding the original design is faster, while ignoring that the modified clock runs 3% faster — a lower CPI does not imply less execution time once the clock period also changes.

Source: Patterson & Hennessy, Computer Organization and Design, 6th ed., Ch.1 §1.6 (performance equation), Ch.4 (the multicycle processor).

Question 5 of 6: Multi-Cycle CPU Performance Comparison

Question 5: Multi-Cycle CPU Performance Comparison (20 marks)