Multicycle CPU Performance Comparison: 25-Comp-A3 Computer Architecture, May 2013, Q5

Question 5: Multicycle 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 per instruction class: LI = 6 (original) / 5 (modified) cycles, ARITHMETIC = 5, MEMORY READ = 5, MEMORY STORE = 4, BRANCH = 4 cycles (unchanged by the modification). Instruction mix: LI 10%, ARITHMETIC 50%, MEMORY READ 20%, MEMORY STORE 10%, BRANCH 10%. The modified design's clock frequency is 5% lower than the original's.

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

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

Average CPI, original design. $$CPI_{orig}=0.10(6)+0.50(5)+0.20(5)+0.10(4)+0.10(4)=0.6+2.5+1.0+0.4+0.4=\boxed{4.9}\ \text{cycles/instr}.$$
Average CPI, modified design (same mix; only LI's own cycle count drops from 6 to 5). $$CPI_{mod}=0.10(5)+0.50(5)+0.20(5)+0.10(4)+0.10(4)=0.5+2.5+1.0+0.4+0.4=\boxed{4.8}\ \text{cycles/instr}.$$
Convert CPI to execution time per instruction. Let $T$ be the original clock's cycle time. The modified clock runs at $0.95\times$ the original frequency, so its cycle time is longer, $T/0.95$: $$t_{orig}=CPI_{orig}\times T=4.9T,\qquad t_{mod}=CPI_{mod}\times\frac{T}{0.95}=\frac{4.8}{0.95}T=5.0526\,T.$$
Compare. Since $4.9T\lt5.0526T$, $$\boxed{t_{orig}\lt t_{mod}\ \Rightarrow\ \text{the ORIGINAL implementation executes programs faster.}}$$ The modified design takes $t_{mod}/t_{orig}=5.0526/4.9=1.0311$ times as long per instruction as the original — i.e. the original is about $\dfrac{t_{mod}-t_{orig}}{t_{mod}}\times100\%\approx3.0\%$ faster (equivalently, the modified design is about $\dfrac{t_{mod}-t_{orig}}{t_{orig}}\times100\%\approx3.1\%$ slower). Lowering LI's own CPI contribution (by $0.1$ cycle, weighted $10\%$) was not enough to offset the 5% clock-frequency penalty that applies to every instruction.

Final results — Question 5
Quantity	Original	Modified
Average CPI	4.9	4.8
Relative cycle time	$T$	$T/0.95=1.0526T$
Avg. execution time/instr.	$4.9T$	$5.0526T$
Faster implementation	Original, by ≈3.0% (modified is ≈3.1% 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 drops by fraction $x$, cycle time scales by $1/(1-x)$.

Why this works: lowering the cycle count for one instruction class only helps overall performance if the resulting CPI improvement outweighs any accompanying clock-period penalty that applies to every single instruction, not just the improved class.

Common pitfall: comparing CPI alone (4.8 < 4.9) and concluding the modified design is faster, while ignoring that its clock runs 5% slower — a smaller 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: Multicycle CPU Performance Comparison

Question 5: Multicycle CPU Performance Comparison (20 marks)