Worked examples — Approximate computing techniques

The scenario matrix

Every problem in this topic is one of these cells. The goal of this page is: one worked example per cell.

Before we compute, we need one symbol nailed down, because it appears everywhere below.

Example 1 — C1: unbiased error over a big sum

Steps.

1. Total error $E={\sum }_{i=1}^N{\epsilon }_i$. Its variance is $\mathrm{Var}(E)=N{\sigma }^2$. Why this step? Independent zero-mean errors add in variance, not in magnitude — the classic $\sqrt{N}$ rule from the parent note.
2. Typical error size $=\sqrt{N{\sigma }^2}=\sqrt{N}\sigma =\sqrt{10000}\times 0.5=100\times 0.5=50$. Why? Standard deviation is $\sqrt{\text{variance}}$; that's the "typical" spread.
3. Signal (the true sum) $=N\mu =10000\times 50=500,000$. Why? We need something to compare the error against — the sum itself.
4. Relative error $=\frac{50}{500000}=0.0001=0.01\%$.

Example 2 — C2: the biased trap

Steps.

1. The bias accumulates linearly: $\mathbb{E}[E]=\sum \mathbb{E}[{\epsilon }_i]=N\times (-0.2)=-2000$. Why this step? Means add directly — there is no cancellation when they all share a sign.
2. Compare to the zero-mean part, which (with $\sigma =0.5$) is only $\pm \sqrt{N}\sigma \approx \pm 50$. Why? To see which term dominates. The $-2000$ swamps the $\pm 50$.
3. Relative error $=\frac{2000}{500000}=0.004=0.4\%$ — 40× worse than Example 1, and it keeps growing with $N$, not shrinking.

Example 3 — C3: truncating a positive product (one-sided sign)

Steps.

1. Exact: $1.375\times 1.625=2.234375$. Why? Ground truth. $A=1.0110_2=1+\frac{1}{4}+\frac{1}{8}$, $B=1.1010_2=1+\frac{1}{2}+\frac{1}{8}$.
2. In binary, $2.234375=10.001111_2$ (6 nonzero fractional bits). Keep 4 fractional bits: $10.0011_2=2.1875$. Why? Truncation chops the tail — hardware simply omits those partial-product columns.
3. Error $=2.234375-2.1875=0.046875$. This is always $\ge 0$: you removed positive weight, so the truncated result is never larger than the true one. Why this matters? Truncation is a biased approximation → it belongs to cell C2's danger family when summed many times.
4. Relative error $=0.046875/2.234375\approx 0.0210=2.1\%$.

Example 4 — C4: rounding instead (symmetric error)

Steps.

1. Tail being dropped is $0.00001100_2=0.046875$, which is more than half of one 4th-bit unit ($2^{-4}=0.0625$; half is $0.03125$). Why this step? Rounding looks at the tail: if $\ge$ half a unit, round up.
2. Since $0.046875>0.03125$, round up: $2.1875+0.0625=2.25$. Why? Rounding pushes to the nearest representable value, not always down.
3. Rounding error $=2.234375-2.25=-0.015625$ (now negative — the sign can go either way). Magnitude $0.015625<0.046875$: 3× smaller than truncation.

Example 5 — C5: precision-scaling energy, the $n^2$ law

Steps.

1. Ratio $=\frac{E_{24}}{E_8}=\frac{24^2}{8^2}=\frac{576}{64}=9$. Why the $n^2$? An array multiplier tiles $\approx n^2$ full-adder cells; each cell burns energy → total scales with area $\propto n^2$.
2. So an INT8 multiply uses about $9\times$ less energy than the FP32 mantissa multiply. Why $24$, not $32$? FP32 is 32 bits stored, but 1 sign + 8 exponent bits do not enter the mantissa multiplier — only the 24-bit significand is multiplied (see Precision and Number Formats (FP32, FP16, INT8)).
3. Degenerate check. Set $n\to n/2$: ratio $=(n/(n/2){)}^2=2^2=4$. That reproduces the parent note's "halving = 4×." Why? The $n^2$ law is scale-free — only the ratio of bit-widths matters.

Example 6 — C6: voltage overscaling, $V^2$ win vs the $V_{th}$ wall

Steps.

1. Power ratio $=\frac{P^{\prime }}{P}=\frac{V^{\prime 2}}{V^2}=\frac{0.8^2}{1.0^2}=0.64$. Why $V^2$ and not $V$? Each switching node charges a capacitor to voltage $V$, storing energy $\frac{1}{2}C{V}^2$ — the square is physics, not a choice.
2. Savings $=1-0.64=0.36=36\%$. Why care? A 20% voltage drop buys a 36% power cut — the "$V^2$ jackpot."
3. But the risk (the $V$ term): gate delay $\propto V/(V-V_{th}{)}^2$ rises as $V\to V_{th}$. At $V_{th}=0.3$ V: delay factor $\propto V/(V-0.3{)}^2$. At $V=1.0$: $1/0.49\approx 2.04$. At $V=0.8$: $0.8/0.25=3.20$ → $1.57\times$ slower. Some paths now miss the clock → timing errors. Why this matters? On the figure, the green power curve falls smoothly while the red delay curve shoots up near $V_{th}$ — that's the error cliff. VOS lives just left of the safe point but right of the cliff.

Example 7 — C7: loop perforation, real image (word problem)

Steps.

1. Speedup: we skip half the rows → $\approx 2\times$ fewer iterations ⇒ ~2× faster. Why acceptable? Adjacent rows in natural images are highly correlated (redundancy → C1-style resilience).
2. PSNR $=10{\log }_{10}\left(\frac{255^2}{6.5}\right)=10{\log }_{10}\left(\frac{65025}{6.5}\right)=10{\log }_{10}(10004)\approx 40.0$ dB. Why this formula? It's the quality ruler from the top of the page — MAX${}^2$ over MSE.
3. $40.0\ge 40$ ✅ — perforation-by-2 just passes. Push to perforate-by-4 and MSE roughly quadruples, dropping PSNR by $\approx 6$ dB → fails. Why $6$ dB per $4\times$? $10{\log }_{10}(4)\approx 6.02$; quadrupling MSE subtracts $6$ dB.

Example 8 — C8: error accumulation in a feedback loop

Steps.

1. Solve the recursion: with $x_0=0$, $x_k=\epsilon \cdot \frac{g^k-1}{g-1}$ (geometric series). Why this step? Feedback multiplies past errors by $g$ each round — the closed form exposes the growth.
2. Plug in: $x_{500}=0.001\cdot \frac{1.01^{500}-1}{0.01}$. Since $1.01^{500}\approx 144.77$, $x_{500}\approx 0.001\times \frac{143.77}{0.01}=0.001\times 14377\approx 14.38$. Why alarming? A $0.001$ per-step error became $\approx 14.4$ — amplified $\sim 14000\times$.
3. Contrast with $g=1$ (no amplification): $x_{500}=500\times 0.001=0.5$. Same per-step error, 28× smaller. Why? When $g>1$ the loop is unstable and errors compound; the $\sqrt{N}$/linear intuitions do NOT apply.

Example 9 — C9: exam twist — approximate the RIGHT field

Steps.

1. Mantissa LSB flip. FP16 mantissa has 10 bits; the LSB is worth $2^{-10}$ of the mantissa. New value $\approx (1.5\pm 2^{-10})\times 2^4=24\pm 0.015625$. Relative error $\approx 0.065\%$. Why small? The mantissa only sets fine detail — this is exactly the LSB we're allowed to approximate.
2. Exponent flip (say $2^4\to 2^{4+8}=2^{12}$ by flipping a high exponent bit): value $\to 1.5\times 2^{12}=6144$. That's $256\times$ too big — magnitude destroyed. Why catastrophic? The exponent controls scale; one wrong bit multiplies the whole number by a power of two.
3. Rule. Approximate mantissa LSBs and pixel data; never exponents, pointers, loop bounds, or control flow. Protect those critical bits with Error-Correcting Codes. Why? A mantissa slip is a $0.065\%$ nuisance; an exponent slip is a $256\times$ (or worse) disaster — the two live on completely different scales, so the field you approximate matters more than how many bits you touch.

Example 10 — C10: zero & degenerate limits (sanity boundary)

Steps.

1. $N=1$ (no aggregation). Relative error $=\frac{\sigma }{\mu \sqrt{1}}=\frac{\sigma }{\mu }$ — the full single-op error, no rescue. With $\sigma =0.5,\mu =50$: $0.01=1\%$. Why? Aggregation needs many terms; one term cancels nothing. The $1/\sqrt{N}$ shrink only starts helping for $N>1$.
2. Error $=0$ (exact op). Then $\sigma =0$ → relative error $=0$ for all $N$. The exact machine is the $\sigma \to 0$ limit of the approximate one — consistent, no discontinuity. Why check? An approximate model must reduce to the exact model when you dial the error to zero.
3. $n=0$ bits (degenerate precision). Multiplier energy $\propto n^2=0$: zero bits ⇒ zero hardware ⇒ zero energy and zero information. The $n^2$ law limits cleanly to the trivial machine at $n=0$. Why check? A good formula must not explode or contradict itself at its boundary.

Active-recall