# Understanding error propagation in fixed-point multiplication

I'm writing a test suite that checks the correctness of a fixed-point arithmetic library that I wrote. Specifically, it deals with Q4.4 numbers, i.e. 4 integer bits and 4 fractional, so its precision is 1/16.

The test suite produces two rational numbers and computes their product, then transforms these rational numbers to Q4.4 numbers and computes their Q4.4 product, then checks that the Q4.4 product is within the expected absolute error of the correct product.

Let $$a$$, $$b$$ be the true numbers (assumed for simplicity to be nonzero), let $$\Delta x$$ be the absolute error, let $$\epsilon_x$$ be the relative error. I computed the absolute error of $$ab$$ to be $$\Delta ab = \epsilon_{ab} * |ab| = (\epsilon_a + \epsilon_b) * |ab| = (|\frac{\Delta a}{a}| + |\frac{\Delta b}{b}|) * |ab|$$, with an upper bound being $$\epsilon_{ab} \leq (|\frac{1/32}{a}| + |\frac{1/32}{b}|)*|ab|$$ since $$\Delta a$$ can be at most half the precision.

This reasoning seemingly fails when computing $$1.3488 * 3.2238 = 4.34826144$$. When expressed with Q4.4 numbers this multiplication gives $$1.375 * 3.250 = 4.500$$, and its actual absolute error is $$\Delta ab = 0.15173856$$; however, the expected value per the formula above is $$4.34826144 * (\frac{1/32}{1.3488} + \frac{1/32}{3.2238}) \approx 4.34826144 * (0.02316874 + 0.00969353) = 4.34826144 * 0.03286227 \approx 0.14289374$$.

To make it clear: the actual value is $$0.152...$$, the expected value is $$0.143...$$

Where am I wrong?

Following on from Paul's answer above, I believe you also need to account for the quantisation error in $$c$$ (the product $$ab$$ might not be representable in your fixed point format; in your case, $$1.375\times 3.25 = 4.46875$$, but Q4.4 quantises this to $$4.5$$), so your starting point is:

$$(a\pm\Delta)(b\pm\Delta) = (c\pm\Delta\pm\delta)$$

where $$\delta$$ is the error you'd expect from propagating through the errors in the multiplication of $$a$$ and $$b$$. Your relative error is then:

$$\epsilon_{ab} = \frac{\pm\delta\pm\Delta}{c} = \epsilon_a + \epsilon_b + \epsilon_a\epsilon_b$$

You are going wrong in forgetting that one of your error formulas is only approximate, not exact.

If we multiply $$(a +\Delta a) \times (b + \Delta b)$$, the result is $$ab + a\Delta b + b\Delta a + \Delta a\Delta b$$ so the relative error is $$\epsilon_{ab} = \frac{a\Delta b + b\Delta a + \Delta a\Delta b}{ab} = \frac {\Delta a}a + \frac {\Delta b}b + \frac{\Delta a}a \frac{\Delta b}b = \epsilon_a + \epsilon_b + \epsilon_a\epsilon_b$$

Now normally, relative errors are $$\ll 1$$, so the product makes a negligible contribution, so we usually just ignore it and pretend that $$\epsilon_{ab} = \epsilon_a + \epsilon_b$$. But in your case the relative errors are big enough that this term can make a measurable contribution.