2.13.md

2.13

Question

Show that under the assumption of small percentage tolerances there is a simple formula for the approximate percentage tolerance of the product of two intervals in terms of the tolerances of the factors. You may simplify the problem by assuming that all numbers are positive.

Answer

We see the percentage of the product is approximately the sum of the percentages:

$$ \begin{align*} p_{x * y} &= 2(y_2 x_2 - x_1 y_2) / (x_1 y_1 + x_2 y_2) \\ &= 2(y_2 x_2 - x_1 y_2) / (x_1 y_1 + x_2 y_2) \\ &= 2 ((\mu_x \mu_y) (p_x + p_y)) / (2 \mu_x \mu_y) \\ &= p_x + p_y \\ \end{align*} $$

For

$$ \begin{align*} \mu_x = \text{center}(x) &= (x_1 + x_2) / 2 \\ p_x = \text{percent}(x) &= 2 (x_2 - x_1) / (x_1 + x_2) \\ x * y= \text{product}(x, y) &= [x_1 y_1, x_2 y_2] \text{ (assuming positive)} \\ \end{align*} $$

Where

$$ \begin{align*} x_1 &= x_1 \mu_x / \mu_x \\ &= x_1 ((x_1 + x_2) / 2) / ((x_1 + x_2) / 2) \\ &= (2 x_1) ((x_1 + x_2) / 2) / (x_1 + x_2) \\ &= ((x_1 + x_2) / 2) (2 x_1) / (x_1 + x_2) \\ &= ((x_1 + x_2) / 2) (2 x_1) / (x_1 + x_2) \\ &= ((x_1 + x_2) / 2) (2 x_1 + x_2 - x_2) / (x_1 + x_2) \\ &= ((x_1 + x_2) / 2) ((x_1 + x_2) - (x_2 - x_1)) / (x_1 + x_2) \\ &= ((x_1 + x_2) / 2) (1 - (x_2 - x_1) / (x_1 + x_2)) \\ &= (\mu_x) (1 - p_x / 2) \\ &= \mu_x - (\mu_x p_x) / 2 \\ x_2 &= \mu_x + (\mu_x p_x) / 2 \text{ (by similar logic)} \implies \\ \\ x_1 y_1 &= (\mu_x - \mu_x p_x / 2)(\mu_y - \mu_y p_y / 2) \\ &= \mu_x \mu_y - \mu_y (\mu_x p_x / 2) - \mu_x (\mu_y p_y / 2) + (p_x p_y) (\mu_x \mu_y) / 4 \\ &\approx \mu_x \mu_y - \mu_y (\mu_x p_x / 2) - \mu_x (\mu_y p_y / 2) + (0) (\mu_x \mu_y) / 4 \text{ (assuming $p_x p_y \approx 0$)}\\ &= \mu_x \mu_y - \mu_y (\mu_x p_x / 2) - \mu_x (\mu_y p_y / 2) \\ &= \mu_x \mu_y (1 - p_x / 2 - p_y / 2) \\ &= \mu_x \mu_y (1 - (p_x + p_y) / 2) \\ x_2 y_2 &= \mu_x \mu_y (1 + (p_x + p_y) / 2) \text{ (by similar logic)} \implies \\ \\ x_2 y_2 - x_1 y_1 &= \mu_x \mu_y (1 + (p_x + p_y) / 2) - \mu_x \mu_y (1 - (p_x + p_y) / 2) \\ &= \mu_x \mu_y (1 + (p_x + p_y) / 2 - 1 + (p_x + p_y) / 2) \\ &= (\mu_x \mu_y) (p_x + p_y) \\ x_1 y_1 + x_2 y_2 &= \mu_x \mu_y (1 - (p_x + p_y) / 2) + \mu_x \mu_y (1 + (p_x + p_y) / 2) \\ &= \mu_x \mu_y (1 - (p_x + p_y) / 2 + 1 + (p_x + p_y) / 2) \\ &= 2 \mu_x \mu_y \\ \end{align*} $$