Percentage difference between loaded miles and empty miles

Question

I dispatch for a trucking company. I'm trying to find a formula to work out the percentage difference between loaded miles and empty miles. For example: One driver drives 200 miles. 150 loaded and 50 empty. That's 75% loaded. I need a formula to figure out that same situation but with numbers that aren't quite so round. I tried $$\left(\frac{x-y}{\frac{x+y}{2}}\right)\cdot 100$$ but it didn't work out the way I'd hoped. Any help would be great. Thanks.

Answer 1

Let $L=$Loaded Miles and $U=$Unloaded Miles.

The percentage of loaded to unloaded would be $$100 \times \frac{L}{U+L}$$

For example if (as in your example) $L=150, U=50$ then percentage of loaded miles is \begin{eqnarray} 100 \times \frac{150}{150+50} &=& 100 \times \frac{150}{200} \\ &=& 100 \times 0.75 \\ &=& 75\% \end{eqnarray} This works on the premise you know at least two values from Total Miles, $U$ and $L$.

Answer 2

I'm not sure you're asking this. But anyways: Let x represent the loaded miles, y the empty miles. Then $$\frac{100y}{(x+y)}=z,$$ where $z$ is the percentage.