@@ -120,20 +120,20 @@ Calculating the product of a list of numbers can result in very high numbers, wh

We mitigate this problem by utilizing an equivalent formula of the weighted product, that directly calculates the ratio of the weighted product of two given assignments.

To understand the problem, let $x =\tuple{p_1, p_2, p_3, q_1}$ and $y =\tuple{P_1, P_2, P_3, Q_1}$ be the qualities of two assignments, where $p$ and $P$ are more-is-better criteria, and $q$ and $Q$ are less-is-better.

For this example, we take all weights to be 1.

Let $\tuple{w_1, w_2, w_3, w_4}$ be the weights.

We could calculate their scores individually and then take the ratio of the scores, as follows.

The result $r$ is the factor by how much better $x$ is than $y$.

If $r$ is greater than 1, then the score of $x$ is $r$ times better than the score of $y$.

If $r$ is less than 1, then $y$ is $r$ times better than $x$.

This calculation has the subexpressions$p_1 p_2 p_3$ and $P_1 P_2 P_3$, which can result in an intermediate result large enough to cause information loss.

This calculation has the subexpression $p_1^{w_1} p_2^{w_2} p_3^{w_3}$, and similar for the P's, which can result in an intermediate result large enough to cause information loss.

By using elementary laws of multiplication and division, we can alternatively calculate $r$ as follows.

\begin{IEEEeqnarray*}{c}

r = \frac{p_1}{P_1}\times\frac{p_2}{P_2}\times\frac{p_3}{P_3}\times\frac{Q_1}{q_1}

r = (\frac{p_1}{P_1})^{w_1}\times(\frac{p_2}{P_2})^{w_2}\times(\frac{p_3}{P_3})^{w_3}\times(\frac{Q_1}{q_1})^{w_4}

\end{IEEEeqnarray*}

We know that the $p_i$ and $P_i$ are objectives of the same capability requirement, so they are most likely in the same order of magnitude.

This results in smaller intermediate numbers, which decreases the chance for information loss.