Weighted Sum Model

Weighted Sum Model

In a MOO and MCDA Problems there are a set of the “best” (Pareto-optimal) solutions. How to decide which one to take?

One possible approach is Weighted Sum

Suppose we have the following table:

• $f_1, …, f_q$ are criteria
• $a_1, …, a_n$ are alternatives
• each criteria is assigned some weight $w_i$

| | $f_1$ | $f_2$ | … | $f_q$ | $a_1$ | $f_1(a_1)$ | $f_2(a_1)$ | … | $f_q(a_1)$ || $a_2$ | $f_1(a_2)$ | $f_2(a_2)$ | … | $f_q(a_2)$ || … | … | … | … | … || $a_n$ | $f_1(a_n)$ | $f_2(a_n)$ | … | $f_q(a_1)$ || | $w_1$ | $w_2$ | … | $w_q$ | | We rank alternatives based on the following score:

• $V(a) = \sum_{i = 1}^{q} w_i f_i (a)$
• $a_i \ P \ a_j \iff V(a_i) > V(a_j)$
• $a_i$ is preferred to $a_j$ if $V(a_i) > V(a_j)$

Downsides

This approach is very simple, but introduces some effects on the decision

Bad Use Of Information

We consider only one aggregated value, and don’t see all the data

• cannot identify weak and strong points

Example:

| | $f_1$ | $f_2$ | $f_3$ | $f_4$ | $a$ | 5 | 5 | 5 | 2 || $b$ | 4 | 4 | 4 | 4 || $w$ | 0.25 | 0.25 | 0.25 | 0.25 | |- $V(a) = 4.25, V(b) = 4$

• $b$ is never the best one
• $a$ is the best almost at all criteria, except for $f_4$ (say he’s in IT and this is a communication skill)
• cannot identify that by only looking at the $V(a)$ and $V(b)$ scores

Conflicts

| | $f_1$ | $f_2$ | $a$ | 5 | 5 || $b$ | 10 | 0 || $c$ | 0 | 10 || $d$ | 5 | 5 || | 0.5 | 0.5 | |$V(a) = V(b) = V(c) = V(d) = 0.5$

• but inside they are very different| | |

  Scale is Important

• suppose that we say that production $p$ is more important then quality $q$
• i.e. $w_p = 2/3, w_q = 1/3$
• $p$ - production per month

| | $p$ | $q$ | Score | $a$ | 100 | 100 | 100 || $b$ | 120 | 80 | 106.6 || $w$ | 2/3 | 1/3 | | In this case $b$ is better

But suppose now we want to use $p$ - production per week | | $p$ | $q$ | Score | $a$ | 25 | 100 | 50 || $b$ | 30 | 80 | 46 || $w$ | 2/3 | 1/3 | | Now all of a sudden $a$ becomes better

• because the scale changed
• need normalization

Not All Solutions

With weighted sum not all the solutions can be found

• With weighted sum approach we cannot find all the efficient solutions just by maximizing the sum

• (a) this solution can be found: this frontier is convex
• (b) cannot be found: only the convex part is discovered

Sources

• Decision Engineering (ULB)