I have a question regarding the Hypervolume indicator in multi-objective optimization:
As far as I know, the Hypervolume is the volume of the trade-off space covered by the Pareto-front. I.e., finding the best possible front implies maximizing the Hypervolume.
Does this definition imply that we want to minimize our multiple objectives? What happens to the sign of the Hypervolume when we are maximizing one or more of the objectives? In other words, are we always trying to maximize the Hypervolume, or are there cases where we want to minimize?
Dear Thomas,
In case if any objective is of maximization type, convert it into minimization type by multiplying it with -1 and then optimize the problem as usual. This wouldn't conflict with aim of maximizing the hypervolume.
Hi Thomas,
In general, I agree with the answers above. The hypervolume indicator, also known in the literature as Lebesgue measure or S-metric, is one of the most popular indicators in Multi Objective Optimization (MOO) problems.
In its essence, the Hypervolume indicator must determine what is the portion of the objective space that is weakly dominated by the an approximation set. So, in that sense we always want to maximize that volume!
However, the way you decide to compute the Hypervolume might depend on two things: the nature of your optimization problem (if it is a maximization or a minimization problem) and the assumptions made by the algorithms that you use to compute the Hypervolume indicator (each algorithm might assume different configurations, such as, the position of the nadir point, the scaling of the objective space values, etc.). For instance, the Quick Hypervolume (QHV) [1] algorithm assumes that the objective space is a unitary hypervolume (i.e., all objective are within the range [0, 1]) and that the nadir point is placed in the origin (i.e., (0, 0, ..., 0)). The maximum value for the Hypervolume indicator is 1 and the minimum value is 0. They study a maximization problem. If you wanted to compute the hypervolume indicator for a minimization problem, I would say the easiest way is to follow Abhinav Kumar Sharma's advice.
Particularly, for the QHV, because the maximum value of the hypervolume will always be 1, to address a minimization problem you could always minimize the value of (1 - QHV) without the need to change the sign of your objectives. In the end, I think it really depends on the assumptions each algorithm makes and on your problem's properties. The sign multiplication seems to be the most flexible, modular, and most correct approach in terms of implementation.
[1] Russo, L. M. S., & Francisco, A. P. (2014). Extending quick hypervolume. Journal of Heuristics, 22(3), 245–271. https://doi.org/10.1007/s10732-016-9309-6
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