Multi-objective optimization: min sum vs min max

The issue

In many models, we observe that an objective that minimizes a (weighted) sum, delivers undesirable results. Here are some examples:

Assignment under preferences

In [1], we formulated a MIP model that tackles the problem of creating groups taking into account the preferences of people. When we just maximize the total sum of preferences that are honored by the assignment, we can see that some participants get a terrible assignment. I.e., taking one for the team. It requires some thought to develop a model that produces a more equitable solution. In a sense, we are looking at trading-off overall efficiency for fairness.

Job scheduling

Typically in job scheduling models, we minimize (1) the total make-span: do things as quickly as possible and (2) the sum of tardiness of jobs. Sometimes we use a weighting scheme for tardy jobs: jobs from valuable clients may get a higher weight to make it less likely they are chosen to be tardy. There are some terms in the objective we can add, for instance:

• The number of jobs that are tardy. The sum may distribute the pain over many jobs. If we want to reduce the number of tardy jobs we can add a term that minimizes the number of tardy jobs.
• The maximum tardiness of a job. Adding such a term will try to prevent the situation where a single job will take the brunt of the pain.

These are examples where just minimizing the sum can lead to unpalatable solutions. It may require some effort to design a more sophisticated objective that takes into account more aspects than just overall efficiency.

Example model

In [2] I discussed a MIP formulation for a matching problem:

Given \(N=2M\) points, find a matching that minimizes the sum of the cost (represented by lengths between the matched points). 

A. Single solution

A.1. MIN SUM Matching 

MIP Model for Min-Max Matching
\[\begin{align}\min\>& \color{darkred}z\\ & \color{darkred}z \ge \color{darkblue}d_{i,j} \color{darkred}x_{i,j} && \forall  i \lt j\\ & \sum_{j|j \gt i} \color{darkred}x_{i,j} + \sum_{j|j \lt i} \color{darkred}x_{j,i} = 1 && \forall i\\ & \color{darkred}x_{i,j} \in \{0,1\} \end{align} \]

The results look like:

Optimal MIN MAX solution

We see we do much better than the maximum length of the MIN SUM problem.

One issue with the MIN MAX problem is that the solution is not unique. Any configuration with lengths less than the maximum length is optimal.

A.3. MIN MAX followed by MIN SUM 

Multi-objective Model
\[\begin{align}\min \>& \color{darkred}z_1 = \max_{i,j|i \lt j} \color{darkblue}d_{i,j} \color{darkred}x_{i,j}\\ \min\>& \color{darkred}z_2 = \sum_{i,j|i \lt j} \color{darkblue}d_{i,j} \color{darkred}x_{i,j}\\ & \sum_{j|j \gt i} \color{darkred}x_{i,j} + \sum_{j|j \lt i} \color{darkred}x_{j,i} = 1 && \forall i\\ & \color{darkred}x_{i,j} \in \{0,1\} \end{align} \]

Of course we need to reformulate the \(\max\) function to make it linear. The multi-objective structure was implemented as:

1. Solve problem with the MIN MAX objective\[\begin{align} \min\>&z_1\\ &z_1 \ge d_{i,j} x_{i,j} &&\forall i\lt j \\ & \sum_{j|j \gt i} x_{i,j} + \sum_{j|j \lt i} x_{j,i} = 1 \\ &x_{i,j} \in \{0,1\} \end{align}\] Let \(z_1^*\) be the optimal objective.
2. Solve problem with the MIN SUM, subject to MIN MAX objective is not deteriorating (i.e. the length of each segment is less than the maximum length found in the previous step). \[\begin{align}\min\>& z_2 = \sum_{i,j|i \lt j} d_{i,j} x_{i,j} \\ & d_{i,j} x_{i,j} \le z_1^* && \forall i\lt j\\ & \sum_{j|j \gt i} x_{i,j} + \sum_{j|j \lt i} x_{j,i} = 1 && \forall i\\ & x_{i,j} \in \{0,1\}\end{align}\] 

This is typically called the lexicographic method or preemptive method.

The results look like:

Optimal multi-objective solution

We see a much better distribution of the lengths with this approach. The underlying reason is that the MIN MAX solution is not unique. There are a ton of solutions with all lengths less than the MIN-MAX objective (at least 50k: I stopped after reaching this number). With the second objective, we choose the "best" of all these solutions.

Here is a summary of the results:

Model	Objective	Sum	Max
MIN SUM	241.267	241.267	38.768
MIN MAX	22.954	403.265	22.954
MULTI-OBJECTIVE (second solve)	247.382	247.382	22.954

We can see that:

• The largest segment is very long for the MIN SUM model
• The sum is very large for the MIN MAX model.
• The MULTI model has a sum only a bit larger than from the MIN SUM model, and the largest length is identical to the MIN MAX model. It prevents the disadvantages of the MIN SUM and MIN MAX models.

A.4. MIN SUM followed by MIN MAX

Multi-objective Model via Weighted Sum

\[\begin{align}\min \>&  \color{darkblue}w_1 \color{darkred}z_1 +  \color{darkblue}w_2 \color{darkred}z_2 \\ & \color{darkred}z_1 \ge   \color{darkblue}d_{i,j} \color{darkred}x_{i,j} && \forall i,j| i \lt j \\ & \color{darkred}z_2 = \sum_{i,j|i \lt j} \color{darkblue}d_{i,j} \color{darkred}x_{i,j}\\ & \sum_{j|j \gt i} \color{darkred}x_{i,j} + \sum_{j|j \lt i} \color{darkred}x_{j,i} = 1 && \forall i\\ & \color{darkred}x_{i,j} \in \{0,1\} \end{align} \]

The convention is we use \(w_i\gt 0\) and \(\sum_i w_i=1\). If we use\[\begin{cases} w_1 = 0.01 \\ w_2 = 0.99\end{cases}\] we get the same solution as in section A.3 MIN ABS followed by MIN SUM.


If we solve our weighted sum model for these values, we get the following report:




This is a bit disappointing: we only observe two different solutions. And we saw these before.

We may miss some efficient (i.e., non-dominated) solutions. In general, the weighted sum method suffers from two problems:

1. We may miss some efficient solutions. Although each point found with the weighted sum method is non-dominated, the other way around is not true: there may be efficient solutions we cannot see with this method.
2. If this method finds several efficient points, these points may not be nicely distributed.