Proportional Fairness for Combinatorial Optimization

Abstract

Proportional fairness (PF) is a widely studied concept in the literature, particularly in telecommunications, network design, resource allocation, and social choice. It aims to distribute utilities to ensure fairness and equity among users while optimizing system performance. Under convexity, PF is equivalent to the Nash bargaining solution, a well-known notion from cooperative game theory, and it can be obtained by maximizing the product of the utilities. In contrast, when dealing with non-convex optimization, PF is not guaranteed to exist; when it exists, it is also the Nash bargaining solution. Consequently, finding PF under non-convexity remains challenging since it is not equivalent to any known optimization problem.

This paper deals with PF in the context of combinatorial optimization, where the feasible set is discrete, finite, and non-convex. For this purpose, we consider a general Max-Max Bi-Objective Combinatorial Optimization (Max-Max BOCO) problem where its two objectives to be simultaneously maximized take only positive values. Then, we seek to find the solution to achieving PF between two objectives, referred to as proportional fair solution (PF solution).

We first show that the PF solution, when it exists, can be obtained by maximizing a suitable linear combination of two objectives. Subsequently, our main contribution lies in presenting an exact algorithm that converges within a logarithmic number of iterations to determine the PF solution. Finally, we provide computational results on the bi-objective spanning tree problem, a specific example of Max-Max BOCO.

Appendix

Proposition 1

If $(P^{PF},Q^{PF}) \in \mathcal {S}$ is a PF solution for Max-Max BOCO, then it is the unique solution that maximizes the product PQ.

Proof

Suppose that $(P^{PF},Q^{PF}) \in \mathcal {S}$ is a PF solution for Max-Max BOCO. We have

$$\begin{aligned} \dfrac{P}{P^{PF}} + \dfrac{Q}{Q^{PF}} \le 2, \; \forall (P,Q) \in \mathcal {S}, \end{aligned}$$

Using Cauchy-Schwarz inequality, we obtain

$$\begin{aligned} 2 \ge \frac{P}{P^{PF}} + \frac{Q}{Q^{PF}} \ge 2 \sqrt{\frac{PQ}{P^{PF}Q^{PF}}}, \end{aligned}$$

Thus, $P^{PF}Q^{PF} \ge PQ, \forall (P,Q) \in \mathcal {S}$.

Now suppose that there exists another PF solution $(P^{*},Q^{*}) \in \mathcal {S}$ such that $P^{*}Q^{*} = P^{PF}Q^{PF}$. We also have

$$\begin{aligned} 2 \ge \frac{P^{*}}{P^{PF}} + \frac{Q^{*}}{Q^{PF}} \ge 2 \sqrt{\frac{P^{*}Q^{*}}{P^{PF}Q^{PF}}} = 2, \end{aligned}$$

Thus, the equality in the Cauchy-Schwarz inequality above must hold, which implies $P^{*} = P^{PF}$ and $Q^{*} = Q^{PF}$. $\square$

Theorem 1

$(P^{PF},Q^{PF}) \in \mathcal {S}$ is the PF solution if and only if $(P^{PF},Q^{PF})$ is a solution of $\mathcal {F}(\alpha ^{PF})$ with $\alpha ^{PF} = P^{PF}/Q^{PF}$.

Proof

$\implies$ Let $(P^{PF},Q^{PF})$ be the PF solution and $\alpha ^{PF} = P^{PF}/Q^{PF}$. We have

$$\begin{aligned} \frac{P}{P^{PF}}+\frac{Q}{Q^{PF}} \le 2, \; \forall (P,Q) \in \mathcal {S}, \end{aligned}$$

(9)

Multiplying (9) by $P^{PF} > 0$ and replacing $P^{PF}/Q^{PF}$ by $\alpha ^{PF}$, we obtain

$$\begin{aligned} & P^{PF} + \alpha ^{PF} Q^{PF} \ge P + \alpha ^{PF}Q, \; \forall (P,Q) \in \mathcal {S}, \end{aligned}$$

Hence, $(P^{PF},Q^{PF})$ is a solution of $\mathcal {F}(\alpha ^{PF})$.

$\Longleftarrow$ Let $(P^{PF},Q^{PF})$ be a solution of $\mathcal {F}(\alpha ^{PF})$ with $\alpha ^{PF} = P^{PF}/Q^{PF}$. We have

$$\begin{aligned} & P + \alpha ^{PF}Q \le P^{PF} + \alpha ^{PF}Q^{PF}, \; \forall (P,Q) \in \mathcal {S}, \end{aligned}$$

Replacing $\alpha ^{PF}$ by $P^{PF}/Q^{PF}$, we obtain (2) which implies $(P^{PF},Q^{PF})$ is the PF solution. $\square$