Solution approaches for equitable multiobjective integer programming problems

Abstract

We consider multi-objective optimization problems where the decision maker (DM) has equity concerns. We assume that the preference model of the DM satisfies properties related to inequity-aversion, hence we focus on finding nondominated solutions in line with the properties of inequity-averse preferences, namely the equitably nondominated solutions. We discuss two algorithms for finding good subsets of equitably nondominated solutions. The first approach is an extension of an interactive approach developed for finding the most preferred nondominated solution when the utility function is assumed to be quasiconcave. We find the most preferred equitably nondominated solution when the utility function is assumed to be symmetric quasiconcave. In the second approach we generate an evenly distributed subset of the set of equitably nondominated solutions to be considered further by the DM. We show the computational feasibility of the two algorithms on equitable multi-objective knapsack problem, in which projects in different categories are to be funded subject to a limited budget. We perform experiments to show and discuss the performances of the algorithms.

Appendices

Appendices

Proof of Theorem 6

Proof

PART 1: We will show that if $z \in \bar{{CD_{Symm}}}(z^{m},z^{k})$ (it is equitably dominated by $C(\overrightarrow{z^{m}};\overrightarrow{z^{k}}))$ then Eqs. 7 and 8 hold. Let z be cone dominated, i.e. $\exists$ $\mu \ge 0: z^{\prime }=\overrightarrow{z^k}+\mu (\overrightarrow{z^k}-\overrightarrow{z^m})$ and $z \preceq _{e} z'$. Then by Theorem 5, $\sum _{i=1}^{t}\overrightarrow{z_{i}} \le \sum _{j \in a}z^{\prime }_j \quad \forall t, \forall a \in K^t$. That is:

$$\begin{aligned} \sum _{i=1}^{t}\overrightarrow{z_{i}} \le \sum _{j \in a}( \overrightarrow{z_j^k}+\mu (\overrightarrow{z_j^k}-\overrightarrow{z_j^m})) \quad \forall t, \forall a \in K^t \end{aligned}$$

(31)

Note that $K^t=K^t_{km} \cup K^t_{mk}$ and $K^t_{km} \cap K^t_{mk}=\emptyset \quad \forall t$. Then:

$$\begin{aligned}&\sum _{i=1}^{t_1}\overrightarrow{z_{i}} \le \sum _{j \in b}( \overrightarrow{z_j^k}+\mu (\overrightarrow{z_j^k}-\overrightarrow{z_j^m})) \quad \forall t_1 \in P, \forall b \in K^{t_1}_{mk} \end{aligned}$$

(32)

$$\begin{aligned}&\sum _{i=1}^{t_2}\overrightarrow{z_{i}} \le \sum _{l \in c}( \overrightarrow{z_l^k}+\mu (\overrightarrow{z_l^k}-\overrightarrow{z_l^m})) \quad \forall t_2 \in P, \forall c \in K^{t_2}_{km} \end{aligned}$$

(33)

Since $\sum _{j \in b} (\overrightarrow{z_j^k}-\overrightarrow{z_j^m}) \le 0$ by definition, Eq. 32 can be rewritten as:

$$\begin{aligned} \mu \sum _{j \in b}(\overrightarrow{z_j^m}-\overrightarrow{z_j^k}) \le \sum _{j \in b}\overrightarrow{z_j^k}-\sum _{i=1}^{t_1}\overrightarrow{z_{i}}\text { } \forall t_1 \in P, \forall b \in K^{t_1}_{mk} \end{aligned}$$

(34)

Since $\mu \ge 0$ and $\sum _{j \in b} (\overrightarrow{z_j^m}-\overrightarrow{z_j^k}) \ge 0$, $\sum _{j \in b}\overrightarrow{z_j^k}-\sum _{i=1}^{t_1}\overrightarrow{z_{i}} \ge 0 \quad \forall t_1\in P\text {, }\forall b \in K^{t_1}_{mk}$ should hold. That is, Eq. 7 holds.

Equation 33 can be rewritten as:

$$\begin{aligned} \sum _{i=1}^{t_2}\overrightarrow{z_{i}}-\sum _{l \in c}\overrightarrow{z_l^k} \le \mu \sum _{l \in c} (\overrightarrow{z_l^k}-\overrightarrow{z_l^m}) \text { } \forall t_2 \in P, \forall c \in K^{t_2}_{km} \end{aligned}$$

(35)

Since $\sum _{l \in c} (\overrightarrow{z_l^k}-\overrightarrow{z_l^m}) > 0$ and $\sum _{j \in b} (\overrightarrow{z_j^m}-\overrightarrow{z_j^k}) \ge 0$, from Eqs. 34 and 35 we have the following (by multiplying Eq. 34 by $\sum _{l \in c} (\overrightarrow{z_l^k}-\overrightarrow{z_l^m})$ and Eq. 35 by $\sum _{j \in b} (\overrightarrow{z_j^m}-\overrightarrow{z_j^k})$):

$(\sum _{i=1}^{t_2}\overrightarrow{z_{i}}-\sum _{l \in c}\overrightarrow{z_l^k}) (\sum _{j \in b}(\overrightarrow{z_j^m}-\overrightarrow{z_j^k}) ) \le \mu (\sum _{l \in c} (\overrightarrow{z_l^k}-\overrightarrow{z_l^m}))(\sum _{j \in b} (\overrightarrow{z_j^m}-\overrightarrow{z_j^k})) \le (\sum _{j \in b}\overrightarrow{z_j^k}-\sum _{i=1}^{t_1}\overrightarrow{z_{i}})(\sum _{l \in c} (\overrightarrow{z_l^k}-\overrightarrow{z_l^m})) \forall t_1 \in P, \forall t_2 \in P, \forall b \in K^{t_1}_{mk}, \forall c \in K^{t_2}_{km}$

PART 2: Now suppose that Eqs. 7 and 8 hold. We will show that z is cone dominated, i.e. $\exists \mu \ge 0 : z \preceq _{e} ( \overrightarrow{z^k}+\mu (\overrightarrow{z^k}-\overrightarrow{z^m}))$. Note that $K^{t_1}_{mk}$ consists of two subsets $K^{t_1}_{strict\_mk}$ and $K^{t_1}_{equal\_mk}$ as follows: $\sum _{j \in b} \overrightarrow{z^m_{j}} > \sum _{j \in b} \overrightarrow{z^k_{j}}$ for all $b \in K^{t_1}_{strict\_mk}$ and $\sum _{j \in b} \overrightarrow{z^m_{j}} = \sum _{j \in b} \overrightarrow{z^k_{j}}$ for all $b \in K^{t_1}_{equal\_mk}$.

Since for $b\in K^{t_1}_{strict\_mk}$ we have $\sum _{j \in b}(\overrightarrow{z^m_j} - \overrightarrow{z^k_j} )>0$ and for $c\in K^{t_2}_{km}$ $\sum _{l \in c}(\overrightarrow{z^k_l} - \overrightarrow{z^m_l} )>0$ , Eq. 8 implies the following:

$$\begin{aligned} \frac{(\sum _{i=1}^{t_2}\overrightarrow{z_i}-\sum _{l \in c}\overrightarrow{z^k_l})}{\sum _{l \in c} (\overrightarrow{z^k_l} - \overrightarrow{z^m_l} )} \le \frac{(\sum _{j \in b}\overrightarrow{z^k_j}-\sum _{i=1}^{t_1}\overrightarrow{z_i})}{ \sum _{j \in b}( \overrightarrow{z^m_j} -\overrightarrow{z^k_j} )} \text { } \forall t_1\in P,\forall t_2\in P, b\in K^{t_1}_{strict\_mk} \text {, } c \in K^{t_2}_{km} \end{aligned}$$

(36)

We will find a $\mu \ge 0$ that makes z equitably dominated by $\overrightarrow{z^k}+\mu (\overrightarrow{z^k}-\overrightarrow{z^m}))$. One can define

$$\begin{aligned} {\bar{\mu }}=\min _{t_1 \in P, b\in K^{t_1}_{strict\_mk}} \frac{(\sum _{j \in b}\overrightarrow{z^k_j}-\sum _{i=1}^{t_1}\overrightarrow{z_i})}{\sum _{j \in b}( \overrightarrow{z^m_j} -\overrightarrow{z^k_j} )}. \end{aligned}$$

(37)

Since $(\sum _{j \in b}\overrightarrow{z^k_j}-\sum _{i=1}^t\overrightarrow{z_i}) \ge 0$ (see Eq. 7) and $\sum _{j \in b} (\overrightarrow{z^m_j} -\overrightarrow{z^k_j} )>0$ for all $b\in K_{strict\_mk}$ , ${\bar{\mu }}\ge 0$ holds.

Note that the following holds for this ${\bar{\mu }}$:

$$\begin{aligned}&\frac{(\sum _{i=1}^{t_2}\overrightarrow{z_i}-\sum _{l \in c}\overrightarrow{z^k_l})}{\sum _{l \in c} (\overrightarrow{z^k_l} - \overrightarrow{z^m_l} )} \le {\bar{\mu }} \le \frac{(\sum _{j \in b}\overrightarrow{z^k_j}-\sum _{i=1}^{t_1}\overrightarrow{z_i})}{ \sum _{j \in b}( \overrightarrow{z^m_j} -\overrightarrow{z^k_j} )}\nonumber \\&\quad \forall t_1\in P, \forall t_2\in P, b\in K^{t_1}_{strict\_mk} \text {, } c \in K^{t_2}_{km} \end{aligned}$$

(38)

$\sum _{i=1}^{t_2}\overrightarrow{z_i}\le \sum _{l \in c}\overrightarrow{z^k_l}+{\bar{\mu }}\sum _{l \in c} (\overrightarrow{z^k_l} - \overrightarrow{z^m_l} ) \text { } \forall t_2 \in P \text {, } \forall c\in K^{t_2}_{km}$(From the left side of Eq. 38)

$\sum _{i=1}^{t_1}\overrightarrow{z_i} \le \sum _{j \in b}\overrightarrow{z^k_j}+{\bar{\mu }} \sum _{j \in b} (\overrightarrow{z^k_j} - \overrightarrow{z^m_j} )\text { } \forall t_1\in P \text {, } \forall b\in K^{t_1}_{strict\_mk}$(From the right side of Eq. 38).

$\sum _{i=1}^t\overrightarrow{z_i} \le \sum _{j \in b}\overrightarrow{z^k_j}+{\bar{\mu }} \sum _{j \in b} (\overrightarrow{z^k_j} -\overrightarrow{z^m_j})=\sum _{j \in b}\overrightarrow{z^k_j} \text { } \forall t\in P \text {, } \forall b\in K^t_{equal\_mk}$ (From condition 7).

Note that $K^t= K^t_{strict\_mk}\cup K^t_{equal\_mk} \cup K^t_{km}$. Therefore the conditions of Theorem 5 are satisfied, making z equitably dominated by $\overrightarrow{z^k}+{\bar{\mu }}(\overrightarrow{z^k}-\overrightarrow{z^m})$. $\square$

Proof of Proposition 7

For $P=\lbrace 1,2,\cdots ,p\rbrace$:

$$\begin{aligned}&y_1=\overrightarrow{z}_1\\&y_2=\overrightarrow{z}_1+\overrightarrow{z}_2\\&\vdots \\&y_p=\overrightarrow{z}_1+\overrightarrow{z}_2+\cdots +\overrightarrow{z}_p \end{aligned}$$

and $\overrightarrow{z}_1 \le \overrightarrow{z}_2\le \cdots \le \overrightarrow{z}_p$

We will prove Proposition 7 by induction. The base case is at $p=2$ where $P=\lbrace 1,2\rbrace$. To show that Proposition 7 holds for the base case, we need to show that $y_2\ge 2y_1$.

Since by definition $y_1=\overrightarrow{z}_1$ and $y_2=\overrightarrow{z}_1+\overrightarrow{z}_2$ where $\overrightarrow{z}_2 \ge \overrightarrow{z}_1$, then $y_2=\overrightarrow{z}_1+\overrightarrow{z}_1+\epsilon \ge 2\overrightarrow{z}_1=2y_1$ where $\epsilon \ge 0$. Hence $y_2\ge 2y_1$.

Hypothesis 1

Assume that Proposition 7 holds for $p=s$.

To complete the proof, we need to show that Proposition 7 holds for $p=s+1$. Due to Hypothesis 1, we just have to show that Proposition 7 holds for all $(j,s+1): 1 \le j \le s$.

$jy_s\ge sy_j$ for all $1\le j \le s$ (Due to Hypothesis 1)

$$\begin{aligned}&\overrightarrow{z}_{s+1} \ge \overrightarrow{z}_j=y_j-y_{j-1}\\&jy_s+j\overrightarrow{z}_{s+1}\ge sy_j+j\overrightarrow{z}_j \\&jy_s+j\overrightarrow{z}_{s+1}\ge sy_j+j(y_j-y_{j-1}).\\&j(y_s+\overrightarrow{z}_{s+1})\ge (s+1)y_j+\underbrace{(j-1)y_j-jy_{j-1}}_{\ge 0, \text { Due to Hypothesis }1}.\\&\hbox {Hence }jy_{s+1}\ge (s+1)y_j. \end{aligned}$$

Therefore, Proposition 7 holds for any p.