Min max min robust (relative) regret combinatorial optimization

Abstract

We consider combinatorial optimization problems with uncertainty in the cost vector. Recently, a novel approach was developed to deal with such uncertainties: instead of a single one robust solution, obtained by solving a min max problem, the authors consider a set of solutions obtained by solving a min max min problem. In this new approach, the set of solutions is computed once and we can choose the best one in real time each time a cost vector occurs yielding better solutions compared to the min max approach. In this paper, we apply the new approach to the absolute and relative regret cases. Algorithms to solve the min max min robust (relative) regret problems are presented with computational experiments.

Appendices

Appendix A: Proof of Proposition 1

• (i)  Let $\mathbf {c} \in \varOmega$ and let $\mathbf {h},\mathbf {x} \in X$.

  • Let $I_{1,0} = \lbrace i: \mathbf {h}_i = 1,\mathbf {x}_i = 0,\;\;1 \le i \le n \rbrace$

  • Let $I_{0,1} = \left\{ i: \mathbf {h}_i = 0,\mathbf {x}_i = 1,\;\;1 \le i \le n \right\}$

  then:

  $$\begin{aligned} \mathbf {c}^t \mathbf {h} - \mathbf {c}^t \mathbf {x} = \sum _{i \in I_{1,0}} \mathbf {c}_i - \sum _{i \in I_{0,1}} \mathbf {c}_i \le \sum _{i \in I_{1,0}} \mathbf {U}_i - \sum _{i \in I_{0,1}} \mathbf {L}_i = {\mathbf {c}^{+}(\mathbf {x})}^t \mathbf {h} - {\mathbf {c}^{+}(\mathbf {x})}^t \mathbf {x} \end{aligned}$$

• (ii) Let $\mathbf {x} \in X$ and let $\mathbf {c} \in \varOmega$. Let $\mathbf {h}$ be an optimal solution for $P(\mathbf {c}^+(\mathbf {x}))$.

  • Let $I_{1,0} = \lbrace i: \mathbf {h}_i = 1,\mathbf {x}_i = 0,\;\;1 \le i \le n \rbrace$

  • Let $I_{0,1} = \left\{ i: \mathbf {h}_i = 0,\mathbf {x}_i = 1,\;\;1 \le i \le n \right\}$

  then:

  $$\begin{aligned}&0 < {\mathbf {c}^+(\mathbf {x})}^t \mathbf {x} - v(P(\mathbf {c}^+(\mathbf {x}))) = {\mathbf {c}^+(\mathbf {x})}^t \mathbf {x} - {\mathbf {c}^+(\mathbf {x})}^t\mathbf {h}\\&\quad = \sum _{i \in I_{0,1}} \mathbf {L}_i - \sum _{i \in I_{1,0}} \mathbf {U}_i\le \sum _{i \in I_{0,1}} \mathbf {c}_i - \sum _{i \in I_{1,0}} \mathbf {c}_i = \mathbf {c}^t \mathbf {x} - \mathbf {c}^t \mathbf {h} \end{aligned}$$

  therefore: $v(P(\mathbf {c})) \le \mathbf {c}^t \mathbf {h} < \mathbf {c}^t \mathbf {x}$.

• (iii) Let $\mathbf {x} \in X$ and let $\mathbf {c} \in \varOmega$, then: $\mathbf {c}^{+}(\mathbf {x})^t \mathbf {x} \overset{(1)}{=} \mathbf {L}^t \mathbf {x} \overset{(2)}{\le } \mathbf {c}^t \mathbf {x}$ where (1) From the definition of $\mathbf {c}^+(\mathbf {x})$ and (2) Since $\mathbf {L} \le \mathbf {c}\;\;\bullet$.

Appendix B: Proof of Proposition 2

Let $X \subseteq {\lbrace 0,1 \rbrace }^n$. Let $f:X \longmapsto \mathbb {R}$ and $g:X \longmapsto \mathbb {R}$ with $f(\mathbf {x}) > 0$ and $g(\mathbf {x}) > 0$ for all $\mathbf {x} \in X$. Let P1 be a problem in ($\mathbf {x}$) defined as:

$$\begin{aligned} (P1):\;\;\underset{\mathbf {x} \in X}{\max }\;\; \frac{f(\mathbf {x})}{g(\mathbf {x})} \end{aligned}$$

Let $\mu \ge 0$ and let $P2(\mu )$ be a problem in ($\mathbf {x}$) defined as;

$$\begin{aligned} (P2(\mu )):\;\;\underset{\mathbf {x} \in X}{\max }\;\; f(\mathbf {x}) - \mu g(\mathbf {x}) \end{aligned}$$

then:

If $v(P2(\mu )) = 0$ then $f(\mathbf {x}) - \mu g(\mathbf {x}) \le 0$ for all $\mathbf {x} \in X$, therefore $\frac{f(\mathbf {x})}{g(\mathbf {x})} \le \mu$ for all $\mathbf {x} \in X$ and we have $v(P1) \le \mu$. If $v(P2(\mu )) = 0$ and $v(P1) < \mu$ we have $\frac{f(\mathbf {x})}{g(\mathbf {x})} < \mu$ for all $\mathbf {{x} \in X}$ and then $f(\mathbf {x}) - \mu g(\mathbf {x}) < 0$ for all $\mathbf {x} \in X$ and we have a contradiction.

If $\mu = v(P1)$ and $v(P2(\mu )) > 0$ then we have $\frac{f(x)}{g(x)} > \mu = v(P1)$ for some $\mathbf {x} \in X$ and we have a contradiction, therefore $v(P2(\mu )) \le 0$. If $\mathbf {x}^*$ is an optimal solution for P1 then we have $f(\mathbf {x}^*) - \mu g(\mathbf {x}^*) = 0$ and then $v(P2(\mu )) \ge 0$ and finally $v(P2(\mu )) = 0$.

Hence we have:

• (1) $v(P2(\mu )) = 0$ if and only if $\mu = v(P1)$.

  If $v(P2(\mu )) = 0$ and $\mathbf {x}^*$ is an optimal solution for $P2(\mu )$ we have $v(P2(\mu )) = f(\mathbf {x}^*) - \mu g(\mathbf {x}^*) = f(\mathbf {x}^*) - v(P1) g(\mathbf {x}^*) = 0$ and then $v(P1) = \frac{f(\mathbf {x}^*)}{g(\mathbf {x}^*)}$, therefore $\mathbf {x}^*$ is an optimal solution for P1. Hence we have:

• (2) If $v(P2(\mu )) = 0$ and $\mathbf {x}^*$ is an optimal solution for $P2(\mu )$ then $\mathbf {x}^*$ is an optimal solution for P1.

Let $X = \lbrace \mathbf {x}^{(1)},\ldots ,\mathbf {x}^{(L)} \rbrace$ then

$v(P2(\mu )) = \max \lbrace f(\mathbf {x}^{(1)}) - \mu g(\mathbf {x}^{(1)}),\ldots ,f(\mathbf {x}^{(L)}) - \mu g(\mathbf {x}^{(L)}) \rbrace$.

Since $g(\mathbf {x}) > 0$ for all $\mathbf {x} \in X$ then we have:

• (3) $v(P2(\mu ))$ is a piecewise linear and decreasing function in $\mu$.

We have:

$$\begin{aligned} v(QX_r(H)) = \underset{\mathbf {x} \in X}{\max }\;\; \frac{ \underset{ \mathbf {h} \in H}{\min }\;\; \mathbf {c}^{+}(\mathbf {x})^t \mathbf {h} - \mathbf {c}^{+}(\mathbf {x})^t \mathbf {x} }{\mathbf {c}^{+}(\mathbf {x})^t \mathbf {x}} = v(RX_r(H))-1 \end{aligned}$$

Let $f(\mathbf {x}) = \underset{ \mathbf {h} \in H}{\min }\;\; \mathbf {c}^{+}(\mathbf {x})^t \mathbf {h}$ and let $g(\mathbf {x}) = \mathbf {c}^{+}(\mathbf {x})^t \mathbf {x}$. We use (1),(2) and (3) to find:

• (i) $v(RX_r(\mu ,H)) = 0$ if and only if $\mu = v(QX_r(H)) + 1$.

• (ii) If $v(RX_r(\mu ,H)) = 0$ and $\mathbf {x}^*$ is an optimal solution for $RX_r(\mu ,H)$ then $\mathbf {x}^*$ is an optimal solution for $QX_r(H)$.

• (iii) $v(RX_r(\mu ,H))$ is a piecewise linear and decreasing convex function in $\mu \;\;\bullet$

Appendix C: Proof of Proposition 3

Let $X = \lbrace \mathbf {x}^{(1)},\ldots ,\mathbf {x}^{(L)} \rbrace$, let $\phi _j = \underset{\mathbf {h} \in H}{\min }\;\; {\mathbf {c}^+(\mathbf {x}^{(j)})}^t \mathbf {h}$ and let $\delta _j = {\mathbf {c}^+(\mathbf {x}^{(j)})}^t \mathbf {x}^{(j)}$ for $j=1,\ldots ,L$.

The algorithm may be rewritten as follows:

Let $\mu _1 = 1$.

Algorithm Find-$\mu ^*$

• 1. $i=1$.

• 2. Solve $R_r(\mu _i,H)$. Let $\mathbf {x}^{(j_i)}$ be an optimal solution.

• 3. If $v(R_r(\mu _i,H)) = 0$ let $\mu ^* = \mu _i$, let $\mathbf {x}^* = \mathbf {x}^{(j_i)}$ and stop.

• 4. Let $\mu _{i+1} = \frac{\phi _{j_i}}{\delta _{j_i}}$, let $i=i+1$ and return to Step 2.

If $v(R_r(\mu _1,H)) > 0$ then $\mu _2 = \frac{\phi _{j_1}}{\delta _{j_1}}$ and $\phi _{j_1} - \mu _1 \delta _{j_1} > 0$. Hence $\mu _2 > 1 = \mu _1$.

If $v(R_r(\mu _2,H)) > 0$ then $\mu _3 = \frac{\phi _{j_2}}{\delta _{j_2}}$ and $\phi _{j_2} - \mu _2 \delta _{j_2} > 0$. Hence $\mu _3 > \mu _2$.

In general we have that $\mu _i < \mu _{i+1}$ for all i. Since X is a finite set then the sequence $\mu _1,\ldots ,\mu _s,\ldots$ generated by the algorithm must be finite and then $v(R_r(\mu _i,H)) = 0$ for some i (otherwise the algorithm generates an infinite sequence of $\mu$-values).

Appendix D

See Tables 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 and 15.

Table 4 SP problems with mesh topology. Q-greedy, T-greedy and S-Q algorithms for the MSR(k) problem. Times, failures and iterations. 10 problems in each set

Table 5 SP problems with mesh topology. Q-greedy, T-greedy and S-Q algorithms for the MSR(k) problem. Reduction percentage for the regret values. 10 problems in each set

Table 6 SP problems with euclidean topology. Q-greedy, T-greedy and S-Q algorithms for the MSR(k) problem. Times and iterations. 10 problems in each set

Table 7 SP problems with euclidean topology. Q-greedy, T-greedy and S-Q algorithms for the MSR(k) problem . Reduction percentage for the regret values. 10 problems in each set

Table 8 p-M problems.Q-greedy, T-greedy and S-Q algorithms for the MSR(k) problem. Times, failures and iterations. 10 problems in each set

Table 9 p-M problems. Q-greedy, T-greedy and S-Q algorithms for the MSR(k) problem. Reduction percentage for the regret values. 10 problems in each set

Table 10 SP problems with mesh topology. Qr-greedy, Tr-greedy and Sr-Qr algorithms for the MrelSR(k) problem. Times, failures and iterations. 10 problems in each set

Table 11 SP problems with mesh topology. Qr-greedy, Tr-greedy and Sr-Qr algorithms for the MrelSR(k) problem . Reduction percentage for the regret values. 10 problems in each set

Table 12 SP problems with euclidean topology. Qr-greedy, Tr-greedy and Sr-Qr algorithms for the MrelSR(k) problem. Times, failures and iterations. 10 problems in each set

Table 13 SP problems with euclidean topology. Qr-greedy, Tr-greedy and Sr-Qr algorithms for the MrelSR(k) problem. Reduction percentage for the regret values. 10 problems in each set

Table 14 p-M problems.Qr-greedy, Tr-greedy and Sr-Qr algorithms for the MrelSR(k) problem. Times, failures and iterations. 10 problems in each set

Table 15 p-M problems. Qr-greedy, Tr-greedy and Sr-Qr algorithms for the MrelSR(k) problem . Reduction percentage for the regret values. 10 problems in each set