A new dual-based cutting plane algorithm for nonlinear adjustable robust optimization

Abstract

This paper explores a class of nonlinear Adjustable Robust Optimization (ARO) problems, containing here-and-now and wait-and-see variables, with uncertainty in the objective function and constraints. By applying Fenchel’s duality on the wait-and-see variables, we obtain an equivalent dual reformulation, which is a nonlinear static robust optimization problem. Using the dual formulation, we provide conditions under which the ARO problem is convex on the here-and-now decision. Furthermore, since the dual formulation contains a non-concave maximization on the uncertain parameter, we use perspective relaxation and an alternating method to handle the non-concavity. By employing the perspective relaxation, we obtain an upper bound, which we show is the same as the static relaxation of the considered problem. Moreover, invoking the alternating method, we design a new dual-based cutting plane algorithm that is able to find a reasonable lower bound for the optimal objective value of the considered nonlinear ARO model. In addition to sketching and establishing the theoretical features of the algorithms, including convergence analysis, by numerical experiments we reveal the abilities of our cutting plane algorithm in producing locally robust solutions with an acceptable optimality gap.

Appendices

Appendices

The appendix of this work is divided into two sections. The first section contains the proof of several points mentioned in the main text. In the second section, we provide a proof for Theorem 5.

Appendix 1 Additional results

In this appendix, first, we provide more details on a point for perspective function mentioned after Remark 1.

Proposition 1

If g is a proper, closed, and convex function, then

$$\begin{aligned} \sup _{t>0,x\in \mathbb {R}^{n_x}}g^{per} (x, t)= \sup _{t\ge 0,x\in \mathbb {R}^{n_x}}g^{per} (x, t), \end{aligned}$$

and

$$\begin{aligned} \inf _{t>0,x\in \mathbb {R}^{n_x}}g^{per} (x, t)= \inf _{t\ge 0,x\in \mathbb {R}^{n_x}}g^{per} (x, t). \end{aligned}$$

Proof

Let $x^0\in \mathbb {R}^{n_x}$. We have

$$\begin{aligned} g^{per} (x^0, t_0=0)&=\displaystyle \liminf _{(x^i,t_i)\rightarrow (x^0,0)}g^{per} (x^i, t_i>0)\\ {}&\le \displaystyle \sup _{(x^i,t_i)\rightarrow (x^0,0)}g^{per} (x^i, t_i>0)\\ {}&\le \displaystyle \sup _{t>0,x\in \mathbb {R}^{n_x}}g^{per} (x, t). \end{aligned}$$

So, $\sup _{t>0,x\in \mathbb {R}^{n_x}}g^{per} (x, t)= \sup _{t\ge 0,x\in \mathbb {R}^{n_x}}g^{per} (x, t)$.

As $\inf _{t>0,x}g^{per} (x, t)\ge \inf _{t\ge 0,x}g^{per} (x, t)$, let $\ell \in \{g^{per} (x, t)| t\ge 0,x\in \mathbb {R}^{n_x}\}$. We want to show $\ell \ge \inf _{t>0,x}g^{per} (x, t)$.

• 1. If $\ell =g^{per} (x^0, t_0)$ for some $x^0\in \mathbb {R}^{n_x}$ and $t_0>0$, then $\ell \ge \inf _{t>0,x}g^{per} (x, t)$.

• 2. If $\ell =g^{per} (x^0,0)$ for some $x^0\in \mathbb {R}^{n_x}$, then

  $$\begin{aligned} \ell =g^{per} (x^0,0)&=\displaystyle \liminf _{(x^i,t_i)\rightarrow (x^0,0)}g^{per} (x^i, t_i>0)\\&\ge \displaystyle \inf _{(x^i,t_i)\rightarrow (x^0,0)}g^{per} (x^i, t_i)\\&\ge \displaystyle \inf _{t>0,x\in \mathbb {R}^{n_x}}g^{per} (x, t). \end{aligned}$$

  The proof is complete.

$\square$

As a consequence of the above proposition, we have

$$\begin{aligned} \sup _{t>0,x}-g^{per} (x, t)=\sup _{t\ge 0,x}-g^{per} (x, t). \end{aligned}$$

The next proposition proves the convexity of the set ${\mathcal {V}}$ and the concavity of the function G introduced in the beginning of Sect. 5.

Proposition 2

The set ${\mathcal {V}}$ is convex, and G is a concave function on ${\mathcal {V}}$.

Proof

We consider two points $\bar{v}=\begin{pmatrix} {\bar{\lambda }} \\ \{{\bar{w}}^j\}_{j=0}^m \end{pmatrix}, {\tilde{v}}=\begin{pmatrix} {\tilde{\lambda }} \\ \{{{\tilde{w}}}^j\}_{j=0}^m \end{pmatrix}\in {\mathcal {V}}$ and $\ell \in [0,1]$. Since $\frac{{\bar{w}}^j}{{\bar{\lambda }}_j},\frac{\tilde{w}^j}{{\tilde{\lambda }}_j}\in dom (g_j^*)$, and $\lambda _{j} {g_j}^*( \tfrac{w^j}{\lambda _j})$ for each j is jointly convex in $(w^j,\lambda _{j})$, we have the following possible cases:

Case 1. $\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j>0:$ In this case,

$$\begin{aligned}&(\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j) g_j^*\left( \frac{\ell {\bar{w}}^j+(1-\ell )\tilde{w}^j}{\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j}\right) \le \ell {\bar{\lambda }}_j g_j^*( \frac{\bar{w}^j}{{\bar{\lambda }}_j})+(1-\ell ){\tilde{\lambda }}_j g_j^*(\frac{\tilde{w}^j}{{\tilde{\lambda }}_j}) < \infty \\ {}&\Rightarrow \frac{\ell \bar{w}^j+(1-\ell )\tilde{w}^j}{\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j}\in dom (g_j^*) \end{aligned}$$

Case 2. $\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j=0:$ In this case, if $0<\ell <1$, then ${\bar{\lambda }}_j=0={\tilde{\lambda }}_j$, and so

$$\begin{aligned} (\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j) g_j^*\left( \frac{\ell {\bar{w}}^j+(1-\ell )\tilde{w}^j}{\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j}\right)&=\delta ^*_{dom(g_j)}(\ell {\bar{w}}^j+(1-\ell ){{\tilde{w}}}^j)\\&\le \delta ^*_{dom(g_j)}(\ell \bar{w}^j)+\delta ^*_{dom(g_j)}((1-\ell ){{\tilde{w}}}^j)\\&< \infty . \end{aligned}$$

If $\ell =0$, then ${\tilde{\lambda }}_j=0$, and hence

$$\begin{aligned} \hspace{-1.12cm} (\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j) g_j^*\left( \frac{\ell {\bar{w}}^j+(1-\ell )\tilde{w}^j}{\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j}\right)&=\delta ^*_{dom(g_j)}(\ell {\bar{w}}^j+(1-\ell ){{\tilde{w}}}^j)\\&=\delta ^*_{dom(g_j)}({{\tilde{w}}}^j)< \infty . \end{aligned}$$

If $\ell =1$, then ${\bar{\lambda }}_j=0$, and thus

$$\begin{aligned} (\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j) g_j^*\left( \frac{\ell {\bar{w}}^j+(1-\ell )\tilde{w}^j}{\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j}\right)&=\delta ^*_{dom(g_j)}(\ell {\bar{w}}^j+(1-\ell ){{\tilde{w}}}^j)\\&=\delta ^*_{dom(g_j)}({\bar{w}}^j)< \infty \end{aligned}$$

So, in all above three cases, we get

$$\begin{aligned}&(\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j) g_j^*\left( \frac{\ell {\bar{w}}^j+(1-\ell )\tilde{w}^j}{\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j}\right) =\delta ^*_{dom(g_j)}(\ell {\bar{w}}^j+(1-\ell ){{\tilde{w}}}^j) < \infty \\&\Rightarrow \frac{\ell {\bar{w}}^j+(1-\ell ){{\tilde{w}}}^j}{\ell {\bar{\lambda }}_j+(1-\ell ){\tilde{\lambda }}_j}\in dom (g_j^*). \end{aligned}$$

Convexity in all other constraints of ${\mathcal {V}}$ obviously holds. So, $\ell \bar{v}+(1-\ell ){\tilde{v}}\in {\mathcal {V}}$ which shows that ${\mathcal {V}}$ is a convex set. The function G on the convex set ${\mathcal {V}}$ is a concave function due to the concavity of each $-\lambda _{j} {g_j}^*( \tfrac{w^j}{\lambda _j})$. $\square$

Appendix 2 Proof of Theorem 5

We first recall optimality condition for a constrained differentiable problem (for more details see e.g., [4]). Consider a (non-convex) problem of the form

$$\begin{aligned} \displaystyle \sup _{y} \left\{ g(y)|~y\in {\mathcal {S}}\right\} , \end{aligned}$$

(21)

where g is a real-valued continuously differentiable function, and ${\mathcal {S}}$ is a nonempty closed convex set. A vector $y^*\in {\mathcal {S}}$ is called a stationary point of problem (21) if

$$\begin{aligned} \nabla g(y^*)^\top (y-y^*)\le 0,~~~\forall y\in {\mathcal {S}}, \end{aligned}$$

where $\nabla g(y^*)$ is the gradient of g at $y^*$.

Lemma 1

Let g be a real-valued continuously differentiable function defined on the Cartesian product of two closed convex sets $C_1\subseteq \mathbb {R}^{n_1}$, $C_2\subseteq \mathbb {R}^{n_2}$. Suppose that $\bar{y}=(\bar{y}^1,\bar{y}^2)\in C_1\times C_2$. Then

$$\begin{aligned} \nabla g(\bar{y})^\top (y-\bar{y})\le 0,~~~\forall y\in C_1\times C_2, \end{aligned}$$

(22)

if and only if the following properties hold:

• (i) $~\nabla _1 g(\bar{y})^\top (y^1-\bar{y}^1)\le 0,~~~\forall y^1\in C_1,$

• (ii) $~\nabla _2 g(\bar{y})^\top (y^2-\bar{y}^2)\le 0,~~~\forall y^2\in C_2,$

where the vector y is partitioned into two component vectors $y^1\in \mathbb {R}^{n_1}$, $y^2\in \mathbb {R}^{n_2}$, as $y\equiv (y^1,y^2)$, and $\nabla _1 g(\bar{y})=\left( \tfrac{\partial g}{\partial y^1}(\bar{y})\right)$, and $\nabla _2 g(\bar{y})=\left( \tfrac{\partial g}{\partial y^2}(\bar{y})\right)$ denote the corresponding gradient vectors.

Proof

$(\Rightarrow )$ Let $y=(y^1,y^2)\in C_1\times C_2$. By setting $y:=(y^1,\bar{y}^2)$ and $y:=(\bar{y}^1,y^2)$ in inequality (22), inequalities (i) and (ii) are derived.

$(\Leftarrow )$ Clearly, (i) and (ii) lead (22). $\square$

Now we are ready to prove Theorem 5. The main line of reasoning can be found in [19] but given here for completeness.

Proof of Theorem 5

Suppose that $z^*=(z^{1*},z^{2*})$ is a limit point of the sequence $\{z^k\}_{k\ge 0}$. Without loss of generality, we assume that $z^k=(u^k,v^k)\rightarrow (z^{1*},z^{2*})$. Our goal is to show that for any $\zeta =(\zeta ^1,\zeta ^2)\in {\mathcal {U}}\times {\mathcal {V}}$, we have

$$\begin{aligned} \nabla {\mathcal {L}}_{{\bar{x}}}(z^*)^\top (\zeta -z^*)\le 0. \end{aligned}$$

According to Lemma 1, the above inequality is equivalent to

$$\begin{aligned}&\nabla _1 {\mathcal {L}}_{{\bar{x}}}(z^{*})^\top (\zeta ^1-z^{1*})\le 0,~~~\forall \zeta ^1\in {\mathcal {U}}, \end{aligned}$$

(23)

$$\begin{aligned}&\nabla _2 {\mathcal {L}}_{{\bar{x}}}(z^{*})^\top (\zeta ^2-z^{2*})\le 0,~~~\forall \zeta ^2\in {\mathcal {V}}, \end{aligned}$$

(24)

where $\nabla {\mathcal {L}}_{{\bar{x}}}(z^*)=\left( \nabla _1 {\mathcal {L}}_{{\bar{x}}}(z^{*})^\top ,\nabla _2 {\mathcal {L}}_{\bar{x}}(z^{*})^\top \right) ^\top$ is the gradient of ${\mathcal {L}}_{\bar{x}}$ at $z^*$. By contradiction, suppose that there exists a vector $\tilde{\zeta }^2\in {\mathcal {V}}$, such that

$$\begin{aligned} \nabla _2 {\mathcal {L}}_{{\bar{x}}}(z^{*})^\top (\tilde{\zeta }^2-z^{2*})>0. \end{aligned}$$

(25)

Set $r^k:=\tilde{\zeta }^2-v^k$. As the sequence $\{v^k\}_{k\ge 0}$ converges to $z^{2*}$, the sequence $\{r^k\}_{k\ge 0}$ converges to $\tilde{\zeta }^2-z^{2*}$. Thus, due to the continuity of the gradient, there exists $N>0$ such that for all $k>N$ we have

$$\begin{aligned} \nabla _2 {\mathcal {L}}_{{\bar{x}}}(z^k)^\top r^k>0. \end{aligned}$$

So, $d^k:=({\textbf {0}}^\top ,(r^k)^\top )^\top$ is an ascent direction of ${\mathcal {L}}_{{\bar{x}}}$ at $z^k$. By backtracking line search [4, Lemma 4.3], for given parameter $\alpha \in (0,1)$, there exists a step size $t_k\in (0,1)$ such that

$$\begin{aligned} {\mathcal {L}}_{{\bar{x}}}(z^k+t_k d^k)- {\mathcal {L}}_{{\bar{x}}}(z^k)\ge \alpha t_k \nabla {\mathcal {L}}_{{\bar{x}}}(z^k)^\top d^k, ~~~\forall k>N. \end{aligned}$$

Therefore

$$\begin{aligned} {\mathcal {L}}_{{\bar{x}}}(u^k,v^k+t_k r^k)- {\mathcal {L}}_{\bar{x}}(u^k,v^k)\ge \alpha t_k \nabla _2 {\mathcal {L}}_{{\bar{x}}}(z^k)^\top r^k>0, ~~~\forall k>N. \end{aligned}$$

(26)

Since ${\mathcal {V}}$ is convex, we have

$$\begin{aligned} v^k+t_k r^k=(1-t_k) v^k+t_k{\tilde{\zeta }}^2\in {\mathcal {V}}, ~~~\forall k>N. \end{aligned}$$

Hence,

$$\begin{aligned} {\mathcal {L}}_{{\bar{x}}}(u^{k+1},v^{k+1})\ge {\mathcal {L}}_{\bar{x}}(u^{k},v^{k+1})\ge {\mathcal {L}}_{{\bar{x}}}(u^k,v^k+t_kr^k)> {\mathcal {L}}_{{\bar{x}}}(u^k,v^k), ~~~\forall k>N. \end{aligned}$$

So, the sequence of function values $\left\{ {\mathcal {L}}_{\bar{x}}(u^k,v^k)\right\}$ is non-decreasing and also bounded above. Therefore, it is convergent. The last inequality and the convergence of $\left\{ {\mathcal {L}}_{{\bar{x}}}(u^k,v^k)\right\}$ implies

$$\begin{aligned} \displaystyle \lim _{k\rightarrow \infty } {\mathcal {L}}_{\bar{x}}(u^k,v^k+t_k r^k)-{\mathcal {L}}_{{\bar{x}}}(u^{k},v^{k})=0. \end{aligned}$$

The above equation and (26) gives

$$\begin{aligned} \nabla _2 {\mathcal {L}}_{{\bar{x}}}(z^*)^\top (\tilde{\zeta }^2-z^{2*})=0, \end{aligned}$$

which contradicts (25). This prove (24). The inequality (23) can be proved similarly. $\square$