A Trust Region Method for the Solution of the Surrogate Dual in Integer Programming

Abstract

We propose an algorithm for solving the surrogate dual of a mixed integer program. The algorithm uses a trust region method based on a piecewise affine model of the dual surrogate value function. A new and much more flexible way of updating bounds on the surrogate dual’s value is proposed, in which numerical experiments prove to be advantageous. A proof of convergence is given and numerical tests show that the method performance is better than a state of the art subgradient solver. Incorporation of the surrogate dual value as a cut added to the integer program is shown to greatly reduce solution times of a standard commercial solver on a specific class of problems.

Appendix

In the following, we will denote the support functions to a closed convex set $A$ by $\delta ^{*}\left( A\right) \left( u\right) :=\sup \left\{ au : a\in A\right\} .$ We use epi-limits and Attouch’s theorem for which the reader may consult [14] for details.

Proof

(of Lemma 4.1) Suppose $V_{\alpha }(u+\gamma d) < V_{\alpha }(u) <+\infty$, for some sufficiently small $\gamma >0$ and $x\in \arg \max \left\{ u\left( Ax^{\prime }-b\right) : x^{\prime }\in X\left( \alpha \right) \right\}$. Then, $x\in X\left( \alpha \right)$ and so

$$\begin{aligned} \gamma d\left( Ax-b\right)&\le \max \left\{ \left( u+\gamma d\right) \left( Ax^{\prime }-b\right) : x^{\prime }\in X\left( \alpha \right) \right\} -u\left( Ax-b\right) \\&= V_{\alpha }(u+\gamma d ) - V_{\alpha }(u) <0, \end{aligned}$$

implying $d\left( Ax-b\right) <0$ holds for all $x\in \arg \max \left\{ u\left( Ax^{\prime }-b\right) : x^{\prime }\in X\left( \alpha \right) \right\}$. Consequently, this is also true for all $s$ defined as in (5).

Conversely, for all $x_{\gamma }\in M\left( \gamma \right) :=\arg \max \left\{ \left( u+\gamma d\right) \left( Ax^{\prime }-b\right) : x^{\prime }\in X\left( \alpha \right) \right\} ,$

$$\begin{aligned} V_{\alpha } (u+\gamma d)&=\left( u+\gamma d\right) \left( Ax_{\gamma }-b\right) \\&=u\left( Ax_{\gamma }-b\right) +\gamma d\left( Ax_{\gamma }-b\right) \le V_{\alpha }( u) +\gamma d\left( Ax_{\gamma }-b\right) \text {.} \end{aligned}$$

Note that since $M(0)$ is bounded and $g_{\gamma }( x) := (u+\gamma d) (Ax-b) +\delta _{X(\alpha ) }(x)$ is a sequence of level set bounded, proper convex functions, epi-converging to $g(x) := u( Ax-b) +\delta _{X( \alpha ) }(x)$, we may invoke [14], Exercise 7.32 (c) and Theorem 7.33 to deduce $\limsup _{\gamma \downarrow 0}M(\gamma ) \subseteq M( 0)$. When we assume $d(Ax-b) <0$ for all $x\in M(0)$, we have $d( Ax_{\gamma }-b) <0$ for $\gamma$ sufficiently small. If we assume otherwise, (i.e., assuming there exists $x_{\gamma _{m}}\in M(\gamma _{m})$ for $\gamma _{m}\downarrow 0$ with $d(Ax_{\gamma _{m}}-b) \ge 0$), the following contradiction follows. As $V_{\alpha }(\cdot )$ is a finite, convex function, it is locally Lipschitz and so $\partial V_{\alpha }(\cdot )$ is locally, uniformly bounded, implying local boundedness of $s_{\gamma }:= \,( Ax_{\gamma }-b)$. This in turn implies local boundedness of $\{ x_{\gamma _{m}}\}$, and on taking any convergent subsequence $x_{\gamma _{m_{k}}}\rightarrow x\in M(0)$, we find that the assumption $d(Ax_{\gamma _{m_{k}} }-b) \ge 0$ implies the contradiction $d(Ax-b) \ge 0$ for some $x\in M(0)$. Thus, $d( Ax_{\gamma }-b) <0$ and $V_{ \alpha }(u+\gamma d ) \le V_{\alpha } (u) +\gamma d(Ax_{\gamma }-b) < V_{\alpha }(u)$, for $\gamma$ small. If there exists a sequence $x_{\gamma _{m}}\in M(\gamma _{m})$ such that $x_{\gamma _{m} }\rightarrow x\in M( 0)$ with $d(Ax-b) <0$, then the same argument implies $d$ is a descent direction. Indeed, the presumption that there exists a further subsequence such that $d( Ax_{\gamma _{m_{k}}}-b) \ge 0$ for all $k$, implies the contradiction $d(Ax-b) \ge 0$. Thus, $\gamma _{m}d(Ax_{\gamma _{m} }-b) <0$, for $m$ large, implying $V_{ \alpha }( u+\gamma _{m}d ) < V_{\alpha }(u)$.

Finally, when $x_{\gamma }\in M( \gamma )$ we have $s_{\gamma }:= ( Ax_{\gamma }-b) \in \partial V_{\alpha } ( u+\gamma d)$, and as $V_{\alpha }(\cdot )$ is a finite, convex function, it is also semi-smooth. Consequently, any convergent subsequence $s_{\gamma _{m}}:= ( Ax_{\gamma _{m}}-b) \rightarrow s= ( Ax-b) \in \partial _{d} V_{\alpha } (u)$, for some $x\in M(0)$. Assuming that $d( Ax-b) <0$ for $s= (Ax-b) \in \partial _{d} V_{\alpha } (u)$, we may argue as above that for $\gamma$ sufficiently small that $d(Ax_{\gamma }-b) <0$. Hence $V_{\alpha }(u+\gamma d) < V_{\alpha }(u)$. $\square$

Proof

(of Lemma 5.1) We assume we have a sequence of trust regions of diameter $\Delta _{k}\downarrow 0$, and an associated sequence of $u_{k+1}$, generated in the evaluation of $V_{\alpha _{k}} (u_{k+1} )$ when calculating $\rho _{k+1}$. In Step 3 of Algorithm SDTR, we add some $s_{k+1}\in \partial V_{\alpha _{k}} \left( u_{k+1}\right)$ by choosing $s_{k+1}=Az_{k+1}-b$.

By assumption, $\left\| u_{k+1}-u_{k}\right\| _{\infty }\downarrow 0$. As $u_{k}\rightarrow u$, we have $\left\| u_{k+1}-u\right\| _{\infty }\rightarrow 0$. Also by assumption, there exists a subsequence of $\left\{ u_{k}\right\} _{k=0}^{\infty }$, denoted by $\left\{ u_{p}\right\} _{p=0}^{\infty }$, with $u_{p} \rightarrow _dd$ such that $v^{\prime }(P(\alpha ,\cdot )) ( u,d) <0$. Denote $t_{p}:=\left\| u_{k_{p}+1}-u_{k_{p}}\right\| _{\infty }$. Then $u_{k_{p}+1}-u_{k_{p}}=t_{p}d_{p}$. Note that in Step 3(b), just before returning to do Step 3 again, we add $z_{k_{p}+1}\in \arg \max \left\{ u_{k_{p}+1}\left( Ax-b\right) : x\in X\left( \alpha _{k_{p}}\right) \right\}$ found while calculating $\rho _{{k_p}+1}$. This implies $\left( Az_{k_{p}+1}-b\right)$ must satisfy the following:

$$\begin{aligned} V_{\alpha _{k_{p}}} ( u_{k_{p}}+t_{p}d_{p} )&=\left( Az_{k_{p}+1}-b\right) \cdot \left( u_{k_{p}}+t_{p}d_{p}\right) \nonumber \\&\ge s\cdot \left( u_{k_{p}}+t_{p}d_{p}\right) ,\quad \forall s=\left( Ax-b\right) \text {, }x\in X_{k_{p}}\left( \alpha _{k_{p}}\right) . \end{aligned}$$

(10)

Take $y_{p}\in X\left( \alpha _{k_{p}}\right)$ satisfying $\beta _{k_{p} +1}=u_{k_{p}+1}\left( Ay_{p}-b\right)$ in the problem (3) from which we obtain $\left( u_{k_{p}+1},\beta _{k_{p}+1}\right)$. Due to (10) and $y_{p}\in X_{k_{p}}( \alpha _{k_{p}})$, we have

$$\begin{aligned} V_{\alpha _{k_{p}}} ( u_{k_{p}}+t_{p}d_{p} )\ge \left( Ay_{p}-b\right) \cdot \left( u_{k_{p}}+t_{p}d_{p}\right) =\beta _{k_{p}+1}. \end{aligned}$$

(11)

When (7) holds, we have $0> V_{\alpha _{k_{p}}} ( u_{k_{p}+1} ) - V_{\alpha _{k_{p}}} ( u_{k_{p}} )\ge \beta _{k_{p}+1}- V_{\alpha _{k_{p}}} ( u_{k_{p} } )$ and so $\rho _{k_{p}+1}\le 1$ . Note that (7) holds in either of the cases: $\beta _{k+1}<0$ or $\beta _{k+1}< \underline{V}_k ( u_{k})$.

Now we wish to invoke Attouch’s theorem, ([14], Theorem 12.35), which states that an epi-convergent family of convex functions has their subdifferentials graphically converging. First we note that as $\alpha _{k_{p}}\downarrow \alpha$, and $X( \alpha _{k_{p}})$ are eventually bounded, the finite sequence of convex functions $V_{\alpha _{k_{p}}} (\cdot )$ is monotonically non-increasing, pointwise convergent and consequently also epi-converges to $V_{\alpha } ( \cdot )$. Thus, $\partial V_{\alpha _{k_{p}}} ( \cdot )$ graphically converges to $\partial V_{\alpha } (\cdot )$.

We need to estimate the magnitude of

$$\begin{aligned} \rho _{k_{p}+1}&\ge \left( \frac{ V_{\alpha _{k_{p}}} \left( u_{k_{p}}+t_{p}d_{p}\right) - V_{\alpha _{k_{m_{p}}}} ( u_{k_{m_{p}}}) }{t_{p}}\right) \left( \frac{\beta _{k_{p}}-V_{ \alpha _{k_{p}}} \left( u_{k_{p}}\right) }{t_{p}}\right) ^{-1} \end{aligned}$$

(12)

As $z_{k_{p}}$ was added to the model in Step 3(b) at iteration $k_{p}-1$, and by assumption is not dropped, we have $z_{k_{p}}\in X_{k_{p}}\left( \alpha _{k_{p}}\right)$ when solving (3) at iteration $k_{p}$. Thus, at the iteration $k_p$, a subgradient $s_{k_{p}}:=Az_{k_{p}}-b$ satisfies $V_{\alpha _{k_{p}-1}} \left( u_{k_{p}}\right) =u_{k_{p}}\left( Az_{k_{p} }-b\right)$. As $z_{k_{p}}\in X_{k_{p}}\left( \alpha _{k_{p}}\right)$, it follows that $\beta _{k_{p}+1}\ge u_{k_{p}+1}s_{k_{p}}=s_{k_{p}}\left( u_{k_{p} }+t_{p}d_{l_{p}}\right)$. Since $s_{k_{p}}\in \partial V_{\alpha _{k_{p}-1}} \left( u_{k_{p}}\right)$, taking a further subsequence, on re-numbering, we may assume we have such a sequence of subgradients with $s_{k_{p}}\rightarrow s = (Ax-b) \in \partial V_{\alpha } (\bar{u})$, for some $x\in X(\alpha )$. (Note that local Lipshitzness of $u\mapsto V_{\alpha } (u)$ ensures $\partial V_{\alpha }( \bar{u})$ is bounded. Then by the graphical convergence of subdifferentials, the sequence $\{ s_{p}\}$ is locally bounded, see [14] Exercise 5.34(b)). As (7) holds for all $p$, we have

$$\begin{aligned} 0&>\frac{ V_{\alpha _{k_{p}}} \left( u_{k_{p}}+t_{p}d_{p}\right) - V_{\alpha _{k_{p}}} \left( u_{k_{p}}\right) }{t_{p}}\ge \frac{\beta _{k_{p}+1}- V_{\alpha _{k_{p}}} \left( u_{k_{p}}\right) }{t_{p}} \\&\ge \frac{s_{k_{p}}\left( u_{k_{p}}+t_{p}d_{p}\right) - V_{\alpha _{k_{p}}} \left( u_{k_{p}}\right) }{t_{p}}\ge \frac{s_{k_{p}}\left( u_{k_{p}}+t_{p}d_{p}\right) -V_{\alpha _{k_{p}-1}} \left( u_{k_{p}}\right) \ }{t_{p}}=s_{k_{p}}d_{p}, \end{aligned}$$

using (11) and monotonicity. The last equality holds as $s_{k_{p}}\in \partial V_{\alpha _{k_{p}-1}} \left( u_{k_{p}}\right)$ implies $V_{\alpha _{k_{p}-1}} ( u_{k_{p}} ) =s_{k_{p}}u_{k_{p}}$. Due to Attouch’s theorem, $s_{k_{p}}\rightarrow s\in \partial V_{\alpha } \left( u\right)$, and so $s_{k_{p}}d_{p}\rightarrow sd\le V^{\prime }_{\alpha } \left( u,d\right)$.

As graphical convergence of sets implies the epi-convergence of the associated support functions, (see [14], Corollary 11.36), we have for all $w_p \rightarrow u$,

$$\begin{aligned} \limsup _{\,_{\begin{array}{c} w_{p}\rightarrow u\\ p\rightarrow \infty \end{array}}}\inf _{d^{\prime }\rightarrow d} V^{\prime }_{\alpha _{k_{p}}} \left( w_{p},d^{\prime }\right)&=\lim _{\,_{\begin{array}{c} w_{p}\rightarrow u\\ p\rightarrow \infty \end{array}}}\inf _{d^{\prime }\rightarrow d}\delta ^{*}\left( \partial V_{\alpha _{k_{p}} } \left( w_{p}\right) \right) \left( d^{\prime }\right) =V^{\prime }_{\alpha } \left( u,d\right) \text {.} \end{aligned}$$

Uniform–local Lipschitzness of $d^{\prime }\mapsto \delta ^{*} ( \partial V_{\alpha _{k_{p}}} ( w_{p}) ) ( d^{\prime })$ follows from the locally uniform boundedness of $\partial V_{\alpha _{k_{p}}} \left( w_{p}\right)$. Hence

$$\begin{aligned} \lim _{\,_{\begin{array}{c} w_{p}\rightarrow u\\ p\rightarrow \infty \end{array}}} V^{\prime }_{\alpha _{k_{p}}} \left( w_{p},d\right) \!=\! \limsup _{\,_{\begin{array}{c} w_{p}\rightarrow u\\ p\rightarrow \infty \end{array}}}\delta ^{*}\left( \partial V_{\alpha _{k_{p}}}\left( w_{p}\right) \right) \left( d\right) \!=\! \limsup _{\,_{\begin{array}{c} w_{p}\rightarrow u\\ p\rightarrow \infty \end{array}}}\inf _{d^{\prime }\rightarrow d} V^{\prime }_{\alpha _{k_{p}}} \left( w_{p},d^{\prime }\right) \!<\!0\text {.} \end{aligned}$$

(13)

Consequently, for $p$ large we have $V^{\prime }_{\alpha _{k_{p}}} \left( w_{p},d\right) <0$, for any $w_{p}\rightarrow u$.

By the mean value theorem for convex functions there exists $\gamma _{p}\in ]0,1[$ such that $V^{\prime }_{\alpha _{k_{p}}} \left( u_{k_{p}}+t_{p}\gamma _{p}d,d\right) \ge v_{p}d_{p}=\frac{V_{\alpha _{k_{p}}} \left( u_{k_{p}}+t_{p}d\right) -V_{\alpha _{k_{p}}} \left( u_{k_{p}}\right) }{t_{p}}$, for some element $v_{p}\in \partial V_{\alpha _{k_{p}}} \left( u_{k_{p}}+t_{p}\gamma _{p}d_{p}\right)$. Hence, for $p$ large, $V^{\prime }_{\alpha _{k_{p}}} \left( u_{k_{p}}+t_{p}\gamma _{p}d,d\right) <0$. Using (12), and observing both quantities in the quotient for $\rho _{k_{p}+1}$ are negative, gives, using the Lipschitzness of $V_{\alpha _{k_{p}}} \left( \cdot \right)$, a lower bound on $\limsup _{p\rightarrow \infty }\rho _{k_{p}+1}$ of

$$\begin{aligned}&\left( \limsup _{\begin{array}{c} t_{p}\downarrow 0\\ u_{k_{p}}\rightarrow u \end{array}}\frac{ V_{\alpha _{k_{p}}} \left( u_{k_{p}}+t_{p}d\right) - V_{\alpha _{k_{p}}} \left( u_{k_{p}}\right) }{t_{p}}\right) \left( s_{k_{p}}d_{p}\right) ^{-1} \\&\ge \limsup _{_{p\rightarrow \infty }} V^{\prime }_{\alpha _{k_{p}}} \left( u_{k_{p}}+t_{p}\gamma _{p}d,d\right) \left( sd\right) ^{-1}\ge \limsup _{\begin{array}{c} w_{p}\rightarrow u\\ p\rightarrow \infty \end{array}} V^{\prime }_{\alpha _{k_{p}}} \left( w_{p},d\right) \left( sd\right) ^{-1}. \end{aligned}$$

Thus $\limsup _{k\rightarrow \infty }\rho _{k+1}\ge V^{\prime }_{\alpha } \left( u,d\right) \times \left( sd\right) ^{-1}\ge 1$. $\square$

Proof

(of Lemma 5.2) The argument given above, in the proof of Lemma 5.1, may be applied with $\alpha _{k}$ fixed. Take a subsequence $\{d_{l_p} \}_{p=0}^\infty$ of $\{ d_{l} := \frac{u_{k,l} - u_k}{\Vert u_{k,l} - u_k \Vert }\}_{p=0}^\infty$ converging to $d$. With the subsequence $u_{k,l_{p}}\rightarrow u_{k}$, we may associate a subsequence of diameters $\Delta _{k,l_{p}}\downarrow 0$. Place $u_{k,l_{p}}=u_{k}+t_{p}d_{l_{p}}$, and note that we continue to reject an update of $u_{k}$. As $u\mapsto V_{\alpha _{k}} \left( u\right)$ is convex, it is also semi-smooth, and so we may assume $s_{l_{p}}\rightarrow s \in \partial _{d} V_{\alpha _{k}} \left( u_{k}\right) :=\left\{ s^{\prime }\in \partial V_{\alpha _{k}} \left( u_{k}\right) : s^{\prime }\cdot d=V^{\prime }_{\alpha _{k}} \left( u_{k} ,d \right) \right\}$. As $z_{k,l_{p}}$ is added to $X_{k,l_{p}}\left( \alpha _{k}\right)$ and retained, we have $z_{k,l_{p-1}}\in X_{k,l_{p-1}}\left( \alpha _{k}\right)$ when solving (3) at iteration $(k,l_{p})$ (and $\alpha _{k}$ is never updated).

Consequently, $\beta _{k,l_{p}}\ge u_{k,l_{p}}\cdot s_{l_{p}-1}=s_{l_{p}-1}\cdot \left( u_{k}+t_{p}d_{p}\right)$. Using (11) replacing $\alpha _{k_{p}}=\alpha _{k}$, and applying (7) to each problem in this sequence yields

$$\begin{aligned} 0&>\frac{V_{\alpha _{k}} \left( u_{k}+t_{p}d_{l_{p}}\right) -V_{\alpha _{k}} \left( u_{k}\right) }{t_{p}} \ge \frac{\beta _{k,l_{p}} - V_{\alpha _{k}} \left( u_{k}\right) }{t_{p}}\nonumber \\&\ge \frac{s_{l_{p}-1}\cdot \left( u_{k}+t_{p}d_{p}\right) - V_{ \alpha _{k}} \left( u_{k}\right) }{t_{p}} = \left[ \left( s_{l_{p}-1}\cdot d_{l_{p}}\right) -\frac{t_{p-1}}{t_{p}}\left( s_{l_{p}-1}\cdot d_{p-1}\right) \right] \nonumber \\&\quad +\left( \frac{t_{p-1}}{t_{p}}\right) \left( \frac{ V_{\alpha _{k}} \left( u_{k}+t_{p-1}d_{l_{p-1}}\right) - V_{\alpha _{k}} \left( u_{k}\right) }{t_{p-1}}\right) , \end{aligned}$$

(14)

as $s_{l_{p}-1}\in \partial V_{\alpha _{k}} \left( u_{k}+t_{p-1}d_{p-1}\right)$. First suppose that there exists a subsequence so that $\left\{ \frac{t_{p_{m}-1}}{t_{p_{m}}}\right\} \rightarrow \lambda \ge 0$. By semi-smoothness, we have the limiting values $s_{l_{p}-1}\cdot d_{p}\rightarrow s\cdot d=V^{\prime }_{\alpha _{k}} \left( u_{k},d\right)$ and $s_{p-1}\cdot d_{p-1}\rightarrow V^{\prime }_{\alpha _{k}} \left( u_{k},d\right)$, so (14) converges to the weighted sum $\left[ 1-\lambda \right] V^{\prime }_{\alpha _{k}} \left( u_{k},d\right) +\lambda V^{\prime }_{\alpha _{k}} \left( u_{k},d\right) =V^{\prime }_{\alpha _{k}} \left( u_{k},d\right)$. Hence, using (12) and noting both quantities in the quotient are negative, we obtain

$$\begin{aligned} \limsup _{p\rightarrow \infty }\rho _{k+l_{p}}&\ge \limsup _{p\rightarrow \infty }\left( \frac{V_{\alpha _{k}} \left( u_{k}+t_{p}d_{l_{p}}\right) - V_{\alpha _{k}} \left( u_{k}\right) }{t_{p}}\right) \left( \frac{\beta _{k,l_{p}}- V_{\alpha _{k}} \left( u_{k}\right) }{t_{p}}\right) ^{-1}\\&\ge \left( \lim _{t_{p}\downarrow 0,d_{l_{p}}\rightarrow d}\frac{V_{\alpha _{k}} \left( u_{k}+t_{p}d_{l_{p}}\right) -V_{\alpha _{k}} \left( u_{k}\right) }{t_{p}}\right) \left( V^{\prime }_{\alpha _{k}} \left( u_{k},d\right) \right) ^{-1}=1\text {.} \end{aligned}$$

In the alternative case, $\left\{ \frac{t_{p-1}}{t_{p}}\right\}$ is unbounded (and $\Delta _{k,l_{p}+1}=\gamma \Delta _{k},_{l_{p}}$ using a $\gamma \in ]0,1[$). Then $u_{k}+t_{p-1}d_{_{p-1}}\in \mathrm{int }B_{k,l_{p}}$, placing us in Step 3(a), as described in Lemma 4.2, resulting in an update of $u_{k}$, contrary to assumption. $\square$

Proof

(of Corollary 5.1) We have $\liminf _{k\rightarrow \infty }\rho _{k+1}$ given by

$$\begin{aligned} \min \left\{ \lim _{m\rightarrow \infty }\rho _{k_{m}+1} : \left\{ \rho _{k_{m}+1}\right\} _{m=0}^{\infty }\text { is a convergent subsequence of }\left\{ \rho _{k+1}\right\} _{k=0}^{\infty }\right\} \text {.} \end{aligned}$$

For part 1, apply Lemma 5.2 to $\left\{ \rho _{k_{m}+1}\right\} _{m=0}^{\infty }$ to obtain $\left\{ \rho _{k_{p}+1}\right\} _{p=0}^{\infty }$ with

$$\begin{aligned} \lim _{m\rightarrow \infty }\rho _{k_{m}+1}=\limsup _{p\rightarrow \infty }\rho _{k_{p}+1}\ge 1. \end{aligned}$$

(15)

For part 2, we use $\rho _{k+1}\ge \xi$, and so

$$\begin{aligned}&\frac{ V_{\alpha _{k_{p}}} \left( u_{k_{p}}+t_{p}d_{p}\right) - V_{\alpha _{k_{p}}} \left( u_{k_{p}}\right) }{t_{p}}\le \xi \left[ \frac{\beta _{k_{p}}- V_{\alpha _{k_{p}}} \left( u_{k_{p}}\right) }{t_{p}}\right] \nonumber \\&=\xi \left[ \frac{\underline{V}_{k_{p}+1}\left( u_{k_{p}}+t_{p}d_{p}\right) - \underline{V}_{k_{p}}\left( u_{k_{p}}\right) }{t_{p}}\right] .\ \end{aligned}$$

(16)

As $u_{k}\rightarrow u$ $\notin \arg \min \left\{ V_{\alpha } \left( w \right) : w\in S^{n}\right\}$, for $k$ large, there exists a descent direction for $V_{\alpha } \left( \cdot \right)$ at $u_{k}$. As $\underline{V}_{k}\left( u_{k}\right) = V_{\alpha } \left( u_{k} \right)$ and $\underline{V}_{k}\left( \cdot \right)$ minorizes $V_{\alpha } \left( \cdot \right)$, there must exist a greater descent in the same direction for $\underline{V}_{k}\left( \cdot \right)$ at $u_{k}$. As $u_{k_{p}+1}$ solves (3), we have $d_{k_{p}}:=\frac{u_{k_{p} +1}-u_{k_{p}}}{ \Vert u_{k_{p}+1}-u_{k_{p}} \Vert }$ in the direction of maximal descent of $\underline{V}_{k_{p}}\left( \cdot \right)$ at $u_{k_{p}}$, and so there exists $\delta >0$ such that we have $\underline{V}_{k_{p}+1}\left( u_{k_{p}}+t_{p}d_{p}\right) - \underline{V}_{k_{p}}\left( u_{k_{p}}\right) \le -\delta t_{p}$, for $p$ sufficiently large. Using (16) and the subgradient inequality for $V_{\alpha _{k_{p}}} \left( \cdot \right)$ at $u_{k_{p}}$ gives (for $p$ large)

$$\begin{aligned} V^{\prime }_{\alpha _{k_{p}}} \left( u_{k_{p}},d_{p}\right) \le \frac{ V_{\alpha _{k_{p}}} \left( u_{k_{p}}+t_{p}d_{p}\right) - V_{\alpha _{k_{p}}} \left( u_{k_{p}}\right) }{t_{p}}\le -\xi \delta <0. \end{aligned}$$

Using (13), we have $V^{\prime }_{\alpha } \left( u,d\right) =\limsup _{p} V^{\prime }_{\alpha _{k_{p}}}\left( u_{k_{p}},d_{p}\right) -\xi \delta <0.$ We may now apply Lemma 5.1 to get (15) again. $\square$

Proof

(of Proposition 5.1) In all cases, we have $V_{\alpha _{k}} \left( u_{k}\right) \ge 0$. Suppose that $\rho _{k,l}<\xi$ for all $l$. Then, applying Lemma 5.4 recursively, we have indices $l_{p}$ with

$$\begin{aligned} V_{\alpha _{k}} \left( u_{k}\right) - \underline{V}_{k,l_{k,p}}\left( u_{k,l_{p}}\right)&\le \eta \left[ V_{\alpha _{k}} \left( u_{k}\right) - \underline{V}_{k,l_{p-1}}\left( u_{k,l_{p-1}}\right) \right] \cdots \nonumber \\&\le \eta ^{p}\left[ V_{\alpha _{k}} \left( u_{k}\right) - \underline{V}_{k,l_{0}}\left( u_{k,l_{0}}\right) \right] \rightarrow _{p\rightarrow \infty }0\text {.} \end{aligned}$$

(17)

When $u_{k}\notin S_{\alpha _{k}}$, we have some $\varepsilon >0$ such that $\left\| u_{k}-p_{\alpha _{k}}\left( u_{k}\right) \right\| _{\infty }\ge \varepsilon$ and $V_{\alpha _{k}} \left( u_{k}\right) -v_{k}^{*}\ge$ $\varepsilon$. By (8), $\min \left( \frac{\Delta _{k,l_{p}}}{ \Vert u_{k}-p_{\alpha _{k}}\left( u_{k}\right) \Vert _{\infty }},1 \right) \left( V_{\alpha _{k}} \left( u_{k}\right) -v_{k}^{*}\right) \rightarrow _{p\rightarrow \infty }0$ implying $\Delta _{k,l_{p}}\rightarrow 0$. Apply Corollary 5.1, part 1 to get a subsequence of $\left\{ \rho _{k,l_{p}}\right\} _{p=0}^{\infty }$ that tends to $1$. Thus, there exists a $p$ with $\rho _{k,l_{p}}\ge \xi >0$, a contradiction.

In the case that $u_{k}\in S_{\alpha _{k}}$, there does not exist any descent for $V_{\alpha _{k}} \left( \cdot \right)$ at $u=u_{k}$. In particular, we have $v_{k}^{*}= V_{\alpha _{k}} \left( u_{k}\right) \ge 0$. Consequently, (3) cannot generate a descent that passes the test $\rho _{k}\ge \xi$. Using (17), we observe that

$$\begin{aligned} v_{k}^{*} - \underline{V}_{k,l_{k,p}}\left( u_{k,l_{p}}\right) \rightarrow _{p}0\text {.} \end{aligned}$$

(18)

As we add a new subgradient each time when solving $V_{\alpha _{k}} (u_{k,l})$, to calculate $\rho _{k,l}$, the model function $u\mapsto \underline{V}_{k,l_{k}}(u)$ is monotonically increasing, and hence convergent to a finite, convex function. As the trust region size is monotonically non-increasing, we have $\left\{ \underline{V}_{k,l_{k}}\left( u_{k,l_{k}}\right) \right\} _{k}$ monotonically non-decreasing. Consequently, using (18), we deduce that we have a subsequence converging to zero. So for the whole sequence, $\underline{V}_{k,l_{k}}\left( u_{k,l_{k}}\right) \uparrow v_{k}^{*}$. When $V_{\alpha _{k}} \left( u_{k}\right) =v_{k}^{*}>0$, (or equivalently $v\left( SD\right) <\alpha _{k}$, the case $v\left( SD\right) \ne \alpha _{k}$), we find after a finite number of iterations that $\beta _{k,l_{k}}= \underline{V}_{k,l_{k,p}}\left( u_{k,l_{k}}\right) >0$. Thus, we are in case 3(a), of Algorithm SDTR, after a finite number of iterations. Alternatively, $V_{\alpha _{k}} \left( u_{k}\right) =v_{k}^{*}=0$, so $\alpha _{k}=v\left( SD\right)$ and $\underline{V}_{k,l_{k}}\left( u_{k,l_{k}}\right) \le 0$. This establishes part 2b of the proposition.

Suppose we have a pure IP and we add a new subgradient each time when solving $V_{\alpha _{k}} \left( u_{k,l}\right)$ when calculating $\rho _{k,l}$. As $u\mapsto V_{\alpha _{k}} \left( u\right)$ is polyhedral, ($X\left( \alpha _{k}\right)$ contains a finite set of points), we add all extremal subgradients after a finite number of iterations $l$. As $u_{k}\in S_{\alpha _{k}}$ we have $0\in \partial V_{\alpha _{k}} \left( u_{k}\right)$, or equivalently, $0 \in \mathrm{co }\left\{ u_{k}\left( Ax_{j}-b\right) :x_{j}\in X_{k,l}\right\}$. But $\underline{V}_{k,l}\left( u\right) =\max \left\{ u\left( Ax_{j}-b\right) : x_{j}\in X_{k,l}\right\}$, so we also have $0\in \partial \underline{V}_{k,l} \left( u_{k}\right) =\mathrm{co }\left\{ u_{k}\left( Ax_{j}-b\right) : x_{j}\in X_{k,l}\right\} ,$ showing that $u_{k}$ is a local (and hence global) minimum of $u\mapsto \underline{V}_{k,l}\left( u\right)$. Then we have the inequalities $0\le \underline{V}_{k,l}\left( u_{k}\right) =u_{k}\left( Ax_{k}-b\right) \le \min _{u\in S^{n}\cap B} \underline{V}_{k,l}\left( u\right) =\beta _{k,l},$ using the inequality $u_{k}\left( Ax_{k}-b\right) = V_{\alpha _{k}} \left( u_{k}\right) \ge 0$. Hence $\underline{V}_{k,l}\left( u_{k}\right) =0$ finitely. $\square$

Proof

(of Theorem 5.1) Whenever $[\overline{\alpha },\underline{\alpha }]$ is updated, we decrease the length of this interval of uncertainty by at least a constant factor. Thus, finite termination is assured unless we have an infinite cycle in the trust region loop, with fixed interval $[\overline{\alpha },\underline{\alpha }]$. Proposition 5.1 indicates that this can only occur in two ways. Via a sequence of decreasing $\left\{ \alpha _{k}\right\}$, and a sequence of acceptable descents (i.e. $\rho _{k,l}\ge \xi )$ which does not terminate, or that for some $k$ we have $\alpha _{k}=v\left( SD\right)$ and $\underline{V}_{k,l_{k}}\left( u_{k,l_{k}}\right) \rightarrow 0= V_{\alpha _{k}} \left( u_{k}\right)$, monotonically. This latter case cannot occur when we have a pure IP.

Consider the first case of acceptable descents. Then there exists a sequence $\left\{ l_{k}\right\}$ such that $u_{k+1}=u_{k,l_{k}}$ and between iteration $\left( k,1\right)$ and $\left( k,l_{k}\right)$, we have $\alpha _{k+l}=\alpha _{k}$ fixed. Thus, by Lemma 5.3,

$$\begin{aligned} \underline{V}_{k,1}\left( u_{k}\right)&- \underline{V}_{k,l_{k}}\left( u_{k,l_{k}}\right) =\sum _{l=1}^{l_{k}} \underline{V}_{k,l}\left( u_{k}\right) -\underline{V}_{k,l}\left( u_{k,l}\right) \\&\ge \sum _{l=1}^{l_{k}}\min \left( \frac{\Delta _{k,l}}{\left\| u_{k}-p_{\alpha _{k}}\left( u_{k}\right) \right\| _{\infty }},1\right) \left( V_{\alpha _{k}} \left( u_{k}\right) -v_{k}^{*}\right) , \end{aligned}$$

where $u_{k+1}=u_{k,l_{k}}$ and $\underline{V}_{k+1,0}\left( u_{k+1}\right) = \underline{V}_{k,l_{k}}\left( u_{k,l_{k}}\right) \ge 0$ (as we are in Step 3 of STDR). The model function $\underline{V}_{k,l_{k}}\left( \cdot \right)$ is not carried to the next stage, but $\alpha _{k}$ is decreased to $\alpha _{k+1}$, and we prune subgradients which decrease the new model function to the new initial model $\underline{V}_{k+1,1}$. Hence, we have the inequality $\underline{V}_{k,l_{k}}\left( u_{k,l_{k}}\right) - \underline{V}_{k+1,1}\left( u_{k+1}\right) \ge 0$. Thus $\underline{V}_{k,1}\left( u_{k}\right) - \underline{V}_{k+1,1}\left( u_{k+1}\right)$ equals

$$\begin{aligned}&\underline{V}_{k,1}\left( u_{k}\right) - \underline{V}_{k,l_{k}}\left( u_{k,l_{k}}\right) + \underline{V}_{k,l_{k}}\left( u_{k,l_{k}}\right) - \underline{V}_{k+1,1}\left( u_{k+1}\right) \\&\quad \ge \sum _{l=1}^{l_{k}}\min \left( \frac{\Delta _{k+l}}{\left\| u_{k}-p_{\alpha _{k}}\left( u_{k}\right) \right\| _{\infty }},1\right) \left( V_{\alpha _{k}} \left( u_{k}\right) -v_{k}^{*}\right) \text {.} \end{aligned}$$

For $M=\sum _{k=0}^{K}l_{k}$, (as we remain in Step 3 of SDTR with $\underline{V}_{K,1}\left( u_{K}\right) \ge 0$),

$$\begin{aligned} +\infty&> \underline{V}_{0,1}\left( u_{0}\right) \ge \underline{V}_{0,1}\left( u_{0}\right) - \underline{V}_{K,1}\left( u_{K}\right) =\sum _{k=0}^{K-1}\left[ \underline{V}_{k,1}\left( u_{k}\right) - \underline{V}_{k+1,1}\left( u_{k+1}\right) \right] \nonumber \\&\ge \sum _{k=0}^{K-1}\sum _{l=0}^{l_{k}}\min \left( \frac{\Delta _{k+l}}{\left\| u_{k}-p_{\alpha _{k}}\left( u_{k}\right) \right\| _{\infty }},1\right) \left( V_{\alpha _{k}} \left( u_{k}\right) -v_{k}^{*}\right) \nonumber \\&=\sum _{j=0}^{M}\min \left( \frac{\Delta _{j}}{\left\| u_{j}-p_{\alpha _{j}}\left( u_{j}\right) \right\| _{\infty }},1\right) \left( V_{\alpha _{j}} \left( u_{j}\right) -v_{j}^{*}\right) , \end{aligned}$$

(19)

where $p_{\alpha _{j}}\left( u\right)$ denotes the projection of $u$ onto the solution set $S_{\alpha _{j}}\subseteq S^{n}$ of minimizers of $u\mapsto V_{\alpha _{j}} \left( u\right)$ and $v_{j}^{*} := \min _{u\in S^{n}} V_{\alpha _{j}} \left( u\right)$. The convergence of the series implies the terms in (19) converge to zero. Next, note that each time we accept a subgradient, (as we have obtained sufficient descent), we decrease the interval $[\underline{\alpha },\alpha _{k}].$ Thus, we have $\alpha _{j}\downarrow \underline{\alpha }$ as $j\rightarrow \infty$. As $\min _{u\in S^{n}} V_{\underline{\alpha }} \left( u\right) <0$, and we assume we remain in Step 3 of SDTR, we have $V_{\alpha _{j}} \left( u_{j}\right) \ge 0$ and hence: $\left\| u_{j}-p_{\alpha _{j}}\left( u_{j}\right) \right\| _{\infty }\ge \delta >0$ for some $\delta$. Thus, there is some $\epsilon >0$ such that $V_{\alpha _{j}} \left( u_{j}\right) -v_{j}^{*} \ge \epsilon >0$. Consequently, we must have $\Delta _{j}\rightarrow 0.$

Consider the first case when we have a MIP. Note that each time we accept a subgradient, (as we have obtained sufficient descent), we reduce the interval $[\underline{\alpha },\alpha _{k}].$ Thus, we have $\alpha _{j}\downarrow \underline{\alpha }$ as $j\rightarrow \infty$ and we may apply Corollary 5.1 to deduce that $\liminf _{j}\rho _{j}\ge 1$. But this implies that there exists a $J$ for which $\rho _{j}\ge \frac{3}{4}$ for $j\ge J$, forcing $\Delta _{j+l}=\min \left\{ 2\Delta _{j+l-1},\overline{\Delta }\right\}$. Eventually, we must have $\Delta _{j}=\overline{\Delta }>0$, a contradiction.

In the case of a pure IP, (no continuous variable), we note that as $\alpha _{j}\downarrow 0$, and $X\left( \alpha _{j}\right) :=\left\{ x\in X : cx\le \alpha _{j}\right\}$, for $j$ sufficiently large, the discrete components of $X\left( \alpha _{j}\right)$ do not change. As we have a pure IP, the function $V_{\alpha _{j}} \left( \cdot \right)$ must be constant and equal to $V_{\underline{\alpha }} \left( \cdot \right)$, for $j$ sufficiently large. Hence for $j$ sufficiently large $V_{\alpha _{j}} \left( \cdot \right) = V_{\underline{\alpha }} \left( \cdot \right)$. We may now apply Corollary 5.1, with $\alpha _{k}=\underline{\alpha }$ constant, to deduce that $\liminf _{j}\rho _{j}\ge 1$.

Thus, we cannot have an infinite loop of acceptable descents without an update of a lower or upper bound. After a finite number of iteration, we must have $\left| \overline{\alpha }-\underline{\alpha }\right| \le \varepsilon$. $\square$