On branching-point selection for trilinear monomials in spatial branch-and-bound: the hull relaxation

Abstract

In Speakman and Lee (Math Oper Res 42(4):1230–1253, 2017), we analytically developed the idea of using volume as a measure for comparing relaxations in the context of spatial branch-and-bound. Specifically, for trilinear monomials, we analytically compared the three possible “double-McCormick relaxations” with the tight convex-hull relaxation. Here, again using volume as a measure, for the convex-hull relaxation of trilinear monomials, we establish simple rules for determining the optimal branching variable and optimal branching point. Additionally, we compare our results with current software practice.

Appendix: technical propositions and lemmas

In this section, we provide the technical propositions and lemmas used for our analysis.

Proposition 1

Given that the upper- and lower-bound parameters respect the labeling $\Omega$, and $\frac{b_1a_2}{b_2} \le \frac{a_1b_2}{a_2}$,

$$\begin{aligned} V_1(q_1) \le V_2\left( \frac{a_1b_2}{a_2}\right) = V_3\left( \frac{a_1b_2}{a_2}\right) . \end{aligned}$$

Proof

It is easy to check that $V_2\left( \frac{a_1b_2}{a_2}\right) = V_3\left( \frac{a_1b_2}{a_2}\right)$.

$$\begin{aligned} V_2\left( \frac{a_1b_2}{a_2}\right) - V_1(q_1) = \frac{(b_3-a_3)(b_2-a_2)}{48(4b_2b_3-a_2b_3-3a_2a_3)a_2^2} \times \left( pa_1^2 + qa_1 + r \right) , \end{aligned}$$

where

$$\begin{aligned} p&=\,\left( -3a_2a_3-a_2b_3+b_2a_3+3b_2b_3\right) \\&\quad \times \Big (-3a_2^3a_3-a_2^3b_3+13a_2^2b_2a_3 +7a_2^2b_2b_3-12a_2b_2^2a_3-20a_2b_2^2b_3+16b_2^3b_3\Big )\\&=\,\Big (3(b_2b_3-a_2a_3)+b_2a_3-a_2b_3\Big ) \\&\quad \times \left( (-3a_2^3+13a_2^2b_2-12a_2b_2^2)a_3 + (-a_2^3+7a_2^2b_2-20a_2b_2^2+16b_2^3)b_3\right) ,\\ q&=\,4a_2b_1(2a_2^2a_3-3a_2b_2a_3-3a_2b_2b_3+4b_2^2b_3)\times (3a_2a_3+a_2b_3-b_2a_3-3b_2b_3),\\ r&=\,4a_2^2b_1^2(a_2a_3+a_2b_3-2b_2b_3)^2. \end{aligned}$$

To show that $V_2\left( \frac{a_1b_2}{a_2}\right) - V_1(q_1)$ is non-negative for all parameters satisfying $\Omega$, we will show that $pa_1^2 + qa_1 + r \ge 0$ for all parameters satisfying $\Omega$.

We observe:

$$\begin{aligned} \Big ((-a_2^3+7a_2^2b_2-20a_2b_2^2+16b_2^3)b_3 + (-3a_2^3+13a_2^2b_2-12a_2b_2^2)a_3\Big ) =:b_3Y + a_3Z, \end{aligned}$$

where

$$\begin{aligned} Y + Z = 4(b_2-a_2)(2b_2-a_2)^2 \ge 0, \end{aligned}$$

and

$$\begin{aligned} Y = \bigg (b_2-a_2\bigg )\bigg (4b_2(b_2-a_2)+12b_2^2+a_2^2\bigg ) + 2a_2^2b_2 \ge 0. \end{aligned}$$

Therefore, by Lemma 6 we have that $b_3Y + a_3Z$ is non-negative and so p is non-negative (Lemma 5). From this we know that $pa_1^2 + qa_1 + r$ is a convex function in $a_1$ and we can find the minimizer by setting the derivative to zero and solving for $a_1$. The minimum occurs at

$$\begin{aligned} a_1 = \frac{2b_1a_2(2a_2^2a_3-3a_2b_2a_3-3a_2b_2b_3+4b_2^2b_3)}{(-3a_2^3a_3-a_2^3b_3+13a_2^2b_2a_3+7a_2^2b_2b_3-12a_2b_2^2a_3 -20a_2b_2^2b_3+16b_2^3b_3)}. \end{aligned}$$

Substituting this in to $pa_1^2 + qa_1 + r$, we obtain that the minimum value of this quadratic is:

$$\begin{aligned} \frac{4a_2^2b_1^2(b_3-a_3)(b_2-a_2)^3(3a_2a_3+a_2b_3-4b_2b_3)^2}{(-3a_2^3a_3-a_2^3b_3+13a_2^2b_2a_3+7a_2^2b_2b_3-12a_2b_2^2a_3 -20a_2b_2^2b_3+16b_2^3b_3)}. \end{aligned}$$

In demonstrating the non-negativity of p, we have already shown that the denominator is non-negative, and it is easy to see that the numerator is non-negative for all values of the parameters satisfying $\Omega$. Therefore $pa_1^2 + qa_1 + r \ge 0$, and consequently, $V_2\left( \frac{a_1b_2}{a_2}\right) - V_1(q_1) \ge 0$ as required. $\square$

Lemma 1

Given that the upper- and lower-bound parameters respect the labeling $\Omega$, and $\frac{b_1a_2}{b_2} \le \frac{a_1b_2}{a_2}$,

$$\begin{aligned} V_1\left( \frac{b_1a_2}{b_2}\right) =V_2\left( \frac{b_1a_2}{b_2}\right) \ge V_2\left( \frac{a_1b_2}{a_2}\right) = V_3\left( \frac{a_1b_2}{a_2}\right) \end{aligned}$$

Proof

It is easy to check that $V_1\left( \frac{b_1a_2}{b_2}\right) =V_2\left( \frac{b_1a_2}{b_2}\right)$ and $V_2\left( \frac{a_1b_2}{a_2}\right) = V_3\left( \frac{a_1b_2}{a_2}\right)$.

Furthermore,

$$\begin{aligned}&V_2\left( \frac{b_1a_2}{b_2}\right) -V_2\left( \frac{a_1b_2}{a_2}\right) \\&\quad =\frac{(b_3-a_3)(b_2-a_2)^2(b_1a_2-a_1b_2)(a_1b_2^2-a_2^2b_1)(3(b_2b_3-a_2a_3) +b_2a_3-a_2b_3)}{12a_2^2b_2^2}\\&\quad \ge 0, \end{aligned}$$

as required. $\square$

Proposition 2

Given that the upper- and lower-bound parameters respect the labeling $\Omega$, and $\frac{b_1a_2}{b_2} > \frac{a_1b_2}{a_2}$,

$$\begin{aligned} V_1(q_1) \le V_4\left( \frac{b_1a_2}{b_2}\right) = V_3\left( \frac{b_1a_2}{b_2}\right) . \end{aligned}$$

Proof

It is easy to check that $V_4\left( \frac{b_1a_2}{b_2}\right) = V_3\left( \frac{b_1a_2}{b_2}\right)$.

$$\begin{aligned} V_4\left( \frac{b_1a_2}{b_2}\right) - V_1(q_1) = \frac{(b_3-a_3)(b_2-a_2)}{48(4b_2b_3-a_2b_3-3a_2a_3)b_2^2} \times \left( pa_1^2 + qa_1 + r \right) , \end{aligned}$$

where

$$\begin{aligned} p&=b_2^2\left( 5b_2b_3-b_2a_3-a_2b_3-3a_2a_3\right) ^2,\\ q&=8b_1b_2\left( 6a_2^2a_3+2a_2^2b_3-3a_2b_2a_3 -9a_2b_2b_3+b_2^2a_3+3b_2^2b_3\right) (b_2b_3-a_2a_3),\\ r&=16b_1^2\left( -3a_2^3a_3-a_2^3b_3+3a_2^2b_2a_3 +5a_2^2b_2b_3-a_2b_2^2a_3-4a_2b_2^2b_3+b_2^3b_3\right) (b_2b_3-a_2a_3). \end{aligned}$$

To show this is non-negative for all parameters satisfying $\Omega$, we will show $pa_1^2 + qa_1 + r \ge 0$ for all parameters satisfying $\Omega$.

Firstly, we observe that

$$\begin{aligned} p=b_2^2(5b_2b_3-b_2a_3-a_2b_3-3a_2a_3)^2 \ge 0. \end{aligned}$$

From this we know that $pa_1^2 + qa_1 + r$ is a convex function in $a_1$, and we can find the minimizer by setting the derivative to zero and solving for $a_1$. The minimum occurs at

$$\begin{aligned} a_1 = \frac{4b_1(6a_2^2a_3+2a_2^2b_3-3a_2b_2a_3-9a_2b_2b_3 +b_2^2a_3+3b_2^2b_3)(a_2a_3-b_2b_3)}{b_2(3a_2a_3+a_2b_3+b_2a_3-5b_2b_3)^2}. \end{aligned}$$

Substituting this in to $pa_1^2 + qa_1 + r$, we obtain that the minimum value of this quadratic is:

$$\begin{aligned} \frac{16b_1^2(b_3-a_3)(b_2-a_2)^3(b_2b_3-a_2a_3)(3a_2a_3+a_2b_3-4b_2b_3)^2}{(3a_2a_3+a_2b_3+b_2a_3-5b_2b_3)^2}, \end{aligned}$$

which is non-negative for all parameters satisfying $\Omega$. Therefore $pa_1^2 + qa_1 + r \ge 0$, and consequently, $V_4\left( \frac{b_1a_2}{b_2}\right) - V_1(q_1) \ge 0$, as required. $\square$

Lemma 2

Given that the upper- and lower-bound parameters respect the labeling $\Omega$, and $\frac{b_1a_2}{b_2} > \frac{a_1b_2}{a_2}$,

$$\begin{aligned} V_1\left( \frac{a_1b_2}{a_2}\right) =V_4\left( \frac{a_1b_2}{a_2}\right) \ge V_4\left( \frac{b_1a_2}{b_2}\right) = V_3\left( \frac{b_1a_2}{b_2}\right) . \end{aligned}$$

Proof

It is easy to check that $V_1\left( \frac{a_1b_2}{a_2}\right) =V_4\left( \frac{a_1b_2}{a_2}\right)$ and $V_4\left( \frac{b_1a_2}{b_2}\right) = V_3\left( \frac{b_1a_2}{b_2}\right)$.

Furthermore,

$$\begin{aligned}&V_4\left( \frac{a_1b_2}{a_2}\right) -V_4\left( \frac{b_1a_2}{b_2}\right) \\&\quad = \frac{(b_3-a_3)(b_2-a_2)^2(b_1a_2^2-a_1b_2^2)(b_1a_2-a_1b_2)(b_2b_3-a_2a_3)}{3a_2^2b_2^2} \ge 0, \end{aligned}$$

as required. $\square$

Lemma 3

Given that the parameters satisfy the conditions $\Omega$, and furthermore, $\frac{b_1a_2}{b_2} \le \frac{a_1b_2}{a_2}$, we have

$$\begin{aligned} q_1 \ge \frac{b_1a_2}{b_2}. \end{aligned}$$

Proof

From the proof of Theorem 4, we know that the midpoint, $q_2$, cannot be less than both $\frac{a_1b_2}{b_1}$ and $\frac{b_1a_2}{b_2}$. Therefore we have:

$$\begin{aligned} q_2 \ge \min \left\{ \frac{a_1b_2}{b_1},\frac{b_1a_2}{b_2}\right\} , \end{aligned}$$

and because we saw in (10) that $q_1 \ge q_2$ we also have

$$\begin{aligned} q_1 \ge \min \left\{ \frac{a_1b_2}{b_1},\frac{b_1a_2}{b_2}\right\} . \end{aligned}$$

Therefore, under the conditions of the lemma, $q_1 \ge \frac{b_1a_2}{b_2}$ as required. $\square$

Lemma 4

Given that the parameters satisfy the conditions $\Omega$, and furthermore, $\frac{b_1a_2}{b_2} \ge \frac{a_1b_2}{a_2}$, we have

$$\begin{aligned} q_1 \ge \frac{a_1b_2}{a_2}. \end{aligned}$$

Proof

We saw in the proof of Lemma 3 that

$$\begin{aligned} q_1 \ge \min \left\{ \frac{a_1b_2}{b_1},\frac{b_1a_2}{b_2}\right\} . \end{aligned}$$

Therefore, under the conditions of the lemma, $q_1 \ge \frac{a_1b_2}{a_2}$ as required. $\square$

For completeness, we state and give proofs of two very simple lemmas (from [24]) which we used several times.

Lemma 5

(Lemma 10.1 in [24] ) For all choices of parameters $0\le a_i < b_i$ satisfying $\Omega$, we have: $b_1a_2-a_1b_2\ge 0$, $b_1a_3-a_1b_3 \ge 0$ and $b_2a_3-a_2b_3 \ge 0$.

Proof

$(b_3-a_3)(b_1a_2-a_1b_2) = b_1a_2b_3 + a_1b_2a_3 - a_1b_2b_3 -b_1a_2a_3 \ge 0$ by $\Omega$. This implies $b_1a_2-a_1b_2 \ge 0$, because $b_3-a_3 > 0$. $b_1a_3-a_1b_3 \ge 0$ and $b_2a_3-a_2b_3 \ge 0$ follow from $\Omega$ in a similar way. $\square$

Lemma 6

(Lemma 10.4 in [24] ) Let $A,B,C,D \in \mathbb {R}$ with $A \ge B\ge 0$, $C+D \ge 0$, $C \ge 0$. Then $AC+BD \ge 0$.

Proof

$AC+BD \ge B(C+D) \ge 0$. $\square$