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Global optimization with spline constraints: a new branch-and-bound method based on B-splines

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Abstract

This paper discusses the use of splines as constraints in mathematical programming. By combining the mature theory of the B-spline and the widely used branch-and-bound framework a novel spatial branch-and-bound (sBB) method is obtained. The method solves nonconvex mixed-integer nonlinear programming (MINLP) problems with spline constraints to global optimality. A broad applicability follows from the fact that a spline may represent any (piecewise) polynomial and accurately approximate other nonlinear functions. The method relies on a reformulation–convexification technique which results in lifted polyhedral relaxations that are efficiently solved by an LP solver. The method has been implemented in the sBB solver Convex ENvelopes for Spline Optimization (CENSO). In this paper CENSO is compared to several state-of-the-art MINLP solvers on a set of polynomially constrained NLP problems. To further display the versatility of the method a realistic pump synthesis problem of class MINLP is solved with exact and approximated pump characteristics.

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Notes

  1. Division by zero is handled by a ‘\(0/0 = 0\)’ convention.

  2. A knot vector satisfying the conditions in Definition 1 is also said to be (\(p+1\))-regular or clamped.

  3. \(\le \mathbf{p}\) is here meant as element-wise inequality. Note that a multivariate spline or polynomial of degree \(\mathbf{p}\) may have terms like \(x_{1}^{p_1}\ldots x_{d}^{p_d}\), so in the conventional sense its degree is \(\sum _j p_j\).

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Acknowledgments

This work was supported by the Center for Integrated Operations in the Petroleum Industry, Trondheim, Norway.

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  1. Department of Engineering Cybernetics, Norwegian University of Science and Technology (NTNU), O.S. Bragstads Plass 2D, 7491, Trondheim, Norway

    Bjarne Grimstad & Anders Sandnes

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  1. Bjarne Grimstad
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Appendices

Appendix 1: Test problems

This appendix holds a collection of nonconvex NLP problems used in the computational study in this paper. Most of the problems can be found in the GLOBALLib library [52] and in the test problem handbooks of Floudas [20, 21]. The same problem set was used in [59].

Problem 1

[21, Chap. 4.10].

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&f(\mathbf{x}) = -x_1 -x_2 \\ \text {subject to}\quad&x_2 \le 2 + 8x_{1}^{2} - 8x_{1}^{3} + 2x_{1}^{4} \\&x_2 \le 36 - 96x_{1} + 88x_{1}^{2} - 32x_{1}^{3} + 4x_{1}^{4} \\&x_1 \in [0,3], x_2 \in [0,4] \end{aligned}$$
(P1)

The global optimum of (P1) is at \(\mathbf{x}^{*} = [2.3295, 3.1785]^\textsf {T}\) with \(f(\mathbf{x}^*) = -5.5080\).

Problem 2

Problem G6 in [40].

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&f(\mathbf{x}) = (x_1 - 10)^3 + (x_2 - 20)^3 \\ \text {subject to}\quad&100 - (x_1 - 5)^2 - (x_2 - 5)^2 \le 0 \\&-82.81 + (x_1 - 6)^2 + (x_2 - 5)^2 \le 0 \\&x_1 \in [13,100], x_2 \in [0,100] \end{aligned}$$
(P2)

Global optimum at \(\mathbf{x}^{*} = [14.0950, 0.8430]^\textsf {T}\) with \(f(\mathbf{x}^*) = -6961.815\).

Problem 3

[42].

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&f(\mathbf{x}) = x_1 \\ \text {subject to}\quad&x_{1}^{2} - x_2 \le 0 \\&x_2 - x_{1}^{2}(x_1 - 2) + 10^{-5} \le 0 \\&\mathbf{x} \in [-10,10]^2 \end{aligned}$$
(P3)

Global optimum at \(\mathbf{x}^{*} = [3.0, 9.00001]^\textsf {T}\) with \(f(\mathbf{x}^*) = 3\).

Problem 4

[21, Chap. 3.5]. Problem ex3.1.4 in GlobalLib.

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&f(\mathbf{x}) = -2x_1 + x_2 - x_3 \\ \text {subject to}\quad&\mathbf{x}^\textsf {T}{\mathsf {A}}^\textsf {T}{\mathsf {A}} \mathbf{x} - 2 \mathbf{y}^\textsf {T}{\mathsf {A}} \mathbf{x} + ||\mathbf{y}||^2 - 0.25 ||\mathbf{b} - \mathbf{z}||^2 \ge 0 \\&x_1 + x_2 + x_3 - 4 \le 0 \\&3x_2 + x_3 - 6 \le 0 \\&x_1 \in [0,2], x_2 \in [0,10], x_3 \in [0,3] \\ \text {with data}\quad&\\ {\mathsf {A}}&= \begin{bmatrix} 0&\quad 0&\quad 1 \\ 0&\quad -1&\quad 0 \\ -2&\quad 1&\quad -1 \end{bmatrix} \\ \mathbf{b}&= [3,~0,~-4]^\textsf {T}\\ \mathbf{y}&= [1.5,~-0.5,~-5]^\textsf {T}\\ \mathbf{z}&= [0,~-1,~-6]^\textsf {T}\end{aligned}$$
(P4)

The global optimum of P4 is at \(\mathbf{x}^{*} = [0.5, 0,3]^\textsf {T}\) with \(f(\mathbf{x}^*) = -4\).

Problem 5

Himmelblau problem from [21]. Problem ex14.1.1 in GlobalLib.

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&f(\mathbf{x}) = x_3 \\ \text {subject to}\quad&2x_{1}^{2} + 4x_{1}x_{2} - 42x_{1} + 4x_{1}^{3} - x_{3} \le 14 \\&-2x_{1}^{2} - 4x_{1}x_{2} + 42x_{1} - 4x_{1}^{3} - x_{3} \le -14 \\&2x_{1}^{2} + 4x_{1}x_{2} - 26x_{2} + 4x_{2}^{3} - x_{3} \le 22 \\&-2x_{1}^{2} - 4x_{1}x_{2} + 26x_{2} - 4x_{2}^{3} - x_{3} \le -22 \\&\mathbf{x} \in [-5,5]^3 \end{aligned}$$
(P5)

This is a root finding problem with \(f(\mathbf{x^*}) = 0\). \(\mathbf{x}^* = [-0.3050690, -0.9133455, 0]^\textsf {T}\) is a known solution to P5.

Problem 6

An optimal design problem for a pressure vessel [43, 59].

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&0.6224 x_{3}x_{4} + 1.7781x_{2}x_{3}^{2} + 3.1661 x_{1}^{2}x_{4} + 19.84 x_{1}^{2}x_{3}\\ \text {subject to}\quad&-x_1 + 0.0193x_3 \le 0 \\&-x_2 + 0.00954x_3 \le 0 \\&-\pi x_{3}^{2} x_4 - (4/3)\pi x_{3}^3 + 750.1728 \le 0 \\&-240 + x_4 \le 0 \\&x_1 \in [1,1.375], x_2 \in [0.625,1], \\&x_3 \in [47.5, 52.5], x_4 \in [90,112] \end{aligned}$$
(P6)

Best known solution is \(\mathbf{x}^* = [1, 0.625, 47.5, 90]^\textsf {T}\) with \(f(\mathbf{x}^*) = 6395.5\).

Problem 7

[21]. Problem ex7.3.2 in GlobalLib.

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&f(\mathbf{x}) = x_4 \\ \text {subject to}\quad&x_{1}^{4} x_{2}^{4} - x_{1}^{4} - x_{2}^{4} x_{3} = 0 \\&1.4 - x_1 - 0.25 x_4 \le 0 \\&-1.4 + x_1 - 0.25 x_4 \le 0 \\&1.5 - x_2 - 0.2 x_4 \le 0 \\&-1.5 + x_2 - 0.2 x_4 \le 0 \\&0.8 - x_3 - 0.2 x_4 \le 0 \\&-0.8 + x_3 - 0.2 x_4 \le 0 \\&\mathbf{x} \in [0,5]^4 \end{aligned}$$
(P7)

The global optimum of P7 is at \(\mathbf{x}^{*} = [1.1275, 1.2820, 1.0179, 1.0899]^\textsf {T}\) with \(f(\mathbf{x}^*) = 1.0899\).

Problem 8

Mechanical design problem from [84].

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&f(\mathbf{x}) = 27.264(2 x_2 x_4 + x_1 x_3 - 2 x_3 x_4) \\ \text {subject to}\quad&61.01627586 - I(\mathbf{x}) \le 0 \\&8 x_1 - I(\mathbf{x}) \le 0 \\&x_1 x_2 x_4 - x_2 x_{4}^{2} + x_{1}^{2} x_3 + x_3 x_{4}^{2} - 2 x_1 x_3 x_4 - 3.5 x_3 I(\mathbf{x}) \le 0 \\&x_1 - 3 x_2 \le 0 \\&2 x_2 - x_1 \le 0 \\&x_3 - 1.5 x_4 \le 0 \\&0.5 x_4 - x_3 \le 0 \\&x_1 \in [3,20], x_2 \in [2,15], x_3 \in [0.125,0.75], x_4 \in [0.25, 1.25] \end{aligned}$$
(P8)

where \(I(\mathbf{x}) = 6x_{1}^{2}x_{2}x_{3} - 12x_{1}x_{2}x_{3}^{2} + 8x_{2}x_{3}^{3} + x_{1}^{3}x_{4} - 6x_{1}^{2}x_{3}x_{4} + 12x_{1}x_{3}^{2}x_{4} - 8x_{3}^{3}x_{4}\). The global optimum is attained at \(\mathbf{x}^{*} = [4.9542, 2, 0.125, 0.25]^\textsf {T}\) with \(f(\mathbf{x}^*) = 42.444\).

Problem 9

Test problem 1 in [20, Chap. 2.2.1]. Problem ex2.1.1 in GlobalLib.

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&f(\mathbf{x}) = \mathbf{c}^{\textsf {T}}\mathbf{x} - \mathbf{x}^{\textsf {T}}\mathbf{Q}\mathbf{x} \\ \text {subject to}\quad&20x_1 + 12x_2 + 11x_3 + 7x_4 + 4x_5 \le 40 \\&\mathbf{x} \in [0,1]^5 \end{aligned}$$
(P9)

where \(\mathbf{c} = [42,44,45,47,47.5]^\textsf {T}\) and \(\mathbf{Q} = 50\mathbf{I}\) (\(\mathbf{I}\) is the identity matrix). The global optimum is attained at \(\mathbf{x}^{*} = [1,1,0,1,0]^\textsf {T}\) with \(f(\mathbf{x}^*) = -17\).

Problem 10

Test problem 2 in [20, Chap. 2.2.1]. Problem ex2.1.2 in GlobalLib

$$\begin{aligned} \underset{\mathbf{x},y}{\text {minimize}} \quad&f(\mathbf{x},y) = \mathbf{c}^{\textsf {T}}\mathbf{x} - 0.5\mathbf{x}^{\textsf {T}}\mathbf{Q}\mathbf{x} - 10y\\ \text {subject to}\quad&6x_1 + 3x_2 + 3x_3 + 2x_4 + x_5 \le 6.5 \\&10x_1 + 10x_3 + y \le 20 \\&y \ge 0 \\&\mathbf{x} \in [0,1]^5 \end{aligned}$$
(P10)

where \(\mathbf{c} = -[10.5,7.5,3.5,2.5,1.5]^\textsf {T}\) and \(\mathbf{Q} = \mathbf{I}\) (\(\mathbf{I}\) is the identity matrix). The global optimum is attained at \(\mathbf{x}^{*} = [0,1,0,1,1]^\textsf {T}\) and \(y^* = 20\) with \(f(\mathbf{x}^*,y^*) = -213\).

Problem 11

[20, Chap. 3.3.1].

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&f(\mathbf{x}) = -25(x_1 - 2)^2 - (x_2 - 2)^2 - (x_3 - 1)^2 \\& -(x_4 - 4)^2 - (x_5 - 1)^2 - (x_6 - 4)^2 \\ \text {subject to}\quad&(x_3 - 3)^2 + x_4 \ge 4 \\&(x_5 - 3)^2 + x_6 \ge 4 \\&x_1 - 3x_2 \le 2 \\&-x_1 + x_2 \le 2 \\&x_1 + x_2 \le 6 \\&x_1 + x_2 \ge 2 \\&x_1 \in [0,6], x_2 \in [0,6], x_3 \in [1,5], \\&x_4 \in [0,6], x_5 \in [1,5], x_6 \in [0,10] \end{aligned}$$
(P11)

The global optimum of P11 is at \(\mathbf{x}^{*} = [5,1,5,0,5,10]^\textsf {T}\) with \(f(\mathbf{x}^*) = -310\).

Problem 12

[21, Chap. 5.2.4].

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&f(\mathbf{x}) = -x_{4}(9 - 6x_{1} - 16x_{2} - 15x_{3}) \\&\qquad \qquad -x_{5}(15 - 6x_{1} - 16x_{2} - 15x_{3}) + x_{6} - 5x_{7} \\ \text {subject to}\quad&x_{3}x_{4} + x_{3}x_{5} \le 50 \\&x_4 + x_6 \le 100 \\&x_5 + x_7 \le 200 \\&x_{4}(3x_1 + x_2 + x_3 - 2.5) - 0.5x_6 \le 0 \\&x_{5}(3x_1 + x_2 + x_3 - 1.5) + 0.5x_7 \le 0 \\&x_1 + x_2 + x_3 = 1 \\&x_1 \in [0,1], x_2 \in [0,1], x_3 \in [0,1], \\&x_4 \in [0,100], x_5 \in [0,200], \\&x_6 \in [0,100], x_7 \in [0,200] \end{aligned}$$
(P12)

The global optimum of P12 is at \(\mathbf{x}^{*} = [0,0.5,0.5,0,100,0,100]^\textsf {T}\) with \(f(\mathbf{x}^*) = -450\).

Problem 13

[20, Chap. 11.3.1].

$$\begin{aligned} \underset{\mathbf{x}}{\text {minimize}} \quad&f(\mathbf{x}) = 0.7854 x_1 x_{2}^{2} \left( 3.3333 x_{3}^{2} + 14.9334 x_3 - 43.0934\right) \\& -1.508 x_1 \left( x_{6}^{2} + x_{7}^{2}\right) + 7.477\left( x_{6}^{3} + x_{7}^{3}\right) \\& + 0.7854\left( x_{4}x_{6}^{2} + x_{5}x_{7}^{2}\right) \\ \text {subject to}\quad&x_{1}x_{2}^{2}x_{3} \ge 27 \\&x_{1}x_{2}^{2}x_{3}^{2} \ge 397.5 \\&x_{2}x_{6}^{4}x_{3}x_{4}^{-3} \ge 1.93 \\&x_{2}x_{7}^{4}x_{3}x_{5}^{-3} \ge 1.93 \\&\left[ \left( 745x_{4}x_{2}^{-1}x_{3}^{-1}\right) ^2 + 16.911\times 10^6\right] ^{0.5}\Bigg /\left( 0.1x_{6}^{3}\right) \le 1100 \\&\left[ \left( 745x_{5}x_{2}^{-1}x_{3}^{-1}\right) ^2 + 157.51\times 10^6\right] ^{0.5}\Bigg /(0.1x_{7}^{3}) \le 850 \\&x_{2}x_{3} \le 40 \\&x_{1}/x_{2} \ge 5 \\&x_{1}/x_{2} \le 12 \\&1.5x_6 - x_4 \le -1.9 \\&1.1x_7 - x_5 \le -1.9 \\&x_1 \in [2.6,3.6], x_2 \in [0.7,0.8], x_3 \in [17,28], \\&x_4 \in [7.3,8.3], x_5 \in [7.3,8.3], x_6 \in [2.9,3.9] \\&x_7 \in [5,5.5] \end{aligned}$$
(P13)

The best known solution for P13 is the point \(\mathbf{x}^{*} = [3.5,0.7,17,7.3,7.71,3.35,5.287]^\textsf {T}\) with \(f(\mathbf{x}^*) = 2994.47\). The problem can be written as a polynomially constrained problem by multiplying to remove all fractional terms in the constraints. This is possible because all variables are positively bounded.

Appendix 2: Proofs

Proposition 1

(Relaxation of bilinear terms) Consider the bilinear term \(y = x_1 x_2\), for \(x_1 \in [x_{1}^{l}, x_{1}^{u}]\) and \(x_2 \in [x_{2}^{l}, x_{2}^{u}]\). Let f be a B-spline representing the bilinear term, i.e. \(f = y\). Then, the convex combination relaxation (24) of f is equivalent to McCormick’s linear relaxation of bilinear terms (see [4, 50]).

Proof

(Proposition 1) Let \(\mathbf{x}_{1,1} = [1, x_1]^\textsf {T}\) and \(\mathbf{x}_{2,1} = [1, x_2]^\textsf {T}\) be the first degree power bases of \(x_1\) and \(x_2\). The the bilinear term can be written as \(y = \varvec{\lambda }^\textsf {T}(\mathbf{x}_{1,1} \otimes \mathbf{x}_{2,1}) = \varvec{\lambda }^\textsf {T}[1, x_1, x_2, x_1 x_2]^\textsf {T}= x_1 x_2\), for \(\varvec{\lambda }^\textsf {T}= [0, 0, 0, 1]\). Using the procedure in Sect. 2.5 one obtains the B-spline form f of y, which has four control points

$$\begin{aligned} {\mathsf {P}} = \begin{bmatrix} x_1^l&\quad x_1^l&\quad x_1^u&\quad x_1^u \\ x_2^l&\quad x_2^u&\quad x_2^l&\quad x_2^u \\ x_1^l x_2^l&\quad x_1^l x_2^u&\quad x_1^u x_2^l&\quad x_1^u x_2^u \end{bmatrix}. \end{aligned}$$
(33)

The relaxation in (24) requires four variables \(\varvec{\lambda } = [\lambda _1, \lambda _2, \lambda _3, \lambda _4]^\textsf {T}\), and is given by the equations

$$\begin{aligned} \underbrace{\begin{bmatrix} x_1^l&\quad x_1^l&\quad x_1^u&\quad x_1^u \\ x_2^l&\quad x_2^u&\quad x_2^l&\quad x_2^u \\ x_1^l x_2^l&\quad x_1^l x_2^u&\quad x_1^u x_2^l&\quad x_1^u x_2^u \\ 1&\quad 1&\quad 1&\quad 1 \end{bmatrix}}_{{\mathsf {A}}} \varvec{\lambda } = \underbrace{\begin{bmatrix} x_1 \\ x_2 \\ y \\ 1 \end{bmatrix}}_{\mathbf{b}}, \varvec{\lambda } \ge \mathbf{0}. \end{aligned}$$
(34)

\({\mathsf {A}}\) is a square matrix of full rank as long as \(x_1^l < x_1^u\) and \(x_2^l < x_2^u\), and it is possible to solve \(\varvec{\lambda } = {\mathsf {A}}^{-1}\mathbf{b}\) analytically. This yields

$$\begin{aligned} \begin{aligned} \lambda _1&= \frac{1}{\gamma } \left( y - x_2^u x_1 - x_1^u x_2 + x_1^u x_2^u \right) , \\ \lambda _2&= \frac{1}{\gamma } \left( -y + x_2^l x_1 + x_1^u x_2 - x_1^u x_2^l \right) , \\ \lambda _3&= \frac{1}{\gamma } \left( -y + x_2^u x_1 + x_1^l x_2 - x_1^l x_2^u \right) , \\ \lambda _4&= \frac{1}{\gamma } \left( y - x_2^l x_1 - x_1^l x_2 + x_1^l x_2^l \right) , \end{aligned} \end{aligned}$$
(35)

where \(\gamma = (x_1^u - x_1^l)(x_2^u - x_2^l)\). Utilizing \(\varvec{\lambda } \ge \mathbf{0}\), and the fact that \(\gamma > 0\), one obtains

$$\begin{aligned} \begin{aligned} y&\ge x_2^u x_1 + x_1^u x_2 - x_1^u x_2^u, \\ y&\le x_2^l x_1 + x_1^u x_2 - x_1^u x_2^l, \\ y&\le x_2^u x_1 + x_1^l x_2 - x_1^l x_2^u, \\ y&\ge x_2^l x_1 + x_1^l x_2 - x_1^l x_2^l, \end{aligned} \end{aligned}$$
(36)

which are precisely the linear constraints of the McCormick relaxation of \(y = x_1 x_2\). \(\square \)

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Grimstad, B., Sandnes, A. Global optimization with spline constraints: a new branch-and-bound method based on B-splines. J Glob Optim 65, 401–439 (2016). https://doi.org/10.1007/s10898-015-0358-4

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