Approximated perspective relaxations: a project and lift approach

Abstract

The perspective reformulation (PR) of a Mixed-Integer NonLinear Program with semi-continuous variables is obtained by replacing each term in the (separable) objective function with its convex envelope. Solving the corresponding continuous relaxation requires appropriate techniques. Under some rather restrictive assumptions, the Projected PR ($\mathrm{P}^2\mathrm{R}$) can be defined where the integer variables are eliminated by projecting the solution set onto the space of the continuous variables only. This approach produces a simple piecewise-convex problem with the same structure as the original one; however, this prevents the use of general-purpose solvers, in that some variables are then only implicitly represented in the formulation. We show how to construct an Approximated Projected PR ($\mathrm{AP}^2\mathrm{R}$) whereby the projected formulation is “lifted” back to the original variable space, with each integer variable expressing one piece of the obtained piecewise-convex function. In some cases, this produces a reformulation of the original problem with exactly the same size and structure as the standard continuous relaxation, but providing substantially improved bounds. In the process we also substantially extend the approach beyond the original $\mathrm{P}^2\mathrm{R}$ development by relaxing the requirement that the objective function be quadratic and the left endpoint of the domain of the variables be non-negative. While the $\mathrm{AP}^2\mathrm{R}$ bound can be weaker than that of the PR, this approach can be applied in many more cases and allows direct use of off-the-shelf MINLP software; this is shown to be competitive with previously proposed approaches in some applications.

Appendix

In this appendix we show that $\mathrm{P}^2\mathrm{R}$ and $\mathrm{AP}^2\mathrm{R}$ can be extended to the case $p_{min} < 0$, albeit at the cost of slightly larger formulations. We first prove an analogous result to Proposition 1.

Proposition 2

If $p_{min} < 0$ then z(p) defined in (19) has the form

$$\begin{aligned} z(p) = \left\{ \begin{array}{l@{\quad }l} \, z_2(p) = f(p) + c &{} \text{ if } \quad p_{min} \le p \le p^-_{int} \\ z^-_1(p) = \big ( f( p^-_{int} ) / p^-_{int} + c / p^-_{int} \big ) p &{} \text{ if } \quad p^-_{int} \le p \le 0 \\ z^+_1(p) = \big ( f( p^+_{int} ) / p^+_{int} + c / p^+_{int} \big ) p &{} \text{ if } \quad 0 \le p \le p^+_{int} \\ \, z_2(p) = f(p) + c &{} \text{ if } \quad p^+_{int} \le p \le p_{max} \end{array}\right. \end{aligned}$$

(32)

where $p^-_{int} \in \{p_{min},1/g^-,0\}$ and $p^+_{int} \in \{0,1/g^+,p_{max}\}$.

Proof

In this case, the form (13) of the constraints in (8) is no longer valid; indeed, $u p_{min} \le p$ rather gives $u \ge p/p_{min}$, and therefore one obtains

$$\begin{aligned} \max \left\{ \, \frac{p}{p_{max}} ,\frac{p}{p_{min}} \, \right\} \le u \le 1. \end{aligned}$$

(33)

Yet, the result of the leftmost “$\max$” only depends on the sign of p; in particular

$$\begin{aligned} p\ge & {} 0 \Longrightarrow \max \{ \, p/p_{max} ,p/p_{min} \, \} = p/p_{max} \\ p\le & {} 0 \Longrightarrow \max \{ \, p/p_{max} ,p/p_{min} \, \} = p/p_{min} . \end{aligned}$$

Therefore, we can proceed by cases, mirroring the previous development with the necessary changes:

• a. If (9) has no solution, the global minimum in (8) is one of the bounds in (33), and there are two sub cases:

  • a.1. The derivative is always negative, and therefore $u^*(p) = 1\Longrightarrow$ (15) holds (i.e., $p_{min} = p_{max} = 0$).

  • a.2. The derivative is always positive, and therefore for $p < 0$, $u^*(p) = p / p_{min}\Longrightarrow$ (14) holds, for $p \ge 0$, $u^*(p) = p / p_{max}\Longrightarrow$ (17) holds.

  All in all, in this case

  $$\begin{aligned} z(p) = \left\{ \begin{array}{ll} \big ( f( p_{min} ) / p_{min} + c / p_{min} \big ) p &{} \text{ if } p < 0 \\ \big ( f( p_{max} ) / p_{max} + c / p_{max} \big ) p &{} \text{ if } p \ge 0 \end{array}\right. . \end{aligned}$$

  (34)

• b. If, instead, the only solution to (9) is (11), one has to separately consider $[p_{min}, 0]$ and $[0, p_{max}]$, since $u^*(p) = \tilde{u}(p)$ if

  $$\begin{aligned} p \in [p_{min}, 0]&\quad \quad \Longrightarrow \quad \quad&p/p_{min} \le \tilde{u}(p) = - p g^- \le 1 \\ p \in [0, p_{max}]&\quad \quad \Longrightarrow \quad \quad&p/p_{max} \le \tilde{u}(p) = \,\, p g^+ \le 1 \end{aligned}$$

  That is, exactly two of the following four cases hold:

  • b.1. $p \ge 0$ and $\tilde{u}(p) \le p / p_{max}\Longleftrightarrow p_{max} \le 1/g^+\Longrightarrow u^*(p) = p / p_{max}\Longrightarrow$ (17) holds.

  • b.2. $p \ge 0$ and $\tilde{u}(p) \ge p / p_{max}\Longleftrightarrow p_{max} \ge 1/g^+$; two further sub cases arise:

    • b.2.1. $(p_{max} \ge ) \, p \ge 1/g^+ \, (\ge 0)\Longrightarrow \tilde{u}(p) \ge 1\Longrightarrow u^*(p) = 1\Longrightarrow$ (15) holds.

    • b.2.2. $(0 \le ) \, p \le 1/g^+ \, (\le p_{max})\Longrightarrow \tilde{u}(p) \le 1\Longrightarrow u^*(p) = \tilde{u}(p)\Longrightarrow$ (18) holds.

    This again gives (19).

  • b.3. $p \le 0$ and $\tilde{u}(p) \le p / p_{min}\Longleftrightarrow (0 >) \, p_{min} \ge - 1/g^-\Longrightarrow u^*(p) = p / p_{min}\Longrightarrow$ (14).

  • b.4. $p \le 0$ and $\tilde{u}(p) \ge p / p_{min}\Longleftrightarrow p_{min} \le - 1/g^- \, (< 0)$; two further subcases arise:

    • b.4.1. $- 1/g^- \le p \le 0\Longleftrightarrow \tilde{u}(p) \le 1\Longrightarrow u^*(p) = \tilde{u}(p)\Longrightarrow$

      $$\begin{aligned} z(p) = \big ( {-}g^- f( - 1 / g^- ) - c g^- \big ) p \end{aligned}$$

      (35)

    • b.4.2. $p_{min} \le p \le -1/g^- \, (< 0)\Longleftrightarrow \tilde{u}(p) \ge 1\Longrightarrow u^*(p) = 1\Longrightarrow$ (15).

    All this gives

    $$\begin{aligned} z(p) = \left\{ \begin{array}{l@{\quad }l} f(p) + c &{} \text{ if } \quad p_{min} \le p \le - 1/g^- \\ \big ( - g^- f( - 1 / g^- ) - c g^- \big ) p &{} \text{ if } \quad - 1/g^- \le p \le 0 \end{array}\right. \end{aligned}$$

    (36)

  To summarize, z(p) is the convex function with at most 4 pieces

  $$\begin{aligned} z(p) = \left\{ \begin{array}{l@{\quad }l} f(p) + c &{} \text{ if } \quad p_{min} \le p \le - 1/g^- \\ \big ( - g^- f( - 1 / g^- ) - c g^- \big ) p &{} \text{ if } \quad - 1/g^- \le p \le 0 \\ \big ( g^+ f( 1 / g^+ ) + c g^+ \big ) p &{} \text{ if } \quad 0 \le p \le 1 / g^+ \\ f(p) + c &{} \text{ if } \quad 1 / g^+ \le p \le p_{max} \end{array}\right. \end{aligned}$$

  (37)

  Under condition b.1, the two rightmost pieces are substituted with the linear piece (17) $(f( p_{max} ) / p_{max} + c / p_{max}) p$ for $0 \le p \le p_{max}$ and/or, under condition b.3, the two leftmost pieces are substituted with the linear piece (14) $(f( p_{min} ) / p_{min} + c / p_{min}) p$ for $p_{min} \le p \le 0$, yielding a 3- or 2-piecewise convex function (piecewise-linear in the latter case as in (34)). $\square$

Example 7

We can extend the rational exponent case of §2.1. For instance, if $k = 4$ (even case), $h=3$, $a=3$, $c=1$, $p_{min} = -2$, and $p_{max} = 2$, one has

$$\begin{aligned} g^{\pm } = \left( \frac{1}{3} \frac{3}{1} \right) ^{\frac{3}{4}} = 1 \qquad \text{ and } \text{ then }\qquad z(p) = \left\{ \begin{array}{l@{\quad }l} 3p^{4/3} + 1 &{} \text{ if } \quad -2 \le p \le -1 \\ -4 p &{} \text{ if } \quad -1 \le p \le 0 \\ 4 p&{} \text{ if } \quad 0 \le p \le 1 \\ 3p^{4/3} + 1 &{} \text{ if } \quad 1 \le p \le 2 \end{array}\right. . \end{aligned}$$

We now prove that also (32) can be reformulated as a compact NLP, thus extending the result of Theorem 1 and the $\mathrm{AP}^2\mathrm{R}$ technique to the case $p_{min} < 0$.

Theorem 2

For z(p) defined in (32) and

$$\begin{aligned} \bar{z}(p) = \left\{ \begin{array}{ll} \min _{u^+ , u^- , q^+ , q^-} &{} h( u^+ , u^- , q^+ , q^- ) \\ &{} - p_{int}^+ u^+ \le q^+ \le ( p_{max} - p_{int}^+ ) u^+ \\ &{} ( p_{min} - p_{int}^- ) u^- \le q^- \le - p_{int}^- u^- \\ &{} p = p^+_{int} u^+ + q^+ + p^-_{int} u^- + q^- \\ &{} u^+ + u^- \le 1,\quad u^+ \in [0, 1],\quad u^- \in [0, 1] \end{array}\right. , \end{aligned}$$

(38)

where

$$\begin{aligned} h( u^+ , u^- , q^+ , q^- )= & {} u^+ z^+_1( p_{int}^+ ) + z_2( q^+ + p^+_{int} ) - z^+_1( p^+_{int} ) + u^- z^-_1( p_{int}^- ) \nonumber \\&+\, z_2( q^- + p^-_{int} ) - z^-_1( p^-_{int} ), \end{aligned}$$

we have $z(p) = \bar{z}(p)$ for all $p \in [p_{min}, p_{max}]$.

Proof

As in Theorem 1, the first step is to bring (32) in the form (21). Here $k = 4$, and using a slightly nonstandard numbering (to better highlight the fundamental symmetry of the function) we have $\alpha _{-2} = p_{min}$, $\alpha _{-1} = p^-_{int}$, $\alpha _0 = 0$, $\alpha _1 = p^+_{int}$, $\alpha _2 = p_{max}$, $z_{-2} = z_2$, $z_{-1} = z^-_1$, $z_1 = z^+_1$. Applying (21) to (32) gives

$$\begin{aligned} z(p) = \left\{ \begin{array}{ll} \min _{p_{-2},p_{-1},p_1,p_2} &{} z_2( p_{-2} + p_{min} ) + z^-_1( p_{-1} + p^-_{int} ) - z^-_1( p^-_{int} ) +\\ &{} z^+_1( p_1 ) + z_2( p_2 + p^+_{int} ) - z( p^+_{int} )\\ &{} 0 \le p_{-2} \le p^-_{int} - p_{min},\quad 0 \le p_{-1} \le - p^-_{int} \\ &{} 0 \le p_1 \le p^+_{int},\quad 0 \le p_2 \le p_{max} - p^+_{int} \\ &{} p = p_{min} + p_{-2} + p_{-1} + p_1 + p_2 \end{array}\right. \end{aligned}$$

(39)

(remember that $z^+_1( 0 ) = 0$), and we want to prove that z(p) given in (39) is equivalent to $\bar{z}(p)$ given in (38) for all $p \in [p_{min}, p_{max}]$. To do that, we start by identifying

$$\begin{aligned} p_{-2} + p_{min} = q^- + p^-_{int} ,\quad p_{-1} = p^-_{int} ( u^- - 1 ) ,\quad p_1 = p^+_{int} u^+ ,\quad p_2 = q^+ \end{aligned}$$

to recover the objective function and most of the constraints in (38), as some simple but somewhat tedious algebra shows. Then, the general result about (21) can be applied to the optimal solution $(p^*_{-2}, p^*_{-1}, p^*_1, p^*_2)$ (or, equivalently, $(\hat{q}^-,\hat{u}^-,\hat{u}^+,\hat{q}^+)$) of (23) for any fixed p, yielding

p	$p^*_{-2}$	$p^*_{-1}$	$p^*_1$	$p^*_2$	$\hat{q}^-$	$\hat{u}^-$	$\hat{u}^+$	$\hat{q}^+$
${[}p_{min}, p^-_{int}]$	$\ge 0$	0	0	0	$\le 0$	1	0	0
${[}p^-_{int}, 0]$	$p^-_{int} - p_{min}$	$\ge 0$	0	0	0	$\in [0, 1]$	0	0
${[}0, p^+_{int}]$	$p^-_{int} - p_{min}$	$- p^-_{int}$	$\ge 0$	0	0	0	$\in [0, 1]$	0
${[}p^+_{int}, p_{max}]$	$p^-_{int} - p_{min}$	$- p^-_{int}$	$p^+_{int}$	$\ge 0$	0	0	1	$\ge 0$

This shows that the constraints

$$\begin{aligned}&- p_{int}^+ u^+ \le q^+ \le ( p_{max} - p_{int}^+ ) u^+ ,\quad ( p_{min} - p_{int}^- ) u^- \le q^- \le - p_{int}^- u^- ,\\&u^- + u^+ \le 1 \end{aligned}$$

are satisfied by $(\hat{q}^-,\hat{u}^-,\hat{u}^+,\hat{q}^+)$ for each value of p. The issue is that $- p_{int}^+ u^+ \le q^+$ is weaker than $0 \le q^+$ and $q^- \le - p_{int}^- u^-$ is weaker than $q^- \le 0$ ($p_{int}^- \le 0 \le p_{int}^+$). However, reasoning as in Theorem 1 one easily shows that relaxing the constraints in this way does not change the optimal solution to (39). $\square$

Once again, the choice of (38) is motivated by the fact that, imposing integrality constraints $u^+ \in \{0, 1\}$, $u^- \in \{0, 1\}$ and with the identification $u = u^+ + u^-$, one obtains a reformulation of the original MINLP whose continuous relaxation is equivalent to PR if $\mathcal {O}$ does not contain constraints linking the u variables, and weaker otherwise. This formulation has twice the number of continuous and binary variables than the ordinary formulation (counting the semi-continuous variables only), but possibly provides (much) stronger bounds.