Efficient solution of ordinary differential equations with a parametric lexicographic linear program embedded

Abstract

This work analyzes the initial value problem in ordinary differential equations with a parametric lexicographic linear program (LP) embedded. The LP is said to be embedded since the dynamics depend on the solution of the LP, which is in turn parameterized by the dynamic states. This problem formulation finds application in dynamic flux balance analysis, which serves as a modeling framework for industrial fermentation reactions. It is shown that the problem formulation can be intractable numerically, which arises from the fact that the LP induces an effective domain that may not be open. A numerical method is developed which reformulates the system so that it is defined on an open set. The result is a system of semi-explicit index-one differential algebraic equations, which can be solved with efficient and accurate methods. It is shown that this method addresses many of the issues stemming from the original problem’s intractability. The application of the method to examples of industrial fermentation processes demonstrates its effectiveness and efficiency.

Appendices

Appendix A: Supporting results for lexicographic optimization

This section presents some definitions, results, and discussion to support the proof of Theorem 2, which deals with finding a specific optimal basis for the lexicographic LP.

Definition 1

Equivalence.

• 1. Two sets $S_1 \subset \mathbb {R}^{n_1}$ and $S_2 \subset \mathbb {R}^{n_2}$ with $n_1 \le n_2$ are equivalent if $n_1 < n_2$ and $S_2 = S_1 \times \{ {\mathbf {0}} \}$, or $n_1 = n_2$ and $S_2 = S_1$.

• 2. Two linear programs are equivalent if their solution sets are equivalent.

Intuitive results regarding equivalence follow.

Lemma 1

• 1. Assume $n_1 \le n_2 \le n_3.$ Let $F_i \subset \mathbb {R}^{n_i}$ for $i \in \{1, 2, 3\}.$ If sets $F_1$ and $F_2$ are equivalent and $F_2$ and $F_3$ are equivalent,  then $F_1$ and $F_3$ are equivalent.

• 2. If two sets $F_1 \in \mathbb {R}^{n_1}$ and $F_2 \in \mathbb {R}^{n_2}$ with $n_1 \le n_2$ are equivalent,  then the linear programs

  $$\begin{aligned} \min \{ {\mathbf {c}} ^{\mathrm{T}}{\mathbf {v}}: {\mathbf {v}}\in F_1 \} \quad \text {and} \quad \min \{ \widehat{{\mathbf {c}}} ^{\mathrm{T}}{\mathbf {v}}: {\mathbf {v}}\in F_2 \} \end{aligned}$$

  are equivalent,  where $\widehat{{\mathbf {c}}} = ({\mathbf {c}}, \widetilde{{\mathbf {c}}})$ for any ${\mathbf {c}} \in \mathbb {R}^{n_1}$ and $\widetilde{{\mathbf {c}}} \in \mathbb {R}^{n_2 - n_1}.$

For the next two results refer to the lexicographic LP

$$\begin{aligned} q({\mathbf {d}})= & {} \inf \{ {\mathbf {c}} ^{\mathrm{T}}{\mathbf {v}}: {\mathbf {M}} {\mathbf {v}}= {\mathbf {d}}, {\mathbf {v}}\ge {\mathbf {0}}\}, \end{aligned}$$

(27)

$$\begin{aligned} {\widehat{q}}({\mathbf {d}})= & {} \inf \{ \widehat{{\mathbf {c}}} ^{\mathrm{T}}{\mathbf {v}}: {\mathbf {M}} {\mathbf {v}}= {\mathbf {d}}, {\mathbf {c}} ^{\mathrm{T}}{\mathbf {v}}= {q}({\mathbf {d}}), {\mathbf {v}}\ge {\mathbf {0}}\}. \end{aligned}$$

(28)

The next result establishes the form of the simplex tableau for the two-level lexicographic LP. Strictly speaking, tableau (29) below is missing the “zeroth” row of reduced costs for the second-level LP (28); for simplicity it is omitted.

Lemma 2

Consider the lexicographic LP (27)–(28). Let B be a dual feasible basis for the first-level LP (27),  and assume that the jth reduced cost is positive $(c_{j} - {\mathbf {c}}_{B}^{\mathrm{T}}{\mathbf {M}}_B^{-1} {\mathbf {m}}_j > 0).$ For all ${\mathbf {d}}$ such that B is optimal for the first-level LP,  the simplex tableau for the second-level LP (28) resulting from the basis $\widehat{B} = \{ j \} \cup B$ is

$$\begin{aligned} \left[ \begin{array}{cc} c_{j} &{}\quad {\mathbf {c}}_{B} ^{\mathrm{T}}\\ {\mathbf {m}}_j &{}\quad {\mathbf {M}}_B \end{array}\right] ^{-1} \left[ \begin{array}{cc} q({\mathbf {d}}) &{}\quad {\mathbf {c}}^{\mathrm{T}}\\ {\mathbf {d}} &{}\quad {\mathbf {M}} \end{array}\right] =\left[ \begin{array}{cc} 0 &{} \quad \frac{ {\mathbf {c}}^{\mathrm{T}}- {\mathbf {c}}_{B}^{\mathrm{T}}{\mathbf {M}}_B^{-1} {\mathbf {M}} }{ c_{j} - {\mathbf {c}}_{B}^{\mathrm{T}}{\mathbf {M}}_B^{-1}{\mathbf {m}}_j } \\ {\mathbf {M}}_B^{-1} {\mathbf {d}} &{}\quad {\mathbf {M}}_B^{-1} \left( {\mathbf {M}} - {\mathbf {m}}_j \frac{{\mathbf {c}}^{\mathrm{T}}- {\mathbf {c}}_{B}^{\mathrm{T}}{\mathbf {M}}_B^{-1} {\mathbf {M}} }{c_{j} - {\mathbf {c}}_{B}^{\mathrm{T}}{\mathbf {M}}_B^{-1}{\mathbf {m}}_j} \right) \end{array}\right] .\quad \end{aligned}$$

(29)

Proof

The proof proceeds by using Schur complements to form the inverse of $\left[ \begin{array}{cc} c_{j} &{}\quad {\mathbf {c}}_{B} ^{\mathrm{T}}\\ {\mathbf {m}}_j &{}\quad {\mathbf {M}}_B \end{array}\right] ,$ performing the matrix multiplication, and simplifying, noting that ${\mathbf {c}}_B ^{\mathrm{T}}{\mathbf {M}}_B^{-1} {\mathbf {d}} = q({\mathbf {d}})$ for all ${\mathbf {d}}$ such that ${\mathbf {M}}_B^{-1} {\mathbf {d}} \ge {\mathbf {0}}$. $\square$

The concept of a “null variable” is important to the proof of Theorem 2, which is defined as a variable which is zero everywhere in the feasible set of an LP. It is clear that removing a null variable and the corresponding parameters (components of the cost vector and columns of the constraint matrix) yields an equivalent LP. The next result states a way to identify null variables in the second-level LP (28).

Lemma 3

Consider the lexicographic LP (27)–(28). Let B be a dual feasible basis for the first-level LP (27),  and assume that the jth reduced cost is positive $(c_{j} - {\mathbf {c}}_{B}^{\mathrm{T}}{\mathbf {M}}_B^{-1} {\mathbf {m}}_j > 0).$ For all ${\mathbf {d}}$ such that B is optimal for the first-level LP,  $v_j = 0$ for all ${\mathbf {v}}$ feasible in the second-level LP (28).

Proof

The result follows from the “null variable theorem” in §4.7 of [24]; this states that $v_j$ is a null variable for a general standard-form LP (27) if and only if there exists a nonzero ${\mathbf {p}}$ such that ${\mathbf {p}} ^{\mathrm{T}}{\mathbf {d}} = 0$, ${\mathbf {p}} ^{\mathrm{T}}{\mathbf {M}} \ge {\mathbf {0}} ^{\mathrm{T}}$, and the jth component of ${\mathbf {p}} ^{\mathrm{T}}{\mathbf {M}}$ is strictly greater than zero. Applying this result to the second-level LP (28), the result follows from inspection of the tableau (29); the first row of $\left[ \begin{array}{cc} c_{j} &{}\quad {\mathbf {c}}_{B} ^{\mathrm{T}}\\ {\mathbf {m}}_j &{}\quad {\mathbf {M}}_B \end{array}\right] ^{-1}$ serves as the appropriate ${\mathbf {p}}$. $\square$

Finally, some aspects of the primal simplex algorithm are noted. If one has a optimal basis already, but a different optimal basis is sought, a pivot could be forced in the sense that, while the ith reduced cost is zero, the ith column is chosen as the pivot column. If the ith reduced cost is zero, but the pivot operation is carried out in the standard way, a new primal feasible basis is obtained for which the reduced costs are the same as the old basis, and so the new basis is also optimal. This is because the reduced costs are updated in a pivot operation by adding a multiple of the pivot row to the reduced costs (the zeroth row of the tableau) so that the ith entry of the zeroth row is zero. But, if the ith reduced cost is already zero, no changes to the zeroth row are made, and so the reduced costs retain the same values.

Appendix B: Domain issues and time-stepping

This section presents an example demonstrating that time-stepping methods such as those mentioned in Sect. 3.2 still suffer from domain issues. More generally, this example shows that implicit integration methods also suffer from domain issues.

Consider an example similar to the one in Sect. 3.1:

$$\begin{aligned} {\mathbf {x}}(0)={\mathbf {0}}, \quad \dot{{\mathbf {x}}}(t) = {\mathbf {f}} \left( {\mathbf {x}}(t),q({\mathbf {x}}(t)) \right) = \left[ \begin{array}{c} 1 \\ {x_2(t)} {q({\mathbf {x}}(t))} - {(x_2(t))}^2 + {2}{x_1(t)} \end{array}\right] , \end{aligned}$$

(30)

where $q(\mathbf {z})= \min \{ v : z_2 \le v \le z_1^2\}.$

The embedded LP is feasible only if ${\mathbf {x}}\in K = \{ \mathbf {z}: z_2 \le z_1^2 \}$, a closed, nonconvex set. Note that ${\mathbf {x}}(t) = (t, t^2)$ is a solution. Letting ${\mathbf {b}}(\mathbf {z}) =(z_1^2,-z_2)$ and rewriting q in terms of the embedded LP’s dual, one has

$$\begin{aligned} q(\mathbf {z}) = \max \{ {\mathbf {b}}(\mathbf {z}) ^{\mathrm{T}}\mathbf {w}: \mathbf {w}\le 0 , w_1 - w_2 = 1 \}. \end{aligned}$$

Letting W denote the feasible set of the above (dual) LP, this is equivalent to finding $\mathbf {w}^* \in W$ such that $(\mathbf {w}- \mathbf {w}^*)^{\mathrm{T}}(-{\mathbf {b}}(\mathbf {z}) ) \ge 0$, $\forall \mathbf {w}\in W$. This is a parametric variational inequality and is denoted $\mathrm {VI}(W, -{\mathbf {b}}(\mathbf {z}) )$. This requires the dynamics to be rewritten as

$$\begin{aligned}&{\mathbf {f}} \left( {\mathbf {x}}(t),q({\mathbf {x}}(t)) \right) \\&\quad =\widehat{{\mathbf {f}}} ({\mathbf {x}}(t),\mathbf {u}(t))= \left[ \begin{array}{c} 1 \\ {x_2(t)} \left( {(x_1(t))^2}{u_1(t)} - {x_2(t)}{u_2(t)} \right) - {(x_2(t))}^2 + {2}{x_1(t)} \end{array} \right] , \end{aligned}$$

where $\mathbf {u}(t)$ is the solution of $\mathrm {VI}(W, -{\mathbf {b}}({\mathbf {x}}(t)))$. Given $h > 0$, an implicit time-stepping scheme takes the form

$$\begin{aligned}&\widetilde{{\mathbf {x}}}(t + h)=\widetilde{{\mathbf {x}}}(t) + h \widehat{{\mathbf {f}}} \left( \widetilde{{\mathbf {x}}}(t+h), \widetilde{\mathbf {u}}(t+h) \right) , \nonumber \\&\quad \widetilde{\mathbf {u}}(t+h)\text { solves }\mathrm {VI}(W, -{\mathbf {b}}(\widetilde{{\mathbf {x}}}(t+h))). \end{aligned}$$

(31)

Typically this implicit system is solved as the equivalent variational inequality $\mathrm {VI}(\mathbb {R}^2 \times W, \mathbf {g}^t )$ where

$$\begin{aligned} \mathbf {g}^t(\mathbf {z},{\mathbf {v}}) = \left[ \begin{array}{c} z_1 - \widetilde{x}_1(t) - h \\ z_2 - \widetilde{x}_2(t) - h( {z_1^2} {z_2} {v_1} - {z_2^2} {v_2} - {z_2^2} + 2{z_1} ) \\ -z_1^2 \\ z_2 \end{array} \right] \end{aligned}$$

(see for instance [30]). However, again letting $\widetilde{{\mathbf {x}}}(0) = {\mathbf {x}}(0) = {\mathbf {0}}$, the initial variational inequality $\mathrm {VI}(\mathbb {R}^2 \times W, \mathbf {g}^0)$ does not have a solution for any choice of h.

To see this, assume, for a contradiction, that a solution exists. Then there is a $(\mathbf {z}^*,{\mathbf {v}}^*) \in \mathbb {R}^2 \times W$ such that

$$\begin{aligned}&(z_1 - z_1^*) (z_1^* - h) +(z_2 - z_2^*) \left( z_2^* - h ( (z_1^*)^2 {z_2^*} {v_1^*} - (z_2^*)^2 {v_2^*} - (z_2^*)^2 + 2{z_1^*} ) \right) \nonumber \\&\quad +(v_1 - v_1^*) ( -(z_1^*)^2 ) + (v_2 - v_2^*) (z_2^*) \ge 0, \end{aligned}$$

(32)

for all $(\mathbf {z},{\mathbf {v}}) \in \mathbb {R}^2 \times W$. First note that $z_1^* = h$, otherwise one could always find a $z_1 \in \mathbb {R}$ such that the inequality (32) did not hold. Similarly, one must have

$$\begin{aligned} z_2^* = {h} \left( (z_1^*)^2{z_2^*}{v_1^*} - (z_2^*)^2{v_2^*} - (z_2^*)^2 + {2}{z_1^*} \right) . \end{aligned}$$

(33)

Using this in inequality (32), one obtains

$$\begin{aligned} (v_1 - v_1^*) ( -h^2 ) + (v_2 - v_2^*) (z_2^*) \ge 0, \end{aligned}$$

for all $(\mathbf {z},{\mathbf {v}}) \in \mathbb {R}^2 \times W$. For any ${\mathbf {v}}\in W$, one can write $v_2 = v_1 - 1$. Then one gets

$$\begin{aligned} (v_1 - v_1^*) ( -h^2 ) + (v_1 - 1 - (v_1^* - 1) ) (z_2^*) \ge 0, \end{aligned}$$

which yields

$$\begin{aligned} (v_1 - v_1^*) ( z_2^* - h^2 ) \ge 0, \end{aligned}$$

for all $v_1 \le 0$, where $v_1^* \le 0$ and $z_2^* \in \mathbb {R}$ satisfy

$$\begin{aligned} {h}{v_1^*}(z_2^*)^2 + (1 - {h^3}{v_1^*}){z_2^*} - {2}{h^2} = 0 \end{aligned}$$

(34)

(which is obtained from Eq. (33) via the substitutions $z_1^* = h$ and $v_2^* = v_1^* - 1$).

One can now analyze three cases:

• 1. $z_2^* > h^2$: However, if this was the case, then whatever the value of $v_1^*$, one could always find a $v_1' < v_1^*$ which then implies $(v_1' - v_1^*) (z_2^* - h^2) < 0$, which is a contradiction.

• 2. $z_2^* < h^2$: However, if this was the case, one must have $v_1^* = 0$, otherwise there exists a $v_1'$ such that $v_1^* < v_1' \le 0$ which then implies that $(v_1' - v_1^*) (z_2^* - h^2) < 0$. Thus, assuming $v_1^* = 0$, use Eq. (34) to check the value of $z_2^*$. However, that yields $z_2^* = 2h^2$, which contradicts $z_2^* < h^2$.

• 3. $z_2^* = h^2$: However, if this was the case, one can use Eq. (34) to check the consistency of values. This yields

  $$\begin{aligned} {h^5} {v_1^*} + h^2 - {h^5} {v_1^*} - {2} {h^2} = 0 \implies -h^2 = 0, \end{aligned}$$

  which contradicts $h > 0$.

Thus, it follows that there does not exist a point $(\mathbf {z}^*,{\mathbf {v}}^*) \in \mathbb {R}^2 \times W$ which solves $\mathrm {VI}(\mathbb {R}^2 \times W,\mathbf {g}^0)$.

Note that the implicit time-stepping scheme (31) is equivalent to the direct method applying an implicit Euler step to the original system (30). Thus, the failure of the system (31) to have a solution indicates that the direct method, even with an implicit integration routine, also fails.