A polyhedral study of production ramping

Abstract

We give strong formulations of ramping constraints—used to model the maximum change in production level for a generator or machine from one time period to the next—and production limits. For the two-period case, we give a complete description of the convex hull of the feasible solutions. The two-period inequalities can be readily used to strengthen ramping formulations without the need for separation. For the general case, we define exponential classes of multi-period variable upper bound and multi-period ramping inequalities, and give conditions under which these inequalities define facets of ramping polyhedra. Finally, we present exact polynomial separation algorithms for the inequalities and report computational experiments on using them in a branch-and-cut algorithm to solve unit commitment problems in power generation.

Appendices

Appendix 1: Proof of convex hull of two-period ramping polytope without start variables

Corollary 1

For $\bar{u} \le \ell + \delta$, $conv(\mathcal {UNS}_t^2) = \{(p,x) \in \mathbb {R}^{2n}: (7){-}(15)\}$, and for $\bar{u} > \ell + \delta$, $conv(\mathcal {UNS}_t^2) = \{(p,x) \in \mathbb {R}^{2n}: (7){-}(13), (16), (17)\}$.

Proof

We use Fourier–Motzkin elimination (see [32] and [22]) of variable $s_{t+1}$ from the convex hull of the feasible solutions to the formulation with start-up variables given by inequalities (5a)–(5i). Inequalities (5d), (5e) and (5h) continue to be facets [given by inequalities (7)–(9)] because these inequalities do not include variable $s_{t+1}$ in their description. Similarly, if $\bar{u}=\ell +\delta$, then inequality (5f) is also a facet, because the coefficient of $s_{t+1}$ is equal to 0. In this case, inequality (5f) is equivalent to inequalities (14) and (15). If $u=\bar{u}$, then inequality (5g) is also a facet, because the coefficient of $s_{t+1}$ is equal to 0. In this case, inequality (5g) reduces to (12).

We need to consider all possible cross products of inequalities (5b), (5c), and (5f) (if $\bar{u} > \ell +\delta$) that provide a lower bound on $s_{t+1}$ with inequalities (5a), (5g) (if $u>\bar{u}$), (5i), and (5f) (if $\bar{u} < \ell +\delta$) that provide an upper bound on $s_{t+1}$.

We first consider the cross product of lower-bounding inequalities (5b) and (5c) with upper-bounding inequalities defined by (5a), (5g) (if $u >\bar{u}$) and (5i).

• The pair of inequalities (5b) and (5a) gives $x_{t} \ge 0$, which is dominated by inequalities (5e) and (5h).

• The pair of inequalities (5b) and (5g) gives inequality (10) (if $u >\bar{u}$).

• The pair of inequalities (5b) and (5i) gives inequality (11).

• The pair of inequalities (5c) and (5a) gives $x_{t+1} \ge 0$, which is dominated by inequalities (5d) and (12).

• The pair of inequalities (5c) and (5g) gives inequality (12) (if $u >\bar{u}$).

• The pair of inequalities (5c) and (5i) gives inequality (13).

Note that depending on whether the coefficient of $s_{t+1}$ in inequality (5f) is positive or negative, we get a lower-bounding or upper-bounding inequality for $s_{t+1}$, respectively. Therefore, we consider the following two cases:

Case 1 :

If $\bar{u} < \ell +\delta$, then inequality (5f) is an upper-bounding inequality given by

$$\begin{aligned} s_{t+1} \le \frac{(\ell +\delta )x_{t+1} - \ell x_{t} - p_{t+1} + p_{t}}{(\ell + \delta - \bar{u})}. \end{aligned}$$

(27)

• The pair of inequalities (5b) and (27) gives inequality (14).

• The pair of inequalities (5c) and (27) gives inequality (15).

Case 2 :

If $\bar{u} > \ell +\delta$, then inequality (5f) is a lower-bounding inequality given by

$$\begin{aligned} s_{t+1} \ge \frac{-(\ell +\delta )x_{t+1} + \ell x_{t} + p_{t+1} - p_{t}}{(\bar{u}- \ell - \delta )}. \end{aligned}$$

(28)

• The pair of inequalities (5a) and (28) gives

  $$\begin{aligned} p_{t+1} - p_{t} \le \bar{u} x_{t+1} - \ell x_{t}, \end{aligned}$$

  (29)

  which is dominated by inequalities (17) and (8). To see this, note that multiplying inequality (8) by $-(\bar{u}-\ell -\delta )$ and adding to inequality (17) gives inequality (29) multiplied by $(u-\ell -\delta )$.

• The pair of inequalities (5g) (when $u >\bar{u}$) and (28) gives

  $$\begin{aligned} \frac{u x_{t+1} - p_{t+1}}{(u - \bar{u})} \ge \frac{-(\ell +\delta )x_{t+1} + \ell x_{t} + p_{t+1} - p_{t}}{(\bar{u}- \ell - \delta )}. \end{aligned}$$

  Rearranging terms we get inequality (17).

• The pair of inequalities (5i) and (28) gives inequality (16).

$\square$

Appendix 2: Facet-definition proof of type-I multi-period ramp-up inequality

Proposition 10

Type-I multi-period ramp-up inequality (20) defines a facet of $conv(\mathcal {U})$ if and only if $\ell + j\delta < u$.

Proof

Necessity: For contradiction assume that $\ell + j\delta \ge u$. From validity of inequality (20) we have $\ell + j\delta \le u$, so $\ell + j\delta = u$. Then inequality (20) can be written as $p_{t+j}-p_t\le (\ell +j\delta )x_{t+j} - \ell x_{t} = u x_{t+j} - \ell x_{t}$, and it is dominated by inequalities (1a) and (1b).

Sufficiency: We use the technique in Theorem 3.6 of §$I.4.3$ in Nemhauser and Wolsey [22]. We show that inequality (20) is the only inequality that is satisfied at equality by all points $(p,x,s) \in \mathcal {U}$ that are tight at (20), i.e., we show that if all points of $\mathcal {U}$ at which inequality (20) is tight satisfy

$$\begin{aligned} \sum _{k=1}^{n}\alpha _{k}p_{k} + \sum _{k=1}^{n}\beta _{k}x_{k} + \sum _{k=2}^{n}\gamma _{k}s_{k} = \alpha _{0}, \end{aligned}$$

(30)

then

• 1. $\alpha _{0}=0$,

• 2. $\alpha _{k} = 0$, $k \in [1,n] {\setminus } \{t,t+j\}$,

• 3. $\alpha _{t} = -\bar{\alpha }$, $\alpha _{t+j} = \bar{\alpha }$,

• 4. $\beta _{k} = 0$, $k \in [1,n] {\setminus } \{t,t+j\}$,

• 5. $\beta _{t} = \bar{\alpha }\ell$,

• 6. $\beta _{t+j} = -\bar{\alpha }(\ell + j\delta )$,

• 7. $\gamma _{k}=0$, $k \in [2,t] \cup [t+j+1,n]$,

• 8. $\gamma _{t+i'} = -\bar{\alpha }\min \{(\bar{u}-\ell -i'\delta ), (u - \ell - j \delta )\}$, $i' \in [1,j]$.

In order to establish the values of the coefficients $\alpha _{k}$, $\beta _{k}$, $\gamma _{k}$ and $\alpha _{0}$, we construct a feasible solution to $\mathcal {U}$ on the face defined by (20). Then a small change in the solution is made to obtain another feasible solution which is on the face defined by inequality (20). Comparing the resulting expressions, the possible values of a set of coefficients are obtained. Also note that from the validity assumption and (A2), $\bar{u}\ge \ell +\delta >\ell$. We start by describing several points feasible to $\mathcal {U}$ that will be used throughout the facet proof. We assume that $k\ge 2$ if we set the value of $s_{k}$. In the following feasible solutions [except for the zero vector (S1)] if the value of a variable is not given, then its value is equal to zero. Let $k_1, k_2 \in [2,n]$ be two periods such that $k_1 < k_2$. Let $\epsilon$ be a very small number greater than zero.

Note that points (S3), (S6), (S7), (S8), and (S9) are feasible because $\bar{u} > \ell$ and $\ell + j\delta < u$.

Next we show the values of the coefficients $\alpha _{k}$, $\beta _{k}$, $k \in [1,n]$, $\gamma _{k}$, $k \in [2,n]$ and $\alpha _{0}$.

• 1. $\alpha _{0}=0$.

  Consider solution (S1). Clearly, this point satisfies inequality (20) at equality because both the left- and the right-hand sides of the inequality are equal to zero. Hence, $\alpha _{0}=0$.

• 2. $\alpha _{k} = 0$, $k \in [1,n] {\setminus } \{t,t+j\}$.

  Consider the following two cases:

  • (a) $k \in [1,t-1] \cup [t+j+1,n]$.

    Consider solution (S2) with $k_1=k$. Clearly, this point satisfies inequality (20) at equality because both the left- and the right-hand sides of the inequality are equal to zero. Now, consider solution (S3) with $k_1=k$. This point also satisfies inequality (20) at equality and is a valid point because $\bar{u} > \ell$ by assumption. Then evaluating (30) at both solutions we get $\alpha _{k} \ell =\alpha _{k} (\ell + \epsilon )$. Hence, $\alpha _{k}=0$.

  • (b) $k \in [t+1,t+j-1]$.

    Consider solution (S5) with $k_1=t$ and $k_2=k$. This point satisfies inequality (20) at equality because both the left- and the right-hand sides of the inequality are equal to $-\ell$. Now, consider solution (S6) with $k_1=t$ and $k_2=k$ (this point also satisfies inequality (20) at equality). Then evaluating (30) at both solutions we get $\alpha _{k} \ell =\alpha _{k} (\ell + \epsilon )$. Hence, $\alpha _{k}=0$.

• 3. $\alpha _{t} = -\bar{\alpha }$, $\alpha _{t+j} = \bar{\alpha }$.

  Consider solution (S7). This point satisfies inequality (20) at equality because both the left- and the right-hand sides of the inequality are equal to $j\delta$. Now, consider solution (S8). This point also satisfies inequality (20) at equality. Because we showed that $\alpha _{k}=0$, $k \in [1,n] {\setminus } \{t,t+j\}$ in part 2, evaluating (30) at both solutions we get $\alpha _{t} \epsilon = -\alpha _{t+j}\epsilon$. Let $\bar{\alpha }:=-\alpha _{t}=\alpha _{t+j}$.

• 4. $\beta _{k} = 0$, $k \in [1,n] {\setminus } \{t,t+j\}$.

  Consider the following two cases:

  • (a) $k > t+j$.

    Consider a solution to $\mathcal {U}$ with $x_{r}=1$, $r \in [t,k]$, $s_{t}=1$, $p_{t+i} = \ell + i\delta$, $i \in [0,j]$, $p_{r} = \ell +j\delta$, $r \in [t+j+1,k]$ and all other variables are equal to zero. This point satisfies inequality (20) at equality because both the left- and the right-hand sides of the inequality are equal to $j\delta$. Now, consider the same solution except we set $x_{k}=0=p_{k}$ (this solution is on the face defined by inequality (20)). Evaluating (30) at both solutions we get $\alpha _{k}(\ell +j\delta ) +\beta _{k} = 0$. Because we showed that $\alpha _{k}=0$ in part 2 we get $\beta _{k}=0$.

  • (b) $t < k < t+j$.

    Consider solution (S5) with $k_1=t$ and $k_2=k$. This point satisfies inequality (20) at equality because both the left- and the right-hand sides of the inequality are equal to $-\ell$. Now, consider solution (S5) with $k_1=t$ and $k_2=k-1$ if $t<k-1$, and solution (S2) with $k_1=t$ if $t=k-1$. Both of the points satisfy inequality (20) at equality. Note that if $t=k-1$ we use solution (S2) because both $k_1=t$ and $k_2=k-1=t$ and we define $k_1 < k_2$ in solution (S5). Evaluating (30) at the described solutions we get $\alpha _{k}\ell +\beta _{k} = 0$. Because we showed that $\alpha _{k}=0$ in part 2 we get $\beta _{k}=0$.

  • (c) $k \le t-1$ for $t \ge 2$.

    Consider the following two cases:

    • i. $k=1$.

      Consider solution (S4). This point is on the face defined by inequality (20) because both the left- and the right-hand sides of the inequality are equal to zero. Evaluating (30) at this solution we get $\alpha _{k}\ell +\beta _{k}=\alpha _{0}=0$ and since $\alpha _{k}=0$ (from part 2) we get $\beta _{k}=0$.

    • ii. $k\ge 2$.

      Consider solution (S10) with $k_1=k$. This point satisfies inequality (20) at equality because both the left- and the right-hand sides of the inequality are equal to zero. Now, consider solution (S2) with $k_1=k-1$. Then evaluating (30) at both solutions we get $\alpha _{k}\ell +\beta _{k}=0$, and because $\alpha _{k}=0$ (from part 2) we get $\beta _{k}=0$.

• 5. $\beta _{t} = \bar{\alpha }\ell$.

  If $t=1$, then we use solution (S4). Because $\alpha _{t} =\alpha _{1} =-\bar{\alpha }$ (from part 3) evaluating this solution at equality (30) we get $\alpha _{1}\ell + \beta _{1}=0$ so $\beta _{1} = \bar{\alpha }\ell$. For $t\ge 2$ consider solution (S5) with $k_1=1$ and $k_2=t$. This point satisfies inequality (20) at equality because both the left- and the right-hand sides of the inequality are equal to $-\ell$. Now consider solution (S5) with $k_1=1$ and $k_2=t-1$. This point is on the face defined by inequality (20) because both the left- and the right-hand sides of the inequality are equal to zero. Then evaluating (30) at both solutions we obtain $\alpha _{t}\ell +\beta _{t}=0$. Because $\alpha _{t}=-\bar{\alpha }$ (from part 3) we get $\beta _{t}=\bar{\alpha }\ell$.

• 6. $\beta _{t+j} = -\bar{\alpha }(\ell + j\delta )$.

  Consider solution (S9). This point satisfies inequality (20) at equality because both the left- and the right-hand sides of the inequality are equal to $j\delta$. Consider the same solution except now we set $x_{t+j}=p_{t+j}=0$. This point is on the face defined by inequality (20) because both the left- and the right-hand sides of the inequality are equal to $-\ell$. Then evaluating (30) at both solutions we obtain $\alpha _{t+j}(\ell + j\delta ) +\beta _{t+j}=0$. Because $\alpha _{t+j}=\bar{\alpha }$ (from part 3) we get $\beta _{t+j}=-\bar{\alpha }(\ell + j\delta )$.

• 7. $\gamma _{k}=0$, $k \in [2,t] \cup [t+j+1,n]$.

  Consider solution (S2) with $k_1=k$. This point satisfies inequality (20) at equality because both the left-and the right-hand sides of the inequality are equal to zero unless $k=t$. If $k=t$, then both the left- and the right-hand sides of the inequality are equal to $-\ell$. Evaluating (30) at this solution we obtain $\alpha _{k}\ell +\beta _{k} + \gamma _{k}=0$. If $k \not = t$, then we have $\alpha _{k}=\beta _{k}=0$ (from parts 2 and 4) so we get $\gamma _{k}=0$. If $k=t$, then because $\alpha _{t}\ell = -\bar{\alpha }\ell$ and $\beta _{t}=\bar{\alpha }\ell$ (from parts 3 and 5), we get $\gamma _{t}=0$.

• 8. $\gamma _{t+i'} = -\bar{\alpha }\min \{(\bar{u}-\ell -i'\delta ), (u - \ell - j \delta )\}$, $i' \in [1,j]$.

  Consider a solution to $\mathcal {U}$ with $x_{t+i'}=1$, $p_{t+i'}=\bar{u}$, $s_{t+i'}=1$, $x_{t+i}=1$, $p_{t+i}=\min \{\bar{u} + (i-i')\delta , u\}$, $i \in [i'+1, j]$ and all other variables are equal to zero. This point satisfies inequality (20) at equality because either both the left- and the right-hand sides of the inequality are equal to $\bar{u}+(j-i')\delta$ or u depending on the value of $p_{t+j}$. Evaluating (30) at this solution we obtain,

  $$\begin{aligned} \alpha _{t+i'}\bar{u} + \beta _{t+i'} + \gamma _{t+i'} + \sum _{i=i'+1}^{j-1}(\alpha _{t+i}p_{t+i} + \beta _{t+i}) + \alpha _{t+j}p_{t+j} +\beta _{t+j} = 0. \end{aligned}$$

  From parts 1–4 and 6 we obtain $\gamma _{t+i'} = -\bar{\alpha }(p_{t+j} - \ell - j\delta )$. Furthermore, because $p_{t+j}$ is either $\bar{u}+(j-i')\delta$ or u, the proof is complete. $\square$