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 twoperiod case, we give a complete description of the convex hull of the feasible solutions. The twoperiod inequalities can be readily used to strengthen ramping formulations without the need for separation. For the general case, we define exponential classes of multiperiod variable upper bound and multiperiod 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 branchandcut algorithm to solve unit commitment problems in power generation.
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Acknowledgments
Pelin DamcıKurt and Simge Küçükyavuz are supported, in part, by the National Science Foundation Grant #1055668, and an allocation of computing time from the Ohio Supercomputer Center. Deepak Rajan’s work is performed under the auspices of the U.S. Department of Energy by Lawrence Livermore National Laboratory under contract DEAC5207NA27344. Alper Atamtürk is supported, in part, by the Office of Assistant Secretary of Defense for Research and Engineering and the National Science Foundation Grant #0970180.
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Appendices
Appendix 1: Proof of convex hull of twoperiod 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 startup 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 lowerbounding inequalities (5b) and (5c) with upperbounding 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 lowerbounding or upperbounding 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 upperbounding 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 lowerbounding 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: Facetdefinition proof of typeI multiperiod rampup inequality
Proposition 10
TypeI multiperiod rampup 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
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 righthand 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)
\(k \in [1,t1] \cup [t+j+1,n]\).
Consider solution (S2) with \(k_1=k\). Clearly, this point satisfies inequality (20) at equality because both the left and the righthand 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)
\(k \in [t+1,t+j1]\).
Consider solution (S5) with \(k_1=t\) and \(k_2=k\). This point satisfies inequality (20) at equality because both the left and the righthand 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, \(\alpha _{k}=0\).

(a)

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 righthand 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 righthand 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 righthand sides of the inequality are equal to \(\ell \). Now, consider solution (S5) with \(k_1=t\) and \(k_2=k1\) if \(t<k1\), and solution (S2) with \(k_1=t\) if \(t=k1\). Both of the points satisfy inequality (20) at equality. Note that if \(t=k1\) we use solution (S2) because both \(k_1=t\) and \(k_2=k1=t\) 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)
\(k \le t1\) 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 righthand 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 righthand sides of the inequality are equal to zero. Now, consider solution (S2) with \(k_1=k1\). Then evaluating (30) at both solutions we get \(\alpha _{k}\ell +\beta _{k}=0\), and because \(\alpha _{k}=0\) (from part 2) we get \(\beta _{k}=0\).

i.

(a)

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 righthand sides of the inequality are equal to \(\ell \). Now consider solution (S5) with \(k_1=1\) and \(k_2=t1\). This point is on the face defined by inequality (20) because both the left and the righthand 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 righthand 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 righthand 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 leftand the righthand sides of the inequality are equal to zero unless \(k=t\). If \(k=t\), then both the left and the righthand 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 \(x_{t+i'}=1\), \(p_{t+i'}=\bar{u}\), \(s_{t+i'}=1\), \(x_{t+i}=1\), \(p_{t+i}=\min \{\bar{u} + (ii')\delta , u\}\), \(i \in [i'+1, j]\) and all other variables are equal to zero. This point satisfies inequality (20) at equality because either both the left and the righthand sides of the inequality are equal to \(\bar{u}+(ji')\delta \) or u depending on the value of \(p_{t+j}\). Evaluating (30) at this solution we obtain,
$$\begin{aligned} \alpha _{t+i'}\bar{u} + \beta _{t+i'} + \gamma _{t+i'} + \sum _{i=i'+1}^{j1}(\alpha _{t+i}p_{t+i} + \beta _{t+i}) + \alpha _{t+j}p_{t+j} +\beta _{t+j} = 0. \end{aligned}$$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 \(\bar{u}+(ji')\delta \) or u, the proof is complete. \(\square \)
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DamcıKurt, P., Küçükyavuz, S., Rajan, D. et al. A polyhedral study of production ramping. Math. Program. 158, 175–205 (2016). https://doi.org/10.1007/s1010701509199
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DOI: https://doi.org/10.1007/s1010701509199
Keywords
 Ramping
 Unit commitment
 Cogeneration
 Production smoothing
 Convex hull
 Polytope
 Valid inequalities
 Facets
 Computation
Mathematics Subject Classification
 90C11
 90C57