# Tight MIP formulations for bounded length cyclic sequences

We study cyclic binary strings with bounds on the lengths of the intervals of consecutive ones and zeros. This is motivated by scheduling problems where such binary strings can be used to represent the state (on/off) of a machine. In this context the bounds correspond to minimum and maximum lengths of on- or off-intervals, and cyclic strings can be used to model periodic schedules. Extending results for non-cyclic strings is not straight forward. We present a non-trivial tight compact extended network flow formulation, as well as valid inequalities in the space of the state and start-up variables some of which are shown to be facet-defining. Applying a result from disjunctive programming, we also convert the extended network flow formulation into an extended formulation over the space of the state and start-up variables.

## Authors

• 11 publications
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## 1. Introduction

In scheduling problems it is often natural to use time-indexed binary variables to model the availability of resources, such as the state of machines (on/off) or roster patterns for the workforce. In these contexts there are often bounds on the lengths of on- and off-intervals, and there is a significant literature on mixed integer programming formulations for this

[Frangioni2006, Frangioni2009, Gentile2017, Hedman2009, Lee2004, Malkin2003, QueyranneWolsey2017]. In particular, Malkin2003 showed that for lower bounds on the lengths of on- and off-intervals, the valid inequalities that can be found in [Wolsey1998] are sufficient to describe the convex hull in the space of the state and start-up variables. PochetWolsey2006 give the convex hull for the case of constant upper and lower bounds, and this was generalized by QueyranneWolsey2017 who considered upper and lower bounds, and allowed these bounds to vary over time. They present a tight extended network formulation, and obtain the convex hull in the space of the state and start-up variables via a projection from a different path formulation.

Our work is motivated by applications in the scheduling of railway maintenance [Liden2015], where it is required in some situations that schedules are cyclic. For this reason, we let the sequence of state variables “wrap around” the time horizon and apply the bounds on the lengths of on- or off-intervals also to intervals that start in the end of the time horizon and continue in the beginning. A more formal problem description is provided in Section 2. In Section 3, we follow the approach from [QueyranneWolsey2017] to derive a compact extended network flow formulation. It turns out that the straightforward cyclic variant of the network formulation from[QueyranneWolsey2017] does not lead to an integral polytope in the space of the flow variables, but we can obtain an integral network flow formulation by considering a larger network that arises from exploiting a simple disjunction. In Section 4 we study a cyclic variant of the Queyranne/Wolsey formulation in the space of the state and start-up variables. We prove that it is a valid formulation, but in contrast to the non-cyclic case the polytope is not integral. For the case that the bounds on the interval lengths are constant over time we provide some valid inequalities, and give sufficient conditions for them to be facet-defining. We also use a result from disjunctive programming to derive an extended formulation for the convex hull in the space of the state and start-up variables. Finally, in Section 5 we describe some directions for further investigations.

## 2. Problem description

Throughout this paper, we denote the set for integers by . Let the time horizon be indexed by with the convention that time is added modulo , that is, is the time period after . For integers and with representing time periods we let the interval wrap around in the natural way, that is, .

As in [QueyranneWolsey2017], we consider parameters that impose bounds on the length of on- and off-intervals in the following way:

• is a lower bound on the length of an on-interval starting in period ,

• is an upper bound on the length of an on-interval starting in period ,

• is a lower bound on the length of an off-interval starting in period ,

• is an upper bound on the length of an off-interval starting in period .

In particular, we require that there are at least one on-period and at least one off-period (otherwise there is an on-interval of length or an off-interval of length , and no matter where we let this start the upper bound on the length of the corresponding interval will be violated). We define binary state variables for to be

 yt={1if period t is an on-period,0if period t is an off-period.

The set of feasible state sequences is characterized by the following implications:

 yt−yt−1=1 ⟹yt+i=1 for all i∈[0,αt−1] t∈[0,n−1], (1) yt−yt−1=1 ⟹yt+i=0 for some i∈[αt,βt] t∈[0,n−1], (2) yt−1−yt=1 ⟹yt+i=0 for all i∈[0,γt−1] t∈[0,n−1], (3) yt−1−yt=1 ⟹yt+i=1 for some i∈[γt,δt] t∈[0,n−1]. (4)

We define the binary start-up variables for to be

 zt=1 ⟺yt−1=0 and yt=1, (5)

and define the set

 Z(n,α,β,γ,δ)={(y,z)∈{0,1}2n: (???), (???), (???), (???), (???), 1⩽y0+⋯+yn−1⩽n−1}.

We are interested in tight linear formulations for , and our approach is to adapt the arguments used in [QueyranneWolsey2017]. Before studying the general case we derive a simple feasibility criterion for the constant bound case in which the bounds for do not change over time. If for all then the number of start-up periods is an integer between and . The following proposition states that for every integer in this range there exists a feasible solution with start-up periods.

###### Proposition 1.

If all , then

 {z0+⋯+zn−1:(y,z)∈Z(n,α,β,γ,δ)}={k∈Z:n/(β+δ)⩽k⩽n/(α+γ)}.

In particular, if and only if for some integer .

###### Proof.

Let . We have to show that there exists with if and only if . First, suppose , set and let denote the indices with for . Then, for every , with and for all . Summing over , we obtain , hence , which implies . For the converse, start with any . Then , hence and . This implies that we can choose and such that . Continuing this way, we obtain with and . Then

 y =11…1p100…0q111…1p200…0q2 … 11…1pk00…0qk, z =100…0p1+q1−1100…0p2+q2−1 … 100…0pk+qk−1

defines a vector

satisfying . ∎

## 3. An extended network formulation

We consider a directed graph with node set , and arc set

 A ={((0,t),(1,l)) : l∈[t+αt,t+βt]}∪{((1,t),(0,l)) : l∈[t+γt,t+δt]}.

Figure 1 illustrates this graph for and for all .

In terms of switching sequences, an arc corresponds to switching on in period and switching off in period , and an arc corresponds to switching off in period and switching on in period . Feasible switching sequences correspond to directed cycles of length where the length of an arc for is . As in [QueyranneWolsey2017] we can use the flow interpretation to obtain a formulation for in the following way. For every node , let and denote the sets of arcs entering and leaving , respectively. For convenience, we will omit one pair of brackets, whenever a node appears as an argument, that is, we will write instead of . For , we define

 Aoff(t) ={((1,r),(0,l))∈A:t∈[r,l−1]}}, Aon(t) ={((0,r),(1,l))∈A:t∈[r,l−1]}}.

If is a cycle of length , then for every , contains exactly one arc from , and in the correspondence between cycles and vectors , we have

 yt ={0if C contains an arc a∈A% off(t)1if C contains an arc a∈Aon(t) t∈[0,n−1], zt =1⟺C contains an arc a∈Aout(0,t) t∈[0,n−1].

Let be the polytope defined by the constraints

 ∑a∈Aoff(0)∪Aon(0)xa =1, (6) ∑a∈Ain(v)xa−∑a∈Aout(v)xa =0 v∈V, (7) yt =∑a∈Aon(t)xa t∈[0,n−1], (8) zt =∑a∈Aout(0,t)xa t∈[0,n−1], (9) xa ⩾0 a∈A. (10)
###### Proposition 2.

The polytope is an extended formulation for , that is, .

###### Proof.

For every we have a corresponding cycle of length . Let us define as . This provides a point , and shows . For the converse inclusion we start with an arbitrary , and fix a vector with . We need to verify that satisfies (1) through (5). For this purpose the following observations are useful:

 ∑a∈Aoff(t)∪Aon(t)xa =1 t∈[0,n−1], (11) ∑a∈Aout(0,t)xa ={1if yt−1=0 and yt=10otherwise. t∈[0,n−1], (12) ∑a∈Aout(1,t)xa ={1if yt−1=1 and yt=00otherwise. t∈[0,n−1]. (13)

These observations can be seen as follows:

(11):

Note that

 =Aout(0,t)∪Aout(1,t), (Aoff(t−1)∪Aon(t−1))∖(Aoff(t)∪Aon(t)) =Ain(0,t)∪Ain(1,t),

and therefore,

 ∑a∈Aoff(t)∪Aon(t)xa=∑a∈Aoff(t−1)∪Aon(t−1)xa+∑a∈Aout(0,t)∪Aout(1,t)xa−∑a∈Ain(0,t)∪A%in(1,t)xa∑a∈Aoff(t−1)∪Aon(t−1)xa.

Together with (6) and induction on this implies (11).

(12):

With we obtain

 ∑a∈Aout(0,t)xa=∑a∈Aon(t)∖Aon(t−1)xa⩽∑a∈Aon(t)xayt,

which implies if . If and , then

 1=yt−yt−1∑a∈Aon(t)xa−∑a∈Aon(t−1)xa⩽∑a∈Aon(t)∖Aon(t−1)xa=∑a∈Aout(0,t)xa

and consequently, . Finally, for we note that and therefore

 ∑a∈Aout(0,t)xa∑a∈Ain(0,t)xa⩽∑a∈Aoff(t−1)xa1−∑a∈Aon(t−1)xa1−yt−1=0.
(13):

With we obtain

 ∑a∈Aout(1,t)xa=∑a∈Aoff(t)∖Aoff(t−1)xa⩽∑a∈Aoff(t)xa1−∑a∈Aon(t)xa1−yt,

which implies if . If and , then

and consequently, . Finally, for we note that and therefore

 ∑a∈Aout(1,t)xa∑a∈Ain(1,t)xa⩽∑a∈Aon(t−1)xayt−1=0.

After establishing (11), (12) and (13), we can now proceed to verify (1) through (5).

(1):

Suppose , that is, and , and fix . Using , we obtain

 yt+i∑a∈Aon(t+i)xa⩾∑a∈Aout(0,t)xa1.

Now (11) implies , and we conclude , as required.

(2):

Suppose , that is, and . Then (12) implies . In particular, for some . Fix an such that for , and note that . Then

 yt+i∑a∈Aon(t+i)xa1−∑a∈Aoff(t+i)xa⩽1−xa∗<1,

and by integrality we conclude , as required.

(3):

Suppose , that is, and , and fix . Using , we obtain

 yt+i∑a∈Aon(t+i)xa1−∑a∈Aoff(t+i)xa⩽1−∑a∈Aout(1,t)xa0.

Now is a consequence of (10) and (8), and we conclude , as required.

(4):

Suppose , that is, and . Then (12) implies . In particular, for some . Fix an such that for , and note that . Then

 yt+i∑a∈Aon(t+i)xa⩾xa∗>0,

and by integrality we conclude , as required.

(5):

This follows immediately from (12) and (9).∎

In the non-cyclic case, the polytope corresponding to is integral and and its projection onto the space gives ([QueyranneWolsey2017, Theorem 1]). Unfortunately, this breaks down in the cyclic case, as the following example shows.

###### Example 1.

Let , , for all . Then is not integral, as the cycle , shown in Figure, with coefficient corresponds to an extreme point of . This point projects to , , which is not contained in .

It is still possible to obtain an extended formulation for as a flow problem in a network of size polynomial in . For this purpose we make copies of the original network: one for every node such that at least one arc in “wraps around”. In other words, there is a copy for node if , and there is a copy for node if . We also add an origin node and a destination node . The underlying idea is that --paths through the copy of the network for node when represent cycles using an arc of the form with , and --paths through the copy for node when represent cycles using an arc of the form with . More formally, with and the extended network has node set

 V′={O,D}∪{(i,t,j,τ):i∈{0,1},t∈[0,n−1],j∈{0,1},τ∈Tj},

and arc set , where

 A′1 ={(O,(1,t,0,τ)) : τ∈T0, t=τ+p−n for some p∈[ατ,βτ]}, A′2 ={(O,(0,t,1,τ)) : τ∈T1, t=τ+q−n for some q∈[γτ,δτ]}, A′3 ={((0,τ,0,τ),D) : τ∈T0}∪{((1,τ,1,τ),D) : τ∈T1}, A′4 ={((i,t,0,τ),(1−i,l,0,τ)) : ((i,t),(1−i,l))∈A,t

The network for , for all , which implies , is shown in Figure 3.

We define to be the set of arcs corresponding to :

 A′on(t)={(O,(1,l,0,τ))∈A′1:l>t}∪{((0,τ,0,τ),D)∈A′3:τ⩽t}∪{((0,k,i,τ),(1,l,i,τ))∈A′4∪A′5:k⩽t

and then we define the polytope by the following constraints:

 ∑a∈Aout(O)x′a =1, (14)