CORE DISCUSSION PAPER9720

THE 0-1 KNAPSACK PROBLEM WITH A SINGLECONTINUOUS VARIABLE

Hugues Marchand1, and Laurence A.Wolsey2

March 1997

AbstractConstraints arising in practice often contain many 0-1 variables

and one or a small number of continuous variables. Existing knapsackseparation routines cannot be used on such constraints. Here we studysuch constraint sets, and derive valid inequalities that can be used ascuts for such sets, as well for more general mixed 0-1 constraints.

Specifically we investigate the polyhedral structure of the knapsackproblem with a single continuous variable, called the continuous 0-1knapsack problem. First different classes of facet-defining inequalitiesare derived based on projection and lifting. The order of lifting, partic-ularly of the continuous variable, plays an important role. Secondly weshow that the flow cover inequalities derived for the single node flowset, consisting of arc flows into and out of a single node with binaryvariable lower and upper bounds on each arc, can be obtained fromvalid inequalities for the continuous 0-1 knapsack problem. Thus theseparation heuristic we derive for continuous knapsack sets can alsobe used to derive cuts for more general mixed 0-1 constraints. Initialcomputational results on a variety of problems are presented.Keywords: continuous 0-1 knapsacks, valid inequalities, lifting andprojection

This text presents research results of the Belgian Program on InteruniversityPoles of Attraction initiated by the Belgian State, Prime Minister’s Office,Science Policy Programming. The scientific responsability is assumed bythe authors.

1Columbia Business School, New York. Supported by a CIM Doctoral Fellowship.2CORE and INMA, Universite Catholique de Louvain.

1 Introduction

Here we study the polyhedral structure of one of the simplest mixed 0-1 sets,a binary knapsack set with a single unbounded continuous variable (calledthe continuous 0-1 knapsack set):

Y = {(y, s) ∈ Bn ×R1+ :

∑j∈N

ajyj ≤ b+ s}

where aj > 0 for j ∈ N = {1, . . . , n}, and b ≥ 0.When s = 0, Y reduces to a binary knapsack set for which a large

number of facet-defining inequalities have been derived [1] [11] [20] [21].For such sets, the concepts of projection, lifting and cover have played animportant role. Using separation heuristics, the resulting inequalities havebeen successfully used as cutting planes in various branch-and-bound andbranch-and-cut systems [6] for problems containing constraints involvingonly binary variables.

In mixed 0-1 programming, so-called single node flow sets, of the form

Z = {(x, y) ∈ Rn+×Bn :∑j∈N+

xj−∑j∈N−

xj ≤ b, ljyj ≤ xj ≤ ujyj j ∈ N+∪N−},

where n =| N+∪N− |, have been used to derive flow cover inequalities [15],[18] which have also been integrated in branch-and-bound [19] and branch-and-cut [16], [5] systems.

The set Y studied here is in many ways simpler than the mixed integerset Z. It also turns out to have a richer structure than the binary knapsackset, and to be a useful relaxation of both Z and mixed 0-1 sets K={(x, y) ∈Rp+ × Bn :

∑p=1 gjxj +

∑n=1 ajyj ≤ b, xj ≤ uj j = 1, . . . , p}. An outline

of the paper now follows. In Section 2 we introduce an example used asmotivation in the later sections, and make some basic observations about theset Y . In Section 3 we derive two families of facet-defining inequalities fromthe underlying knapsack problem and the complemented knapsack problemrespectively. Here the continuous variable is introduced (lifted) last. InSection 4 we develop three more families of valid inequalities. We essentiallyproject out all but one of the binary variables, leaving just one binary andone continuous variable. The resulting set has just one nontrivial facet whichis then lifted by reintroducing the other variables. It turns out that if allvariables projected to one are lifted before all variables projected to zero, or

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vice versa, the lifting coefficients, and hence the resulting inequalities, calledcontinuous cover and continuous reverse cover inequalities, are unique. Withother lifting sequences, the facets obtained are sequence dependent. For thesmall example, the five classes suffice to completely define the convex hull.

In Section 5 we show that the set Y can be viewed as a relaxation of themixed integer set Z, and that certain continuous cover inequalities for Yproduce flow cover inequalities for Z. By varying the relaxation and usinglifting, new variants of the flow cover inequalities are obtained. In Section6 we present computational results showing the behaviour of a separationheuristic for Y producing inequalities of the type derived in Sections 3 and4. The instances tackled include continuous knapsack sets, single node flowsets and mixed 0-1 problems from the MIPLIB library [3]. The resultsare comparable with those obtained using flow-cover separation routines.Finally we terminate with a discussion of possible extensions.

2 The Continuous Knapsack Problem

Consider the set:

Y = {(y, s) ∈ Bn ×R+ :∑j∈N

ajyj ≤ b+ s}

where aj > 0 for j ∈ N = {1, . . . , n}, b ≥ 0, and we assume throughout that∑j∈N aj > b.

First we observe that conv(Y ) is full-dimensional. The following charac-terisation of the trivial facets is straightforward.

Proposition 1 i) The inequality∑j∈N ajyj ≤ b + s defines a facet of

conv(Y ) if aj ≤∑i∈N ai − b for all j ∈ N .

ii) The inequality s ≥ 0 defines a facet of conv(Y ) if aj ≤ b for all j ∈ N .iii) The inequality yj ≥ 0 defines a facet of conv(Y ).iv) The inequality yj ≤ 1 defines a facet of conv(Y ).

We now introduce an example to which we will refer in the followingsections.

Example. Let

Y = {(y, s) ∈ B5 ×R1+ : 7y1 + 6y2 + 5y3 + 3y4 + 2y5 ≤ 11 + s}.

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A complete description of the convex hull is given by the trivial facets yj ≥ 0and yj ≤ 1 for all j = {1 . . . 5}, s ≥ 0, the defining inequality 7y1 + 6y2 +5y3 + 3y4 + 2y5 ≤ 11 + s, and the following system of inequalities:

2y1 + 2y2 ≤ 2 + s (1)y1 + y3 ≤ 1 + s (2)y1 + y4 + y5 ≤ 2 + s (3)

3y1 + 3y2 + 3y3 + 3y4 ≤ 6 + s (4)2y1 + 2y2 + 2y3 + 2y5 ≤ 4 + s (5)3y1 + 2y2 + 2y3 + y4 + y5 ≤ 4 + s (6)6y1 + 5y2 + 5y3 + 3y4 + 2y5 ≤ 10 + s (7)5y1 + 4y2 + 3y3 + y4 + 2y5 ≤ 7 + s (8)6y1 + 5y2 + 4y3 + 3y4 + y5 ≤ 9 + s (9)4y1 + 4y2 + 2y3 + 2y5 ≤ 6 + s (10)5y1 + 5y2 + 3y3 + 3y4 ≤ 8 + s (11)3y1 + 2y2 + y3 ≤ 3 + s (12)4y1 + 3y2 + 4y3 + 3y4 ≤ 7 + s (13)3y1 + 2y2 + 3y3 + 2y5 ≤ 5 + s (14)4y1 + 3y2 + 2y3 + y4 + y5 ≤ 5 + s (15)5y1 + 4y2 + 4y3 + 3y4 + y5 ≤ 8 + s (16)4y1 + 3y2 + 3y3 + y4 + 2y5 ≤ 6 + s (17)

This description was obtained using the code of [7]. 2

3 Knapsack and Complemented Knapsack Facets

One way to generate facets of conv(Y ) is to adapt facets of the underlyingknapsack polytope, and reintroduce the continuous variable s by a liftingprocedure. Let N = {j ∈ N : aj ≤ b} and Y0 = {(y, s) ∈ Y : s = 0, yj =0, j ∈ N \ N}.

Proposition 2 If∑j∈N πjyj ≤ π0 with π ≥ 0, π0 > 0 defines a non trivial

facet of conv(Y0), then

4

∑j∈N

πjyj ≤ π0 +s

β(18)

defines a facet of conv(Y ), where β = mins>0s

η(s)−π0, η(s) = max{∑j∈N πjyj :∑

j∈N ajyj ≤ b+ s, y ∈ B|N |} and πj = π0 + aj−bβ for j ∈ N \ N .

Proof. By [4], we observe that∑j∈N πjyj ≤ π0 + s

β is a facet of theconvex hull of the set Y where yj = 0 for j ∈ N \ N . Moreover βπ0 ≤ b andβπj ≤ aj for all j ∈ N .

Now suppose that there is a point (y, s) ∈ Y with yk = 1 for somek ∈ N \ N and

∑j∈N βπjyj > βπ0 + s. Note that we also have βπj =

βπ0+(aj−b) ≤ aj for j ∈ N\N as βπ0 ≤ b. Therefore s <∑j∈N βπjyj−βπ0

≤ ∑j∈N\{k} ajyj + (βπ0 + (ak − b))yk − βπ0 =

∑j∈N ajyj − b and then

(y, s) 6∈ Y , a contradiction.Hence (18) is a valid inequality for Y . Moreover as the points yk = 1, s =

ak − b for k ∈ N \ N are tight, it defines a facet of conv(Y ). 2

Thus each nontrivial facet of the knapsack set gives rise to a distinctfacet of conv(Y ).

Example (continued) The underlying knapsack set is

Y0 = {y ∈ B5 : 7y1 + 6y2 + 5y3 + 3y4 + 2y5 ≤ 11}A complete description of the convex hull is given by the trivial facets yj ≥ 0for all j = {1 . . . 5}, y4 ≤ 1, y5 ≤ 1 and the following inequalities.

y1 + y2 ≤ 1 (19)y1 + y3 ≤ 1 (20)y1 + y4 + y5 ≤ 2 (21)y1 + y2 + y3 + y4 ≤ 2 (22)y1 + y2 + y3 + y5 ≤ 2 (23)

3y1 + 2y2 + 2y3 + y4 + y5 ≤ 4 (24)

Note that whereas conv(Y ) has 17 nontrivial facets, conv(Y0) has only 6.

For the knapsack facet (19), β = 2 and so by Proposition 2 the correspondingfacet-defining inequality for the continuous knapsack set Y is 2y1 + 2y2 ≤

5

2 + s. In this way the six inequalities (19)− (24) of the knapsack set Y0 leadto the six facet-defining inequalities (1)− (6) for conv(Y ). 2

To derive another class of facets, we introduce the slack variable s =b + s −∑j∈N ajyj . Complementing the variables using yj = 1 − yj leadsto the equation

∑j∈N aj yj + s = (

∑j∈N aj − b) + s, and after dropping the

nonnegative variable s, we obtain the set

Y = {(y, s) ∈ Bn ×R1+ :

∑j∈N

aj yj ≤ (∑j∈N

aj − b) + s}.

Now we can apply the same lifting approach to Y . Again let N ={j ∈ N : aj ≤

∑k∈N ak − b} and Y0 = {(y, s) ∈ Y : s = 0, yj = 0, j ∈

N \ N}. If∑j∈N µj yj ≤ µO is a facet-defining inequality for conv(Y0),

we obtain a facet-defining inequality∑j∈N µj yj ≤ µ0 + s/β for conv(Y ).

Recomplementing variables leads to a facet-defining inequality∑j∈N

(aj − βµj)yj ≤ b+ β(µ0 −∑j∈N

µj) + s

for conv(Y ).

Example (continued) In the complemented space (y, s), the underlyingcontinuous knapsack set is

Y = {(y, s) ∈ B5 ×R1+ : 7y1 + 6y2 + 5y3 + 3y4 + 2y5 ≤ 12 + s}

A complete description of the convex hull of the associated knapsack setY0 is given by the trivial facets yj ≥ 0 for all j = {1 . . . 5}, yj ≤ 1 for allj = {3 . . . 5} and the following system of inequalities,

y1 + y2 ≤ 1 (25)y1 + y2 + y3 + y4 ≤ 2 (26)y1 + y2 + y3 + y5 ≤ 2 (27)

For the facet (25), µ = (1, 1, 0, 0, 0), µ0 = 1 and β = 1. So in the originalspace of variables we obtain

(7− 1)y1 + (6− 1)y2 + (5− 0)y3 + (3− 0)y4 + (2− 0)y5 ≤ 11 + 1(1− 2) + s

6

or 6y1 + 5y2 + 5y3 + 3y4 + 2y5 ≤ 10 + s, facet-defining inequality (7) ofconv(Y ).

In the same way, using Proposition 2, the three inequalities (25)− (27)lead to the three facet-defining inequalities (7)− (9) of conv(Y ). 2

4 Continuous Cover Inequalities

Here we examine facets of conv(Y ) obtained by first projecting out all butone of the binary variables, then deriving a facet-defining inequality in thetwo-dimensional space consisting of one binary variable and the continuousvariable s, and finally lifting back the binary variables.

We start by reviewing some results on sequential lifting. For a morethorough introduction and basic references on 0-1 lifting and superadditivity,see [13]. Let

W (d) = {(z, s) ∈ B|N+|+|N−| ×R1+ :

∑j∈N+

ajzj +∑j∈N−

ajzj ≤ d+ s}

with aj > 0 for j ∈ N+, aj < 0 for j ∈ N− and N = N+ ∪ N−. Nowsuppose that M+ ⊆ N+,M− ⊆ N−,M = M+ ∪M− and

∑j∈M

πjzj ≤ π0 + s (28)

is a tight valid inequality for the set WM (d) = W (d) ∩ {z : zj = 0 j /∈M}.The lifting question is then to determine values {αj}j∈N\M so that∑

j∈Mπjzj +

∑j∈N\M

αjzj ≤ π0 + s

is valid for W (d).One way to lift variables is to introduce them one at a time (sequential

lifting). Suppose that we start with some variable zk with k ∈ N \M .Let GM (u) = max{∑j∈M πjzj − s : (z, s) ∈ WM (u)}, and φM (u) =

GM (d)−GM (d− u) for u ∈ R1. As inequality (28) is tight, π0 = GM (d).

Proposition 3 The inequality

αkzk +∑j∈M

πjzj ≤ π0 + s (29)

is valid for WM∪{k}(d) if and only if αk ≤ φM (ak). If (28) is facet-definingfor WM (d), then (29) is facet-defining for WM∪{k}(d) if αk = φM (ak).

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Now to sequentially lift in a second variable, it is necessary to updatethe functions GM and φM , so as to calculate the next lifting coefficient.Specifically if k ∈ N \M was introduced first with lifting coefficient αk =φM (ak), then

GM∪{k}(u) = max[GM (u), GM (u− ak) + φM (ak)],

so in general the function GM can increase as more variables are liftedin. However GM (d) does not change, and so φM can decrease. In thenext two Propositions we explore two cases in which these functions remainunchanged over an important part of their domains.

Definition. A function F : D ⊆ Rm → R1 is superadditive on D if F (d1) +F (d2) ≤ F (d1 + d2) for all d1, d2, d1 + d2 ∈ D.

We distinguish whether the variable lifted has a positive coefficient k ∈N+ \M+ or negative coefficient k ∈ N− \M− in the description of W (d).

Proposition 4 [22] i) φM (u) = minv≤d[GM (v)−GM (v − u)] for u ∈ R1+

if and only ifii) φM is superadditive on R1

+.If i) and ii) hold, theniii) for any k ∈ N+ \M+, φM (u) = φM∪{k}(u) for u ∈ R1

+, andiv) the inequality

∑j∈M πjzj +

∑j∈N+\M+ φM (aj)zj ≤ π0 + s is valid

for WM∪(N+\M+)(d), and it is facet-defining for WM∪(N+\M+)(d) if (28) isfacet-defining for WM (d).

Note that a slightly stronger statement can be made: if iii) holds for allpossible values of ak ∈ R1

+, or if iv) holds for all values of aj ∈ R1+ with

j ∈ N+ \M+ and | N+ \M+ |≥ 2, then i) and ii) hold.In a similar fashion, we obtain

Proposition 5 The following are equivalenti) φM (u) = minv≥d[GM (v)−GM (v − u)] for u ∈ R1

−ii) φM is superadditive on R1

−If i) and ii) hold and k ∈ N− \M−, theniii) φM (u) = φM∪{k}(u) for u ∈ R1

−iv) the inequality

∑j∈M πjzj +

∑j∈N−\M− φM (aj)zj ≤ π0 + s is valid

for WM∪(N−\M−)(d), and it is facet-defining for WM∪(N−\M−)(d) if (28) isfacet-defining for WM (d).

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We now apply the above results to derive facet-defining inequalities forY . One pair of definitions is needed.

Definition. A pair (k,C) is a k-cover for Y if k ∈ C ⊆ N ,∑j∈C aj = b+ λ

with λ > 0, and∑j∈C\{k} aj < b.

A pair (k, T ) is a k-reverse-cover for Y if k ∈ T ⊆ N ,∑j∈T aj =

∑j∈N aj −

b+ µ with µ > 0, and∑j∈T\{k} aj <

∑j∈N aj − b.

Note that if (k,C) is a k-cover for Y , then (k, T ) is a k-reverse-cover forY with T = (N \ C) ∪ {k}, and vice versa, and ak = λ + µ. The resultingpair of k-covers, denoted (k,C, T ) is called a k-cover-pair.

4.1 Projection

Here we see that, given a k-cover-pair (k,C, T ), there is a natural projectionof all the binary variables except for k. The set Y can be rewritten as

Y = {(y, s) ∈ Bn ×R1+ : −akyk −

∑j∈C\{k}

ajyj +∑

j∈T\{k}ajyj ≤ −λ+ s},

or alternatively as

Y = {(y, s) ∈ Bn ×R1+ : akyk −

∑j∈C\{k}

ajyj +∑

j∈T\{k}ajyj ≤ µ+ s}.

Now we project down to the space of (yk, s) by setting yj = 1 for j ∈ C \{k}and yj = 0 for j ∈ T \ {k}. The resulting set is

Q = {(yk, s) ∈ B1×R1+ : −akyk ≤ −λ+s} = {(yk, s) ∈ B1×R1

+ : akyk ≤ µ+s}.

It is easily checked that the unique nontrivial facet-defing inequality for Qis the Gomory mixed integer [8] or MIR [14] inequality−λyk ≤ −λ+ s, or (ak − µ)yk ≤ s.Now we lift back the variables for j ∈ N \ {k}. Different orderings will leadto different facets.

4.2 Continuous Cover Inequalities

Here we start by lifting in the variables that have been set to yj = 1 (oryj = 0) for j ∈ C \ {k}. Let YC = {(y, s) ∈ Y : yj = 0 j ∈ N \ C} andC = {j ∈ C : aj > λ}, so k ∈ C.

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Proposition 6 If (k,C) is a k-cover for Y , the inequality∑j∈C

λyj +∑

j∈C\Cajyj ≤ (| C | −1)λ+

∑j∈C\C

aj + s (30)

is facet-defining for conv(YC).

Proof. Given the facet-defining inequality

−λyk ≤ −λ+ s

forQ = {(yk, s) ∈ B1 ×R1

+ : −akyk ≤ −λ+ s},we sequentially lift the variables yj j ∈ C \ {k}. We have that G{k}(u) =max{−λyk−s : −akyk−s ≤ u, (yk, s) ∈ B1×R1

+}. ThusG{k}(u) = min(u, 0)for u ≥ −λ, and φ{k}(u) = G{k}(−λ) − G{k}(−λ − u) = max[−λ, u] foru ∈ R1

−. As φ{k} is superadditive on R1−, we obtain from Proposition 5 that

the lifted inequality

−∑j∈C

λyj −∑

j∈C\Cajyj ≤ −λ+ s

is facet-defining for YC . After complementation, the claim follows. 2

Next we lift the variables from yj = 0 for j ∈ T \ {k} = N \ C. NowGC(u) =

max{−∑j∈C

λyj −∑

j∈C\Cajyj − s : −

∑j∈C

ajyj − s ≤ u, (y, s) ∈ B|C| ×R1+}.

Suppose WLOG that C = {1, . . . , r} with a1 ≥ . . . ≥ ar > λ. Let Aj =∑ji=1 ai for j ≤ r, and A0 = 0. It can be verified that

GC(u) =

0 if u ≥ 0−jλ+ u+Aj if −Aj − λ ≤ u ≤ −Aj j = 0, . . . , r − 1−jλ if −Aj < u ≤ −Aj−1 − λ j = 1, . . . , r−rλ+ u−Ar if u ≤ −Ar

and φC(u) = GC(−λ)−GC(−λ− u) is given on R1+ by

φC(u) =

(j − 1)λ if Aj−1 ≤ u ≤ Aj − λ j = 1, . . . , r(j − 1)λ+ [u− (Aj − λ)] if Aj − λ ≤ u ≤ Aj j = 1, . . . , r − 1(r − 1)λ+ [u− (Ar − λ)] if Ar − λ ≤ u.

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Proposition 7 If (k,C) is a k-cover for Y , then the inequality∑j∈C

λyj +∑

j∈C\Cajyj +

∑j∈N\C

φC(aj)yj ≤ (| C | −1)λ+∑

j∈C\Caj + s (31)

is facet-defining for conv(Y ).

Proof. Function φC is superadditive on R1+ and thus by Proposition 4

−∑j∈C

λyj −∑

j∈C\Cajyj +

∑j∈N\C

φ(aj)yj ≤ −λ+ s

is facet-defining for conv(Y ). Complementing variables, the claim follows.2

Example (continued)

C = {1, 2, 4} is a 1-cover with λ = 5. By Proposition 6 we obtain the validinequality

5y1 + 5y2 + 3y4 ≤ 8 + s

for YC .The function φC is given by

φC(u) =

0 0 ≤ u ≤ 2u− 2 2 ≤ u ≤ 75 7 ≤ u ≤ 8u− 3 8 ≤ u

Therefore φC(a3) = φC(5) = 3, φC(a5) = φC(2) = 0 and, by Proposition 7,the inequality

5y1 + 5y2 + 3y3 + 3y4 ≤ 8 + s

defines a facet of conv(Y ).Proposition 7 produces the following facets: (1) (C = {1, 2}), (2) (C =

{1, 3}), (3) (C = {1, 4, 5}), (4) (C = {2, 3, 4}), (5) (C = {2, 3, 5}), (7)(C = {2, 3, 4, 5}), (10) (C = {1, 2, 5}), and (11) (C = {1, 2, 4}). 2

We can derive facets of the form (31) for the complemented problem Yand re-express the inequalities in the initial space of variables. However wechoose to derive this new class directly.

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4.3 Continuous Reverse Cover Inequalities

Here we start by lifting in the variables that have been set to yj = 0 forj ∈ T \ {k}. Let YT = {(y, s) ∈ Y : yj = 1 j ∈ N \ T}. Suppose thatT = {1 . . . p}, order {aj}j∈T s.t. aj ≥ aj+1, j ∈ {1 . . . p − 1} and definer=min {j ∈ T : aj > µ}. Note that r ≥ 1 as ak > µ.

Proposition 8 If (k, T ) is a k-reverse-cover for Y , the inequality∑j∈T

(aj − µ)+yj ≤ s

is facet-defining for conv(YT ).

Proof. Given the facet-defining inequality (ak−µ)yk ≤ s for Q = {(yk, s) ∈B1×R1

+ : akyk ≤ µ+s}, we sequentially lift the variables yj j ∈ T \{k}. Wehave that G{k}(u) = max{(ak − µ)yk − s : akyk − s ≤ u, (yk, s) ∈ B1 ×R1

+}.Thus G{k}(u) = min(0, u) for u ≤ µ, and φ{k}(u) = G{k}(µ)−G{k}(µ−u) =(u − µ)+ for u ≥ 0. As φ{k} is superadditive on R1

+, the claim followsimmediately from Proposition 4. 2

Next we lift the variables yj for j ∈ C \{k} = N \T with coefficient −aj .GT (u) = max{∑j∈T (aj −µ)+yj − s :

∑j∈T ajyj − s ≤ u, (y, s) ∈ B|T |×R1

+}and φT (u) = GT (µ)−GT (µ−u). It is straightforward to verify that φT (u) =−ψT (−u) on R1

− where ψT is defined on R1+ as follows:

ψT (u) =

u− iµ if Ai ≤ u ≤ Ai+1 − µ for i = 0 . . . r − 1Ai − iµ if Ai − µ ≤ u ≤ Ai for i = 1 . . . r − 1Ar − rµ if u ≥ Ar − µ

where A0 = 0 and Ai =∑ij=1 aj for i = 1 . . . r.

Proposition 9 If (k, T ) is a k-reverse-cover for Y , then the inequality∑j∈T

(aj − µ)+yj +∑

j∈N\TψT (aj)yj ≤

∑j∈N\T

ψT (aj) + s (32)

is facet-defining for Y .

Proof. The function φT is superadditive on R1− and hence by Propositions

5 and 8, ∑j∈T

(aj − µ)+yj +∑

j∈N\TφT (−aj)yj ≤ s

12

is facet-defining for Y . The claim follows after complementation and sub-stitution. 2

Example (continued)

T = {2, 3, 5} is a 2-reverse cover with a2 = 6 > µ = 1. Therefore byProposition 8,

5y2 + 4y3 + 1y5 ≤ sis facet-defining for YT . Now

ψT (d) =

d 0 ≤ d ≤ 55 5 ≤ d ≤ 6d− 1 6 ≤ d ≤ 109 10 ≤ d ≤ 11d− 2 11 ≤ d ≤ 1210 12 ≤ d

So ψT (a1) = ψT (7) = 6, ψT (a4) = 3 and so the inequality

6y1 + 5y2 + 4y3 + 3y4 + y5 ≤ 9 + s

obtained by Proposition 9 defines a facet of conv(Y ).Proposition 9 produces the following facets: (1) T = {1, 3, 4, 5}, (2)

T = {1, 2, 4, 5}, (3) T = {1, 2, 3}, (9) T = {2, 3, 5}, (7) T = {1, 2}, (8)T = {2, 3, 4}, (10) T = {1, 3, 4}, (11) T = {1, 3, 5}, (12) T = {2, 3, 4, 5}, (13)T = {1, 2, 5}, (14) T = {1, 2, 4}, (10) T = {1, 3, 4} and (15) T = {1, 4, 5}. 2

4.4 Projected Lifted Continuous Cover Inequalities

Again let (k,C, T ) be a k-cover pair, and suppose now that the lifting orderis arbitrary. We start from Q = {(y, s) ∈ B1 ×R1

+ : akyk ≤ µ+ s}, and liftterms ajyj for j ∈ T \ {k} and −ajyj for j ∈ C \ {k}.

Proposition 10 For each lifting sequence, a facet-defining inequality of theform

(ak − µ)yk +∑

j∈C\{k}γjyj +

∑j∈T\{k}

γjyj ≤∑

j∈C\{k}γj + s

is obtained, with 0 ≤ γj ≤ (aj−µ)+ for j ∈ T \{k}, and aj ≥ γj ≥ min[λ, aj ]for j ∈ C \ {k}.

13

Proof. From Proposition 3, φ{k}(u) provides an upper bound on the liftingcoefficient. Another result from [22] states that γ{k}(u) = minv∈R1 [G{k}(v)−G{k}(v−u)] always produces a feasible value and hence lower bound for thelifting coefficent. As G{k}(u) = min{u,min[(u − µ)+, λ]}, it follows thatγ{k}(u) = min(u, 0). Therefore for j ∈ T \ {k}, u = aj > 0, so γ{k}(u) =0 ≤ γj ≤ φ{k}(u) = (aj − µ)+, and for j ∈ C \ {k}, u = −aj < 0, soγ{k}(u) = −aj ≤ −γj ≤ φ{k}(u) = max[−λ,−aj ] = −min[λ, aj ]. 2

Example (continued) ({5}, {1, 4, 5}, {2, 3, 5}) is a 5-cover-pair with λ =

µ = 1. The set Y now can be written as (y, s) ∈ B5 ×R1+ satisfying

−7y1 + 6y2 + 5y3 − 3y4 + 2y5 ≤ 1 + s.

Projecting onto y1 = y4 = 1, y2 = y3 = 0, we obtain Q = {(y5, s) ∈ B1×R1+ :

2y5 ≤ 1+s} with facet-defining inequality y5 ≤ s. Lifting in the variables inthe order y4, y3, y2, y1, and calculating the coefficients one by one by repeateduse of Proposition 3, we obtain

−4y1 + 3y2 + 2y3 − y4 + y5 ≤ s, or

4y1 + 3y2 + 2y3 + y4 + y5 ≤ 5 + s.

Starting from the same k-cover-pair, and lifting in the order y3, y1, y4, y2

leads to facet-defining inequality (16).Finally taking the k-cover-pair (4, {1, 4, 5}, {2, 3, 4}) and lifting in the

order y3, y1, y5, y2 leads to (17).We have explained all the facets of conv(Y ) in our example. 2

5 The Continuous Knapsack Problem and the FlowModel

Here we show how valid inequalities for Y can be used to derive valid in-equalities for single node flow sets, and in particular generate a large classof flow cover inequalities.

The continuous knapsack set Y can be seen as a relaxation of the flowmodel

Z = {(x, y) ∈ Rn+×Bn :∑j∈N+

xj−∑j∈N−

xj ≤ b, ljyj ≤ xj ≤ ujyj j ∈ N+∪N−},

14

with lj , uj ≥ 0 for j ∈ N+ ∪ N−. This means that valid inequalities for Yprovide valid inequalities for the flow model Z. We first present a simpleexample.

Example 2. Consider an instance of the flow model Z consisting of thepoints (x, y) ∈ R5

+ ×B5 satisfying:

x1 + x2 + x3 + x4 + x5 ≤ 11x1 ≤ 7y1, x2 ≤ 6y2, x3 ≥ 5y3, x4 ≥ 3y4, x5 ≥ 2y5

After substituting xj + tj = ujyj with tj ≥ 0 for j = 1, 2, and xj ≥ ljyj forj = 3, 4, 5, we obtain

7y1 + 6y2 + 5y3 + 3y4 + 2y5 ≤ 11 + (t1 + t2)

of the form Y . The valid inequality (1) for Y gives 2y1 + 2y2 ≤ 2 + (t1 + t2)which after eliminating t1, t2 gives

x1 + x2 − 5y1 − 4y2 ≤ 2

which is a basic flow cover inequality for Z, see [15]. 2

We now describe a general procedure:

i) Let (N+l , N

+u , N

+b ) be a partition of N+ and (N−l , N

−u , N

−b ) a partition of

N−.

ii) For j ∈ N+l ∪N−l , substitute xj = ljyj + sj with sj ≥ 0.

iii) For j ∈ N+u ∪N−u , substitute xj = ujyj − tj with tj ≥ 0.

The resulting relaxation is:

Z = {(x, y, s, t) :∑j∈N+

l

ljyj +∑j∈N+

u

ujyj −∑j∈N−

l

ljyj −∑j∈N−u

ujyj

+∑j∈N+

b

xj +∑j∈N+

l

sj +∑j∈N−u

tj

≤ b+∑j∈N+

u

tj +∑j∈N−

l

sj +∑j∈N−

b

xj ,

15

yj ∈ {0, 1} j ∈ N−l ∪N+l ∪N−u ∪N+

u , x, s, t ≥ 0}.

iv) Complement the binary variables with negative coefficients, set b′ =b +

∑j∈N−

llj +

∑j∈N−u uj , define a nonnegative variable s′ =

∑j∈N+

utj +∑

j∈N−lsj +

∑j∈N−

bxj , and define aj = lj for j ∈ N+

l ∪N−l and aj = uj forj ∈ N+

u ∪N−u .

The final relaxation is

Y = {(y, s′) :∑

j∈N+l∪N+

u

ajyj +∑

j∈N−l∪N−u

ajyj ≤ b′ + s′,

yj ∈ {0, 1} j ∈ N−l ∪N+l ∪N−u ∪N+

u , s′ ≥ 0}

Example 3. Consider a set Z consisting of points (x, y) ∈ R7+ × B7 satis-

fyingx1 + x2 + x3 − x4 − x5 − x6 − x7 ≤ 1

7y1 ≤ x1 ≤ 12y1, x2 ≤ 6y2, x3 ≤ 5y3, 5y4 ≤ x4 ≤ 9y4,

1y5 ≤ x5 ≤ 3y5, 4y6 ≤ x6 ≤ 9y6, x7 = 2y7

Setting N+l = {1}, N+

u = {2}, N+b = {3}, N−l = {4}, N−u = {5, 7} and

N−b = {6}, we obtain after substitution and complementation

7y1 + 6y2 + 5y4 + 3y5 + 2y7 + s1 + x3 + t5 + t7 ≤ 11 + t2 + s4 + x6

and after further relaxation

7y1 + 6y2 + 5y4 + 3y5 + 2y7 ≤ 11 + s′.

The continuous knapsack inequality (11): 5y1 + 5y2 + 3y4 + 3y5 ≤ 8 + s′

gives after substituting back the inequality

5y1 + (x2 − y2) ≤ 2 + (x4 − 2y4) + 3y5 + x6

that is valid for Z. 2

Having the relaxation Y , we can in particular derive continuous cover(31) and continuous reverse cover (32) inequalities for Y , and then convertthem back into valid inequalities for Z. In the next part of this section weexpress both inequalities in the original space Z.

16

A k-cover C for Z is a set of the form C = C+l ∪ C+

u ∪ C−l ∪ C−u 6= ∅where C−l ⊆ N−l , etc.,∑

j∈C+l

lj +∑j∈C+

u

uj −∑

j∈N−l\C−

l

lj −∑

j∈N−u \C−u

uj − b = λ > 0

and either k ∈ C+l ∪ C−l with lk > λ or k ∈ C+

u ∪ C−u with uk > λ.Now (k,C) is a k-cover for Y , and φC can be defined as before to generate

the continuous cover inequality (31).

Proposition 11 If C is a k-cover for Z, then∑j∈C+

u

[xj − (uj − λ)+yj ] +∑j∈C+

l

min{lj , λ}yj

+∑

j∈N+u \C+

u

[xj − (uj − φC(uj))yj ] +∑

j∈N+l\C+

l

φC(lj)yj

≤∑j∈C+

u

min{uj , λ}+∑j∈C+

l

min{lj , λ}−λ+∑j∈C−

l

[xj−(lj−λ)+yj ]+∑j∈C−u

min{uj , λ}yj

+∑

j∈N−l\C−

l

[xj−ljyj−φC(lj)(1−yj)]−∑

j∈N−u \C−u

φC(uj)(1−yj)+∑j∈N−

b

xj (33)

is a valid inequality for Z.

Proof. Remembering that aj = lj for j ∈ C+l ∪ C−l and aj = uj for

j ∈ C+u ∪ C−u , the inequality (31) gives:∑

j∈C+u

[min(λ, uj)yj − tj ] +∑j∈C+

l

[min(λ, lj)yj ] +∑

j∈N+u \C+

u

[φC(uj)yj − tj ]

+∑

j∈N+l\C+

l

φC(lj)yj +∑j∈C−u

min(λ, uj)yj +∑j∈C−

l

[min(λ, lj)yj − sj ]

+∑

j∈N−u \C−u

φC(uj)yj +∑

j∈N−l\C−

l

[φC(lj)yj − sj ]

≤∑

j∈C+u ∪C−u

min(λ, uj) +∑

j∈C+l∪C−

l

min(λ, lj)− λ+∑j∈N−

b

xj .

The result follows by substituting back for sj , tj and complementing thevariables yj . 2

17

Similarly a k-reverse-cover T for Z is a set of the form T = T+l ∪ T+

u ∪T−l ∪ T−u 6= 0 where T−l ⊆ N−l , etc.,

b+∑j∈T−

l

lj +∑j∈T−u

uj −∑

j∈N+l\C+

l

lj −∑

j∈N+u \T+

u

uj = µ > 0

and either k ∈ T+l ∪ T−l with lk > µ or k ∈ T+

u ∪ T−u with uk > µ.Now (k, T ) is a k-reverse-cover for Y and ψT can also be defined as

before.

Proposition 12 If T is a k-reverse-cover for Z, then∑j∈T+

l

(lj − µ)+yj +∑j∈T+

u

[xj − (uj − (uj − µ)+)yj ]

−∑

j∈N+l\T+l

ψT (lj)(1− yj) +∑

j∈N+u \T+

u

[xj − ψT (uj)(1− yj)− ujyj ]

≤ −∑j∈T−u

(uj − µ)+(1− yj) +∑j∈T−

l

[xj − (ljyj + (lj − µ)+(1− yj))]

+∑

j∈N−l\T−l

[xj − (lj − ψT (lj))yj ] +∑

j∈N−u \T−u

ψT (uj)yj +∑j∈N−

b

xj (34)

is a valid inequality for Z.

In many cases, valid inequalities (33) and (34) define faces of high dimensionsof conv(Z). In particular we can prove,

Proposition 13 Suppose N− = ∅, uj ≤ b and lj = 0 for all j ∈ N+. Theninequalities (33) and (34) define facets of conv(Z).

More generally we observe that the inequalities (33) and (34) are comple-mentary in exactly the same way as the inequalities of Van Roy and Wolsey[18] and Stallaert [17] respectively. Under certain conditions inequality (33)dominates the Van Roy-Wolsey inequality, but in general neither dominates.By complementarity, the same remark applies to (34) and the inequality ofStallaert.

18

6 Computational Aspects

Separation heuristics have been developed for the knapsack and continuouscover inequalities described in Sections 3 and 4. Also by applying these rou-tines to the complementary set Y , complemented knapsack and continuousreverse cover inequalites are also obtained. The heuristics follow standardlines. The cover C is generated greedily from the linear programming relax-ation of the continuous knapsack constraint, or by greedily choosing elementswith the smallest coefficient aj .

To apply the heuristics to flow models of the form Z, it is necessaryto obtain a set of the form Y as in Section 5. Therefore one needs tochoose the sets N+

b , N+u , N

+l , N

−b , N

−u , N

−l . The approach is as described in

[4]. Two candidate sets Y are generated. The idea of the first is to try tocreate a pure surrogate knapsack constraint. Suppose (x∗, y∗, s∗) is the linearprogramming solution. If x∗j = ujy

∗j , then put j ∈ N+

u or N−u as appropriate.If x∗j = ljy

∗j , then put j ∈ N+

l or N−l . Otherwise if j ∈ N+, put j ∈ N+l , and

if j ∈ N−, put j ∈ N−u . The second is based on minimizing the differencebetween each continuous variable and its simple or variable bounds i.e. ifljy∗j + s∗j = x∗j = ujy

∗j − t∗j , and min{s∗j , t∗j} = s∗j or min{s∗j , t∗j} = t∗j , then

we put j ∈ Nl or j ∈ Nu respectively.Initial tests have been carried out on two problem sets using the MIPO

[2] code running on a Sun Sparc Ultra 1 with the separation routine addedto generate cuts at the top node only.

The first set consists of randomly generated mixed integer knapsackproblems involving optimization over sets of the form

K = {(x, y) ∈ Rp+ ×Bn :p∑

=1

gjxj +n∑

=1

ajyj ≤ b, xj ≤ uj j = 1, . . . , p}.

Adding cuts as described above, the problems with 30 variables generatedas in [23] are all solved at the top node. To generate some more difficultinstances, the objective coefficients have been correlated with those of theconstraint, the number of variables increased and the value of b modified.Results for six more difficult instances are presented.

In Table 1 results are given using pure branch-and-bound (BB) withoutcuts, and cut-and-branch (CB) with the continuous knapsack separationroutines called only at the top node. By default MIPO includes a simpleprimal heuristic at the top node, uses reduced cost fixing in the tree, anduses a cut pool. This means that cuts added at the top node and later

19

discarded, may be added again later at some node of the tree. A pure bestbound strategy is used to choose the next node. LP, XLP and IP are thevalues of the linear programming relaxation, the value after adding cuts, andthe optimal value respectively. The instance names begin with the numbersn and p of 0− 1 and continuous variables. Thus 25.5k2 is instance number2 with n = 25 and p = 5. (The data for the two largest instances are givenin the appendix of [12]).

Inst LP XLP IP BBnod BBSec CBnod CBSec Cuts25.5k1 553358.2 553283.2 553283.2 93 0.13 1 0.11 1325.5k2 553624.1 553479.1 553479.1 215 0.22 1 0.15 1450.5k1 1106620.2 1106524.4 1106524.4 653 0.78 1 0.19 1550.5k2 1106691.7 1106589.4 1106575.5 1269 1.53 7 0.57 30100.5k1 1911261.1 1911164.0 1911164.0 447 0.95 1 0.23 15100.5k2 1108425.9 1108272.7 1108187.4 +100000 ****** 5323 35.63 55

Table 1: Mixed Integer Knapsacks-Cut and Branch with MIPO

The second set are standard mixed integer problems from MIPLIB [3]that have been preprocessed by MINTO [16]. The results with MIPO areagain obtained running the cut routine only at the top node. Comparativeresults using flow cover separation routines can be found in [10].

7 Comments

We have shown that the continuous knapsack model Y is somewhat richer instructure than the pure knapsack model, and that it also allows us to deriveinequalities for the more general flow model Z studied again recently in [9].In the example treated in detail most of the inequalities found belong tomore than one family, but it is clear that in general each of the five familiescontributes new facet-defining inequalities.

The inequalities we have derived are very basic in that we have onlyexamined sequential lifting in which the continuous variable is essentiallyintroduced last or first. The great advantage of the superadditivity is thatone avoids solving knapsack problems to calculate each lifting coefficient.Using other sequential lifting orders and non-sequential lifting are also the-oretically possible, but are computationally more complicated.

In practice we have shown that four classes of inequalities can be sep-arated and used efficiently, and for mixed 0-1 rows such as the flow model

20

instance lp xlp ip nb cut CB node CB timeegout 511.62 567.29 568.1 35 5 0.35

fixnet3 5413.77 51973 51973 156 1 6.34fixnet4 7703.48 8675.61 8936 350 67 33.04fixnet6 3192.04 3624.35 3983 299 1661 131.28fxch.3 152.01 187.89 197.98 408 667 22.43

mod013 256.02 270.57 280.95 52 183 2.07modglob 20430947.6 20681347.1 20740508.1 361 1789 61.73khb05250 95919464 106741250.4 106940226 205 31 6.81

rgn 48.80 68.00 82.20 72 2337 19.29set1al 11651.63 15867.25 15869.75 400 35 8.66set1ch 35118.11 40680.22 54537.75 306 +5000 ****set1cl 2183.57 6484.25 6484.25 400 1 5.54

utrans.1 158.66 177.86 183.337 93 111 2.35utrans.2 207.74 233.68 239.217 116 155 3.93utrans.3 388.31 416.39 432.285 135 645 13.15

vpm1 16.43 19.50 20 96 243 5.94

Table 2: MIPLIB problems-Cut and Branch with MIPO

Z, the use of the continuous knapsack model combined with appropriatechoices of substitutions gives results comparable with those obtained withflow cover inequality separation routines.

To extend the results on facet-defining inequalities from Y to the moregeneral mixed 0-1 set K, we can first, by complementing variables if neces-sary, assume that K is in canonical form in which for all variables xj withgj > 0, uj = ∞. Now two natural sets to consider are Y + = {(y, s, s+) ∈Bn × R1

+ × R1+ :

∑j∈N ajyj + s+ ≤ b+ s} and Y (u) = Y ∩ {(y, s) : s ≤ u}.

It is easily seen that all tight valid inequalities for Y + have a zero coef-ficient for s+, and so the nontrivial facet-defining inequalities of Y + andY are the same. This indicates that we can essentially ignore terms gjxjwith gj > 0 in the description of K. Y (u) is more interesting. LettingY0(u) denote the knapsack set {y ∈ Bn :

∑j∈N ajyj ≤ b + u}, it follows

from the observation that {y ∈ Rn : ∃s with (y, s) ∈ Y (u)} = Y0(u) thatthe facet-defining inequalities of conv(Y (u)) in which s has a coefficient ofzero are exactly the facet-defining inequalities of conv(Y0(u)). It is alsoimmediate that the facet-defining inequalities of conv(Y ) in which s has anon-zero coefficient are valid, but not necessarily facet-defining, for Y (u).

21

Now if P = {j ∈ {1, . . . , p} : gj < 0} and P1 ⊆ P , all facets of K ofthe form

∑nj=1 πjyj ≤ π0 + α[

∑j∈P1

(−gjxj)] can be obtained by examiningrelaxations of K of the form

K = {(y, s′) ∈ Bn ×R1+ :

n∑j=1

ajyj ≤ b′ + s′, s′ ≤ u′}

where b′ = [b+∑j∈P\P1

(−gj)uj ], s′ =∑j∈P1

(−gj)xj and u′ =∑j∈P1

(−gj)ujwhich is precisely of the form Y (u).

The approach of projecting and lifting can also be used to obtain strongvalid inequalities for 0-1 knapsacks with GUBs, and integer knapsacks, andalso allows one to generate flow cover inequalities for the correspondingsingle node generalizations of these models.

Acknowledgement. The authors are grateful to Sebastian Ceria and YvesPochet for their helpful comments.

22

References

[1] E. Balas, Facets of the Knapsack Polytope, Mathematical Programming8, 146-164 (1975)

[2] E. Balas, S. Ceria and G. Cornuejols, Mixed 0–1 Programming by Lift-and-Project in a Branch-and-cut Framework, Management Science 42,1229-1246 (1996).

[3] Bixby R.E., E.A. Boyd and R.R. Indovina MIPLIB ”A test set of MixedInteger Programming Problems”, SIAM News, March (1992), 16.

[4] S. Ceria, C. Cordier, H. Marchand and L.A. Wolsey, Cutting Planesfor Integer Programs with General Integer Variables, CORE DP9575,Universite Catholique de Louvain, Louvain-la-Neuve (1995).

[5] C. Cordier, H. Marchand, R. Laundy and L.A. Wolsey, bc − opt: abranch-and-cut code for mixed integer programs, CORE DP9778, Uni-versite Catholique de Louvain, Louvain-la-Neuve (1997).

[6] H. Crowder, E.L. Johnson and M.W. Padberg, Solving Large ScaleZero-One Linear Programming Problems, Operations Research 31, 803-834 (1983).

[7] K. Fukuda and A. Prodon, Double Description Method Revisited, in’Combinatorics and Computer Science’, Lecture Notes in ComputerScience no 1120, Springer Verlag, 1996.

[8] R. E. Gomory, Solving Linear Programming Problems in Integers, inCombinatorial Analysis, R. E. Bellman and M. Hall, Jr., eds., AmericanMathematical Society, (1960), pp. 211-216.

[9] Z.Gu, G.L. Nemhauser and M.W.P. Savelsbergh, Sequence IndependentLifting, LEC-95-08, Georgia Institute of Technology (1995).

[10] Z.Gu, G.L. Nemhauser and M.W.P. Savelsbergh, Lifted Flow CoverInequalities for Mixed 0-1 Integer Programs, LEC-96-05, Georgia Insti-tute of Technology (1996).

[11] P.L. Hammer, E.L. Johnson and U.N. Peled, Facets of Regular 0-1Polytopes, Mathematical Programming 8, 179-206 (1975).

23

[12] H. Marchand and L.A. Wolsey, The 0-1 Knapsack Problem with a Sin-gle Continuous Variable, CORE DP9720, Universite Catholique de Lou-vain, Louvain-la-Neuve, March 1997.

[13] G.L. Nemhauser and L.A. Wolsey, Integer and Combinatorial Opti-mization, Wiley (1988).

[14] G.L. Nemhauser and L.A. Wolsey, A Recursive Procedure for Gener-ating all Cuts for 0-1 Mixed Integer Programs, Mathematical Program-ming 46, 379-390 (1990).

[15] M.W. Padberg, T.J. Van Roy and L.A. Wolsey, Valid Linear Inequal-ities for Fixed Charge Problems, Operations Research 33, 842-861(1985).

[16] M.W.P Savelsbergh and G.L. Nemhauser, Functional description ofMINTO, a Mixed INTeger Optimizer, Report COC-91-03A, GeorgiaInstitute of Technology, Atlanta, Georgia (1993).

[17] J. Stallaert, On the Polyhedral Structure of Capacited Fixed ChargeNetwork Problems, Ph.D. Thesis Univ. of California (1992).

[18] T.J. Van Roy and L.A. Wolsey, Valid Inequalities for Mixed 0-1 Pro-grams, Discrete Applied Mathematics 4, 199-213 (1986).

[19] T.J. Van Roy and L.A. Wolsey, Solving Mixed 0-1 Problems by Auto-matic Reformulation, Operations Research 35, 45-57 (1987).

[20] R. Weismantel, On the 0-1 Knapsack Polytope, Mathematical Program-ming 77, 49-68 (1997).

[21] L.A. Wolsey, Faces for a Linear Inequality in 0-1 Variables, Mathemat-ical Programming 8, 165-178 (1975).

[22] L.A. Wolsey, Valid Inequalities and Superadditivity for 0/1 Integer Pro-grams, Mathematics of Operations Research 2, 66-77 (1977).

[23] X-G Yao and E.A. Boyd, Cutting Planes for Mixed-Integer KnapsackPolyhedra, Texas A&M University, April 1995.

24

Appendix

100.5k1-----------

cx 4270 3527 4754 1150 2073cy 51751 51860 55705 56130 53085 57530 51747 56499 56534

56514 55599 57267 53441 52824 52672 52773 53033 5942550290 55967 55577 55756 50694 58317 53499 58089 5389950908 58589 56702 56630 57328 51454 58707 59472 5012158891 52142 56351 55054 56422 55533 59803 51223 5290757942 53226 52646 53288 59609 51551 53452 53773 5842656584 51711 56555 51163 54787 54486 49720 53091 5717157935 53135 56948 56521 50688 50654 54104 53171 5929952633 56701 59115 54617 52092 56615 51114 52080 5593056961 52813 52496 54802 56010 50518 58912 51143 5872459619 54345 53613 59256 51273 50315 59290 56079 4987951539

ax 4.282 3.499 4.736 1.162 2.092ay 51.862 52.307 55.521 55.920 52.906 57.631 52.035 56.296

56.773 56.183 55.170 57.323 53.624 52.726 52.779 52.80052.667 59.279 50.029 55.826 55.283 55.734 50.851 57.97453.161 58.544 53.521 50.833 58.209 57.046 56.198 57.65251.393 58.261 59.648 50.303 58.835 52.063 55.984 54.78256.374 55.483 59.933 51.273 52.602 57.496 53.083 52.23653.757 59.816 52.034 53.433 53.895 58.387 56.986 51.74756.685 51.136 54.308 54.813 50.117 52.935 57.668 57.80253.254 57.260 56.207 50.269 50.576 54.249 52.956 59.23652.314 56.816 59.384 54.674 51.793 56.606 51.520 52.08055.998 56.952 52.529 52.291 54.648 56.020 50.723 59.17651.456 58.270 59.751 54.065 53.758 59.635 51.451 50.53458.832 56.197 50.044 51.100

<= 1900.000

ux 4.888 3.695 3.180 3.053 4.620

25

100.5k2

cx 3757 4142 2151 3100 591cy 55248 54343 60328 55011 55107 59119 52853 57237 53645

58842 57345 52098 59247 52568 56959 55957 59319 5254053960 56709 50493 56986 56692 53381 54780 57353 5379552920 50598 59702 57115 51625 59410 51300 57719 5838757958 54085 56325 56071 53622 55972 53967 50787 5520953946 60321 58904 56785 51960 59582 56125 51415 5271758034 56988 50510 58293 59050 51810 53909 55773 5677057249 58073 57573 53767 53591 51522 53810 59026 5358159199 55797 56781 55561 58979 56813 50170 53779 5907157574 57288 58368 53504 53619 51367 54393 58928 5965952932 56499 52616 56985 59852 51427 59172 57323 5366356180

ax 3.737 4.172 2.183 3.107 0.587ay 54.996 54.840 59.894 54.643 54.840 59.453 52.711 56.891

53.485 58.595 57.820 52.204 58.753 52.071 57.105 56.30159.077 52.856 54.340 56.979 50.422 57.370 56.983 53.27154.478 57.294 53.531 52.518 50.575 59.206 57.374 52.05458.996 50.952 57.706 58.183 57.792 54.416 56.724 55.67153.699 56.264 53.570 50.544 55.250 53.933 59.858 58.46256.981 51.958 59.680 55.736 51.872 52.767 58.008 56.51950.374 57.892 58.801 52.151 53.966 55.691 56.852 57.62258.033 57.254 53.718 53.250 51.788 54.209 58.834 53.72459.269 55.503 56.558 55.948 59.147 57.270 50.375 53.74459.383 57.838 57.703 58.437 53.357 53.199 50.996 54.57458.751 59.863 53.291 56.350 52.224 56.523 59.932 51.60058.695 57.510 53.420 55.887

<= 1100.000

ux 2.129 3.412 3.012 4.893 3.517

26