Class Bfs 121109084059495
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Transcript of Class Bfs 121109084059495
Lecture 6: Cutting Plane Methodand Strong Valid Inequalities
(3 units)
Outline
I Valid inequality
I Gomory’s cutting plane method
I Polyhedra, faces and facets
I Cover inequalities for 0-1 knapsack set
I Lifting and separation of cover inequalities
I Branch-and-cut algorithm
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Valid Inequality
I Let X = {x | Ax ≤ b, x ∈ Zn+}. Consider the following linear
program:
(CIP) min cT x
s.t. x ∈ conv(X ).
Then (IP) is equivalent to (CIP).
I Question: How to describe or approximateconv(X ) = {x | Ax ≤ b, x ≥ 0}?
I An inequality πT x ≤ π0 is said to be a valid inequality for Xif πT x ≤ π0 for all x ∈ X .
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I Question:(1) How to find good or useful valid inequality?(2) How to use valid inequality to solve an integer program?
I Chvatal-Gomory procedure for VI. X = {x ∈ Zn+ | Ax ≤ b},
A = (a1, . . . , an).(i) (surrogate)
∑nj=1 µTajxj ≤ µTb, µ ≥ 0.
(ii) (round off)∑n
j=1bµTajcxj ≤ µTb.
(iii)∑n
j=1bµTajcxj ≤ bµTbc.I Theorem: every valid inequality of X can be obtained by
applying Chvatal-Gomory procedure for a finite number oftimes.
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Gomory’s Cutting Plane Procedure for IP
I Integer programming:
min{cT x | x ∈ Zn+,Ax = b}.
I Create the valid inequalities (cutting planes) directly from thesimplex tableau
I Given an (optimal) LP basis B, write the (pure) IP as
min cTB B−1b +
∑
j∈NB
cjxj
s.t. (xB)i +∑
j∈NB
aijxj = bi , i = 1, 2, ...m,
xj ∈ Z1+, j = 1, 2, ...n,
NB is the set of nonbasic variables.
I cj ≥ 0, j ∈ NB, bi ≥ 0, i = 1, . . . ,m.
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I If the LP solution is not integral, then there exists some row iwith bi 6∈ Z.
I The C-G cut for row i is
(xB)i +∑
j∈NB
baijcxj ≤ bbic.
I Substitute for (xB)i to get
∑
j∈NB
(aij − baijc)xj ≥ bi − bbic.
I Let fij = aij − baijc, fj = bi − bbic.∑
j∈NB
fijxj ≥ fi .
I Since the LP optimal solution x∗j = 0 for j ∈ NB, and0 ≤ fij < 1, 0 < fi < 1, this inequality cut off x∗!
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I Example 1
min−5x1 − 8x2
s.t. x1 + x2 ≤ 6
5x1 + 9x2 ≤ 45
x1, x2 ∈ Z+
I The feasible region is
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I Optimal Simplex Tableau:
x1 x2 x3 x4 b
1 0 2.25 −0.25 2.250 1 −1.25 0.25 3.750 0 1.25 0.75 41.25
I The Gomory cut from the second row of the tableau is:
0.75x3 + 0.25x4 ≥ 0.75
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I The modified feasible set is
I The fractional optimal solution x = (2.35, 3.75) violates thecutting plane and is removed from the new feasible set.
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I Example 2
min−4x1 + x2
s.t. 7x1 − 2x2 ≤ 14
x2 ≤ 3
2x1 − 2x2 ≤ 3
x1, x2 ∈ Z+
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I Optimal Simplex Tableau:
x1 x2 x3 x4 x5 b
1 0 17
27 0 20
70 1 0 1 0 30 0 −2
7107 1 23
7
0 0 47
17 0 −59
7
Cut from the first row of the tableau is:
1
7x3 +
2
7x4 ≥ 6
7.
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I Reoptimization:
x1 x2 x3 x4 x5 x6 b
1 0 0 0 0 1 20 1 0 0 −1
2 1 12
0 0 1 0 −1 −5 10 0 0 1 1
2 6 52
0 0 0 0 12 3 −15
2
Cut from the second row of the tableau is:
1
2x5 ≥ 1
2.
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I Reoptimization:
x1 x2 x3 x4 x5 x6 x7 b
1 0 0 0 0 1 0 20 1 0 0 0 1 −1 10 0 1 0 0 −5 −2 20 0 0 1 0 6 1 20 0 0 0 1 0 −1 1
0 0 0 0 0 3 1 −7
Done! Optimal solution x∗ = (2, 1)T .
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Linear algebra and convex analysis review
I Affine independence: A finite collection of vectors x1, . . . ,xk ∈ Rn is affinely independent if the unique solution to
k∑
i=1
αixi = 0,
k∑
i=1
αi = 0,
is αi = 0, i = 1, 2, ..., k.
I Linear independence implies affine independence, but not viceversa.
I The following statements are equivalent:
1. x1, . . . , xk ∈ Rn are affinely independent.2. x2 − x1, . . . , xk − x1 are linearly independent.3. (x1, 1), . . . , (xk , 1) ∈ Rn+1 are linearly independent.
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I A polyhedron is a set of the form{x ∈ Rn | Ax ≤ b} = {x ∈ Rn | aix ≤ bi , i ∈ M}, whereA ∈ Rm@n and b ∈ Rm.
I Dimension of Polyhedra: A polyhedron P is of dimension k,denoted dim(P) = k, if the maximum number of affinelyindependent points in P is k + 1. A polyhedron P ∈ Rn isfull-dimensional if dim(P) = n. LetM = {1, ...,m},M= = {i ∈ M | aix = bi ,∀x ∈ P}, (the equality set),M≤ = M \M= (the inequality set).Let (A=, b=), (A≤, b≤) be the corresponding rows of (A, b).If P ∈ Rn, then dim(P) + rank(A=, b=) = n.
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Valid Inequalities and Faces
I The inequality denoted by (π, π0) is called a valid inequalityfor P if πx ≤ π0,∀x ∈ P.
I Note that (π, π0) is a valid inequality if and only if P lies inthe half-space {x ∈ Rn | πx ≤ π0}.
I If (π, π0) is a valid inequality for P andF = {x ∈ P | πx = π0}, F is called a face of P and we saythat (π, π0) represents or defines F .
I A face is said to be proper if F 6= ∅ and F 6= P. Note that aface has multiple representations.
I The face represented by (π, π0) is nonempty if and only ifmax{πx | x ∈ P} = π0.
I If the face F is nonempty, we say it supports P. Note that theset of optimal solutions to an LP is always a face of thefeasible region.
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Facets
I Let P be a polyhedron with equality set M=. IfF = {x ∈ P | πx = π0} is nonempty, then F is a polyhedron.We can get the polyhedron F by taking some of theinequalities of P and making them equalities.
I The number of distinct faces of P is finite.
I A face F is said to be a facet of P if dim(F ) = dim(P)− 1.The inequality corresponding to a facet is called a strong validinequality.
I If F is a facet of P, then in any description of P, there existssome inequality representing F . (By setting the inequality toequality, we get F ).
I Every inequality that represents a face that is not a facet isunnecessary in the description of P.
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An example
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0-1 Knapsack Inequalities
I Valid Inequalities for the Knapsack Problem. We areinterested in valid inequalities for the knapsack set
S = {x ∈ {0, 1}n |N∑
j=1
ajxj ≤ b}.
N = {1, 2, . . . , .n}. Assume that aj > 0, j ∈ N, aj < b, j ∈ N.We are interested in finding facets of conv(S).
I Simple facets. What is dim(conv(S))? 0, ej , j ∈ N are n + 1affinely independent points in conv(S), so dim(conv(S)) = n.
I xk ≥ 0 is a facet of conv(S).
I Proof. 0, ej , j ∈ N \ {k} are n affinely independent pointsthat satisfy xk = 0.
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I xk ≤ 1 is a facet of conv(S) if aj + ak ≤ b, ∀j ∈ N \ {k}.I Proof. ek , ej + ek , j ∈ N \ {k} are n affinely independent
points that satisfy xk = 1.
I A set C ⊆ N is a cover if∑
j∈C aj > b. A cover C is aminimal cover if C \ {j} is not a cover ∀j ∈ C .
I If C ⊆ N is a cover, then the cover inequality
∑
j∈C
xj ≤ |C | − 1
is a valid inequality for S .
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I Example:
S = {x ∈ B7 | 11x1+6x2+6x3+5x4+5x5+4x6+x7 ≤ 19}.I Minimal covers:
C = {1, 2, 3},C = {1, 2, 6},C = {1, 5, 6},C = {3, 4, 5, 6}.
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Can we do better?
I Are these inequalities the strongest ones we can come upwith? What does strongest mean? We all know that facetsare the “strongest”, but can we say anything else?
I If πx ≤ π0 and µx ≤ µ0 are two valid inequalities for P ∈ Rn+,
we say that πx ≤ π0 dominates πx ≤ µ0 if ∃u ≥ 0 such thatπ ≥ uµ, π0 ≤ uµ0 and (π, π0) 6= u(µ, µ0).
I If πx ≤ π0 dominates πx ≤ µ0, then
{x ∈ Rn+ | πx ≤ π0} ⊆ {x ∈ Rn
+ | µx ≤ µ0}.
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I Strengthening cover inequalities. If C ⊆ N is a minimal cover,the extended cover E (C ) is defined asE (C ) = C ∪ {j ∈ N | aj ≥ ai ,∀i ∈ C}. If E (C ) is an extendedcover for S , then the extended cover inequality
∑
j∈E(C)
xj ≤ |C | − 1
is a valid inequality for S .
I The cover inequality x3 + x4 + x5 + x6 ≤ 3 is dominated bythe extended cover inequality x1 + x2 + x3 + x4 + x5 + x6 ≤ 3.
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I Let C be a minimal cover. If C = N, then
∑
j∈C
xj ≤ |C | − 1
is a facet of conv(S).
I Proof Rk = C \ {k}, ∀k ∈ C . xRk satisfies
∑
j∈C
xRkj = |C | − 1.
Also, xR1 , . . . , xR|C | are affinely independent. Since C = N,there are n affinely independent vectors satisfy∑
j∈C xRkj = |C | − 1 at equality.
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I Let C = {j1, . . . , jr} be a minimal cover. Letp = min{j | j ∈ N \ E (C )}. If C = E (C ), and∑
j∈C\{j1} aj + ap ≤ b, then∑
j∈C xj ≤ |C | − 1 is a facet ofconv(S).
I Proof. Tk = C \ {j1} ∪ {k}, ∀k ∈ N \ E (C ).|Tk ∩ E (C )| = |C | − 1 and
∑
j∈Tk∩E(C)
xTkj = |C | − 1.
|Rk |+ |Tk | = N. xRk , k ∈ C , xTk , k ∈ N \ C , are n affinelyindependent vectors.
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I Example:
S =∑
{x ∈ {0, 1}5 | 79x1+53x2+53x3+45x4+45x5 ≤ 178}.
I Consider minimal cover C = {1, 2, 3}. The valid inequality is:
x1 + x2 + x3 ≤ 2.
C = E (C ). p = 4, C \ {1} ∪ {4} = {2, 3, 4}.53 + 53 + 45 = 151 < 178. So x1 + x2 + x3 = 2 gives a facetof conv(S).
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Lifting Cover Inequalities
I Question: Can we find the the a valid inequality as strong aspossible?
I Example: C = {3, 4, 5, 6}, the valid inequality for C is:
x3 + x4 + x5 + x6 ≤ 3.
I Setting x1 = x2 = x7 = 0, the cover inequalitiesx3 + x4 + x5 + x6 ≤ 3 is valid for{x ∈ {0, 1}4 | 6x3 + 5x4 + 5x5 + 4x6 ≤ 19}.
I If x1 is not fixed at 0, can we strengthen the inequality? Forwhat values of α1 is the inequality
α1x1 + x3 + x4 + x5 + x6 ≤ 3
valid forP2,7 = {x ∈ {0, 1}5 | 11x1 + 6x3 + 5x4 + 5x5 + 4x6 ≤ 19}.
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I ⇔ α1 + x3 + x4 + x5 + x6 ≤ 3 is valid for all x ∈ {0, 1}4
satisfying 6x3 + 5x4 + 5x5 + 4x6 ≤ 19− 11;⇔ α1 + ζ ≤ 3, where
ζ = max{x3 + x4 + x5 + x6 | 6x3 + 5x4 + 5x5 + 4x6 ≤ 8}.
I ζ = 1 ⇒ α1 ≤ 2. Thus α1 = 2 gives the strongest inequality.
I How to find the best value αj , j ∈ N \ C such that
∑
j∈C
xj +∑
j∈N\Cαjxj ≤ |C | − 1
is valid for S?
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I Lifting ProcedureI Let j1, . . . , jr be an ordering of N \ C . Set t = 1.I The valid inequality
t−1∑
i=1
αji xji +∑
j∈C
xj ≤ |C | − 1
is given. Solve the following knapsack problem:
ζt = maxt−1∑
i=1
αji xji +∑
j∈C
xj
s.t.t−1∑
i=1
aji xji +∑
j∈C
ajxj ≤ b − ajt
x ∈ {0, 1}|C |+t−1.
I Set αjt = |C | − 1− ζt . Stop if t = r .
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I Example: C = {3, 4, 5, 6}, j1 = 1, j2 = 2, j3 = 7. α1 = 2.Consider x2, we have
ζ2 = max 2x1 + x3 + x4 + x5 + x6
s.t. 11x1 + 6x3 + 5x4 + 5x5 + 4x6 ≤ 19− 6 = 13,
x ∈ {0, 1}5.
So ζ2 = 2 and αj2 = α2 = 3− 2 = 1.Consider x7 now, we have
ζ3 = max 2x1 + x2 + x3 + x4 + x5 + x6
s.t. 11x1 + 6x2 + 6x3 + 5x4 + 5x5 + 4x6 ≤ 19− 1 = 18,
x ∈ {0, 1}6.
So ζ3 = 3 and αj3 = α7 = 3− 3 = 0. We obtain a validinequality:
2x1 + x2 + x3 + x4 + x5 + x6 + x7 ≤ 3.
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Separation of Cover Inequalities
I We often try to solve problems that have knapsack rows withlots more variables than that... Obviously I do not want toadd all of those facets at once. What to do?
I Given some P, find an inequality of the form∑j∈C xj ≤ |C | − 1 such that
∑j∈C x∗j > |C | − 1. This is
called a separation problem.
I Note that∑
j∈C xj ≤ |C | − 1 can be rewritten as
∑
j∈C
(1− xj) ≥ 1.
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I Separation Problem: Given a fractional LP solution x∗, does ∃cover C ⊆ N such that
∑j∈C (1− x∗j ) < 1? or is
γ = minC⊆N
{∑
j∈C
(1− xj) |∑
j∈C
aj > b} < 1?
I If γ ≥ 1, then x∗ satisfies all the cover inequalities.
I If γ < 1 with optimal solution zR , then∑
j∈R xj ≤ |R| − 1 is aviolated cover inequality.
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Branch-and-Cut Method
I Branch and cut is an LP-based branch and bound scheme inwhich the linear programming relaxations combined with bycutting plane method.
I The valid inequalities are generated dynamically usingseparation procedures.
I At each node of the search tree, cuts are generated and usedto improve the LP relaxation.
I Branch-and-cut method is very efficient for some hard integerprogramming problems.
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