Transcript CS6234: Lecture 4  Linear Programming  LP and Simplex Algorithm [PS82]-Ch2  Duality [PS82]-Ch3  Primal-Dual Algorithm [PS82]-Ch5  Additional topics: Reading/Presentation by students Lecture notes adapted from.

CS6234: Lecture 4
 Linear Programming
 LP and Simplex Algorithm [PS82]-Ch2
 Duality
[PS82]-Ch3
 Primal-Dual Algorithm
[PS82]-Ch5
 Additional topics:
Reading/Presentation by students
Lecture notes adapted from Comb Opt course by
Jorisk, Math Dept, Maashricht Univ,
Hon Wai Leong, NUS
(CS6234, Spring 2009) Page 1
Copyright © 2009 by Leong Hon Wai
Combinatorial Optimization
Chapter 5 [PS82]
The primal dual algorithm
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Combinatorial Optimization
Masters OR
Primal Dual Method
found x, succeed!
primal
restricted
primal
dual
restricted
dual
y
z
Construct a better dual
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Combinatorial Optimization
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Source: CUHK, CS5160
LP in standard form + Dual
min
s.t.
c’ x
Ax
x
=
≥
b (≥ 0 )
0
(P)
Primal
max
s.t.
π’ b
π’ A ≤
c
π’
free
Dual
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(D)
Definition of the Dual
Definition 3.1: Given an LP in general form, called
the primal, the dual is defined as follows
Primal
Dual
Min c’x
Max π’b
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a’i x = bi
iεM
πi free
xj ≥ 0
jεN
π’ Ai ≤ ci
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Complementary slackness
Theorem 3.4 A pair, x, π, respectively
feasible in a primal-dual pair is optimal if
and only if:
ui = πi (a’ix - bi) = 0 for all i
vj = (cj - π’Aj) xj = 0 for all j.
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(1)
(2)
Idea of a Primal-Dual Algorithm
Suppose we have a dual feasible solution π.
If we can find a primal feasible solution x such that
xj = 0 whenever cj – π’Aj > 0 then:
Ai x = bi for all i and hence
(1) holds because x is a feasible solution to the primal
(2) holds because xj = 0 whenever cj – π’Aj = 0.
Thus the complementary slackness relations hold, and
hence x and π are optimal solutions to the primal
problem (P) and the dual problem (D) resp.
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Outline of the primal dual
algorithm
P
x
D
RP
ρ
π
Adjustment to π
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DRP
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Primal Dual Method
found x, succeed!
primal
restricted
primal
dual
restricted
dual
y
z
Construct a better dual
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Source: CUHK, CS5160
Getting started: finding a dual feasible π
If c ≥ 0, π = 0 is a dual feasible solution.
When cj < 0 for some j, introduce variable xm+1,
and the additional constraint
x1 + x2 + … + xm+1 = bm+1.
(bm+1 large enough)
The dual (D) then becomes
max
s.t.
π’ b + πm+1 bm+1
π’ A + πm+1
πm+1
≤ cj
≤ 0.
for all j,
The solution πi = 0, i=1…m, and
πm+1 = minj cj < 0 is feasible for the dual (D).
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Given a dual feasible solution π
Thus, we assume we have a dual feasible solution π.
Consider the set
J = { j : π’j Aj = cj }.
A solution x to the primal (P) is optimal iff
xj=0 for all j not in J.
Hence, we aim for an x feasible in
Σj=1n aij xj
= bi
i =1,..,m
xj
≥ 0,
for all j in J
xj
= 0,
for all j not in J
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Restricted Primal (RP)
Min ξ = Σj=1m xja
Σj=1n aijxj +
xj
xj
xja
xja
= bi
≥ 0,
= 0,
≥ 0,
i =1,..,m
for all j in J
for all j not in J
i =1,..,m
(RP)
If ξopt = 0, the corresponding optimal solution to (RP) yields an
x which satisfies (together with π) the complementary
slackness conditions.
Thus, in the remainder we consider the case where ξopt > 0.
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The dual of the restricted primal:
Min
Σj=1m xja
s.t. ΣjεJ aij xj +
xja
xj
xj
xja
Max
s.t.
π’ b
π’ Aj
πi
πi
≤0
≤ 1,
free
= bi
≥ 0,
= 0,
≥ 0,
i =1,..,m
for all j in J
for all j not in J
i =1,..,m
for all j in J
i=1…m,
i=1…m,
We denote by ρ the optimal solution of this (DRP).
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DRP
(RP)
Comparison of (D) and (DRP)
(D)
Max
s.t.
(DRP)
π’ b
π’ Aj
≤ cj
j=1…n
πi
free
i=1…m.
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Max
s.t.
π’ b
π’ Aj
πi
πi
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≤0
≤ 1,
free
for all j in J
i=1…m,
i=1…m.
A Primal Dual iteration….
Let θ ε R, and consider π* = π + θρ.
Assume θ is such that π* is feasible in (D). Then π*’
b = π’b+ θρ’b.
Obviously ρ’b = ξopt > 0, since otherwise x together
with π satisfies the complementary slackness
conditions. Thus, by choosing θ > 0, π*’ b > π’b,
yielding an improved solution of the dual (D).
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A Primal Dual iteration
For π* to be feasible in (D), it must hold that:
π*’Aj = π’Aj + θρ’Aj ≤ cj , j=1…n .
From the definition of (DRP) it holds that
ρ’Aj ≤ 0 for all j in J. Thus if ρ’Aj ≤ 0 for all j, we
can choose θ = +∞.But then (D) is unbounded,
and hence (P) is infeasible. (Theorem 5.1)
Thus, we assume that ρ’Aj > 0 for some j not in J.
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A Primal Dual iteration
We are going to allow j to ‘enter the basis’ of (D)…
Let θ1 to be the maximum value such that
π’ Aj+ θρ’Aj ≤ cj for all j not in J and ρ’Aj > 0
Dual constraint j is satisfied at equality, and then by
complementary slackness, primal variable xj can have
nonnegative value
Define π* = π + θ1 ρ. Then π*’b = π’ b+ θ1 ρ’ b > π*’b.
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The Primal Dual Algorithm
Input: Feasible solution π for (D).
Output: Optimal solution x for (P) if it exists.
Infeasible  false; opt  false;
While not (infeasible or opt) begin
J  {j: π’ Aj =cj }
Solve (RP) giving solution x.
If ξopt = 0
then opt  true
else if ρ’A ≤ 0
then infeasible  true
else π π + θ1 ρ
End {while}
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Admissible columns
Definition Given a solution π to the dual
(D), let J = { j : π’ Aj =cj }. Then any column
Aj , jεJ is called an admissable columns.
Theorem 5.3. A column which is in the
optimal basis of (RP) and admissable in
some iteration of the Primal Dual algorithm
remains admissable at the start of the next
iteration.
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Admissable columns
Proof. If Aj is in the optimal basis of (RP) then, by
complementary slackness dj - ρ’ Aj = 0, where dj
is the cost coefficient in the objective function of
the (RP). Hence dj = 0, and thus ρ’ Aj = 0.
This in turn implies that
π*’ Aj =
π’ Aj+ θρ’Aj
=
π’ Aj
=
cj
Since j was an admissable column. Thus j remains
an admissable column.
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Consequence
In each iteration, the optimal solution x of (RP), is
also a basic feasible solution of the (RP) in the
next iteration.
Thus subsequent (RP)’s can be solved taking the
optimal solution of the previous one as a starting
point for a simplex iteration. Moreover, using an
anti cycling rule, the Primal Dual algorithm is
finite (see Theorem 5.4).
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A Primal Dual method for Shortest
Path
node-arc incidence matrix A:
1
if arc k leaves node i
aij = {
-1
if arc k enters node i
0
otherwise
Primal (P):
min
s.t.
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c’ f
Σk Ak fk
Σk Ak fk
f
=0
=1
≥0
for all i ≠ s,t.
i = s.
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Dual (D) of the shortest path
problem
max πs
s.t. πi π
πt
πj
≤ cij,
free,
= 0.
Admissable arcs J = { arcs (i,j) : πi - πj = cij }.
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Restricted Primal (RP) of Shortest
Path
min
Σi x i a
s.t.
ΣkεJ
Ak fk + xia
Σ{j:(s,j)εJ} A(s,j)f(s,j) + xsa
fk
fk
x ia
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=0
=1
≥0
=0
≥0
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for all i ≠ s,t.
for all k,
for all k not in J,
for all i.
Dual of the Restricted Primal
(DRP)
max πs
s.t. πi πi
πt
π
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πj
≤ 0,
≤ 1,
= 0,
free,
for all arcs (i,j) in J.
for all i.
Combinatorial Optimization
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Solving (DRP)
Obviously πs ≤ 1, and since we aim to
maximize πs we try πs = 1. But then all
nodes i reachable by an admissable arc
from s must also have πi = 1. This
argument applies recursively.
Similarly since πt = 0, all nodes i
(recursively) reachable from the sink, must
have πi = 0.
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Optimal solution ρ to (DRP)
t
s
1
1
0
If the source can be reached from the sink by a path
consisting of admissable arcs πs = 1 end hence (RP) has
optimal value zero and we are done.
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Solving (DRP)
Let θ1 to be the maximum value such that
π’ Ak+ θρ’Ak≤ ck for all k not in J and ρ’Ak > 0.
Thus we must consider (i,j) such that ρi -ρj =1.
Therefore θ1 is the minimum over the
aforementioned (i,j) of cij – (πi - πj ).
Interpretation on next slide…
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Interpretation of Primal Dual
iteration
We assume cij > 0 for i,j = 1…m.
Then the solution πi = 0 for i=1…m is
feasible in (D), and selected as the starting
solution.
We select a non admissable arc from a
green or red node i (which has ρi = 1 to a
yellow node j, which has ρi = 0. Since arc
(i,j) is non admissable it must hold that
πi – πj < cij.
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Example
3
1
3
2
2
t
3
s
2
5
1
2
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1
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4
Initial dual feasible solution
π1=0
3
1
π3=0
3
2
2
πt=0
t
πs=0
3
s
2
5
1
2
1
π2=0
Admissable arcs: Ø
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4
π4=0
Initial solution of (RP)
xsa=1, xia=0, for all i ≠ s,
no variables fk for admissable columns Ak.
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First DRP
ρ1=1
3
1
ρ3=1
3
2
2
ρt=0
t
ρs=1
3
s
2
5
1
2
1
ρ2=1
θ1 = c3t + πt - π3 = 2
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4
ρ4=1
Next solution of (RP)
xsa=1, xia=0, for all i ≠ s,
variables fk for admissable columns Ak from
node 3 to t, but primal variables remain
unchanged. (and will remain unchanged,
until some arc leaving s is admissable…
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Next dual feasible solution
π1=2
3
1
π3=2
3
2
2
πt=0
t
πs=2
3
s
2
5
1
2
1
π2=2
Admissable arcs: (3,t)
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4
π4=2
Next DRP
ρ1=1
3
1
ρ3=0
3
2
2
ρt=0
t
ρs=1
3
s
2
5
1
2
1
ρ2=1
θ1 = c43 + π3 - π4 = 2
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4
ρ4=1
Next dual feasible solution
π1=4
3
1
π3=2
3
2
2
πt=0
t
πs=4
3
s
2
5
1
2
1
π2=4
Admissable arcs: (3,t), (4,3)
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4
π4=4
Next DRP
ρ1=1
3
1
ρ3=0
3
2
2
ρt=0
t
ρs=1
3
s
2
5
1
2
1
ρ2=1
4
ρ4=0
θ1 = c24 + π4 - π2 = 1 = c13 + π3 - π1 = 1
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Next dual feasible solution
π1=5
3
1
π3=2
3
2
2
πt=0
t
πs=5
3
s
2
5
1
2
1
π2=5
Admissable arcs: (3,t), (4,3), (1,3), (2,4)
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4
π4=4
Next DRP
ρ1=0
3
1
ρ3=0
3
2
2
ρt=0
t
ρs=1
3
s
2
5
1
2
1
ρ2=0
θ1 = cs2 + π2 - πs = 1
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4
ρ4=0
Next dual feasible solution
π1=5
3
1
π3=2
3
2
2
πt=0
t
πs=6
3
s
2
5
1
2
1
π2=5
Admissable arcs: (3,t), (4,3), (1,3), (2,4),(s,2)
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4
π4=4
Next solution of (RP)
xia=0, for all i,
fs2=1, f24=1, f43=1, f3t=1, all other fij = 0.
Solution value 0  f is feasible in primal (P),
and since c ’x = π’b optimal in (P)!
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Combinatorialization…
(P) has an integer cost vector c, and (0,1) right hand side. (RP)
has a (0,1) cost vector and right hand side b. Therefore in (RP)
the complexity is not from the numbers but from the
combinatorics.
Similarly (D) has a (0,1) vector b, and integer right hand side c.
(DRP) has a (0,1) cost vector and right hand side. Therefore in
(DRP) the complexity is not from the numbers but from the
combinatorics.
The primal dual algorithm solves a problem with ‘numerical’
complexity, by repeatedly solving a problem which has only
‘combinatorial’ complexity.
This concept, which Papadimitriou calls ‘combinatorialization’ is
frequently encountered in combinatorial optimization.
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Additional notes from [CUHK]
Hon Wai Leong, NUS
(CS6234, Spring 2009) Page 44
Copyright © 2009 by Leong Hon Wai
Primal Dual Method
found x, succeed!
primal
restricted
primal
dual
restricted
dual
y
z
Construct a better dual
Lecture 20: March 28
Primal Dual Program
Primal Program
Dual Program
If there is a feasible primal solution x and a feasible dual solution y,
then both are optimal solutions.
Primal-Dual Method: An algorithm to construct such a pair of solutions.
Optimality Condition
Suppose there is a feasible primal solution x, and a feasible dual solution y.
How do we check that they are optimal solutions?
Avoid strict inequality
Optimality Condition
Avoid strict inequality
Complementary Slackness Conditions
Primal complementary slackness condition:
Dual complementary slackness condition:
Primal Dual Method
Start from a feasible dual solution.
Search for a feasible primal solution
satisfying complementary slackness conditions.
repeat
If not, improve the objective value of dual solution.
Restricted Primal
Given a feasible dual solution y, how do we search for a feasible
primal solution x that satisfies complementary slackness conditions?
Formulate this as an LP itself!
Primal complementary slackness condition:
Dual complementary slackness condition:
Restricted Primal
If j not in J, then we need x(j) to be zero.
If i not in I, then we need
If zero, we are done
Restricted primal
Restricted Dual
Suppose the objective of the restricted primal is not zero, what do we do?
Then we want to find a better dual solution.
Restricted primal
nonzero
Restricted dual
Restricted Dual
Consider y+єz as the new dual solution.
Dual Program
nonzero
Larger value
Still feasible
Restricted dual
General Framework
found x, succeed!
primal
restricted
primal
dual
y
restricted
dual
z
Construct a better dual
Bipartite Matching
Primal complementary slackness condition:
Hungarian Method
Start from a feasible vertex cover
Find a perfect matching using tight edges
non-zero
Consider y- єz, a better dual
Remarks
It is not a polynomial time method.
It reduces the weighted problem to the unweighted problem,
so that the restricted primal linear program is easier to solve,
and often there are combinatorial algorithms to solve it.
Many combinatorial algorithms, like max-flow, matching,
min-cost flow, shortest path, spanning tree, …, can be derived
within this framework.
Approximation Algorithm
How do we adapt the primal-dual method for approximation algorithms?
We want to construct a primal feasible solution x and
a dual feasible solution y so that cx and by are “close”.
Avoid losing too much
Approximate Optimality Condition
Avoid strict inequality
Approximate Complementary Slackness Condition
Primal complementary slackness condition:
Dual complementary slackness condition:
Only a sufficient condition
Vertex Cover
Primal complementary slackness condition:
Dual complementary slackness condition:
Approximate Optimality Conditions
Primal complementary slackness condition:
Just focus on this!
Pick only vertices that go tight.
Dual complementary slackness condition:
Pick only edge with one vertex in the vertex cover.
This is nothing!
This would imply a 2-approximation.
Algorithm
Pick only vertices that go tight.
Algorithm (2-approximation for vertex cover)
Initially, x=0, y=0
When there is an uncovered edge
Pick an uncovered edge, and raise y(e) until some vertices go tight.
Add all tight vertices to the vertex cover.
Output the vertex cover x.
Familiar?
This is the greedy matching 2-approximation
when every vertex has the same cost.
Thank you.
Q &A
Hon Wai Leong, NUS
(CS6234, Spring 2009) Page 65
Copyright © 2009 by Leong Hon Wai