6.859J/15.083J Integer Programming and
Combinatorial Optimization
Professors: Dimitris Bertsimas,
Andreas Schulz,
1 Structure of Class
Formulations, complexity and relaxations, Lec. 1-9
Slide 1
Robust Discrete Optimization, Lec. 10-11
Algebra and geometry of IO, Lec. 12-15
Algorithms for IO, Lec. 16-23
Mixed Integer Optimization, Lec. 24-25
2 Requirements
Slide 2
Homeworks: 30%
Midterm Exam: 30%
Final Exam: 40%
Contributions to class: An important tie breaker
Use of CPLEX for solving IO problems
3 Todays Lecture
Slide 3
Modeling with integer variables
What is a good formulation?
Theme: The Power of Formulations
4 Integer Optimization
4.1 Mixed IO
(MIO) max
s.t.
c x+hy
Ax + nBy b
x 2 Z+ (x 0 x integer)
y 2 Rm+ (y 0)
0
0
1
Slide 4
4.2 Pure IO
cx
Ax b
x 2 Z+n
(IO) max
s.t.
0
Slide 5
Important special case: Binary Optimization
(BO) max c x
s.t. Ax b
x 2 f0 1gn
0
4.3 LO
cx
By b
y 2 Rn+
(LO) max
s.t.
0
Slide 6
5 Modeling with Binary Variables
5.1 Binary Choice
if event occurs
x 2 1
0 otherwise
Example 1: IO formulation of the knapsack problem
n : projects, total budget b
aj : cost of project j
cj : value of project j
if project j is selected.
xj = 1
0
otherwise.
n
P
max
cj xj
jP
�1
s.t.
aj xj b
xj 2 f0 1g
Slide 7
Slide 8
5.2 Modeling relations
At most one event occurs
X
j
Slide 9
xj 1
Neither or both events occur
x2 ; x1 = 0
2
If one event occurs then, another occurs
0 x2 x1
If x = 0, then y = 0 if x = 1, then y is uncontrained
0 y Ux
x 2 f0 1g
5.3 The assignment problem
Slide 10
n people
m jobs
cij : cost of assigning person j to job i:
xij = 10 person j is assigned to job i
P
min
cij xij
n
P
s.t.
xij = 1 each job is assigned
j�1
m
P
xij 1 each person can do at most one job.
i�1
xij 2 f0 1g
5.4 Multiple optimal solutions
Slide 11
Generate all optimal solutions to a BOP.
max c x
s.t. Ax b
x 2 f0 1gn
0
Generate third best?
Extensions to MIO?
5.5 Modeling nonconvex objective functions
How to model min c(x), where c(x) is piecewise linear but not convex?
Slide 12
6 What is a good formulation?
6.1 Facility Location
Slide 13
Data
N = f1 : : :ng potential facility locations
I = f1 : : :mg set of clients
cj : cost of facility placed at j
hij : cost of satisfying client i from facility j:
3
Decision variables
a facility is placed at location j
xj = 1
0
otherwise
yij = fraction of demand of client i
satised by facility j:
Slide 14
n
m P
n
P
P
IZ1 = min
cj xj +
hij yij
j �1
i�1 j �1
n
P
s.t.
yij = 1
j �1
yij xj
xj 2 f0 1g 0 yij 1:
Consider an alternative formulation.
n
m P
n
P
P
IZ = min
cx +
h y
Slide 15
2
j �1
n
P
s.t.
j �1
m
P
j j
yij = 1
i�1 j �1
ij ij
yij m xj
i�1
xj 2 f0 1g 0 yij 1:
Are both valid?
Which one is preferable?
6.2 Observations
IZ1 = IZ2 , since the integer points both formulations dene are the same.
Slide 16
n
X
xj 1
P1 = f(x y) : yij = 1 yij xj 00 yij 1
j �1
P2 = f(x y) :
n
X
j �1
yij = 1
0 xj 1
0 yij 1
Let
Z1 = min cx + hy
(x y) 2 P1
Z2 Z1 IZ1 = IZ2
4
m
X
i�1
yij m xj Slide 17
Z2 = min cx + hy
(x y) 2 P2
6.3 Implications
Slide 18
Finding IZ1 (= IZ2 ) is dicult.
Solving to nd Z1 Z2 is a LOP. Since Z1 is closer to IZ1 several methods
(branch and bound) would work better (actually much better).
Suppose that if we solve min cx + hy (x y) 2 P1 we nd an integral
solution. Have we solved the facility location problem?
Slide 19
Formulation 1 is better than Formulation 2. (Despite the fact that 1 has
a larger number of constraints than 2.)
What is then the criterion?
6.4 Ideal Formulations
Slide 20
Let P be a linear relaxation for a problem
Let
H = f(x y) : x 2 f0 1gng \ P
Consider Convex Hull (H)
X
X
= fx : x = i xi i = 1 i 0 xi 2 H g
i
i
Slide 21
The extreme points of CH (H) have f0 1g coordinates.
So, if we know CH (H) explicitly, then by solving min cx + hy (x y) 2
CH (H) we solve the problem.
Message: Quality of formulation is judged by closeness to CH (H).
CH (H) P1 P2
7 Minimum Spanning
Tree (MST)
How do telephone companies bill you?
It used to be that rate/minute: Boston ! LA proportional to distance in
MST
Other applications: Telecommunications, Transportation (good lower bound
for TSP)
Slide 22
5
Slide 23
Given a graph G = (V E ) undirected and Costs ce e 2 E.
Find a tree of minimum cost spanning all the nodes.
1 if edge e is included in the tree
Decision variables xe =
0 otherwise
Slide 24
The tree should be connected. How can you model this requirement?
Let S be a set of vertices. Then S and V n S should be connected
Let (S) = fe = (i j) 2 E : ij 22 SV n S
Then,
X
xe 1
e2(S )
What is the number of edges in a tree?
P
Then, xe = n ; 1
e2E
7.1 Formulation
P
IZM ST = min cexe
8 ePE x 1
>
< eP(S ) e
H >
x = n;1
>eE e
:
xe 2 f0 1g:
Is this a good formulation?
Slide 25
2
2
8 S V S 6= V
2
Pcut = fx 2 R E : 0 x e
X
xe = n ; 1
e E
X
xe 1 8 S V S 6= V g
j
j
2
Is Pcut the CH (H)?
e2(S )
6
Slide 26
7.2 What is C H (H)?
Let
X
Slide 27
Psub = fx 2 R E : xe = n ; 1
e E
X
xe jS j ; 1 8 S V S 6= V g
j
j
2
e2E (S )
i
2S
E(S) = e = (i j) : j 2 S
Why is this a valid IO formulation?
Slide 28
Theorem: Psub = CH (H).
) Psub is the best possible formulation.
MESSAGE: Good formulations can have an exponential number of con
straints.
8 The Traveling Salesman
Problem
Given G = (V E ) an undirected graph. V = f1 : : : ng, costs ce 8 e 2 E. Find
a tour that minimizes total length.
8.1 Formulation I
xe =
1 if edge e is included in the tour.
0
otherwise.
P
min
ce xe
eP
E
s:t:
xe 2 S E
eP
(S )
xe = 2 i 2 V
e (i)
xe 2 f0 1g
Slide 29
Slide 30
2
2
2
8.2 Formulation II
min
s:t:
Pc x
Pe ex jS j ; 1 S E
e
eP
E (S )
x = 2 i 2 V
Slide 31
2
e
e2(i)
xe 2 f0 1g
7
Slide 32
TSP =
Pcut
fx 2 RjE j
P
e2(S )
xe 2
0 xe 1g P
fx 2 R E xe = 2
e (i)
P
xe jS j ; 1
e (S )
0 xe 1g
TSP =
Psub
P
e2(i)
xe = 2
j
j
2
2
TS P
Pcut
Slide 33
TSP 6 CH (H)
= Psub
Theorem:
Nobody knows CH (H) for the TSP
9 Minimum Matching
Slide 34
Given G = (V E ) ce costs on e 2 E. Find a matching of minimum cost.
Formulation:
P
min Pce xe
s:t:
xe = 1 i 2 V
e (i)
xe 2 f0 1g
2
Is the linear relaxation CH (H)?
Let
Slide 35
P x =1
PM AT = fx 2 R E :
e
e (i)
P
xe 1 jS j = 2k + 1 S =
6 e (S )
xe 0g
Theorem: PM AT = CH (H)
j
j
2
2
10 Observations
Slide 36
For MST, Matching there are ecient algorithms. CH (H) is known.
For TSP 6 9 ecient algorithm. TSP is an NP ; hard problem. CH (H)
is not known.
Conjuecture: The convex hull of problems that are polynomially solvable
are explicitly known.
8
11 Summary
1. Modeling with binary variables allows a lot of modelling power.
2. An IO formulation is better than another one if the polyhedra of their
linear relaxations are closer to the convex hull of the IO.
3. A good formulation may have an exponential number of constraints.
4. Conjecture: Formulations characterize the complexity of problems. If a
problem is solvable in polynomial time, then the convex hull of solutions
is known.
9
Slide 37
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15.083J / 6.859J Integer Programming and Combinatorial Optimization
Fall 2009
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