Linear Programming (Optimization)

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Transcript Linear Programming (Optimization)

IE 531 Linear Programming
Spring 2010
박성수
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Course Objectives
 Why need to study LP?
Important tool by itself
Theoretical basis for later developments (IP, Network, Graph, Nonlinear,
scheduling, Sets, Coding, Game, … )
Formulation + package is not enough for advanced applications and
interpretation of results
 Objectives of the class:
Understand the theory of linear optimization
Preparation for more in-depth optimization theory
Modeling capabilities
Introduction to use of software (Xpress-MP and/or CPLEX)
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 Prerequisite: basic linear algebra/calculus,
earlier exposure to LP/OR helpful,
mathematical maturity (reading proofs, logical thinking)
 Be steady in studying.
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Chapter 1 Introduction
 Mathematical Programming Problem:
min/max
f(x)
subject to
gi(x)  0, i = 1, ..., m,
(hj(x) = 0, j = 1, ..., k,)
( x  X  R n)
f, gi, hj : Rn  R
 If f, gi, hj linear (affine) function  linear programming problem
If f, gi, hj (or part of them) nonlinear function  nonlinear programming
problem
If solution set restricted to be integer points  integer programming
problem
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 Linear programming: problem of optimizing (maximize or minimize) a
linear (objective) function subject to linear inequality constraints.
 General form:
{max, min} c'x
subject to
ai'x  bi , iM1
ai'x  bi , iM2
ai'x = bi , iM3
xj  0, jN1 ,
xj  0, jN2
c, ai , x Rn
(There may exist variables unrestricted in sign)
 inner product of two column vectors x, y  Rn :
x’y = i = 1n xiyi
If x’y = 0, x, y  0, then x, y are said to be orthogonal. In 3-D, the angle
between the two vectors is 90 degrees.
( vectors are column vectors unless specified otherwise)
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 Big difference from systems of linear equations is the existence of
objective function and linear inequalities (instead of equalities)
 Much deeper theoretical results and applicability than systems of
linear equations.
 x1, x2, …, xn : (decision) variables
bi : right-hand-side
ai'x { , ,  } bi : i th constraint
xj { ,  } 0 : nonnegativity (nonpositivity) constraint
c'x : objective function
 Other terminology:
feasible solution, feasible set (region), free (unrestricted) variable,
optimal (feasible) solution, optimal cost, unbounded
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Important submatrix multiplications
 Interpretation of constraints:
A: mn
  a1 '    |
  A
A

  1
  am '    |
| 
An 

| 
Ax  nj1 A j x j  im1 ai ' xei , where ei is i-th unit vector
y' A  im1 yi ai '  nj1 y' A j e j '
denote constraints as Ax { , ,  } b
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 Any LP can be expressed as min c'x, Ax  b
max c'x  min (-c'x) and take negative of the optimal cost
ai'x  bi  -ai'x  -bi
ai'x = bi  ai'x  bi , -ai'x  -bi
nonnegativity (nonpositivity) are special cases of inequalities which will
be handled separately in the algorithms.
Feasible solution set of LP can always be expressed as Ax  b (or Ax
 b) (called polyhedron, a set which can be described as a solution
set of finitely many linear inequalities)
 We may sometimes use max c'x, Ax  b form (especially, when we
study polyhedron)
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Brief History of LP (or Optimization)
 Gauss: Gaussian elimination to solve systems of equations
Fourier(early 19C) and Motzkin(20C) : solving systems of linear
inequalities
Farkas, Minkowski, Weyl, Caratheodory, … (19-20C):
Mathematical structures related to LP (polyhedron, systems of
alternatives, polarity)
Kantorovich (1930s) : efficient allocation of resources
(Nobel prize in 1975 with Koopmans)
Dantzig (1947) : Simplex method
1950s : emergence of Network theory, Integer and combinatorial
optimization, development of computer
1960s : more developments
1970s : disappointment, NP-completeness, more realistic
expectations
Khachian (1979) : ellipsoid method for LP
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1980s : personal computer, easy access to data, willingness to use
models
Karmarkar (1984) : Interior point method
1990s : improved theory and software, powerful computers
software add-ins to spreadsheets, modeling languages,
large scale optimization, more intermixing of O.R. and A.I.
Markowitz (1990) : Nobel prize for portfolio selection (quadratic
programming)
Nash (1994) : Nobel prize for game theory
21C (?) : Lots of opportunities
more accurate and timely data available
more theoretical developments
better software and computer
need for more automated decision making for complex systems
need for coordination for efficient use of resources (e.g.
supply chain management, APS, traditional engineering problems,
bio...)
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Application Areas of Optimization
 Operations Managements
Production Planning
Scheduling (production, personnel, ..)
Transportation Planning, Logistics
Energy
Military
Finance
Marketing
E-business
Telecommunications
Games
Engineering Optimization (mechanical, electrical, bioinformatics, ... )
System Design
…
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Resources
 Societies:
INFORMS (the Institute for Operations Research and Management
Sciences) : www.informs.org
MPS (The Mathematical Programming Society) : www.mathprog.org
Korean Institute of Industrial Engineers : kiie.org
Korean Operations Research Society : www.korms.or.kr
 Journals:
Operations Research, Management Science, Mathematical
Programming, Networks, European Journal of Operational Research,
Naval Research Logistics, Journal of the Operational Research
Society, Interfaces, …
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Standard form problems
 Standard form : min c'x, Ax = b, x  0
Ax  nj1 A j x j  im1 ai ' xei
Find optimal weights (nonnegative) from possible nonnegative linear
combinations of columns of A to obtain b vector
Find optimal solution that satisfies linear equations and nonnegativity
 Reduction to standard form
Free (unrestricted) variable xj  xj+ - xj- , xj+, xj-  0
j aijxij  bi  j aijxij + si = bi , si  0 (slack variable)
j aijxij  bi  j aijxij - si = bi , si  0 (surplus variable)
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 Any (practical) algorithm can solve the LP problem in equality form
only (except nonnegativity)
 Modified form of the simplex method can solve the problem with free
variables directly (w/o using difference of two variables).
It gives more sensible interpretation of the behavior of the algorithm.
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1.2 Formulation examples
 Minimum cost network flow problem
Directed network G=(N, A), (|N| = n )
arc capacity uij , (i, j) A, unit flow cost cij , (i, j) A
bi : net supply at node i (bi > 0: supply node, bi < 0: demand node),
(i bi = 0)
Find min cost transportation plan that satisfies supply, demand at each
node and arc capacities.
minimize
subject to
(i, j)A cijxij
{j : (i, j)A} xij - {j : (j, i)A} xji = bi , i = 1, …, n
(out flow - in flow = net flow at node i)
(some people use, in flow – out flow = net flow)
xij  uij ,
(i, j)A
xij  0 ,
(i, j)A
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 Choosing paths in a communication network ( (fractional)
multicommodity flow problem)
 Multicommodity flow problem: Several commodities share the
network. For each commodity, it is min cost network flow problem.
But the commodities must share the capacities of the arcs.
Generalization of min cost network flow problem. Many
applications in communication, distribution / transportation systems
Several commodities case
Actually one commodity. But there are multiple origin and destination
pairs of nodes (telecom, logistics, ..)
 Given telecommunication network (directed) with arc set A, arc
capacity uij bits/sec, (i, j) A, unit flow cost cij /bit , (i, j) A,
demand bkl bits/sec for traffic from node k to node l.
Data can be sent using more than one path.
Find paths to direct demands with min cost.
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Decision variables:
xijkl : amount of data with origin k and destination l that
traverses link (i, j) A
Let bikl = bkl if i = k
-bkl if i = l
0
otherwise
 Formulation (flow formulation)
minimize
subject to
(i, j)A k l cijxijkl
{j : (i, j)A} xijkl -  {j : (j, i)A} xjikl = bikl ,
i, k, l = 1, …, n
(out flow - in flow = net flow at node i for
commodity from node k to node l)
k l xijkl  uij ,
(i, j)A
(The sum of all commodities should not exceed the
capacity of link (i, j) )
xijkl  0 ,
(i, j)A,
k, l =1, …, n
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 Alternative formulation (path formulation)
Let K: set of origin-destination pairs (commodities)
P(k): set of all possible paths for sending commodity k  K
P(k;e): set of paths in P(k) that traverses arc e  A
E(p): set of links contained in path p
Decision variables:
ypk : fraction of commodity k sent on path p
kK pP(k) wpkypk
pP(k) ypk = 1, for all kK
kK pP(k; e) bkypk  ue ,
for all eA
0  ypk  1, for all p  P(k), k  K
where wpk = bkeE(p) ce
minimize
subject to
 If ypk  {0, 1}, it is a single path routing problem (path selection,
integer multicommodity flow).
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 path formulation has smaller number of constraints, but enormous
number of variables.
can be solved easily by column generation technique (later).
Integer version is more difficult to solve.
 Extensions: Network design - also determine the number and type of
facilities to be installed on the links (and/or nodes) together with routing
of traffic.
 Variations: Integer flow. Bifurcation of traffic may not be allowed.
Determine capacities and routing considering rerouting of traffic in case
of network failure, Robust network design (data uncertainty), ...
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 Pattern classification (Linear classifier)
Given m objects with feature vector ai Rn , i = 1, …, m.
Objects belong to one of two classes. We know the class to which
each sample object belongs.
We want to design a criterion to determine the class of a new object
using the feature vector.
Want to find a vector (x, xn+1) Rn+1 with x Rn such that, if i S, then
ai'x  xn+1, and if i S, then ai'x < xn+1. (if it is possible)
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 Find a feasible solution (x, xn+1) that satisfies
ai'x  xn+1, i S
ai'x < xn+1. i S
for all sample objects i
Is this a linear programming problem?
( no objective function, strict inequality in constraints)
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 Is strict inequality allowed in LP?
consider min x, x > 0  no minimum point. only infimum of
objective value exists
 If the system has a feasible solution (x, xn+1), we can make the
difference of the rhs and lhs large by using solution M(x, xn+1) for M > 0
and large. Hence there exists a solution that makes the difference at
least 1 if the system has a solution.
Remedy: Use ai'x  xn+1,
i S
ai'x  xn+1-1, i S
 Important problem in data mining with applications in target marketing,
bankruptcy prediction, medical diagnosis, process monitoring, …
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 Variations
What if there are many choices of hyperplanes? any reasonable criteria?
What if there is no hyperplane separating the two classes?
Do we have to use only one hyperplane?
Use of nonlinear function possible? How to solve them?
• SVM (support vector machine), convex optimization
More than two classes?
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1.3 Piecewise linear convex obj. functions
 Some problems involving nonlinear functions can be modeled as LP.
 Def: Function f : Rn  R is called a convex function if for all x, y Rn
and all   [0, 1]
f(x + (1- )y)  f(x) + (1- )f(y).
( the domain may be restricted)
f called concave if -f is convex
(picture: the line segment joining (x, f(x)) and (y, f(y)) in Rn+1 is not
below the locus of f(x) )
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 Def: x, y Rn, 1, 2  0, 1+ 2 = 1
Then 1x + 2y is said to be a convex combination of x, y.
Generally, i=1k ixi , where i=1k i = 1 and i  0, i = 1, ..., k is a convex
combination of the points x1, ..., xk
 Def: A set S  Rn is convex if for any x, y S, we have x + (1 - ) y S
for any  [ 0, 1 ].
Picture:
1x + 2y = 1x + (1 - 1) y, 0  1  1
= y + 1 (x – y)
(line segment joining x, y lies in S)
x (1 = 1)
(x-y)
y (1 = 0)
(x-y)
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Picture
( x, f ( x))  R n 1
f (x)
( y, f ( y ))
(x  (1   ) y, f ( x)  (1   ) f ( y))
 f ( x)  (1   ) f ( y)
f ( x  (1   ) y)
x
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 x  (1   ) y
y
x  Rn
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 relation between convex function and convex set
 Def: f: Rn  R. Define epigraph of f as epi(f) = { (x, )  Rn+1 :   f(x) }
 Then previous definition of convex function is equivalent to epi(f) being
a convex set. When dealing with convex functions, we frequently
consider epi(f) to exploit the properties of convex sets.
 Consider operations on functions that preserve convexity and
operations on sets that preserve convexity.
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 Example:
Consider f(x) = max i = 1, …, m (ci'x + di), ci Rn, di R
(maximum of affine functions, called a piecewise linear convex
function.)
f (x)
c1'x+d1
c2'x+d2
c3'x+d3
x  Rn
x
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 Thm: Let f1, …, fm : Rn  R be convex functions. Then
f(x) = max i = 1, …, m fi(x) is also convex.
pf)
f(x + (1- )y) = max i=1, …, m fi(x + (1- )y )
 max i=1, …, m (fi(x) + (1- )fi(y) )
 max i=1, …, m fi(x) + max i=1, …, m (1- )fi(y)
= f(x) + (1- )f(y)
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 Min of piecewise linear convex functions
Minimize max I=1, …, m (ci'x + di)
Subject to
Ax  b
Minimize
Subject to
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z
z  ci'x + di ,
Ax  b
i = 1, …, m
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 Q: What can we do about finding max of a piecewise linear convex
function?
maximum of a piecewise linear concave function (can be obtained as
min of affine functions)?
Min of a piecewise linear concave function?
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 Convex function has a nice property such that a local min point is a
global min point. (when domain is Rn or convex set) (HW later)
Hence finding min of a convex function defined over a convex set is
usually easy. But finding a max of a convex function is difficult to
solve. Basically, we need to examine all local max points.
Similarly, finding a max of concave function is easy, but finding min of
a concave function is difficult.
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 In constraints, f(x)  h
where f(x) is piecewise linear convex function f(x) = max i=1, …, m (fi'x +
gi).
 fi'x + gi  h,
i = 1, … , m
Q: What about constraints f(x)  h ? Can it be modeled as LP?
 Def: f: Rn  R, convex function,   R
The set C = { x: f(x)   } is called the level set of f
 level set of a convex function is a convex set. (HW later)
solution set of LP is convex (easy)  non-convex solution set can’t be
modeled as LP.
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Problems involving absolute values
 Minimize i = 1, …, n ci |xi|
subject to
Ax  b
(assume ci  0)
More direct formulations than piecewise linear convex function is
possible.
(1)
Min i ci zi
subject to
Ax  b
xi  zi , i = 1, …, n
-xi  zi , i = 1, …, n
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(2)
Min i ci (xi+ + xi-)
subject to
Ax+ - Ax-  b
x+ , x-  0
(want xi+ = xi if xi  0, xi- = -xi if xi
< 0 and xi+xi- = 0, i.e., at most one of
xi+, xi- is positive in an optimal
solution.
ci  0 guarantees that.)
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Data Fitting
 Regression analysis using absolute value function
Given m data points (ai , bi ), i = 1, …, m, ai Rn , bi R.
Want to find x Rn that predicts results b given a with function b = a'x
Want x that minimizes prediction error | bi - ai'x | for all i.
minimize
subject to
z
bi - ai'x  z, i = 1, … , m
-bi + ai'x  z, i = 1, … , m
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 Alternative criterion
minimize i = 1, …, m | bi - ai'x |
minimize
subject to
z1 + … + z m
bi - ai'x  zi , i = 1, … , m
-bi + ai'x  zi , i = 1, … , m
Quadratic error function can't be modeled as LP, but need calculus
method (closed form solution)
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 Special case of piecewise linear objective function : separable
piecewise linear objective function.
function f: Rn  R, is called separable if f(x) = f1(x1) + f2(x2) + … +
fn(xn)
fi(xi)
c1 < c2 < c3 < c4
c4
c3
slope: ci
c2
c1
0
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x1i
a1
x2i
a2
x3i
a3
xi
x4i
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 Express xi in the constraints as xi  x1i + x2i + x3i + x4i , where
0  x1i  a1, 0  x2i  a2 - a1 , 0  x3i  a3 - a2, 0  x4i
In the objective function, use :
min c1x1i + c2x2i + c3x3i + c4x4i
Since we solve min problem, it is guaranteed that we get
xki > 0 in an optimal solution implies xji , j < k have values at their
upper bounds.
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1.4 Graphical representation and solution
 Let a  Rn, b R.
Geometric intuition for the solution sets of
{ x : a’x = 0 }
{ x : a’x  0 }
{ x : a’x  0 }
{ x : a’x = b }
{ x : a’x  b }
{ x : a’x  b }
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 Geometry in 2-D
{ x : a’x  0 }
a
0
{ x : a’x  0 }
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{ x : a’x = 0 }
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 Let z be a (any) point satisfying a’x = b. Then
{ x : a’x = b } = { x : a’x = a’z } = { x : a’(x – z) = 0 }
Hence x – z = y, where y is any solution to a’y = 0, or x = y + z.
Similarly, for { x : a’x  b }, { x : a’x  b }.
{ x : a’x  b }
a
z
{ x : a’x  b }
0
{ x : a’x = b }
{ x : a’x = 0 }
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 min c1x1 + c2x2
s.t.
-x1 + x2  1,
x1  0, x2  0
x2
c=(1, 0)
c=(-1, -1)
c=(1, 1)
c=(0, 1)
x1
{x: x1 + x2 = 0}
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{x: x1 + x2 = z}
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 Representing complex solution set in 2-D
( n variables, m equations (coefficient vectors are linearly
independent), nonnegativity, and n – m = 2 )
x3
x1 = 0
x2
x2 = 0
x3 = 0
x1
 See text sec. 1.5, 1.6 for more backgrounds
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