Presentation on theme: "Geometric Interpretation of Linear Programs"— Presentation transcript:
1Geometric Interpretation of Linear Programs Chris Osborn, Alan Ray, Carl Bussema, and Chad Meiners16 March 2005
2IntroductionVisualizing algebraic concepts geometrically can give new insight and understandingUnderstand properties of LP in terms of geometryUse geometry as aid to solve LPModel geometric problems as LPSome concepts from earlier chapters; some new
4Region of FeasibilityGraphical region describing all feasible solutions to a linear programming problemIn 2-space: polygon, each edge a constraintIn 3-space: polyhedron, each face a constraint
5Feasibility in 2-Space 2x1 + x2 ≤ 4 In an LP environment, restrict to Quadrant I since x1, x2 ≥ 0
6Feasibility in 3-SpaceFive total constraints; therefore 5 faces to the polyhedron
7Simplex Method Every time a new dictionary is generated: Simplex moves from one vertex to another vertex along an edge of polyhedronAnalogous to increasing value of a non-basic variable until bounded by basic constraintEach such point is a feasible solution
9Simplex Illustrated: First Pivot Current solution:x1 = 0x2 = 0x3 = 5z=3x1+2x2+5x3=25
10Simplex Illustrated: Second Pivot Current solution:x1 = 2x2 = 0x3 = 5z=3x1+2x2+5x3=31
11Simplex Illustrated: Final Pivot Final solution (optimal):x1 = 0x2 = 4x3 = 5z=3x1+2x2+5x3=33
12Simplex Review and Analysis Simplex pivoting represents traveling along polyhedron edgesEach vertex reached tightens one constraint (and if needed, loosens another)May take a longer path to reach final vertex than needed
13Simplex Weaknesses: Exponential Iterations: Klee-Minty Reviewed Cases with high complexity (2n-1 iterations)Normal complexity is O(m3)How was this problem solved?
14Geometric Interpretation & Klee-Minty Saw non-optimal solution earlierHow can we represent the Klee-Minty problem class graphically?
15Step 1: Constructing a Shape Start with a cube.What characteristics do we want the cube to have?What is the worst case to maximize z?
16Step 1: Constructing a Shape Goal 1: Create a shape with a long series of increasing facetsGoal 2: Create an LP problem that forces this route to be taken
18Step 3: Achieving 2n-1 Iterations: Altering the Algebra LetConvertto
19The Final SolutionMost desirable:Least desirable:
20Simplex Weaknesses: Degeneration: A Graphic Example
21Simplex Weaknesses: Degeneration: Summary How does the degeneracy of this problem impact the graphical solution?Degenerate solutions express the same vertex in a different way.How have we dealt with degeneracy previously?
22Different Facets, One Point We shift the multiple facets into two, separate ones[0, 0, 1][0, 0, e2][e - e2, 0, e2]
23Non-Graphic ExampleWhere is the fourth colliding facet in this example:Sometimes, degeneracy occurs without visible fourth facets (as above)
24The Graphic MethodUse geometry to quickly solve LP problems in 2 variablesPlot all restrictions in 2D plane (x1, x2)Result plus axes forms polyhedronRegion of feasible solutionsDraw any line with same slope as objective function through polyhedron“Move” line until leaving feasible regioni.e., Find parallel tangent
29Graphic Method Discussion Pro:Works for any number of constraintsFast, especially with graphing toolGives visual representation of tradeoff between variablesCon:Only works well in 2D (feasible but difficult in 3D)For very large number of constraints, could be annoying to plotFor large range / ratio of coefficients, plot size limits precision and ability to quickly find tangent
30Second Graphic Method Example max 4x1 + 6x2subject to:x1 – 3x2 ≤ 32x1 + 3x2 ≤ 12-2x1 + 7x2 ≤ 21x1,x2 ≥ 0Same constraints; new objective. What changes?
33Geometric Duality Continued Think of dual variables as coefficients for primal constraints (α = y1, = y2, = y3): (α) (x1 – 3x2) ≤ (α) 3 () (2x1 + 3x2) ≤ () 12 () (-2x1 + 7x2) ≤ () 21Resulting sum is linear combination: (α+2-2)x1 + (-3α+3+7)x2 ≤ 3α+12+21We can graph this for various choices of α, , and
34Geometric Duality: Linear Combinations Graphed (α+2-2)x1 + (-3α+3+7)x2 ≤ 3α+12+21Three examples shownα = = 1, = 0α = = = 1α = 0, = = 1Pink line is parallel tangentNotice: primal solution two vertices away from originTwo constraints matter; third irrelevant here.Duality implication: = 0
35Geometric Duality Graphed Again (α+2)x1 + (-3α+3)x2 ≤ 3α+12Gives a line always passing through (5,2/3), the primal solutionIf primal solution optimal, there must exist some α, such that resulting line matches parallel tangent. Why?Duality theorem guarantees
36Convex Set and Hulls Convex Sets Convex Hulls Applying LP Theorem to Convex Sets and Hulls
37Convex Sets Two of these sets are not like the others S1 and S4 are convexS2 and S3 are notA set S Rnis convex iffGiven a,b SFor all 0 ≤ t ≤ 1ta + (1-t)b SS2S3S4
38Property of Convex Sets The intersection of two convex sets results in a convex setEvery set has a minimal convex set that contains it
39Convex Hulls Given a set S Rn Convex Hull H Contains S Is convex Is contained by all convex sets containing S (i.e. it is minimal)
40Convex Hulls as Linear Equations Given a set S RnFor each point z in HThere are k pointsz1,…,zk in Spositive variables t1,…,tkSuch thatz = ti zi1 = tiz1zz = ti zi Represents k equations!!!z3z2
41LP TheoremIf a system of m linear equations has a nonnegative solution, then it has a solution with at most m positive variablesSo givenv=1…n aivxv = b (for i = 1…m)xv ≥ 0At most m of the variables x1…xn are positive
42Implications Upon Convex Hulls For a space S RnWe have at most n+1 points that define a hull pointSo for R2, every point z in H is defined by at most 3 points in SWhy?Hull points are represented by n+1 linear equationsThus we have at most n+1 positive scaling variables tizz3z1z2
43Convex Hulls as Linear Equations For a set S R2Point z is in a hull of S iffThere are three points in SThe weighted sum of these three points equal Szz3z1z2Redundant
44Some More Observations Every half-space is convexEvery polyhedron is convexThe convex hull of a finite set of points is a polyhedron
45LP TheoremEvery unsolvable system of linear inequalities for n variables contains a unsolvable subsystem of at most n+1 inequalitiesWe use this theorem for the common point theorem
46Common Point TheoremLet F be a finite family of at least n + 1 convex sets in RnSuch that every n+1 sets in F have a point in commonAll sets in F have a point in common
47Common Point Theorem (continued) Note that we can’t make guarantees without every n+1 sets in F having a point in common
48Common Point Theorem (Why) The intersection of each n+1 sets is a system of n+1 linear inequialitiesTherefore the whole system cannot have an unsolvable subsystem of n+1 linear inequalitiesThus we have a point in common for the family of convex sets
49LP TheoremA system of linear inequalities is inconsistent iff it is unsolvableThis means that unsolvable linear inequalities must have inconsistent constraintse.g. x1 = 1 and x1 = 2Likewise inconsistent constraint make linear inequalities unsolvable
50Separation Theorem for Polyhedra For every pair of disjoint polyhedraThere exists a pair of disjoint half-spacesSuch that each half-space contains a polyhedron
51Conclusions Geometry useful for: Questions? Understanding properties of linear programsSolving (some) linear programsModeling linear programs visuallyAnd geometric problems can be modeled with linear programsQuestions?