Geometric Interpretation of Linear Programs Chris Osborn, Alan Ray, Carl Bussema, and Chad Meiners 16 March 2005

Introduction Visualizing algebraic concepts geometrically can give new insight and understanding Understand properties of LP in terms of geometry Use geometry as aid to solve LP Model geometric problems as LP Some concepts from earlier chapters; some new

Overview Feasibility Simplex Method Simplex Weaknesses Graphic Method Exponential Iterations Degeneracy Graphic Method Duality Convex Sets and Hulls

Region of Feasibility Graphical region describing all feasible solutions to a linear programming problem In 2-space: polygon, each edge a constraint In 3-space: polyhedron, each face a constraint

Feasibility in 2-Space 2x1 + x2 ≤ 4 In an LP environment, restrict to Quadrant I since x1, x2 ≥ 0

Feasibility in 3-Space Five total constraints; therefore 5 faces to the polyhedron

Simplex Method Every time a new dictionary is generated: Simplex moves from one vertex to another vertex along an edge of polyhedron Analogous to increasing value of a non-basic variable until bounded by basic constraint Each such point is a feasible solution

Simplex Illustrated: Initial Dictionary Current solution: x1 = 0 x2 = 0 x3 = 0 z=3x1+2x2+5x3=0

Simplex Illustrated: First Pivot Current solution: x1 = 0 x2 = 0 x3 = 5 z=3x1+2x2+5x3=25

Simplex Illustrated: Second Pivot Current solution: x1 = 2 x2 = 0 x3 = 5 z=3x1+2x2+5x3=31

Simplex Illustrated: Final Pivot Final solution (optimal): x1 = 0 x2 = 4 x3 = 5 z=3x1+2x2+5x3=33

Simplex Review and Analysis Simplex pivoting represents traveling along polyhedron edges Each vertex reached tightens one constraint (and if needed, loosens another) May take a longer path to reach final vertex than needed

Simplex Weaknesses: Exponential Iterations: Klee-Minty Reviewed Cases with high complexity (2n-1 iterations) Normal complexity is O(m3) How was this problem solved?

Geometric Interpretation & Klee-Minty Saw non-optimal solution earlier How can we represent the Klee-Minty problem class graphically?

Step 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?

Step 1: Constructing a Shape Goal 1: Create a shape with a long series of increasing facets Goal 2: Create an LP problem that forces this route to be taken

Step 2: Increasing Objective Function: Modifying the Cube [0, 1, 0.8] [0, 1, 0.82] [1, 0.8, 0] [0, 1, 0] [1, 0, 0.98] [0, 0, 1] [1, 0, 0] [0, 0, 0] Squash the cube New dictionary

Step 3: Achieving 2n-1 Iterations: Altering the Algebra Let Convert to

The Final Solution Most desirable: Least desirable:

Simplex Weaknesses: Degeneration: A Graphic Example

Simplex 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?

Different Facets, One Point We shift the multiple facets into two, separate ones [0, 0, 1] [0, 0, 1+0.25e2] [e - e2, 0, 1 + 0.25e2]

Non-Graphic Example Where is the fourth colliding facet in this example: Sometimes, degeneracy occurs without visible fourth facets (as above)

The Graphic Method Use geometry to quickly solve LP problems in 2 variables Plot all restrictions in 2D plane (x1, x2) Result plus axes forms polyhedron Region of feasible solutions Draw any line with same slope as objective function through polyhedron “Move” line until leaving feasible region i.e., Find parallel tangent

Graphic Method Example: Step 1: Plot Boundary Conditions max 5x1 + 4x2 subject to: x1 – 3x2 ≤ 3 2x1 + 3x2 ≤ 12 -2x1 + 7x2 ≤ 21 x1,x2 ≥ 0

Graphic Method Example: Step 2: Determine Feasibility max 5x1 + 4x2 subject to: x1 – 3x2 ≤ 3 2x1 + 3x2 ≤ 12 -2x1 + 7x2 ≤ 21 x1,x2 ≥ 0 Based only on this, where might the optimal solution be?

Graphic Method Example: Step 3: Plot Objective = c max 5x1 + 4x2 subject to: x1 – 3x2 ≤ 3 2x1 + 3x2 ≤ 12 -2x1 + 7x2 ≤ 21 x1,x2 ≥ 0

Graphic Method Example: Step 4: Find Parallel Tangent max 5x1 + 4x2 subject to: x1 – 3x2 ≤ 3 2x1 + 3x2 ≤ 12 -2x1 + 7x2 ≤ 21 x1,x2 ≥ 0 Optimal solution: x1=5, x2=2/3, z=83/3

Graphic Method Discussion Pro: Works for any number of constraints Fast, especially with graphing tool Gives visual representation of tradeoff between variables Con: Only works well in 2D (feasible but difficult in 3D) For very large number of constraints, could be annoying to plot For large range / ratio of coefficients, plot size limits precision and ability to quickly find tangent

Second Graphic Method Example max 4x1 + 6x2 subject to: x1 – 3x2 ≤ 3 2x1 + 3x2 ≤ 12 -2x1 + 7x2 ≤ 21 x1,x2 ≥ 0 Same constraints; new objective. What changes?

Second Graphic Method Example: No Tangent Exists max 4x1 + 6x2 subject to: x1 – 3x2 ≤ 3 2x1 + 3x2 ≤ 12 -2x1 + 7x2 ≤ 21 x1,x2 ≥ 0 Optimal solution: 1.05 ≤ x1 ≤ 5, 2x1 + 3x2 = 12, z=24

Geometric Interpretation of Duality Consider earlier problem: max 5x1 + 4x2 subject to: x1 – 3x2 ≤ 3 2x1 + 3x2 ≤ 12 -2x1 + 7x2 ≤ 21 x1,x2 ≥ 0 Optimal: x1*=5, x2*=2/3 Prove optimal if equal to corresponding dual solution min 3y1 + 12y2 + 21y3 subject to: y1 + 2y2 – 2y3 ≥ 5 -3y1 + 3y2 + 7y3 ≥ 4 y1, y2 ≥ 0

Geometric 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) ≤ () 21 Resulting sum is linear combination: (α+2-2)x1 + (-3α+3+7)x2 ≤ 3α+12+21 We can graph this for various choices of α, , and 

Geometric Duality: Linear Combinations Graphed (α+2-2)x1 + (-3α+3+7)x2 ≤ 3α+12+21 Three examples shown α =  = 1,  = 0 α =  =  = 1 α = 0,  =  = 1 Pink line is parallel tangent Notice: primal solution two vertices away from origin Two constraints matter; third irrelevant here. Duality implication:  = 0

Geometric Duality Graphed Again (α+2)x1 + (-3α+3)x2 ≤ 3α+12 Gives a line always passing through (5,2/3), the primal solution If primal solution optimal, there must exist some α,  such that resulting line matches parallel tangent. Why? Duality theorem guarantees

Convex Set and Hulls Convex Sets Convex Hulls Applying LP Theorem to Convex Sets and Hulls

Convex Sets Two of these sets are not like the others S1 and S4 are convex S2 and S3 are not A set S  Rn is convex iff Given a,b  S For all 0 ≤ t ≤ 1 ta + (1-t)b  S S2 S3 S4

Property of Convex Sets The intersection of two convex sets results in a convex set Every set has a minimal convex set that contains it

Convex 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)

Convex Hulls as Linear Equations Given a set S  Rn For each point z in H There are k points z1,…,zk in S positive variables t1,…,tk Such that z =  ti zi 1 =  ti z1 z z =  ti zi Represents k equations!!! z3 z2

LP Theorem If a system of m linear equations has a nonnegative solution, then it has a solution with at most m positive variables So given v=1…n aivxv = b (for i = 1…m) xv ≥ 0 At most m of the variables x1…xn are positive

Implications Upon Convex Hulls For a space S  Rn We have at most n+1 points that define a hull point So for R2, every point z in H is defined by at most 3 points in S Why? Hull points are represented by n+1 linear equations Thus we have at most n+1 positive scaling variables ti z z3 z1 z2

Convex Hulls as Linear Equations For a set S  R2 Point z is in a hull of S iff There are three points in S The weighted sum of these three points equal S z z3 z1 z2 Redundant

Some More Observations Every half-space is convex Every polyhedron is convex The convex hull of a finite set of points is a polyhedron

LP Theorem Every unsolvable system of linear inequalities for n variables contains a unsolvable subsystem of at most n+1 inequalities We use this theorem for the common point theorem

Common Point Theorem Let F be a finite family of at least n + 1 convex sets in Rn Such that every n+1 sets in F have a point in common All sets in F have a point in common

Common Point Theorem (continued) Note that we can’t make guarantees without every n+1 sets in F having a point in common

Common Point Theorem (Why) The intersection of each n+1 sets is a system of n+1 linear inequialities Therefore the whole system cannot have an unsolvable subsystem of n+1 linear inequalities Thus we have a point in common for the family of convex sets

LP Theorem A system of linear inequalities is inconsistent iff it is unsolvable This means that unsolvable linear inequalities must have inconsistent constraints e.g. x1 = 1 and x1 = 2 Likewise inconsistent constraint make linear inequalities unsolvable

Separation Theorem for Polyhedra For every pair of disjoint polyhedra There exists a pair of disjoint half-spaces Such that each half-space contains a polyhedron

Conclusions Geometry useful for: Questions? Understanding properties of linear programs Solving (some) linear programs Modeling linear programs visually And geometric problems can be modeled with linear programs Questions?