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

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Presentation on theme: "Geometric Interpretation of Linear Programs Chris Osborn, Alan Ray, Carl Bussema, and Chad Meiners 16 March 2005."— Presentation transcript:

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

2 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

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

4 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

5 Feasibility in 2-Space  2x 1 + x 2 ≤ 4  In an LP environment, restrict to Quadrant I since x 1, x 2 ≥ 0

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

7 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

8 Simplex Illustrated: Initial Dictionary Current solution: x 1 = 0 x 2 = 0 x 3 = 0 z=3x1+2x2+5x3=0

9 Simplex Illustrated: First Pivot Current solution: x 1 = 0 x 2 = 0 x 3 = 5 z=3x1+2x2+5x3=25

10 Simplex Illustrated: Second Pivot Current solution: x 1 = 2 x 2 = 0 x 3 = 5 z=3x1+2x2+5x3=31

11 Simplex Illustrated: Final Pivot Final solution (optimal): x 1 = 0 x 2 = 4 x 3 = 5 z=3x1+2x2+5x3=33

12 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

13 Simplex Weaknesses: Exponential Iterations: Klee-Minty Reviewed  Cases with high complexity (2 n -1 iterations)  Normal complexity is O(m 3 )  How was this problem solved?

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

15 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?

16 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

17 Step 2: Increasing Objective Function: Modifying the Cube  Squash the cube  New dictionary [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]

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

19 The Final Solution  Most desirable:  Least desirable:

20 Simplex Weaknesses: Degeneration: A Graphic Example

21 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?

22 Different Facets, One Point  We shift the multiple facets into two, separate ones [0, 0, 1] [0, 0,  2 ] [  -  2, 0,  2 ]

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

24 The Graphic Method  Use geometry to quickly solve LP problems in 2 variables  Plot all restrictions in 2D plane (x 1, x 2 )  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

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

26 Graphic Method Example: Step 2: Determine Feasibility max 5x 1 + 4x 2 subject to: x 1 – 3x 2 ≤ 3 2x 1 + 3x 2 ≤ 12 -2x 1 + 7x 2 ≤ 21 x 1,x 2 ≥ 0 Based only on this, where might the optimal solution be?

27 Graphic Method Example: Step 3: Plot Objective = c max 5x 1 + 4x 2 subject to: x 1 – 3x 2 ≤ 3 2x 1 + 3x 2 ≤ 12 -2x 1 + 7x 2 ≤ 21 x 1,x 2 ≥ 0

28 Graphic Method Example: Step 4: Find Parallel Tangent max 5x 1 + 4x 2 subject to: x 1 – 3x 2 ≤ 3 2x 1 + 3x 2 ≤ 12 -2x 1 + 7x 2 ≤ 21 x 1,x 2 ≥ 0 Optimal solution: x 1 =5, x 2 =2/3, z=83/3

29 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

30 Second Graphic Method Example max 4x 1 + 6x 2 subject to: x 1 – 3x 2 ≤ 3 2x 1 + 3x 2 ≤ 12 -2x 1 + 7x 2 ≤ 21 x 1,x 2 ≥ 0 Same constraints; new objective. What changes?

31 Second Graphic Method Example: No Tangent Exists max 4x 1 + 6x 2 subject to: x 1 – 3x 2 ≤ 3 2x 1 + 3x 2 ≤ 12 -2x 1 + 7x 2 ≤ 21 x 1,x 2 ≥ 0 Optimal solution: 1.05 ≤ x 1 ≤ 5, 2x 1 + 3x 2 = 12, z=24

32 Geometric Interpretation of Duality  Consider earlier problem: max 5x 1 + 4x 2 subject to: x 1 – 3x 2 ≤ 3 2x 1 + 3x 2 ≤ 12 -2x 1 + 7x 2 ≤ 21 x 1,x 2 ≥ 0 Optimal: x 1 * =5, x 2 * =2/3  Prove optimal if equal to corresponding dual solution min 3y y y 3 subject to: y 1 + 2y 2 – 2y 3 ≥ 5 -3y 1 + 3y 2 + 7y 3 ≥ 4 y 1, y 2 ≥ 0

33 Geometric Duality Continued  Think of dual variables as coefficients for primal constraints (α = y 1,  = y 2,  = y 3 ): (α) (x 1 – 3x 2 ) ≤ (α) 3 (  ) (2x 1 + 3x 2 ) ≤ (  ) 12 (  ) ( -2x 1 + 7x 2 ) ≤ (  ) 21  Resulting sum is linear combination: (α+2  -2  )x 1 + (-3α+3  +7  )x 2 ≤ 3α+12  +21   We can graph this for various choices of α, , and 

34 Geometric Duality: Linear Combinations Graphed  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 (α+2  -2  )x 1 + (-3α+3  +7  )x 2 ≤ 3α+12  +21 

35 Geometric Duality Graphed Again  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 (α+2  )x 1 + (-3α+3  )x 2 ≤ 3α+12 

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

37 Convex Sets  Two of these sets are not like the others  S 1 and S 4 are convex  S 2 and S 3 are not  A set S  R n is convex iff Given a,b  S For all 0 ≤ t ≤ 1 ta + (1-t)b  S S1S1 S2S2 S3S3 S4S4

38 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

39 Convex Hulls  Given a set S  R n  Convex Hull H Contains S Is convex Is contained by all convex sets containing S (i.e. it is minimal)

40 Convex Hulls as Linear Equations  Given a set S  R n  For each point z in H  There are k points z 1,…,z k in S  positive variables t 1,…,t k  Such that z =  t i z i 1 =  t i z2z2 z z3z3 z1z1

41 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 a iv x v = b (for i = 1…m) x v ≥ 0  At most m of the variables x 1 …x n are positive

42 Implications Upon Convex Hulls  For a space S  R n  We have at most n+1 points that define a hull point  So for R 2, 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 t i z z3z3 z1z1 z2z2

43 Convex Hulls as Linear Equations  For a set S  R 2  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 z3z3 z1z1 z2z2

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

45 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

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

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

48 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

49 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. x 1 = 1 and x 1 = 2 Likewise inconsistent constraint make linear inequalities unsolvable

50 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

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


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