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 Graphic Method
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 2x1 + x2 ≤ 4
In an LP environment, restrict to Quadrant I since x1, x2 ≥ 0

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

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:
x1 = 0
x2 = 0
x3 = 0
z=3x1+2x2+5x3=0

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

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

11. Simplex Illustrated: Final Pivot
Final solution (optimal):
x1 = 0
x2 = 4
x3 = 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 (2n-1 iterations)
Normal complexity is O(m3)
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
[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

18. Step 3: Achieving 2n-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, e2]
[e - e2, 0, e2]

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 (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

25. 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

26. 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?

27. 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

28. 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

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

31. 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

32. 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

33. 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 

34. 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

35. 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

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
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

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  Rn 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  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

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 aivxv = b (for i = 1…m)
xv ≥ 0
At most m of the variables x1…xn are positive

42. 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

43. 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

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 Rn
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. x1 = 1 and x1 = 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: Questions?
Understanding properties of linear programs
Solving (some) linear programs
Modeling linear programs visually
And geometric problems can be modeled with linear programs
Questions?