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Geometric Interpretation of Linear Programs
Chris Osborn, Alan Ray, Carl Bussema, and Chad Meiners
16 March 2005
Five total constraints; therefore 5 faces to the polyhedron
Feasibility in 3-Space
0,,
5
42subject to
523maximize
321
3
21
321
xxx
x
xx
xxx
Simplex Illustrated: Initial Dictionary
321
35
214
523
5
24
xxxz
xx
xxx
Current solution:
x1 = 0x2 = 0x3 = 0z=3x1+2x2+5x3=0
Simplex Illustrated: First Pivot
521
53
214
52325
5
24
xxxz
xx
xxx
Current solution:
x1 = 0x2 = 0x3 = 5z=3x1+2x2+5x3=25
Simplex Illustrated: Second Pivot
5423
221
53
421
221
1
531
5
2
xxxz
xx
xxx
Current solution:
x1 = 2x2 = 0x3 = 5z=3x1+2x2+5x3=31
Simplex Illustrated: Final Pivot
541
53
412
52733
5
24
xxxz
xx
xxx
Final solution (optimal):
x1 = 0x2 = 4x3 = 5z=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
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 Conditionsmax 5x1 + 4x2
subject to:
x1 – 3x2 ≤ 3
2x1 + 3x2 ≤ 12
-2x1 + 7x2 ≤ 21
x1,x2 ≥ 0
Graphic Method Example:Step 2: Determine Feasibilitymax 5x1 + 4x2
subject to:
x1 – 3x2 ≤ 3
2x1 + 3x2 ≤ 12
-2x1 + 7x2 ≤ 21
x1,x2 ≥ 0Based 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 Tangentmax 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
Second Graphic Method Examplemax 4x1 + 6x2
subject to:
x1 – 3x2 ≤ 3
2x1 + 3x2 ≤ 12
-2x1 + 7x2 ≤ 21
x1,x2 ≥ 0Same 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 ≥ 0Optimal 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
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)x1 + (-3α+3+7)x2 ≤ 3α+12+21
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)x1 + (-3α+3)x2 ≤ 3α+12
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
S1
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 thatz = ti zi
1 = ti
z2
z
z3
z1
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 givenv=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
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
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:
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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