Consider the simplex method for finding a better BFS. Then the following are
Our new B = B remove r add k will also be a basis.
When our BFS is non-degenerate, our new z-value gets smaller in the
Theorem: Consider an LP that is in standard form. The LP is feasible iff it has a
BFS. (ie it will have an extreme point when it is feasible!)
(Remember, standard form forces Rank(A) = k = number of rows)
We first start by choosing a
Theorem: Consider an LP in standard form. All Extreme points are BFS.
We start with a BFS x* and show that it is an extreme point.
We will want to take advantage of the following theorem.
Corollary: Let S be a set of the form S cfw_x
Theorem: Consider an LP that has an extreme point and has an optimal solution
(ie not unbounded). Then one of the extreme points must be an optimal
We will assume once more that our LP is in semi-FM form:
s.t Ax b
Theorem: Let S be a set of the form S cfw_x R n | Ax b . Let x* be a feasible solution, and
let A= represent the matrix where x* holds as equality from A. x* is an extreme
point iff Rank(A=) = n.
First we start with proving the forwa
Consider an LP problem P and its dual problem D. Let the objective function
for P be cTx and the objective function for D be bTy. Then the following hold
cT x bT y
When x and y are feasible in their respective
Theorem: The intersection of two convex sets is convex.
Let A and B be two convex sets, and let C be the intersection of A and B.
Consider any two points in C : x and y.
Let z be any choice of lambda such that: z = x + (1-)y where
Fundamental Theorem for LPs:
Any linear program falls in one of the following 3 categories:
The LP is unbounded, and has no solution.
The LP is infeasible, and has no solution.
The LP has an optimal solution.
To prove this theorem,
Theorem: Consider the LP given:
c1 x1 c2 x2 . cn xn c
c1,1 x1 c1, 2 x2 . c1,n xn b1
cm,1 x1 cm, 2 x2 . cm ,n xn bm
Solving the above LP will result in the exact same solution when solving the LP below
c1 x1 c2 x2 . cn
A system of linear inequalities is feasible if and only if the Fourier-Motzkin Elimination
Algorithm does not produce any contradictory inequalities.
If a system of inequalities is feasible
There is an x* that satisfies all
Theorem: Solving the Auxiliary Problem using the Simplex Method will
have two outcomes:
The optimal solution will be 0 (thus all of the s variables are 0) and the
resulting BFS (not including the s-variables) will be a BFS in the origina
1. G is a tree iff it is connected and has |A| =|N| - 1.
2. If G is a tree, then for each a,b in N, there is a unique path from a to b.
3. If G is a tree, then adding 1 arc will produce 1 unique cycle.
2. We will prove 2 tru
Instructions: You may work in groups of up to 4 people answering the questions below. The tutorial is marked
out of 1 (either 0 or 1). If you get at least half of the answers correct, you will re
Test #1: (
Name (print): _
Part A: Short Answer (1 Mark Each)
Question 1: Which of the following inequalities can be implied by
None of the above
Question 2: The variables in the dual correspond to _ in the primal pro
Given any node incidence matrix of a connected network whose b-vector sums to
0, then we can remove any row and the resulting matrix (and remove the
corresponding b-value from b) and the new LP will have an A with full row rank and