# Lecture 3 Simplex Method

```ISM 206
Lecture 3
The Simplex Method
Announcements
Outline
• LP so far
• Why we can look only at basic feasible solutions
• Optimality conditions
• The simplex method
• The step from one bfs to the next
• Tableu method
• Phase I: Finding an initial BFS
LP so far
• Formulated LP’s in various contexts
• Transform any LP into a standard form LP
• Intuition of simplex method: Find the best corner
point feasible solution
• Math required:
– Corner point Feasible or basic feasible solutions
correspond to a set of n active constraints
– Any set of active constraints corresponds to a basis
from the matrix A
– The basis is a set of linearly independent columns
Standard Form
max c1 x1  c2 x2  ...  cN xN
subject to
a11x1  a12 x2  ...  a1N xN  b1
a21 x1  a22 x2  ...  a2 N x N  b2
...
aM 1 x1  aM 2 x2  ...  aMN x N  bM
x j  0, j  1..N
max cx
Concise version:
subject to Ax  b
x0
A is an m by n matrix: n variables, m constraints
Standard Form to Augmented Form
max c1 x1  c2 x2  ...  cN xN
max c1 x1  c2 x2  ...  c N x N
subject to
subject to
a11x1  a12 x2  ...  a1N xN  b1
a21 x1  a22 x2  ...  a2 N x N  b2
...
aM 1 x1  aM 2 x2  ...  aMN x N  bM
x j  0, j  1..N
max cx
subject to Ax  b
x0
A is an m by n matrix: n variables, m constraints

a11x1  a12 x2  ...  a1N xN  x1  b1

a21x1  a22 x2  ...  a2 N x N  x2  b2
...

aM 1 x1  aM 2 x2  ...  aMN x N  xN  bM

x j , xi  0, j  1..N , i  1..m
max cx
 x
subject to [ A, I ]    b
x
x0
Solutions, Extreme points and
bases
• Key fact:
– If a LP has an optimal solution, then it has an optimal
extreme point solution (proved today)
• Basic Feasible solution (Corner Point Feasible):
– The vector x is an extreme point of the solution space
iff it is a bfs of Ax=b, x&gt;=0
• If A is of full rank then there is at least one basis
B of A
– B is set of linearly independent columns of A
• B gives us a basic solution
– If this is feasible then it is called a basic feasible
solution (bfs) or corner point feasible (cpf)
Optimal basis theorem
Theorem
If a LP in standard form has a finite optimal
solution then it has an optimal basic
feasible solution
Proof
Requires the representation theorem…
Simplex Method
• Checks the corner points
• Gets better solution at each iteration
1. Find a starting solution
2. Test for optimality
– If optimal then stop
3. Perform one iteration to new CPF (BFS)
solution. Back to 2.
Simplex Method: basis change
• One basic variable is replaced by another
• The optimality test identifies a non-basic
variable to enter the basis
– The entering variable is increased until one of
the other basic variables becomes zero
– This is found using the minimum ratio test
– That variable departs the basis
Idea of simplex iterations
• New matrix A is an m by n matrix
– m constraints
– n variables
• Z = objective value (another variable)
• m+1 constraints, n variables
– We can rearrange m+1 equations, trying to
• Maximize Z
• Keeping x &gt;=0
• Consider basic solutions
– m x-variables are nonzero
– All others are zero
The simplex method
• Example
• Table version
Questions and Break
A basic feasible solution
• B=basis of A.
• Write LP in terms of basis
A  [B N ]
Ax  Bx B  NxN  b
xB   m , x N   n  m
xN  0, xB  B 1b
X is a basic solution of the LP
X is a basic feasible solution if it is feasible!
(example)
max c' x
s.t. Ax  b
x0
Optimality of a basis
We want to test of a basic feasible solution is optimal
Use the basic notation from before
Bx B  NxN  b
xB  B 1 (b  NxN )
c' x  cB ' xB  c N ' x N
 cB ' B 1 (b  NxN )  c N ' x N
 cB ' B 1b  cB ' B 1 NxN  c N ' x N
 z0   ( z j  c j )x j
j
z0  cB ' B 1b
z j  cB ' B 1 A. j j  N
max cB ' xB  cN ' xN
s.t.
Bx B  NxN  b
xB , x N  0
Optimality test
max c' x  z0   ( z j  c j )x j
j
Test :
z j  c j  0 j  N
Finding an initial bfs
• The ‘phase 1’ approach
• The ‘big M’ method
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