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MSIS 685: Linear Programming
Lecture 2
9/22/98
Scribe: Nawei chen
The Simplex Method
Two forms of linear programming problems:
The standard form
the Canonical form
max CT X
s.t. AX b
X0
mxn
AR
min CT X
s.t. AX =b
X0
AR
mxn
m
m
bR
bR
n
n
X R
X R
X3
C
We use canonical form in our course.
X2
The simplex strategy (Geometrically):
X1
E
Figure 1
We use figure 1,a geometrical expression of a LP problem, as example
A) Find an extreme point E (any one of vertices).
B) Test if E is optimal (that is check if objective function value at E is better than all of
its neighbors).
If optimal, stop. Else go to C)
C) Move to a neighbor of E that has a better objective function value.
D) Go to A)
Convex Sets and Polyhedron
Definition Hyper plane : a set of points satisfy
R
Fx I
|
G
J
HS
: a x ........a x
G
J
|THx K
1
.
.
1 1
n
n
n
n
XR
, XT= (x1,
U
|
 bV
|W
x2,…… xn )
Definition Half space : a set of points satisfy
R
Fx I
|
G
J
HS
: a x ........a x
G
J
|THx K
1
.
.
1 1
n
n
n
U
|
 bV
|W
In LP, every constraint is half space. Feasible set is the intersection of all half space .
Definition CRn is convex
If for any pair of points X1, X2  C then every point of Z on the line segment from X1
to X2 is also in C. Z can be expressed as Z= X1 + (1-) X2, 01
Example:
Figure 1
figure 2
Figure 1 is convex and figure 2 is not convex
If Z=1X1+…nXn and 0i1
Then Z is a convex combination of X1, X2…Xn.
The convex hull of X1, X2…Xn is the set
conv(X1, X2……Xn)=1X1+…….nXn , 0i1,i=1}
The convex hull of a set of vectors is the smallest convex set containing the set.
Definition Z==1 X1 +2 X2+…+m Xm is a strict convex combination of X1, X2,…..
Xm if at least one of 0<i<1, otherwise it is non-strict convex combination.
Definition if C is a convex set and X  C, then X is an extreme point if it can’t be
expressed as a strict convex combination of other points of C.
Figure a
figure b
Example : figure a) has infinite extreme points. All the vertices and points on curve on
Figure b) are extreme points.
Example . a half space is convex.
Proof :
R
Fx I
|
G
J
HS
: a x ........a x
G
|THx J
K
1
.
.
n
1 1
n
n
U
|
 bV
|W
Let XT=(x1, x2,…….. xn) H
YT=(y1, y2,…….. yn) H
ZT =(z1,z2,…….. zn)
Z = X +(1-) Y, 01, so
a1z1 + ……. anzn = a1[x1 +(1-) y1]+……… an[ xn +(1-) yn]
= [ a1 x1+…….. an xn] +(1-)[ a1 y1+…….. an yn]
  b+(1-) b =b
then Z must in H..
Theorem: if C1, C2 are convex set, then is C1  C2.
Proof: let X1, X2  C1  C2, Z = X1 +(1-) X2, 01
X1, X2  C1, so Z C1
X1, X2  C2, so Z C2
So, Z C1 C2
The feasible set of LP is convex since it is intersection of half spaces.
Definition A convex polyhedron is the intersection of a finite number of half-spaces.
Note: the feasible set of a canonical linear program is a polyhedron.
Definition A bounded convex polyhedron is called a polytope.
Fact: a polyhedron has a finite number of extreme points.
Max CTX
Subject to X C
Where C is a convex set, if it exists, is attained at an extreme point of C.
Fact: The optimum of
An extreme point of a polyhedron in Rn is always a point where at least n of the
inequalities are satisfied with equalities.
Note: in a LP, an extreme point is always the solution of a system of n equations.
Definition Two extreme points X1 and X2 are neighbors if their corresponding system
of equations differ only in a equation.
We use an example to illustrate the steps of simplex method.
Max
5x1 + 3 x2 + 4x3
Subject to
x1 + x2 + x3  1
2 x1 - x2 + x3  5
x1 + 3 x2 - x3  4
x1 , x2, x3  0
Step 1. Introduce n variable to make n inequalities into equalities.
Then, we introduce three slack variable x4,
x5, x6, and the constraints become
x1 + x2 + x3 + x4
=1
2 x1 - x2 + x3
+ x5
=5
x1 + 3 x2 - x3
+ x6 = 4
x1,2,3 ,4,5,6  0
step 2.
Find an initial solution (extreme point) and construct a dictionary.
Group non basic variable to the right side and basic variable to the left side.
When x1 = x2 = x3 = 0 ,all constraints are satisfied. we get the initial solution.
We construct the dictionary:
Z=
5 x1 + 3 x2 + 4 x3
x4 = 1 - x1 - x2 - x3
x5= 5 - 2 x1 + x2 - x3
x6= 4 - x1 - 3 x2 + x3
basic var
x1,2,3 ,4,5,6  0
non-basic var
step 3. Move from one point to its neighbor. Since only one equation differs , we will
change one basic variable with one non-basic variable.
Choose x1 as the entering variable, since the coefficient of x1 is positive.
x1 should satisfy x1 1, x1  5/2, x1  4 to make x4,5,6  0,then min(1,5/2,4)=1,
we choose x4
as out variable.
Step 4 . Construct new dictionary .
We use x1 as entering variable, x4 as out variable and get new dictionary:
Z = 5 -2 x2 - x3 - 5 x4
x1 = 1 - x2 - x3 - x4
x5 = 3 + 3 x2 + x3
x6= 3 - 2 x2 +2 x3 + x4
step 5. If coefficients of variable function in objective function is positive , we will go
to step 3. Else, when coefficients is negative, we get the optimal solution.
The coefficients in step 4. Objective functions are all negative, we get the
optimal solution. Max Z= 5 at the extreme point (1,0,0,0,3,3).