Simplex Method Continued …
from a presentation at the Fuqua School of Business
MIT and James Orlin
© 2003
Rev. 2.71 by M. Miccio on December 15, 2014
file Simplex2_AMII_05b_gr
MIT and James Orlin © 2003
1
Today’s Lecture
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



Review of the simplex algorithm
Alternative Optimal Solutions
Unbounded Solution
Degeneracy
Is the simplex algorithm finite? (Answer, yes,
but only if we are careful)
MIT and James Orlin © 2003
2
Today’s Lecture





Review of the simplex algorithm
Alternative Optimal Solutions
Unbounded Solution
Degeneracy
Is the simplex algorithm finite? (Answer, yes,
but only if we are careful)
MIT and James Orlin © 2003
3
Review of the Notation







n
number of variables
m
number of constraints
c
index of entering variable
r
index of pivot row
– note: the rth basic variable leaves the basis
The original data is cj, aij, bi
After performing pivots we represent the revised
coefficients as cj, aij, bi
The bar − indicates that it is the coefficient after
some pivots
MIT and James Orlin © 2003
4
Tableau
Basic Variables
Non-basic Variables
-z
x1
x2
xr
xm xm+1
xs
xn
1
0
0
0
0
cm+1
cs
cn
=
-z0
0
1
0
0
0 a1,m+1
a1,s
a1,n
=
b1
0
0
1
0
0 a2,m+1
a2,s
a2,n
=
b2
0
0
0
1
0 ar,m+1
ar,s
ar,n
=
br
0
0
0
0
1 am,m+1
am,s
am,n
=
bm
.....
.....
CV
m constraints, n variables
MIT and James Orlin © 2003
5
The basic feasible solution

The current values are all non-negative.
– This is needed for canonical form

There is a basic variable associated with each
constraint.
– in this Tableau, the basic variable associated with
constraint i is xi.

There are (n-m) nonbasic variables.
– In this Tableau, the nonbasic variables are
xm+1, … xn.

This bfs is as follows: x1 =b1, … , xm =bm.
All other variables are 0.
MIT and James Orlin © 2003
6
Today’s Lecture






Review of the simplex algorithm.
Alternative Optimal Solutions
Unbounded Solution
Degeneracy
Degenerate Solutions
Proving finiteness and optimality under no
degeneracy
MIT and James Orlin © 2003
7
Alternative Optima (maximization)
Initial Tableau
z = 0x1 -4x2 + 8
max!
How much can x1 be
increased?
-z
x1
x2
x3
x4
1
0
-4
0
0
=
-8
0
1
3
1
0
=
6
0
-1
2
0
1
=
2
This basic feasible solution is optimal! Is
there another optimal solution?
Suppose that x1 entered the basis.
What would leave?
MIT and James Orlin © 2003
8
Alternative Optima (maximization)
Initial Tableau
-z
x1
x2
x3
x4
1
c
01
c
-42
0
0
=
-8
0
1
3
1
0
=
6
0
-1
2
0
1
=
2
Use the min ratio rule to select the pivot row and to
determine which variable leaves the basis.
MIT and James Orlin © 2003
9
Alternative Optima (maximization)
Tableau No.1
-z
x1
x2
x3
x4
1
0
-4
0
0
=
-8
0
1
3
1
0
=
6
0
-1
0
52
10
1
=
82
Let x1 enter the basis. Let the basic variable in
constraint 1 leave the basis.
Note: the solution is different, but the objective function
value is the same.
MIT and James Orlin © 2003
10
Alternative Optima
Test for infinite solutions
 c'j  0
for any non - basic column j ?
There are alternative optima
ifcj = 0 for some j where xj is non-basic
MIT and James Orlin © 2003
11
Today’s Lecture






Review of the simplex algorithm.
Alternative Optimal Solutions
Unbounded Solution
Degeneracy
Degenerate Solutions
Proving finiteness and optimality under no
degeneracy
MIT and James Orlin © 2003
12
Unboundedness
z = 1x1+ 0x2 + 2 max!
-z
x1
x2
x3
x4
1
1
0
0
-1
=
-2
0
-3
0
1
-1.5
=
3
0
-2
1
0
.5
=
1
x1 = D
0
= 11 − (-2)D
x2 =
= 33 − (-3)D
x3 =
x4 = 0
z == 2 + D
The entry variable is x1. Set x1 = D>0 while x4 = 0.
D can be as large as we like !
If all the coefficients in the entering column are  0,
then the solution is unbounded from above
MIT and James Orlin © 2003
13
Unboundedness
Test for Unbounded Solution
i, a  0
'
ic
If aic  0 for all i, then the solution is unbounded.
MIT and James Orlin © 2003
14
Simplex Algorithm step by step (Search for Max)
Step 0. The problem is in canonical form withb  0.
Step 1. If c  0 then stop. The solution is optimal. If there exists
some cj>0, then we continue.
Step 2. Choose any non-basic variable to pivot in with cs > 0, e.g.,
cs = maxj {cj | cj > 0 }. If ais  0 for all i, then stop; the LP is
unbounded. If we continue, then there exists some ais > 0.
Step 3. Pivot out the basic variable in row r, where r is chosen by
the min ratio rule, that is r = argmini (bi/ais : ais > 0 ).
Step 4. Replace the basic variable in row r with variable xs and reestablish canonical form (i.e., pivot on the coefficient ars)
Step 5. Go back to Step 1.
MIT and James Orlin © 2003
15
Today’s Lecture





Review of the simplex algorithm.
Alternative Optimal Solutions
Unbounded Solution
Degeneracy and Degenerate Solutions
Proving finiteness and optimality under no
degeneracy
MIT and James Orlin © 2003
16
Degeneracy
Initial Tableau
BV
-z
x1
x2
x3
x4
-z
1
-3
2
0
0
=
-2
x3
0
-3
3
1
0
=
6
x4
0
-4
2
0
1
=
02
A bfs is degenerate if the Tableau revealsbi = 0 for some i.
Otherwise, it is non-degenerate.
This bfs is degenerate.
MIT and James Orlin © 2003
17
Degeneracy
BV
-z
x1
x2
x3
x4
-z
1
-3
2
0
0
=
-2
x3
0
-3
3
1
0
=
6
x4
0
-4
2
0
1
=
32
A bfs is degenerate if the Tableau revealsbi = 0 for some i.
Otherwise, it is non-degenerate.
This bfs is non-degenerate.
MIT and James Orlin © 2003
18
Degeneracy
BV
-z
x1
x2
x3
x4
-z
1
-3
2
0
0
=
-2
x3
0
-3
3
1
0
=
6
x4
0
-4
2
0
1
=
02
x1 = 0
x2 = D
x3 = 6 - 3D
x4 = 0 - 2D
z = 2 + 2D
Suppose that variable x2 will enter the basis.
Degenerate  the solution may stay the same
 z stays the same
MIT and James Orlin © 2003
19
A degenerate pivoting
Tableau No.1
BV
-z
x1
x2
x3
x4
-z
1
-3
1
02
0
-1
0
=
-2
x3
0
-3
3
03
1
-3/2
0
=
6
x4
0
-4
-2
12
0
1/2
1
=
0
x1 = 0
x2 = 0
x3 = 6
x4 = 0
z= 2
The entering variable is x2, which takes a zero value.
The exiting variable is the one in constraint 2.
In this case the objective function did not change.
(But the tableau did !)
MIT and James Orlin © 2003
20
Cycling
Problem:
– if one of the basic variables is zero: Xi = 0 + …
a pivot could be performed which does not change
the corresponding value of the objective function
 danger of pivoting back (“cycling”)
Solution:
– use e.g. Bland’s anti-cycling rule:
• always select the candidate pivot column with smallest
index, i.e., the leftmost in Tableau (e.g., x2 instead of x4)
• if several rows have the same minimum ratio, choose for
pivoting the first one (i.e., topmost) of them
– this is alternative to the pivot column selection rule
that was previously discussed
The Simplex Algorithm
11 November, 2002
21
Degeneracy
Test for Degeneracy
 2 or more rows for which :
'
r
'
i
b
b
 min i ' > 0
'
a rc
a ic
The next bfs will be degenerate !
The Simplex Algorithm
11 November, 2002
22
Non-degeneracy leads to strict improvement
BV
-z
x1
x2
x3
x4
-z
1
-3
2
0
0
=
-2
x3
0
-3
3
1
0
=
6
x4
0
-4
2
0
1
=
2
x1 = 0
x2 = D
x3 = 6 - 3D
x4 = 2 - 2D
z = 2+ 2D
If the bfs is non-degenerate, then the entering
variable can strictly increase, and the objective value
will strictly improve.
MIT and James Orlin © 2003
23
Cycling
Alternative to Bland’s anti-cycling rule

Perturb the RHS just a little, in just the
right way
– No basis is degenerate
– Every bfs to the perturbed problem is also a
bfs for the original problem
– The optimal basis for the perturbed problem is
optimal for the original problem
MIT and James Orlin © 2003
24
An LP before perturbation
Initial Tableau
BV
-z
x1
x2
x3
x4
-z
1
-3
2
0
0
=
0
x3
0
-3
3
1
0
=
6.
x4
0
-4
2
0
1
=
2.
Goal: modify the problem by changing the RHS just a little
Perturb it to avoid degeneracy
Perturb it so little that solving the perturbed problem also solves
the original problem.
MIT and James Orlin © 2003
25
An example of a perturbed initial bfs
BV
-z
x1
x2
x3
x4
-z
1
-3
2
0
0
=
x3
0
-3
3
1
0
=
6.00000000000013
x4
0
-4
2
0
1
=
2.00000000000041
0
In theory perturbations are even smaller
In practice, the simplex method works fine without perturbations,
that is, it really obtains the optimum basic feasible solution.
MIT and James Orlin © 2003
26
Today’s Lecture






Review of the simplex algorithm.
Alternative Optimal Solutions
Unbounded Solution
Degeneracy
Degenerate Solutions
Proving finiteness and optimality under no
degeneracy
MIT and James Orlin © 2003
27
Theorem: If every feasible basic
solution is non-degenerate, the simplex
method is finite.
1.
The number of feasible basic solutions is at most
n! / (n-m)! m!
which is the number of ways of selecting m basic
variables
2.
Each feasible basic solution is different
• Each has a better cost than the last, assuming non-degeneracy
Therefore, the simplex method is finite
3.
•
Typical number of iterations is between m and 3m.
Number of operations per iteration: O(nm)
MIT and James Orlin © 2003
28
Some Remarks on Degeneracy and
Alternative Optima
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
It might seem that degeneracy would be rare. After all,
why should we expect that the RHS would be 0 for
some variable?
In reality, degeneracy is incredibly common.
The simplex method is not necessarily finite unless
care is taken in the pivot rule. In reality, almost no
one takes care, and the simplex method is not only
finite, but it is incredibly efficient.
The issue of alternative optima arises frequently, and
is of importance in practice.
MIT and James Orlin © 2003
29
Software for Simplex implementation and solution
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Demonstration Example
Maximize the Objective Function (P)
P = 15 x1 + 10 x2
subject to:
L1:
0.25 x1 + 1.0 x2 <= 65
L2:
1.25 x1 + 0.5 x2 <= 90
L3:
1.0 x1 + 1.0 x2 <= 85
x1 >= 0; x2 >= 0
Solved step-by-step on
Operations Research-Linear Programming-Simplex Algorithm
by Elmer G. Wiens
http://www.egwald.ca/operationsresearch/lpsimplex.php
MIT and James Orlin © 2003
31
Demonstration Simplex solver
Solved step-by-step Tableau
and
Final Solution
on
Finite mathematics utility: simplex method tool
http://www.zweigmedia.com/RealWorld/simplex.html
MIT and James Orlin © 2003
32
SIMPLEX with MatLab®
LINEAR PROGRAMMING WITH MATLAB
course by Edward Neuman
Department of Mathematics
Southern Illinois University at Carbondale
function simplex2p(type, c, A, rel, b)
% The Two-Phase Method for solving the LP problem
% implements the Bland’s rule to avoid cycling in case of degeneracy
%
% min (max) z = c*x
% subject to Ax rel b
%
x >= 0
...
% input data
type = ‘min’
(OR ‘max’)
% simplex2p always searches for MIN!
c = [c1 c2 ... ck] % is a row vector
A = [a11 a12 … a1k; ... ; am1 am2 … amk]
rel = ‘<<< … >>> … ===‘
b = [b1; b2; ... bm] % must be a column vector
33
Exercise 5 (for home)
with contraints of type ≤
z = 3x1 + 2x2 - x3 + x4
x1 + 2x2 + x3 - x4  5
2x1 + 4x2 + x3 + x4  1
x1  0, x2  0, x3  0
x4
max !
inequality
inequality
non-negativity
uncostrained in sign
34