1. Simplex Method Continued …
from a presentation at the Fuqua School of Business
MIT and James Orlin
© 2003
Rev by M. Miccio on December 15, 2014
MIT and James Orlin © 2003
file Simplex2_AMII_05b_gr 1

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 2

3. 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

4. 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

5. m constraints, n variables
Tableau
Basic Variables
Non-basic Variables -z x1 x2 xr xm
xm+1 xs xn CV 1
cm+1
cs
cn
-z0 = 1
a1,m+1
a1,s
a1,n
b1 = 1
a2,m+1
a2,s
a2,n
b2 = 1
ar,m+1
ar,s
ar,n
br = 1
am,m+1
am,s
am,n
bm =
m constraints, n variables
MIT and James Orlin © 2003 5

6. 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

7. 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

8. Alternative Optima (maximization) Initial Tableau
z = 0x1 -4x2 + 8 max!
How much can x1 be increased? -z x1 x2 x3 x4 1 -4 -8 = 1 3 1 6 = -1 2 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

9. Alternative Optima (maximization) Initial Tableau1
c1
c2 -4 -8 = 1 3 1 6 = -1 2 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

10. Alternative Optima (maximization) Tableau No.1-4 -4 -8 -8 = 1 1 3 3 1 1 6 6 = -1 2 5 1 1 1 8 2
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

11. Test for infinite solutions
Alternative Optima
Test for infinite solutions
There are alternative optima
ifcj = 0 for some j where xj is non-basic
MIT and James Orlin © 2003 11

12. 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

13. Unboundedness z = 1x1+ 0x2 + 2 max! -z x1 x2 x3 x4 1 -2 1 -3 1 -1 1-1 1 -2 3 =
x1 = 0 x2 = 1 x3 = 3 x4 = 0 z = 2
x1 = D x2 = 1 − (-2)D x3 = 3 − (-3)D x4 = 0 z = 2 + D -3 1
-1.5 = -2 1 .5
Verif. Cosa succede per a21 = +2
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

14. Unboundedness Test for Unbounded Solution
If aic  0 for all i, then the solution is unbounded.
MIT and James Orlin © 2003 14

15. 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 re-establish canonical form (i.e., pivot on the coefficient ars)
Step 5. Go back to Step 1.
MIT and James Orlin © 2003 15

16. 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

17. Degeneracy Initial TableauBV -z x1 x2 x3 x4 -z 1 -3 2 -2 = x3 -3 3 1 1 6 = x4 -4 2 1 2
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

18. Degeneracy 3 BV -z x1 x2 x3 x4 -z 1 -3 2 -2 = x3 -3 3 1 1 6 = x4 -4 2-2 = x3 -3 3 1 1 6 = x4 -4 2 1 3 3 2
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

19. Degeneracy BV -z x1 x2 x3 x4 -z 1 -3 2 -2 =
x1 = 0 x2 = D x3 = 6 - 3D x4 = 0 - 2D z = D x3 -3 3 1 1 6 = x4 -4 2 1 2
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

20. A degenerate pivoting Tableau No.1BV -z x1 x2 x3 x4 -z 1 -3 1 2 -1 -2 -2 =
x1 = 0 x2 = 0 x3 = 6 x4 = 0 z = 2 x3 -3 3 3 1 1 1
-3/2 6 6 = x4 -2 -4 1 2 1
1/2
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

21. Cycling Problem: Solution:
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

22. Degeneracy Test for Degeneracy The next bfs will be degenerate !
The Simplex Algorithm
11 November, 2002 22

23. Non-degeneracy leads to strict improvementBV -z x1 x2 x3 x4 -z 1 -3 2 -2 =
x1 = 0 x2 = D x3 = 6 - 3D x4 = 2 - 2D z = 2+ 2D x3 -3 3 1 1 6 = x4 -4 2 1 2
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

24. 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

25. An LP before perturbation Initial TableauBV -z x1 x2 x3 x4 -z 1 -3 2 = x3 -3 3 1 1 6. = x4 -4 2 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

26. An example of a perturbed initial bfsBV -z x1 x2 x3 x4 -z 1 -3 2 = x3 -3 3 1 1 = x4 -4 2 1
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

27. 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

28. Theorem: If every feasible basic solution is non-degenerate, the simplex method is finite.
The number of feasible basic solutions is at most n! / (n-m)! m! which is the number of ways of selecting m basic variables
Each feasible basic solution is different
Each has a better cost than the last, assuming non-degeneracy
Therefore, the simplex method is finite
Typical number of iterations is between m and 3m. Number of operations per iteration: O(nm)
MIT and James Orlin © 2003 28

29. Some Remarks on Degeneracy and Alternative Optima
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

30. Software for Simplex implementation and solution

31. Demonstration Example
Maximize the Objective Function (P)
P = 15 x x2
subject to:
L1: x x2 <= 65 L2: x x2 <= 90 L3: 1.0 x x2 <= 85
      x1 >= 0; x2 >= 0
Solved step-by-step on
Operations Research-Linear Programming-Simplex Algorithm by Elmer G. Wiens
MIT and James Orlin © 2003 31

32. Demonstration Simplex solver
Solved step-by-step Tableau
and
Final Solution on
Finite mathematics utility: simplex method tool
MIT and James Orlin © 2003 32

33. SIMPLEX with MatLab® function simplex2p(type, c, A, rel, b)
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

34. Exercise 5 (for home) with contraints of type ≤
z = 3x1 + 2x2 - x3 + x4 max !
x1 + 2x2 + x3 - x4  5 inequality
2x1 + 4x2 + x3 + x4  1 inequality
x1  0, x2  0, x3  0 non-negativity
x4 uncostrained in sign