Transcript Degeneracy and the Convergence of the Simplex Algorithm

The big M method
LI Xiao-lei
The big M method



The simplex algorithm requires a starting
bfs. We found a starting bfs by using the
slack variables as our basic variables.
If an LP has any ≥ or equality constraints,
a starting bfs may not be readily
apparent.
The big M method (or the two-phase
simplex method) may be used.
Example 4

Bevco manufactures an orange-flavored soft
drink called Oranj by combining orange soda
and orange juice.
Each ounce of orange soda contains 0.5 oz of
sugar and 1 mg of vitamin C. each ounce of
orange juice contains 0.25 oz of sugar and 3
mg of vitamin C.
It costs Bevco 2¢ to produce an ounce of
orange soda and 3¢ to produce an ounce of
orange juice.
Example 4

Bevco’s marketing department has
decided that each 10-oz bottle of Oranj
must contain at least 20 mg of vitamin C
and at most 4 oz of sugar.
Use LP to determine how Bevco can
meet the marketing department’s
requirements at minimum cost.
Example 4

Solution
Let
x1=number of ounces of orange soda in a bottle of Oranj
x2=number of ounces of orange juice in a bottle of Oranj
Then the appropriate LP is
min z=2x1+3x2
s.t. 1/2x1+1/4x2≤4 (sugar constraint)
x1+ 3x2≥20 (vitamin C constraint)
(17)
x1+ x2=10 (10 oz in bottle of Oranj)
x1,x2≥0
Example 4

Put (17) into standard form
z -2x1 -3x2
=0
1/2 x1+1/4x2+s1 =4
x1+ 3x2
-e2=20
x1+ x2
=10
All variables nonnegative
Example 4

Search for a bfs
In row 1, s1=4 could be used as a basic variable
In row 2, e2=-20 violates the sign restriction.
In row 3, there is no readily apparent basic
variable.

Artificial variables
To remedy this problem, we simply “invent” a
basic feasible variable for each constraint that
needs one. These variables are created by us
and are not real variables, we call them
artificial variables.
If an artificial variable is added to row i, we label
it ai.
Example 4

In the current problem, we need to add
an artificial variable a2 to row 2 and an
artificial variable a3 to row 3.
z -2x1 -3x2
=0
1/2 x1+1/4x2+s1
=4
x1+ 3x2
-e2+a2
=20
(18)
x1+ x2
+a3 =10
We now have a bfs: z=0,s1=4,a2=20,a3=10


There is no guarantee that the optimal solution
to (18) will be the same as the optimal solution
to (17).
In a min problem, we can ensure that all the
artificial variables will be zero by adding a term
Mai to the objective function for each artificial
variable ai. (in a max problem, add a term –Mai
to the objective function.)
Here M represents a “very large” positive
number.
Example 4

In (18), we could change our objective function
to
min z=2x1+3x2+Ma2+Ma3
The row 0 will change to
z-2x1-3x2-Ma2-Ma3=0


With this modified objective function, it seems
reasonable that the optimal solution to (18) will
have a2=a3=0. in this case, the optimal solution
to (18) will solve the original problem (17).
Some of the artificial variables may assume
positive values in the optimal solution. If this
occurs, the original problem has no feasible
solution.
Description of big M method

Step 1
Modify the constraints so that the right-hand
side of each constraint is nonnegative.
Each constraint with a negative right-hand side
be multiplied through by -1. if you multiply an
inequality by any negative number, the
direction of the inequality is reversed.
For example,
x1+x2≥-1 into –x1-x2≤1
x1-x2≤-2 into -x1+x2≥2
Description of big M method

Step 1’
Identify each constraint that is now an = or ≥
constraint. Prepare for step 3 – which
constraint will the artificial variable to be
added.

Step 2
Convert each inequality constraint to standard
form.
For ≤ constraint i, we add a slack variable si;
For ≥ constraint i, we add an excess variable ei;
Description of big M method

Step 3
If constraint i is a ≥ or = constraint, add an
artificial variable ai, also add the sign
restriction ai≥0.

Step 4
Let M denote a vary large positive number.
If the LP is a min problem, add (for each
artificial variable) Mai to the objective function.
If the LP is a max problem, add (for each
artificial variable) -Mai to the objective function.
Description of big M method

Step 5
Since each artificial variable will be in the
starting basis, all artificial variables must be
eliminated from row 0 before beginning the
simplex. This ensures that we begin with a
canonical form. In choosing the entering
variable, remember that M is a very large
positive number.
Description of big M method
For example, 4M-2 is more positive than
3M+900, and -6M-5 is more negative than 5M-40.

Now solve the transformed problem by
the simplex.
If all artificial variables are equal to zero in the
optimal solution, we have found the optimal
solution to the original problem.
If any artificial variables are positive in the
optimal solution, the original problem is
infeasible.
Example 4 (continued)
Solution
 Step 1
Since none of the constraints has a negative
right-hand side, we don’t have to multiply any
constraint through by -1.

Step 1’
Constraints 2 and 3 will require artificial
variables.
Example 4 (continued)

Step 2
Add a slack variable s1 to row 1 and subtract an
excess variable e2 from row 2.
min z= 2x1+ 3x2
row 1: 1/2x1+1/4x2+s1 =4
row 2:
x1+ 3x2 -e2=20
row 3:
x1+ x2
=10
Example 4 (continued)

Step 3
Add an artificial variable a2 to row 2 and an
artificial variable a3 to row 3.
min z= 2x1+ 3x2
row 1: 1/2x1+1/4x2+s1
=4
row 2:
x1+ 3x2 -e2+a2=20
row 3:
x1+ x2
+a3=10
From this tableau, the initial bfs will be
s1=4,a2=20 and a3=10.
Example 4 (continued)

Step 4
For the min problem, we add Ma2+Ma3 to the
objective function.
The objective function is now
min z=2x1+3x2+Ma2+Ma3
Example 4 (continued)

Step 5
Row 0 is now z-2x1-3x2-Ma2-Ma3=0
To eliminate a2 and a3 from row 0, simply
replace row 0 by row 0+M(row 2)+M(row 3).
This yields
Row 0:
z2x1 -3x2
-Ma2-Ma3=0
M(row 2):
mx1+3Mx2-Me2+Ma2
=20M
M(row 3):
Mx1 +Mx2
+Ma3=10M
New row 0: z+(2M-2)x1+(4M-3)x2-Me2
=30M
Example 4 (continued)
Initial tableau for Bevco
z
x1
x2
s1
e2
a2
a3
rhs
Basic
variable
1
2M-2
4M-3
0
-M
0
0
30M
z=30M
0
½
¼
1
0
0
0
4
s1=4
16
0
1
③
0
-1
1
0
20
a2=20
20/3*
0
1
1
0
0
0
1
10
a3=10
10
ratio
Since we are solving a min problem, the variable with the most
positive coefficient in row 0 should enter the basis.
Since 4M-3>2M-2, variable x2 should enter the basis. The ratio test
indicates that x2 should enter the basis in row 2.
Example 4 (continued)

Replace row 2 by 1/3(row 2). The new
row 2 is
1/3x1+x2-1/3e2+1/3a2=20/3

Eliminate x2 from row 0 by adding –(4M3)(new row 2) to row 0. the new row 0 is
z+(2M-3)/3x1+(M-3)/3e2+(3-4M)/3a2=(60+10M)/3

Using ero’s to eliminate x2 from row 1
and 3.
Example 4 (continued)
First tableau for Bevco
a3
rhs
Basic
variable
z
x1
x2
s1
e2
a2
ratio
1
(2M3)/3
0
0
(M3)/3
(30
4M)/3
(60+1 z=(60+10
0M)/3 M)/3
0
5/12
0
1
1/12
-1/12
0
7/3
s1=7/3
28/5
0
1/3
1
0
-1/3
1/3
0
20/3
x2=20/3
20
0
2/3
0
0
1/3
-1/3
1
10/3
a3=10/3
5*
The ratio test indicates that x1 should enter the basis in
the third row of the current tableau.
Example 4 (continued)

Replace row 3 by 3/2(row 3). The new
row 3 is
x1+1/2e2-1/2a2+3/2a3=5

Eliminate x1 from row 0 by adding (32M)/3 (new row 3) to row 0. the new row
0 is
z-1/2e2+(1-2M)/2a2+(3-2M)/2a3=25

Using ero’s to eliminate x1 from row 1
and 2.
Example 4 (continued)
Optimal tableau for Bevco
a3
rhs
Basic
variable
z
x1
x2
s1
e2
a2
1
0
0
0
-1/2
(1(325
2M)/2 2M)/2
z=25
0
0
0
1
-1/8
1/8
-5/8
¼
s1=1/4
0
0
1
0
-1/2
½
-1/2
5
x2=5
0
1
0
0
1/2
-1/2
3/2
5
x1=5
Since all variables in row 0 have nonpositive coefficients, this
is an optimal tableau.
Since all artificial variables are equal to zero, we have found
the optimal solution to the Bevco problem.
The optimal solution to the Bevco problem: z=25,x1=x2=5,
s1=1/4,e2=0.
Example 4 (continued)
Note
 The a2 column could have been dropped
after a2 left the basis (at the conclusion
of the first pivot), and the a3 column
could have been dropped after a3 left the
basis( at the conclusion of the second
pivot).
summary


If any artificial variable is positive in the
optimal Big M tableau, the original LP
has no feasible solution.
When the Big M method is used, it is
difficult to determine how large M should
be. Generally, M is chosen to be at least
100 times larger than the largest
coefficient in the original objective
function.