Week_4: simplex method II
1
1.introduction
LPs in which all the constraints are (≤) with nonnegative right-hand sides offer a
convenient all-slack starting basic feasible solution. Models involving (=) and/or (≥)
constraints do not.
The procedure for starting "ill-behaved" LPs with (=) and (≥) constraints is to use
artificial variables that play the role of slacks at the first iteration, and then dispose of
them legitimately at a later iteration. Two closely related methods are introduced here:
the M-method and the two-phase method.
2.M-method
The M-method starts with the LP in equation form, If equation i does not have a slack (or
a variable that can play the role of a slack), an artificial variable, Ri, is added to form a
starting solution similar to the convenient all-slack basic solution, However, because the
artificial variables are not part of the original LP model, they are assigned a very high
penalty in the objective function, thus forcing them (eventually) to equal zero in the
optimum solution, This will always be the case if the problem has a feasible solution, The
following rule shows how the penalty is assigned in the cases of maximization and
minimization:
Penalty Rule for Artificial Variables.
Given M, a sufficiently large positive value (mathematically, M
∞ ), the objective
coefficient of an artificial variable represents an appropriate penalty if:
𝑎𝑟𝑡𝑖𝑓𝑖𝑐𝑖𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑐𝑜𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 =
−𝑀 𝑖𝑛 𝑚𝑎𝑥𝑖𝑚𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑠
𝑀 𝑖𝑛 𝑚𝑖𝑛𝑖𝑚𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑠
Example:
Minimize
z=4x1+x2
Subject to
3x1+x2=3
4x1+3x2≥6
x1+2x2≤4
x1,x2≥0
Using x3 as a surplus in the second constraint and X4 as a slack in the third constraint, the equation form
of the problem is given as:
minimize z=4x1+x2
subject to
2
3x1+x2
=3
4x1+3x2 -x3
=6
x1+2x2
+x4 =4
X1,x2,x3,x4≥0
The third equation has its slack variable, X4, but the first and second equations do not. Thus, we add
the artificial variables RI and Rz in the first two equations and penalize them in the objective function
with M R1 + M R2 (because we are minimizing). The resulting LP is given as
minimize z=4x1+x2+MR1+MR2
subject to
3x1+x2
+R1
=3
4x1+3x2-x3
+R2=6
x1+2x2
+x4
=4
x1, x2, x3, x4, R1, R2 ≥ 0
The associated starting basic solution is now given by (R1, R2, x4) = (3,6,4).
M must assume a numeric value, M must be sufficiently large relative to the original objective
coefficients so it will act as a penalty that forces the artificial variables to zero level in the optimal
solution. Using M = 100, the starting simplex tableau is given as follows (for convenience, the zcolumn is eliminated because it does not change in aI1 the iterations):
First objective function written as
Minimize z-4x1-x2-0x3-100R1-100R2
And then simplex table as:
basic
Z
R1
R2
X4
X1
-4
3
4
1
X2
-1
1
3
2
X3
0
0
-1
0
R1
-100
1
0
0
R2
-100
0
1
0
X4
0
0
0
1
solution
0
3
6
4
Before proceeding with the simplex method computations, we need to make the z-row
consistent with the rest of the tableau. Specifically, in the tableau, X1=X2=X3= 0, which
yields the starting basic solution Rl =3, R2=6 , and X4 = 4. This solution yields
z = 100 x 3 + 100 x 6 = 900 (instead of 0, as the right-hand side of the z-row currently
shows). This inconsistency stems from the fact that R1 and R2 have nonzero coefficients
(-100, -100) in the z-row (compare with the all-slack starting solution in example,
where the z-row coefficients of the slacks are zero).
We can eliminate this inconsistency by substituting out Rj and R2 in the z-row using the
appropriate constraint equations. In particular, notice the highlighted elements (= 1) in the
R1-row and the Rz-row. Multiplying each of R1-row and R2-row by 100 and adding the
sum to
the z-row will substitute out R1 and R2 in the objective row-that is,
3
New z-row = Old z-row + (100 X R1-row + 100 x R2-row) The modified
tableau thus becomes
leave
basic
Z
R1
R2
X4
inter
X1
696
3
4
1
Pivot
column
X2
399
1
3
2
X3
-100
0
-1
0
R1
0
1
0
0
R2
0
0
1
0
X4
0
0
0
1
solution
900
3
6
4
Pivot row
Because we are minimizing the objective function, the variable x1 having the most positive
coefficient in the z-row (= 696) enters the solution
To determine leave variable we compute ratio of solution on coefficient of pivot column
and choose minimize ratio.
Basic
R1
R2
X4
Inter variable
Solution
ratio
x1
3
3
X1=3/3=1
4
6
X1=6/4=3/2=1.5
1
4
X1=4/1=4
Conclusion: inter x1 variable and leave R1 variable
minimize
the new tableau can be computed by using the familiar Gauss-Jordan operations.
1. Replace Sl in the Basic column with XI:
New xl-row = Current R1-row ÷ 3
= (1/3)(3 1 0 1 0 0 3)
= (1 1/3 0 1/3 0 0 1)
2. new z-row=current z-row-(696)*new x1-row
=(696 399 -100 0 0 0 900)-(696)*( 1 1/3 0 1/3 0 0 1)
=(0 399-232 -100 -232 0 0 900-696)=(0 167 -100 -232 0 0 204)
3.new R2-row =current R2-row –(4)*new x1-row
=(4 3 -1 0 1 0 6)-(4)*( 1 1/3 0 1/3 0 0 1)
=(0 3-4/3 -1 -4/3 1 0 2 )=(0 5/3 -1 -4/3 1 0 2)
4-new x4-row=current x4-row –(1)*new x1- row
=(1 2 0 0 0 1 4)-(1)( 1 1/3 0 1/3 0 0 1)
=(0 2-1/3 0 -1/3 0 1 3)=(0 5/3 0 -1/3 0 1 3 )
4
leave
basic
Z
x1
R2
X4
X1
0
1
0
0
inter
X2
167
1/3
5/3
5/3
Pivot
column
X3
-100
0
-1
0
R1
-232
1/3
-4/3
-1/3
R2
0
0
1
0
X4
0
0
0
1
solution
204
1
2
3
Pivot row
Because we are minimizing the objective function, the variable x2 having the most positive
coefficient in the z-row (=167) enters the solution
To determine leave variable we compute ratio of solution on coefficient of pivot column
and choose minimize ratio.
Basic
Inter variable
Solution
ratio
x2
1/3
1
X2=1/(1/3)=3
5/3
2
X2=2/(5/3)=6/5
5/3
3
X2=3/(5/3)=9/5
Conclusion: inter x2 variable and leave R2 variable
X1
R2
X4
minimize
The last tableau shows that x2 and R2 are the entering and leaving variables,
respectively.
Continuing with the simplex computations, two more iterations are needed to reach the
optimum:
X1=2/5
X2=9/5
Z=17/5
3. Alternative Optima
the objective function can assume the same optimal value at more than one solution
point, thus giving rise to alternative optima. The next example shows that there is an
infinite number of such solutions. It also demonstrates the practical significance of
encountering such solutions.
Example:
Maximize
z=2x1+4x2
Subject to
x1+2x2≤ 5
5
x1+x2≤ 4
x1,x2 ≥0
solution:
step 1: convert inequality to equality form.
maximize
z-2x1-4x2=0
subject to
x1+2x2+x3
=5
x1+x2+
x4=4
x1,x2,x3,x4x≥ 0
step 2: determine start basic feasible solution by set x1,x2=0 (nonbasic variable)
x3=5
x4=4
Basic
variable
leave
X3
X4
Z
X1
1
0
0
-2
1
1
X2
inter
-4
2
1
Pivot
column
X3
X4
solution
0
1
0
0
0
1
0
5
4
Pivot -row
Step 3: Select an entering variable (which represent large negative value of objective function
), here x2=-4
Step 4: leave variable (which represent minimum ratio value of solution on coefficient of
pivot column )
Basic
X3
X4
Entering
X2
Solution
2
5
1
4
Conclusion: x2 enters and x3 leaves
6
Ratio
(or intercept)
X3=5/2=2.5
X1=4/1=4
Minimum
Step 5: Determine the new basic solution by using the appropriate Gauss-Jordan
Computations
These computations are applied to the preceding tableau in the following manner:
1. Replace x2 in the Basic column with x3:
New x2-row = Current x3-row ÷ 2
= (1/2)( 0 1 2 1 0 5)
= (0 1/2 1 1/2 0 2.5)
2. new z-row=current z-row-(-4)*new x2-row
=(1 -2 -4 0 0 0 )-(-4)*( 0 1/2 1 1/2 0 2.5 )
=( 1 0 0 2 0 10)
3.new x4-row =current x4-row –(1)*new x2-row
=( 0 1 1 0 1 4) -(1)*( 0 1/2 1 1/2 0 2.5)
=( 0 1/2 0 -1/2 1 1.5)
Basic
variable
leave
X2
X4
Z
X1
inter
0
1/2
1/2
1
0
0
X2
X3
X4
solution
0
1
0
2
1/2
-1/2
0
0
1
10
2.5
1.5
Pivot
column
and then applied same steps
Basic
variable
Z
X2
X1
Z
X1
X2
X3
X4
solution
1
0
0
0
0
1
0
1
0
2
1
-1
0
-1
2
10
1
3
Final result
1- At point B (x1=0, x2=1/2)
2- At point C (x1=3, x2=1)
7
Pivot
row
8