INDR 262 Introduction to Optimization Methods
STARTING SOLUTION AND CONVERGENCE
Simplex Method: Simple method starts from an initial basic feasible solution.
min z = cx
st. Ax ≤ b
x≥0
If b ≥ 0
Add slacks xs to each constraint:
Ax + xs = b
x, xs ≥ 0
Therefore, an initial basic feasible solution can be easily obtained when we have only Ax ≤ b in
the set of constraints.
Question: What happens if b is not nonnegative Ax≥b and x≥0, b must be nonnegative for a
feasible solution.
Add artificial variables, xa to negative rows (or rows without a designated basic
variable).
How to eliminate artificial variables?
Artificial variables should be 0.
Two methods for solving problems without an initial basic feasible solution:
1. The big-M method
2. Two – Phase method
1. The big-M Method
max cx
s.t. Ax=b
x≥0
If an initial basic feasible solution is unknown, introduce the artificial variable xa:
Ax+xa=b
x, xa≥0
In order to eliminate artificial variables from the basis, introduce sufficiently large M (bigM) as cost coefficient of the artificial variables in the objective
max z = cx - Mxa
s.t.
Ax + xa=b
x,
xa ≥ 0
2. The Two-Phase Method
Phase I: Formulate the constraint of the LP problem as in the big-M method to secure an
initial basic feasible solution. Next, find a basic solution of the resulting
equation that minimizes the sum of the artificial variables.

1
Metin Turkay
INDR 262 Introduction to Optimization Methods
min z=xa
s.t. Ax+xa=b
x, xa ≥ 0
If the optimal solution is positive, the LP problem has no feasible solution. Otherwise, go
to phase II.
Phase II: Use the feasible solution obtained in Phase I as the initial basic feasible
solution. Solve the LP problem with simplex method.
Example:
min
z=
subject to
3x1
+2x2 +4x3
2x1
3x1
x1
+ x2 +3x3 = 60
+3x2 +5x3 ≥ 120
≥ 0
x2
≥ 0
x3 ≥ 0
Convert the minimization objective to maximization:
min z = 3x1 +2x2 +4x3
max -z = -3x1 -2x2 -4x3
a)
Solution by the Big M method.
Rewrite the problem introducing slack (x4) and artificial variables (x5, x6):
max -z=
(0)
-z
+3x1 +2x2 +4x3
+Mx5 +Mx6 = 0
(1)
2x1 + x2 +3x3
+ x5
= 60
(2)
3x1 +3x2 +5x3 -x4
+ x6 = 120
Since the artificial variables x5 and x6 are basic, we must reorganize row 0 to have 0
reduced costs for these variables:
New row 0
= Old row 0 – M*(row 1) – M*(row 2)
= [3 2 4 0 M M  0] – M*[2 1 3 0 1 0  60] – M*[3 3 5 -1 0 1  120]
= [-5M+3 -4M+2 -8M+4 M 0 0  -180M]
Iter
0
1
2
B.V.
Eqn
z
x1
x2
x3
x4
x5
x6
RHS
Z
0
-1
-5M+3
-4M+2
-8M+4
M
0
0
-180M
x5
1
0
2
1
3
0
1
0
60
20
x6
2
0
3
3
5
-1
0
1
120
24
z
0
-1
⅓M+⅓
-4/3M+⅔
0
M
/3M-4/3
0
-20M-80
x3
1
0
2/3
1/3
1
0
1/3
0
20
60
x6
2
0
-1/3
4/3
0
-1
-5/3
1
20
15
z
0
-1
½
0
0
½
M- ½
M- ½
-90
x3
1
0
¾
0
1
¼
¾
-¼
15
x2
2
0
-¼
1
0
-¾
-5/4
¾
15
8
Ratio

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Metin Turkay
INDR 262 Introduction to Optimization Methods
The optimal solution: x1* = 0, x2* = 15, x3* = 15, z* = 90
b) Solution by the Two-Phase method.
Phase I: min z =
x5 +x6
subject to
2x1 + x2 +3x3
+x5
3x1 +3x2 +5x3 -x4
+x6
x1,
x2, x3, x4,
x5, x6
Convert the minimization objective to maximization:
min
z=
x5
+x6
max
-z=
-x5
-x6
(0)
(1)
(2)
-z
2x1
3x1
+ x2
+3x2
+3x3
+5x3
+x5
+x5
= 60
= 120
≥ 0
+x6
-x4
= 0
= 60
= 120
+x6
Since the artificial variables x5 and x6 are basic, we must reorganize row 0 to have 0
reduced costs for these variables:
New row 0
= Old row 0 – 1*(row 1) – 1*(row 2)
= [0 0 0 0 1 1  0] – 1*[2 1 3 0 1 0  60] – 1*[3 3 5 -1 0 1  120]
= [-5 -4 -8 1 0 0  -180]
Iter
0
1
2
B.V.
Eqn
z
x1
x2
x3
x4
x5
x6
RHS
Ratio
z
0
-1
-5
-4
-8
1
0
0
-180
x5
1
0
2
1
3
0
1
0
60
20
x6
2
0
3
3
5
-1
0
1
120
24
z
0
-1
⅓
-4/3
0
1
8
/3
0
-20
x3
1
0
2/3
1/3
1
0
1/3
0
20
60
x6
2
0
-1/3
4/3
0
-1
-5/3
1
20
15
z
0
-1
0
0
0
0
1
1
0
x3
1
0
¾
0
1
¼
¾
-¼
15
x2
2
0
-¼
1
0
-¾
-5/4
¾
15
Phase II: Rewrite the optimization model by,
• dropping the artificial variables x5 and x6
• using the original objective function
min z = 3x1 +2x2 +4x3
s.t.
¾x1
+x3 +¼x4 =15
-¼x1 + x2
-¾x4 =15
x1,
x2,
x3,
x4 ≥ 0

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Metin Turkay
INDR 262 Introduction to Optimization Methods
Convert the minimization objective to maximization:
min z= 3x1 +2x2 +4x3
max -z= -3x1 -2x2 -4x3
max -z=
s.t.
-3x1 -2x2 -4x3
¾x1
+ x3 +¼x4 =15
-¼x1 + x2
-¾x4 =15
x1, x2, x3, x4 ≥ 0
(0)
(1)
(2)
-z
+3x1 +2x2 +4x3
= 0
¾x1
+ x3 +¼x4 =15
-¼x1 + x2
-¾x4 =15
Since the variables x2 and x3 are basic, we must reorganize row 0 to have 0 reduced costs
for these variables:
New row 0
= Old row 0 – 4*(row 1) – 2*(row 2)
= [3 2 4 0  0] – 4*[¾ 0 1 ¼  15] – 2*[-¼ 1 0 –¾  15]
= [½ 0 0 ½ -90]
Iter
0
B.V.
Eqn
z
x1
x2
x3
x4
RHS
z
0
-1
½
0
0
½
-90
x3
1
0
¾
0
1
¼
15
x2
2
0
-¼
1
0
-¾
15
The optimal solution: x1* = 0, x2* = 15, x3* = 15, z* = 90

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Metin Turkay
INDR 262 Introduction to Optimization Methods
Analysis of the big-M Method
max z = cx - Mxa
s.t.
Ax + xa = b
x,
xa ≥ 0
Optimal solution is found
x*≥0, xa*=0
Artificial values have all
0 values. The solution is
optimal.
x*≥0, xa*≠0
Some of the artificial
variables are nonzero. The
solution is infeasible
The problem is unbounded.
zk–ck = min (zj–cj)>0, yk≤0
x*≥0, xa*=0
Unbounded
x*≥ 0, xa*≠0
Infeasible

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Metin Turkay
INDR 262 Introduction to Optimization Methods
Degeneracy
When there is a tie in the minimum ratio test for determining the leaving variable, one or
more of the basic variables will be zero in the next iteration, the new solution is called a
degenerate solution.
→ The model has at least one redundant constraint.
Ex : max z = 3x1 + 9x2
s.t.
x1 + 4x2 ≤ 8
x1 + 2x2 ≤ 4
x1, x2 ≥ 0
Initial Simplex Tableau:
xB
z
x3
x4
x1
-3
1
1
x2
-9
4
2
x3
0
1
0
x4
0
0
1
RHS
0
8
4
x3 leaves x2 enters
xB
z
x2
x4
z
0
0
0
x1
-3/4
1/4
1/2
xB
z
x2
x1
z
1
0
0
x1
0
0
1
x2
0
1
0
x3
9/4
1/4
-1/2
x4
0
0
1
RHS
18
2
0
x4 leaves x1 enters
Implications of degenerate solutions:
x2
0
1
0
x3
3/2
1/2
-1
x4
3/2
-1/2
2
RHS
18
2
0
a) cycling
b) circling

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Metin Turkay
INDR 262 Introduction to Optimization Methods
Unbounded Solutions:
In some LP models, the values of the variables may be increased indefinitely without
violating any of the constraint. This condition is called unsoundness.
Ex: max
s.t.
z = x1 + 2x2
x1 - x2 ≤ 10
2x1
≤ 40
x1, x2 ≥ 0
xB
z
x3
x4
z
1
0
0
x1
-1
1
2
x2
-2
-1
0
x3
0
1
0
x4
0
0
1
RHS
0
10
40
Observations:
• x2 is selected as the entering basic variable since it has the most negative row-0
coefficient
• x3 cannot leave the basis because a12 is -1 indicating that the value of the entering
basic variable, x2, should decrease which is not possible since the current value of x2
which is a nonnegative variable is 0; decreasing x2 further will make the solution
infeasible
• x4 cannot leave the basis because a12 is 0 indicating that the value of the entering basic
variable, x2, can increase to infinity without violating this constraint
Therefore, this problem is unbounded.

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Metin Turkay
INDR 262 Introduction to Optimization Methods
Example 1:
min z= 4x1
s.t.
3x1
4x1
x1
x1,
+ x2
+ x2
+3x2
+2x2
x2
=3
≥6
≤4
≥0
First introduce the slack variables:
min z= 4x1 + x2
s.t.
3x1 + x2
4x1 +3x2 –x3
x1 +2x2
+x4
x1,
x2, x3, x4
=3
=6
=4
≥0
No initial basis, introduce artificial variables.
a) big-M Method:
min
s.t.
z=
4x1
3x1
4x1
x1
x1,
+ x2
+Mx5 +Mx6
+ x2
+ x5
=3
+3x2 –x3
+ x6 = 6
+2x2
+x4
=4
x2, x3, x4,
x5, x6 ≥ 0
xB
z
x5
x6
x4
Eqn
0
1
2
3
z
1
0
0
0
x1
-4
3
4
1
x2
-1
1
3
2
x3
0
0
-1
0
x5
-M
1
0
0
x6
-M
0
1
0
x4
0
0
0
1
RHS
0
3
6
4
Eliminate inconsistency in basic variable x5 and x6.
Multiply each row whose basic variable is an artificial variable by the negative of its row-0
entry and add to row-0:
New row 0:
Old row 0:
+M*row 1:
+M*ow 2:
New row 0=
xB
z
x5
x6
x4
Old row 0 +M*row 1
-4
-1
0
3M
M
0
4M
3M -M
-4+7M -1+ 4M -M
Eqn
0
1
2
3
z
1
0
0
0
x1
7M-4
3
4
1
+M*row 2
-M -M 0
M
0 0
0
M 0
0
0 0
x2
4M-1
1
3
2
x3
-M
0
-1
0
| 0
|3M
|6M
|9M
x5
0
1
0
0
x6
0
0
1
0
x4
0
0
0
1
RHS
9M
3
6
4
Ratio
1
3/2
4

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Metin Turkay
INDR 262 Introduction to Optimization Methods
Entering Variable: x1 has the most positive coefficient in row-0.
Leaving Variable: minimum ratio test → x5 is the leaving basic variable
xB
z
x1
x6
x4
Eqn
0
1
2
3
z
1
0
0
0
x1
0
1
0
0
x2
1/3(5M+1)
1/3
5/3
5/3
x3
-M
0
-1
0
x5
-1/3(7M+4)
1/3
-4/3
-1/3
x6
0
0
1
0
x4
0
0
0
1
RHS
2M+4
1
2
3
Ratio
3
6/5
9/5
Entering Variable: x2 has the most positive coefficient in row 0.
Leaving Variable: minimum ratio test → x6 is the leaving basic variable
xB
z
x1
x2
x4
Eqn
0
1
2
3
z
1
0
0
0
x1
0
1
0
0
x2
0
0
1
0
x3
-1/5
1/5
-3/5
1
x5
-M+8/5
3/5
-4/5
1
x6
-M-1/5
-1/5
3/5
-1
x4
0
0
0
1
RHS
18/5
3/5
6/5
1
b) Two-Phase Method:
Phase I: min z0 =
x5 +x6
s.t.
3x1 + x2
+x5
=3
4x1 +3x2 –x3
+x6 = 6
x1 +2x2
+x4
=4
x1, x2, x3, x4, x5, x6 ≥ 0
xB
z0
x5
x6
x4
Old z0–row : 0
+1*x5–row : 3
+1*x6–row : 4
New z0 – row : 7
0
1
3
4
xB
z0
x5
x6
x4
z0
1
0
0
0
0
0
-1
-1
z0
1
0
0
0
x1
0
3
4
1
-1
1
0
x2
0
1
3
2
-1
0
1
0
x1
7
3
4
1
x3
0
0
-1
0
0 |
0 |
0 |
0
x2
4
1
3
2
x5
-1
1
0
0
x6
-1
0
1
0
x4
0
0
0
1
RHS
0
3
6
4
x6
0
0
1
0
x4
0
0
0
1
RHS
9
3
6
4
0
3
6
0 | 9
x3 x5
-1
0
0
1
-1
0
0
0

9
Metin Turkay
INDR 262 Introduction to Optimization Methods
→ Simplex Method: Optimal Solution for the Phase I problem
xB
z0
x1
x2
x4
z0
0
0
0
0
x1
0
1
0
0
x2
0
0
1
0
x3
0
1/5
-3/5
1
x5
-1
3/5
-4/5
1
x6
-1
-1/5
3/5
-1
x4
0
0
0
1
RHS
0
3/5
6/5
1
→ Optimal Solution with z0 = 0
→ Continue with Phase II.
Phase II:
min
s.t.
Delete artificial variables and use the optimal solution in Phase I as the LP problem:
z= 4x1 +x2
x1
+1/5x3
= 3/5
x2 -3/5x3
= 6/5
x3 +x4 = 1
x1, x2, x3, x4 ≥ 0
xB
z
x1
x2
x4
z
1
0
0
0
x1
-4
1
0
0
x2
-1
0
1
0
x3
0
1/5
-3/5
1
x4
0
0
0
1
RHS
0
3/5
6/5
1
x1 and x2 have nonzero coefficients in the row-0, they must be substituted out.
Old row-0 : -4
+4*x1–row : 4
+1*x2–row : 0
New row – 0 : 0
-1
0
1
0
4/5
-3/5
0
1/5
0 | 0
0 | 12/5
0 | 6/5
0
| 18/5
z
1
0
0
0
x1
0
1
0
0
→ Initial simplex tableau:
xB
z
x1
x2
x4
x2
0
0
1
0
x3
1/5
1/5
-3/5
1
x4
0
0
0
1
RHS
18/5
3/5
6/5
1
Continue until finding the optimal solution!

10
Metin Turkay