OPERATIONS RESEARCH
Michał Kulej
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Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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Artificial Starting Solution
The Big M Method
The linear programming problem in which all constraints are (≤) with
nonnegative right-hand sides offers a all-slack starting basic feasible
solution. Problems with (=) and/or (≥) constraints need to use artificial
variables to the beginning of simplex algorithm. There are two methods: the M-method(also called the Big M-method) and the Two-Phase
method.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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2 / 41
Artificial Starting Solution
The Big M Method
Example
max Z = 2x1 + x2 − 3x3
x1 + x2 + x3 ≥ 6
2x1 + x2 = 14
x1 , x2 , x3 ≥ 0
After converting to standard form we have:
max Z = 2x1 + x2 − 3x3
x1 + x2 + x3 − s1 = 6
2x1 + x2 = 14
x1 , x2 , x3 , s1 ≥ 0
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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3 / 41
Artificial Starting Solution
The Big M Method
Adding Artificial Variables
The above system of equations is not in basic form - there is not a
basic variable in each equation.
If the equation i does not have a slack (or a variable that can play the
role of a slack), an artificial variable, ai , is added to form a starting
solution similar to the all-slack basic solution. However, because the
artificial variables are not part of the original linear model, they are assigned a very high penalty in the objective function, thus forcing them
(eventually) to equal zero in the optimal solution.
In the considering example we add two artificial variables a1 and a2 .
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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4 / 41
Artificial Starting Solution
The Big M Method
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 objective coefficient =
−M, in maximization problems
M, in minimization problems.
So we have:
max Z = 2x1 + x2 − 3x3 − Ma1 − Ma2
a1
+x1 + x2 + x3 − s1 = 6
a2 +2x1 + x2 = 14
x1 , x2 , x3 , s1 , a1 , a2 ≥ 0
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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5 / 41
Artificial Starting Solution
The Big M Method
The Big M Method Calculations
First and second simplex tableau are as follows:
cB
BV
a1
a2
x~1
x2
x3
s1
Solution
−M
−M
a~1
a2
Z
1
0
0
0
1
0
1
2
−3M − 2
1
1
−2M − 1
1
0
−M + 3
−1
0
M
6
14
−20M
2
x1
1
0
1
1
1
−1
6
−M
a~2
Z
−2
3M + 2
1
0
0
0
−1
M+1
−2
2M − 1
2
−2M − 2
2
−2M + 12
6/1 = 6
14/2 = 7
−
2/2 = 1
The optimal simplex tableau:
cB BV
2 x1
0 s1
Z
a1
a2
x1 x2
1
1
1
0
2
2
1
−1
0 − 21
2
M M +1 0
0
x3 s1 Solution
0 0
7
1
−1 1
3 0
14
Optimal solution is: x1 = 7, s1 = 1, x2 = x3 = 0 with Z = 14.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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6 / 41
Artificial Starting Solution
The Big M Method
No Feasible Solution - an Example
The use of the penalty M will not force an artificial variable to zero level
in the final simplex iteration if LPP does not have a feasible solution
(i.e. the constraints are not consistent). In this case, the final simplex
tableau will include at least one artificial variable at positive level.
Solve the problem :
max Z = 2x1 + 2x2
6x1 + 4x2 ≤ 24
x1 ≥ 5
x1 , x2 ≥ 0
The standard form:
max Z = 2x1 + 2x2
s1 +6x1 + 4x2 = 24
x1 − s2 = 5
x1 , x2 , s1 , s2 ≥ 0
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
7 / 41
Artificial Starting Solution
The Big M Method
Using The Big M Method to Solve the Problem
It is enough to add only one artificial variable :
max Z = 2x1 + 2x2 − Ma2
s1
+6x1 + 4x2 = 24
a2 +x1 − s2 = 5
x1 , x2 , s1 , s2 ≥ 0
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
8 / 41
Artificial Starting Solution
The Big M Method
The Big M Method Calculations
cB BV
0
s~1
−M a2
Z
x~1
x2 s2 Solution
s1 a2
1 0
6
4
0
24
24/6 = 4
0 1
1
0 −1
5
5/1 = 5
0 0 −M − 2 −2 M
−5M
cB BV
2
x1
−M a2
Z
s1
1
6
− 16
1
6M
+
1
3
a2 x1
0 1
1 0
0 0
x2
2
3
− 23
2
3M
−
2
3
s2 Solution
0
4
−1
1
M −M + 8
The last simplex tableau is optimal - all Z - row coefficients are
nonnegative. The artificial variable a2 is basic variable with positive
value so the problem is not consistent - it has no feasible solutions.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
9 / 41
Artificial Starting Solution
The Two-Phase Method
The Essence of The Two-Phase Method
This method solves the linear programming problem in two phases:
Phase I attempts to find a starting feasible solution and, if one is found,
Phase II is used to solve the original problem.
Phase I Put the problem in equation form, and add the necessary
artificial variables to the constraints to get starting feasible
basic solution. Next, find a basic solution of the resulting
equations that, regardless of whether the problem is
maximization or minimization, always minimizes the sum
of the artificial variables. If minimum value of the sum is
positive, the problem has no feasible solution, which ends
the process. Otherwise, proceed to Phase II.
Phase II Use the feasible solution from Phase I as starting feasible
basic solution for original problem.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
10 / 41
Artificial Starting Solution
The Two-Phase Method
Example
We use this method to solve the following problem:
max Z = 4x1 + x2
3x1 + x2 = 3
4x1 + 3x2 ≥ 6
x1 + 2x2 ≤ 4
x1 , x2 ≥ 0
Using s2 as a surplus in the second constraint and s3 as a slack in the
third constraint, the standard form of the problem is given as:
max Z = 4x1 + x2
3x1 + x2 = 3
4x1 + 3x2 − s2 = 6
x1 + 2x2 + s3 = 4
x1 , x2 , s2 , s3 ≥ 0
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
11 / 41
Artificial Starting Solution
The Two-Phase Method
The Model with Arificial Variables - Phase I
The third equation has its basic variable (s3 ) but the first and second do
not. Thus we add the artificial variables a1 and a2 and we minimize the
sum of the artificial variables a1 + a2 . The resulting model is given as:
min Z a = a1 + a2
+a1
−s2
+a2
3x1 + x2
4x1 + 3x2
x1 + 2x2
x1 , x2 , s2 , s3 , a1 , a2 ≥ 0
+s3
= 3
= 6
= 4
Taking a1 , a2 , s3 as a basic variables and using formula (2) for
computing coefficients of Z a - row we obtain the first simplex tableau:
cB BV
1 a1
1 a2
0 s3
Za
Michał Kulej (Wrocł Univ. of Techn.)
x1 x2 s2 a1 a2 s3 Solution
3 1 0
1 0 0
3
4 3 −1 0 1 0
6
1 2 0
0 0 1
4
7 4 −1 0 0 0
9
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Artificial Starting Solution
The Two-Phase Method
The Optimal Simplex Tableau of Phase I
Now we use simplex algorithm for minimization problem and receive
(after two iterations) the following optimum tableau:
cB BV
0 x1
0 x2
0 s3
Za
x1 x2 s2
1
1 0
5
0 1 − 53
0 0
1
0 0
0
a1
s3 Solution
3
0
5
3
6
0
5
5
1 −1 1
1
−1 −1 0
0
3
5
− 45
a2
− 15
Because minimum Z a = 0. Phase I produces the basic feasible
solution x1 = 53 , x2 = 56 , s3 = 1. We can eliminate columns for a1 and
a2 from tableau and move to Phase II.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
13 / 41
Artificial Starting Solution
The Two-Phase Method
Phase II
After deleting the artificial columns, we write the original problem as
max Z = 4x1 + x2
= 53
x1
+ 15 s2
x2 − 35 s2
= 56
s2 +s3 = 1
x1 , x2 , s2 , s3 ≥ 0
Now we can begin the simplex algorithm with the following simplex
tableau:
cB BV x1 x2 s2 s3 Solution
1
3
0
4 x1 1 0
5
5
6
3
1 x2 0 1 − 5 0
5
0 s2 0 0
1
1
1
1
18
0
Z
0 0
5
5
Because we are maximizing, this is the optimal tableau. The optimal
solution is x1 = 53 , x2 = 65 and Z = 18
5 .
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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14 / 41
Artificial Starting Solution
The Two-Phase Method
Removal of Artificial Variables after Phase I
The removal of the artificial variables and their columns at the end of Phase I
can take place only when they are all nonbasic. If one or more artificial
variables are basic (at zero level) at the end of Phase I, then the following
additional steps must be undertaken to remove them prior to start of Phase II.
Step 1. Select a zero artificial variable to leave the basic solution
(choosing the leaving variable) and designate its row as the pivot
row. The entering variable can be any nonbasic variable with a
nonzero(positive or negative) coefficient in the pivot row. Perform the
associated simplex iteration.
Step 2. Remove the column of the (just-leaving) artificial variable from
the tableau. If all the zero artificial variables have been removed, go to
Phase II. Otherwise, go back to Step 1.
Remark All commercial packages use the two-phase method to solve
linear programming problems.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
15 / 41
Sensitivity Analysis
Changes in Objective Coefficients
Changes in Objective Coefficients
The sensitivity analysis is concerned with how changes in an parameters of the linear programming model affect the optimal solution.
Let us consider the following problem (described in the basic form) of
manufacturing two products W1 and W2 from two row materials S1 , S2 :
max Z = 3x1 + 2x2
s1
+2x1 + x2 = 100
s2
+x1 + x2 = 80
s3 +x1 = 40
x1 , x2 , s1 , s2 , s3 ≥ 0
[Maximizing the revenue]
[Limit on row material S1 ]
[Limit on row material S2 ]
[Demand on W1 ]
The optimal solution of the model is x1 = 20, x2 = 60 with Z = 180.
We want to determine the range of values of parameter (i.e. coefficient
of objective function - unit price of product W1 in our example ) over
which the optimal solution remain unchanged.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
16 / 41
Sensitivity Analysis
Changes in Objective Coefficients
The Optimal Simplex Tableau of the Problem
cB BV
3 x1
2 x2
0 s3
Z
s1 s2 s3 x1 x2 Solution
1 -1 0 1 0
20
-1 2 0 0 1
60
-1 1 1 0 0
20
1 1 0 0 0
180
Let us assumed that unit price of W 1 equals 3 + δ. Then we have:
cB
BV
3 + δ x1
2
x2
0
s3
Z
Michał Kulej (Wrocł Univ. of Techn.)
s1
s2
s3 x1 x2 Solution
1
−1
0 1 0
20
−1
2
0 0 1
60
−1
1
1 0 0
20
1 + δ −δ + 1 0 0 0
180
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Sensitivity Analysis
Changes in Objective Coefficients
Possibly Changes of Values of c1
The optimal solution remains optimal as long as all the Z − row
coefficients are nonnegative, so we get:
1+δ ≥0
−δ + 1 ≥ 0
Solving these system inequalities we obtain that δ ∈ [−1, 1]. Thus
optimal solution will remain optimal for values of price of W1 belonging
to the interval [2,4].
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
18 / 41
Sensitivity Analysis
Changes in Right-Hand Side
Changes in Right-Hand Side
Now we examine how the optimal solution to linear problem will
changes if the right-hand side of a constraint is changed. For example
we consider the constraint for row material S1 (coefficient b1 = 100).
Changing in bi will leave Z - row of simplex tableau unchanged but will
change the the solution column of the simplex tableau. If at least one
coefficient becomes negative, the current solution is no longer feasible.
So we look for the value of δ for which the following system equalities
is consistent:

2x1 + x2 = 100 + δ



x1 + x2 = 80
 s3 + x1 = 40


x1 , x2 , s1 , s2 , s3 ≥ 0
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
19 / 41
Sensitivity Analysis
Changes in Right-Hand Side
Matrix Form of System Equations
We describe this system in matrix form:


 

2 1 0
x1
100 + δ
 1 1 0   x2  =  80

s3
1 0 1
40
Multiplying both sides of this equations by the inverse matrix of the
coefficients matrix we get:
  
−1 
 
0
100 + δ
2 1 0
x1
≥ 0 
 x2  =  1 1 0   80
0
40
1 0 1
s3

Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
20 / 41
Sensitivity Analysis
Changes in Right-Hand Side
The Inverse Matrix of Basis
The inverse matrix we could obtain from the optimal simplex tableau. It
is consisting of the columns in the optimal tableau that correspond to
the initial basic variables BV = {s1 , s2 , s3 }(taking in the same order):

−1 
1 −1 0
2 1 0
 1 1 0  =  −1 2 0 
−1 1 1
1 0 1

Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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21 / 41
Sensitivity Analysis
Changes in Right-Hand Side
Hence we get:
  

0
100 + δ
1 −1 0
≥ 0 
 −1 2 0   80
0
40
−1 1 1

For the current set of basic variable (basis) to remain optimal we
require that the following system of inequalities holds:
20 + δ ≥ 0
60 − δ ≥ 0
20 − δ ≥ 0
The solution remains optimal (basic variable x1 , x2 , s3 ) as long as
δ ∈ [−20, 20] or if between 80 and 120 row material S1 (b1 ∈ [80, 120],
where the current amount is b1 = 100) are available.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
22 / 41
Dual Problem
Definition Of the Dual Problem
Associated with any linear programming problem is another linear
problem, called the dual problem (Dual in short). Now we explain how
to find the dual problem to the given LPP, we discuss the economic
interpretation of the dual problem and we discuss the relation that exist
between an LPP (called Primal) and its dual problem. We consider the
LPP with normal (canonical) form :
max Z = c1 x1 + c2 x2 + · · · + cn xn
a11 x1
a21 x1
..
.
+
+
a12 x2
a22 x2
..
.
+ ···
+ ···
am1 x1 + am2 x2 + · · ·
+
+
a1n xn
a2n xn
..
.
≤
≤
b1
b2
+ amn xn ≤ bm
xj ≥ 0 (j = 1, 2, . . . , n)
The original problem is called the primal.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
23 / 41
Dual Problem
Definition Of the Dual Problem
Dual Problem of LPP if Primal is in Canonical Form
The dual problem is defined as follows:
min W = b1 y1 + b2 y2 + · · · + bm ym
a11 y1 + a21 y2 + · · ·
a12 y1 + a22 y2 + · · ·
..
..
.
.
a1n y1 + a2n y2 + · · ·
+ am1 ym ≥ c1
+ am2 ym ≥ c2
..
.
+ amn ym ≥ cn
yi ≥ 0 (i = 1, 2, . . . , m)
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
24 / 41
Dual Problem
Definition Of the Dual Problem
The Construction of the Dual Problem from the Primal
Problem
min W
(y1 ≥ 0)
(y2 ≥ 0)
..
.
y1
y2
..
.
(ym ≥ 0) ym
Michał Kulej (Wrocł Univ. of Techn.)
max Z
(x1 ≥ 0) (x2 ≥ 0)
x1
x2
a11
a12
a21
a22
..
..
.
.
am1
am2
≥ c1
≥ c2
OPERATIONS RESEARCH
···
···
···
···
(xn ≥ 0)
xn
a1n
a2n
..
.
amn
≥ cn
≤ b1
≤ b2
..
.
≤ bm
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25 / 41
Dual Problem
Definition Of the Dual Problem
Example
The Furniture Company STYLE manufactures tables and chairs. A
table sells for $160, and chair sells for $200. The demand for tables
and chairs is unlimited. The manufacture of each type of furniture
require labor, lumber, and inventory space. The amount of each
resource needed to make tables and chairs and daily limit of available
resources is given in the following table:
Resource
Labor(hours)
Lumber((feet)3 )
Inventory space((feet)2 )
Resources needed
Table
Chair
2
4
18
18
24
12
Amount of
resource available(hours)
40
216
240
STYLE wants to maximize total revenue.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
26 / 41
Dual Problem
Definition Of the Dual Problem
Primal and Dual of the Example Problem
Primal problem
max z = 160x1 + 200x2
2x1 + 4x2 ≤ 40
18x1 + 18x2 ≤ 216
24x1 + 12x2 ≤ 240
x1 , x2 ≥ 0.
Dual problem
min w = 40y1 + 216y2 + 40y3
2y1 + 18y2 + 24y3 ≥ 160
4y1 + 18y2 + 12y3 ≥ 200
y1 , y2 , y3 ≥ 0.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
27 / 41
Dual Problem
Economic Interpretation of Duality
Interpretation of Dual for STYLE
Suppose there is an entrpreneur who wants to purchase all of Style’s
resourses i.e. 40 hours of labor, 210 (feet)3 of lumber and 240(feet)2
of inventory space. Then he must determine the price he is willing to
pay for a unit of each of STYLE’s resources. Let y1 , y2 i y3 be the price
for one hour of labor, one cubic feet of lumber and one square feet of
inventory space. We show that the resource prices should be
determine by solving the dual problem. The total price the
entrepreneur must pay for the resources is 40y1 + 216y2 + 240y3 and
since he wish to minimize the cost of purchasing the resources he
wants to:
minimize W = 40y1 + 216y2 + 240y3 .
But he must set resource prices high enough to induce STYLE to sell
its resources.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
28 / 41
Dual Problem
Economic Interpretation of Duality
Interpretation of Dual for STYLE
For example, he must offer STYLE at lest $160 for a combination of
resources that includes 2 hours of labor, 18 cubic feet of lumber and
24 square feet of inventory space, because STYLE could use these
resources to manufacture table that can be sold for $160. Since he is
offering 2y1 + 18y2 + 24y3 for the resources used to produce table, he
must choose y1 , y2 , y3 to satisfy
2y1 + 18y2 + 24y3 ≥ 160.
Similar reasoning for the chair gives:
4y1 + 18y2 + 12y3 ≥ 200.
The sign restrictions y1 , y2 , y3 ≥ 0 must also hold. So we see that the
solution to the dual of STYLE problem does yield prices for labor,
lumber and inventory space.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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29 / 41
Dual Problem
Economic Interpretation of Duality
Finding the Dual to any LPP
We illustrate how to find the dual problem on the following example:
max z = 2x1 + x2
x1 + x2 = 2
2x1 − x2 ≥ 3
x1 − x2 ≤ 1
x1 ≥ 0
x2 − Unrestricted
Michał Kulej (Wrocł Univ. of Techn.)
←→
←→
←→
←→
←→
min w = 2y1 + 3y2 + y3
y1 - Unrestricted
y2 ≤ 0
y3 ≥ 0
y1 + 2y2 + y3 ≥ 2
y1 − y2 − y3 = 1.
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Dual Problem
Economic Interpretation of Duality
Rules for Construction of Dual
The general conclusion from the preceding example is that the
variables and constraints in the primal and dual problems are defined
by rules in the following table:
Maximization problem
Constraints
≥
≤
=
Variables
≥0
≤0
Unrestricted
↔
↔
↔
↔
↔
Minimization problem
Variables
≤0
≥0
Unrestricted
Constraints
≥
≤
=
Table: Rules for construction the Dual Problem
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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31 / 41
Dual Problem
Primal-Dual Relationships
The Key Relationships between the Primal and Dual
Theorem
The dual of the dual problem yields the original primal problem.
Theorem (Weak duality property)
If we choose any feasible solution to the primal and any feasible
solution to the dual, the W − value for the feasible dual solution will be
at least as large as the Z − value for the feasible primal solution. Let
x = [x1 , x2 , . . . , xn ]T be any feasible solution to the primal and
y = [y1 , y2 , . . . , ym ] be any feasible solution to the dual. Then
(Z − value for x) ≤ (W − value for y)
From this theorem results two properties:
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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32 / 41
Dual Problem
Primal-Dual Relationships
Property
If x̄ = (x̄1 , x̄2 , . . . , x̄n ) and ȳ = (ȳ1 , ȳ2 , . . . , ȳm ) are feasible solutions of
primal problem and dual problem respectively such that
Z = c1 x̄1 + c2 x̄2 + . . . + cn x̄n = b1 ȳ1 + b2 ȳ2 + . . . + bm ȳm = W , then x̄
must be optimal solution for primal problem and ȳ must be optimal
solution for dual problem.
Property
If the primal(dual) is unbounded, then the dual(primal) problem is
infeasible.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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33 / 41
Dual Problem
Primal-Dual Relationships
Duality Theorem
Theorem
The following are only possible relations between the primal and dual
problems:
1
If one problem has feasible solutions and a bounded objective
function(and so has optimal solution), then so does the other
problem.
2
If one problem has feasible solutions and an unbounded objective
function(and so no optimal solution), then the other problem has
no feasible solutions.
3
If one problem has no feasible solutions, then the other problem
has either no feasible solution or an unbounded objective function.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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34 / 41
Dual Problem
Primal-Dual Relationships
Interpretation of Optimal Dual Decision Variables
Now we can give an interpretation of the dual problem for
maximization problem. We know that for optimal solutions x̄ of primal
problem and ȳ of dual problem the following equality holds:
Z = c1 x̄1 + c2 x̄2 + . . . + cn x̄n = b1 ȳ1 + b2 ȳ2 + . . . + bm ȳm = W
So each bi ȳi can be interpreted as the contribution to profit by having
bi units of resource i available for the primal problem. Thus, the
optimal dual variable ȳi (it is called shadow price) is interpreted as the
contribution to profit per unit of resource i(i = 1, 2, . . . , m).
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
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35 / 41
Dual Problem
Reding the Optimal Dual Solution if Primal is a Maximum Problem
Reading Optimal Dual from Z-row of Optimal Simplex
Tableau for STYLE
After solving primal problem by simplex method we could read the
optimal dual solution from the optimal simplex tableau. Let us consider
the LPP for firm STYLE. The optimal simplex tableau is as follows:
Table: Optimal simplex tableau
s1
200
160
0
x2
x1
s3
Z
1
2
− 12
6
20
s2
1
− 18
1
9
-2
20
3
s3
0
0
1
0
x1
0
1
0
0
x2
1
0
0
0
8
4
48
2240
The 
basic variablesare ZB = {x2 , x1 , s3 } and the basis
 is1

1
4 2 0
2 − 18 0
1

B =  18 18 0 . Inverse matrix for B is: B−1 =  − 21
9 0
12 24 1
6 −2 1
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
36 / 41
Dual Problem
Reding the Optimal Dual Solution if Primal is a Maximum Problem
The Optimal Solution of Dual Problem
(y 1 , y 2 , y 3 ) we can compute using matrix B−1 as follows:

(y 1 , y 2 , y 3 ) = cB B−1 = (200, 160, 0) 
1
2
− 12
6
1
− 18

0
1
 = (20, 20 , 0).
9 0
3
−2 1
where the vector cB contains the coefficients of objective function
corresponding the basic variables. The optimal dual solution are
coefficients of variables s1 , s2 , s3 of Z − row optimal simplex tableau.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
37 / 41
Dual Problem
Reding the Optimal Dual Solution if Primal is a Maximum Problem
Reding Dual Solution from Optimal Simplex Tableau
If the primal problem is any form , then the optimal dual solution may
be read from Z − row optimal simplex tableau by using the following
rules:
Optimal value of dual variable yi if constraint i is a (≤) equal to
coefficient of si in Z − row optimal simplex tableau.
Optimal value of dual variable yi if constraint i is a (≥) equal to
-(coefficient of ei in Z − row optimal simplex tableau), where ei is
surplus variable.
Optimal value of dual variable yi if constraint i is an equality (=)
equal to (coefficient of ai in Z − row optimal simplex tableau)-M,
where ai is artificial variable.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
38 / 41
Dual Problem
Reding the Optimal Dual Solution if Primal is a Maximum Problem
Example
max z = 3x1 + 2x2 + 5x5
x1 + 3x2 + 2x3 ≤ 15
2x2 − x3 ≥ 5
2x1 + x2 − 5x3 = 10
x1 , x2 , x3 ≥ 0.
To solve the problem we use the M method:
max Z = 3x1 + 2x2 + 5x5 − Ma2 − Ma3
x1 + 3x2 + 2x3 + s1 = 15
2x2 − x3 − e2 + a2 = 5
2x1 + x2 − 5x3 + a3 = 10
x1 , x2 , x3 , s1 , a2 , a3 ≥ 0.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
39 / 41
Dual Problem
Reding the Optimal Dual Solution if Primal is a Maximum Problem
The Big-M Method, the Last Simplex Tableau
5 x3
2 x2
3 x1
Z
x1 x2 x3 s1 e2
5
4
0 0 1 23
23
9
2
− 23
0 1 0 23
17
9
1 0 0 23
23
58
0 0 0 51
23
23
a2
5
− 23
9
23
17
− 23
58
M − 23
a3
2
− 23
1
− 23
7
23
M+
9
23
15
23
65
23
120
23
565
23
The dual problem has the following form:
min W = 15y1 + 5y2 + 10y3
y1 + 2y3 ≥ 3
3y1 + 2y2 + y3 ≥ 2
2y1 − y2 − 5y3 ≥ 5
y1 ≥ 0, y2 ≤ 0, y3 − Unrestricted.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
40 / 41
Dual Problem
Reding the Optimal Dual Solution if Primal is a Maximum Problem
Optimal Dual Solution
Reading optimal dual solution from optimal simplex tableau we get:
The first constraint is inequality ‘≤’ so y1 =
row for column s1 ).
51
23
(coefficient of Z −
58
The second constraint is inequality ‘≥’ so y2 = − 23
(- coefficient of
Z − row for column e2 ).
The third constraint is equality ’=’ so y3 =
row for column a3 minus M).
9
23
(coefficient of Z −
Optimal value of objective function of dual problem is
9
565
51
+ 5(− 58
W = 15 23
23 ) + 10 23 = 23 and equals optimal value of
objective function of the primal problem.
Michał Kulej (Wrocł Univ. of Techn.)
OPERATIONS RESEARCH
BIS
41 / 41