Week_7: Duality
1
1- Define dual problem
The dual problem is an LP defined directly and systematically from the primal (or
original) LP model. The two problems are so closely related that the optimal solution
of one problem automatically provides the optimal solution to the other.
The dual is defined for various forms of the primal depending on the sense of
optimization (maximization or minimization), types of constraints (≥, ≤, or =), and
orientation of the variables (nonnegative or unrestricted).
define the primal in equation form as follows:
maximize or minimize 𝑧 =
subject to
𝑛
𝑗 =1 𝑐𝑗
𝑛
𝑗 =1 𝑎𝑖𝑗 𝑥𝑖𝑗
𝑥𝑗 ≥0
𝑥𝑗
= 𝑏𝑖𝑗
𝑖 = 1,2,3 … … … 𝑚
𝑗 = 1,2,3 … . . 𝑛
The variables xi, j = 1, 2, ... , n, include the surplus, slack, and artificial variables.
the dual problem is constructed from the primal. Effectively, we have
1. A dual variable is defined for each primal (constraint) equation.
2. A dual constraint is defined for each primal variable.
1. The constraint (column) coefficients of a primal variable define the left-hand
side coefficients of the dual constraint and its objective coefficient define the
right-hand side.
3. The objective coefficients of the dual equal the right-hand side of the primal
constraint equations.
dual
variable
y1
y2
.
.
ym
primal variable
…….
xj
…… xn
……
cj
……. cn
x1
c1
x2
c2
a11
a21
.
.
am1
a12
a22
a1j
a2j
a1n
a2n
b1
b2
am2
amj
jth dual
constrian
amn
bm
dual objective
coficient
right hand side
The rules for determining the sense of optimization (maximization or minimization),
the type of the constraint or =, and the sign of the dual variables are summarized
in Table below:
2
Objective
Primal
problem(objective)
Maximization
Minimization
Minimization
maximization
Dual problem
Constraints type
Variable sign
≥
≤
Unrestricted
unrestricted
Example:
Primal
primal in equation form
dual variable
Maximize z=5x1+12x2+4x3
maximize z=5x1+12x2+4x3+0x4
Subject to
subject to
x1+2x2+x3≤10
x1+2x2+x3+x4=10
y1
2x1-x2-3x3=8
2x1-x2+3x3+0x4=8
y2
x1,x2,x3≥0
x1,x2,x3x4≥0
----------------------------------------------------------------------------------------------------------------Dual Problem
Minimize w=10y1+8y2
Subject to
y1+2y2≥5
2y1-y2≥12
y1+3y2≥4
y1+0y2≥0
→)y1≥0,y2 unrestricted)
y1,y2 unrestricted
}
Example:
Primal
primal in equation form
dual variable
Minimize z=15x1+12x2
minimize z=15x1+12x2+0x3+0x4
Subject to
subject to
x1+2x2≥3
x1+2x2-x3+0x4=3
y1
2x1-4x2≤5
2x1-4x2+0x3+x4=5
y2
x1,x2≥0
x1,x2,x3,x4≥0
----------------------------------------------------------------------------------------------------------------Dual Problem
Maximize w=3y1+5y2
Subject to
y1+2y2≤15
2y1-4y2≤12
-y1
≤0
y2≤0
→)y1≥0,y2≤0)
y1,y2 unrestricted
}
3
Summary of the Rules for Constructing the Dual. The general conclusion from the
preceding examples is that the variables and constraints in the primal and dual
problems are defined by the rules in Table below. It is a good exercise to verify that
these explicit rules are subsumed by the general rules in Table below
Table: rules for constructing the dual problem
Maximization problem
Constraints
≥
≤
=
Variables
≥0
≤0
Unrestricted
minimize problem
Variables
≤0
≥0
unrestricted
Constraints
≥
≤
=
2.Optimal Dual Solution
The primal and dual solutions are so closely related that the optimal solution of either
problem directly yields (with little additional computation) the optimal solution to the
other. Thus, in an LP model in which the number of variables is considerably smaller
than the number of constraints, computational savings may be realized by solving the
dual, from which the primal solution is determined automatically
there are two methods for determining the dual values. Note that the dual of the dual
is itself the primal, which means that the dual solution can also be used to yield the
optimal primal solution automatically.
Method 1.
𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓
=
𝑑𝑢𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒
𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑝𝑟𝑖𝑚𝑎𝑙 𝑧 𝑐𝑜𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑥𝑖
+
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥𝑖
4
Method 2:
𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑠
=
𝑜𝑓 𝑑𝑢𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒
𝑟𝑜𝑤 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓
𝑜𝑝𝑡𝑖𝑚𝑎𝑙
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠
× 𝑝𝑟𝑖𝑚𝑎𝑙
𝑜𝑓 𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑝𝑟𝑖𝑚𝑎𝑙 𝑏𝑎𝑠𝑖𝑐 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
𝑖𝑛𝑣𝑒𝑟𝑠𝑒
Example:
Consider the following LP:
Maximize z=5x1+12x2+4x3
Subject to
x1+2x2+x3≤10
2x1-x2+3x3=8
x1,x2,x3≥0
To prepare the problem for solution by the simplex method, we add a slack x4 in the
first constraint and an artificial R in the second. The resulting primal and the
associated dual problems are thus defined as follows:
Primal
Dual
-
----------------------------------------------------------------------------------------------------------------maximize z=5x1+12x2+4x3-MR
minimize w=10y1+8y2
subject to
subject to
x1+2x2+x3+x4
2x1-x2+3x3
y1+2y2≥5
=10
+ R=8
2y1-y2≥12
x1,x2,x3,x4,R≥0
y1+3y2≥4
y1
≥0
y2≥-M(→y2 unrestricted)
table below provides the optimal primal tableau.
Basic
Z
X2
X1
X1
0
0
1
X2
0
1
0
X3
3/5
-1/5
7/5
5
X4
29/5
2/5
1/5
R
-2/5+M
-1/5
2/5
solution
54.8
12/5
26/5
Method 1: In Table above, the starting primal variables x4 and R uniquely
correspond to the dual variables yl and y2, respectively. Thus, we determine the
optimum dual solution as follows:
Start primal basic variables
z-equation coefficients
Original objective
coefficient
Dual variables
Optimal dual values
X4
29/5
R
-2/5+M
0
-M
y1
29/5+0=29/5
y2
-2/5+M+(-M)=-2/5
Method 2_ The optimal inverse matrix, highlighted under the starting variables X4
and R, is given in Table as
2 −1
𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 = 5 5
1 2
5 5
First, we note that the optimal primal variables are listed in the tableau in row order
as x2 and then x1 This means that the elements of the original objective coefficients
for the two variables must appear in the same order-namely,
(Original objective coefficients) = (Coefficient of x2, coefficient of x1 )
= (12,5)
Thus, the optimal dual values are computed as
𝑦1, 𝑦2 =
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒
× 𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑖𝑛𝑣𝑒𝑟𝑠𝑒
𝑐𝑜𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑥2, 𝑥1
= 12
5
2 −1
5 5
1 2
5 5
6
=
29
5
−2
5