Duality Theory
LI Xiaolei
Finding the dual of an LP
• When taking the dual of a given LP, we refer to
the given LP as the primal. If the primal is a max
problem, the dual will be a min problem, and vice
versa.
• For convenience, we define the variables for the
max problem to be z, x1,x2,…,xn and the variables
for the min problem to be w,y1,y2,…,ym.
Finding the dual of an LP
• A normal max problem may be written as
max z  c1 x1  c2 x2    cn xn
s.t.
a11 x1  a12 x2    a1n xn  b1
a21 x1  a22 x2    a2 n xn  b2




am1 x1  am 2 x2    amn xn  bm
x j  0(i  1,2 , ,n)
(16)
Finding the dual of an LP
• The dual of a normal max problem is defined
to be,
min w  b1 y1  b2 y2    bm ym
s.t.
a11 y1  a21 y2    am1 ym  c1
a12 y1  a22 y2    am 2 ym  c2




a1n y1  a2 n y2    amn ym  cn
yi  0(i  1,2, ,m)
(17)
Finding the dual of an LP
• A min problem like (17) that has all ≥constraints
and all variables nonnegative is called a normal
min problem.
• If the primal is a normal min problem like (17), we
define the dual of (17) to be (16).
• A tabular approach makes it easy to find the dual
of an LP. A normal min problem is found by
reading down; the dual is found by reading
across in the table.
Finding the dual of an LP
max z
x1  0 x2  0
min w
xn  0
x1
x2

xn
y1  0
y1
a11
a12

a1n
 b1
y2  0
y2
a21
a22

a2 n
 b2






ym  0
ym
am1
am 2

amn
 bm
 c1
 c2

 cn
Finding the dual of an LP
• Illustrate by the Dakota problem,
max z  60 x1  30 x2  20 x3
s.t.
8 x1  6 x2  x3  48
4 x1  2 x2  1.5 x3  20
2 x1  1.5 x2  0.5 x3  8
x j  0(i  1,2 ,3 )
Finding the dual of an LP
max z
x1  0 x2  0 x3  0
min w
x1
x2
x3
y1  0
y1
8
6
1
 48
y2  0
y2
4
2
1.5
 20
y3  0
y3
2
1.5
0.5
8
 60
 30
 20
Finding the dual of an LP
• Then, reading down, we find the Dakota
dual to be
min w  48 y1  20 y2  8 y3
s.t.
8 y1
 4 y2  2 y3  60
6 y1
 2 y2  1.5 y3  30
y1  1.5 y2  0.5 y3  20
x j  0(i  1,2 ,3 )
Finding the dual of a nonnormal LP
• Many LPs are not normal max or min problem.
For example,
min w  2 y1  4 y2  6 y3
max z  2 x1  x2
s.t.
y1  2 y2  y3  2
s.t.
x1  x2  2
y1
 y3  1 (19)
2 x1  x2  3 (18)
y 2  y3  1
x1  x2  1
2 y1  y2
3
x1  0, x2 urs
y1 urs, y 2 , y3  0
Finding the dual of a nonnormal LP
• An LP can be transformed into normal form.
• To place a max problem into normal form, we
proceed as follows:
– Step 1 multiply each ≥ constraint by -1, converting it into
a ≤ constraint.
– Step 2 replace each equality constraint by two inequality
constraints (a ≤ constraint and a ≥ constraint). Then
convert the ≥ constraint to a ≤ constraint.
– Step 3 replace each urs variable xi by xi=xi’-xi’’, where xi’
≥0 and xi’’ ≥0 .
Finding the dual of a nonnormal LP
• (18) has been transformed onto the
following LP:
max z  2 x1  x2  x2
s.t.
x1  x2  x  2
- x1  x2  x  2
- 2 x1  x2  x  3
x1  x2  x  1
x1 , x2 , x2  0
Finding the dual of a nonnormal LP
• Transform a nonnormal min problem into a
normal min problem:
– Step 1 multiply each ≤ constraint by -1, converting it into
a ≥ constraint.
– Step 2 replace each equality constraint by two inequality
constraints (a ≤ constraint and a ≥ constraint). Then
convert the ≥ constraint to a ≤ constraint.
– Step 3 replace each urs variable yi by yi=yi’-yi’’, where yi’
≥0 and yi’’ ≥0 .
Finding the dual of a nonnormal max problem
• Find the dual of a nonnormal max LP without going through
the transformations,
– Step 1 fill in table so that the primal can be read across.
– Step 2 making the following changes, (a) if the ith primal
constraint is a ≥ constraint, the corresponding dual
variable yi must satisfy yi≤0. (b) if the ith primal
constraint is an equality constraint, the dual variable yi is
now unrestricted in sign. (c) if the ith primal variable is
urs, the ith dual constraint will be an equality constraint.
– Then the dual can be read down in the usual fashion.
Finding the dual of a nonnormal max problem
• For example,
max z  2 x1  x2
s.t.
max z
( x1  0) ( x2 urs )*
min w
x1  x2  2
x1
x2
2 x1  x2  3
y1
1
1
 2*
x1  x2  1
y2
2
1
 3*
y3
1
1
1
2
1
x1  0, x2 urs
( y3  0)
Finding the dual of a nonnormal max problem
max z
( x1  0) ( x2 urs )
min w
x1
x2
( y1 urs )
y1
1
1
2
( y 2  0)
y2
2
1
3
( y 3  0)
y3
1
1
1
2
1
Finding the dual of a nonnormal min LP
– Step 1 write out the primal table.
– Step 2 making the following changes, (a) if the ith primal
constraint is a ≤ constraint, the corresponding dual
variable xi must satisfy xi ≤0. (b) if the ith primal
constraint is an equality constraint, the dual variable xi is
now unrestricted in sign. (c) if the ith primal variable is
urs, the ith dual constraint will be an equality constraint.
– Then the dual can be read down in the usual fashion.
Economic interpretation of the dual problem
• The dual of the Dakota problem is,
min w  48 y1  20 y2  8 y3
s.t.
8 y1  4 y2  2 y3  60 (desk constraint )
6 y1  2 y2  1.5 y3  30 ( table constraint )
y1  1.5 y2  0.5 y3  20 (chair constraint )
y1 , y2 , y3  0
Economic interpretation of the dual problem
• Suppose an entrepreneur wants to purchase all
of Dakota’s resources. Then the entrepreneur
must determine the price he or she is willing to
pay for a unit of each of Dakota’s resources,
then
– y1=price paid for 1 board ft of lumber
– y2=price paid for 1 finishing hour
– y3=price paid for 1 carpentry hour
Economic interpretation of the dual problem
• The total price that should be paid for these resources is w.
the cost of purchasing the resources is to be minimized.
• In setting resource prices, what constraints does the
entrepreneur face?
For example, the entrepreneur must offer Dakota at least
$60 for a combination of resources that includes 8 board
feet of lumber, 4 finishing hours, and 2 carpentry hours,
because Dakota could use these resources to produce
a desk that can be sold for $60. the same reason shows
the other two constraints.
The dual theorem and its consequences
• The dual theorem states that the primal and dual
have equal optimal objective function values.
The dual theorem and its consequences
• Primal problem
max z  c1 x1  c2 x2    cn xn
s.t.
a11 x1  a12 x2    a1n xn  b1
a21 x1  a22 x2    a2 n xn  b2




am1 x1  am 2 x2    amn xn  bm
x j  0(i  1,2 , ,n)
(22)
The dual theorem and its consequences
• Dual problem
min w  b1 y1  b2 y2    bm ym
s.t.
a11 y1  a21 y2    am1 ym  c1
a12 y1  a22 y2    am 2 ym  c2




a1n y1  a2 n y2    amn ym  cn
yi  0(i  1,2, ,m)
(23)
The dual theorem and its consequences
• Weak duality
If we choose any feasible solution to the primal
and any feasible solution to the dual, the wvalue for the feasible dual solution will be at
least as large as the z-value for the feasible
primal solution.
The dual theorem and its consequences
• Lemma 1
Let
 x1 
x 
x   2
 
 
 xn 
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).
The dual theorem and its consequences
• If a feasible solution to either the primal or the
dual is readily available, weak duality can be
used to obtain a bound on the optimal objective
function value for the other problem.
The dual theorem and its consequences
• Lemma 2
Let
 x1 
x 
x   2
 
 
 xn 
Be a feasible solution to the primal and y=[y1
y2 … ym] be a feasible solution to the dual. If
cx=yb, then x is optimal for the primal and y is
optimal for the dual.
The dual theorem and its consequences
• Lemma 3
If the primal is unbounded, the dual problem is
infeasible.
• Lemma 4
If the dual is unbounded, the primal is infeasible.
The dual theorem
• Theorem 1
The dual theorem
Suppose BV is an optimal basis for the primal.
Then cBVB-1 is an optimal solution to the dual.
Also,
zw
The dual theorem
• Remarks
– A basis BV that is feasible for the primal is
optimal if and only if cBVB-1 is dual feasible.
– When we find the optimal solution to the
primal by using the simplex algorithm, we
have also found the optimal solution to the
dual.
The dual theorem
• How to read the optimal dual solution from row
0 of the optimal tableau if the primal is a max
problem
Optimal value of dual variable yi
if constraint i is a ≤ constraint
=coefficient of si in optimal row 0
Optimal value of dual variable yi
if constraint i is a ≥ constraint
=-(coefficient of ei in optimal row 0)
Optimal value of dual variable yi
if constraint i is a = constraint
=(coefficient of ai in optimal row 0)-M
Example 10
• To solve the following LP,
max z  3x1  2 x2  5 x3
s.t.
x1  3x2  2 x3  15
 2 x2
 x3  5
2 x1  x2  5 x3  10
x1 , x2 , x3  0
Example 10
• The optimal tableau,
z
x1
x2
x3
s1
e2
1
0
0
0
0
0
0
1
0
0
1
0
0
1
0
0
51
23
4
23
2
23
9
23
58
23
5
23
9
23
17
23

a2
a3
rhs
M  58
23
M  239
 235
 232
9
23
17
23
 231
565
23
15
23
65
23
120
23

7
23
BV
z
565
23
15
23
65
23
120
23
x3 
x2 
x1 
Example 10
• Solution
To find the dual from the tableau,
max z
( x1  0) ( x2  0) ( x3  0)
min w
x1
x2
x3
( y1  0)
y1
1
3
2
 15
( y2  0)
y2
0
2
1
 5*
( y3 urs )
y3
2
1
5
 10 *
3
2
5
Example 10
From the optimal primal tableau, we can find the
optimal solution to dual as follows:
Since the first primal constraint is a ≤ constraint,
y1=coefficient of s1 in optimal row 0=51/23.
Since the second primal constraint is a ≥
constraint, y2=-(coefficient of e2 in optimal row
0)=-58/23.
Since the third constraint is an equality constraint,
y3=(coefficient of a3 in optimal row 0)-M=9/23.
Example 10
By the dual theorem, the optimal dual objective
function value w must equal 565/23.
In summary, the optimal dual solution is
w  565
23 , y1 
51
23
, y2   58
23 , y3 
9
23
The dual simplex method
• When we use the simplex method to solve a max
problem (primal) , we begin with a primal feasible
solution. Since at least one variable in row 0 of the
initial tableau has a negative coefficient, our initial
primal solution is not dual feasible.
• Through a sequence of simplex pivots, we
maintain primal feasibility and obtain an optimal
solution when dual feasibility is attained.
The dual simplex method
• In many situations, it is easier to solve an LP by
beginning with a tableau in which each variable
in row 0 has a nonnegative coefficient (so the
tableau is dual feasible) and at least one
constraint has a negative right-hand side (so the
tableau is primal infeasible).
• The dual simplex method maintains a
nonnegative row 0 (dual feasibility) and
eventually obtains a tableau in which each righthand side is nonnegative (primal feasibility).
Dual simplex method for a max problem
• Step1
Is the right-hand side of each constraint nonnegative? If so,
an optimal solution has been found; if not, at least one
constraint has a negative right-hand side, and we go to
step 2.
Dual simplex method for a max problem
• Step2
Choose the most negative basic variable as the variable to
leave the basis. The row in which the variable is basic will be
the pivot row. To select the variable that enters the basis, we
compute the following ratio for each variable xj that has a
negative coefficient in the pivot row:
Coefficient of xj in row 0
Coefficient of xj in pivot row
Choose the variable with the smallest ratio as the entering
variable.
This form of the ratio test maintains a dual feasible tableau
(all variables in row 0 have nonnegative coefficients). Now
use ero’s to make the entering variable a basic variable in
the pivot row.
Dual simplex method for a max problem
• Step 3
If there is any constraint in which the right-hand side is
negative and each variable has a nonnegative coefficient,
the LP has no feasible solution. If no constraint indicating
infeasibility is found, return to step 1.
Dual simplex method for a max problem
• To illustrate the case of an infeasible LP, suppose the dual
simplex method yielded a constraint such as x1+2x2+x3=-5.
since x1,x2,x3≥0, then x1+2x2+x3≥0, the constraint cannot be
satisfied. In this case, the original LP must be infeasible.
Dual simplex method
• Three uses of the dual simplex follow:
– Finding the new optimal solution after a constraint is added
to an LP
– Finding the new optimal solution after changing a right-hand
side of an LP
– Solving a normal min problem
Finding the new optimal solution after a
constraint is added to an LP
• The dual simplex method is often used to find the new
optimal solution to an LP after a constraint is added.
When a constraint is added, one of the following three
cases will occur:
– Case 1 the current optimal solution satisfies the new
constraint.
– Case 2 the current optimal solution does not satisfy
the new constraint, but the LP still has a feasible
solution.
– Case 3 the additional constraint causes the LP to have
no feasible solution.
Finding the new optimal solution after a
constraint is added to an LP
• Suppose we have added the constraint
x1+x2+x3≤11 to the Dakota problem.
The current optimal solution (z=280, x1=2, x2=0, x3=8)
satisfies this constraint.
Adding a constraint to an LP either leaves the feasible
region unchanged or eliminates points from the
feasible region. So, it either reduces the optimal zvalue or leaves it unchanged.
Since the current solution is still feasible and has z=280,
it must still be optimal.
Finding the new optimal solution after a
constraint is added to an LP
• Suppose that in the Dakota problem, adds the
constraint x2≥1
Since the current optimal solution has x2=0, it is no
longer feasible and cannot be optimal.
Appending the constraint –x2+e4=-1 to the optimal
Dakota tableau.
Finding the new optimal solution after a
constraint is added to an LP
BV
 5 x2
z
 2 x2
 2 x2
x1
 54 x2
 x2
 s1
 x3
 10s2
 10s3
 280 z  280
 2 s2
 8s3
 24
s1  24
 2 s2
 4s3
8
x3  8
 12 s2
 32 s3
2
x1  2
 1
e4  1
 e4
The variable e4=-1 is the most negative basic variable, so
e4 will exit from the basis, and row 4 will be the pivot row.
Since x2 is the only variable with a negative coefficient in
row 4, x2 must enter into the basis.
Finding the new optimal solution after a
constraint is added to an LP
BV
z
 s1
 x3
x1
 10s2
 10s3
 5e4
 275 z  275
 2 s2
 8s3
 2e4
 26
s1  26
 2 s2
 4 s3
 2e4
 10
x3  10
 12 s2
 32 s3
 54 e4

x1 
x2
This is an optimal tableau.
 e4
3
4
1
3
4
x2  1
Finding the new optimal solution after a
constraint is added to an LP
• Suppose we add the constraint x1+x2≥12 to the
Dakota problem.
Appending the constraint –x1-x2+e4=-12 to the optimal
Dakota tableau yields,
BV
 5 x2
z
 2 x2
 2 x2
x1
 54 x2
 x1
 x2
 s1
 x3
 10s2
 10 s3
 280
z  280
 2 s2
 8s3
 24
s1  24
 2 s2
 4s3
8
x3  8
 12 s2
 32 s3
2
x1  2
 e4
 12 e4  12
Finding the new optimal solution after a
constraint is added to an LP
To eliminate x1 from the new constraint,
BV
 5 x2
z
 10s2
 10 s3
 280
z  280
 2 s2
 8s3
 24
s1  24
 2 s2
 4 s3
8
x3  8
 54 x2
 12 s2
 32 s3
2
x1  2
0.25 x2
 0.5s2
 1.5s3
 2 x2
 2 x2
x1
 s1
 x3
 e4
 10 e4  10
Since e4=-10 is the most negative basic variable, e4 will leave the
basis and row 4 will be the pivot row. The variable s2 is the only
one with a negative coefficient in row 4, so s2 enters the basis.
Finding the new optimal solution after a
constraint is added to an LP
BV
 10 x2
z
 x2
 x2
x1
 s1
 x3
 40s3
 20e4
 80
z  280
 2s3
 4e4
 16
s1  16
 2s3
 4e4
 32 x3  32
 x2
 0.5 x2
s2
 3s3
 e4
 12
x2  12
 2e4
 20
s2  20
Now x3 must leave the basis, and row 2 will be
the pivot. Since x2 is the only variable in row 2
with a negative coefficient, x2 now enters the
basis.
Finding the new optimal solution after a
constraint is added to an LP
BV
 10 x3
z
 x3
x2
x1
 60s3
 s1
 60e4
 4 s3
 240 z  240
 16
s1  16
 x3
 2 s3
 4e4
 32
x2  32
 x3
 2s3
 3e4
 20
x1  20
 4 s3
 4e4
 36
s2  36
 0.5 x3
 s2
Row 3 cannot be satisfied. Hence the Dakota
problem with the additional constraint x1+x2≥12
has no feasible solution.
Finding the new optimal solution after
changing a right-hand side
• If the right-hand side of a constraint is changed
and the current basis becomes infeasible, the
dual simplex can be used to find the new
optimal solution.
• To illustrate, suppose that 30 finishing hours are
now available in Dakota problem. Then it
changed the current optimal tableau to that
shown ->
Finding the new optimal solution after
changing a right-hand side
z
+5x2
+10s2+10s3=380
-2x2 +s1+ 2s2 -8s3=44
-2x2+x3
+2s2 -4s3=28
x1+1.25x2
-0.5s2+1.5s3=-3
Basic
variables
z=380
s1=44
x3=28
x1=-3
x1 is the most negative one, so x1 must leave the
basis, and row 3 will be the pivot row. Since s2 has
the only negative coefficient in row 3, s2 will enter
the basis.
Finding the new optimal solution after
changing a right-hand side
Basic
variables
z+20x1+30x2
+40s3=320 z=320
4x1 +3x2 +s1
-2s3=32 s1=32
4x1 +3x2+x3
+2s3=16 x3=16
-2x1-2.5x2
+s2 -3s3=6
s2=6
This is an optimal tableau.
Solving a normal min problem
• To solve the following LP:
min z  x1  2 x2
s.t.
x1 - 2 x2  x3  4
2 x1  x2  x3  6
x1 , x2 , x3  0
Convert the LP to a max problem with
objective function z’=-x1-2x2.
Solving a normal min problem
Subtracting excess variables e1 and e2 from the
two constraints, and multiply each constraint
through by -1, we can use e1 and e2 as basic
variables.
z
 x1
 2 x2
0
 x1
 2 x2
 x3
 2 x1
 x2
 x3
 e1
BV
z  0
 4 e1  4
 e2
 6 e2  6
Solving a normal min problem
At least one constraint has a negative right-hand
side, so this is not an optimal tableau.
We choose the most negative basic variable e2 to
leave the basis. Since e2 is basic in row 2, row 2
will be the pivot row.
To determine the entering variable, we find the
following ratios:
x1 ratio=1/2=0.5
x2 ratio=2/1=2
The smaller ratio is the x1 ratio, so x1 enters the
basis in row 2.
Solving a normal min problem
z
x1
 32 x2
 12 x3
 52 x2
 32 x3
 12 x2
 12 x3
 e1
 12 e2
BV
 3 z  3
 12 e2
 1 e1  1
 12 e2
3
Since there is no constraint indicating
infeasibility, we return to step 1.
e2  3
Solving a normal min problem
The first constraint has a negative right-hand side,
so the tableau is not optimal.
Since e1=-1 is the only negative basic variable, e1
will exit from the basis, and row 1 will be the
pivot row.
The ratios are
x3 ratio=(1/2)/(3/2)=1/3
e2 ratio=(1/2)/(1/2)=1
The smallest ratio is 1/3, so x3 will enter the basis
in row 1.
Solving a normal min problem
BV
z
 73 x2
 53 x2
x1
 13 x2
 x3
 13 e1
 13 e2
  103
 23 e1
 13 e2

 13 e1
 13 e2

2
3
10
3
z   103
e1 
e2 
2
3
10
3
Since each right-hand side is nonnegative, this
is an optimal tableau.
The original problem was a min problem, so
the optimal solution to the original problem is
z=10/3,x1=10/3, x3=2/3 and x2=0.