Dual problem

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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
min w
y1  0
y2  0

ym  0
y1
y2

ym
max z
x1  0 x2  0
x2
x1
a12
a11
a22
a21


am 2
am1
 c2
 c1





xn  0
xn
 b1
a1n
 b2
a2 n


 bm
amn
 cn
Finding the dual of an LP

Illustrate by the Dakota problem,
max z  60 x1  30x2  20x3
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
x1
x2
x3
min w
y1  0
y2  0
y3  0
y1
y2
y3
8
4
2
 60
6
2
1.5
 30
1
1.5
0.5
 20
 48
 20
8
Finding the dual of an LP

Then, reading down, we find the Dakota dual to
be
max 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,
(18) max z
s.t.
 2 x1  x2
(19) min w  2 y1  4 y2  6 y3
x1  x2  2
2 x1  x2  3
x1  x2  1
x1  0, x2 urs
s.t.
y1  2 y2  y3  2
 y3  1
y1
y 2  y3  1
2 y1  y2
3
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 fashing.
Finding the dual of a nonnormal max
problem

For example,
max z  2 x1  x2
s.t.
x1  x2  2
max z
2 x1  x2  3
x1  x2  1
x1  0, x2 urs
( x1  0) ( x2 urs)*
x1
x2
min w
( y 3  0)
y1
y2
y3
1
2
1
2
1
1
1
1
 2*
 3*
1
Finding the dual of a nonnormal max
problem
max z
( x1  0) ( x2 urs)
x1
x2
min w
( y1 urs)
( y 2  0)
( y3  0)
y1
y2
y3
1
2
1
2
1
1
1
1
2
3
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 fashing.
Economic interpretation of the dual
problem

The dual of the Dakota problem is,
min w  48y1  20 y2  8 y3
s.t.
8 y1  4 y2  2 y3  60 (desk const raint)
6 y1  2 y2  1.5 y3  30 ( t able const raint)
y1  1.5 y2  0.5 y3  20 (chair const raint)
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 w-value 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 
2

x
 
 
 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 =coefficient of si in optimal row 0
if constraint i is a ≤ constraint
Optimal value of dual variable yi =-(coefficient of ei in optimal row 0)
if constraint i is a ≥ constraint
Optimal value of dual variable yi =(coefficient of ai in optimal row 0)-M
if constraint i is a = constraint
Example 10

To solve the following LP,
max z  3 x1  2 x2  5 x3
s.t.
x1  3 x2  2 x3  15
 2 x2
 x3  5
2 x1  x2  5 x3  10
x1 , x2 , x3  0
Example 10

The optimal tableau,
z
1
x1
0
x2
0
x3
0
0
0
0
1
0
0
1
0
0
1
0
0
s1
e2
51
23
4
23
2
23
9
23
58
23
5
23
9
23
17
23

a2
M  58
23
a3
M  239
 235
 232
9
23
17
23
 231

7
23
rhs
565
23
15
23
65
23
120
23
BV
z  565
23
x3  15
23
x2 
x1 
65
23
120
23
Example 10

Solution
To find the dual from the tableau,
max z
( x1  0) ( x2  0) ( x3  0)
x1
x2
x3
min w
( y1  0)
( y 2  0)
( y3 urs)
y1
y2
y3
1
0
2
3
3
2
1
2
2
1
5
5
 15
 5*
 10*
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
51
58
9
w  565
,
y

,
y


,
y

1
2
3
23
23
23
23
The dual theorem

How to read the optimal dual solution from row 0
of the optimal tableau if the primal is a min
problem
Optimal value of dual variable xi =coefficient of si in optimal row 0
if constraint i is a ≤ constraint
Optimal value of dual variable xi =-(coefficient of ei in optimal row 0)
if constraint i is a ≥ constraint
Optimal value of dual variable xi =(coefficient of ai in optimal row 0)+M
if constraint i is a = constraint
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 right-hand 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 righthand 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 z-value 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
 4 s3
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
 10s3
 280
z  280
 2 s2
 8s3
 24
s1  24
 2 s2
 4 s3
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
 10s3
 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
0.25x2
 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
 10x2
z
 x2
 x2
x1
 s1
 x3
 40s3
 20e4
 80
 2 s3
 4e4
 16 s1  16
 2 s3
 4e4
 32 x3  32
 x2
 0.5 x2
s2
 3s3
z  280
 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
 10x3
z
 x3
x2
x1
 60s3
 s1
 60e4
 4 s3
 240 z  240
 16
s1  16
 x3
 2 s3
 4e4
 32
x2  32
 x3
 2 s3
 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
 x1  2 x2
 2 x1  x2
 x3
 x3
 e1
 e2
BV
0
z  0
 4 e1  4
 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
 52 x2
 12 x2
 12 x3
 32 x3
 12 x3
 e1
 12 e2
 12 e2
 12 e2
BV
 3 z  3
 1 e1  1
 3 e2  3
Since there is no constraint indicating infeasibility,
we return to step 1.
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
z
x1
 73 x2
 53 x2
 13 x2
 x3
 13 e1  13 e2
 23 e1  13 e2
 13 e1  13 e2
  103
 23
 103
BV
z   103
e1  23
e2  103
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.
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