duality theory li xiaolei. finding the dual of an lp when taking the dual of a given lp, we refer...

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,…,y

m.


A normal max problem may be written as

(16)

,n),,(ix

bxaxaxa

bxaxaxa

bxaxaxats

xcxcxcz

j

mnmnmm

nn

nn

nn

210

..

max

2211

22222121

11212111

2211


The dual of a normal max problem is defined to be,

(17)

,m),,(iy

cyayaya

cyayaya

cyayayats

ybybybw

i

nmmnnn

mm

mm

mm

210

..

min

2211

22222112

11221111

2211

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.


n

mmnmmmm

n

n

n

n

ccc

baaayy

baaayy

baaayy

xxx

xxxw

z

21

21

22222122

11121111

21

21

0

0

0

000min

max


Illustrate by the Dakota problem,

),,(ix

xxx

xxx

xxxts

xxxz

j 3210

85.05.12

205.124

4868 ..

203006 max

321

321

321

321


203060

85.05.120

205.1240

481680

000min

max

33

22

11

321

321

yy

yy

yy

xxx

xxxw

z


Then, reading down, we find the Dakota dual to be

),,(ix

yyy

yyy

yyyts

yyyw

j 3210

205.05.1

305.12 6

602 4 8 ..

8 2084 max

321

321

321

321

Finding the dual of a nonnormal LP

Many LPs are not normal max or min problem. For example,

ursxx

xx

xx

xxts

xxz

,0

1

32

2 ..

2max

21

21

21

21

21

0,y urs,

3 2

1

1

2 2 ..

642min

321

21

32

31

321

321

y y

yy

yy

yy

yyyts

yyyw(18) (19)


An LP can be transformed into normal form. To place a max problem into normal form, we proc

eed as follows: Step 1 multiply each ≥ constraint by -1, converting it into

a ≤ constraint. Step 2 replace each equality constraint by two inequalit

y 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 .


(18) has been transformed onto the following LP:

0,,

1

32-

2-

2 ..

2max

221

21

21

21

21

221

xxx

xxx

xxx

xxx

xxxts

xxxz


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 inequalit

y 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 y

i’ ≥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.


For example,

12

111)0(

312

211

) ()0(min

max

33

*2

*1

21

*21

yy

y

y

xx

ursxxw

z

ursxx

xx

xx

xxts

xxz

,0

1

32

2 ..

2max

21

21

21

21

21


12

111)0(

312)0(

211) (

) ()0(min

max

33

22

11

21

21

yy

yy

yursy

xx

ursxxw

z

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,

0,,

)constraintchair ( 205.05.1

)constraint table( 305.12 6

)constraintdesk ( 602 4 8 ..

8 2048min

321

321

321

321

321

yyy

yyy

yyy

yyyts

yyyw


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


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.


Primal problem

(22)

,n),,(ix

bxaxaxa

bxaxaxa

bxaxaxats

xcxcxcz

j

mnmnmm

nn

nn

nn

210

..

max

2211

22222121

11212111

2211


Dual problem

(23)

,m),,(iy

cyayaya

cyayaya

cyayayats

ybybybw

i

nmmnnn

mm

mm

mm

210

..

min

2211

22222112

11221111

2211


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.


Lemma 1

Let

Be any feasible solution to the primal and y=[y1 y

2 … ym] be any feasible solution to the dual, then (z-value for x)≤(w-value for y).

nx

x

x

x

2

1


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.


Lemma 2 Let

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.

nx

x

x

x

2

1


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, wz

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


if constraint i is a ≥ constraint

=-(coefficient of ei in optimal row 0)


if constraint i is a = constraint

=(coefficient of ai in optimal row 0)-M

Example 10

To solve the following LP,

0,,

105 2

5 2

152 3 ..

5 23max

321

321

32

321

321

xxx

xxx

xx

xxxts

xxxz

Example 10

The optimal tableau,

23120

123120

237

2317

2317

239

2365

22365

231

239

239

232

2315

32315

232

235

235

234

23565

23565

239

2358

2358

2351

3221321

0010

0100

1000

0001

BV

x

x

x

zMM

rhsaaesxxxz

Example 10

Solution

To find the dual from the tableau,

523

*10512) (

*5120)0(

15231)0(

)0()0()0(min

max

33

22

11

321

321

yursy

yy

yy

xxx

xxxw

z

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

239

32358

22351

123565 ,,, yyyw

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

if constraint i is a ≤ constraint

=coefficient of si in optimal row 0


if constraint i is a ≥ constraint

=-(coefficient of ei in optimal row 0)


if constraint i is a = constraint

=(coefficient of ai in optimal row 0)+M

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.


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 0Coefficient 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.


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.


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.


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.


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.


11

22

88422

2424822

28028010105

442

1323

221

245

1

33232

13212

322

eex

xssxx

xssxx

ssssx

zssxz

BV

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.


11

1010242

2626282

27527551010

242

43

143

445

323

221

1

34323

14321

432

xex

xessx

xessx

sesss

zessz

BV

This is an optimal tableau.


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,

1212

22

88422

2424822

28028010105

4421

1323

221

245

1

33232

13212

322

eexx

xssxx

xssxx

ssssx

zssxz

BV


To eliminate x1 from the new constraint,

10105.15.025.0

22

88422

2424822

28028010105

44322

1323

221

245

1

33232

13212

322

eessx

xssxx

xssxx

ssssx

zssxz

BV

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.


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.

2020235.0

1212

323242

161642

28080204010

24322

2421

34332

14312

432

sessx

xexx

xesxx

sessx

zesxz

BV


Row 3 cannot be satisfied. Hence the Dakota problem with the additional constraint x1+x2≥12 has no feasible solution.

3636445.0

202032

323242

16164

240240606010

24323

14331

24332

1313

433

sessx

xesxx

xesxx

sssx

zesxz

BV

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 ->


Basic

variables

z +5x2 +10s2+10s3=380 z=380

-2x2 +s1+ 2s2 -8s3=44 s1=44

-2x2+x3 +2s2 -4s3=28 x3=28

x1+1.25x2 -0.5s2+1.5s3=-3 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.


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:

Convert the LP to a max problem with objective function z’=-x1-2x2.

0,,

6 2

4 2- ..

2min

321

321

321

21

xxx

xxx

xxxts

xxz


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.

662

442

002

22321

11321

21

eexxx

eexxx

zxxz

BV


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=2The smaller ratio is the x1 ratio, so x1 enters the basis in

row 2.


Since there is no constraint indicating infeasibility, we return to step 1.

33

11

33

2221

321

221

1

1221

1323

225

221

321

223

eexxx

eeexx

zexxz

BV


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)=1The smallest ratio is 1/3, so x3 will enter the basis in

row 1.


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.

310

2310

231

131

231

1

32

132

231

132

3235

310

310

231

131

237

eeexx

eeexx

zeexz

BV

duality theory li xiaolei. finding the dual of an lp when taking the dual of a given lp, we refer...

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