[PPT]Linear Programming (Optimization) - SOLABsolab.kaist.ac.kr/files/LP/LP2015/LP_Chap_04.pptx · Web viewLinear Programming 2015 Chapter 4. Duality Theory Given min 𝑐′𝑥,

Linear Programming 2015 1

Chapter 4. Duality Theory Given min , s.t. , (called primal problem)

there exists another LP derived from the primal problem using the same data, but with different structure (called the dual prob-lem, 쌍대문제 ).

The relation between the primal and the dual problem consti-tutes very important basis for understanding the deeper struc-ture of LP (compared to systems of linear equations).It provides numerous insights in theory and algorithms and im-portant ingredients in the theory of LP.

Objective value of the dual problem of a feasible dual solution provides lower bound on the optimal primal objective value and the dual problem can be derived for this purpose. However, the text derives it as a special case of the Lagrangian dual problem.

Consider an optimization problem (OPT) , . A relaxation of OPT is any min-imization problem

(RP) with the following two properties: (R1) (R2) for

Prop: If RP is infeasible, so is OPT. If OPT is feasible, then . Pf) From (R1), if , so is OPT. Now suppose that is finite, and let be an optimal solution to OPT. Then , where the first inequality follows from (R2) and the second one follows from (R1). Finallly, if , (R1) and (R2) imply that . Relaxation for a maximization problem also can be defined similarly.



Given min , s.t. , (called primal problem)consider a relaxed problem in which the constraints is elimi-nated and instead included in the objective with penalty , where is a price vector of the same dimension as .

Lagrangian function : The problem becomes min

s.t. (called Lagrangian relaxation)

Optimal value of this problem for fixed is denoted as .Suppose is an optimal solution to the LP, then

since is a feasible solution to LP. gives a lower bound on the optimal value of the LP. Want close lower bound.


Lagrangian dual problem :max s.t. no constraints on ,

where

Hence the dual problem ismax s.t. ( , )


Remark :(1) If LP has inequality constraints

dual constraints are

Or the dual can be derived directly min , s.t. ( Let )

max , s.t.

Hence the dual problem ismax , s.t.


(2) If are free variables, then

dual constraints are


4.2 The dual problem

min max s.t. s.t.


Table 4.1 : Relation between primal and dual variables and con-straints

Dual of a maximization problem can be obtained by converting it into an equivalent min problem, and then take its dual.

PRIMAL minimize maximize DUAL

constraintsfree

variables

variablesfree

constraints


Vector notationmin max s.t. s.t.

min max s.t. s.t.

Thm 4.1 : If we transform the dual into an equivalent minimiza-tion problem and then form its dual, we obtain a problem equiva-lent to the original problem.( the dual of the dual is primal, involution property)

For simplicity, we call the min form as primal and max form as dual. But any form can be considered as primal and the corre-sponding dual can be defined.


Dual

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freexx

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xx

3

2

1

3

321

21

00

463253

.. ts

321 465max ppp

.. ts

33 2 3120

0

32

21

21

3

2

1

pppppp

pp

freep

321 465min xxx .. ts

33- 2312

00

32

21

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1

xxxxxx

xx

freexDual

321 32max ppp

freepp

ppppp

pp

3

2

1

3

321

21

00

46325 3

.. ts

(Ex 4.1)


Ex 4.2 : Duals of equivalent LP’s are equivalent.

min max s.t. s.t. free

min max s.t. s.t. free free,

min max s.t. s.t.


Ex 4.3: Redundant equation can be ignored.

Considermin max s.t. (feasible) s.t.

Suppose .For feasible , get . dual constraints are .

dual objective is .Let . max

s.t.


Thm 4.2 : If we use the following transformations, the corre-sponding duals are equivalent, i. e., they are either both infea-sible or they have the same optimal cost.(a) free variable difference of two nonnegative variables(b) inequality equality by using nonnegative slack var.(c) If LP in standard form (feasible) has redundant equality constraints, eliminate them.


4.3 The duality theorem Thm 4.3 : ( Weak duality )

If is feasible to primal, is feasible to dual, then .pf) Let , , If feasible to P and D respectively, then

Cor 4.1 : If any one of primal, dual is unbounded the other infeasible

Cor 4.2 : If feasible and , then optimal.pf) for all primal feasible . Hence is optimal to the primal problem. Similarly for .


Thm 4.4 : (Strong duality)If a LP has an optimal solution, so does its dual, and the respec-tive optimal costs are equal.

pf) Consider standard form first.Get optimal dual solution from the optimal basis.Suppose have min , , full row rank, and this LP has optimal so-lution.Use simplex to find optimal basis with .Let , then dual feasibleAlso , hence is an optimal solution to the dual and .For general LP, convert it to standard LP with full row rank and apply the result, then convert the dual to the dual for the original LP.


D1 1

2 D2

Fig 4.1 : Proof of the duality theorem for general LP

Duals of equivalent prob-lems are equivalent Equivalent

Duality for standard form problems


Table 4.2 : different possibilities for the primal and the dual

Finite opti-mum

Unbounded Infeasible

Finite opti-mum

Possible Impossible Impossible

Unbounded Impossible Impossible Possible

Infeasible Impossible Possible Possible

(P)

(D)


Note:(1) Later, we will show strong duality without using simplex method. (see example 4.4 later)

(2) Optimal dual solution provides “certificate of optimality”certificate of optimality is (short) information that can be used to check the optimality of a given solution in polynomial time.( Two view points : 1. Finding the optimal solution.

2. Proving that a given solution is optimal.For computational complexity of a problem, the two view points usually give the same complexity (though not proven yet, P = NP ?). Hence, researchers were almost sure that LP can be solved in polynomial time even before the discovery of a polynomial time algorithm for LP.)

(3) The nine possibilities in Table 4.2 can be used importantly in determining the status of primal or dual problem.


Thm 4.5 : (Complementary slackness)Let and be feasible solutions to primal and dual, then and are optimal solutions to the respective problems iff

for all , for all .

pf) Define , Then for feasible .

From strong duality, if optimal, we have .Hence Conversely, if , hence optimal.


Note :(1) CS theorem provides a tool to prove the optimality of a given primal solution. Given a primal solution, we may identify a dual solution that satisfies the CS condition. If the dual solution is feasible, both x and y are feasible solutions that satisfy CS con-ditions, hence optimal (If nondegenerate primal solution, then the system of equations has a unique solution. See ex. 4.6).

(2) CS theorem does not need that and be basic solutions.

(3) CS theorem can be used to design algorithms for special types of LP (e.g. network problems). Also an interior point algo-rithm tries to find a system of nonlinear equations similar to the CS conditions.

(4) See the strict complementary slackness in exercise 4.20.


Geometric view of optimal dual solution

(P) min s.t. , , ( assume span )

(D) max s.t.

Let , such that are linearly independent. has unique solution which is a basic solution. Assume that is nondegenerate, i.e., , for .Let be dual vector.


The conditions that and are optimal are

(a) , ( primal feasibility )(b) ( complementary slackness )(c) ( dual feasibility )

( unique solution )(d) ( dual feasibility )


Figure 4.3

𝑐 𝑎5

𝑎4

𝑎3

𝑎2

𝑎1𝐷

𝐶

𝐵𝐴

𝑐

𝑎2

𝑎1

𝑐𝑎3

𝑎1

𝑐𝑎4 𝑎1

𝑐𝑎5

𝑎1


Figure 4.4 degenerate basic feasible solution

𝑐𝑎3

𝑎2

𝑎1

𝑎3 𝑎2𝑎1

𝑥∗


4.4 Optimal dual var as marginal costs

Suppose standard LP has nondegenerate optimal b.f.s. and is the optimal basis, then for small , . not affected, hence remains optimal basis.Objective value is ( )So is marginal cost (shadow price) of requirement .

Another interpretation of duality, for standard form problems: ex) diet problem (ex. 1.3 in sec.1.1)

Have different foods and different nutrients. Nutrient re-quirement vector , . amount of nutrient in one unit of food.min ()

, nutrient vector of one unit of food.Interpret as the “fair” price of per unit of the nutrient.Cost of one unit of food is but it also has a value of if priced at the nutri-ent market.CS conditions asserts that every food that is used () to synthesize the ideal food, should be consistently priced at the two markets (when ). The value of the ideal food is , which is equal to in the nutrient market.

Also can explain the logic of simplex method. With current basis , have cur-rent value of nutrients (. If have nonbasic with , cost of food is cheaper than the value of nutrients obtained from buying one unit of food. Hence it is better to buy food (try to make , or basic).



4.5 Dual simplex method

For standard problem, a basis gives primal solution and dual solution .At optimal basis, have ( )and , hence optimal.( have primal feasible, dual feasible solutions and objective values are the same).Sometimes, it is easy to find dual feasible basis. Then, start-ing from a dual feasible basis, try to find the basis which sat-isfies the primal feasibility.(Text gives algorithm for tableau form, but revised dual sim-plex algorithm also possible.)


Given tableau, for all , for some .

(1) Find row with . is component of .

(2) For with , find such that

(3) Perform pivot ( enters, leaves basis )( dual feasibility maintained)


Ex 4.7 :

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2/102/3122/1215060301014354321 xxxxx

4x5x

4x2x


Note :(1) row 0 row 0 + ( row )

( from the choice of )For , add (nonnegative number) to row 0 For , have . Hence dual feasibility maintained.

Objective value increases by . ( note that )

If , objective value strictly increases.


(2) If in all iterations, objective value strictly increases, hence fi-nite termination. (need lexicographic pivoting rule for general case)At terminationcase a) , optimal solution.case b) entries of row are all , then dual is unbounded. Hence primal is infeasible.Reasoning : (a) Find unbounded dual solution. Let be the current dual feasible solution ( ).Want such that as .In other words, .Suppose . Let (negative of row of ), then and .Hence is dual feasible and as .(Is an extreme ray of dual polyhedron?)

(continued)(b) Current row : Since , and for feasibility, but no feasible solution to primal exists. hence dual unbounded.



The geometry of the dual simplex method For standard LP, a basis gives a basic solution (not necessarily

feasible) .The same basis provides a dual solution by , i.e. .The dual solution is dual feasible if .So, in the dual simplex, we search for dual b.f.s. while corre-sponding primal basic solutions are infeasible until primal feasi-bility (hence optimality) is attained.( see Example 4.8 )

See example 4.9 for the cases when degeneracy exists. Different bases may lead to the same basic solution for the pri-mal, but to different basic solutions for the dual.


4.6 Farkas’ lemma and linear inequalities Thm 4.6 : Exactly one of the following holds:

(a) There exists some such that .(b) There exists some such that and .

(Note that we used (I) (II) earlier. Rows of are considered as generators of a cone. But, here, columns of are considered as generators of a cone.)pf) not both : ( i.e. one holds the other not hold)Text shows (a) ~(b) which is equivalent to (b) ~(a).~(a) (b) : Consider

(P) max (D) min

~(a) primal infeasible dual infeasible or un-bounded

but is dual feasible dual unbounded with and .


Other expression for Farkas’ lemma.

Cor 4.3 ) Suppose that any vector satisfying also satisfy .Then such that .

See ‘Applications of Farkas’ lemma to asset pricing’ and separat-ing hyperplane theorem used for proving Farkas’ lemma.


The duality theorem revisited Proving strong duality using Farkas’ lemma (not using simplex

method as earlier)

(P) min (D) max s.t.

Suppose optimal to (P). Let Then that satisfies , must satisfy ( i. e., is infeasible )( Note that above statement is nothing but saying, if optimal, then there does not exists a descent and feasible direction at . Necessary condition for optimality of . )


(continued)Otherwise, let .Then

for small Hence feasible for small and .Contradiction to optimality of .By Farkas’, such that .Let for . such that and . By weak duality, is an optimal dual solution.


Figure 4.2 strong duality theorem

𝑐

𝑎3

𝑎2𝑎1

𝑝2𝑎2 𝑝1𝑎1

{𝑑 :𝑐 ′𝑑<0 }

{𝑑 :𝑎𝑖 ′ 𝑑≥0 , 𝑖∈𝐼 }

𝑥∗

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[PPT]Linear Programming (Optimization) - SOLABsolab.kaist.ac.kr/files/LP/LP2015/LP_Chap_04.pptx · Web viewLinear Programming 2015 Chapter 4. Duality Theory Given min 𝑐′𝑥,