4. Duality and Sensitivity

Transcription

1 4. Duality and Sensitivity For every instance of an LP, there is an associated LP known as the dual problem. The original problem is known as the primal problem. There are two de nitions of the dual pair of problems, which are equivalent. Duality can also be shown to be a `symmetric relationship (property of involution) in that the dual of the dual is the primal problem. 4. Primal Dual Formulations CANONICAL FORM P: Minimize c T x subject to Ax b x 0 D: Maximize y T b or b T y subject to y T A c T or A T y c y 0 STANDARD FORM P Minimize c T x subject to Ax = b x 0 D Maximize y T b or b T y subject to y T A c T (or A T y c) y free variables (u.r.s.) Note that if the number of variables in the primal problem is n and the number of constraints is m then the the number of variables in the dual problem is m and the number of constraints is n: In fact variables in the dual problem correspond to constraints in the primal problem and vice-versa. In other words, if xr n and A is (m n) then y R m and A T y c represents a system of n inequalities (the dual constraints) in m new unknown dual variables (y ; :::; y m ) : Example Primal:

2 Minimize 0x + x + 4x subject to x + x x 4x + x 0 x ; x ; x 0 Dual: Maximize y + 0y subject to y + 4y 0 y y + y 4 y ; y 0 Example Primal: Minimize 6x + 8x subject to x + x s = 4 x + x s = 7 x ; x ; s ; s 0 Dual: Maximize 4w + 7w subject to w + w 6 w + w 8 w ; w 0 The primal in this example is stated in standard form P. However the variables s ; s play the part of non-negative surplus variables since they do not appear in the objective function and it is easy to write down an equivalent primal problem in the canonical form P. The dual problem is the same, whether written in the form of D or as D. Economic interpretation of the dual

3 The dual of the diet problem introduced earlier min x s.t. c T x Ax r x 0 is max y s.t. y T b y T A c T y 0 We can provide an interesting economic interpretation of the dual in this case: Recall that a ij represents the amount of nutrient i contained in unit amount of the j th food. Consider a pill maker (an alternative supplier of nutrients) who wants to set a price for each nutrient. Let i be the price per unit of the i th nutrient (i = ; :::; m) forming a price vector. The pill maker s price to supply the required amount of each nutrient in the diet is mx i r i : i= The price of all nutrients needed to make up a unit amount of the j th food should be competitive, therefore not be more than the price of unit amount of the j th food. Hence mx i a ij c j i= j = ; :::; n Identifying y with we see that the dual problem maximizes the income of the pill maker subject to a competitiveness constraint. 4. Duality - basic properties 4.. Equivalence of dual forms The duality relationships P-D, P-D are equivalent. Proof We rst take the dual pair in standard form P-D as the de nition and nd the dual of a problem in the canonical form P: Minimize c T x subject to Ax b x 0

4 Introduce a vector of m surplus variables s 0 so the constraints can be written or Ax s = b x A I m = b s which is of the form A 0 x 0 = b where A 0 and x 0 are partitioned matrices. Applying the duality result for the primal in standard form P, the dual problem is Maximize subject to b T y A T I m y c 0 In this dual problem, although the D dual variables are not explicitly unrestricted in sign (free variables), the constraints imply that A T y c and y 0, which provides the dual problem D. We can also take the dual pair in canonical form P-D as the de nition and nd the dual of a problem in the canonical form P. (Exercise) 4.. Involution property The dual of the dual is the primal. Proof Consider the dual in canonical form D (yr m ; A(m n)) Maximize subject to b T y A T y c y 0 We may write this in the form of a canonical P primal problem Minimize subject to b T y A T y c y 0 The corresponding canonical dual problem in variables wr n is Maximize c T w subject to A T T w b T w 0 which is equivalent to the primal problem P Minimize subject to c T w Aw b w 0 4

5 4... Mixed Constraints The dual of LP problems that are neither in standard form nor in canonical form can be written down using rules summarized in the following table. Min Problem Max problem Type of Constraint Type of Variable a T i x b i y i 0 (P, D) a T i x b i y i 0 a T i x = b i y i u.r.s. (P, D) Type of Variable Type of Constraint x j 0 y T A j c j (either case) x j 0 y T A j c j x j u.r.s. y T A j = c j Notes:. u.r.s. denotes unrestricted in sign or free variable. a T i denotes i th primal constraint coe cients i.e. the i th row of A-matrix. A j denotes j th dual constraint coe cients i.e. the j th column of A-matrix Example (mixed constraints) Primal: Dual: Maximize z = x + 6x subject to x + x = x + x 4x + 7x 8 x u.r.s., x 0 Minimize w = y + y + 8y subject to y y + 4y = y + y + 7y 6 y u.r.s., y 0; y 0 Use of the Table: m = ; n = so dual has variables y ; y ; y and constraints Since primal is a maximization, refer to Table nd column for primal properties - read o dual properties from Table st column.

6 Variable x is u.r.s. )st dual constraint is = Variable x 0 ) nd dual constraint is (for minimization) st primal constraint is =, ) y is u.r.s. nd primal constraint is ) y 0 rd primal constraint is ) y 0 4. Duality Theorems The primal and dual LP problems are so intimately related that solving either problem provides the solution to the other. For the following results we assume the primal is a minimization in standard form P with dual D. 4.. Weak duality lemma If x, y are feasible for the primal and dual problems respectively, then) c T x y T b (4.) Proof Let x, y be feasible for problems P, D respectively. Then ) y T A c T (4.a) y T A c T 0 (4.b) y T A c T x 0 (4.c) (4.c) follows from (4.b) because x 0 (feasible for P) so P n i= y T A c T x i 0. i Now Ax = b; again because x is feasible for P, so y T Ax c T x 0 y T b c T x 0 y T b c T x as required. Corollary If x, y are feasible for P, D respectively and c T x = y T b then x and y are optimal for their respective problems. i.e. If a pair of feasible solutions for the primal and dual problems have the same objective function value, then they are optimal for their respective problems. 6

7 Corollary If either problem is unbounded (optimal solution is at in nity) the other problem in infeasible. [ NB The converse is not true as both problems can be infeasible ] Example In Example of Section 4. the primal and dual objective functions are z = 6x + 8x and w = 4y + 7y : It can easily be veri ed (e.g. graphically) that z min = w max = 4 x = 7 ; 0 T and y = 0; 6 T. and these optima are achieved at Let us solve the dual problem through simplex iterations. In standard form constraints are y + y + v = 6 y + y + v = 8 A basis of dual slacks is available: Max y y v 6 v 8 z After one pivot, the optimal tableau is y v z y v We may con rm the elements of the tableau as follows: B = 0 ; N = 0 B = 0 B N = = = (y ij ) 7

8 b = B b = 0 (z j c j ) = c T B y j c j = 6 8 ; 7 = 6 8 y v z 0 = c T B b = 7; = 4 The matrix form of the reduced tableau is Non-basic variables where z j c j are the elements of c T B B N c T N and z 0 = c T B b. Basic vars Y = B N = (y ij ) b z fz j c j g z 0 (4.) Simplex multipliers (dual variables) The components of the m-vector y T 0 = c T B B (4.4) known as simplex multipliers are the dual variables associated with a particular basis B: The dual vector is generally infeasible for the dual problem during intermediate tableau iterations. Only when the bottom row satis es primal optimality does y T 0 become dual feasible. We will show below that. the dual variables y 0 corresponding to an optimal basis are feasible. the dual objective w 0 = y T 0 b equals z 0, the minimum value of the primal. Therefore by Corollary to the weak duality lemma we have constructed an optimal dual vector y Strong duality theorem If either the primal or the dual has a nite optimal solution then so has the other and the corresponding objective function values are equal. If either problem is unbounded the other problem is infeasible. Proof Let y T 0 = c T B B where B is an optimal basis. y T 0 A = y T 0 h = h h c T B c T B i h i B N = c T BB B N i c T B B N (4.a) i = c T (4.b) c T N 8

9 so that y T 0 is dual feasible, with objective value w 0 = y T 0 b = c T BB b = c T Bx B = z 0 Notice that in going from (4.a) to (4.b) we have assumed that c T B B N c T N. This follows from the optimality conditions z j c j 0 8j, which are simply the components of c T B B N c T N. Example (continued) For the above example y T 0 = c T BB = 7; 0 = = x ; x 7 ; 0 : 0 Hence the primal optimal solution is x = 7 ; x = 0: 4.. Complementary slackness The complementary slackness (C-S) conditions, which are a consequence of duality, relate the values of the primal and dual vectors at a common optimum. They can be stated for either the asymmetric (P,D) or the symmetric (P D) primal dual pair. Theorem (Complementary slackness - asymmetric case) Let x; y be feasible for P, D respectively. A necessary and su cient condition (NSC) that they both be optimal is that for each j (j = ; :::; n) i) If x j > 0 then y T A j = c j ii) If y T A j < c j then x j = 0 Proof. Since x; y are feasible for their respective problems, we have Ax = b, x 0 y T A c T 9

10 Thus y T A + v T = c T with v T 0. Postmultiply by x and apply Ax = b y T Ax + v T x = c T x y T b + v T x = c T x c T x y T b = v T x (4.6) By corollary to the Weak Duality theorem, x; y are optimal if and only if r.h.s.= 0. Since v T x = P n j= v jx j with v j ; x j 0; r.h.s.= 0 if and only if v j x j = 0 for each j. So i) x j > 0 ) v j = 0 and ii) v j > 0 ) x j = 0 as required: [In fact ii) follows from i) and feasibility of y ] Theorem (Complementary slackness - symmetric case) Let x; y be feasible for P, D respectively. A necessary and su cient condition (NSC) that they both be optimal is that i) x j v j = 0 8j ii) s i y i = 0 8i where v j is the j th dual slack (corresponding to x j ) and s i is the i th primal slack so Comment: By analogy with the asymmetric case, we can write each equation as a pair of implications, 8j; x j > 0 ) v j = 0; v j > 0 ) x j = 0 (4.7a) 8i; s i > 0 ) y i = 0; y i > 0 ) s i = 0 (4.7b) Proof. Since x; y are feasible for their respective problems, we have Ax b, x 0 y T A c T, y 0 or, introducing slack variables s 0, v T 0 Ax s = b, x; s 0 y T A + v T = c T, y; v 0 Then y T Ax y T s = y T b y T Ax + v T x = c T x so c T x y T b = v T x + y T s (4.8) The result follows by analogy to the previous (asymmetric) case. 0

11 Example (continued) Recall the primal was Minimize 6x + 8x subject to x + x 4 x + x 7 x ; x 0 and the dual Maximize 4w + 7w The optimal solution to the primal is subject to w + w 6 (x ; x ; s ; s ) = w + w 8 w ; w 0 7 ; 0; ; 0 From the C-S conditions, the optimal solution to the dual takes the form (v ; v ; w ; w ) = (0; ; 0; ) i.e. the st dual variable is zero at the optimum (w = 0). Hence = 6 from the st dual constraint which is active (satis ed with equality) since v = 0: Hence (w ; w ) = 0; The Dual Simplex Algorithm Suppose we have a tableau that satis es the bottom row optimality conditions but which corresponds to an infeasible solution. i.e. we have non-basic variables x B Y =(y ij ) b z-row z j c j s z 0 with z j c j 0; each j (for minimization) and b i < 0 for some i. Such a tableau is said to be dual feasible because the associated dual vector y T 0 = c T B B satis es y T 0 A=y T 0 [BjN] c T The dual simplex algorithm is applicable. The rules for selecting a pivot element are:

12 . If b 0; stop; the current basis is optimal. Otherwise choose pivot row p by min i bi = b p (b p < 0). If y pj 0 8j stop; the problem is infeasible (because dual is unbounded). Otherwise, choose pivot column q by the minimum ratio rule z j min j y pj c j : y pj < 0 = z q c q y pq (4.9). Pivot on the element y pq (< 0) according to usual Simplex rules and return to Step. Notes i) The dual simplex algorthm is in fact the primal simplex algorithm applied to the dual problem. ii) The dual simplex algorithm can be useful (a) in avoiding a Phase I calculation when an infeasible but optimal tableau is available by inspection, (b) in sensitivity analysis when the current optimal solution becomes infeasible through the addition of an extra constraint or a small change in the right hand side vector b: in integer programming by Gomory s cutting planes. Example (Dual Simplex) Minimize x +x +4x subject to x +x + x x x +x 4 x ; x ; x 0 An initial (infeasible) tableau using s ; s as basic variables is After st dual simplex pivot x x x s s " After nd dual simplex pivot s x s x x 4 4 "

13 x x s s x Optimal solution is (x ; x ; x ; s ; s ) = ; ; 0; 0; 0 Optimal solution to dual (w; w) = 8 ; can be con rmed using C-S conditions (Ex.) N.B.. The OF value z is monotonic increasing in successive tableaux (min problem).. Since I m appears in the the initial A matrix and y T 0 = c T B B = c T B B ( I m ) w; w also appear negatively as the z j c j values for surplus variables s ; s. 4. Sensitivity Analysis,How is an optimal solution (of a LP in standard form) in uenced by. changes to the objective function coe cients c. changes to the rhs. of the constraints b. changes to the constraint matrix A; in particular (a) adding a new constraint a T m+x = b m+ (b) adding a new variable (or activity) x m+? From a knowledge of the simplex tableau at optimality non-basic variables x B Y =B N b = B b z-row c T B B N c T N z 0 = c T B B b (4.0) we see that. small changes to c a ect only the bottom row. small changes to b a ect only the rhs.. changes to A can potentially in uence all sections of the tableau (we will only consider in detail adding a new constraint)

14 Example (Furniture manufacturer) A manufacturer of furniture makes three products: desks, tables and chairs. Each item of furniture made consumes amounts of three resources: timber, nishing time and carpentry time. To manufacture one desk requires 8 ft. of timber, 4 hours nishing time and hours carpentry time A table requires 6 ft. of timber, hours nishing time and. hours carpentry time A chair requires ft. of timber,. hours nishing time and 0. hours carpentry time The total weekly availability of timber is 48 ft., of nishing time is 0 hrs and of carpentry time is 8 hrs. The unit pro t from selling a desk is $60, from a table is $0, from a chair is $0. The LP formulation to maximize the manufacturers pro t subject to the given resource constraints is as follows: Let x ; x ; x = no. of units of each product (desks, tables and chairs respectively) to make per week. Maximize 60x +0x +0x subject to 8x +6x + x 48 4x +x +:x 0 x +:x 0:x 8 x ; x ; x 0 resulting in the following tableaux Initial Tableau x x x s s 4 0 s Final Tableau s x s s 8 4 x 4 8 x The optimal solution is to manufacture desks and 8 chairs resulting in a pro t of $80. There is a surplus of 4 ft. of timber, but all available hours of nishing time, carpentry time are used up. Often the data assumed in the calculation are imprecise or subject to change. We now ask Q. What range of variation in unit pro t leaves the current solution optimal? Q. What is the optimal product mix if the unit pro t for a table increases to $40? Q. What if the total available nishing time increases to 0 hours? These questions can be answered through a sensitivity analysis. 4

15 . Changes to c From (4.0) we see these a ect the optimality conditions z j c j 0 (for max) each j, and the optimal value z 0 = c T Bb (though not of course the polytope representing the feasible region). Consider increasing a single component of c B ; say c by an amount, i.e. c nal tableau is c +: The modi ed s x s s 8 4 x 4 8 x The condition for the current solution to remain optimal is which need to be satis ed simultaneously. Hence or equivalently 6 c 80 For example, if c = 6 ( = 4) we nd an alternative optimal manufacturing plan is to stop manufacturing desks and instead to produce.6 tables and. desks per week s x s s 8 4 x 4 8 x s x s 6 s (7:) 6 x (:) 8 x (:6) NB Non-integer production values can still have meaning in a continuous production setting, although integer programming (IP) techniques (see Section ) will provide ways of nding optimal integer solutions. Similarly we can establish ranges for c and c (leaving the other components unchanged) that ensure the current solution remains optimal.. Suppose c = 40 ( = 0 in c ). As x is nonbasic in the optimal tableau, only that column s z j c j

16 value is changed. It becomes negative ( ) and after one pivot optimality is restored s x s s 8 4 x 4 8 x s x s 6 s (7:) 6 x (:) 8 x (:6) This is the same BFS as before.. Changes to b These a ect the rhs. of an optimal tableau (4.0). Consider an increase of to the second resource limitation b, b b + : We consider the e ect on b of the rhs. becoming b 0 = b + b with b = (0; ; 0) T e = (0; ; 0), the nd column of the identity matrix I : = e where b = B b b 0 = B (b + b) = b +B b = b +B e Notice that the product B e picks out the second column of B which appears in the nal tableau under s (which corresponded to e in the initial tableau). Hence The rhs. remains feasible while b = ; 8 + 0; 0 i.e. while 4 4 () 6 b 4 e.g. if b = 0 ( = 0) the nal tableau becomes infeasible. In this example feasibility (hence the optimal solution) is obtained after one iteration of the dual simplex algorithm s x s s 8 44 x 4 8 x s x x s x 6 s