College of Management, NCTU Operation Research I Fall, 2008

Jin Y. Wang 6-1

Chap6 Duality Theory and Sensitivity Analysis

The rationale of duality theory Max 4x1 + x2 + 5x3 + 3x4 S.T. x1 x2 x3 + 3x4 1 5x1 + x2 + 3x3 + 8x4 55 x1 + 2x2 + 3x3 5x4 3

x1~x4 0 9 If we multiply a certain number (y0) on both sides of any constraint, the

result wont be affected. 9 The arbitrary y is called the multiplier. y1, y2, and y3 are the multipliers of

constraints 1, 2, and 3, respectively. 9 How can we tell the range of the optimal objective value? (Let Z* be the

optimal value) Consider 4x1 + x2 + 5x3 + 3x4 5/3 (5x1 + x2 + 3x3 + 8x4) Z* 5/3(55) = 275/3 Similarly 4x1 + x2 + 5x3 + 3x4 constraint 2 + constraint 3 = (5x1+ x2 + 3x3 + 8x4) + (x1 + 2x2 + 3x3 5x4) Z* < 55 + 3 = 58

9 That is (1) Z* 0b1 + 5/3b2 + 0b3 (2) Z* 0b1 + 1 b2 + 1b3

9 In general, we want to determine the values of yi such that the following statement holds. Z* 4x1 + x2 + 5x3 + 3x4 y1(x1 x2 x3 + 3x4) + y2(5x1 + x2 + 3x3 + 8x4) +

y3(x1 + 2x2 + 3x3 5x4) y1 + 55y2 + 3y3

9 Issue: which is the best estimate of the optimal value?

College of Management, NCTU Operation Research I Fall, 2008

Jin Y. Wang 6-2

9 So, we can conclude that (1) y1, y2, y3 0 (2) 4 y1 + 5y2 y3

1 y1 + y2 + 2y3 5 y1 + 3y2 + 3y3 3 3y1 + 8y2 5y3

(3) Since we want to estimate Z* as best as possible, i.e. we want to have y1 + 55y2 + 3y3 as less as possible. Therefore, we have Min y1 + 55y2 + 3y3

9 The above formulation is called the Dual problem. 9 The original formulation is called the Primal problem. 9 Note: (1) Primal: coefficients of the objective function Dual: RHS (2) Primal: RHS Dual: coefficients of the objective function (3) Primal: coefficients of a variable in the functional constraint

Dual: coefficients in a functional constraint (4) Max Min (5)

Example of finding the dual problem Max 5x1 + 4x2 + 3x3 S.T. 2x1 + 3x2 + x3 5

4x1 + x2 + 2x3 11 3x1 + 4x2 + 2x3 8 x1, x2, x3 0

Matrix representation of the primal and dual problems

College of Management, NCTU Operation Research I Fall, 2008

Jin Y. Wang 6-3

9 Primal Problem (P) Max == nj jj xcZ 1 Max Z = cx S.T. = nj ijij bxa1 , for i = 1, 2, , m S.T. Ax b xj 0, for j = 1, 2, , n x 0

9 Dual Problem (D) Min == mj ii ybW 1 Min W = yb S.T. = mi jiij cya1 , for j = 1, 2, , n S.T. yA c

yi 0, for i = 1, 2, , m y 0 Weak duality property

If x is a feasible primal solution, and y is a feasible dual solution, then cx yb

9 Corollary: If the primal is feasible and unbounded, the dual is infeasible and

vice versa.

Strong duality property If x* is an optimal solution for the primal problem and y* is an optimal solution for the dual problem, then cx* = y*b.

A corollary for weak and strong duality property 9 These two properties implies that cx < yb for feasible solutions if one or both

of them are not optimal for their respective problems, whereas equality holds when both are optimal.

Complementary solutions property

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9 At each iteration, the simplex method simultaneously identifies a CPF solution x for the primal problem and a complementary solution y for the dual problem (in tabular form, the coefficients of the slack variables in row 0), where cx = yb.

9 If x is not optimal for the primal problem, then y is not feasible for the dual problem. P: Max Z = 5x1 + 4x2 + 3x3 S.T. 2x1 + 3x2 + x3 5

4x1 + x2 + 2x3 11 3x1 + 4x2 + 2x3 8

x1, x2 x3 0 D: Min W = 5y1 +11y2 + 8y3

S.T. 2y1 + 4y2 + 3y3 5 3y1 + y2 + 4y3 4 y1 + 2y2 + 2y3 3 y1, y2, y3 0

Z - 5x1 - 4x2 - 3x3 = 0

2x1 + 3x2 + x3 + x4 = 5 4x1 + x2 + 2x3 + x5 = 11 3x1 4x2 + 2x3 + x6 = 8

y1 = , y2 = , y3 = Z - 1/2 x1 + 2x2 + 3/2x6 = 12

1/2 x1 + x2 + x4 + 1/2x6 = 1 x1 3x2 + x5 + x6 = 3

3/2x1 + 2x2 + x3 + 1/2x6 = 4 y1 = , y2 = , y3 =

Z + 3x2 + x4 + x6 = 13

x1 + 2x2 + 2x4 x6 = 2 5x2 2x4+ x5 = 1

x2 + x3 3x4 + 2x6 = 1 y1 = , y2 = , y3 =

9 Recall that

College of Management, NCTU Operation Research I Fall, 2008

Jin Y. Wang 6-5

Coefficient of: Iteration Basic Variable

Z Original Variable Slack Variables Right Side

Any Z xB

1 0

cBB-1A c B-1A

cBB-1 B-1

cBB-1b B-1b

9 Let y (dual variables) = cBB-1 (the shadow price). Obviously, Z (=cx) = yb. 9 The coefficients of original and slack variables in the objective row are yAc

and y, respectively. 9 Note that yAc represents the value of surplus variables in the dual. 9 If the optimality hasnt yet reached, at least one coefficient in the objective

row is < 0. Thus, at lease one constraint of the dual is violated.

9 This also explains the strong duality property in some sense.

Complementary optimal solutions property 9 At the final iteration, the simplex method simultaneously identifies an optimal

solution x* for the primal problem and a complementary optimal solution y* for the dual problem (the coefficients of the slack variables in row 0), where cx* = y*b

9 In fact, y* = CBB-1 are the shadow prices. Symmetry property 9 For any primal problem and its dual problem, all relationships between them

must be symmetric because the dual of this dual problem is this primal problem.

Duality theorem All possible relationships between the primal and dual problems 9 If one problem has feasible solutions and a bounded objective function (and so

has an optimal solution), then so does the other problem. 9 If one problem has feasible solutions and an unbounded objective function

(and so no optimal solution), then the other problem has no feasible solutions.

College of Management, NCTU Operation Research I Fall, 2008

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9 If one problem has no feasible solutions, then the other problem has either no feasible solutions or an unbounded objective function.

Dual Optimal Infeasible Unbounded

Optimal Infeasible

Primal

Unbounded Example for primal infeasibledual infeasible Max 2x1 x2 Min y1 2y2 S.T. x1 x2 1 S.T. y1 y2 2

x1 + x2 2 y1 + y2 1 x1, x2 0 y1, y2 0

Applications of dual problem 9 Dual problem can be solved directly by the simplex method.

The number of functional constraints affects the computational effort of the simplex method far more than the number of variables does.

9 It is useful for evaluating a proposed solution for the primal problem.

Economic interpretation of duality 9 Primal Problem (P)

Max == nj jj xcZ 1 S.T. = nj ijij bxa1 , for i = 1, 2, , m xj 0, for j = 1, 2, , n

9 Dual Problem (D) Min == mj ii ybW 1 S.T. = mi jiij cya1 , for j = 1, 2, , n yi 0, for i = 1, 2, , m

College of Management, NCTU Operation Research I Fall, 2008

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9 If resource i is not used for production, we rent it out with unit rental of yi. 9 =mi iij ya1 means the rental income if we reduce the production amount of

product j by one.

9 = mi jiij cya1 means the rental income be greater than the sales income. 9 The objective of dual is to minimize the rental income in order to obtain the

reasonable rental price.

Primal-Dual relationships 9 Dual is a LP problem and also has corner-point solutions. 9 By using the augmented form, we can express these corner-point solutions as

basic solutions. 9 Because the functional constraints have the form, this augmented form is

obtained by subtracting the surplus from the left-hand side of each constraint.

Complementary basic solutions property 9 Each basic solution in the primal problem has a complementary basic solution

in the dual problem, where their respective objective function values are equal. 9 Recall again:

Coefficient of: Iteration Basic Variable

Z Original Variable Slack Variables Right Side

Any Z 1 cBB-1A c cBB-1 cBB-1b

9 Given row 0 (objective row) of the simplex tableau for the primal basic

solution, the complementary dual basic solution can be easily found. Primal Variable Associated Dual Variables

(Original variable) xj (Surplus Variable) ym+j (j = 1, 2, , n) (Slack Variable) xn+i (Original Variable) yi (i = 1, 2, ., m)

College of Management, NCTU Operation Research I Fall, 2008

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9 The m basic variables for the primal problem are required to have a coefficient of zero in the objective row, which thereby requires the m associated dual variables to be zero, i.e., nonbasic variables for the dual problem.

9 The values of the remaining n (basic) variables will be the simultaneous

solution to the system of equations. 9 The dual solution read from row 0 must also be a basic solution.

Complementary slackness property Primal Variable Associated Dual Variable

Basic Nonbasic (m variables) Nonbasic Basic (n variables)

9 If one of (primal) variables is a basic variable (> 0), then the corresponding one (dual) must be a nonbasic variable (= 0).

9 Wyndor Example: Max Z = 3x1+ 5x2 Min W = 4y1 + 12y2 + 18y3 S.T. x1 4 S.T. y1 + 3y3 3

2x2 12 2y2 + 2y3 5 3x1 + 2x2 18 x1, x2 0 y1, y2, y3 0

Max Z = 3x1+ 5x2 Min W = 4y1 + 12y2 + 18y3 S.T. x1 + x3 = 4 S.T. y1 + 3y3 y4 = 3

2x2 + x4 = 12 2y2 + 2y3 y5 = 5 3x1 + 2x2 + x5 = 18 Given x = (4, 6, 0, 0, -6), we know that x1, x2, and x5 are basic variables.

The coefficients of these variables in row 0 are zero. Thus, y4, y5, and y3 are nonbasic variables (=0). y1 = 3

2y2 = 5

College of Management, NCTU Operation Research I Fall, 2008

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Given x = (0, 6, 4, 0, 6), we have

9 Let x1*~xn* be primal feasible solution, y1*~ym* be dual feasible solution.

Then, they are both optimal if and only if

(1) xj* > 0 implies =

=m

ijiij cya

1

* .

(2) =