Summary of the simplex method

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1 MVE165/MMG630, The simplex method; degeneracy; unbounded solutions; infeasibility; starting solutions; duality; interpretation Ann-Brith Strömberg

2 Summary of the simplex method Optimality condition: The entering variable in a maximization (minimization) problem should have the largest positive (negative) marginal value (reduced cost). The entering variable determines a direction in which the objective value increases (decreases). If all reduced costs are negative (positive), the current basis is optimal. Feasibility condition: The leaving variable is the one with smallest nonnegative quotient. Corresponds to the constraint that is reached first

3 Simplex search for linear (minimization) programs (Ch. 4.6) 1. Initialization: Choose any feasible basis, construct the corresponding basic solution x 0, let t = 0 2. Step direction: Select a variable to enter the basis using the optimality condition (negative marginal value). Stop if no entering variable exists 3. Step length: Select a leaving variable using the feasibility condition (smallest non-negative quotient) 4. New iterate: Compute the new basic solution x t+1 by performing matrix operations. 5. Let t := t + 1 and repeat from 2

4 Degeneracy (Ch. 4.10) If the smallest nonnegative quotient is zero, the value of a basic variable will become zero in the next iteration The solution is degenerate The objective value will not improve in this iteration Risk: cycling around (non-optimal) bases Reason: a redundant constraint touches the feasible set Example: x 1 + x 2 6 x 2 3 x 1 + 2x 2 9 x 1, x 2 0

5 Degeneracy Typical objective function progress of the simplex method objective value iteration number Computation rules to prevent from infinite cycling: careful choices of leaving and entering variables In modern software: perturb the right hand side (b i + b i ), solve, reduce the perturbation and resolve starting from the current basis. Repeat until b i = 0.

6 Unbounded solutions (Ch. 4.4, 4.6) If all quotients are negative, the value of the variable entering the basis may increase infinitely The feasible set is unbounded In a real application this would probably be due to some incorrect assumption Example: minimize z = x 1 2x 2 subject to x 1 +x 2 2 2x 1 +x 2 1 x 1, x 2 0 Draw graph!!

7 Unbounded solutions (Ch. 4.4, 4.6) A feasible basis is given by x 1 = 1, x 2 = 3, with corresponding tableau: Homework: Find this basis using the simplex method. basis z x 1 x 2 s 1 s 2 RHS z x x Entering variable is s 2 Row 1: x 1 = 1 + s 2 0 s 2 1 Row 2: x 2 = 3 + s 2 0 s 2 3 No leaving variable can be found, since no constraint will prevent s 2 from increasing infinitely

8 Starting solution finding an initial basis (Ch. 4.9) Example: minimize z = 2x 1 +3x 2 subject to 3x 1 +2x 2 = 14 2x 1 4x 2 2 Draw graph!! 4x 1 +3x 2 19 x 1,x 2 0 Add slack and surplus variables minimize z = 2x 1 +3x 2 subject to 3x 1 +2x 2 = 14 2x 1 4x 2 s 1 = 2 4x 1 +3x 2 +s 2 = 19 x 1,x 2,s 1,s 2 0 How finding an initial basis? Only s 2 is obvious!

9 Artificial variables Add artificial variables a 1 and a 2 to the first and second constraints, respectively Solve an artificial problem: minimize a 1 + a 2 minimize w = a 1 +a 2 subject to 3x 1 +2x 2 +a 1 = 14 2x 1 4x 2 s 1 +a 2 = 2 4x 1 +3x 2 +s 2 = 19 x 1,x 2,s 1,s 2,a 1,a 2 0 The phase one problem An initial basis is given by a 1 = 14, a 2 = 2, and s 2 = 19: basis w x 1 x 2 s 1 s 2 a 1 a 2 RHS w a a s

10 Find an initial solution using artificial variables x 1 enters a 2 leaves (then x 2 s 2, then s 1 a 1 ) basis w x 1 x 2 s 1 s 2 a 1 a 2 RHS w a a s w a x s w a x x w s x x A feasible basis is given by x 1 = 4, x 2 = 1, and s 1 = 2

11 Infeasible linear programs (Ch. 4.9) If the solution to the phase one problem has optimal value = 0, a feasible basis has been found Start optimizing the original objective function z from this basis (homework) If the solution to the phase one problem has optimal value w > 0, no feasible solutions exist What would this mean in a real application? Alternative: Big-M method: Add the artificial variables to the original objective with a large coefficient Example: minimize z = 2x 1 + 3x 2 minimize z a = 2x 1 + 3x 2 + Ma 1 + Ma 2

12 Alternative optimal solutions (Ch. 4.6) Example: maximize z = 2x 1 +4x 2 subject to x 1 +2x 2 5 x 1 +x 2 4 Draw graph!! x 1,x 2 0 The extreme points (0, 5 2 ) and (3,1) have the same optimal value z = 10 All solutions that are positive linear (convex) combinations of these are optimal: (x 1,x 2 ) = α (0, 5 ) + (1 α) (3,1), 0 α 1 2

13 A general linear program in standard form A linear program with n non-negative variables, m equality constraints (m < n), and non-negative right hand sides: maximize z = subject to n c j x j j=1 n a ij x j j=1 On matrix form it is written as: = b i, i = 1,...,m, x j 0, j = 1,...,n. maximize z = c T x, subject to Ax = b, x 0 n, where x R n, A R m n, b R m + (b 0 m ), and c R n.

14 General derivation of the simplex method (Ch. 4.8) B = set of basic variables, N = set of non-basic variables B = m and N = n m Partition matrix/vectors: A=(B,N), x=(x B,x N ), c=(c B,c N ) The matrix B (N) contains the columns of A corresponding to the index set B (N) Analogously for x and c Rewrite the linear program: maximize z = c T x maximize z = c T B x B + c T N x N subject to Ax = b, = subject to Bx B + Nx N = b, x 0 n x B 0 m, x N 0 n m Substitute: x B = B 1 b B 1 Nx N = maximize z = c T B B 1 b + [c T N ct B B 1 N]x N subject to B 1 b B 1 Nx N 0 m, x N 0 n m

15 Optimality and feasibility Optimality condition (for maximization) The basis B is optimal if c T N ct B B 1 N 0 n m (marginal values = reduced costs 0) If not, choose as entering variable j N the one with the largest value of the reduced cost c j c T B B 1 A j Feasibility condition For all i B it holds that x i = (B 1 b) i (B 1 A j ) i x j Choose the leaving variable i B according to { (B i 1 } b) i = arg min i B (B 1 A j ) i (B 1 A j ) i > 0

16 In the simplex tableau, we have basis z x B x N s RHS z 1 0 c T N ct B B 1 N c T B B 1 c T B B 1 b x B 0 I B 1 N B 1 B 1 b s denotes possible slack variables (columns for s are copies of certain columns for (x B,x N )) The computations performed by the simplex algorithm involve matrix inversions and updates of these A non-basic (basic) variable enters (leaves) the basis one column, A j, of B is replaced by another, A k Row operations Updates of B 1 (and B 1 N, B 1 b, and c T B B 1 ) Efficient numerical computations are crucial for the performance of the simplex algorithm

17 Intuitive derivation of duality (Ch. 6.1) A linear program with optimal value z maximize z := 20x 1 +18x 2 weights subject to 7x 1 +10x (1) v 1 16x 1 +12x (2) v 2 x 1,x 2 0 How large can z be? Compute upper estimates of z, e.g. Multiply (1) by 3 21x1 + 30x z Multiply (2) by x1 + 18x z 8100 Combine: 0.6 (1)+1 (2) 20.2x1 +18x z 7560 Do better than guess compute optimal weights! Value of estimate: w = 3600v v 2 min 7v v 2 20 Constraints on weights: 10v v 2 18 v 1,v 2 0

18 The best (lowest) possible upper estimate of z minimize w := 3600v v 2 subject to 7v v v v 2 18 v 1,v 2 0 A linear program! It is called the dual of the original linear program

19 The lego model the market problem Consider the lego problem maximize z = 1600x x 2 subject to 2x 1 + x 2 6 2x 1 + 2x 2 8 x 1, x 2 0 Option: Sell bricks instead of making furniture v 1 (v 2 ) = price of a large (small) brick Market wish to minimize payment: minimize 6v 1 + 8v 2 I sell if prices are high enough: 2v1 + 2v otherwise better to make tables v1 + 2v otherwise better to make chairs v1, v 2 0 prices are naturally non-negative

20 Linear programming duality To each primal linear program corresponds a dual linear program [Primal] minimize z = c T x, subject to Ax = b, x 0 n, [Dual] maximize w = b T y, subject to A T y c. On component form: [Primal] minimize z = n j=1 c jx j subject to n j=1 a ijx j = b i, i = 1,...,m, x j 0, j = 1,...,n, [Dual] maximize w = n j=1 b iy i subject to m i=1 a ijy i c j, j = 1,...,n.

21 An example A primal linear program minimize z = 2x 1 +3x 2 subject to 3x 1 +2x 2 = 14 2x 1 4x 2 2 4x 1 +3x 2 19 x 1,x 2 0 The corresponding dual linear program maximize w = 14y 1 +2y 2 +19y 3 subject to 3y 1 +2y 2 +4y 3 2 2y 1 4y 2 +3y 3 3 y 1 free, y 2 0, y 3 0

22 Rules for constructing the dual program (Ch. 6.2) maximization minimization dual program primal program primal program dual program constraints variables 0 0 = free variables constraints 0 0 free = The dual of the dual of any linear program equals the primal

23 Duality properties (Ch. 6.3) Weak duality: Let x be a feasible point in the primal (minimization) and y be a feasible point in the dual (maximization). Then, z = c T x b T y = w Strong duality: In a pair of primal and dual linear programs, if one of them has an optimal solution, so does the other, and their optimal values are equal. Complementary slackness: If x is optimal in the primal and y is optimal in the dual, then x T (c A T y) = y T (b Ax) = 0. If x is feasible in the primal, y is feasible in the dual, and x T (c A T y) = y T (b Ax) = 0, then x and y are optimal for their respective problems.

24 Relations between primal and dual optimal solutions primal (dual) problem dual (primal) problem unique and unique and non-degenerate solution non-degenerate solution unbounded solution = no feasible solutions no feasible solutions = unbounded solution or no feasible solutions degenerate solution alternative solutions

25 Exercises on duality Homework! Formulate and solve graphically the dual of: minimize z = 6x 1 +3x 2 +x 3 subject to 6x 1 3x 2 +x 3 2 3x 1 +4x 2 +x 3 5 x 1,x 2,x 3 0 Then find the optimal primal solution Verify that the dual of the dual equals the primal

MVE165/MMG631 Linear and integer optimization with applications Lecture 5 Linear programming duality and sensitivity analysis

MVE165/MMG631 Linear and integer optimization with applications Lecture 5 Linear programming duality and sensitivity analysis Ann-Brith Strömberg 2017 03 29 Lecture 4 Linear and integer optimization with