Summary of the simplex method

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1 MVE165/MMG631,Linear and integer optimization with applications The simplex method: degeneracy; unbounded solutions; starting solutions; infeasibility; alternative optimal solutions Ann-Brith Strömberg

2 Summary of the simplex method Optimality condition: For a minimization [maximization] problem, the variable entering the basis should have the largest negative [positive] reduced cost [marginal value] The entering variable determines a direction (along an edge of the polyhedron) in which the objective value decreases [increases] If all reduced costs are positive [negative] then the current basis is optimal Feasibility condition: The variable leaving the basis is the one with the smallest quota between the corresponding right hand side and the (non-negative) constraint coefficient of the entering variable (search direction) The variable leaving the basis corresponds to the constraint that is reached first A negative (zero) coefficient of the entering variable: the corresponding (in)equality is not constraining

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 reduced cost). If no entering variable exists: Stop, the solution is optimal 3. Step length: Select a leaving variable using the feasibility condition (smallest non-negative quota). If all coefficients 0: Stop, the problem has an unbounded solution 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 Special case: Degeneracy (Ch. 4.10) If the smallest non-negative 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 Special case: Degeneracy example minimize z = x 1 2x 2 x 1 + x 2 6 x 2 3 x 1 + 2x 2 9 x 1, x 2 0 basis z x 1 x 2 x 3 x 4 x 5 RHS z x x B = (x 3, x 4,x 5) = (6,3, 9) x x z x x B = (x 3, x 2,x 5) = (3,3, 3) x x z x x B = (x 3, x 2,x 1) = (0,3, 3) x x z x x B = (x 4, x 2,x 1) = (0,3, 3) x x z x x B = (x 5, x 2,x 1) = (0,3, 3) x x This last iteration will not appear in the simplex course. Why?

6 Special case: 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 + b), solve, decrease b and resolve starting from the current basis. Repeat until b = 0

7 Unbounded solutions (Ch. 4.4, 4.6) If all quotas 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 modelling 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!!

8 Unbounded solutions (Ch. 4.4, 4.6) A feasible basis is given by x 1 = 1, x 2 = 3, with corresponding tableau: 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 Homework: Find the above basis using the simplex method

9 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 to find an initial basis? Only s 2 is obvious!

10 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

11 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

12 Find an initial solution illustration

13 Infeasible linear programs (Ch. 4.9) If the solution to the phase one problem has an optimal value w = 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 an 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

14 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

15 Next lecture on Thursday, April 11 Sensitivity analysis (Ch. 5) Linear optimization duality theory (Ch. 6)

Ann-Brith Strömberg. Lecture 4 Linear and Integer Optimization with Applications 1/10

Ann-Brith Strömberg. Lecture 4 Linear and Integer Optimization with Applications 1/10 MVE165/MMG631 Linear and Integer Optimization with Applications Lecture 4 Linear programming: degeneracy; unbounded solution; infeasibility; starting solutions Ann-Brith Strömberg 2017 03 28 Lecture 4