IEOR 4004 Midterm Review (part I)

1. IEOR 4004 Midterm Review(part I) March 10, 2014

2. Summary • Modeling optimization problems • Linear program • Simplex method • Algorithm • Initialization • Variants • Sensitivity analysis (in part II) • Duality (in part II) • Upper bounds and Shadow prices • Complementary slackness

3. Mathematical modeling • Simplified (idealized) formulation • Limitations • Only as good as our assumptions/input data • Cannot make predictions beyond the assumptions We need more maps

4. Mathematical modeling Problem simplification Model formulation Model Algorithm selection Numeric calculation Problem Solution Interpretation Sensitivity analysis

5. Mathematical modeling • Deterministic= values known with certainty • Stochastic= involves chance, uncertainty • Linear, non-linear, convex, semi-definite

6. Formulating an optimization problem Linear program Linear objective • Decision variables • Objective • Constraints • Domains x1, x2, x3, x4 x1 (1 − 2x2)2 Minimize + 2x2 + 3x3 (x1)2 + (x2)2 + (x3)2 + (x4)2 ≤ 2 x1 − 2x2 + x3 −3x4≤ 1 − x1 + 3x2 + 2x3 + x4≥ − 2 − 2x1x3 = 1 Linear constraints x3 in {0,1} x4 in [0,1] x1≥ 0 x2 ≠ 0.5 Sign restriction

7. Linear program • solution= values of variables that together satisfy all constraints • Feasiblesolution = satisfies also sign constraints • Infeasible solution = not feasible • Basis= set of m variables (where m = # of equations) • Variable in basis = basicvariable, all other non-basic • Basic solution= set all non-basic variables to zero {x1, x2} x1 = − 1 x2= −1 x3= 0 x4= 0 x1 = 4/3 x2= 0 x3= −1/3 x4= 0 x1 − 2x2 + x3 −3x4 = 1 − x1 + 3x2 + 2x3 + x4= − 2 x1≥ 0

8. Linear program • Feasible region = set of all feasible solutions • LP is Infeasibleif feasible region empty • LP is Unboundedif the objective function is unbounded over the feasible region objective function grows beyond bounds objective function growth is limited

9. Simplex algorithm • Converting to standard form Add slack variables Substitute negative variables Substitute unrestricted variables Dictionary x5 = 3 − x1 + 2x2 − x3’ + 3x4+− 3x4− x6= 2 − x1 + 3x2 + 2x3’ + x4+ − x4− − 3x4+ + 3x4− + x4+− x4− − x3’ − 2x3’ − 3x4 + x4 x1 − 2x2 − x1 + 3x2 + x3 + 2x3 + x5 = − x6 = ≤ ≥ 3 − 2 x1, x2 ≥ 0 x5,x6 ≥ 0 z = − 2x2 + 3x3’ x4+, x4− ≥ 0 x3’ ≥ 0 x3 ≤ 0

10. Simplex method (maximization) x5 = 3 − x1 + 2x2 − x3’ + 3x4+− 3x4− x6= 2 − x1 + 3x2 + 2x3’ + x4+ − x4− START z = − 2x2 + 3x3’ Initial feasible solution xj Pivot xj to the basis Optimality test Are all coefficients in znon-positive ? Ratio test Find min b/a a = coeff of xi inxj b = value of xj a, b opposite signs xi Variable selection NO YES END xj does not exist LP is unbounded optimal solution found Alternative solutions?

11. Simplex algorithm Ratio test: • Anatomy of a dictionary x5: 3/1 = 3 x6: no constraint (different sign) (same sign) x5 = 3 − x1 + 2x2 − x3’ + 3x4+− 3x4− x6= 2 − x1 + 3x2 + 2x3’ + x4+ − x4− basis objective(s) z = − 2x2 + 3x3’ negative coeff positive coeff zero coeff x5 leaves x3’ enters (min ratio)

12. Simplex/Graphical method x3, x4are slack variables • Illustration 150 profit: 180 0 80 max 3x1 +2x2 x1 + x2 ≤ 80 2x1 + x2 ≤ 100 x1, x2 ≥ 0 max 3x1 +2x2 x1 + x2 + x3 = 80 2x1 + x2 + x4 = 100 x1, x2, x3, x4 ≥ 0 z start: x1=0, x2=0, x3=80, x4=100 x1 increases until x4=0 60 x2 increases until x3=0 optimal 40 x3 x1 x4 20 isoprofit line x2 0 20 60 80 3x1 +2x2 = profit 40

13. Initialization - Phase I • Goal: make all original variables non-basic (=0) • If all-0 doesn’t satisfy a constraint, add artificialvariable ai: • add +ai if the right-hand side is > 0 • add –ai if the right-hand side < 0 • Add slackvariables (where needed) • New objective: minimize the sum of artificial variables • Startingbasis: all artificial + slack variables from equations without artificial variables If optimal value negative, then LP Infeasible Else dropartificial variables and introducez s1 = 1 − x1 + 2x2 − x3 + 3x4 a2= 2 − x1 + 3x2 + 2x3+ x4 a3= 1 − x2− x3− 2x4+ e3 z = 2x2− 3x3 Maximize w = − a2 − a3 x1 = 5/2 − (7/2)x3 − (5/4)s1 + (7/4)e3 x2 = 0 − 2x3 − (1/2)s1 + (1/2)e3 x4= 1/2 + (1/2)x3 + (1/4)s1 + (1/4)e3 x1 = 5/2 − (7/2)x3+ (1/4)a2− (7/4)a3 − (5/4)s1 + (7/4)e3 x2 = 0 − 2x3 + (1/2)a2− (1/2)a3 − (1/2)s1 + (1/2)e3 x4= 1/2 + (1/2)x3 − (1/4) a2 − (1/4)a3 + (1/4)s1 + (1/4)e3 x1 − 2x2 + x3 −3x4 − x1 + 3x2 + 2x3 + x4 x2 + x3 +2x4 ≤ 1 = − 2 ≥ 1 + s1 = 1 = − 2 = 1 − a2 − e3 + a3 w = − a2 − a3 w = −3 + x1 − 4x2− 3x3− 3x4 − e3 z = 2x2− 3x3 z = − 7x3 − s1+ e3 w = − a2 − a3

14. Degeneracy x1=20 x2=60 x3=x4=x5=0 • Basic solution is degenerateif some basic variable is zero • multiple dictionaries 80 60 max 3x1 +2x2 x1 + x2+ x3 = 80 2x1 + x2+ + x4 = 100 6x1+ x2 + +x5 = 180 x1, x2, x3, x4, x5≥ 0 40 20 20 40 60 80 x1 = 20 + x3 − x4 x2 = 60 − 2x3 + x4 x5 = − 4x3 + 5x4 z = 180 − x3 − x4 x1 = 20 + 0.2x3 − 0.2x5 x2 = 60 − 1.2x3 + 0.2x5 x4 = 0.8x3 + 0.2x5 z = 180 − 1.8x3 − 0.2x5 x1 = 20 + 0.25x4 − 0.25x5 x2 = 60 − 1.5x4 + 0.5x5 x3 = 1.25x4− 0.25x5 z = 180 − 2.25x4 + 0.25x5 optimal optimal ???

15. Upper bounded Simplex Ratio test: • Anatomy of a dictionary 1. (different sign) x5: 3/1 = 3 2. (entering variable) x3: 2 3. (same sign) x6: (4−1)/2 = 1.5 (different sign) x5 = 3 − x1 + 2x2 − x3 + 3x4 x6= 1 − x1 + 3x2 + 2x3 + x4 basis objective(s) z = − 2x2 + 3x3 (same sign) 0 ≤ x1, x2 , x4, x5 0 ≤ x3 ≤ 2 0 ≤ x6≤ 4 x3enters bounds x6 leaves 0 ≤ x6’ ≤ 4 substitute x6 = 4 − x6’ (min ratio)

16. Dual Simplex Ratio test: • Anatomy of a dictionary x1: no constraint x2: 5/2 = 2.5 x3: no constraint x4: 7/4 = 1.75 (positive) negative value infeasible! x5 = − 3 − x1 + 2x2 − x3 + 4x4 x6= 1 − x1 + 3x2+ 2x3 + x4 basis objective(s) z = − 5x2− 3x3 − 7x4 dually feasible (all coefficients non-positive) x5leaves x4 enters (min ratio)