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**IEOR 4004 Midterm Review (part I)**

March 10, 2014

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**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

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**Mathematical modeling**

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

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**Mathematical modeling**

Problem simplification Model formulation Model Algorithm selection Numeric calculation Problem Solution Interpretation Sensitivity analysis

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**Mathematical modeling**

Deterministic = values known with certainty Stochastic = involves chance, uncertainty Linear, non-linear, convex, semi-definite

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**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 x1 ≥ 0 x3 in {0,1} x4 in [0,1] x2 ≠ 0.5 Sign restriction

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

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**Linear program Feasible region = set of all feasible solutions**

LP is Infeasible if feasible region empty LP is Unbounded if the objective function is unbounded over the feasible region objective function grows beyond bounds objective function growth is limited

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**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− x1 − 2x2 − x1 + 3x2 − x3’ − 2x3’ + x3 + 2x3 + x5 = − x6 = − 3x4 + x4 ≤ ≥ 3 − 2 z = − 2x2 + 3x3’ x1, x2 ≥ 0 x5, x6 ≥ 0 x3 ≤ 0 x3’ ≥ 0 x4+, x4− ≥ 0

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**Simplex method (maximization)**

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

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**Simplex algorithm Anatomy of a dictionary Ratio test: 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’ zero coeff negative coeff positive coeff x3’ enters x5 leaves (min ratio)

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**Simplex/Graphical method**

x3, x4 are slack variables Illustration profit: 180 150 80 max 3x1 +2x2 x1 + x2 + x3 = 80 2x1 + x2 + x4 = 100 x1, x2, x3, x4 ≥ 0 max 3x1 +2x2 x1 + x2 ≤ 80 2x1 + x2 ≤ 100 x1, x2 ≥ 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 isoprofit line 20 x2 3x1 +2x2 = profit 20 40 60 80

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**Initialization - Phase I**

Goal: make all original variables non-basic (=0) If all-0 doesn’t satisfy a constraint, add artificial variable ai: add +ai if the right-hand side is > 0 add –ai if the right-hand side < 0 Add slack variables (where needed) New objective: minimize the sum of artificial variables Starting basis: all artificial + slack variables from equations without artificial variables If optimal value negative, then LP Infeasible Else drop artificial variables and introduce z Maximize w = − a2 − a3 z = 2x2 − 3x3 s1 = 1 − x1 + 2x2 − x3 + 3x4 a2 = 2 − x1 + 3x2 + 2x3 + x4 a3 = 1 − x2 − x3 − 2x4 + e3 x1 = 5/2 − (7/2)x3 + (1/4)a2 − (7/4)a3 − (5/4)s1 + (7/4)e3 x2 = − x3 + (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 = 5/2 − (7/2)x3 − (5/4)s1 + (7/4)e3 x2 = − x3 − (1/2)s1 + (1/2)e3 x4 = 1/2 + (1/2)x3 + (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 w = − a2 − a3 w = −3 + x1 − 4x2 − 3x3 − 3x4 − e3 + a3 w = − a2 − a3 z = − x3 − s e3 z = 2x2 − 3x3

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**Degeneracy Basic solution is degenerate if some basic variable is zero**

x1=20 x2=60 x3=x4=x5=0 Basic solution is degenerate if some basic variable is zero multiple dictionaries 80 60 max 3x1 +2x2 x1 + x2 + x3 = 80 2x1 + x x4 = 100 6x1 + x 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 = x3 − 0.2x5 x2 = 60 − 1.2x x5 x4 = 0.8x x5 z = 180 − 1.8x3 − 0.2x5 x1 = x4 − 0.25x5 x2 = 60 − 1.5x x5 x3 = 1.25x4 − 0.25x5 z = 180 − 2.25x x5 optimal optimal ???

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**Upper bounded Simplex Anatomy of a dictionary Ratio test:**

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 + 2x x4 basis objective(s) z = − 2x2 + 3x3 (same sign) 0 ≤ x1, x2 , x4, x5 0 ≤ x3 ≤ 2 0 ≤ x6 ≤ 4 x3 enters bounds 0 ≤ x6’ ≤ 4 substitute x6 = 4 − x6’ x6 leaves (min ratio)

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**(all coefficients non-positive)**

Dual Simplex Anatomy of a dictionary Ratio test: 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 + 2x x4 basis objective(s) z = − 5x2 − 3x3 − 7x4 dually feasible (all coefficients non-positive) x5 leaves x4 enters (min ratio)

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