Math 5593 Linear Programming Weeks 12/13

Math 5593 Linear Programming Weeks 12/13 University of Colorado Denver, Fall 2013, Prof. Engau 1 Introduction 2 Polyhedral Theory 3 LP and Lagrangean Relaxation 4 Computational Methods and Algorithms

Integer Programming and Combinatorial Optimization In practice, many decision variables must be limited to integers (people, products, etc.) or binary values (on / off, yes / no, etc.): var var name integer / binary; (Binary) Integer Program (IP/BIP): max c T x : Ax b, x integer / binary Mixed Integer (Linear) Program (MIP): max c T x + d T y : Ax + By b, y integer / binary Combinatorial Optimization Problems (COP): Given a finite set N = 1, 2,..., n, weights c j for all j N, and a family F P(N) of subsets of N: max or min j S c j : S F. Solving IPs, BIPs, or MIPs as LP and rounding fractional values to integers does not often work well and may destroy feasibility!

Review: Example Problems and Modeling Techniques Many classic OR problems and models are inherently discrete: Easy : Transportation, assignment, other network flows Difficult : TSP, scheduling, facility location, knapsack In addition, integer or binary variables have model applications: Cardinality constraints: at least k (set covering constraint), at most k (set packing/budget), exactly k (set partitioning) Zero or minimum/maximum/range constraints: ly x uy Either-or disjunctions: A 1 x b 1 + My, A 2 x b 2 + M(1 y) If-then conditionals: A 1 x > b 1 My, A 2 x b 2 + M(1 y) Fixed costs (in objective): min cvar T x + cfix T y where x My Piecewise linear function/approximation: L i y i x i L i y i 1 Linearization of binary quadratic programs: x 0, 1 n x 2 i = x i, x i x j = x ij, maxx i + x j 1, 0 x ij minx i, x j

IP Basics: Polyhedral Theory and Formulations Goal: Given a discrete (integer) set X Z n, find a continuous (polyhedral) formulation P R n for X such that X = P Z n : X = x Z n : Ax b = x R n : Ax b Z n = P Z n Ex: X = x Z 2 : 7x 1 + 10x 2 56, 0 x 1 6, 0 x 2 4 P = x R 2 : 7x 1 + 10x 2 56, 0 x 1 6, 0 x 2 4 x 2 4 2 0 0 2 4 6 x 1 P is a trivial formulation for X: P is the gray shaded area. X are the grid points in P. Is P a good formulation for X? Perfect for box constraints. Poor for 7x 1 + 10x 2 56.

Improving P: Better and Ideal Formulations Def: A formulation P 1 for X Z n is better than P 2 if P 1 P 2. Say P 1 strengthens or tightens or cuts off parts of P 2. Example: X = 7x 1 + 10x 2 56, 0 x 1 6, 0 x 2 4 Z 2 x 2 4 2 Idea: Improve a formulation by adding supporting constraints! Add x 1 + x 2 7 Add 2x 1 + 3x 2 16 But P must stay convex! 0 0 2 4 6 x Best possible: convex hull! 1 The ideal formulation for X Z n is its convex hull P = conv(x)! All extreme points of P are integer ( integer polyhedron ). Can solve IP maxc T x : x X as LP maxc T x : x P. Unfortunately, finding conv(x) is often (at least) as hard as IP!

When is Solving IP as Easy as LP: Integer Polyhedra! Let A, b integer and consider IP max c T x : Ax = b, x Z n +. Question: When does LP (x R n +) have integer solution? Answer: If P = x R n + : Ax = b is an integer polyhedron! Equivalent Answer: If the extreme points (bfs) of P are integer: (x B, x N ) = (B 1 b, 0) Z n + for all (the optimal) basis matrix B of A. Cramer s Rule: If Ax = b is nonsingular, then x i = det(a i ) det(a) where A i is the matrix formed by replacing the ith column of A with b. We can apply Cramer s Rule to the basic system Bx B = b Clear that det(bi ) is an integer, but need det(b) 1, 1 Inverse Rule: B 1 = Adj(B) det(b) where Adj(B) is adjugate (adjunct) matrix (transpose of cofactor matrix C where C ij = ( 1) i+j B ij ) Again clear that Adj(B) is integer, so need det(b) ±1. Result: If B is optimal and det B ±1, 0, then LP solves IP.

Integer Polyhedra and Totally Unimodular Matrices Definition: A matrix A is called totally unimodular (TU) if every square submatrix B of A has its determinant det(b) ±1, 0. [ ] 1 1 A = has det(a) = 1, but det(b) = 2 = 2 (not TU). 1 2 [ ] [ ] 1 1 1 1 A = has det(a) = 0 and is TU, but is not. 1 1 1 1 1 0 1 1 [ ] How about A = 1 1 0 1 1 0? Not TU: find! 1 1 1 1 0 1 1 0 1 1 How about A = 0 1 1 0? TU! (good exercise) 1 1 0 1 Result: P = x R n + : Ax = b is an integer polyhedron for all integer values of b for which it is bounded if and only if A is TU.

TU Matrices: Conditions and Examples Verifying whether a matrix is TU is a challenging undertaking! There is an exponential number of submatrices to check. However, it is easy to show that a matrix is not TU (if we already found or know a suitable certificate submatrix) For insiders: yes, that sounds a lot like NP-complete However, some conditions and special cases that might help: If A is TU, then all entries of A must satisfy a ij ±1, 0. A is TU if and only if A T is TU if and only if [ A I ] is TU. A is TU if a ij 0, 1, 1, a ij ±1 for at most two i in each column j, and there is a partition of rows I = I 1 I 2 such that for all columns with exactly two nonzero entries: a ij a ij = 0. i I 1 i I 2 Examples: [ ] 1 0 1 1 0 1 1 0 (not TU), 1 1 0 1 [ ] 1 0 1 1 0 1 1 0 (I 1 = I, I 2 = ) 1 1 0 1

TU Matrices: Conditions and Examples (Continued) A special class of TU matrices are node-arc incidence matrices for graphs G = (V, E) with node or vertex set V = 1, 2,..., n and arc or edge set E = e 1,..., e m : For every e j = (i 1, i 2 ) let 1 if i = i 1 (arc e j originates at node i 1 ) a ij = 1 if i = i 2 (arc e j terminates at node i 2 ) 0 otherwise Example: A = e 1 e 2 e 3 1 0 1 0 1 1 1 1 0 1 2 3 1 e 1 2 3 Network matrices satisfy partition condition (I 1 = I, I 2 = ). Discrete network flow problems without mean constraints (assignment, transportation, shortest-path, maximum-flow) are easy. Covering, packing, or TSP constraints are mean! e 3 e 2

Improving P (Continued): Valid Inequalities Let P = x R n : Ax b be a formulation for X = P Z n. If A is TU, then P = conv(x) is integer: ideal formulation. The converse is not true, in general! (A is hardly ever TU). Example: X = 7x 1 + 10x 2 56, 0 x 1 6, 0 x 2 4 Z 2 conv(x) = x 1 +x 2 7, 2x 1 +3x 2 16, 0 x 1 6, 0 x 2 4 Why drop 7x 1 + 10x 2 56? Valid but dominated redundant! An inequality (a, b) (or a T x b) is a valid inequality (vi) if a T x b for all x X. A vi (a 1, b 1 ) dominates (a 2, b 2 ) if there is λ > 0 such that a 1 λa 2, b 1 λb 2, and (a 1, b 1 ) λ(a 2, b 2 ). Note that two vis (a 1, b 1 ) = λ(a 2, b 2 ) are the same if λ > 0. A vi is redundant if it is dominated by a linear combination of other vis (Note: 1 (1, 1, 7) + 3 (2, 3, 16) = (7, 10, 55))

What are Good Valid Inequalities: Faces and Facets Result: A vi (a 1, b 1 ) dominates another vi (a 2, b 2 ) if and only if x R n : a1 T x b 1 x R n : a2 T x b 2. Result: Given a formulation P = x R n + : Ax b, a vi (c, d) is redundant for P iff there is y such that A T y c and b T y d. Proof: LP Duality Theorem or Farkas Lemma (as Exercise) Clear that good vis are non-dominated and non-redundant. Def: Given a vi (c, d) for P = x R n : Ax b, a face is a set F = x P : c T x = d (a face is proper if F ). A face of dimension dim(p) 1 (usually n 1) is called a facet. Intersections of P with a supporting hyperplane are faces. Face-defining vis are non-dominated, could be redundant. Facet-defining vis are non-dominated and non-redundant. Optimal representation only needs facet-defining inequalities!

3D Geometric Illustration: Faces, Edges, and Vertices For n = 3, we distinguish three types of k-faces (k = 0, 1, 2): 2-faces: faces of max dimension n 1 (polyhedral facets) 1-faces: edges (can be facets for 2-dimensional polygons) 0-faces: vertices (can still be facets for 1-dimensional lines) This is part of Polyhedral Combinatorics!

Finding Good Valid Inequalities: Preprocessing Consider the following IP and try to strengthen its formulation: max 2x 1 + 3x 2 + x 3 s.t. 8x 1 + 4x 2 3x 3 21, 3x 1 2x 2 + 4x 3 16 0 x 1 5, 1 x 2 4, 0 x 3 3, x Z 3. Idea: Use the box constraints to tighten lower or upper bounds! Solve first inequality for x 1 and use bounds from x 2 and x 3 : 8x 1 21 4x 2 + 3x 3 21 4(1) + 3(3) = 26 So x 1 26 8 = 3.25 and thus x 1 3 (because x 1 is integer). Repeat for x 2 and x 3 (paying attenting to inequality signs): 4x 2 21 8x 1 + 3x 3 21 8(0) + 3(3) = 30 3x 3 8x 1 + 4x 2 21 8(0) + 4(1) 21 = 17 So x 2 30 4 = 7 and x3 17 3 = 5 does not improve!

Finding Good Valid Inequalities: Preprocessing (Cont.) After preprocessing the first inequality, our new formulation is P 1 = 8x 1 + 4x 2 3x 3 21, 3x 1 2x 2 + 4x 3 16, 0 x 1 3, 1 x 2 4, 0 x 3 3. Give up? No way! Let s try the same for the second constraint: 3x 1 16 + 2x 2 4x 3 16 + 2(1) 4(3) = 6 2x 2 3x 1 + 4x 3 16 3(3) + 4(3) 16 = 5 4x 3 16 3x 1 + 2x 2 16 3(3) + 2(1) = 9 So x 1 6 3 = 2, x 2 5 2 = 2, x3 9 4 = 3 and we improved to P 2 = 8x 1 + 4x 2 3x 3 21, 3x 1 2x 2 + 4x 3 16, 2 x 1 3, 1 x 2 2, 3 x 3 3. Only four solutions to check: (2, 1, 3), (2, 2, 3), (3, 1, 3), (3, 2, 3).

Finding More Valid Inequalities: Preprocessing (Cont.) Preprocessing also helps for formulations of binary programs: X = x R 4 + : 0 (x 1, x 2, x 3, x 4 ) 1, 2x 1 3x 2 + x 3 x 4 0, 2x 1 + 5x 2 + 4x 3 + x 4 6, 3x 2 + 2x 3 + 4x 4 5 0, 1 4 Idea: Fix one variable value and look for hidden relationships! Suppose that x 1 = 1 and solve the first inequality for x 2 : 3x 2 2x 1 + x 3 x 4 2(1) + 0 1 1 So x 1 = 1 implies x 2 1 3 = 1 (if-then!) and thus x1 x 2. Suppose that x 2 = 1 and solve second inequality for x 3 : 4x 3 6 + 2x 1 5x 2 x 4 6 + 2(1) 5(1) 0 = 3 So x 2 = 1 implies x 3 3 4 = 0 and thus x2 + x 3 1. Similarly, the third inequality implies x 2 + x 3 + x 4 2. Together with x 2 + x 3 1, then x 4 = 1 and x 2 + x 3 = 1. Only three solutions to check: (1, 1, 0, 1), (0, 1, 0, 1), (0, 0, 1, 1).

Finding All (!) Valid Inequalities: Integer Rounding To keep things simple, consider a single knapsack constraint: 2x 1 + 5x 2 + 4x 3 + x 4 7 where x Z 4 + or 0, 1 4 Idea: We can multiply inequality by arbitrary positive numbers! Example: Multiply by 1 2 4, then 4 x 1 + 5 4 x 2 + x 3 + 1 4 x 4 7 4. Constraint remains valid if we round coefficients downward: 2 4 x1 + 5 4 x2 + x 3 + 1 4 x4 = x 1 + x 2 + x 3 7 4 Now LHS is integer so also round down RHS to 7 4 = 1: x 1 + x 2 + x 3 1 where x Z 4 + or 0, 1 4 Is the new constraint better than 2x 1 + 5x 2 + 4x 3 + x 4 7? No: (1, 1, 1, 1) is infeasible but satisfies the new constraint. But: ( 7 2, 0, 0, 0), (0, 7 5, 0, 0), (0, 0, 7 4, 0) were feasible and are now cut off the new constraint tightens formulation!

Gomory Cuts and the Chvátal-Gomory Procedure Consider a formulation P = x R n + : Ax b for X = P Z n +. 1 Let (a i, b i ) be one or a lin. comb. of the inequalities in P. 2 Let λ 0, then λai T x = n λa ijx j λb i same as (a i, b i ). j=1 3 Since x 0, then n λaij xj λb i is vi for P and X. j=1 4 Since LHS is integer, n j=1 λaij xj λb i is valid for X. 5 New vi may be invalid for P and improve formulation for X! Result: This simple procedure is sufficient to generate all valid inequalities, in principle. Generated vis are called Gomory cuts. Given an arbitrary first formulation P for a set X = P Z n +, every valid inequality for X can be generated by applying the Chvátal-Gomory procedure a finite (!) number of times. Proofs: Gomory 1958 (proof of concept), Chvátal 1973 (for bounded sets), Schrijver 1980 (for unbounded sets)

Example: Gomory Cuts for Rolling Mill Formulation Box, polyhedron, and convex hull: B = x R 2 + : x 1 6, x 2 4 P = B x : 7x 1 + 10x 2 56 C = B x : 2x 1 + 3x 2 16, x 1 + x 2 7 0 0 2 4 6 x 1 Preprocessing does not help: x 1 56 7 = 8, x2 56 10 = 5 Try Gomoroy: 7 3 x1 + 10 3 x2 = 2x 1 + 3x 2 56 3 = 18 ( 3 ) 10 7 x1 + ( ) 3 10 10 x2 = 2x 1 + 3x 2 ( 3 10) 56 = 16 Exercise: Generate x 1 + x 2 7. Need linear combination! ( 710 ) ( 10 ) ( 810 ) + = x 1 + 10 8 x2 = x 1 + x 2 62 8 = 7 56 6 62 x 2 4 2

Binary Programs: Covers and Extended Formulation Consider P = x R n + : Ax b for binary X = P 0, 1 n. Let a i be row of A and consider inequality n j=1 a ijx j b i. Find a cover C 1, 2,..., n such that j C a ij > b i. Then j C x j C 1 is called a cover inequality for X. Ex: 3x 1 + 2x 2 + 4x 3 5 has covers 1, 2, 3, 1, 3, 2, 3 x 1 + x 2 + x 3 2, x 1 + x 3 1, x 2 + x 3 1. Extended Formulation: rather than adding new constraints, add new variables and lift problem into a higher-dimensional space! Relaxation-Linearization Technique (Adams-Sherali 1990): Let x k 0, 1 and multiply inequalities by x k and (1 x k ) n j=1 a ijx j x k b i x k n i=1 a ijx j (1 x k ) b i (1 x k ). Add x jk = x j x k and max0, x j + x k 1 x jk minx j, x k. Result: Repeated RLT lift-and-project finds the convex hull!

IP Optimality Conditions and Relaxation Gaps Question: Given IP max c T x : Ax b, x Z n + and a feasible (integer) solution x, how do we know whether x is optimal? If z LP = max c T x : Ax b, x R n + and z LP = z IP = ct x (or c T x = z LP if c is integer), then x is optimal solution. Otherwise, we don t: no easy-to-check optimality condition! In practice, we use some sort of optimality (relaxation) gap. Definition: If (P) maxf (x): x X, then (R) maxg(x): x Y is a relaxation for (P) if X Y and f (x) g(x) for all x X. If (R) is infeasible/bounded, then (P) is infeasible/bounded. If (R) is optimal at x X and f (x ) = g(x ), then x is also optimal for (P) and there is no relaxation gap: z R z P = 0. In general, the relaxation gap of (R) for (P) is z R z P 0. Note: An ideal relaxation for IP is LP max c T x : x conv(x).

Linear Programming Relaxation Consider zip = max x Z n + c T x : Ax b and its LP relaxation: zlp c = max T x : Ax b, x R n +. Relaxation clear: larger (relaxed) feasible set, same objective. When is it tight (no gap)? If A is TU, b integer, or we lucky. Otherwise, try to improve relaxation using valid inequalities. Example: Consider a knapsack problem max c T x : a T x b : z LP(λ) = max c T x : n λa i T x λb. i=1 Drawback of LP Relaxations: finding zlp requires optimization! Although just LPs solving many LPs is not very practical. A way to more quickly get good upper bounds is desirable. Duality! Dual feasible solutions give primal upper bounds.

Primal-Dual Bounds and Duality Gaps Def: Problems (P) maxf (x): x X and (D) ming(y): y Y are a (weak) primal-dual pair if f (x) g(y) for all x X, y Y. If one problem is unbounded, then the other is infeasible. If one problem is feasible, then the other is bounded: f (x) zp z D g(y) for all x X and y Y. Duality gap is zd z P 0 (if no gap, then pair is strong). Note: get bounds from feasible solutions without optimization! Lots of examples: LP duality (strong), max flows/min cuts in integer networks (strong), max cardinality matchings/min node cover in bipartitite graphs (strong, König s Theorem) For IP max c T x : Ax b, x Z n can use weak (LP) dual min b T y : A T y c, y R m +. Unfortunately, bounds from LP duals are often quite weak.

Lagrangean Relaxation and Dual Problem Consider max f (x): g(x) 0, h(x) = 0 and relax constraints: L(λ, µ) = max f (x) λ T g(x) µ T h(x) Exercise: Show that this is a relaxation for λ > 0 (and any µ). L(λ, µ) is Lagrangean relaxation with L-multipliers (λ, µ). The Lagrangean Dual Problem is the min-max problem min L(λ, µ) = min max λ>0,µ λ>0,µ x f (x) λ T g(x) µ T h(x) For LP/IP: max c T x : Ax b, x X (X is R n + or Z n +): zl = min L(λ) = min max c T x + λ T (b Ax). λ>0 λ>0 x X. Easy to get bounds: compute objective with λ > 0, x X. Not easy to optimize: uses subgradiant and NLP methods.

Generic (not Genetic!) Integer Programming Algorithm Let X = x Z n : Ax b and consider IP max c T x : x X. 0 Start with an initial formulation P = x R n : Ax b. 1 Apply preprocessing to strengthen the formulation P. 2 Solve relaxation for x R n and an upper bound z. 3 If x Z n and c T x = z, then stop: x is optimal. 4 If x Z n but c T x < z, then x may be optimal: check dual bounds and their duality gaps (if small enough, stop). 5 If x / Z n, then x is not optimal. Try some of the following: Apply rounding heuristic to find x X and apply Step 4. Find a cutting plane / vi for X that cuts off infeasible x. Select xi / Z and split the problem into two subproblems: P 1 = P x R n : x i x i P 2 = P x R n : x i x i 6 Update the formulation(s) P (P 1, P 2 ) and go back to Step 1. Or use Complete Enumeration: try all x X and pick the best!

Gomory s Fractional Cutting-Plane Method Separation Problem: Given formulation P = x R n + : Ax b for X = P Z n and x P, either show that x conv(x) or find a vi (c, d) such that for all x conv(x): c T x d < c T x. Do relaxations ever have optimal solutions x conv(x) but x / Z n? Yes: if there are multiple optimal solutions along an at least 2-dimensional face (edge) of conv(x). Will we ever encounter such solutions? Yes: if we use an interior-point method. Not when using the simplex method! If x / Z n + is found using a simplex method, then we know that ( ) xb = B 1 b B 1 N x N = b āijx j. j N Take any fractional xi = b i / Z and generate valid Gomory cut: x i + āijx j b i x i + āij xj bi. j N j N New inequality cuts off x because x i = b > bi and x N = 0.

Decomposition / Branch-and-Bound (Vanderbei 23.5) Decomposition: Consider the problem z = maxf (x): x X. If X = i X i and z i = maxf (x): x X i, then z = maxz i. If X i are singletons, then this is complete enumeration. Smart decomposition schemes use partial enumeration. We will discuss Branch-and-Bound on the following example: max 17x 1 + 12x 2 s.t. 10x 1 + 7x 2 40 x 1 + x 2 5 x 1, x 2 Z + Formulation: P 0 = x R 2 + : 10x 1 + 7x 2 40, x 1 + x 2 5 LP Solution: (x 1, x 2 ) = ( 5 3, 10 ) 3 Upper Bound: z 0 = 205 3 = 68.33

Branch-and-Bound Example (First Branching) Note that (x 1, x 2 ) = ( 5 3, 10 ) 3 is fractional, so we can decompose: P 1 = x R 2 + : 10x 1 + 7x 2 40, x 1 + x 2 5, x 1 1 P 2 = x R 2 + : 10x 1 + 7x 2 40, x 1 + x 2 5, x 1 2 First solve formulation P 1 : LP Solution: (x 1, x 2 ) = (1, 4) Objective Value: z 1 = 65 Feasible: incumbent best! Then solve formulation P 2 : LP Solution: (x 1, x 2 ) = ( 2, 20 ) 7 Upper Bound: z 2 = 478 7 = 68.29 Can still beat incumbent best!

Enumeration Tree: Status after First Branching

Branch-and-Bound Example (Second Branching) For P 2, (x 1, x 2 ) = ( 2, 20 ) 7 is fractional so we decompose further: P 3 = x R 2 + : 10x 1 + 7x 2 40, x 1 + x 2 5, x 1 2, x 2 2 P 9 = x R 2 + : 10x 1 + 7x 2 40, x 1 + x 2 5, x 1 2, x 2 3 First solve formulation P 3 : LP Solution: (x 1, x 2 ) = ( 13 5, 2) Upper Bound: z 3 = 341 5 = 68.2 Can still beat incumbent best! Then decide between two options: Breath-first search: solve P 9. Depth-first search: decompose into P 4 (x 1 2) and P 5 (x 1 3).

Enumeration Tree: Status after Second Branching

Branch-and-Bound Example (Third Branching) Applying a depth-first search, we continue to decompose P 3 : P 4 = x R 2 + : 10x 1 + 7x 2 40, x 1 + x 2 5, x 1 = 2, x 2 2 P 5 = x R 2 + : 10x 1 + 7x 2 40, x 1 + x 2 5, x 1 3, x 2 2 First solve formulation P 4 : LP Solution: (x 1, x 2 ) = (2, 2) Objective Value: z 4 = 58 Feasible but not optimal. Continue with formulation P 5 : LP Solution: (x 1, x 2 ) = ( 3, 10 ) 7 Upper Bound: z 4 = 477 7 = 68.14 Can still beat incumbent best!

Enumeration Tree: Status after Third Branching

Branch-and-Bound Example (Fourth Branching) For P 5, (x 1, x 2 ) = ( 3, 10 ) 7 is fractional so again we decompose: P 6 = x R 2 + : 10x 1 + 7x 2 40, x 1 + x 2 5, x 1 3, x 2 1 P 10 = x R 2 + : 10x 1 + 7x 2 40, x 1 + x 2 5, x 1 3, x 2 = 2 First solve formulation P 6 : LP Solution: (x 1, x 2 ) = ( 10 3, 1) Upper Bound: z 4 = 681 10 = 68.1 Can still beat incumbent best! Continue with depth-first search: P 7 (x 1 3) yields x = (3, 1) with z 7 = 63 (feasible but not optimal) P 8 (x 1 4) yields x = (4, 0) with z 8 = 68 (best / optimal solution!)

Full Enumeration Tree with Problem Decomposition

Depth-First Search versus Breadth-First Search Several reasons for depth-first when using branch-and-bound: Integer solutions tend to lie deep in the enumeration tree, and finding (good) solutions quickly can improve method. If our current best solution has objective z and some LP bound z < z, we can stop to decompose (prune) branch. Also: if algorithm crashes - at least you have something Can use recursion to implement method (fun exercise). Easy to restart simplex after adding a new constraint. Example: Optimal dictionary for initial LP relaxation over P 0 is ζ = 205 3 5 3 w 1 1 3 w 2 x 1 = 5 3 1 3 w 1 + 7 3 w 2 x 2 = 10 3 + 1 3 w 1 10 3 w 2 For P 2 (x 1 2), add constraint w 3 = x 1 2 = 1 3 1 3 w 1 + 7 3 w 2.

Restarting Branch-and-Bound using Dual Simplex After adding w 3 = x 1 2 = 1 3 1 3 w 1 + 7 3 w 2, new dictionary is ζ = 205 3 5 3 w 1 1 3 w 2 x 1 = 5 3 1 3 w 1 + 7 3 w 2 x 2 = 10 3 + 1 3 w 1 10 3 w 2 w 3 = 1 3 1 3 w 1 + 7 3 w 2 Because dictionary is primal infeasible but dual feasible, we can use the dual simplex method and pivot rules (check: w 3 w 2 ): ζ = 478 7 12 7 w 1 1 7 w 3 x 1 = 2 + w 3 x 2 = 20 7 1 7 w 1 10 7 w 3 w 2 = 1 7 + 1 7 w 1 + 3 7 w 3 The new dictionary is primal-dual feasible, so (x 1, x 2 ) = ( 2, 20 7 and x 2 = 478 7 are optimal for the LP relaxation over P 2. Quick! )