15.081J/6.251J Introduction to Mathematical Programming Lecture 4: Geometry of Linear Optimization III� � � � � � 1 Outline Slide 1 1. Projections of Polyhedra 2. Fourier-Mo tzkin Elimination Algorithm 3. Optimality Conditions 2 Projections of polyhedra Slide 2 • πk : ℜn �→ ℜk projects x onto its first k coordinates: πk (x) = πk(x1, . . . , xn) = (x1, . . . , xk ). • Πk(S) = πk(x) | x ∈ S ; Equivalently Πk(S) = (x1, . . . , xk) there exist xk+1, . . . , xn s.t. (x1, . . . , xn) ∈ S . x1 x2 x3 P 1( S ) P 2( S ) 2.1 The Elimination Algorithm 2.1.1 By example Slide 3 • Consider the polyhedron x1 + x2 ≥ 1 1� � x1 + x2 + 2x3 ≥ 2 2x1 + 3x3 ≥ 3 x1 − 4x3 ≥ 4 −2x1 + x2 − x3 ≥ 5. • We re w rite these constraints 0 ≥ 1 − x1 − x2 x3 ≥ 1 − (x1/2) − (x2/2) x3 ≥ 1 − (2x1/3) −1 + (x1/4) ≥ x3 −5 − 2x1 + x2 ≥ x3. • Eliminate variable x3, obtaing polyhedron Q 0 ≥ 1 − x1 − x2 −1 + x1/4 ≥ 1 − (x1/2) − (x2/2) −1 + x1/4 ≥ 1 − (2x1/3) −5 − 2x1 + x2 ≥ 1 − (x1/2) − (x2/2) −5 − 2x1 + x2 ≥ 1 − (2x1/3). 2.2 The Elimination Algorithm Slide 4 1. Rewrite jn =1 aij xj ≥ bi in the form n−1 ainxn ≥ − aij xj + bi, i = 1, . . . , m; j=1 if ain =� 0, divide both sides by ain. By letting x = (x1, . . . , xn−1) that P is represented by: ′ xn ≥ di + fix, if ain > 0, ′ dj + fj x ≥ xn, if ajn < 0, ′ 0 ≥ dk + fkx, if akn = 0. 2. Let Q be the p olyhedron in ℜn−1 defined by: ′′ dj + fj x ≥ di + fix, if ain > 0 and ajn < 0, ′ 0 ≥ dk + fkx, if akn = 0. Theorem: The polyhedron Q constructed by the elimination algor ithm is equa l to the projection Πn−1(P ) of P . 2� � � � � 2.3 Implications • Let P ⊂ ℜn+k be a polyhedron. Then, the set x ∈ ℜn � there ex ists y ∈ ℜk such that (x, y) ∈ P is also a polyhedron. • Let P ⊂ ℜn be a polyhedro n and let A be an m × n matrix. Then, the set Q = {Ax | x ∈ P } is also a polyhedr on. • The convex hull of a finite number of vectors is a poly hedron. 2.4 Algorithm for LO ′ • Consider min c x subject to x ∈ P . ′ • Define a new variable x0 and introduce the constraint x0 = c x. • Apply the elimination a lgorithm n times to eliminate the variables x1, . . . , xn • We are left with the set ′Q = x0 | there exists x ∈ P such that x0 = c x , and the optimal cost is equal to the smallest element of Q. 3 Optimality Conditions 3.1 Feasible directions • We a re at x ∈ P and we contemplate moving away from x, in the direction of a vecto r d ∈ ℜn. • We need to consider those choices of d that do not immediately take us outside the feasible set. • A vector d ∈ ℜn is said to be a feasible direction at x, if there exists a positive scalar θ for which x + θd ∈ P . 3 Slide 5 Slide 6 Slide 7 Slide 8� � � � � Slide 9 • x be a BFS to the standard fo rm problem corresponding to a basis B. • xi = 0, i ∈ N, xB = B−1B. • We consider moving away from x, to a new vector x + θd, by selecting a nonbasic variable xj and increasing it to a positive value θ, while keeping the rema ining nonbasic variables at zero. • Algebraically, dj = 1, and di = 0 for every nonbasic index i o ther than j. • The vector xB of basic variables changes to xB + θdB . • Feasibility: A(x + θd) = B ⇒ Ad = 0. • 0 = Ad = �ni=1 Aidi = �mi=1 AB(i)dB(i) + Aj = BdB + Aj ⇒ dB = −B−1Aj . • Nonnegativity constraints? – If x nondegener ate, xB > 0; thus xB + θdB ≥ 0 for θ is sufficiently small. – If xdegenerate, then d is not always a feasible direction. Why? • Effects in cost? ′ Cost change: c ′ d = cj − cB B−1Aj This quantity is called reduced cost cj of the variable xj . 3.2 Theorem Slide 10 • x BFS associated with basis B • c re duced costs Then • If c ≥ 0 ⇒ x optimal • x optimal and non-degenerate ⇒ c ≥ 0 3.3 Proof • y arbitrary feasible solution • d = y − x ⇒ Ax = Ay = b ⇒ Ad = 0 ⇒ BdB + i∈N ⇒ dB = − i∈N ⇒ c ′ d = c ′ B Slide 11 Aidi = 0 B−1Aidi dB + cidi i∈N Slide 12 ′ = (ci − cB B−1Ai)di = cidi i∈Ni∈N 4• Since yi ≥ 0 and xi = 0, i ∈ N, then di = yi − xi ≥ 0, i ∈ N • c ′ d = c ′ (y − x) ≥ 0 ⇒ c ′ y ≥ c ′ x ⇒ x optimal (b) Your turn 5MIT OpenCourseWarehttp://ocw.mit.edu 6.251J / 15.081J Introduction to Mathematical Programming Fall 2009 For information about citing these materials or our Terms of Use, visit: