15.081J/6.251J Introduction to Mathematical Programming Lecture 8: Duality Theory I� � 1 Outline Slide 1 • Motivation of duality • General form of the dual • Weak and strong duality • Relations between pr imal and dual • Economic Interpretation • Complementary Slackness 2 Motivation 2.1 An idea from Lagrange Slide 2 Consider the LOP, called the primal with optimal solution x ∗ ′ min c x s.t. Ax = b x ≥ 0 Relax the constraint g(p) = min c ′ x + p ′ (b − Ax) s.t. x ≥ 0 g(p) ≤ c ′ x ∗ + p ′ (b − Ax∗) = c ′ x ∗ Get the tightest lower bo und, i.e., max g(p) g(p) = min c ′ x + p ′ (b − Ax) x≥0 ′ ′ ′ = p b + min (c − p A)x x≥0 Note that � ′ ′ 0, if c ′ − p ′ A ≥ 0 ′ ,min (c − p A)x = x≥0 −∞, otherwise. Dual max g(p) ⇔ max p ′ b s.t. p ′ A ≤ c ′ 13 General form of the dual Primal ′ min c x s.t. a′i x ≥ bi i ∈ M1 a′i x ≤ bi i ∈ M2 a′i x = bi i ∈ M3 xj ≥ 0 j ∈ N1 xj ≤ 0 j ∈ N2 xj ><0 j ∈ N3 3.1 Example min x1 + 2x2 + 3x3 s.t. −x1 + 3x2 = 5 2x1 − x2 + 3x3 ≥ 6 x3 ≤ 4 x1 ≥ 0 x2 ≤ 0 x3 free, Primal constraints variables min ≥ bi ≤ bi = bi ≥ 0 ≤ 0 > <0 Dual max s.t. max s.t. max ≥ 0 ≤ 0 > <0 ≤ cj ≥ cj = cj min c x max p ′ b s.t. Ax = b s.t. p ′ A ≤ c ′ x ≥ 0 ′ ′ bmin c x max p s.t. Ax ≥ b s.t. p ′ A = c ′ p ≥ 0 4 Weak Duality Slide 7 Theorem: If x is primal feasible and p is dual feasible then p ′ b ≤ c ′ x Proof p ′ b = p ′ Ax ≤ c ′ x 2 Theorem: The dual of the dual is the primal. 3.2 A matrix view ′ Slide 3 p ′ b pi ≥ 0 pi ≤ 0 p > i <0 p ′ Aj ≤ cj p ′ Aj ≥ cj p ′ Aj = cj i ∈ M1 i ∈ M1 i ∈ M3 j ∈ N1 j ∈ N2 j ∈ N3 5p1 + 6p2 + 4p3 p1 free p2 ≥ 0 p3 ≤ 0 −p1 + 2p2 ≤ 1 3p1 − p2 ≥ 2 3p2 + p3 = 3. Slide 4 Slide 5 dual variables constraints Slide 6Corollary: If x is primal feasible, p is dual feasible, and p ′ b = c ′ x, then x is optimal in the primal and p is optimal in the dual. 5 Strong Duality Slide 8 Theorem: If the LOP has optimal so lution, then so does the dual, and optimal costs are equal. Proof: ′min c x s.t. Ax = b x ≥ 0 Apply Simplex; optimal solution x, basis B. Optimality conditions: c ′ − cB ′ B−1A ≥ 0 ′ Slide 9 Define p ′ = c ′ B−1 ⇒ p ′ A ≤ c ′ B ⇒ p dual feasible for max p ′ b s.t. p ′ A ≤ c ′ ′ p ′ b = cB ′ B−1b = cB ′ xB = c x ⇒ x, p are primal and dual optimal 5.1 Intuition Slide 10 x * a1 a2 a3 p1a1p2a2 c . 3� � 6 Relations between primal and dual Slide 11 Finite opt. Unbounded Infeasible Finite opt. * Unbounded * Infeasible * * 7 Economic Interpretation Slide 12 • x optimal nondegenerate solution: B−1b > 0 • Suppose b changes to b + d for some small d • How is the optimal cost affected? • For small d feasibilty unaffected • Optimality conditions unaffected • New cost c ′ B−1(b + d) = p ′ (b + d)B • If resource i changes by di, cos t changes by pidi: “Marginal Price” 8 Complementary slackness 8.1 Theorem Slide 13 Let x primal feasible and p dual fea sible. Then x, p optimal if and only if ′ pi(aix − bi) = 0, ∀i ′ xj (cj − p Aj ) = 0, ∀j 8.2 Proof Slide 14 • ui = pi(ai ′ x − bi) and vj = (cj − p ′ Aj )xj • If x primal feasible and p dual feasible, we have ui ≥ 0 and vj ≥ 0 for all i and j. • Also ′ c x − p ′ b = ui + vj . i j ′ • By the str ong duality theorem, if x and p are optimal, then c x = p ′ b ⇒ ui = vj = 0 for all i, j. • Conversely, if ui = vj = 0 for all i, j, then c ′ x = p ′ b, • ⇒ x and p are optimal. 48.3 Example Slide 15 min 13x1 + 10x2 + 6x3 max 8p1 + 3p2 s.t. 5x1 + x2 + 3x3 = 8 s.t. 5p1 + 3p2 ≤ 13 3x1 + x2 = 3 p1 + p2 ≤ 10 x1 , x2 , x3 ≥ 0 3p1 ≤ 6 Is x ∗ = (1, 0, 1 ) ′ optimal? Slide 16 5p1 + 3p2 = 13, 3p1 = 6 ⇒ p1 = 2, p2 = 1 Objective=19 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: