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Linear Programming Homogenous Self Dual Model, Lecture Notes - Mathematics - Prof. J Vanderbei.pdf, Prof. J Vanderbei, Mathematics, Linear Programming, Homogenous Self Dual Model, Predictor-Corrector Algorithm
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Robert J. Vanderbei
December 14, 2005 ORF 522
Operations Research and Financial Engineering, Princeton University http://www.princeton.edu/∼rvdb
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Primal-Dual Pair
maximize cT^ x subject to Ax ≤ b x ≥ 0
minimize bT^ y subject to AT^ y ≥ c y ≥ 0
Homogeneous Self-Dual Problem
maximize 0 subject to −AT^ y +cφ ≤ 0 Ax −bφ ≤ 0 −cT^ x +bT^ y ≤ 0 x, y, φ ≥ 0
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Theorem. Let (x, y, φ) be a solution to HSD. If φ > 0 , then
Proof.
x∗^ is primal feasible—obvious.
y∗^ is dual feasible—obvious.
Weak duality theorem implies that cT^ x∗^ ≤ bT^ y∗.
3rd HSD constraint implies reverse inequality.
Primal feasibility, plus dual feasibility, plus no gap implies op- timality.
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0 −AT^ c A 0 −b −cT^ bT^0
x y φ
(^) −→ x n + m + 1 −→ n
In New Notation:
maximize 0 subject to Ax + z = 0 x, z ≥ 0
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Solve linearized system for (∆x, ∆z).
Pick step length θ.
Step to a new point:
x ¯ = x + θ∆x, z¯ = z + θ∆z.
Even More Notation
ρ ¯ = ρ(¯x, z¯), μ¯ = μ(¯x, z¯)
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Proof.
ρ ¯ = A(x + θ∆x) + (z + θ∆z) = Ax + z + θ(A∆x + ∆z) = ρ − θ(1 − δ)ρ = (1 − θ + θδ)ρ.
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Neighborhoods of {(x, z) > 0 : x 1 z 1 = x 2 z 2 = · · · = xnzn}
N (β) =
(x, z) > 0 : ‖XZe − μ(x, z)e‖ ≤ βμ(x, z)
Note: β < β′^ implies N (β) ⊂ N (β′).
Predictor-Corrector Algorithm
Odd Iterations–Predictor Step
Assume (x, z) ∈ N (1/4).
Compute (∆x, ∆z) using δ = 0.
Compute θ so that (¯x, z¯) ∈ N (1/2).
Even Iterations–Corrector Step
Assume (x, z) ∈ N (1/2).
Compute (∆x, ∆z) using δ = 1.
Put θ = 1.
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In Complementarity Space
Let uj = xj zj j = 1, 2 ,... , n.
Let u (^) j = x (^) j z (^) j j = 1 , 2 ,... , n.
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u 1
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u (0) u (1)
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u (4) u (6) u (3) u (5)
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Theorem.
Proof.
μ ¯ = (1 − θ)μ since δ = 0.
‖ X¯ Ze¯ − μe¯ ‖ = ‖∆X∆Ze‖ ≤ μ/ 4.
Need to show also that (¯x, z¯) > 0. Intuitively clear (see earlier picture) but proof is tedious. See text.
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Progress toward optimality is controlled by the stepsize θ. Theorem. In predictor steps, θ ≥ 2 √^1 n. Proof. Consider taking a step with step length t ≤ 1 / 2
n:
x(t) = x + t∆x, z(t) = z + t∆z.
From earlier theorems and lemmas,
‖X(t)Z(t)e − μ(t)e‖ ≤ (1 − t)‖XZe − μe‖ + t^2 ‖∆X∆Ze‖ ≤ (1 − t)
μ 4
nμ 2 ≤ (1 − t)
μ 4
μ 8 ≤ (1 − t)
μ 4
μ 4
=
μ(t) 2
Therefore (x(t), z(t)) ∈ N (1/2) which implies that θ ≥ 1 / 2
n.
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Just a final remark: If primal and dual problems are feasible, then algorithm will produce a solution to HSD with φ > 0 from which a solution to original problem can be extracted. See text for details.
