ps9sol_ee550_fall08

Ps9sol_ee550_fall08 - III FLOW OVER 2 PATHS We solve the following optimization problem minimize g f 1,f 2 = -f 1 = 2 e f 1-1 3 e f 2-1 subject to

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UNIVERSITY OF SOUTHERN CALIFORNIA, FALL 2008 1 EE 550: Problem Set # 9 Solution I. MULTI-COMMODITY FLOW Here is an example (numbers in squares are the link capacities): Fig. 1. II. TRANSMISSION RATE ALLOCATION (a) If we have a vector ( μ 1 ,...,μ N ) that satisfies i μ i < μ max , we can form a new vector ( μ 1 + δ,μ 2 ,...,μ N ) that gives queue 1 all the remaining rate δ = μ max - i μ i . This strictly reduces the average delay in queue 1 and hence reduces the overall average delay. So the delay optimal allocation must meet the μ max constraint with equality. (b) Try to match derivatives. We have φ i ( μ i ) = λ i / ( μ i - λ i ) . We want to minimize i φ i ( μ i ) subject to i μ i = μ max and μ i 0 for all i . Thus: φ 0 i ( μ i ) = - λ i / ( μ i - λ i ) 2 = - γ Therefore: p λ i = μ i - λ i Thus: μ i = λ i + θ p λ i for some constant θ (that is, θ = 1 / γ ). But N i =1 μ i = μ max , and so we have: μ * i = λ i + θ p λ i where θ = μ max - N j =1 λ j N j =1 p λ j This equalizes the derivatives and satisfies all constraints, and is hence optimal for this convex optimization problem.
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Unformatted text preview: III. FLOW OVER 2 PATHS We solve the following optimization problem: minimize: g ( f 1 ,f 2 = λ-f 1 ) = 2( e f 1-1) + 3( e f 2-1) subject to: ≤ f 1 ≤ λ, ≤ f 1 ,f 2 ≤ μ. By taking derivative of g ( f 1 ,λ-f 1 ) with respect to f 1 and set it to zero, we get f * 1 = (1 / 2) log(3 / 2) + λ/ 2 = 0 . 203 + λ/ 2 . (a) When λ = 1 , f * 1 = 0 . 703 , f * 2 = 0 . 297 . (b) When λ = 0 . 15 , . 203 + λ/ 2 > λ . Thus f * 1 must be an end point, that is, either . 15 or . In this case, it is easy to verify that . 15 is better. Thus f * 1 = 0 . 15 , f * 2 = 0 ....
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This note was uploaded on 12/21/2010 for the course EE 550 taught by Professor Neely during the Fall '08 term at USC.

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