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ECE 665 Spring 2004 Computer Algorithms with Applications to VLSI CAD

ECE 665 Spring 2004 Computer Algorithms with Applications to VLSI CAD. Linear Programming - Duality Max-Flow/Min-cut. Primal Problem. x3. x7. x1. x4. x0. x0. max x0. st. x2. x5. x8. -x1-x2+x0 = 0. x6. x1-x3-x4 = 0. Max flow. x2-x5-x6 = 0. x3+x5-x7 = 0. x4+x6-x8 = 0.

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ECE 665 Spring 2004 Computer Algorithms with Applications to VLSI CAD

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  1. ECE 665Spring 2004Computer AlgorithmswithApplications to VLSI CAD Linear Programming - Duality Max-Flow/Min-cut

  2. Primal Problem x3 x7 x1 x4 x0 x0 max x0 st x2 x5 x8 -x1-x2+x0 = 0 x6 x1-x3-x4 = 0 Max flow x2-x5-x6 = 0 x3+x5-x7 = 0 x4+x6-x8 = 0 x7+x8-x0 = 0 Edge capacities Flow variables x1 <= 12 1 2 x2 <= 9 x3 <= 10 x4 <= 3 10 x5 <= 4 7 12 4 x6 <= 10 3 4 6 5 x7 <= 7 3 x8 <= 13 9 13 end 10 Example3: Max-Flow Min-Cut ECE 665 - Max-Flow/Min-Cut Duality

  3. Primal Problem - Solution X3=7 X1=10 X7=7 X4=3 X0=19 X0=19 X2=9 X8=12 X5=0 LP OPTIMUM FOUND AT STEP 3 min cut X6=9 OBJECTIVE FUNCTION VALUE 1) 19.000000 VARIABLE VALUE REDUCED COST 1 2 X0 19.000000 0.000000 X1 10.000000 0.000000 X2 9.000000 0.000000 10 7 X3 7.000000 0.000000 12 4 3 5 6 4 X4 3.000000 0.000000 3 X5 0.000000 1.000000 9 13 X6 9.000000 0.000000 10 X7 7.000000 0.000000 X8 12.000000 0.000000 Example3: Max-Flow Min-Cut (2) Saturated edges: x2=9 x4=3 x7=7 Max flow= 19 = min cut of forward saturated eges ECE 665 - Max-Flow/Min-Cut Duality

  4. Dual Problem - formulation x3 x7 x1 x4 x0 x0 max x0 st x2 x5 x8 z1 -x1-x2+x0 = 0 x6 x1-x3-x4 = 0 Node constraints: dual variables z x2-x5-x6 = 0 x3+x5-x7 = 0 x4+x6-x8 = 0 z6 Dual objective vector: [z1,…, z6, w1,…,w8] x7+x8-x0 = 0 w1 x1 <= 12 1 2 Dual objective cost: [ 0, … , 0, c1, …, c8] x2 <= 9 x3 <= 10 Edge constraints: dual variables w x4 <= 3 10 Edge capacities (12,9,10,3,4,10,7,13) x5 <= 4 7 12 4 x6 <= 10 3 4 6 5 x7 <= 7 3 w8 x8 <= 13 9 13 end 10 Example3: Max-Flow Min-Cut (3) Convert primal problem: Dual obj function: Fdual = 0 z1+…+0 z2 + 12 w12 +… + 13 w8 ECE 665 - Max-Flow/Min-Cut Duality

  5. Dual Problem st z1-z6 >= 1 -z1+z2+w1 >= 0 z2 z4 w3 -z1+z3+w2 >= 0 w7 w1 -z2+z4+w3 >= 0 z1 z6 -z2+z5+w4 >= 0 w4 -z3+z4+w5 >= 0 w5 -z3+z5+w6 >= 0 1 2 w2 w8 -z4+z6+w7 >= 0 w6 z5 -z5+z6+w8 >= 0 z3 z= node variables end w = edge variables 3 4 6 5 Example3: Max-Flow Min-Cut (4) min 12w1 +9w2 +10w3 +3w4 +4w5 +10w6 +7w7 +13w8 ECE 665 - Max-Flow/Min-Cut Duality

  6. Dual Problem - solution z4=1 z2=1 min cut w3=0 w1=0 w7=1 z1=1 z6=0 w4=1 w5=0 OBJECTIVE FUNCTION VALUE S T w2=1 1) 19.000000 w8=0 z3=0 w6=0 z5=0 VARIABLE VALUE REDUCED COST W1 0.000000 2.000000 W2 1.000000 0.000000 1 2 W3 0.000000 3.000000 W4 1.000000 0.000000 W5 0.000000 4.000000 zi = 1 if nodei is in set S 0 if it is in set T W6 0.000000 1.000000 W7 1.000000 0.000000 W8 0.000000 1.000000 3 4 5 6 Z1 1.000000 0.000000 Z6 0.000000 0.000000 wi = 1 if edge iis in the min-cut 0 otherwise Z2 1.000000 0.000000 Z3 0.000000 0.000000 Z4 1.000000 0.000000 Z5 0.000000 0.000000 Example3: Max-Flow Min-Cut (5) LP OPTIMUM FOUND AT STEP 9 Interpretation of dual variables ECE 665 - Max-Flow/Min-Cut Duality

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