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ECE 665 Spring 2005 Computer Algorithms with Applications to VLSI CAD

Explore linear programming duality in the longest path problem for VLSI CAD applications. Formulate primal and dual LP, analyze solutions, and interpret results.

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ECE 665 Spring 2005 Computer Algorithms with Applications to VLSI CAD

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  1. ECE 665Spring 2005Computer AlgorithmswithApplications to VLSI CAD Linear Programming Duality – Longest Path Problem

  2. w1 w2 w3 Linear Program Dual – Example1 • Primal LP (original) max100x1 + 80x2 s.to: 10x1 + 5x2  50 5x1 + 5x2  35 5x1 + 15x2  80 Primal solution: x1=3, x2 =4, Fp = 620 maxcT x s.to: A x  b x  0 • Dual LP (original) • min50w1 + 35w2 + 80w3 • s.to: • 10w1 + 5w2 + 5w3  100 • 5w1 + 5w2 + 15w3  80 • Dual solution: • w1=4, w2 =12, w3=0 • Fd = 620 • maxwT b • s.to:wTA  cT • w  0 ECE 665 - Longest Path dual

  3. Formulate it as a Linear Program Assign variables xi to each edge ei xi = 1 if edge eiis in the longest path xi = 0 otherwise Write the objective function Write the constraint set x1 x3 x2 x4 x6 x5 4 2 x7 5 x9 2 x8 1 2 3 2 5 3 3 6 5 4 6 Example 2: Longest Path Problem ECE 665 - Longest Path dual

  4. Primal Problem x1 x3 4 2 x2 5 x4 2 x6 x5 3 2 x7 x9 5 x8 3 1 2 6 3 6 5 4 Example 2: Longest Path Problem (2) max 4x1 + 5x2 + 2x3 + 2x4 + 2x5 + 3x6 + 5x7 + 3x8 + 6x9 st -x1 - x2 - x3 = -1 x1 - x4 - x7 = 0 x2 + x4 - x5 = 0 x3 - x6 - x9 = 0 x5 + x6 - x8 = 0 x7 + x8 + x9 = 1 end ECE 665 - Longest Path dual

  5. Primal Problem - solution x1 max 4x1 + 5x2 + 2x3 + 2x4 + 2x5 + 3x6 + 5x7 + 3x8 + 6x9 x3 st x2 -x1 - x2 - x3 = -1 x1 - x4 - x7 = 0 x4 x2 + x4 - x5 = 0 x3 - x6 - x9 = 0 x5 + x6 - x8 = 0 x6 x5 4 x7 + x8 + x9 = 1 end 2 x7 5 LP OPTIMUM FOUND AT STEP 2 x9 2 x8 1 2 OBJECTIVE FUNCTION VALUE 3 1) 11.000000 2 VARIABLE VALUE REDUCED COST X1 1.000000 0.000000 5 X2 0.000000 1.000000 3 3 4 6 5 6 X3 0.000000 0.000000 X4 1.000000 0.000000 X5 1.000000 0.000000 X6 0.000000 3.000000 X7 0.000000 2.000000 X8 1.000000 0.000000 X9 0.000000 3.000000 Example 2: Longest Path Problem (3) ECE 665 - Longest Path dual

  6. Dual Problem formulation + solution w1=0 min -w1 + w6 4 w2=4 2 st 5 -w1 + w2 > 4 w4=5 -w1 + w3 > 5 2 -w1 + w4 > 2 w3=6 -w2 + w3 > 2 -w3 + w5 > 2 3 2 -w4 + w5 > 3 -w2 + w6 > 5 w5=8 -w5 + w6 > 3 5 -w4 + w6 > 6 3 1 2 6 end LP OPTIMUM FOUND AT STEP 7 w6=11 OBJECTIVE FUNCTION VALUE 1) 11.000000 3 6 5 4 VARIABLE VALUE REDUCED COST W1 0.000000 0.000000 W2 4.000000 0.000000 W3 6.000000 0.000000 W4 5.000000 0.000000 W5 8.000000 0.000000 W6 11.000000 0.000000 Example 2: Longest Path Problem (4) Interpretation of dual variables: wi = distance ofnodeI from source Note: This is still a longest path problem (Critical Path or Schedulingproblem): Find a minimumdistance from sink to source that satisfies all edge-length constraints. ECE 665 - Longest Path dual

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