s02lec09

s02lec09 - PRIMAL PROBLEM: 15.053 Thursday, March 7...

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1 15.053 Thursday, March 7 z Duality 2 The dual problem, in general illustrating duality with 2-person 0-sum game theory Handouts: Lecture Notes PRIMAL PROBLEM: maximize z = 3x 1 + 4x 2 +6x 3 + 8x 4 subject to x 1 + x 2 + x 3 + x 4 = 1 2x 1 + 3x 2 +4x 3 + 5x 4 = 3 x 1 , x 2 , x 3 , x 4 0 Subject to y 1 + 2y 2 3 y 1 + 4y 2 6 y 1 + 3y 2 4 y 1 + 5y 2 8 minimize y 1 + 3y 2 DUAL PROBLEM: Observation 1. The constraint matrix in the primal is the transpose of the constraint matrix in the dual. Observation 2. The RHS coefficients in the primal become the cost coefficients in the dual. PRIMAL PROBLEM: maximize z = 3x 1 + 4x 2 +6x 3 + 8x 4 subject to x 1 + x 2 + x 3 + x 4 = 1 2x 1 + 3x 2 +4x 3 + 5x 4 = 3 x 1 , x 2 , x 3 , x 4 0 Subject to y 1 + 2y 2 3 y + 4y 6 y 1 + 3y 2 4 y 1 + 5y 2 8 minimize y 1 + 3y 2 DUAL PROBLEM: Observation 3. The cost coefficients in the primal become the RHS coefficients in the dual. Observation 4. The primal (in this case) is a max problem with equality constraints and non-negative variables The dual (in this case) is a minimization problem with constraints and variables unconstrained in sign. 4 PRIMAL PROBLEM: maximize z = 3x 1 + 4x 2 +6x 3 + 8x 4 subject to x 1 + x 2 + x 3 + x 4 = 1 2x 1 + 3x 2 +4x 3 + 5x 4 = 3 x 1 , x 2 , x 3 , x 4 0 Subject to y 1 + 2y 2 3 y 1 + 4y 2 6 y 1 + 3y 2 4 y 1 + 5y 2 8 minimize y 1 + 3y 2 DUAL PROBLEM: Question. How does the dual change if we have inequality constraints? Recall that the optimal dual variables are the shadow prices for the primal 5 PRIMAL PROBLEM: maximize z = 3x 1 + 4x 2 +6x 3 + 8x 4 subject to x 1 + x 2 + x 3 + x 4 2x 1 + 3x 2 +4x 3 + 5x 4 x 1 , x 2 , x 3 , x 4 0 Method. Recall that the optimal dual variables are shadow prices. price y 1 y 2 Suppose we replace the 1 by 1 + ∆. Can the optimal objective value go down? Can it go up? 1+ 3+ Suppose we replace the 3 by 3 + ∆. Can the optimal objective value go down? Can it go up? Conclusion: y 1 0. Conclusion: y 2 0. 6 PRIMAL PROBLEM: maximize z = 3x 1 + 4x 2 +6x 3 + 8x 4 subject to x 1 + x 2 + x 3 + x 4 =1 2x 1 + 3x 2 +4x 3 + 5x 4 =3 x 1 , x 2 , x , x 0 subject to y 1 + 2y 2 3 y 1 + 4y 2 6 y 1 + 3y 2 4 y 1 + 5y 2 8 minimize y 1 + 3y 2 DUAL PROBLEM: Now suppose that we permit variables in the primal problem to be either 0 or unconstrained in sign.
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s02lec09 - PRIMAL PROBLEM: 15.053 Thursday, March 7...

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