{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

4120_lecture13-F10

4120_lecture13-F10 - Computational Methods for Management...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Computational Methods for Management and Economics Carla Gomes Lecture 13 Reading: 6.1- 6.4 of textbook
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Outline Duality
Background image of page 2
Duality Every maximization LP problem in the standard form gives rise to a minimization LP problem called the dual problem Every feasible solution in one yields a bound on the optimal value of the other If one of the problems has an optimal solution, so does the other and the two optimal values coincide These results have very interesting economic interpretations
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Bounds on the optimal value Let x1 = the number of doors to produce x2 = the number of windows to produce Maximize Z = 3 x1 + 5 x2 subject to x1 ≤ 4 2 x2 ≤ 12 3 x1 + 2 x2 ≤ 18 and x1 ≥ 0, x2 ≥ 0. Lower bound (LB) on Z* Z* ≥ LB any feasible solution Upper bound (UB) on Z* Z* ≤ UB – how do we come up with upper bounds for Z* for a maximization LP problem?
Background image of page 4
“Guessing” upper bounds Let x1 = the number of doors to produce x2 = the number of windows to produce Maximize Z = 3 x1 + 5 x2 subject to x1 ≤ 4 2 x2 ≤ 12 3 x1 + 2 x2 ≤ 18 and x1 ≥ 0, x2 ≥ 0. 1 – multiplying the 3 rd costraint by 3 9 x1 + 6 x2 ≤ 54 Z = 3 x1 + 5 x2 ≤ 9 x1 + 6 x2 ≤ 54 Z *≤ 54 2 – multiplying the 3 rd constraint by 2.5 7.5 x1 + 5 x2 ≤ 45 Z = 3 x1 + 5 x2 ≤ 7.5 x1 + 5 x2 ≤ 45 Z *≤ 45 3 – multiplying the 2 nd constraint by 2 and add it to the 3 rd 3 x1 + 6 x2 ≤ 42 Z = 3 x1 + 5 x2 ≤ 3 x1 + 6 x2 ≤ 42 Z *≤ 42 4 – multiplying the 2 nd constraint by 1.5 and add it to the 3 rd 3 x1 + 5 x2 ≤ 36 Z = 3 x1 + 5 x2 ≤ 3 x1 + 5 x2 ≤ 36 Z *≤ 36
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
A principled way of finding upper-bounds on Z* Dual Problem Let x1 = the number of doors to produce x2 = the number of windows to produce Maximize Z = 3 x1 + 5 x2 subject to x1 ≤ 4 y1 2 x2 ≤ 12 y2 3 x1 + 2 x2 ≤ 18 y3 and x1 ≥ 0, x2 ≥ 0.
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}