Non-Linnear Programming With Constraints - Lagrange's Method

# Non-Linnear Programming With Constraints - Lagrange's...

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UNC-Wilmington ECN 321 Department of Economics and Finance Dr. Chris Dumas Non-linear Programming with Constraints, “Lagrange’s Method” -- Example Problems Example (1) * Nonlinear programming problem * One choice variable * One constraint Note: When we have a nonlinear problem with one or more constraints, we need to use Lagrange's method. max U = 2*ln(3X) X subject to: 5X ≤ 100 Because this is a non-linear programming problem with a constraint, we will use Lagrange's method as the solution method. Converting the problem into an equivalent Lagrangian problem, we re-write the objective function as a Lagrangian expression, denoted "L." In doing so, we incorporate the constraint into the Lagrangian expression by introducing the new choice variable λ into the problem as a Lagrangian multiplier: max L = 2*ln(3X) + λ (100 - 5X) X, λ subject to: nothing F.O.C.'s: (1) 0 5 3 ) X 3 1 ( 2 X L = λ - = (2) 0 X 5 100 L = - = λ The FOC equations provide us with two equations in two unknowns, X and λ . Solve these two equations for the solution values of X and λ . There are several ways to work the algebra to solve these two equations. One way is illustrated below: Solve FOC (2) for X: 100 - 5X = 0 100 = 5X X * = 100/5 = 20 (Note: we typically use a star "*" superscript to denote a solution value.) Substitute X * back into FOC (1) and solve for λ : 2(1/(3X))3 - 5 λ = 0 2(1/(3·20))3 - 5 λ = 0 1/10 = 5 λ λ * = 1/50 = 0.02 Hence, the solution is: X * = 20 , λ * = 0.02. (Note: Recall that in Lagrangian problems, the solution value of the Lagrangian multiplier, λ * = 0.02, is the shadow value of the constraint .) 1

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UNC-Wilmington ECN 321 Department of Economics and Finance Dr. Chris Dumas Example (2): * Nonlinear programming problem * Two choice variables * No interaction among choice variables in the utility function. * One
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Non-Linnear Programming With Constraints - Lagrange's...

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