week1113 - 8 Nonlinear optimization Linear programs are...

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8 Nonlinear optimization Linear programs are relatively easy to solve. By contrast, as we have seen, integer programs may be much harder. To solve an integer program, we try to capitalize on our ability to solve linear programs quickly, studying lin- ear programming relaxations that in some sense approximate the underlying problem. Even without integrality constraints, optimization problems may be much harder than linear programs, due to nonlinearity . Since we understand linear programs so well, we might naturally study nonlinear optimization problems first by approximating them using linear programs. Approximating nonlinear functions by linear functions is the central idea of differential calculus, so we begin by reviewing some ideas from calculus. Consider the following nonlinear function of one variable, x : f ( x ) = 35 - 12 x 2 + 4 x 3 + 3 x 4 . We plot the graph of this function below. The derivative of this function is f 0 ( x ) = - 24 x + 12 x 2 + 12 x 3 , so using the values f ( - 1) = 22 and f 0 ( - 1) = 24, we arrive at the “linear approximation” to f at - 1: f ( x ) f ( - 1)+ f 0 ( - 1)( x - ( - 1)) = 22+24( x +1) = g ( x ) for all x near - 1 . Calculus provides a crucial tool to help us maximize or minimize nonlinear functions: at any maximizing or minimizing point ¯ x , the graph of the linear 105
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f 0 x ) should be zero, which implies ¯ x = - 2, 0, or 1. From the graph we can see that ¯ x = - 2 is a minimizer : in other words, f ( x ) f ( - 2) for all x . On the other hand, ¯ x = 1 is a local minimizer , meaning f ( x ) f (1) for all x near 1. Sometimes, to emphasize the difference, we refer to minimizers as global minimizers . AMPL allows us to model nonlinear functions, and the solver MINOS in- cludes some capability to solve nonlinear optimization problems. This flexi- bility comes with some pitfalls, however. To illustrate, consider the following example. ampl: var x; ampl: minimize objective: 35 - 12*x^2 + 4*x^3 + 3*x^4; ampl: solve; MINOS 5.5: optimal solution found. 0 iterations, objective 35 Nonlin evals: obj = 3, grad = 2. ampl: display x; x = 0 As we can see from the graph, despite what AMPL tells us, x = 0 is certainly not a solution of our problem: in fact it is a local maximizer ! We can encourage AMPL to rethink by restarting at a different initial point. ampl: let x:=10; ampl: solve; MINOS 5.5: optimal solution found. 4 iterations, objective 3 Nonlin evals: obj = 12, grad = 11. ampl: display x; x = -2 This time, AMPL has indeed solved the problem. But different initial points can produce different output. ampl: let x:=0.5; ampl: solve; MINOS 5.5: optimal solution found. 106
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This note was uploaded on 04/06/2008 for the course ORIE 321 taught by Professor Shmoys/lewis during the Spring '07 term at Cornell University (Engineering School).

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week1113 - 8 Nonlinear optimization Linear programs are...

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