11 Nonlinear Programming - Nonlinear Programming In...

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± ± ± ± ± ± ± ± ± ± ± Nonlinear Programming In previous chapters, we have studied linear programming problems. For an LP, our goal was to maximize or minimize a linear function subject to linear constraints. But in many interesting maximization and minimization problems, the objective function may not be a linear function, or some of the constraints may not be linear constraints. Such an optimization problem is called a nonlinear programming problem (NLP). In this chapter, we discuss techniques used to solve NLPs. We begin with a review of material from differential calculus, which will be needed for our study of nonlinear programming. 11.1 Review of Differential Calculus Limits The idea of a limit is one of the most basic ideas in calculus. DEFINITION The equation lim x a f ( x ) ± c means that as x gets closer to a (but not equal to a ), the value of f ( x ) gets arbitrarily close to c . It is also possible that lim x a f ( x ) may not exist. 1 Show that lim x 2 x 2 ² 2 x ± 2 2 ² 2(2) ± 0. 2 Show that lim x 0 ³ 1 x ³ does not exist. Solution 1 To verify this result, evaluate x 2 ² 2 x for values of x close to, but not equal to, 2. 2 To verify this result, observe that as x gets near 0, ³ 1 x ³ becomes either a very large pos- itive number or a very large negative number. Thus, as x approaches 0, ³ 1 x ³ will not approach any single number. Limits EXAMPLE 1
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11.1 Review of Differential Calculus 611 Continuity DEFINITION A function f ( x ) is continuous at a point a if lim x a f ( x ) ± f ( a ) If f ( x ) is not continuous at x ± a , we say that f ( x ) is discontinuous (or has a discontinuity) at a . Bakeco orders sugar from Sugarco. The per-pound purchase price of the sugar depends on the size of the order (see Table 1). Let x ± number of pounds of sugar purchased by Bakeco f ( x ) ± cost of ordering x pounds of sugar Then f ( x ) ± 25 x for 0 ² x ³ 100 f ( x ) ± 20 x for 100 ² x ² 200 f ( x ) ± 15 x for x ´ 200 For all values of x , determine if x is continuous or discontinuous. Solution From Figure 1, it is clear that lim x 100 f ( x ) and lim x 200 f ( x ) do not exist. Thus, f ( x ) is discontinuous at x ± 100 and x ± 200 and is continuous for all other values of x satisfying x µ 0. Continuous Functions EXAMPLE 2 TABLE 1 Price of Sugar Paid by Bakeco Size of Order Price per Pound (¢) 10 0 ² x ³ 100 25 100 ² x ² 200 20 100 ² x ´ 200 15 f ( x ) x 100 200 300 $10 $20 $25 $30 $40 FIGURE 1 Cost of Purchasing Sugar for Bakeco
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612 CHAPTER 11 Nonlinear Programming Differentiation DEFINITION The derivative of a function f ( x ) at x ± a [written f ² ( a )] is deFned to be lim ³ x 0 ´ f ( a µ³ ³ x x ) f ( a ) ´ If this limit does not exist, then f ( x ) has no derivative at x ± a . We may think of f ² ( a ) as the slope of f ( x ) at x ± a. Thus, if we begin at x ± a and increase x by a small amount ³ ( ³ may be positive or negative), then f ( x ) will increase by an amount approximately equal to ³ f ² ( a ). If f ² ( a ) · 0, then f ( x ) is increasing at x ± a , whereas if f ² ( a ) ¸ 0, then f ( x ) is decreasing at x ± a . The derivatives of many func-
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11 Nonlinear Programming - Nonlinear Programming In...

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