11 Nonlinear Programming

# 11 Nonlinear Programming - Nonlinear Programming In...

This preview shows pages 1–4. Sign up to view the full content.

± ± ± ± ± ± ± ± ± ± ± 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

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

View Full Document
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
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-

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## 11 Nonlinear Programming - Nonlinear Programming In...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online