which means that *i is the negative of the imputed
value of the unit of ai.
Example 8.1 Consider the problem:
Solution.
From the first two equations we get solving this with the last equation yields the quantities
r- component row vector such that
Suppose (y*, *) is a solution of (8.6) and (8.7). Note
that y* depends on a, i.e., y*=y*(a). Now
is the optimum value of the objective function. The Lagrange multipliers satisfy the relation
y: be an n-component column vector, a: be an r-component column vector, b: be an s-component column vector. h: En E1, g: En Er, w: En Es be given functions.
8.1.2 Inequality Constraints
Note that (8.10) is analogous to (8.6). Also (8.11) repeats the inequality constraint (8.3) in the same way that (8.7) repeated the equality constraint (8.2). However, the conditions in (8.12) are new and are particular to the inequality-constrained problem.
Example 8.3 solve the problem:
Solution. The Lagrangian is The necessary conditions are
Case 1: = 0
From (8.16) we obtain x = 4 , which does not satisfy (8.17), thus, infeasible.
Chapter 8 The Maximum Principle: Discrete Time
8.1 Nonlinear Programming Problems We begin by starting a general form of a nonlinear programming problem.
Case 2: x=6
(8.17) holds. From (8.16) we get = 4, so that (8.18)
holds. The optimal solution is Nhomakorabeathen
since it is the only solution satisfying the necessary conditions.
Case 2:
Here from (8.13) we get = - 4, which does not satisfy the inequality 0 in (8.15).
From these two cases we conclude that the optimum solution is x* = 4 and
We assume functions g and w to be column vectors with components r and s , respectively. We consider the nonlinear programming problem:
subject to
8.1.1 Lagrange Multipliers Suppose we want to solve (8.1) without imposing constraint (8.2) or (8.3). The problem is now the classical unconstrained maximization problem of calculus, and the first-order necessary conditions for its solution are
The points satisfying (8.4) are called critical points. With equality constraints, the Lagrangian is
where is an r-component row vector.
The necessary condition for y* to be a (maximum) solution to be (8.1) and (8.2) is that there exists an
Example 8.4 Find the shortest distance between the point (2.2) and the upper half of the semicircle of radius one, whose center is at the origin. In order to simplify the calculation, we minimize h , the square of the distance:
Example 8.2
Solution. We form the Lagrangian The necessary conditions (8.10)-(8.12) become
Case 1: From (8.13) we get x = 4, which also satisfies (8.14). Hence, this solution, which makes h(4)=16, is a possible candidate for the maximum solution.