Chapter 8 The Maximum Principle: Discrete Time

1. Chapter 8 The Maximum Principle: Discrete Time 8.1 Nonlinear Programming Problems We begin by starting a general form of a nonlinear programming problem. 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  Esbe given functions.

2. We assume functions g and w to be column vectors with components r and s , respectively. We consider the nonlinear programming problem: subject to

3. 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.

4. The necessary condition for y* to be a (maximum) solution to be (8.1) and (8.2) is that there exists an 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 which means that *i is the negative of the imputed value of the unit of ai.

5. Example 8.1 Consider the problem: Solution. From the first two equations we get solving this with the last equation yields the quantities

6. 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.

7. Example 8.2 Solution. We form the Lagrangian The necessary conditions (8.10)-(8.12) become

8. 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. 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

9. Example 8.3 solve the problem: Solution. The Lagrangian is The necessary conditions are

10. Case 1: = 0 From (8.16) we obtain x = 4 , which does not satisfy (8.17), thus, infeasible. Case 2:x=6 (8.17) holds. From (8.16) we get  = 4, so that (8.18) holds. The optimal solution is then since it is the only solution satisfying the necessary conditions.

11. 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:

12. The Lagrangian function for this problem is The necessary conditions are From (8.24) we see that either  =0 or x2+y2=1,i.e., we are on the boundary of the semicircle. If  =0, we see from (8.20) that x=2. But x=2 does not satisfy (8.22) for any y , and hence we conclude >0 and x2+y2=1.

13. Figure 8.1 Shortest Distance from a Point to a Semi-Circle

14. From (8.25) we conclude that either or y =0. If , then from (8.20), (8.21) and  >0, we get x = y. Solving the latter with x2+y2=1 , gives If y =0, then solving with x2+y2=1 gives These three points are shown in figure 8.1. Of the three points found that satisfy the necessary conditions, clearly the point found in (a) is the nearest point and solves the closest-point problem. The point (-1,0) in (c) is in fact the farthest point; and the point (1,0) in (b) is neither the closest nor the farthest point.

15. Example 8.5 Consider the problem: subject to The set of points satisfying the constraints is shown shaded in figure 8.2. From the figure it is obvious that the solution point (0,1) maximizes the value of y. Let us see if we can find it using the above procedure.

16. Figure 8.2: Graph of Example 8.5

17. The Lagrangian is The necessary conditions are together with (8.27) and (8.28). From (8.30) we get  =0, since x-1/3 is never 0 in the range -1 x  1. But substitution of  =0 into (8.31) gives  = - 1 < 0, which fails to solve (8.33).

18. Example 8.6 Consider the problem: subject to The constraints are now differentiable, and the optimum solution is (x*,y*)=(0,1) and h*=1. But once again the Kuhn-Tucker method fails, as we will see. The Lagrangian is

19. so that the necessary conditions are together with (8.35) and (8.36). From (8.41) we get, either y =0 or =0. Since y =0 minimizes the objective function, we choose  =0. From (8.38) we get either  =0 or x=0. Since substitution of  =  =0 into (8.39) shows that it is unsolvable, we choose x =0,   0. But then (8.40) gives (1-y)3 = 0 or y =1. However,  =0 and y =1 means that once more there is no solution to (8.39).

20. The reason for failure of the method in Example 8.6 is that the constraints do not satisfy what is called the constraint qualification, which will be discussed in the next section. In order to motivate the definition we illustrate two different situation in figure 8.3. In figure 8.3(a) we show two boundary curves and intersecting the boundary point . The two tangents to these curves are shown, and  is a vector lying between the two tangents. Starting at , there is a differentiable curve drawn so that it lies entirely within the feasible set Y, such that its initial slope is equal to .

21. Figure 8.3: Kuhn-Tucker Constraint Qualifications

22. Whenever such a curve can be drawn from every boundary point in Y and every  contained between the tangent lines, we say that the constraints defining Y satisfy the Kuhn-Tucker constraint qualification. Figure 8.3(b) illustrates a case of a cusp at which the constraint qualification does not hold. Here the two tangents to the graphs and coincide, so that 1 and2are vectors lying between these two tangents. Notice that for vector 1, it is possible to find the differentiable curve satisfying the above condition, but for vector 2no suchcurve exists. Hence, the constraint qualification does not hold for the example in figure 8.3(b).

23. 8.1.3 Theorems from Nonlinear Programming We first state the constraint qualification symbolically. For the problem defined by (8.1), (8.2), and (8.3), let Y be the set of all feasible vectors satisfying (8.2) and (8.3), i.e., Let be any point of Y and let be the vector of tight constraints at point , i.e., z includes all the g constraints in (8.2) and those constraints in (8.3) which are satisfied as equalities. Define the set

24. Then, we shall say that the constraints set Y satisfies the Kuhn-Tucker constraint qualification at if z is differentiable at and if, for every , there exists a differentiable curve defined for such that The Lagrangian function is

25. The Kuhn-Tucker conditions at for this problem are where  and  are row vectors of multipliers to be determined.

26. Theorem 8.1 (Sufficient Conditions). If h,g, and w are differentiable, h is concave, g is affine, w is concave, and solve the conditions (8.44)-(8.47), then is a solution to the maximization problem (8.1)-(8.3). Theorem 8.2 (Necessary Conditions). If h,g, and w are differentiable, and solve the maximization problem, and the constraint qualification holds at , then there exist multipliers and such that satisfy conditions (8.44)-(8.47).

27. 8.1.4 A Discrete-Time Optimal Control Problem Here, the sate xk is assumed to be measured at the beginning of period k and control uk is implemented during period k. this convention is depicted in figure 8.4.

28. We also define continuously differentiable functions f: F: g: S: Then, a discrete-time optimal problem in the Bolza form is subject to the difference equations In (8.49) the term is known as the difference operator.

29. 8.1.5 A Discrete Maximum Principle The Lagrangian function of the problem is We now define the Hamiltonian function Hk to be Using (8.52) we can rewrite (8.51) as

30. If we differentiate (8.53) with respect to xk for k =1,2,…,T-1 , we obtain which upon rearranging terms becomes If we differentiate (8.53) with respect to xT, we get The difference equations (8.54) with terminal boundary conditions (8.55) are called adjoint equations.

31. If we differentiate L with respect to k and state the corresponding Kuhn-Tucker conditions for the multiplies k and constraint(8.50), we have and

32. We note that, if Hk is concave in k, is concave in k, and the constraint qualification holds, then conditions (8.56) and (8.57) are precisely the necessary and sufficient conditions for solving the following Hamiltonian maximization problem:

33. Theorem 8.3. If for every k , Hk in (8.52) and g(uk,k) arc concave in uk , and the constraint qualification holds, then the necessary conditions for uk* , k =0,1,…,T-1, to be an optimal control for the problem (8.48)-(8.50) are

34. Example 8.7 Consider the discrete-time optimal control problem: subject to We shall solve this problem for T=6 and T 7. Solution. The Hamiltonian is

35. Let us assume, as we did in Example 2.3, that k< 0 as long as xk is positive so that k=-1. Given this assumption, (8.61) becomes , whose solution is By differentiating (8.63), we obtain the adjoint equation Let us assume T =6. Substitute (8.65) into (8.66) to obtain

36. From Appendix A.11, we find the solution to be where c is a constant. Since 6 = 0, we can obtain the value of c by setting k =6 in the above equation. Thus, so that A sketch of the value for k and xk appears in figure 8.5. Note that 5 =0, so that the control 4 is singular. However, since x4=1, we choose  4=-1 in order to bring x5 down to 0.

37. Figure 8.5: Sketch of xk and k

38. The solution of the problem for T 7 is carried out in the same way that we solved example 2.3. Namely, observe that x5 =0 and 5= 6=0, so that the control is singular. We simply make k=0 for k  7 so that  k=0 for all k  7 . It is clear without a formal proof That this maximizes (8.60). Example 8.8 Let usconsider a discrete version of the production-inventory example of Section 6.1; see Kleindorfer (1975). Let Ik, Pk and Sk be the inventory, production, and demand at time k , respectively. Let I0 be the initial inventory, let and be the goal

39. levels of inventory and production, and let h and c be inventory and production cost coefficients. The problem is: subject to Form the Hamiltonian where the adjoint variable satisfies

40. To maximize the Hamiltonian, let us differentiate (8.70) to obtain Since production must be nonnegative, we obtain the optimal production as Expressions (8.69), (8.71), and (8.72) determine a two-point boundary value problem. For a given set of data, it can be solved numerically by using a spreadsheet software EXCEL; see Section 2.5. If the constraint Pk 0 is dropped it can be solved analytically by the method of Section 6.1, with difference equations replacing the differential equations used there.

41. 8.2 A General Discrete Maximum Principle subject to Assumptions required are: (i) and are continuously differentiable in xk for every uk and k . (ii) The sets are b-directionally convex for every x and k, where b =(-1,0,…,0). That is, given  and w in and 0   1 , there exists

42. such that and for every x and k. It should be noted that convexity implies b-directional convexity, but not the converse. (iii) satisfies the Kuhn-Tucker constraint qualification.