Now suppose we want to solve the problem defined by (8.1)–(8.3). As before, we define the Lagrangian
L(x, λ, μ) =h(x) +λ[f(x)−a] +μ[g(x)−b]. (8.9) The Kuhn-Tucker necessary conditions for this problem cannot be as easily derived as for the equality-constrained problem in the preceding section. We will write them first, and then give interpretations to make them plausible. The necessary conditions forx∗ to be a solution of (8.1)–
(8.3) are that there exist anr-dimensional vectorλand ans-dimensional row vector μsuch that
Lx =hx+λfx+μgx = 0, (8.10)
f = a, (8.11)
g ≥ b, (8.12)
μ≥0, μ(g−b) = 0. (8.13)
8.1. Nonlinear Programming Problems 263 Note thatg is appended in (8.10) in the same way f is appended in (8.6). Also (8.12) repeats the inequality constraint (8.3) in the same way that (8.11) repeats the equality constraint (8.2). However, the conditions in (8.13) are new and particular to the inequality-constrained problem.
We will see that they include some of the boundary points of the feasible set of points as well as unconstrained maximum solution points, as can- didates for the solution to the maximum problem. This is best brought out by examples.
Example 8.2 Solve the problem:
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⎪⎪
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⎨
⎪⎪
⎪⎪
⎪⎩
max{h(x) = 8x−x2} subject to
x≥2.
Solution We form the Lagrangian
L(x, μ) = 8x−x2+μ(x−2).
The necessary conditions (8.10)–(8.13) become
Lx = 8−2x+μ= 0, (8.14)
x−2≥0, (8.15)
μ≥0, μ(x−2) = 0. (8.16)
Observe that the constraint μ(x−2) = 0 in (8.16) can be phrased as:
either μ= 0 or x= 2.We treat these two cases separately.
Case 1: μ = 0. From (8.14) we get x = 4, which also satisfies (8.15).
Hence, this solution, which makes h(4) = 16,is a possible candidate for the maximum solution.
Case 2: x= 2.Here from (8.14) we getμ=−4,which does not satisfy the inequalityμ≥0 in (8.16).
From these two cases we conclude that the optimum solution isx∗ = 4 and h∗=h(x∗) = 16.
264 8. The Maximum Principal: Discrete Time Example 8.3 Solve the problem:
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⎪⎩
max{h(x) = 8x−x2} subject to
x≥6.
Solution The Lagrangian is
L(x, μ) = 8x−x2+μ(x−6).
The necessary conditions are
Lx= 8−2x+μ = 0, (8.17)
x−6 ≥ 0, (8.18)
μ≥0, μ(x−6) = 0. (8.19)
Again, the condition μ(x−6) = 0 is an either-or relation which gives two cases.
Case 1: μ = 0. From (8.17) we obtain x = 4, which does not satisfy (8.18), so this case is infeasible.
Case 2: x = 6. Obviously (8.18) holds. From (8.17) we get μ = 4, so (8.19) holds as well. The optimal solution is then
x∗ = 6, h∗=h(x∗) = 12,
since it is the only solution satisfying the necessary conditions.
The examples above involve only one variable, and are relatively obvious. The next example, which is two-dimensional, will reveal more of the power and the difficulties of applying the Kuhn-Tucker conditions.
Example 8.4 Find the shortest distance between the point (2,2) and the upper half of the semicircle of radius one with its center at the origin, shown as the curve in Fig.8.1. In order to simplify the calculation, we minimizeh,the square of the distance. Hence, the problem can be stated
8.1. Nonlinear Programming Problems 265 as the following nonlinear programming problem:
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⎨
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⎪⎪
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⎪⎩ max-
−h(x, y) =−(x−2)2−(y−2)2. subject to
x2+y2= 1, y≥0.
The Lagrangian function for this problem is
L=−(x−2)2−(y−2)2+λ(x2+y2−1) +μy. (8.20) The necessary conditions are
−2(x−2) + 2λx= 0, (8.21)
−2(y−2) + 2λy+μ= 0, (8.22)
x2+y2−1 = 0, (8.23)
y≥0, (8.24)
μ≥0, μy= 0. (8.25)
First, we conclude that λ = 0, since otherwise λ = 0 would imply x= 2 from (8.21), which would contradict (8.23). Next, from (8.25) we conclude that eitherμ= 0 ory = 0.Ifμ= 0,then from (8.21) and (8.22), we getx=y.Solving the equationx=ytogether withx2+y2= 1 gives:
(a) (1/√ 2,1/√
2) and h=−(9−4√ 2).
If y= 0,then solving with x2+y2 = 1 gives:
(b) (1,0) and h=−5, (c) (−1,0) and h=−13.
These three points are shown in Fig.8.1. Of the three points found that satisfy the necessary conditions, clearly the point (1/√
2,1/√
2) 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. The associated multiplier values can be easily computed, and these are: (a) λ= 1−2√
2, μ = 0;
(b) λ=−1, μ= 4; and (c) λ= 3, μ= 4.
266 8. The Maximum Principal: Discrete Time
(-1,0) (1,0)
Farthest Point
Closest Point
Figure 8.1: Shortest distance from point (2,2) to the semicircle The fact that there are three points satisfying the necessary condi- tions, and only one of them actually solves the problem at hand, empha- sizes that the conditions are only necessary and not sufficient. In every case it is important to check the solutions to the necessary conditions to see which of the solutions provides the optimum.
Next we work two examples that show some technical difficulties that can arise in the application of the Kuhn-Tucker conditions.
Example 8.5 Consider the problem:
max{h(x, y) =y} (8.26)
subject to
(1−y)3−x2 ≥ 0, (8.27)
x ≥ 0 (8.28)
y ≥ 0. (8.29)
The set of points satisfying the constraints is shown shaded in Fig.8.2.
From the figure it is obvious that the solution point (0,1) maximizes the value of y.
Hence, the optimum solution is (x∗, y∗) = (0,1) and h∗ = 1. Let us see if we can find it using the above procedure. The Lagrangian is
L=y+λ[(1−y)3−x2] +μx+νy, (8.30)
8.1. Nonlinear Programming Problems 267
Figure 8.2: Graph of Example8.5 so that the necessary conditions are
Lx = −2xλ+μ= 0, (8.31)
Ly = 1−3λ(1−y)2+ν= 0, (8.32) λ ≥ 0, λ[(1−y)3−x2] = 0, (8.33)
μ ≥ 0, μx= 0, (8.34)
ν ≥ 0, νy= 0, (8.35)
together with (8.27)–(8.29). Let us check if these conditions hold at the point (0,1). At y = 1, the constraint y ≥ 0 is not active, and we have ν = 0.With ν= 0 and y= 1,(8.32) cannot be satisfied.
The reason for failure of the method in Example8.5is that the con- straints do not satisfy what is called theconstraint qualification. A com- plete study of the topic is beyond the scope of this book, but we state in the next section a constraint qualification sufficient for our purposes.
For further information, see Mangasarian (1969).