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120 Gradient Optimization<br />

The matrix algebra rules for differentiation applied to (5.6) produce<br />

g(x)=b+Ax, (5.13)<br />

where g(x) is <strong>of</strong>ten written as VF(x), called grad F. Whichever symbol is used,<br />

g is a vector, like x; g is the gradient <strong>of</strong> F. From the central example in (5.7),<br />

the vector b and matrix A can be identified so that (5.13) can be written as:<br />

g(x) = [ ~~~]+[ -ig -m x . (5.14)<br />

Some readers may be more comfortable differentiating (5.8) with respect to<br />

both XI and x,:<br />

gl=VIF=26xl-lOx,-60, (5.15)<br />

g,=V,F=-lOxl+26x 2<br />

-132. (5.16)<br />

Appendix Program A5-2 evaluates the function value and the gradient elements<br />

(derivatives) for this particular example. The reader is urged to use that<br />

program in conjunction with Figure 5.5. Note that the gradient vectors are<br />

always perpendicular to the level curves and point in the direction <strong>of</strong> steepest<br />

ascent.<br />

Finding the minimum <strong>of</strong> an n-variable quadratic function requires setting<br />

each <strong>of</strong> the n gradient components equal to zero. This means setting (5.13)<br />

equal to vector zero; this is equivalent to setting both (5.15) and (5.16) to zero<br />

for that particular example. When (5.13) equals zero, then<br />

Ax= -b. (5.17)<br />

But the matrix inverse A-I is defined by the relationship<br />

(5.18)<br />

where the unit matrix U has all zero elements, except for l's on the main<br />

diagonal. A property <strong>of</strong> the unit matrix is that when it multiplies a vector, the<br />

result is just that vector. Multiplying both sides <strong>of</strong> (5.17) by A-I yields the x<br />

values where g(x) = 0:<br />

x=-A-Ib. (5.19)<br />

Identifying b and A by comparing (5.13) with the example in (5.14), (5.19)<br />

yields<br />

XI] _ I [26 10] [ 60] _ [5 ]<br />

[ x 2<br />

- 576 10 26 132 - 7 '<br />

(5.20)<br />

where inverse matrix A-I was found by the three conceptual steps for finding<br />

inverses: transpose, form the signed c<strong>of</strong>actors, and divide by the determinant<br />

(see any <strong>book</strong> on matrix algebra, for instance, Noble, 1969). A glance at<br />

Figure 5.5 shows that the center <strong>of</strong> the level curves is indeed at (5,7), the<br />

vector from the origin to the center. The solution (5.19) is the translation <strong>of</strong><br />

the ellipses from the origin, as shown in Figure 5.7. The rotation <strong>of</strong> the ellipse<br />

with respect to the major axes is discussed in the next section, as well as the<br />

issue <strong>of</strong> whether (5.19) determines a true minimum function point.

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