# SSM-Ch22 - Chap 22 Sec 22.1 Unconstrained Optimization...

This preview shows pages 1–3. Sign up to view the full content.

Chap . 22 Unconstrained Optimization . Linear Programming Sec . 22 . 1 Basic Concepts . Unconstrained Optimization Problem Set 22 . 1 . Page 939 3 . Cauchy s method of steepest descent . The given function is f u x U u 3 x 1 2 U 2 x 2 2 u 12 x 1 U 16 x 2 . (A) The given starting value is x 0 u ± 1 1 ² T . Proceed as in Example 1, beginning with the general formulas and using the starting value later. To simplify notations, denote the components of the gradient of f by f 1 and f 2 . The gradient of f is U f u x U u ± f 1 f 2 ² T u ± 6 x 1 u 12 4 x 2 U 16 ² T . In terms of components, f 1 u 6 x 1 u 12, f 2 u 4 x 2 U 16. (B) Furthermore, z u t U u ± z 1 z 2 ² T u x u t U f u x U u ± x 1 u tf 1 x 2 u tf 2 ² T . In terms of components, z 1 u t U u x 1 u tf 1 , z 2 u t U u x 2 u tf 2 . (C) Now obtain g u t U u f u z u t UU from f u x U in (A) by replacing x 1 with z 1 and x 2 with z 2 . This gives g u t U u 3 z 1 2 U 2 z 2 2 u 12 z 1 U 16 z 2 . Calculate the derivative of g u t U with respect to t , obtaining g ± u t U u 6 z 1 z 1 ± U 4 z 2 z 2 ± u 12 z 1 ± U 16 z 2 ± . From (C) you see that z 1 ± u u f 1 and z 2 ± u u f 2 . Substitute this and z 1 and z 2 from (C) into g ± u t U , obtaining g ± u t U u 6 u x 1 u tf 1 U u u f 1 U U 4 u x 2 u tf 2 U u u f 2 U U 12 f 1 u 16 f 2 . Order the terms as follows. Collect the terms containing t and denote their sum by D (suggesting “denominator” in what follows). This gives tD u t u 6 f 1 2 U 4 f 2 2 U . (D) Denote the sum of the other terms by N (suggesting “numerator”), obtaining

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

View Full Document
214 Optimization, Graphs Part F N u u 6 x 1 f 1 u 4 x 2 f 2 U 12 f 1 u 16 f 2 . (E) With these notations you have g ± u t U u tD U N . Solving g ± u t U u 0 for t gives t u u N D . Step 1 . For the given x u x 0 u ± 1 1 ² T you have x 1 u 1, x 2 u 1 and from (B) f 1 u u 6, f 2 u 20, tD u 1816 t , N u u 436, so that t u t 0 u u N / D u 0.240088. From this and (B) and (C) you obtain the next approximation x 1 of the desired solution in the form x 1 u z u t 0 U u ± 1 u t 0 u u 6 U 1 u t 0 U 20 ² T u ± 1 U 6 t 0 1 u 20 t 0 ² T u ± 2.44053 u 3.80176 ² T . This completes the first step. Step 2 . Instead of x 0 now use x 1 , in terms of components, x 1 u 2.44053, x 2 u u 3.80176, and do the computations in Step 2 with these data. And so on. The results for the first six steps are as follows. n x f 0 1.00000 1.00000 9.0000 1 2.44053 u 3.80176 u 43.3392 2 1.98753 u 3.93766 u 43.9918 3 2.00549 u 3.99753 u 44.0000 4 1.99985 u 3.99922 u 43.9999 5 2.00006 u 3.99994 u 44.0000 6 2.00006 u 3.99994 u 44.0000 You can see that the convergence is fast, and you also notice the effect of roundoff in the last few digits. Completing squares you find that f can be written f u x U u 3 u x 1 u 2 U 2 U 2 u x 2 U 4 U 2 u 44. This explains the numeric results. It also shows that the level curves
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

SSM-Ch22 - Chap 22 Sec 22.1 Unconstrained Optimization...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online