292 Nonlinear Programming I: One-Dimensional Minimization Methods
f′(l)
l
l~ 1 *l
~
2 * l~ 3 *
Figure 5.20 Situation whenfA′varies very slowly.
Remarks:
1.The secant method is identical to assuming a linear equation forf′(λ) This.
implies that the original function,f (λ), is approximated by a quadratic equation.
2.In some cases we may encounter a situation where the functionf′(λ) varies
very slowly withλ, as shown in Fig. 5.20. This situation can be identified
by noticing that the pointBremains unaltered for several consecutive refits.
Once such a situation is suspected, the convergence process can be improved
by taking the next value ofλi+ 1 as ( A+B)/2 instead of finding its value from
Eq. (5.74).
Example 5.14 Find the minimum of the function
f (λ)= 0. 65 −
0. 75
1 +λ^2
− 0. 65 λtan−^1
1
λ
using the secant method with an initial step size oft 0 = 0. 1 ,λ 1 = 0. 0 , andε= 0 .01.
SOLUTION λ 1 = A= 0. 0 , t 0 = 0. 1 , f′(A) =− 1 .02102, B=A+t 0 = 0. 1 ,
f′ (B)=− 0 .744832. Sincef′(B) < 0 , we set newA= 0 .1,f′(A) =− 0 .744832,t 0 =
2 ( 0. 1 )= 0. 2 ,B=λ 1 +t 0 = 0. 2 , and computef′(B) =− 0 .490343. Sincef′(B) < 0 ,
we set new A= 0 .2, f′(A) =− 0 .490343, t 0 = 2 ( 0. 2 )= 0. 4 , B=λ 1 +t 0 = 0. 4 ,
and computef′(B) =− 0 .103652. Sincef′(B) < 0 , we set newA= 0 .4,f′(A)=
− 0. 1 03652,t 0 = 2 ( 0. 4 )= 0. 8 ,B=λ 1 +t 0 = 0. 8 , and computef′(B) =+ 0 .180800.
Sincef′(B) > 0 , we proceed to findλ 2.
Iteration 1
SinceA=λ 1 = 0. 4 ,f′(A) =− 0. 103652 , B= 0. 8 , f′(B) =+ 0 .180800, we compute
λ 2 =A−
f′(A)(B −A)
f′(B)−f′(A)
= 0. 545757
Convergence check:|f′(λ 2 ) = |+| 0.0105789|>ε.