6.4 Univariate Method 317Step 3: To find whether the value off decreases alongS 1 or−S 1 , we use the probe
lengthε. Since
f 1 =f(X 1 ) =f( 0 , 0 )= 0 ,
f+ =f(X 1 +εS 1 ) =f(ε, 0 )= 0. 01 − 0 + 2 ( 0. 0001 )
+ 0 + 0 = 0. 0102 >f 1f− =f(X 1 −εS 1 ) =f(−ε, 0 )= − 0. 01 − 0 + 2 ( 0. 0001 )
+ 0 + 0 = − 0. 0098 < f 1 ,−S 1 is the correct direction for minimizingf fromX 1.
Step 4: To find the optimum step lengthλ∗ 1 , we minimize
f(X 1 −λ 1 S 1 ) =f(−λ 1 , 0 )=(−λ 1 )− 0 + 2 (−λ 1 )^2 + 0 + 0 = 2 λ^21 −λ 1As df/dλ 1 = at 0 λ 1 =^14 , we haveλ∗ 1 =^14.
Step 5: Set
X 2 =X 1 −λ∗ 1 S 1 ={
0
0
}
−^14
{
1
0
}
=
{
−^14
0
}
f 2 = f(X 2 ) =f(−^14 , 0 )=−^18.Iterationi= 2
Step 2: Choose the search directionS 2 asS 2 =
{ 0
1}
.
Step 3: Sincef 2 = f(X 2 ) =− 0 .125,
f+ =f(X 2 +εS 2 ) =f(− 0. 25 , 0. 01 )= − 0. 1399 < f 2
f− =f(X 2 +εS 2 ) =f(− 0. 25 ,− 0. 01 )= − 0. 1099 >f 2S 2 is the correct direction for decreasing the value off fromX 2.
Step 4: We minimizef (X 2 +λ 2 S 2 ) o findt λ∗ 2.
Here
f (X 2 +λ 2 S 2 ) =f(− 0. 25 , λ 2 )
=− 0. 25 −λ 2 + 2 ( 0. 25 )^2 − 2 ( 0. 25 )(λ 2 )+λ^22=λ^22 − 1. 5 λ 2 − 0. 125
df
dλ 2= 2 λ 2 − 1. 5 = 0 at λ∗ 2 = 0. 75Step 5: Set
X 3 =X 2 +λ∗ 2 S 2 ={
− 0. 25
0
}
+ 0. 75
{
0
1
}
=
{
− 0. 25
0. 75
}
f 3 = f(X 3 ) =− 0. 6875