334 Nonlinear Programming II: Unconstrained Optimization Techniques
Iteration 2Step 1: Asf (X 1 ) = 80. 0 , f (X 2 ) = 56 .75, andf (X 3 ) = 96 .0,Xh=X 3 ={
4. 0
5. 0
}
and Xl=X 2 ={
2. 0
5. 5
}
Step 2:The centroid isX 0 =
1
2
(X 1 +X 2 )=
1
2
{
4. 0 + 2. 0
4. 0 + 5. 5
}
=
{
3. 0
4. 75
}
f (X 0 ) = 67. 31Step 3:Xr= 2 X 0 −Xh={
6. 0
9. 5
}
−
{
4. 0
5. 0
}
=
{
2. 0
4. 5
}
f(Xr)= 2. 0 − 4. 5 + 2 ( 4. 0 )+ 2 ( 9. 0 )+ 20. 25 = 43. 75Step 4: Asf (Xr) < f (Xl) we find, XeasXe= 2 Xr−X 0 ={
4. 0
9. 0
}
−
{
3. 0
4. 75
}
=
{
1. 0
4. 25
}
f (Xe)= 1. 0 − 4. 25 + 2 ( 1. 0 )+ 2 ( 4. 25 )+ 18. 0625 = 25. 3125Step 5: Asf (Xe) < f (Xl) we replace, XhbyXeand obtain the new vertices asX 1 =
{
4. 0
4. 0
}
, X 2 =
{
2. 0
5. 5
}
, and X 3 ={
1. 0
4. 25
}
Step 6: For convergence, we computeQasQ=
[
( 80. 0 − 67. 31 )^2 + 6 ( 5. 75 − 67. 31 )^2 + 5 ( 2. 3125 − 67. 31 )^2
3
] 1 / 2
= 62. 1
SinceQ>ε, we go to the next iteration.
This procedure can be continued until the specified convergence is satisfied. When
the convergence is satisfied, the centroidX 0 of the latest simplex can be taken as the
optimum point.