6.7 Simplex Method 333SOLUTION
Iteration 1
Step 1: The function value at each of the vertices of the current simplex is given by
f 1 =f(X 1 )= 4. 0 − 4. 0 + 2 ( 16. 0 )+ 2 ( 16. 0 )+ 16. 0 = 80. 0
f 2 =f(X 2 )= 5. 0 − 4. 0 + 2 ( 25. 0 )+ 2 ( 20. 0 )+ 16. 0 = 107. 0
f 3 =f(X 3 )= 4. 0 − 5. 0 + 2 ( 16. 0 )+ 2 ( 20. 0 )+ 25. 0 = 96. 0Therefore,Xh=X 2 ={
5. 0
4. 0
}
, f (Xh) = 107. 0 ,Xl=X 1 ={
4. 0
4. 0
}
, and f (Xl) = 80. 0Step 2: The centroidX 0 is obtained as
X 0 =1
2
(X 1 +X 3 )=
1
2
{
4. 0 + 4. 0
4. 0 + 5. 0
}
=
{
4. 0
4. 5
}
with f (X 0 ) = 87. 75Step 3: The reflection point is found as
Xr= 2 X 0 −Xh={
8. 0
9. 0
}
−
{
5. 0
4. 0
}
=
{
3. 0
5. 0
}
Thenf (Xr)= 3. 0 − 5. 0 + 2 ( 9. 0 )+ 2 ( 15. 0 )+ 25. 0 = 71. 0Step 4: Asf (Xr) < f (Xl) we find, Xeby expansion as
Xe= 2 Xr−X 0 ={
6. 0
10. 0
}
−
{
4. 0
4. 5
}
=
{
2. 0
5. 5
}
Then
f (Xe)= 2. 0 − 5. 5 + 2 ( 4. 0 )+ 2 ( 11. 0 )+ 30. 25 = 56. 75Step 5: Sincef (Xe) < f (Xl) we replace, XhbyXeand obtain the vertices of the new
simplex as
X 1 ={
4. 0
4. 0
}
, X 2 =
{
2. 0
5. 5
}
, and X 3 ={
4. 0
5. 0
}
Step 6: To test for convergence, we compute
Q=
[
( 80. 0 − 87. 75 )^2 + 6 ( 5. 75 − 87. 75 )^2 + 6 ( 9. 0 − 87. 75 )^2
3
] 1 / 2
= 91. 06
As this quantity is not smaller thanε, we go to the next iteration.