744 Practical Aspects of Optimization
Table 14.2ExactY 0 =
0 .116462E− 02
0 .232923E− 02
0 .514654E− 01
− 0 .703216E− 01
Exact(Y 0 +Y)=
0 .970515E− 03
0 .194103E− 02
0 .428879E− 01
− 0 .586014E− 01
Value ofi Yi Yi=Y 0 +∑i
k= 1Yk1
− 0 .232922E− 03
− 0 .465844E− 03
− 0 .102930E− 01
0 .140642E− 01
0 .931695E− 03
0 .186339E− 02
0 .411724E− 01
− 0 .562573E− 01
2
0 .465842E− 04
0 .931683E− 04
0 .205859E− 02
− 0 .281283E− 02
0 .978279E− 03
0 .195656E− 02
0 .432310E− 01
− 0 .590702E− 01
3
− 0 .931678E− 05
− 0 .186335E− 04
− 0 .411716E− 03
0 .562563E− 03
0 .968962E− 03
0 .193792E− 02
0 .428193E− 01
− 0 .585076E− 01
4
0 .186335E− 05
0 .372669E− 05
0 .823429E− 04
− 0 .112512E− 03
0 .970825E− 03
0 .194165E− 02
0 .429016E− 01
− 0 .586201E− 01
If we find the exact solution atr basicdesign vectorsX 1 ,X 2 ,... ,Xr, the corresponding
solutions,Yi, are found by solving the equations[Ki]Yi= P, i= 1 , 2 ,... , r (14.24)where the stiffness matrix, [Ki], is determined at the design vectorXi. If we consider a
new design vector,XN, in the neighborhood of the basic design vectors, the equilib rium
equations atXNcan be expressed as[KN]YN=P (14.25)where [KN] is the stiffness matrix evaluated atXN. By approximatingYNas a linear
combinationof the basic displacement vectorsYi, i = 1 , 2 ,... , r, we haveYN≈c 1 Y 1 +c 2 Y 2 + · · · +crYr=[Y]c (14.26)where [Y]=[Y 1 ,Y 2 , · ·· ,Yr] is an n×r matrix andc= {c 1 , c 2 , · ·· , cr}T is an
r-component column vector. Substitution of Eq. (14.26) into Eq. (14.25) gives[KN][Y]c=P (14.27)