?1.717??0??1.717?T(5)(5) X(5)?X??S?????1.717,3.178????212?3.178??1??3.178?????????(5)(5)F(?)?f(X1??S2)?4?4.5?1.717?4?(3.178??)?1.7172?2?(3.178??)2?2?1.717?(3.178??)?1.7174?2?1.7172?(3.178??)由
dF(?)?0可得??0.155 d?(5)2故X?X??S(5)1(5)2?1.717??0??1.717?T ???0.155??1.717,3.333????1??3.333?3.178??????而函数值：
2242f(X(5)2)?4?4.5?1.717?4?3.333?1.717?2?3.333?2?1.717?3.333?1.717?2?1.717?3.333?1.154函数下降量：
(4)f(X(5)1.154?1.2622)?f(X2)??0.0856 (4)f(X2)1.262
第六轮搜索：
先从X2点出发，沿着S15)因为 X(2??S1(6)(5)(6)(6)?[1,0]T方向进行一维最优化搜索，求X1点，
?1.717??1??1.717???T ??????1.717??,3.333????????3.333??0??3.333?所以
(6)2242F(?)?f(X(5)2??S1)?4?4.5(1.717??)?4?3.333?(1.717??)?2?3.333?2?(1.717??)?3.333?(1.717??)?2?(1.717??)?3.333dF(?)??21.623044?11.332??4(1.717??)3 d?dF(?)由?0可得??0.055
d?(6)(6)(6)则X1?X(5)??2??1S1?1.717??1??1.772?T ?0.055??1.772,3.333????0??3.333?3.333??????而函数值：
(6)f(X1)?4?4.5?1.772?4?3.333?1.7722?2?3.3332?2?1.772?3.333?1.7724?2?1.7722?3.333?1.116再从X1点出发，沿S2(6)(6)?[0,1]T方向进行一维最优化搜索，求X(6)2点：
X(6)2?X(6)1??S(6)2?1.772??0??1.772?T ??????1.772,3.333??????1??3.333???3.333??????(6)(6)F(?)?f(X1??S2)?4?4.5?1.772?4?(3.333??)?1.7722?2?(3.333??)2?2?1.772?(3.333??)?1.7724?2?1.7722?(3.333??)由
dF(?)?0可得??0.123 d?word文档 可自由复制编辑
(6)(6)故X(6)?X??S212???1.772??0??1.772?T ?0.123??1.772,3.456????????3.333??1??3.456?而函数值：
2242f(X(6)2)?4?4.5?1.772?4?3.456?1.772?2?3.456?2?1.772?3.456?1.772?2?1.772?3.456?1.086函数下降量：
(5)f(X(6))?f(X1.086?1.15422)??0.0589 (5)f(X2)1.154
第七轮搜索：
先从X2点出发，沿着S16)因为 X(2??S1(7)(6)(7)(7)?[1,0]T方向进行一维最优化搜索，求X1点，
?1.772??1??1.772???T ??????1.772??,3.456????????3.456??0??3.456?所以
(7)2242F(?)?f(X(6)2??S1)?4?4.5(1.772??)?4?3.456?(1.772??)?2?3.456?2?(1.772??)?3.456?(1.772??)?2?(1.772??)?3.456dF(?)??23.364128?11.824??4(1.772??)3 d?dF(?)由?0可得??0.041
d?(7)(7)(7)则X1?X(6)??2??1S1?1.772??1??1.813?T ?0.041??1.772,3.456????0??3.456?3.456??????而函数值：
(7)f(X1)?4?4.5?1.813?4?3.456?1.8132?2?3.4562?2?1.813?3.456?1.8134?2?1.8132?3.456?1.063再从X1点出发，沿S2(7)(7)?[0,1]T方向进行一维最优化搜索，求X(7)2点：
X(7)2?X(7)1??S(7)2?1.813??0??1.813?T ??????1.813,3.456??????1??3.456???3.456??????(7)(7)F(?)?f(X1??S2)?4?4.5?1.813?4?(3.456??)?1.8132?2?(3.456??)2?2?1.813?(3.456??)?1.8134?2?1.8132?(3.456??)由
dF(?)?0可得??0.094 d?(7)2故X?X(7)1??S(7)2?1.813??0??1.813?T ???0.094??1.813,3.550????1??3.550?3.456??????而函数值：
2242f(X(7)2)?4?4.5?1.813?4?3.550?1.813?2?3.550?2?1.813?3.550?1.813?2?1.813?3.550?1.045word文档 可自由复制编辑
函数下降量：
(6)f(X(7))?f(X1.045?1.08622)??0.0295 (6)f(X2)1.086
第八轮搜索：
先从X2点出发，沿着S17)因为 X(2??S1(8)(7)(8)(8)?[1,0]T方向进行一维最优化搜索，求X1点，
?1.813??1??1.813???T ??????1.813??,3.550????????3.550??0??3.550?所以
(8)2242F(?)?f(X(7)2??S1)?4?4.5(1.813??)?4?3.550?(1.813??)?2?3.550?2?(1.813??)?3.550?(1.813??)?2?(1.813??)?3.550dF(?)??24.718600?12.200??4(1.813??)3 d?dF(?)由?0可得??0.032
d?则X(8)1?X(7)2??S(8)(8)11?1.813??1??1.845?T ???0.032??1.845,3.550????0??3.550?3.550??????而函数值：
(8)f(X1)?4?4.5?1.845?4?3.550?1.8452?2?3.5502?2?1.845?3.550?1.8454?2?1.8452?3.550?1.031(8)(8)T(8)再从X1点出发，沿S2?[0,1]方向进行一维最优化搜索，求X2点：
X(8)2?X??S(8)1(8)2?1.845??0??1.845?T ??????1.845,3.550??????1??3.550???3.550??????(8)(8)F(?)?f(X1??S2)?4?4.5?1.845?4?(3.550??)?1.8452?2?(3.550??)2?2?1.845?(3.550??)?1.8454?2?1.8452?(3.550??)由
dF(?)?0可得??0.075 d?(8)2故X?X??S(8)1(8)2?1.845??0??1.845?T ???0.075??1.845,3.625????1??3.625?3.550??????而函数值：
2242f(X(8)2)?4?4.5?1.845?4?3.625?1.845?2?3.625?2?1.845?3.625?1.845?2?1.845?3.625?1.020函数下降量：
(7)f(X(8)1.020?1.0452)?f(X2)??0.0239
f(X(7))1.0452
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将以上各跌代计算结果整理得如下表格： 迭代次数 1 2 3 4 5 6 7 8 迭代点 [1.296，2.2]T [1.296，2.488]T [1.439，2.488]T [1.439，2.755]T [1.555，2.755]T [1.555，2.987]T [1.646，2.987]T [1.646，3.178]T T[1.717，3.178] [1.717，3.333]T [1.772，3.333]T [1.772，3.456]T [1.813，3.456]T [1.813，3.550]T [1.845，3.550]T [1.845，3.625]T 步长 -0.704 0.288 0.143 0.267 0.116 0.232 0.091 0.191 0.071 0.155 0.055 0.123 0.041 0.094 0.032 0.075 函数值 2.120 1.954 1.798 1.656 1.531 1.424 1.334 1.262 1.202 1.154 1.116 1.086 1.063 1.045 1.031 1.020 函数下降量 0.2598 0.1525 0.1401 0.1138 0.0856 0.0589 0.0295 0.0239 f(X(k))?f(X(k?1))在函数下降量的终止准则?0.03的基础上，结合优化计算结果可知，
f(X(k?1))所求最优点为[1.845，3.625]，对应的函数值为f(X)?1.020。
4-2解：
T
解：已知初始点X0(1)?X(0)?[?24]T，因此f(X0(1))?26。
(1)第一轮搜索方向，取：S1(1)?[10]T,S2(1)?[01]T。从初始点X0(1)?[?24]T出
发，沿S1方向进行一维最优化搜索，求X1点
即求解：f(X0(1)(1)(1)??1S1)?minf(X0（1）(1)(1) ??S1）?minF（?）(1)因为X0??S(1)1=
??2??1?T?4????0??[??24]故 ????(1)F（?）?f(X0(1)2??S1）?1.（5??2）?0.5?16?（??2）?4?（2??2）?1.5?2?12??26(1)(1)dF（?）df(X0??S1）（1）??0得3??12?0，??4??1。所以 令
d?d?word文档 可自由复制编辑
X1=X0(1)(1)??（1）(1)11=
S??2??1??2?(1)f(X)?2 函数值?4?1?4??0??4???????(1)再从X1出发，沿S2方向进行一维最优化搜索，求X2点；
X1(1)(1)(1)??S2(1)?2??0?T=???????[24??] ?4??1?(1)2??S2）?1.5?4?0.5?（4??）?2?（4??）?4?0.5?2?2??2
(1)F（?）?f(X1令
df(X1(1)d?(1)(1)??S2）（1）?0得??2?0 ，???2??2所以
(1)X2=X1?0?（1）(1)?2?(1)??2S2=???2???[22]T函数值f(X2)?0
?4??1?(1)计算X3=2X2(1)?2???2??6?(1)(1)?X0，得X3=2????????，
?2??4??0?(1)(1)(1)f1?f(X0)?26，f2?f(X2)?0，f3?f(X3)?42
f(X0所以 ?m（1）(1))-f(X1)=26-2=24，f(X1)-f(X2)=2-0=2
(1)(1)(1)(1)(1)?f(X0)-f(X1)=26-2=24，m=1.
用判别准则检验：f3?42?f1?26
据此取：S1(2)?S1(1)?[10]T,S2(2)?S2(1)?[01]T。由于f2?0?f3?26，第二轮
搜索的初始点取：X0(2)?X2(1)?[22]T，f(X0(2))?0.
从初始点X0(2)?[22]T出发，沿S1(2)方向进行一维最优化搜索，求X1(2)点。 ?2??2??1??0?(2)X0(2)??S1(2)=????????(2)?2??? ??2?F（?）?f(X02??S1）?1.（52??）?0.5?4?（2??）?2?（22??）?1.5?2?2? (2)令
df(X0(2)??S1）2（2）?0得3??2?0 ，?????1所以
d?3X1(2)=X0(2)42?1?（2）(2)?2???1S1=??????[3?2?3?0?2(2)2]T函数值f(X1)??
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机械优化设计课后习题答案
