1、實驗7 無約束優(yōu)化題目5 某分子由25個原子組成，并且已經(jīng)通過實驗測量得到了其中某些原子對之間的距離（假設(shè)在平面結(jié)構(gòu)上討論），如表7.8所示。請你確定每個原子的位置關(guān)系。表7.8原子對距離原子對距離原子對距離原子對距離（4，1）0.9607(5,4)0.4758(18,8)0.8363(15,13)0.5725（12,1）0.4399(12,4)1.3402(13,9)0.3208(19,13)0.7660(13,1)0.8143(24,4)0.7006(15,9)0.1574(15,14)0.4394(17,1)1.3765(8,6)0.4945(22,9)1.2736(16,14)1.09

2、52(21,1)1.2722(13,6)1.0559(11,10)0.5781(20,16)1.0422(5,2)0.5294(19,6)0.6810(13,10)0.9254(23,16)1.8255(16,2)0.6144(25,6)0.3587(19,10)0.6401(18,17)1.4325(17,2)0.3766(8,7)0.3351(20,10)0.2467(19,17)1.0851(25,2)0.6893(14,7)0.2878(22,10)0.4727(20,19)0.4995(5,3)0.9488(16,7)1.1346(18,11)1.3840(23,19)1.2277(

3、20,3)0.8000(20,7)0.3870(25,11)0.4366(24,19)1.1271(21,3)1.1090(21,7)0.7511(15,12)1.0307(23,21)0.7060(24,3)1.1432(14,8)0.4439(17,12)1.3904(23,22)0.8025【模型建立】設(shè)第i個點所在的位置為，因為所求的是各原子間的位置關(guān)系，可以設(shè)定第一個點坐標(biāo)為，然后再計算其他原子的位置，使它在最大程度上滿足上表中提供的數(shù)據(jù)，即讓達(dá)到最小，其中表示第i個原子和第j個原子之間的距離，數(shù)據(jù)如上表中所示。問題轉(zhuǎn)化為無約束優(yōu)化：【算法設(shè)計】上面是一個無約束優(yōu)化問題，要求可以歸結(jié)

4、到無約束規(guī)劃模型min f(x)，令x=0,x2,x3,x4,x25,0,y2,y3,y25,調(diào)用基本命令fminunc來做，也可以令,調(diào)用lsqnonlin命令來做?！境绦颉糠椒ㄒ?用lsqnonlin命令實現(xiàn)function f=distance(x,d) f(1)=(x(1,3)2+(x(2,3)2-d(1)2; %對應(yīng)條件第4個和第1個原子間距f(2)=(x(1,11)2+(x(2,11)2-d(2)2;f(3)=(x(1,12)2+(x(2,12)2-d(3)2;f(4)=(x(1,16)2+(x(2,16)2-d(4)2;f(5)=(x(1,20)2+(x(2,20)2-d(5)2

5、;f(6)=(x(1,4)-x(1,1)2+(x(2，4)-x(2,1)2-d(6)2;f(7)=(x(1,15)-x(1,1)2+(x(2,15)-x(2,1)2-d(7)2;f(8)=(x(1,16)-x(1,1)2+(x(2,16)-x(2,1)2-d(8)2;f(9)=(x(1,24)-x(1,1)2+(x(2,24)-x(2,1)2-d(9)2;f(10)=(x(1,4)-x(1,2)2+(x(2,4)-x(2,2)2-d(10)2;f(11)=(x(1,19)-x(1,2)2+(x(2,19)-x(2,2)2-d(11)2;f(12)=(x(1,20)-x(1,2)2+(x(2,2

6、0)-x(2,2)2-d(12)2;f(13)=(x(1,23)-x(1,2)2+(x(2,23)-x(2,2)2-d(13)2;f(14)=(x(1,4)-x(1,3)2+(x(2,4)-x(2,3)2-d(14)2;f(15)=(x(1,11)-x(1,3)2+(x(2,11)-x(2,3)2-d(15)2;f(16)=(x(1,23)-x(1,3)2+(x(2,23)-x(2,3)2-d(16)2;f(17)=(x(1,7)-x(1,5)2+(x(2,7)-x(2,5)2-d(17)2;f(18)=(x(1,12)-x(1,5)2+(x(2,12)-x(2,5)2-d(18)2;f(19

7、)=(x(1,18)-x(1,5)2+(x(2,18)-x(2,5)2-d(19)2;f(20)=(x(1,24)-x(1,5)2+(x(2,24)-x(2,5)2-d(20)2;f(21)=(x(1,7)-x(1,6)2+(x(2,7)-x(2,6)2-d(21)2;f(22)=(x(1,13)-x(1,6)2+(x(2,13)-x(2,6)2-d(22)2;f(23)=(x(1,15)-x(1,6)2+(x(2,15)-x(2,6)2-d(23)2;f(24)=(x(1,19)-x(1,6)2+(x(2,19)-x(2,6)2-d(24)2;f(25)=(x(1,20)-x(1,6)2+(

8、x(2,20)-x(2,6)2-d(25)2;f(26)=(x(1,13)-x(1,7)2+(x(2,13)-x(2,7)2-d(26)2;f(27)=(x(1,17)-x(1,7)2+(x(2,17)-x(2,7)2-d(27)2;f(28)=(x(1,12)-x(1,8)2+(x(2,12)-x(2,8)2-d(28)2;f(29)=(x(1,14)-x(1,8)2+(x(2,14)-x(2,8)2-d(29)2;f(30)=(x(1,21)-x(1,8)2+(x(2,21)-x(2,8)2-d(30)2;f(31)=(x(1,10)-x(1,9)2+(x(2,10)-x(2,9)2-d(

9、31)2;f(32)=(x(1,12)-x(1,9)2+(x(2,12)-x(2,9)2-d(32)2;f(33)=(x(1,18)-x(1,9)2+(x(2,18)-x(2,9)2-d(33)2;f(34)=(x(1,19)-x(1,9)2+(x(2,19)-x(2,9)2-d(34)2;f(35)=(x(1,21)-x(1,9)2+(x(2,21)-x(2,9)2-d(35)2;f(36)=(x(1,17)-x(1,10)2+(x(2,17)-x(2,10)2-d(36)2;f(37)=(x(1,24)-x(1,10)2+(x(2,24)-x(2,10)2-d(37)2;f(38)=(x(

10、1,14)-x(1,11)2+(x(2,14)-x(2,11)2-d(38)2;f(39)=(x(1,16)-x(1,11)2+(x(2,16)-x(2,11)2-d(39)2;f(40)=(x(1,14)-x(1,12)2+(x(2,14)-x(2,12)2-d(40)2;f(41)=(x(1,18)-x(1,12)2+(x(2,18)-x(2,12)2-d(41)2;f(42)=(x(1,14)-x(1,13)2+(x(2,14)-x(2,13)2-d(42)2;f(43)=(x(1,15)-x(1,13)2+(x(2,15)-x(2,13)2-d(43)2;f(44)=(x(1,19)-

11、x(1,15)2+(x(2,19)-x(2,15)2-d(44)2;f(45)=(x(1,22)-x(1,15)2+(x(2,22)-x(2,15)2-d(45)2;f(46)=(x(1,17)-x(1,16)2+(x(2,17)-x(2,16)2-d(46)2;f(47)=(x(1,18)-x(1,16)2+(x(2,18)-x(2,16)2-d(47)2;f(48)=(x(1,19)-x(1,18)2+(x(2,19)-x(2,18)2-d(48)2;f(49)=(x(1,22)-x(1,18)2+(x(2,22)-x(2,18)2-d(49)2;f(50)=(x(1,23)-x(1,18

12、)2+(x(2,23)-x(2,18)2-d(50)2;f(51)=(x(1,22)-x(1,20)2+(x(2,22)-x(2,20)2-d(51)2;f(52)=(x(1,22)-x(1,21)2+(x(2,22)-x(2,21)2-d(52)2;clear allx0=zeros(1,24);ones(1,24);d=0.9607,0.4399,0.8143,1.3765,1.2722,0.5294,0.6144,0.3766,0.6893,. 0.9488,0.8000,1.1090,1.1432,0.4758,1.3402,0.7006,0.4945,1.0559,. 0.6810,

13、0.3587,0.3351,0.2878,1.1346,0.3870,0.7511,0.4439,0.8363,. 0.3208,0.1574,1.2736,0.5781,0.9254,0.6401,0.2467,0.4727,1.3840,. 0.4366,1.0307,1.3904,0.5725,0.7660,0.4394,1.0952,1.0422,1.8255,. 1.4325,1.0851,0.4995,1.2277,1.1271,0.7060,0.8052;%設(shè)定初值x,norms=lsqnonlin(distance,x0,d);p=x; %p第一行即為第二個原子的橫、縱坐標(biāo)；p

14、第二行為第三個原子的橫、縱坐標(biāo)a=0,x(1,:);b=0,x(2,:);plot(a,b,*); %畫散點圖表示出原子的位置【輸出結(jié)果】每個原子的位置如下（即第二個原子的位置為（1.2378,0.0746），第三個原子為（1,6142,1,。1211），）：p = 1.2378 0.0746 1.6142 1.1211 0.5485 0.7778 0.9121 0.4836 0.8546 1.1533 0.7009 0.4533 0.9871 0.6573 0.4694 0.2363 0.8326 1.0437 0.4356 0.6211 -0.0650 -0.4184 0.8165 0.1

15、140 0.7351 0.2500 0.3971 0.4986 1.8229 0.2932 1.3241 -0.3717 1.7657 0.9905 1.3514 0.7020 0.8921 0.7787 0.5040 1.1701 1.3106 1.1878 0.8031 1.8060 0.5290 1.4757 0.8946 0.7093為了形象直觀地表示出各點的位置，畫出如下的散點圖：方法二 用fminunc實現(xiàn)function f=distance1(x,d) f=(x(1,3)2+(x(2,3)2-d(1)2; f=f+(x(1,11)2+(x(2,11)2-d(2)2;f=f+(x

16、(1,12)2+(x(2,12)2-d(3)2;f=f+(x(1,16)2+(x(2,16)2-d(4)2;f=f+(x(1,20)2+(x(2,20)2-d(5)2;f=f+(x(1,4)-x(1,1)2+(x(2*5-2)-x(2,1)2-d(6)2;f=f+(x(1,15)-x(1,1)2+(x(2,15)-x(2,1)2-d(7)2;f=f+(x(1,16)-x(1,1)2+(x(2,16)-x(2,1)2-d(8)2;f=f+(x(1,24)-x(1,1)2+(x(2,24)-x(2,1)2-d(9)2;f=f+(x(1,4)-x(1,2)2+(x(2,4)-x(2,2)2-d(10

17、)2;f=f+(x(1,19)-x(1,2)2+(x(2,19)-x(2,2)2-d(11)2;f=f+(x(1,20)-x(1,2)2+(x(2,20)-x(2,2)2-d(12)2;f=f+(x(1,23)-x(1,2)2+(x(2,23)-x(2,2)2-d(13)2;f=f+(x(1,4)-x(1,3)2+(x(2,4)-x(2,3)2-d(14)2;f=f+(x(1,11)-x(1,3)2+(x(2,11)-x(2,3)2-d(15)2;f=f+(x(1,23)-x(1,3)2+(x(2,23)-x(2,3)2-d(16)2;f=f+(x(1,7)-x(1,5)2+(x(2,7)-x

18、(2,5)2-d(17)2;f=f+(x(1,12)-x(1,5)2+(x(2,12)-x(2,5)2-d(18)2;f=f+(x(1,18)-x(1,5)2+(x(2,18)-x(2,5)2-d(19)2;f=f+(x(1,24)-x(1,5)2+(x(2,24)-x(2,5)2-d(20)2;f=f+(x(1,7)-x(1,6)2+(x(2,7)-x(2,6)2-d(21)2;f=f+(x(1,13)-x(1,6)2+(x(2,13)-x(2,6)2-d(22)2;f=f+(x(1,15)-x(1,6)2+(x(2,15)-x(2,6)2-d(23)2;f=f+(x(1,19)-x(1,6

19、)2+(x(2,19)-x(2,6)2-d(24)2;f=f+(x(1,20)-x(1,6)2+(x(2,20)-x(2,6)2-d(25)2;f=f+(x(1,13)-x(1,7)2+(x(2,13)-x(2,7)2-d(26)2;f=f+(x(1,17)-x(1,7)2+(x(2,17)-x(2,7)2-d(27)2;f=f+(x(1,12)-x(1,8)2+(x(2,12)-x(2,8)2-d(28)2;f=f+(x(1,14)-x(1,8)2+(x(2,14)-x(2,8)2-d(29)2;f=f+(x(1,21)-x(1,8)2+(x(2,21)-x(2,8)2-d(30)2;f=f

20、+(x(1,10)-x(1,9)2+(x(2,10)-x(2,9)2-d(31)2;f=f+(x(1,12)-x(1,9)2+(x(2,12)-x(2,9)2-d(32)2;f=f+(x(1,18)-x(1,9)2+(x(2,18)-x(2,9)2-d(33)2;f=f+(x(1,19)-x(1,9)2+(x(2,19)-x(2,9)2-d(34)2;f=f+(x(1,21)-x(1,9)2+(x(2,21)-x(2,9)2-d(35)2;f=f+(x(1,17)-x(1,10)2+(x(2,17)-x(2,10)2-d(36)2;f=f+(x(1,24)-x(1,10)2+(x(2,24)-

21、x(2,10)2-d(37)2;f=f+(x(1,14)-x(1,11)2+(x(2,14)-x(2,11)2-d(38)2;f=f+(x(1,16)-x(1,11)2+(x(2,16)-x(2,11)2-d(39)2;f=f+(x(1,14)-x(1,12)2+(x(2,14)-x(2,12)2-d(40)2;f=f+(x(1,18)-x(1,12)2+(x(2,18)-x(2,12)2-d(41)2;f=f+(x(1,14)-x(1,13)2+(x(2,14)-x(2,13)2-d(42)2;f=f+(x(1,15)-x(1,13)2+(x(2,15)-x(2,13)2-d(43)2;f=

22、f+(x(1,19)-x(1,15)2+(x(2,19)-x(2,15)2-d(44)2;f=f+(x(1,22)-x(1,15)2+(x(2,22)-x(2,15)2-d(45)2;f=f+(x(1,17)-x(1,16)2+(x(2,17)-x(2,16)2-d(46)2;f=f+(x(1,18)-x(1,16)2+(x(2,18)-x(2,16)2-d(47)2;f=f+(x(1,19)-x(1,18)2+(x(2,19)-x(2,18)2-d(48)2;f=f+(x(1,22)-x(1,18)2+(x(2,22)-x(2,18)2-d(49)2;f=f+(x(1,23)-x(1,18)

23、2+(x(2,23)-x(2,18)2-d(50)2;f=f+(x(1,22)-x(1,20)2+(x(2,22)-x(2,20)2-d(51)2;f=f+(x(1,22)-x(1,21)2+(x(2,22)-x(2,21)2-d(52)2;clear all;d=0.9607,0.4399,0.8143,1.3765,1.2722,0.5294,0.6144,0.3766,0.6893,0.9488,0.8000,1.1090,1.1432,0.4758,1.3402,0.7006,0.4945,1.0559,0.6810,0.3587,0.3351,0.2878,1.1346,0.3870

24、,0.7511,0.4439,0.8363,0.3208,0.1574,1.2736,0.5781,0.9254,0.6401,0.2467,0.4727,1.3840,0.4366,1.0307,1.3904,0.5725,0.7660,0.4394,1.0952,1.0422,1.8255,1.4325,1.0851,0.4995,1.2277,1.1271,0.7060,0.8052;x0=zeros(1,26),ones(1,24);opt = optimset(tolx,1e-16,tolf,1e-16);x,z,exit1,out1 = fminunc(distance1,x0,o

25、pt,d);p=0,0;x(1:24),x(25:48);a=0,x(1:24);b=0,x(25:48);plot(a,b,*);%畫散點圖表示原子位置【輸出結(jié)果】每個原子的位置如下（即第二個原子的位置為（-0.6818，0.6856），第二個原子為（-0.3563，0.2298），）：p = 0 0 -0.6818 0.6856 -0.3563 0.2298 -0.0530 0.9715 -0.4866 1.1721 0.0506 1.0700 0.3992 0.6744 0.0547 0.5349 -0.3948 0.3789 -0.1388 1.0752 0.4190 0.9660 -

26、0.3796 -0.3227 -0.7430 0.3687 0.1511 0.9167 -0.2615 0.6897 -0.5621 0.0830 -0.9836 0.9375 -0.2070 -0.2659 -0.0144 0.4168 0.0717 0.9085 -0.0117 1.2804 -0.0890 1.6054 0.6877 1.4172 -0.6661 1.3305 0.0125 0.7379畫出位置散點圖如下：【結(jié)果分析】（1）關(guān)于lsqononlin與fminunc精確性的探討用上面兩種不同的命令所求出來的結(jié)果并不一樣，可能是因為模型建立稍有差別導(dǎo)致算法不同而引起的。為了找

27、到更適合解決本題的模型，在命令窗口中輸入以下命令比較兩種算法的精確性。對于方法一 （用lsqnonlin命令實現(xiàn)） d=0.9607,0.4399,0.8143,1.3765,1.2722,0.5294,0.6144,0.3766,0.6893,. 0.9488,0.8000,1.1090,1.1432,0.4758,1.3402,0.7006,0.4945,1.0559,. 0.6810,0.3587,0.3351,0.2878,1.1346,0.3870,0.7511,0.4439,0.8363,. 0.3208,0.1574,1.2736,0.5781,0.9254,0.6401,0.2

28、467,0.4727,1.3840,. 0.4366,1.0307,1.3904,0.5725,0.7660,0.4394,1.0952,1.0422,1.8255,. 1.4325,1.0851,0.4995,1.2277,1.1271,0.7060,0.8052;x=p;f(1)=(x(1,3)2+(x(2,3)2-d(1)2; %對應(yīng)條件第4個和第1個原子間距f(2)=(x(1,11)2+(x(2,11)2-d(2)2;f(3)=(x(1,12)2+(x(2,12)2-d(3)2;f(4)=(x(1,16)2+(x(2,16)2-d(4)2;f(5)=(x(1,20)2+(x(2,20)

29、2-d(5)2;f(6)=(x(1,4)-x(1,1)2+(x(2*5-2)-x(2,1)2-d(6)2;f(7)=(x(1,15)-x(1,1)2+(x(2,15)-x(2,1)2-d(7)2;f(8)=(x(1,16)-x(1,1)2+(x(2,16)-x(2,1)2-d(8)2;f(9)=(x(1,24)-x(1,1)2+(x(2,24)-x(2,1)2-d(9)2;f(10)=(x(1,4)-x(1,2)2+(x(2,4)-x(2,2)2-d(10)2;f(11)=(x(1,19)-x(1,2)2+(x(2,19)-x(2,2)2-d(11)2;f(12)=(x(1,20)-x(1,2

30、)2+(x(2,20)-x(2,2)2-d(12)2;f(13)=(x(1,23)-x(1,2)2+(x(2,23)-x(2,2)2-d(13)2;f(14)=(x(1,4)-x(1,3)2+(x(2,4)-x(2,3)2-d(14)2;f(15)=(x(1,11)-x(1,3)2+(x(2,11)-x(2,3)2-d(15)2;f(16)=(x(1,23)-x(1,3)2+(x(2,23)-x(2,3)2-d(16)2;f(17)=(x(1,7)-x(1,5)2+(x(2,7)-x(2,5)2-d(17)2;f(18)=(x(1,12)-x(1,5)2+(x(2,12)-x(2,5)2-d(

31、18)2;f(19)=(x(1,18)-x(1,5)2+(x(2,18)-x(2,5)2-d(19)2;f(20)=(x(1,24)-x(1,5)2+(x(2,24)-x(2,5)2-d(20)2;f(21)=(x(1,7)-x(1,6)2+(x(2,7)-x(2,6)2-d(21)2;f(22)=(x(1,13)-x(1,6)2+(x(2,13)-x(2,6)2-d(22)2;f(23)=(x(1,15)-x(1,6)2+(x(2,15)-x(2,6)2-d(23)2;f(24)=(x(1,19)-x(1,6)2+(x(2,19)-x(2,6)2-d(24)2;f(25)=(x(1,20)-

32、x(1,6)2+(x(2,20)-x(2,6)2-d(25)2;f(26)=(x(1,13)-x(1,7)2+(x(2,13)-x(2,7)2-d(26)2;f(27)=(x(1,17)-x(1,7)2+(x(2,17)-x(2,7)2-d(27)2;f(28)=(x(1,12)-x(1,8)2+(x(2,12)-x(2,8)2-d(28)2;f(29)=(x(1,14)-x(1,8)2+(x(2,14)-x(2,8)2-d(29)2;f(30)=(x(1,21)-x(1,8)2+(x(2,21)-x(2,8)2-d(30)2;f(31)=(x(1,10)-x(1,9)2+(x(2,10)-x

33、(2,9)2-d(31)2;f(32)=(x(1,12)-x(1,9)2+(x(2,12)-x(2,9)2-d(32)2;f(33)=(x(1,18)-x(1,9)2+(x(2,18)-x(2,9)2-d(33)2;f(34)=(x(1,19)-x(1,9)2+(x(2,19)-x(2,9)2-d(34)2;f(35)=(x(1,21)-x(1,9)2+(x(2,21)-x(2,9)2-d(35)2;f(36)=(x(1,17)-x(1,10)2+(x(2,17)-x(2,10)2-d(36)2;f(37)=(x(1,24)-x(1,10)2+(x(2,24)-x(2,10)2-d(37)2;

34、f(38)=(x(1,14)-x(1,11)2+(x(2,14)-x(2,11)2-d(38)2;f(39)=(x(1,16)-x(1,11)2+(x(2,16)-x(2,11)2-d(39)2;f(40)=(x(1,14)-x(1,12)2+(x(2,14)-x(2,12)2-d(40)2;f(41)=(x(1,18)-x(1,12)2+(x(2,18)-x(2,12)2-d(41)2;f(42)=(x(1,14)-x(1,13)2+(x(2,14)-x(2,13)2-d(42)2;f(43)=(x(1,15)-x(1,13)2+(x(2,15)-x(2,13)2-d(43)2;f(44)=

35、(x(1,19)-x(1,15)2+(x(2,19)-x(2,15)2-d(44)2;f(45)=(x(1,22)-x(1,15)2+(x(2,22)-x(2,15)2-d(45)2;f(46)=(x(1,17)-x(1,16)2+(x(2,17)-x(2,16)2-d(46)2;f(47)=(x(1,18)-x(1,16)2+(x(2,18)-x(2,16)2-d(47)2;f(48)=(x(1,19)-x(1,18)2+(x(2,19)-x(2,18)2-d(48)2;f(49)=(x(1,22)-x(1,18)2+(x(2,22)-x(2,18)2-d(49)2;f(50)=(x(1,2

36、3)-x(1,18)2+(x(2,23)-x(2,18)2-d(50)2;f(51)=(x(1,22)-x(1,20)2+(x(2,22)-x(2,20)2-d(51)2;f(52)=(x(1,22)-x(1,21)2+(x(2,22)-x(2,21)2-d(52)2; a=0; for i=1:52a=a+f(i);enda運(yùn)行結(jié)果為a =0.2190對于方法二（用fminunc命令實現(xiàn)） f = (x(4-1)2 + x(24+4-1)2 - d(1)2)2;f = f + (x(12-1)2 + x(24+12-1)2 - d(2)2)2;f = f + (x(13-1)2 + x(24

37、+13-1)2 - d(3)2)2;f = f + (x(17-1)2 + x(24+17-1)2 - d(4)2)2;f = f + (x(21-1)2 + x(24+21-1)2 - d(5)2)2; f = f + (x(5-1) - x(2-1)2 + (x(24+5-1) - x(24+2-1)2 - d(6)2)2; f = f + (x(16-1) - x(2-1)2 + (x(24+16-1) - x(24+2-1)2 - d(7)2)2; f = f + (x(17-1) - x(2-1)2 + (x(24+17-1) - x(24+2-1)2 - d(8)2)2; f =

38、f + (x(25-1) - x(2-1)2 + (x(24+25-1) - x(24+2-1)2 - d(9)2)2; f = f + (x(5-1) - x(3-1)2 + (x(24+5-1) - x(24+3-1)2 - d(10)2)2; f = f + (x(20-1) - x(3-1)2 + (x(24+20-1) - x(24+3-1)2 - d(11)2)2; f = f + (x(21-1) - x(3-1)2 + (x(24+21-1) - x(24+3-1)2 - d(12)2)2; f = f + (x(24-1) - x(3-1)2 + (x(24+24-1) -

39、x(24+3-1)2 - d(13)2)2; f = f + (x(5-1) - x(4-1)2 + (x(24+5-1) - x(24+4-1)2 - d(14)2)2; f = f + (x(12-1) - x(4-1)2 + (x(24+12-1) - x(24+4-1)2 - d(15)2)2;f = f + (x(24-1) - x(4-1)2 + (x(24+24-1) - x(24+4-1)2 - d(16)2)2; f = f + (x(8-1) - x(6-1)2 + (x(24+8-1) - x(24+6-1)2 - d(17)2)2; f = f + (x(13-1) -

40、 x(6-1)2 + (x(24+13-1) - x(24+6-1)2 - d(18)2)2; f = f + (x(19-1) - x(6-1)2 + (x(24+19-1) - x(24+6-1)2 - d(19)2)2; f = f + (x(25-1) - x(6-1)2 + (x(24+25-1) - x(24+6-1)2 - d(20)2)2; f = f + (x(8-1) - x(7-1)2 + (x(24+8-1) - x(24+7-1)2 - d(21)2)2; f = f + (x(14-1) - x(7-1)2 + (x(24+14-1) - x(24+7-1)2 -

41、d(22)2)2;f = f + (x(16-1) - x(7-1)2 + (x(24+16-1) - x(24+7-1)2 - d(23)2)2;f = f + (x(20-1) - x(7-1)2 + (x(24+20-1) - x(24+7-1)2 - d(24)2)2; f = f + (x(21-1) - x(7-1)2 + (x(24+21-1) - x(24+7-1)2 - d(25)2)2; f = f + (x(14-1) - x(8-1)2 + (x(24+14-1) - x(24+8-1)2 - d(26)2)2; f = f + (x(18-1) - x(8-1)2 +

42、 (x(24+18-1) - x(24+8-1)2 - d(27)2)2; f = f + (x(13-1) - x(9-1)2 + (x(24+13-1) - x(24+9-1)2 - d(28)2)2; f = f + (x(15-1) - x(9-1)2 + (x(24+15-1) - x(24+9-1)2 - d(29)2)2; f = f + (x(22-1) - x(9-1)2 + (x(24+22-1) - x(24+9-1)2 - d(30)2)2; f = f + (x(11-1) - x(10-1)2 + (x(24+11-1) - x(24+10-1)2 - d(31)2

43、)2; f = f + (x(13-1) - x(10-1)2 + (x(24+13-1) - x(24+10-1)2 - d(32)2)2; f = f + (x(19-1) - x(10-1)2 + (x(24+19-1) - x(24+10-1)2 - d(33)2)2; f = f + (x(20-1) - x(10-1)2 + (x(24+20-1) - x(24+10-1)2 - d(34)2)2; f = f + (x(22-1) - x(10-1)2 + (x(24+22-1) - x(24+10-1)2 - d(35)2)2; f = f + (x(18-1) - x(11-

44、1)2 + (x(24+18-1) - x(24+11-1)2 - d(36)2)2; f = f + (x(25-1) - x(11-1)2 + (x(24+25-1) - x(24+11-1)2 - d(37)2)2; f = f + (x(15-1) - x(12-1)2 + (x(24+15-1) - x(24+12-1)2 - d(38)2)2; f = f + (x(17-1) - x(12-1)2 + (x(24+17-1) - x(24+12-1)2 - d(39)2)2; f = f + (x(15-1) - x(13-1)2 + (x(24+15-1) - x(24+13-

45、1)2 - d(40)2)2; f = f + (x(19-1) - x(13-1)2 + (x(24+19-1) - x(24+13-1)2 - d(41)2)2; f = f + (x(15-1) - x(14-1)2 + (x(24+15-1) - x(24+14-1)2 - d(42)2)2; f = f + (x(16-1) - x(14-1)2 + (x(24+16-1) - x(24+14-1)2 - d(43)2)2; f = f + (x(20-1) - x(16-1)2 + (x(24+20-1) - x(24+16-1)2 - d(44)2)2; f = f + (x(2

46、3-1) - x(16-1)2 + (x(24+23-1) - x(24+16-1)2 - d(45)2)2; f = f + (x(18-1) - x(17-1)2 + (x(24+18-1) - x(24+17-1)2 - d(46)2)2; f = f + (x(19-1) - x(17-1)2 + (x(24+19-1) - x(24+17-1)2 - d(47)2)2; f = f + (x(20-1) - x(19-1)2 + (x(24+20-1) - x(24+19-1)2 - d(48)2)2; f = f + (x(23-1) - x(19-1)2 + (x(24+23-1

47、) - x(24+19-1)2 - d(49)2)2; f = f + (x(24-1) - x(19-1)2 + (x(24+24-1) - x(24+19-1)2 - d(50)2)2; f = f + (x(23-1) - x(21-1)2 + (x(24+23-1) - x(24+21-1)2 - d(51)2)2; f = f + (x(23-1) - x(22-1)2 + (x(24+23-1) - x(24+22-1)2 - d(52)2)2; d=0.9607,0.4399,0.8143,1.3765,1.2722,0.5294,0.6144,0.3766,0.6893,0.9

48、488,0.8000,1.1090,1.1432,0.4758,1.3402,0.7006,0.4945,1.0559,0.6810,0.3587,0.3351,0.2878,1.1346,0.3870,0.7511,0.4439,0.8363,0.3208,0.1574,1.2736,0.5781,0.9254,0.6401,0.2467,0.4727,1.3840,0.4366,1.0307,1.3904,0.5725,0.7660,0.4394,1.0952,1.0422,1.8255,1.4325,1.0851,0.4995,1.2277,1.1271,0.7060,0.8052;x=

49、p(:,1),p(:,2); f運(yùn)行結(jié)果為f =0.0375兩種算法初值的設(shè)定是一樣的。對比兩個運(yùn)行結(jié)果，用fminunc求出的距離差的和小于用lsqonlin求出的距離差的和。由此可見，用fminunc命令求解更為精確。（2）改變初值對結(jié)果的影響鑒于前面的分析，我們采用fminunc命令得到的結(jié)果來繼續(xù)下面的討論。將初值改為x0=zeros(1,2),ones(1,46);運(yùn)行結(jié)果如下： p = 0 0 0.5553 1.5928 0.1561 0.4287 0.7454 0.6616 0.8003 1.1268 1.1061 0.7317 1.1218 0.9560 0.8743 1.14

50、78 0.2187 0.4924 0.9728 0.8388 0.3939 0.8547 -0.4765 0.1503 0.0492 0.8143 0.9413 0.6822 0.5017 0.4869 0.0320 1.2791 0.2655 1.3360 1.6969 1.3201 0.5989 0.2906 0.8971 0.6896 1.2488 0.2158 1.2155 1.2784 1.7608 0.6915 0.7832 1.3882 0.8195 0.9554畫出位置散點圖：對于其他初值的設(shè)定情況也大致相同。由此可見，改變初值，對于最終結(jié)果的影響是很大的。不同的初值，得到原子的位置不一樣。我覺得可能的原因是fminunc求出的是局部極小值，在不同的初值條件下函數(shù)會收斂到不同的極小值，因此會有上述情況發(fā)生?！緦嶒灲Y(jié)論】對于無約束優(yōu)化求極小值的問題，可以采用lsqnonlin和fminunc等不同的命令，得到的結(jié)果雖算法的不同而有所不同。此外，設(shè)置不同的初值，也可以得到不同的結(jié)果，因此說明初值的選取對研究問題其一定的作用。為了得到更準(zhǔn)確的模型，需要根據(jù)實際問題慎重地選取初值。吸收室c1（t）中心室 c（t）k1k題目8 給藥方案設(shè)計需要依據(jù)藥物