[分享][推荐]数学建模培训讲座

宝宝 · 发表于 2004-6-3 03:17:32

<

>时下，已经开始数学建模竞赛报名了，宝宝将自己当是学习的讲座贴出来，希望新手回去好好学习。
<

>对了，斑竹不好意思宝宝不能上传，只好这样贴出来。
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>1. 一维插值



对表格给出的函数，求出没有给出的函数值。


在实际工作中，经常会遇到插值问题。


例1：表1是待加工零件下轮廓线的一组数据，现需要得到x坐标每改变0.1时所对应的y的坐标.


<TABLE cellSpacing=0 cellPadding=0 border=1>

<TR>
<TD vAlign=top width=41>
x

</TD>
<TD vAlign=top width=41>
0

</TD>
<TD vAlign=top width=41>
3

</TD>
<TD vAlign=top width=41>
5

</TD>
<TD vAlign=top width=41>
7

</TD>
<TD vAlign=top width=41>
9

</TD>
<TD vAlign=top width=41>
11

</TD>
<TD vAlign=top width=41>
12

</TD>
<TD vAlign=top width=41>
13

</TD>
<TD vAlign=top width=41>
14

</TD>
<TD vAlign=top width=41>
15

</TD></TR>
<TR>
<TD vAlign=top width=41>
y

</TD>
<TD vAlign=top width=41>
0

</TD>
<TD vAlign=top width=41>
1.2

</TD>
<TD vAlign=top width=41>
1.7

</TD>
<TD vAlign=top width=41>
2.0

</TD>
<TD vAlign=top width=41>
2.1

</TD>
<TD vAlign=top width=41>
2.0

</TD>
<TD vAlign=top width=41>
1.8

</TD>
<TD vAlign=top width=41>
1.2

</TD>
<TD vAlign=top width=41>
1.0

</TD>
<TD vAlign=top width=41>
1.6

</TD></TR></TABLE>
下面是关于插值的两条命令(专门用来解决这类问题)：
y=interp1(x0,y0,x) 分段线性插值
y=spline(x0,y0,x) 三次样条插值
其中x0,y0是已知的节点坐标，是同维向量。y对应于x处的插值。y与x是同维向量。
解决上述问题,我们可分两步:
一 用原始数据绘图作为选用插值方法的参考.
二 确定插值方法进行插值计算
对于上述问题,可键入以下的命令:
x0=[0,3,5,7,9,11,12,13,14,15]';


y0=[0,1.2,1.7,2.0,2.1,2.0,1.8,1.2,1.0,1.6]'


plot(x0,y0) %完成第一步工作
x=0:0.1:15;
y=interp1(x0,y0,x'); %用分段线性插值完成第二步工作


plot(x,y)
y=spline(x0,y0,x'); 
plot(x,y) %用三次样条插值完成第二步工作



练习：对y=1/(1+x2),-5≤x≤5，用n（=11）个节点（等分）作上述两种插值，用m（=21）个插值点（等分）作图，比较结果。
解：键入并运行如下命令
n=11;m=21;x=-5:10/(m-1):5;y=1./(1+x.^2);
xo=-5:10/(n-1):5;yo=1./(1+xo.^2);
y1=interp1(xo,yo,x);
y2=spline(xo,yo,x);
plot(x,y,'r',x,y1,'b',x,y2,'k')
练习：在某处测得海洋不同深度处水温如下：
<TABLE cellSpacing=0 cellPadding=0 border=1>

<TR>
<TD vAlign=top width=95>
深度</TD>
<TD vAlign=top width=95>
446</TD>
<TD vAlign=top width=95>
714</TD>
<TD vAlign=top width=95>
950</TD>
<TD vAlign=top width=95>
1422</TD>
<TD vAlign=top width=95>
1634</TD></TR>
<TR>
<TD vAlign=top width=95>
水温</TD>
<TD vAlign=top width=95>
7.04</TD>
<TD vAlign=top width=95>
4.28</TD>
<TD vAlign=top width=95>
3.40</TD>
<TD vAlign=top width=95>
2.54</TD>
<TD vAlign=top width=95>
2.13</TD></TR></TABLE>
求深度为500、1000、1500米处的水温。
解：输入程序：
D=[446,714,950,1422,1634];
T=[7.04,4.28,3.40,2.54,2.13];
Di=[500,1000,1500];
Ti=interp1(D,T,Di)
MATLAB的命令interp1(X,Y,Xi,’method’)用于一元插值.其中Method可选’nearest’(最近邻插值),’linear’(线性插值),’spline’(三次样条插值),’cubic’(三次多项式插值)

宝宝 · 发表于 2004-6-3 03:18:15

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0cm 0cm 0pt 21.25pt; TEXT-INDENT: 1.45pt; mso-list: l44 level1 lfo47; tab-stops: list 40.7pt">1. 二维插值<

0cm 0cm 0pt 21.25pt">MATLAB中二维插值的命令是：<

0cm 0cm 0pt 21.25pt">z=interp2（x0,y0,z0,x,y,'meth'） 例2：在一个长为5个单位，宽为3个单位的金属薄片上测得15个点的温度值，试求出此薄片的温度分布，并绘出等温线图。（数据如下表）<TABLE medium none; BORDER-TOP: medium none; MARGIN-LEFT: 77.4pt; BORDER-LEFT: medium none; BORDER-BOTTOM: medium none; BORDER-COLLAPSE: collapse; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-table-layout-alt: fixed" cellSpacing=0 cellPadding=0 border=1><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-diagonal-down: .5pt solid windowtext" vAlign=top width=60>y<v:shapetype> <v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path gradientshapeok="t" connecttype="rect" extrusionok="f"></v:path><lock v:ext="edit" aspectratio="t"></lock></v:shapetype><v:shape><v:imagedata></v:imagedata></v:shape> x<v:shape> <v:imagedata></v:imagedata></v:shape></TD><TD windowtext 0.5pt solid; 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PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=60> 84</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 45.6pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=61> 84</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 44.4pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=59> 82</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=60> 85</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=60> 86</TD></TR></TABLE>程序：temps=[82,81,80,82,84;79,63,61,65,87;84,84,82,85,86]; mesh(temps) %根据原始数据绘出温度分布图，可看到此图的粗造度。<v:shape><v:imagedata></v:imagedata></v:shape> %下面开始进行二维函数的三阶插值。 width=1:5; depth=1:3; di=1:0.2:3; wi=1:0.2:5; [WI,DI]=meshgrid(wi,di);%增加了节点数目 ZI=interp2(width,depth,temps,WI,DI,'cubic');% 对数据（width,depth,temps）进 % 行三阶插值拟合。 surfc(WI,DI,ZI)contour(WI,DI,ZI)<v:shape><v:imagedata></v:imagedata></v:shape> <v:shape><v:imagedata></v:imagedata></v:shape>

宝宝 · 发表于 2004-6-3 03:18:43

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0cm 0cm 0pt">3．曲线拟合<

0cm 0cm 0pt">假设一函数g(x)是以表格形式给出的，现要求一函数f(x)，使f(x)在某一准则下与表格函数（数据）最为接近。<

0cm 0cm 0pt">由于与插值的提法不同，所以在数学上理论根据不同，解决问题的方法也不同。此处，我们总假设f(x)是多项式。例3：弹簧在力F的作用下伸长x厘米。F和x在一定的范围内服从虎克定律。试根据下列数据确定弹性系数k,并给出不服从虎克定律时的近似公式。<TABLE medium none; BORDER-TOP: medium none; BORDER-LEFT: medium none; BORDER-BOTTOM: medium none; BORDER-COLLAPSE: collapse; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt" cellSpacing=0 cellPadding=0 border=1><TR 12.95pt"><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 12.95pt; BACKGROUND-COLOR: transparent" vAlign=top width=44>x</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 12.95pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=44>1</TD><TD windowtext 0.5pt solid; 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PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 12.95pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=44>9</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 12.95pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=44>12</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 12.95pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=44>13</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 12.95pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=44>15</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 33pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 12.95pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=44>17</TD></TR><TR 14.7pt"><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 14.7pt; BACKGROUND-COLOR: transparent; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=44>F</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 14.7pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=44>1.5</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 14.7pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=44>3.9</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 14.7pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=44>6.6</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 14.7pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=44>11.7</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 14.7pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=44>15.6</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 14.7pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=44>18.8</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 14.7pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=44>19.6</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 32.95pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 14.7pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=44>20.6</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 33pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 14.7pt; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=44>21.1</TD></TR></TABLE>解题思路：可以用一阶多项式拟合求出k，以及近似公式。在MATLAB中，用以下命令拟合多项式。polyfit（x0,y0,n）一般，也需先观察原始数据的图像，然后再确定拟和成什么曲线。对于上述问题,可键入以下的命令:x=[1,2,4,7,9,12,13,15,17]'; F=[1.5,3.9,6.6,11.7,15.6,18.8,19.6,20.6,21.1]'; plot(x,F,'.')从图像上我们发现：前5个数据应与直线拟合，后5个数据应与二次曲线拟合。于是键入 a=polyfit(x(1:5),F(1:5),1); a=polyfit(x(5:9),F(5:9),2)得注意：有时，面对一个实际问题，究竟是用插值还是用拟合不好确定，还需大家在实际中仔细区分。同时，大家（包括学过计算方法的同学）注意去掌握相应的理论知识。

宝宝 · 发表于 2004-6-3 03:20:54

<

0cm 0pt?>4．数值积分



<

TEXT-INDENT: 0cm 0pt; 15.75pt?>先看一个例子：
<

TEXT-INDENT: 0cm 0pt; 15.75pt?>例4．现要根据瑞士地图计算其国土面积。于是对地图作如下的测量：以西东方向为横轴，以南北方向为纵轴。（选适当的点为原点）将国土最西到最东边界在x轴上的区间划取足够多的分点xi，在每个分点处可测出南北边界点的对应坐标y1 ，y2。用这样的方法得到下表
x 7.0 10.5 13.0 17.5 34.0 40.5 44.5 48.0 56.0
y1 44 45 47 50 50 38 30 30 34
y2 44 59 70 72 93 100 110 110 110
x 61.0 68.5 76.5 80.5 91.0 96.0 101.0 104.0 106.5
y1 36 34 41 45 46 43 37 33 28
y2 117 118 116 118 118 121 124 121 121
x 111.5 118.0 123.5 136.5 142.0 146.0 150.0 157.0 158.0
y1 32 65 55 54 52 50 66 66 68
y2 121 122 116 83 81 82 86 85 68

[此贴子已经被作者于2004-6-2 20:23:50编辑过]

宝宝 · 发表于 2004-6-3 03:20:57

<

0cm 0cm 0pt">4．数值积分<

0cm 0cm 0pt; TEXT-INDENT: 15.75pt">先看一个例子：<

0cm 0cm 0pt; TEXT-INDENT: 15.75pt">例4．现要根据瑞士地图计算其国土面积。于是对地图作如下的测量：以西东方向为横轴，以南北方向为纵轴。（选适当的点为原点）将国土最西到最东边界在x轴上的区间划取足够多的分点xi，在每个分点处可测出南北边界点的对应坐标y1 ，y2。用这样的方法得到下表

宝宝 · 发表于 2004-6-3 03:23:43

<

0cm 0cm 0pt">根据地图比例知18mm相当于40km，试由上表计算瑞士国土的近似面积。（精确值为41288km2）。<

0cm 0cm 0pt">解题思路：数据实际上表示了两条曲线，实际上我们要求由两曲线所围成的图形的面积。<

0cm 0cm 0pt">解此问题的方法是数值积分的方法。具体解时我们遇到两个问题：1。数据如何输入；2。没有现成的命令可用。解：对于第一个问题，我们可把数据考备成M文件（或纯文本文件）。然后，利用数据绘制平面图形。键入load mianji.txtA=mianji';plot(A(:,1),A(:,2),'r',A(:,1),A(:,3),'g')<v:shapetype><v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path gradientshapeok="t" connecttype="rect" extrusionok="f"></v:path><lock v:ext="edit" aspectratio="t"></lock></v:shapetype><v:shape><v:imagedata></v:imagedata></v:shape>接下来可以计算面积。键入：a1=trapz(A(:,1)*40/18,A(:,2)*40/18);a2=trapz(A(:,1)*40/18,A(:,3)*40/18);d=a2-a1d = 4.2414e+004至此，问题可以说得到了解决。之所以说还有问题，是我们觉得误差较大。但计算方法的理论给了我们更精确计算方法。只是MATLAB没有相应的命令。想得到更理想的结果，我们可以自己设计解决问题的方法。（可以编写辛普森数值计算公式的程序，或用拟合的方法求出被积函数，再利用MATLAB的命令quad,quad8）

宝宝 · 发表于 2004-6-3 03:24:12

<

0cm 0cm 0pt; TEXT-INDENT: 21.75pt">5．数值微分<

0cm 0cm 0pt 21pt">实际的例：已知20世纪美国人口统计数据如下，根据数据计算人口增长率。（其实还可以对于后十年人口进行预测）<TABLE medium none; BORDER-TOP: medium none; MARGIN-LEFT: 8.2pt; BORDER-LEFT: medium none; BORDER-BOTTOM: medium none; BORDER-COLLAPSE: collapse; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt" cellSpacing=0 cellPadding=0 border=1><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 45.1pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent" vAlign=top width=60><

0cm 0cm 0pt">年份</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 36.6pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=49>1900</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 36.6pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=49>1910</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=50>1920</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=50>1930</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=50>1940</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=50>1950</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=50>1960</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=50>1970</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=50>1980</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=50>1990</TD></TR><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 45.1pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=60>人口×106</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 36.6pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=49>76.0</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 36.6pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=49>92.0</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=50>106.5</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=50>123.2</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=50>131.7</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=50>150.7</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=50>179.3</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=50>204.0</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=50>226.5</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: #ece9d8; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: #ece9d8; WIDTH: 37.45pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; BACKGROUND-COLOR: transparent; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=50>251.4</TD></TR></TABLE>解题思路：设人口是时间的函数x(t).于是人口的增长率就是x(t)对t的导数.如果计算出人口的相关变化率<v:shapetype> <v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path gradientshapeok="t" connecttype="rect" extrusionok="f"></v:path><lock v:ext="edit" aspectratio="t"></lock></v:shapetype><v:shape><v:imagedata></v:imagedata></v:shape>。那么人口增长满足<v:shape> <v:imagedata></v:imagedata></v:shape>，它在初始条件x(0)=x0下的解为<v:shape> <v:imagedata></v:imagedata></v:shape>.（用以检查计算结果的正确性）解：此问题的特点是以离散变量给出函数x(t),所以就要用差分来表示函数x(t)的导数.<v:shape><v:imagedata></v:imagedata></v:shape>常用后一个公式。（因为，它实际上是用二次插值函数来代替曲线x(t)）即常用三点公式来代替函数在各分点的导数值：<v:shape><v:imagedata></v:imagedata></v:shape>MATLAB用命令diff按两点公式计算差分;此题自编程序用三点公式计算相关变化率.编程如下:for i=1:length(x) if i==1 r(1)=(-3*x(1)+4*x(1+1)-x(1+2))/(20*x(1)); elseif i~=length(x) r(i)=(x(i+1)-x(i-1))/(20*x(i)); else r(length(x))=(x(length(x)-2)-4*x(length(x)-1)+3*x(length(x)))/(20*x(length(x))); endendr=r;保存为diff3.m文件听候调用.再在命令窗内键入X=[1900,1910,1920,1930,1940,1950,1960,1970,1980,1990];x=[76.0, 92.0, 106.5, 123.2, 131.7, 150.7, 179.3, 204.0, 226.5, 251.4];diff3;由于r以离散数据给出,所以要用数值积分计算.键入x(1,1)*exp(trapz(X(1,1:9),r(1:9)))数值积分命令：trapz(x),trapz(x,y),quad(‘fun’,a,b)等.

宝宝 · 发表于 2004-6-3 03:27:28

<

0cm 0cm 0pt; tab-stops: 0cm">6：微分方程数值解（单摆问题）<

0cm 0cm 0pt; TEXT-INDENT: 21pt; tab-stops: 0cm; mso-char-indent-count: 2.0; mso-char-indent-size: 10.5pt">单摆问题的数学模型是<

0cm 0cm 0pt; TEXT-INDENT: 21pt; TEXT-ALIGN: center; tab-stops: 0cm; mso-char-indent-count: 2.0; mso-char-indent-size: 10.5pt" align=center><v:shapetype><v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path gradientshapeok="t" connecttype="rect" extrusionok="f"></v:path><lock v:ext="edit" aspectratio="t"></lock></v:shapetype><v:shape><v:imagedata></v:imagedata></v:shape>在初始角度不大时,问题可以得到很好地解决,但如果初始角较大,此方程无法求出解析解.现问题是当初始角为100和300时,求出其解,画出解的图形进行比较。解:若θ0较小,则原方程可用<v:shape> <v:imagedata></v:imagedata></v:shape>来近似.其解析解为θ(t)= θ0cosωt,<v:shape> <v:imagedata></v:imagedata></v:shape>. 若不用线性方程来近似,那么有两个模型:<v:shape><v:imagedata></v:imagedata></v:shape> 取g=9.8,l=25, 100=0.1745, 300=0.5236.用MATLAB求这两个模型的数值解,先要作如下的处理:令x1=θ,x2=θ’,则模型变为<v:shape><v:imagedata></v:imagedata></v:shape>再编函数文件function xdot=danbai(t,x)xdot=zeros(2,1);xdot(1)=x(2);xdot(2)=-9.8/25*sin(x(1));在命令窗口键入[t,x]=ode45(‘danbai’,[0:0.1:20],[0.1745,0]);[t,y]=ode45(‘danbai’,[0:0.1:20],[0.5236,0]);plot(t,x(:,1),’r’,t,y(:,1),’k’);参考书:数学实验,高等教育出版社

宝宝 · 发表于 2004-6-3 03:27:51

<

0cm 0cm 0pt">7．解优化问题<

0cm 0cm 0pt">线性规划有约束极小问题<

0cm 0cm 0pt">模型<v:shapetype><v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path gradientshapeok="t" connecttype="rect" extrusionok="f"></v:path><lock v:ext="edit" aspectratio="t"></lock></v:shapetype><v:shape><v:imagedata></v:imagedata></v:shape>用命令[x, fval]= linprog(f,A,b,A1,b1,lb,ub)例1：Find x that minimizes f(x)=-5x1-4x2-6x3subject tox1-x2+x3≦203x1+2x2+4x3≦423x1+2x2≦300≦x1, 0≦x2,0≦x3First, enter the coefficients:f = [-5; -4; -6]A = [1 -1 1 3 2 4 3 2 0];b = [20; 42; 30];lb = zeros(3,1);Next, call a linear programming routine:[x,fval,exitflag,output,lambda] = linprog(f,A,b,[],[],lb)；Entering x, fval,lambda.ineqlin, and lambda.lower getsx = 0.0000 15.0000 3.0000fval = -78.0000和其它信息。例2：解问题<v:shape><v:imagedata></v:imagedata></v:shape>把问题极小化并将约束标准化<v:shape><v:imagedata></v:imagedata></v:shape>键入c=[-2,-3,-5];a=[-2,-5,-1];b=-10;a1=[1,1,1];b1=7;LB=[0,0,0];[x，y]=linprog（c,a,b,a1,b1,LB）得当X=(6.4286,0.5714,0.0000)时,z=14.5714最大.例3: 解问题<v:shape><v:imagedata></v:imagedata></v:shape>解:键入c=[-2,-1,1];a=[1,4,-1;2,-2,1];b=[4;12];a1=[1,1,2];b1=6;lb=[0;0;-inf];ub=[inf;inf;5];[x,z]=linprog(c,a,b,a1,b1,lb,ub) 得当X=(4.6667,0.0000,0.6667)时,z=-8.6667最小.

宝宝 · 发表于 2004-6-3 03:28:08

<

0cm 0cm 0pt">非线性规划有约束极小问题<

0cm 0cm 0pt">模型1<

0cm 0cm 0pt; TEXT-ALIGN: center" align=center><v:shapetype><v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path gradientshapeok="t" connecttype="rect" extrusionok="f"></v:path><lock v:ext="edit" aspectratio="t"></lock></v:shapetype><v:shape><v:imagedata></v:imagedata></v:shape>用命令x=constr('f ',x0)。ExamplesFind values of x that minimize f(x)=-x1x2x3, starting at the point x = [10; 10; 10] and subject to the constraints 0≤x1+2x2+2x3≤72.-x1-2x2-2x3≤0,x1+2x2+2x3≤72,第一步：编写M文件function [f,g]=myfun(x)f=-x(1)*x(2)*x(3);g(1)=-x(1)-2*x(2)-2*x(3);g(2)=x(1)+2*x(2)+2*x(3)-72;第二步：求解在MATLAB工作窗中键入x0=[10,10,10];x=constr('myfun',x0)即可模型2：<v:shape><v:imagedata></v:imagedata></v:shape>MATLAB求解此问题的命令是：[x,fval,exitflag,output,lambda,grad,hessian]=fmincon(‘fun’,x0,A,b,A1,b1,LB,UB,’nonlcon’,options,p1,p2,…)fun是目标函数的m_文件名.nonlcon是约束函数C(x)和C1(x)的m_文件名.文件输出为[C,C1].例:求解最优化问题<v:shape><v:imagedata></v:imagedata></v:shape>第1步: 建立目标函数和非线性约束的m_文件.Function y=e1511(x)% 目标函数的m_文件Y=exp(x(1))*(4*x(1)^2+2*x(2)^2+4*x(1)*x(2)+2*x(2)+1); Function [c1,c2]=e1511b(x)% 非线性约束的m_文件C1=[1.5+x(1)*x(2)-x(1)-x(2);-x(1)*x(2)-10];C2=0;第2步: 运行程序.键入x0=[-1,1];a1=[1,1];b1=0;[x,f,exitflab,output]=fmincon(‘e1511’,x0,[],[],a1,b1,[],[],’e1511b’)得结果.输出结果的意义:经过4次迭代(iterations:4)收敛到了(exitfag=1)最优解x(1)=-1.2247,x(2)=1.2247,目标函数最优值为1.8951.

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