认证考试数学建模试验谜底.docx
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认证考试数学建模试验谜底.docx
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认证考试数学建模试验谜底
实验一:
食饵与捕食者的相互依存与制约分析
练习1:
解微分方程组:
Input:
f.m
function,f=f(x,y)
f=[-2,1;998,-999]*y+[2*sin(x);999*(cos(x)-sin(x))];
f2.m
[x,y]=ode23('f',[0,10],[2,3]);
x
pause
y
plot(x,y)
Result:
练习2:
解二阶微分方程
Input:
●zuni.m
function,f=zuni(t,y);
global,a,wo;
A=[0,1;
,-wo^2,-2*a];
f=A*y;
●ex2.m
global,a,wo;
a=5;
wo=1;
[t,y]=ode45('zuni',[0:
0.01:
2*pi],[0,4])
y
plot(t,y(:
1),'b')
hold,on
plot(t,y(:
2),'g')
output:
2.利用matlab实现食饵与捕食者系统的仿真:
Volterra食饵与捕食者模型:
取
Input:
●shier.m
function,dx=shier(t,x)
global,r,d,a,b;
dx=zeros(2,1);
dx
(1)=(r-a*x
(2))*x
(1)
dx
(2)=(-d+b*x
(1))*x
(2)
●shier2.m
global,r,d,a,b;
r=1;d=0.5;a=0.1;b=0.02;
ts=0:
0.1:
15;
x0=[25,2];
[t,x]=ode45('shier',ts,x0);
x
plot(x(:
1),x(:
2))
3.利用matlab实现两种群共生系统的仿真:
仿照上例编写程序。
input:
peace.m
function,dx=peace(t,x)
global,N1,N2,r1,r2,q1,q2;
dx=zeros(2,1);
dx
(1)=r1*x
(1)*(1-x
(1)./N1+q1*x
(2)./N2)
dx
(2)=r2*x
(2)*(-1+q2*x
(1)./N1-x
(2)./N2)
peace2.m
global,N1,N2,r1,r2,q1,q2;
N1=input('种群1最大容量=,');
n1=input('种群1初始数量=,');
N2=input('种群2最大容量=,');
n2=input('种群1初始数量=,');
q1=input('请输入相关系数q1=');
q2=input('请输入相关系数q2=');
r1=0.05;
r2=0.01;
ts=0:
1:
5000;
x0=[n1,n2];
[t,x]=ode45('peace',ts,x0);
%plot(x(:
1),x(:
2))
plot(t,x(:
1),'b')
hold,on
plot(t,x(:
2),'r')
result:
取N1=100,N2=200,n1=10,n2=20,q1=0.3,q2=3
4.,,利用matlab实现两种群的竞争系统的仿真:
仿照上例编写程序。
Input:
●fight.m
function,dx=fight(t,x)
global,N1,N2,r1,r2,q1,q2;
dx=zeros(2,1);
dx
(1)=r1*x
(1)*(1-x
(1)./N1-q1*x
(2)./N2)
dx
(2)=r2*x
(2)*(1-q2*x
(1)./N1-x
(2)./N2)
%dx
(1)=r1*x
(1)*(1-x
(1)./N1)
%dx
(2)=r2*x
(2)*(1-x
(2)./N2)
●fight2.m
global,N1,N2,r1,r2,q1,q2;
N1=input('种群1最大容量=,');
n1=input('种群1初始数量=,');
N2=input('种群2最大容量=,');
n2=input('种群1初始数量=,');
q1=input('请输入相关系数q1=');
q2=input('请输入相关系数q2=');
r1=0.05;
r2=0.01;
ts=0:
1:
5000;
x0=[n1,n2];
[t,x]=ode45('fight',ts,x0);
%plot(x(:
1),x(:
2))
plot(t,x(:
1),'b')
hold,on
plot(t,x(:
2),'r')
result:
取N1=100,N2=200,n1=10,n2=20,q1=0.3,q2=0.5
实验二:
社会经济系统的冲量过程分析
1.熟练matlab的基本函数,掌握矩阵运算的函数使用:
矩阵
input:
matrix_1.m
A=,[1,2,3;
,,,4,5,6;
,,,7,8,9];
B=,[1,1,1;
,,,2,2,2;
,,,3,3,3];
A
B
disp('
(1)矩阵的加减')
C=A+B
D=A-B
disp('
(2)矩阵的乘法')
E=A*B
disp('(3)矩阵的逆运算')
inv(A)
disp('4)矩阵的幂运算')
A^3
disp('(5)矩阵的特征值函数(x为特征向量矩阵,y为特征值矩阵')
F=,[7,3,-2;
,,3,4,-1;
,,-2,-1,3];
F
[x,y]=eig(F)
disp('(6)矩阵的转置')
F'
disp('(6)矩阵的除法')
disp('A右除B')
A/B
disp('A左除B')
A\B
output:
A,=
,,,,1,,,,,2,,,,,3
,,,,4,,,,,5,,,,,6
,,,,7,,,,,8,,,,,9
B,=
,,,,1,,,,,1,,,,,1
,,,,2,,,,,2,,,,,2
,,,,3,,,,,3,,,,,3
(1)矩阵的加减
C,=
,,,,2,,,,,3,,,,,4
,,,,6,,,,,7,,,,,8
,,,10,,,,11,,,,12
D,=
,,,,0,,,,,1,,,,,2
,,,,2,,,,,3,,,,,4
,,,,4,,,,,5,,,,,6
(2)矩阵的乘法
E,=
,,,14,,,,14,,,,14
,,,32,,,,32,,,,32
,,,50,,,,50,,,,50
(3)矩阵的逆运算
Warning:
Matrix,is,close,to,singular,or,badly,scaled.
,,,,,,,,Results,may,be,inaccurate.,RCOND,=,1.541976e-018.
>,In,F:
\MATLAB6p5\work\matrix_1.m,at,line,15
ans,=
,1.0e+016,*
,,-0.4504,,,,0.9007,,,-0.4504
,,,0.9007,,,-1.8014,,,,0.9007
,,-0.4504,,,,0.9007,,,-0.4504
4)矩阵的幂运算
ans,=
,,,,,,,,468,,,,,,,,,576,,,,,,,,,684
,,,,,,,1062,,,,,,,,1305,,,,,,,,1548
,,,,,,,1656,,,,,,,,2034,,,,,,,,2412
(5)矩阵的特征值函数(x为特征向量矩阵,y为特征值矩阵
F,=
,,,,7,,,,,3,,,,-2
,,,,3,,,,,4,,,,-1
,,,-2,,,,-1,,,,,3
x,=
,,,0.5774,,,-0.0988,,,-0.8105
,,-0.5774,,,,0.6525,,,-0.4908
,,,0.5774,,,,0.7513,,,,0.3197
y,=
,,,2.0000,,,,,,,,,0,,,,,,,,,0
,,,,,,,,0,,,,2.3944,,,,,,,,,0
,,,,,,,,0,,,,,,,,,0,,,,9.6056
(6)矩阵的转置
ans,=
,,,,7,,,,,3,,,,-2
,,,,3,,,,,4,,,,-1
,,,-2,,,,-1,,,,,3
(6)矩阵的除法
A右除B
Warning:
Matrix,is,singular,to,working,precision.
(Type,"warning,off,MATLAB:
singularMatrix",to,suppress,this,warning.)
>,In,F:
\MATLAB6p5\work\matrix_1.m,at,line,32
ans,=
,,Inf,,,Inf,,,Inf
,,Inf,,,Inf,,,Inf
,,Inf,,,Inf,,,Inf
A左除B
Warning:
Matrix,is,close,to,singular,or,badly,scaled.
,,,,,,,,Results,may,be,inaccurate.,RCOND,=,1.541976e-018.
(Type,"warning,off,MATLAB:
nearlySingularMatrix",to,suppress,this,warning.)
>,In,F:
\MATLAB6p5\work\matrix_1.m,at,line,34
ans,=
,,-0.3333,,,-0.3333,,,-0.3333
,,,0.6667,,,,0.6667,,,,0.6667
,,,,,,,,0,,,,,,,,,0,,,,,,,,,0
2.分析社会经济系统的演化
第一步,输入能源经济系统有向图的邻接矩阵A;
第二步,计算
;
第三步,分析变化的规律,画出系统中每个变量随时间变化的增减趋势图。
Input:
ex_2.m
A=[0,-1,1,-1,0,0,0;
-1,0,0,,0,0,0,0;
0,-1,0,,0,1,0,0;
0,,0,0,,0,0,0,1;
1,,0,0,,0,0,1,0;
0,,0,0,,0,0,0,1;
1,,0,0,,0,0,0,0]
B=A;
for,i=2:
10
%sprintf('A的%d次幂为',i)
B=cat(4,B,,A^i);
end
B
row=input('要查看变量元素的行数=');
line=input('要查看变量元素的列数=');
temp=zeros(1,10);
for,j=1:
10
C=B(:
:
1,j);
temp(1,j)=C(row,line);
end
temp
disp('变化图形如下')
plot([1:
10],temp)
[x,y]=eig(A);
y
Output:
,,,,察看[1,2]号元素的变化
A,=
,,,,0,,,,-1,,,,,1,,,,-1,,,,,0,,,,,0,,,,,0
,,,-1,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0
,,,,0,,,,-1,,,,,0,,,,,0,,,,,1,,,,,0,,,,,0
,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,1
,,,,1,,,,,0,,,,,0,,,,,0,,,,,0,,,,,1,,,,,0
,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,1
,,,,1,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0
B(:
:
1,1),=
,,,,0,,,,-1,,,,,1,,,,-1,,,,,0,,,,,0,,,,,0
,,,-1,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0
,,,,0,,,,-1,,,,,0,,,,,0,,,,,1,,,,,0,,,,,0
,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,1
,,,,1,,,,,0,,,,,0,,,,,0,,,,,0,,,,,1,,,,,0
,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,1
,,,,1,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0
B(:
:
1,2),=
,,,,1,,,,-1,,,,,0,,,,,0,,,,,1,,,,,0,,,,-1
,,,,0,,,,,1,,,,-1,,,,,1,,,,,0,,,,,0,,,,,0
,,,,2,,,,,0,,,,,0,,,,,0,,,,,0,,,,,1,,,,,0
,,,,1,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0
,,,,0,,,,-1,,,,,1,,,,-1,,,,,0,,,,,0,,,,,1
,,,,1,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0,,,,,0
,,,,0,,,,-1,,,,,1,,,,-1,,,,,0,,,,,0,,,,,0
B(:
:
1,3),=
,,,,1,,,,-1,,,,,1,,,,-1,,,,,0,,,,,1,,,,,0
,,,-1,,,,,1,,,,,0,,,,,0,,,,-1,,,,,0,,,,,1
,,,,0,,,,-2,,,,,2,,,,-2,,,,,0,,,,,0,,,,,1
,,,,0,,,,-1,,,,,1,,,,-1,,,,,0,,,,,0,,,,,0
,,,,2,,,,-1,,,,,0,,,,,0,,,,,1,,,,,0,,,,-1
,,,,0,,,,-1,,,,,1,,,,-1,,,,,0,,,,,0,,,,,0
,,,,1,,,,-1,,,,,0,,,,,0,,,,,1,,,,,0,,,,-1
B(:
:
1,4),=
,,,,1,,,,-2,,,,,1,,,,-1,,,,,1,,,,,0,,,,,0
,,,-1,,,,,1,,,,-1,,,,,1,,,,,0,,,,-1,,,,,0
,,,,3,,,,-2,,,,,0,,,,,0,,,,,2,,,,,0,,,,-2
,,,,1,,,,-1,,,,,0,,,,,0,,,,,1,,,,,0,,,,-1
,,,,1,,,,-2,,,,,2,,,,-2,,,,,0,,,,,1,,,,,0
,,,,1,,,,-1,,,,,0,,,,,0,,,,,1,,,,,0,,,,-1
,,,,1,,,,-1,,,,,1,,,,-1,,,,,0,,,,,1,,,,,0
B(:
:
1,5),=
,,,,3,,,,-2,,,,,1,,,,-1,,,,,1,,,,,1,,,,-1
,,,-1,,,,,2,,,,-1,,,,,1,,,,-1,,,,,0,,,,,0
,,,,2,,,,-3,,,,,3,,,,-3,,,,,0,,,,,2,,,,,0
,,,,1,,,,-1,,,,,1,,,,-1,,,,,0,,,,,1,,,,,0
,,,,2,,,,-3,,,,,1,,,,-1,,,,,2,,,,,0,,,,-1
,,,,1,,,,-1,,,,,1,,,,-1,,,,,0,,,,,1,,,,,0
,,,,1,,,,-2,,,,,1,,,,-1,,,,,1,,,,,0,,,,,0
B(:
:
1,6),=
,,,,2,,,,-4,,,,,3,,,,-3,,,,,1,,,,,1,,,,,0
,,,-3,,,,,2,,,,-1,,,,,1,,,,-1,,,,-1,,,,,1
,,,,3,,,,-5,,,,,2,,,,-2,,,,,3,,,,,0,,,,-1
,,,,1,,,,-2,,,,,1,,,,-1,,,,,1,,,,,0,,,,,0
,,,,4,,,,-3,,,,,2,,,,-2,,,,,1,,,,,2,,,,-1
,,,,1,,,,-2,,,,,1,,,,-1,,,,,1,,,,,0,,,,,0
,,,,3,,,,-2,,,,,1,,,,-1,,,,,1,,,,,1,,,,-1
B(:
:
1,7),=
,,,,5,,,,-5,,,,,2,,,,-2,,,,,3,,,,,1,,,,-2
,,,-2,,,,,4,,,,-3,,,,,3,,,,-1,,,,-1,,,,,0
,,,,7,,,,-5,,,,,3,,,,-3,,,,,2,,,,,3,,,,-2
,,,,3,,,,-2,,,,,1,,,,-1,,,,,1,,,,,1,,,,-1
,,,,3,,,,-6,,,,,4,,,,-4,,,,,2,,,,,1,,,,,0
,,,,3,,,,-2,,,,,1,,,,-1,,,,,1,,,,,1,,,,-1
,,,,2,,,,-4,,,,,3,,,,-3,,,,,1,,,,,1,,,,,0
B(:
:
1,8),=
,,,,6,,,,-7,,,,,5,,,,-5,,,,,2,,,,,3,,,,-1
,,,-5,,,,,5,,,,-2,,,,,2,,,,-3,,,,-1,,,,,2
,,,,5,,,-10,,,,,7,,,,-7,,,,,3,,,,,2,,,,,0
,,,,2,,,,-4,,,,,3,,,,-3,,,,,1,,,,,1,,,,,0
,,,,8,,,,-7,,,,,3,,,,-3,,,,,4,,,,,2,,,,-3
,,,,2,,,,-4,,,,,3,,,,-3,,,,,1,,,,,1,,,,,0
,,,,5,,,,-5,,,,,2,,,,-2,,,,,3,,,,,1,,,,-2
B(:
:
1,9),=
,,,,8,,,-11,,,,,6,,,,-6,,,,,5,,,,,2,,,,-2
,,,-6,,,,,7,,,,-5,,,,,5,,,,-2,,,,-3,,,,,1
,,,13,,,-12,,,,,5,,,,-5,,,,,7,,,,,3,,,,-5
,,,,5,,,,-5,,,,,2,,,,-2,,,,,3,,,,,1,,,,-2
,,,,8,,,-11,,,,,8,,,,-8,,,,,3,,,,,4,,,,-1
,,,,5,,,,-5,,,,,2,,,,-2,,,,,3,,,,,1,,,,-2
,,,,6,,,,-7,,,,,5,,,,-5,,,,,2,,,,,3,,,,-1
B(:
:
1,10),=
,,,14,,,-14,,,,,8,,,,-8,,,,,6,,,,,5,,,,-4
,,,-8,,,,11,,,,-6,,,,,6,,,,-5,,,,-2,,,,,2
,,,14,,,-18,,,,13,,,-13,,,,,5,,,,,7,,,,-2
,,,,6,,,,-7,,,,,5,,,,-5,,,,,2,,,,,3,,,,-1
,,,13,,,-16,,,,,8,,,,-8,,,,,8,,,,,3,,,,-4
,,,,6,,,,-7,,,,,5,,,,-5,,,,,2,,,,,3,,,,-1
,,,,8,,,-11,,,,,6,,,,-6,,,,,5,,,,,2,,,,-2
要查看变量元素的行数=1
要查看变量元素的列数=2
temp,=
,,,-1,,,,-1,,,,-1,,,,-2,,,,-2,,,,-4,,,,-5,,,,-7,,,-11,,,-14
变化图形如下
实验三:
软件开发人员的薪金问题分析
利用统计工具箱,完成软件开发人员的薪金问题的分析
(1)原始模型
Input:
function,wage=wage(X,Y)
A=[1,1,1,0;1,0,0,0;1,1,0,0;1,0,0,1;1,0,0,0;2,1,0,1;2,0,0,1;2,0,1,0;2,0,0,0;3,0,0,1;3,1,1,0;3,1,0,1;3,1,0,0;
,,4,0,1,0;4,1,0,0;4,0,0,0;4,0,0,1;5,0,0,1;5,0,0,0;5,1,1,0;6,0,1,0;6,1,0,0;6,0,0,1;6,1,0,1;7,1,1,0;8,0,0,1;
,,8,1,1,0;8,1,0,0;8,0,1,0;10,0,1,0;10,0,0,1;10,1,0,0;10,1,0,1;11,1,0,1;11,0,1,0;12,0,0,1;
,,12,1,0,0;13,0,1,0;13,1,0,1;14,0,0,1;15,1,0,0;16,1,0,1;16,0,0,1;16,0,1,0;17,0,0,1;20,0,1,0];
X=[ones(46,1),A]
Y=[13876;11608;18701;11283;11767;20872;11772;10535;12195;12313;14975;21371;19800;
,,11417;20263;13231;12884;13245;13677;15965;12366;21352;13839;22884;16978;14803;
,,17404;22184;13548;14467;15942;23174;23780;25410;14861;16882;
,,24170;15990;26330;17949;25685;27837;18838;17483;19207;19346]
[b,bint,r,rint,stats]=regress(Y,X);,,%求回归系数的点估计b和区间估计bint
b,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,%求点估计b
bint,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,%求区间估计bint
stats,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,%stats由,
(1)相关系数R^2,
(2)F值,(3)对应概率
figure
rcoplot(r,rint),,,,,,,,,,,,,,,,,,,,,,,,%画出残差r和置信区间rint
figure
plot(A(:
1),r,'k+'),,,,,,,,,,,,,,,,,,,,,%模型
(1)残差r和x1的关系
output:
b,=
,1.0e+004,*
,,,1.1033
,,,0.0546
,,,0.6883
,,-0.2994
,,,0.0148
bint,=
,1.0e+004,*
,,,1.0258,,,,1.1807
,,,0.0484,,,,0.0608
,,,0.6248,,,,0.7517
,,-0.3826,,,-0.2162
,,-0.0636,,,,0.0931
stats,=
0.9567,,226.4258,,,,,,,,,0
(2)修改模型
Input:
function,wage2=wage2(X,Y)
A=[1,1,1,0;1,0,0,0;1,1,0,0;1,0,0,1;1,0,0,0;2,1,0,1;2,0,0,1;2,0,1,0;2,0,0,0;3,0,0,1;3,1,1,0;3,1,0,1;3,1,0,0;
,,4,0,1,0;4,1,0,0;4,0,0,0;4,0,0,1;5,0,0,1;5,0,
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