Question ideas
The competition questions will be updated as soon as they are released. Follow Digital Analog Kitty: https://blog.csdn.net/alionCUMT?spm=1011.2266.3001.5343
Complete code
The competition questions will be updated as soon as they are released. Follow Digital Analog Kitty: https://blog.csdn.net/alionCUMT?spm=1011.2266.3001.5343
Competition information
The 13th Asia-Pacific College Student Mathematical Modeling Competition in 2023 will be hosted by the Beijing Image and Graphics Society and organized by the Asia-Pacific College Student Mathematical Modeling Competition Organizing Committee. The competition follows the competition charter and relevant regulations. Colleges and universities are warmly welcome to organize students to register for the competition.
In the 12th Asia-Pacific College Student Mathematical Modeling Competition in 2022, there will be 9,700 teams from around the world, covering 969 universities, with more than 27,000 students actively participating. Participating universities include 39 domestic 985 universities and 114 211 universities, such as Peking University, Tsinghua University, Zhejiang University, Tongji University, Shanghai Jiao Tong University, Fudan University, Sichuan University, Dalian University of Technology, etc. In addition, in addition to universities in mainland China, there are also University of California, Berkeley, Johns Hopkins University, and New York University from the United States; Middlesex University, University of Oxford, University of Liverpool, University of Nottingham, and University of Edinburgh from the United Kingdom; Germany RWTH Aachen University and North Hessen University of Applied Sciences; St. Petersburg State University of Architecture and Architecture in Russia; University of Melbourne and University of Sydney in Australia; University of Malaya in Malaysia; Tohoku University in Japan; Pantheon-Assas University in Paris, France ; City University of Macau, Macau University of Science and Technology, Macau Polytechnic, and University of Macau in Macau; Beijing Normal University-Hong Kong Baptist University Joint International College, Chinese University of Hong Kong, Hong Kong University of Science and Technology, and Hong Kong Polytechnic University in Hong Kong; as well as Sino-foreign cooperation Universities such as the University of Nottingham Ningbo, Shenzhen Moscow State University, and Xi'an Jiaotong-Liverpool University are all participating.
At present, the competition has significant influence internationally and is regarded as one of the evaluation competitions among domestic universities as a warm-up competition for the US competition, bonus points for graduate studies, bonus points for comprehensive assessment, and innovative scholarships.
analytic hierarchy process
The problem is boiled down to the determination of the relative importance of the lowest layer relative to the highest layer (goal)
Model solution steps:
1. 建立模型
2. 构造判断矩阵，两两比较
一致性检验:检查传递性，是否在误差区间
3. 构造每一层层次比较矩阵
4. 求最底层对最高层权重。
Premise: The judgment matrix construction is reasonable
You can solve sub-problems instead of using the entire analytic hierarchy process.
Consistency test code: input hierarchical comparison matrix, and output weights and consistency test results at the same time
disp('请输入判断矩阵A(n阶)');
A=input('A=');
[n,n]=size(A);
x=ones(n,100);
y=ones(n,100);
m=zeros(1,100);
m(1)=max(x(:,1));
y(:,1)=x(:,1);
x(:,2)=A*y(:,1);
m(2)=max(x(:,2));
y(:,2)=x(:,2)/m(2);
p=0.0001;i=2;k=abs(m(2)-m(1));
while k>p
i=i+1;
x(:,i)=A*y(:,i-1);
m(i)=max(x(:,i));
y(:,i)=x(:,i)/m(i);
k=abs(m(i)-m(i-1));
end
a=sum(y(:,i));
w=y(:,i)/a;
t=m(i);
disp(w);
%以下是一致性检验
CI=(t-n)/(n-1);RI=[0 0 0.52 0.89 1.12 1.26 1.36 1.41 1.46 1.49 1.52 1.54 1.56 1.58 1.59];
CR=CI/RI(n);
if CR<0.10
disp('此矩阵的一致性可以接受!');
disp('CI=');disp(CI);
disp('CR=');disp(CR);
end
Multi-attribute decision-making model
Methods for sorting solutions using existing decision-making information
Using WAA: Weighted Arithmetic Average Operator Calculation
Step1: Normalization of attribute values (all have fixed formulas):
- Benefit type
- cost type
- Fixed type
- District type
…
Step2:(Same level analysis method) Construct a comparison matrix: Calculate the weight of each attribute
Step3:Normalized byattribute weightx The processed data can obtain the weight of each plan, and just compare the weight of each plan.
Gray forecast
A method of making predictions with a small amount of incomplete information.
GM(1,1) prediction:
function []=greymodel(y)
% 本程序主要用来计算根据灰色理论建立的模型的预测值。
% 应用的数学模型是 GM(1,1)。
% 原始数据的处理方法是一次累加法。
y=input('请输入数据 ');
n=length(y);
yy=ones(n,1);
yy(1)=y(1);
for i=2:n
yy(i)=yy(i-1)+y(i);
end
B=ones(n-1,2);
for i=1:(n-1)
B(i,1)=-(yy(i)+yy(i+1))/2;
B(i,2)=1;
end
BT=B';
for j=1:n-1
YN(j)=y(j+1);
end
YN=YN';
A=inv(BT*B)*BT*YN;
a=A(1);
u=A(2);
t=u/a;
i=1:n+2;
yys(i+1)=(y(1)-t).*exp(-a.*i)+t;
yys(1)=y(1);
for j=n+2:-1:2
ys(j)=yys(j)-yys(j-1);
end
x=1:n;
xs=2:n+2;
yn=ys(2:n+2);
plot(x,y,'^r',xs,yn,'*-b');
det=0;
sum1=0;
sumpe=0;
for i=1:n
sumpe=sumpe+y(i);
end
pe=sumpe/n;
for i=1:n;
sum1=sum1+(y(i)-pe).^2;
end
s1=sqrt(sum1/n);
sumce=0;
for i=2:n
sumce=sumce+(y(i)-yn(i));
end
ce=sumce/(n-1);
sum2=0;
for i=2:n;
sum2=sum2+(y(i)-yn(i)-ce).^2;
end
s2=sqrt(sum2/(n-1));
c=(s2)/(s1);
disp(['后验差比值为：',num2str(c)]);
if c<0.35
disp('系统预测精度好')
else if c<0.5
disp('系统预测精度合格')
else if c<0.65
disp('系统预测精度勉强')
else
disp('系统预测精度不合格')
end
end
end
disp(['下个拟合值为 ',num2str(ys(n+1))]);
disp(['再下个拟合值为',num2str(ys(n+2))]);
dijkstra algorithm
Can find the shortest path from the starting point to all other vertices
Note: The weight in the first file is the weighted adjacency matrix
weight= [0 2 8 1 Inf Inf Inf Inf Inf Inf Inf;
2 0 6 Inf 1 Inf Inf Inf Inf Inf Inf;
8 6 0 7 5 1 2 Inf Inf Inf Inf;
1 Inf 7 0 Inf Inf 9 Inf Inf Inf Inf;
Inf 1 5 Inf 0 3 Inf 2 9 Inf Inf;
Inf Inf 1 Inf 3 0 4 Inf 6 Inf Inf;
Inf Inf 2 9 Inf 4 0 Inf 3 1 Inf;
Inf Inf Inf Inf 2 Inf Inf 0 7 Inf 9;
Inf Inf Inf Inf 9 6 3 7 0 1 2;
Inf Inf Inf Inf Inf Inf 1 Inf 1 0 4;
Inf Inf Inf Inf Inf Inf Inf 9 2 4 0;];
[dis, path]=dijkstra(weight,1, 11)
dijkstra.m
function [min,path]=dijkstra(w,start,terminal)
n=size(w,1); label(start)=0; f(start)=start;
for i=1:n
if i~=start
label(i)=inf;
end, end
s(1)=start; u=start;
while length(s)<n
for i=1:n
ins=0;
for j=1:length(s)
if i==s(j)
ins=1;
end,
end
if ins==0
v=i;
if label(v)>(label(u)+w(u,v))
label(v)=(label(u)+w(u,v));
f(v)=u;
end,
end,
end
v1=0;
k=inf;
for i=1:n
ins=0;
for j=1:length(s)
if i==s(j)
ins=1;
end,
end
if ins==0
v=i;
if k>label(v)
k=label(v); v1=v;
end,
end,
end
s(length(s)+1)=v1;
u=v1;
end
min=label(terminal); path(1)=terminal;
i=1;
while path(i)~=start
path(i+1)=f(path(i));
i=i+1 ;
end
path(i)=start;
L=length(path);
path=path(L:-1:1);
Floyd algorithm
Find the shortest path from the starting point to all other vertices and output the path matrix. You need to look at it manually. Like the results of the dijstra algorithm, the two algorithms can be analyzed at the same time to confirm each other.
a= [ 0,50,inf,40,25,10;
50,0,15,20,inf,25;
inf,15,0,10,20,inf;
40,20,10,0,10,25;
25,inf,20,10,0,55;
10,25,inf,25,55,0];
[D, path]=floyd(a)
function [D,path,min1,path1]=floyd(a,start,terminal)
D=a;n=size(D,1);path=zeros(n,n);
for i=1:n
for j=1:n
if D(i,j)~=inf
path(i,j)=j;
end,
end,
end
for k=1:n
for i=1:n
for j=1:n
if D(i,k)+D(k,j)<D(i,j)
D(i,j)=D(i,k)+D(k,j);
path(i,j)=path(i,k);
end,
end,
end,
end
if nargin==3
min1=D(start,terminal);
m(1)=start;
i=1;
path1=[ ];
while path(m(i),terminal)~=terminal
k=i+1;
m(k)=path(m(i),terminal);
i=i+1;
end
m(i+1)=terminal;
path1=m;
end
simulated annealing algorithm
Mimics the annealing phenomenon in nature. This section discusses the solution to the TSP problem
TSP issues cover:Road transportation, logistics planning, Internet setting nodes
function [ newpath , position ] = swap( oldpath , number )
% 对 oldpath 进 行 互 换 操 作
% number 为 产 生 的 新 路 径 的 个 数
% position 为 对 应 newpath 互 换 的 位 置
m = length( oldpath ) ; % 城 市 的 个 数
newpath = zeros( number , m ) ;
position = sort( randi( m , number , 2 ) , 2 ); % 随 机 产 生 交 换 的 位 置
for i = 1 : number
newpath( i , : ) = oldpath ;
% 交 换 路 径 中 选 中 的 城 市
newpath( i , position( i , 1 ) ) = oldpath( position( i , 2 ) ) ;
newpath( i , position( i , 2 ) ) = oldpath( position( i , 1 ) ) ;
end
function [ objval ] = pathfare( fare , path )
% 计 算 路 径 path 的 代 价 objval
% path 为 1 到 n 的 排 列 ，代 表 城 市 的 访 问 顺 序 ；
% fare 为 代 价 矩 阵 ， 且 为 方 阵 。
[ m , n ] = size( path ) ;
objval = zeros( 1 , m ) ;
for i = 1 : m
for j = 2 : n
objval( i ) = objval( i ) + fare( path( i , j - 1 ) , path( i , j ) ) ;
end
objval( i ) = objval( i ) + fare( path( i , n ) , path( i , 1 ) ) ;
end
function [ fare ] = distance( coord )
% 根 据 各 城 市 的 距 离 坐 标 求 相 互 之 间 的 距 离
% fare 为 各 城 市 的 距 离 ， coord 为 各 城 市 的 坐 标
[ v , m ] = size( coord ) ; % m 为 城 市 的 个 数
fare = zeros( m ) ;
for i = 1 : m % 外 层 为 行
for j = i : m % 内 层 为 列
fare( i , j ) = ( sum( ( coord( : , i ) - coord( : , j ) ) .^ 2 ) ) ^ 0.5 ;
fare( j , i ) = fare( i , j ) ; % 距 离 矩 阵 对 称
end
end
function [ ] = myplot( path , coord , pathfar )
% 做 出 路 径 的 图 形
% path 为 要 做 图 的 路 径 ，coord 为 各 个 城 市 的 坐 标
% pathfar 为 路 径 path 对 应 的 费 用
len = length( path ) ;
clf ;
hold on ;
title( [ '近似最短路径如下，路程为' , num2str( pathfar ) ] ) ;
plot( coord( 1 , : ) , coord( 2 , : ) , 'ok');
pause( 0.4 ) ;
for ii = 2 : len
plot( coord( 1 , path( [ ii - 1 , ii ] ) ) , coord( 2 , path( [ ii - 1 , ii ] ) ) , '-b');
x = sum( coord( 1 , path( [ ii - 1 , ii ] ) ) ) / 2 ;
y = sum( coord( 2 , path( [ ii - 1 , ii ] ) ) ) / 2 ;
text( x , y , [ '(' , num2str( ii - 1 ) , ')' ] ) ;
pause( 0.4 ) ;
end
plot( coord( 1 , path( [ 1 , len ] ) ) , coord( 2 , path( [ 1 , len ] ) ) , '-b' ) ;
x = sum( coord( 1 , path( [ 1 , len ] ) ) ) / 2 ;
y = sum( coord( 2 , path( [ 1 , len ] ) ) ) / 2 ;
text( x , y , [ '(' , num2str( len ) , ')' ] ) ;
pause( 0.4 ) ;
hold off ;
clear;
% 程 序 参 数 设 定
Coord = ... % 城 市 的 坐 标 Coordinates 第一列代表第一个城市的x，y坐标，第二列代表第二个城市的x，y坐标...
[ 0.6683 0.6195 0.4 0.2439 0.1707 0.2293 0.5171 0.8732 0.6878 0.8488 ; ...
0.2536 0.2634 0.4439 0.1463 0.2293 0.761 0.9414 0.6536 0.5219 0.3609 ] ;
t0 = 1 ; % 初 温 t0
iLk = 20 ; % 内 循 环 最 大 迭 代 次 数 iLk
oLk = 50 ; % 外 循 环 最 大 迭 代 次 数 oLk
lam = 0.95 ; % λ lambda
istd = 0.001 ; % 若 内 循 环 函 数 值 方 差 小 于 istd 则 停 止
ostd = 0.001 ; % 若 外 循 环 函 数 值 方 差 小 于 ostd 则 停 止
ilen = 5 ; % 内 循 环 保 存 的 目 标 函 数 值 个 数
olen = 5 ; % 外 循 环 保 存 的 目 标 函 数 值 个 数
% 程 序 主 体
m = length( Coord ) ; % 城 市 的 个 数 m
fare = distance( Coord ) ; % 路 径 费 用 fare
path = 1 : m ; % 初 始 路 径 path
pathfar = pathfare( fare , path ) ; % 路 径 费 用 path fare
ores = zeros( 1 , olen ) ; % 外 循 环 保 存 的 目 标 函 数 值
e0 = pathfar ; % 能 量 初 值 e0
t = t0 ; % 温 度 t
for out = 1 : oLk % 外 循 环 模 拟 退 火 过 程
ires = zeros( 1 , ilen ) ; % 内 循 环 保 存 的 目 标 函 数 值
for in = 1 : iLk % 内 循 环 模 拟 热 平 衡 过 程
[ newpath , v ] = swap( path , 1 ) ; % 产 生 新 状 态
e1 = pathfare( fare , newpath ) ; % 新 状 态 能 量
% Metropolis 抽 样 稳 定 准 则
r = min( 1 , exp( - ( e1 - e0 ) / t ) ) ;
if rand < r
path = newpath ; % 更 新 最 佳 状 态
e0 = e1 ;
end
ires = [ ires( 2 : end ) e0 ] ; % 保 存 新 状 态 能 量
% 内 循 环 终 止 准 则 ：连 续 ilen 个 状 态 能 量 波 动 小 于 istd
if std( ires , 1 ) < istd
break ;
end
end
ores = [ ores( 2 : end ) e0 ] ; % 保 存 新 状 态 能 量
% 外 循 环 终 止 准 则 ：连 续 olen 个 状 态 能 量 波 动 小 于 ostd
if std( ores , 1 ) < ostd
break ;
end
t = lam * t ;
end
pathfar = e0 ;
% 输 入 结 果
fprintf( '近似最优路径为：\n ' )
population competition model
To simulate competition between two populations, just change the initial parameters.
Application: Sales of similar products by different companies...
fun.m:
function dx=fun(t,x,r1,r2,n1,n2,s1,s2)
r1=1;
r2=1;
n1=100;
n2=100;
s1=0.5;
s2=2;
dx=[r1*x(1)*(1-x(1)/n1-s1*x(2)/n2);r2*x(2)*(1-s2*x(1)/n1-x(2)/n2)];
p3.m:
h=0.1;%所取时间点间隔
ts=[0:h:30];%时间区间
x0=[10,10];%初始条件
opt=odeset('reltol',1e-6,'abstol',1e-9);%相对误差1e-6，绝对误差1e-9
[t,x]=ode45(@fun,ts,x0,opt);%使用5级4阶龙格—库塔公式计算
plot(t,x(:,1),'r',t,x(:,2),'b','LineWidth',2),grid;
pause;
plot(x(:,1),x(:,2),'LineWidth',2),grid %作相轨线
queuing theory
Solve the relationship between the number of call lines and phone user calls
Model contains:
-
customer input process
-
Queuing structure and queuing rules
-
Service organizations and service rules
-
Arrival interval and service time distribution
System status parameters:
-
N(t) The total number of customers at time t
-
Transient probability P(t) The probability of system state N(t)=n at time t
-
Steady state probability P=lim P(t)
System operating indicator parameters:
-
Number of customers and Number of customers waiting to be served
-
Total timeandQueue time
-
Busy period and busy period service volume
-
loss rate
-
Service intensity
classic model
By inputting mu and lambda, calculate Lq, Ls, Ws, Wq.
M/M/1 queuing system
For a service desk, the customer source satisfies Poisson distribution, single team, single service desk, first in, first out, and random service time.
M/M/S queuing system
There are S service desks, each service desk is independent of each other, and the others are the same as above.
code
M/M/1 process simulation code
Change simulation time,mu,
Change lambda-> There are more service desks, fewer people come to a certain counter, and the pressure is reduced.
This string of code will generate a simulated image of the queuing process
clear
clc
%*****************************************
%初始化顾客源
%*****************************************
%总仿真时间
Total_time = 10;
%队列最大长度
N = 10000000000;
%到达率与服务率
lambda = 10;
mu = 6;
%平均到达时间与平均服务时间
arr_mean = 1/lambda;
ser_mean = 1/mu;
arr_num = round(Total_time*lambda*2);
events = [];
%按负指数分布产生各顾客达到时间间隔
events(1,:) = exprnd(arr_mean,1,arr_num);
%各顾客的到达时刻等于时间间隔的累积和
events(1,:) = cumsum(events(1,:));
%按负指数分布产生各顾客服务时间
events(2,:) = exprnd(ser_mean,1,arr_num);
%计算仿真顾客个数，即到达时刻在仿真时间内的顾客数
len_sim = sum(events(1,:)<= Total_time);
%*****************************************
%计算第 1个顾客的信息
%*****************************************
%第 1个顾客进入系统后直接接受服务，无需等待
events(3,1) = 0;
%其离开时刻等于其到达时刻与服务时间之和
events(4,1) = events(1,1)+events(2,1);
%其肯定被系统接纳，此时系统内共有
%1个顾客，故标志位置1
events(5,1) = 1;
%其进入系统后，系统内已有成员序号为 1
member = [1];
for i = 2:arr_num
%如果第 i个顾客的到达时间超过了仿真时间，则跳出循环
if events(1,i)>Total_time
break;
else
number = sum(events(4,member) > events(1,i));
%如果系统已满，则系统拒绝第 i个顾客，其标志位置 0
if number >= N+1
events(5,i) = 0;
%如果系统为空，则第 i个顾客直接接受服务
else
if number == 0
%其等待时间为 0
2009.1516
%PROGRAMLANGUAGEPROGRAMLANGUAGE
events(3,i) = 0;
%其离开时刻等于到达时刻与服务时间之和
events(4,i) = events(1,i)+events(2,i);
%其标志位置 1
events(5,i) = 1;
member = [member,i];
%如果系统有顾客正在接受服务，且系统等待队列未满，则 第 i个顾客进入系统
else len_mem = length(member);
%其等待时间等于队列中前一个顾客的离开时刻减去其到 达时刻
events(3,i)=events(4,member(len_mem))-events(1,i);
%其离开时刻等于队列中前一个顾客的离开时刻加上其服
%务时间
events(4,i)=events(4,member(len_mem))+events(2,i);
%标识位表示其进入系统后，系统内共有的顾客数
events(5,i) = number+1;
member = [member,i];
end
end
end
end
%仿真结束时，进入系统的总顾客数
len_mem = length(member);
%*****************************************
%输出结果
%*****************************************
%绘制在仿真时间内，进入系统的所有顾客的到达时刻和离
%开时刻曲线图（stairs：绘制二维阶梯图）
stairs([0 events(1,member)],0:len_mem);
hold on;
stairs([0 events(4,member)],0:len_mem,'.-r');
legend('到达时间 ','离开时间 ');
hold off;
grid on;
%绘制在仿真时间内，进入系统的所有顾客的停留时间和等
%待时间曲线图（plot：绘制二维线性图）
figure;
plot(1:len_mem,events(3,member),'r-*',1: len_mem,events(2,member)+events(3,member),'k-');
legend('等待时间 ','停留时间 ');
grid on;
M/M/S four indicator calculation codes
Change mu and lambda and convert them into number of service people per hour.
This string of code will calculate the four key indicators of M/M/S from the input parameters.
s=2;
mu=4;
lambda=3;
ro=lambda/mu;
ros=ro/s;
sum1=0;
for i=0:(s-1)
sum1=sum1+ro.^i/factorial(i);
end
sum2=ro.^s/factorial(s)/(1-ros);
p0=1/(sum1+sum2);
p=ro.^s.*p0/factorial(s)/(1-ros);
Lq=p.*ros/(1-ros);
L=Lq+ro;
W=L/lambda;
Wq=Lq/lambda;
fprintf('排队等待的平均人数为%5.2f人\n',Lq)
fprintf('系统内的平均人数为%5.2f人\n',L)
fprintf('平均逗留时间为%5.2f分钟\n',W*60)
fprintf('平均等待时间为%5.2f分种\n',Wq*60)
linear programming model
A small question within a larger mathematical modeling problem
Sample code:
max=2*x1+3*x2;
x1+2*x2<=8;
4*x1<=16;
4*x2<=12;
Nonlinear programming and 01 programming
Nonlinear: There are nonlinear components in it
@gin(x1): Find the integer of x1
Code example:
Model:
max=98*x1+277*x2-x1*x1-0.3*x1*x2-2*x2*x2;
x1+x2<100;
x1<=2*x2;
@gin(x1);
@gin(x2);
end
01 Planning: The problem of unknown quantities with values only 0 and 1 (reflected by restrictive conditions)
Code example:
Model:
Min=8*x11+13*x12+18*x13+23*x14+10*x21+14*x22+16*x23+27*x24+2*x31+10*x32+21*x33+26*x34+14*x41+22*x42+26*x43+28*x44;
x11+x12+x13+x14=1;
x21+x22+x23+x24=1;
x31+x32+x33+x34=1;
x41+x42+x43+x44=1;
x11+x21+x31+x41=1;
x12+x22+x32+x42=1;
x13+x23+x33+x43=1;
x14+x24+x34+x44=1;
end
int16
Principal component analysis
The operation is a bit confusing and I don’t understand it very well.
Just follow the video for SPSS operations, involving simple operations in Excel
1.输入数据
2.找到Compent Martix，里面有若干变量F1,F2
3.提取数据到excel
4.将变量Fi的数据除以变量Fi特征值的平方根，算出指标对应的系数，得出新因子Fi与原变量的线性关系
5.以特征值为权重，加权平均写出F表达式
6.写出表达式后，归一化数据，带入表达式，算出最终结果，比较F
Perform multiple variablesLinear transformationSelect a smaller number of important variables
Select the eigenvalue greater than 1 and calculate the index corresponding coefficient.
After selecting important variables, you can guess the actual meaning
Cluster analysis
Birds of a feather flock together, watch the video for details
Look at the tree diagram and icicle diagram, just put the diagram on top and put the classification conclusion.
Multiple regression analysis
Understand whether two or more variables are related, the direction and strength of the correlation
For example: the relationship between income level and education, industry, and type of work