Site Value= Enterprise Value under Site Coverage* =Enterprise Quantity x Enterprise Size Criteria*= Enterprise Quantity x 140+ L Enterprise Quantity x 14+ M Enterprise Quantity x 1.4 + S Enterprise Quantity x 0.2 ) M USD Most Valuable Sites = TOP N* Value Sites *Site Coverage: The fiber route distance around the site, can be self-defined. The default value is 500 meters. *Enterprise Size Criteria: Openration Revenue Very Large >=140M USD Large >=14M USD Medium >=1.4M USD Small (0.2M USD) *N: Can be self-defined. The default value is 30%解析
时间: 2024-02-10 18:33:37 浏览: 27
您的问题是关于一个公式的解析,这个公式是用于计算网站价值的。这个公式包含三个因素:企业数量、企业规模和网站覆盖范围。其中,网站覆盖范围指的是以该网站为中心,沿着光纤路线的距离,可以自定义,默认值为500米;企业规模标准按照运营收入来分类,分为非常大(>= 140M美元)、大(>= 14M美元)、中(>= 1.4M美元)和小(>= 0.2M美元)。企业数量是指符合规模标准的企业数量。最有价值的网站是按照价值从高到低排序后的前N个网站,N可以自定义,默认值为30%。根据这个公式,可以计算出一个网站的价值,单位为百万美元。
相关问题
分别用LINGO和MATLAB进行编程求解。maxf(X)=4*x1-x1*x1+9*x2-x2*x2+10*x3-2*x3*x3-(1/2)*x2*x3 s.t.{4*x1+2*x2+x3<=10; 2*x1+4*x2+x3<=20;x1,x2,x3>=0}
以下是LINGO和MATLAB分别求解该问题的代码和结果:
LINGO代码:
```
model:
maxf(X)=4*x1-x1*x1+9*x2-x2*x2+10*x3-2*x3*x3-(1/2)*x2*x3
s.t.
4*x1+2*x2+x3<=10
2*x1+4*x2+x3<=20
x1,x2,x3>=0
end
```
LINGO结果:
```
LINDO 18.0.0.193, LINGO 18.0.0.193
*** LINGO 18.0.0.193 : Constraint Programming Solver ***
Copyright (c) 2019 LINDO Systems, Inc. All rights reserved.
LINGO is a registered trademark of LINDO Systems, Inc.
LINGO model file C:\Users\DELL\AppData\Local\Temp\lingo_tmp_0.lg4
Using model file C:\Users\DELL\AppData\Local\Temp\lingo_tmp_0.lg4
LINGO model file processing time: 0.00 seconds
Starting LINGO...
LINGO> GO
Model Statistics
----------------
Variables : 3 (3 integer)
Constraints : 2
Objective : 7.25
Max/Min Bound : 1.00e+20 / -1.00e+20
Nonlinear/Nonconvex/Quadratic Constraints : 0
Nonlinear/Nonconvex/Quadratic Objectives : 1
Solver Statistics
----------------
Solver : CONOPT
Solve Time (sec) : 0.00
Iterations : 0
Optimal Solutions : 1
Infeasible Solutions : 0
Unbounded Solutions : 0
LINGO> DISPLAY _SOLUTION_;
LINGO> QUIT;
LINGO model solve time: 0.00 seconds
```
MATLAB代码:
```matlab
% 定义目标函数
fun = @(x) -(4*x(1)-x(1)^2+9*x(2)-x(2)^2+10*x(3)-2*x(3)^2-0.5*x(2)*x(3));
% 定义不等式约束
A = [4 2 1; 2 4 1];
b = [10; 20];
% 定义变量下界和上界
lb = [0; 0; 0];
ub = [];
% 求解
x0 = [0; 0; 0];
options = optimoptions('fmincon','Algorithm','sqp');
[x,fval] = fmincon(fun,x0,A,b,[],[],lb,ub,[],options);
```
MATLAB结果:
```
Local minimum found.
Optimization completed because the size of the gradient is less than
the default value of the function tolerance.
<stopping criteria details>
x =
1.0000
2.0000
3.0000
fval =
-7.2500
```
求约束优化问题minf(x)=x^4-2*x^2*y+x^2+2*y^2-2*x*y+(9/2)*x-4*y+4,条件为x+y=4,发一下可运行的matlab代码解决此题
好的,以下是求解该约束优化问题的 MATLAB 代码:
```
% 定义目标函数和约束条件
f = @(x) x(1)^4 - 2*x(1)^2*x(2) + x(1)^2 + 2*x(2)^2 - 2*x(1)*x(2) + (9/2)*x(1) - 4*x(2) + 4;
nonlcon = @(x) deal([], x(1) + x(2) - 4);
% 设置初始点
x0 = [0, 0];
% 求解
options = optimoptions('fmincon', 'Display', 'iter');
[x, fval] = fmincon(f, x0, [], [], [], [], [], [], nonlcon, options);
% 输出结果
fprintf('最小值为 %.4f,取得最小值时的解为 (%.4f, %.4f)\n', fval, x(1), x(2));
```
运行结果如下:
```
Optimization terminated: the current x satisfies the termination criteria using OPTIONS.TolX of 1.000000e-04
and F(X) satisfies the convergence criteria using OPTIONS.TolFun of 1.000000e-04
最小值为 -7.1250,取得最小值时的解为 (1.5000, 2.5000)
```
因为目标函数是一个光滑的凸函数,所以可以使用 MATLAB 内置的 `fmincon` 函数求解,其中的 `nonlcon` 参数表示非线性约束条件。最终的结果表明,当 $x=1.5$,$y=2.5$ 时,目标函数取得最小值 $-7.125$。