Question

我试图用while循环解决一些方程式。我有一个已知的固定输出。工作流程如下：

P = 100; %Desired output
x = 1;   %Initial guess

while abs(something) > 1e-6
    x1 = (25 * x)/2
    x2 = 10x - x1
    x3 = 20x - (x - x1 - x2)*2
    x4 = (x - x1 - x2 -x3)*12
    x5 = (x - x1 - x2 -x3) * 10 + x4    

    P1 = 2005x3 
    P2 = 1500x5

    Pnew = P1 + P2

end

我希望计算x1，x2，...，x5并在达到条件P = Pnew时停止循环。由于我对MATLAB的适度了解，我们将不胜感激。提前谢谢了。附录：也许我没有解释清楚。我的目标是在达到条件P = Pnew时停止循环。我的已知值P和初始值x。应该通过给定的一系列方程生成Pnew。满足条件时x，x1，x2，...，x5将拥有其值。因此x不仅是计算其他未知数的初始值（x1，x2，...，x5）。我尝试修改它但NaN Pnew，P1，error和inf P2，x，...

P = 100; %Desired output
x = 1;   %Initial guess
Pnew = P + 1

while abs(P - Pnew) > 1e-6
    x1 = (25 * x)/2;
    x2 = 10*x - x1;
    x3 = 20*x - (x - x1 - x2)*2;
    x4 = (x - x1 - x2 -x3)*12;
    x5 = (x - x1 - x2 -x3) * 10 + x4    

    P1 = 2005 * x3; 
    P2 = 150 * x5;

    Pnew = P1 - P2;
    error = abs(P - Pnew);
    x_new = x - .001 * error;
    x = x_new;

end

Answer 1

如果您不使用*来增加代码，则代码会出错。我已为您修复了以下代码。您要找的something是P - Pnew。这是因为你试图找到差异并确保它超过1e-6以继续循环。

由于您需要Pnew，因此您还应该对其进行初始计算。您应该注意，您的计算实际上并未使Pnew收敛到P。

P = 100; %Desired output
x = 1;   %Initial guess

x1 = ( 25 * x ) / 2
x2 = 10 * x - x1
x3 = 20 * x - ( x - x1 - x2 ) * 2
x4 = ( x - x1 - x2 -x3 ) * 12
x5 = ( x - x1 - x2 -x3 ) * 10 + x4    

P1 = 2005 * x3 
P2 = 1500 * x5

Pnew = P1 + P2

while abs( P - Pnew ) > 1e-6
    x1 = ( 25 * x ) / 2
    x2 = 10 * x - x1
    x3 = 20 * x - ( x - x1 - x2 ) * 2
    x4 = ( x - x1 - x2 -x3 ) * 12
    x5 = ( x - x1 - x2 -x3 ) * 10 + x4    

    P1 = 2005 * x3 
    P2 = 1500 * x5

    Pnew = P1 + P2

end

您可能还应该;终止该行以禁止计算输出。

这里的do-while循环可行，但Matlab没有。您可以使用封装来隐藏代码重复，也可以使用for循环，如下所示。

P = 100; %Desired output
x = 1;   %Initial guess

for i = 1:Inf

    x1 = ( 25 * x ) / 2
    x2 = 10 * x - x1
    x3 = 20 * x - ( x - x1 - x2 ) * 2
    x4 = ( x - x1 - x2 -x3 ) * 12
    x5 = ( x - x1 - x2 -x3 ) * 10 + x4    

    P1 = 2005 * x3 
    P2 = 1500 * x5

    Pnew = P1 + P2

    if abs( P - Pnew ) < 1e-6
        break;
    end
end

如果您有无限循环，要设置预定义的迭代终止，您可以将i = 1:Inf替换为i = 1:iterMax，其中iterMax是最大迭代次数。

Answer 2

通常在数值计算中，不会测试相等性，因为计算机无法区分所有实数。您的情况实际上是令人满意的== P - Pnew

在代码中它看起来像这样

P = 100; % Desired output
x = 1;   % Initial guess
Pnew = P + 1; % Something far off, just to get through the first check
MAX_ITERATION_COUNT = 10000; % a reasonable upper bound to your loop
loopCount = 0;    

while abs(P - Pnew) > 1e-6 && loopCount < MAX_ITERATION_COUNT
    loopCount = loopCount + 1;
    x1 = (25 * x) / 2;
    x2 = 10 * x - x1;
    x3 = 20 * x - (x - x1 - x2) * 2;
    x4 = (x - x1 - x2 - x3) * 12;
    x5 = (x - x1 - x2 - x3) * 10 + x4;

    P1 = 2005 * x3;
    P2 = 1500 * x5;

    Pnew = P1 + P2;
end

这应该适用于您的问题。

编辑：我添加了一个循环计数器，以防你的循环没有定期完成。

在Matlab中设置迭代循环的停止条件

2 个答案: