如何用matlab求解logistic模型
t=[0,30,45,76]
p=[2200,4000,8260,10000]
p=L/(1+(L/P0-1)e^(-rt))
希望求解出L,p0,r
并画出预测图形
logistic模型属于非线性问题。所以求解其模型函数的系数可以用nlinfit（）非线性回归分析函数来解决。由于给出的数据偏少，通过有效的插值方法，增加合理的数据点。主要代码：
a0=[-35.287，813.17，0.0098613];
t=0:5:75; %t=[0,30,45,76];
p=[2200 2259.2 2429.3 2699.5 3058.7 3495.8 4000 5228.9 7048.4 8260 8738.6 9152.7 9494 9754.1 9924.9 9997.9]; %p=[2200,4000,8260,10000];
fun=@(a,t)a(1)./(1+(a(1)/a(2)-1)*exp(-a(3).*t));
a= nlinfit(t,p,fun,a0);
运行结果
L=11278.4096；P0=1257.3315；r=0.061172
决定系数R²：0.93268
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第1个回答 2014-05-28
建立m函数文件存为logistic1
function f=logistic1(b)
t=[0,5,10,24,33,48,57,72,96,120,144,168,192,216];y=[0,0.028,0.103,0.336,0.450,0.597,0.716,0.778,0.835,0.849,0.816,0.839,0.811,0.816];
f = y-b(1)./(1+b(2).*exp(-b(3).*t));
b0=[10,2,2];
>> b=leastsq('logistic1',b0)
b =
0.8221 13.9173 0.0818
或者cftool
General model:
f(x) = b/(1+a*exp(-k*x))
Coefficients (with 95% confidence bounds):
a = 13.92 (6.301, 21.53)
b = 0.822 (0.7911, 0.853)
k = 0.08184 (0.06479, 0.0989)
Goodness of fit:
SSE: 0.01404
R-square: 0.9898
Adjusted R-square: 0.9879
RMSE: 0.03572
function f=logistic1(b)
t=[0,5,10,24,33,48,57,72,96,120,144,168,192,216];y=[0,0.028,0.103,0.336,0.450,0.597,0.716,0.778,0.835,0.849,0.816,0.839,0.811,0.816];
f = y-b(1)./(1+b(2).*exp(-b(3).*t));
b0=[10,2,2];
>> b=leastsq('logistic1',b0)
b =
0.8221 13.9173 0.0818
或者cftool
General model:
f(x) = b/(1+a*exp(-k*x))
Coefficients (with 95% confidence bounds):
a = 13.92 (6.301, 21.53)
b = 0.822 (0.7911, 0.853)
k = 0.08184 (0.06479, 0.0989)
Goodness of fit:
SSE: 0.01404
R-square: 0.9898
Adjusted R-square: 0.9879
RMSE: 0.03572
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