灰色理论例题matlab求解

1、建立GM (1, 1)模型对产品销售额预测祁诗阳冯晓凯申静某大型企业1999年至2004年的产品销售额如下表，试建立GM(1,1)预测模型，并预测 2005年的产品销售额。年份199920002001200220032004销售额 (亿元)2.673.133.253.363.563.72有题目知X(0)= (X(0)(1),. X(0)(6) = (2.67,3.13,3.25,3.36,3.56,3.72)构造累加生成序列X (1) = (X (1),. X (6) = (2.67,5.8,9.05,12.41,15.97,19.69)对X (1作紧邻均值生成_ 1，、z (k) = _ (

2、z (k) + z (k -1)2k = 2,.6编程如下：x=2.67 5.8 9.05 12.41 15.97 19.69;z(1)=x(1);for i=2:6z(i)=0.5*(x(i)+x(i-1);endformat long gz结果如下：z =Columns 1 through 42.67 4.235 7.425 10.73Columns 5 through 614.19 17.83因此z (1) = (z (1)(1)，z (5) = (4.235,7.425,10.73,14.19,17.83)于是构造b矩阵和y矩阵如下：-4.235 1、(3.13)-7.425 13.2

3、5B =-10.73 1y =3.36-14.19 13.5617.83 L、3.72 /对参数&进行最小二乘估计，采用matlab编程完成解答如下:B=-4.235 -7.425 -10.73 -14.19 -17.83】，ones(5,1);Y=3.13 3.25 3.36 3.56 3.72;format long ga=inv(B*B)*B*Y结果如下： a =-0.04396098154759662.92561659879905即 a =-0.044, u=2.96- =-66.55d则GM(1,1)白化方程为dxcr0.044x = 2.96 dt预测模型为：X(k +1) = 6

4、9.22e0.044*k - 66.551、关联度检验法：米用matlab编程得到模拟序列for i=1:6X(i)=69.22*exp(0.044*(i-1)-66.55；endformat long gx(1)=X(1);for i=2:6x(i)=X(i)-X(i-1);endX结果如下：x =Columns 1 through 42.673.11367860537808 3.25373920141375 3.40010005288617Columns 5 through 63.553044560121343.71286887145915因此模拟序列为(0)= (0)(1),. (0)(

5、6) = (2.67,3.113,3.253,3.40,3.553,3.712)求模拟序列和原始序列的相关度 (0) = ( (0)(1),. (0)(6) = (2.67,3.13,3.25,3.36,3.56,3.72)初始化，即将该序列所有数据分别除以第一个数据。原始序列变为 x1=(1，1.172，1.217，1.258，1.333，1.393)模拟序列变为 x2=(1，1.166，1.218，1.273，1.331，1.390)序列差=(0，0.006, 0.001，0.002, 0.003)两级差 M=maxmax A =0.006m=minmin A =0计算关联系数取P = 0

6、.50.003门(k)= 一:A(k )| + 0.003n (1) = 1 n (2) = 0.3333) = 0.75 4) = 0.65) = 0.5计算关联度Y = 1 寸 n (k) = 0.6366 0.65k=1因此此模型符合，预测出来的2005年的产品销售额也可信。2、残差检验模拟序列为 (0) = ( (0)(1),. (0)(6) = (2.67,3.113,3.253,3.40,3.553,3.712)原始序列为 (0) = ( (0)(1),. (0)(6) = (2.67,3.13,3.25,3.36,3.56,3.72)由 e(k )= |x(。)(k )一文(。)

7、(k 得，残差序列为：e = (e(1)，e(6) = (0,0.017,0.003,0.04,0.007,0.008)所以，相对相对误差为：rel (k) = M)= (0,0.005,0.001,0.012,0.002,0.002) x(0)(k)平均相对误差为：。=1 寸 rel(k) = 0.0037k=1从上述可得到平均精度为99.63%，所以模型符合，预测结果可信。3、后验差检验模拟序列：x (0) = (x (0)(1),. x (0)(6) = (2.67,3.113,3.253,3.40,3.553,3.712) 原始序列：x (0) = (x (0)(1),. x (0)(

8、6) = (2.67,3.13,3.25,3.36,3.56,3.72)残差序列e = (e(1)，e(6) = (0,0.017,0.003,0.04,0.007,0.008)x = J_ X(0)(k) = 3.28 k=1采用VC编程完成均方差比值C的解答程序：#include#includevoid main() ( int i;double x6=2.67, 3.13, 3.25, 3.36, 3.56, 3.72;/x 为初始序列double y6=2.67, 3.113, 3.253, 3.4, 3.553, 3.712;/y 为模拟序列double b6;double a=0.

9、00,s,c=0.00,d,e=0.00,f,w;/f 为 S2,s 为 s1,w 为均方差比值 Cfor(i=0;i6;i+)a+=(xi-3.28)*(xi-3.28);s=sqrt(a/6);printf(a=%f,s=%fn”,a,s);for(i=0;i6;i+)bi=xi-yi;printf(b%d=%fn”,i,bi);c+=bi;d=c/6;printf(c=%f,d=%fn”,c,d);for(i=0;i6;i+)e+=(bi-d)*(bi-d);f=sqrt(e/6);w=f/s;printf(f=%f,w=%fn”,f,w);S = ：1 寸 X(0)(k)-X=s = 0.8204 1*5k=1S = ；1(k) J = f = 0.03282 6k=1c = % = w = 0.0399(0.5精度为2级，合格1小误差概率：0.5533598)= 1精度为1级好p = p|(k)-e 0.6745S=