灰色理论例题matlab求解

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)=(X⑴(1),...X⑴(6))=(2.67,5.8,9.05,12.41,15.97,19.69)对X(1作紧邻均值生成_1，、、z⑴(k)=_(z⑴(k)+z⑴(k-1))2k=2,.・・・6编程如下：x=[2.675.89.0512.4115.9719.69];z(1)=x(1);fori=2:6z(i)=0.5*(x(i)+x(i-1));endformatlonggz结果如下：z=Columns1through42.674.2357.42510.73Columns5through614.1917.83因此z(1)=(z(1)(1)，…z⑴(5))=(4.235,7.425,10.73,14.19,17.83)于是构造b矩阵和y矩阵如下：'-4.2351、(3.13)-7.42513.25B=-10.731y=3.36-14.1913.56"17.83L、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=[3.133.253.363.563.72]';formatlongga=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dx⑴cr0.044x=2.96dt预测模型为：X⑴(k+1)=69.22e0.044*k-66.551、关联度检验法：米用matlab编程得到模拟序列fori=1:6X(i)=69.22*exp(0.044*(i-1))-66.55；endformatlonggx(1)=X(1);fori=2:6x(i)=X(i)-X(i-1);endX结果如下：x=Columns1through42.673.113678605378083.253739201413753.40010005288617Columns5through63.553044560121343.71286887145915因此模拟序列为£(0)=(£(0)(1),...£(0)(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=maxmaxA=0.006m=minminA=0计算关联系数取P=0.50.003门(k)=一:A(k)|+0.003n(1)=1n(2)=0.333"3)=0.75"4)=0.6"5)=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(。)(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)(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)=3.28k=1采用VC编程完成均方差比值C的解答程序：#include<stdio.h>#include<math.h>voidmain()(inti;doublex[6]={2.67,3.13,3.25,3.36,3.56,3.72};//x为初始序列doubley[6]={2.67,3.113,3.253,3.4,3.553,3.712};//y为模拟序列doubleb[6];doublea=0.00,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;i<6;i++){a+=(x[i]-3.28)*(x[i]-3.28);}s=sqrt(a/6);printf("a=%f,s=%f\n”,a,s);for(i=0;i<6;i++)

{b[i]=x[i]-y[i];printf("b[%d]=%f\n”,i,b[i]);c+=b[i];}d=c/6;printf("c=%f,d=%f\n”,c,d);for(i=0;i<6;i++){e+=(b[i]-d)*(b[i]-d);}f=sqrt(e/6);w=f/s;printf("f=%f,w=%f\n”,f,w);S=：1寸X(0)(k)-X」=s=0.82041*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]｝=L^(