例题matlab求解.docx

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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(o)(l),.x(o)(6) = (267,3.13,3.25336,3.56,3.72)构造累加生成序列X二(X ,X (6) = (2.67,5.8,9.05,12.41,15.97,19.69)对X作紧邻均值生成Z (Z (k) + d-l)k 2,.6编程如下：x=2.67 5

2、.8 9.05 12.41 15.97 19.69;z(l)=x(l);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 二(Z ,Z )=(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 3.13、-7.425 13.25B =-10.73 1Y =3.36-14.19 13.56117.83 1；3.72,对参数今进行最小二乘估

3、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 二-0.04396098154759662.92561659879905即 H =-0. 044, u=2. 96 W =-66. 55则GM(1, 1)白化方程为公().044x = 2.96dt预测模型为：x(1)伏 +1) = 69.22e04*k 一 66.551、关联度检验法： 采用matlab编程得到模

4、拟序列 for i=l:6X(i)=69.22*exp(0.044*(i-1 )-66.55: end format long gX(I)=X(I); for i=2:6 x(i)=X(i)-X(i-l); endX 结果如下：X =Columns 1 through 42.673.11367860537808 3.25373920141375 3.40010005288617Columns 5 through 6 3.553044560121343.71286887145915因此模拟序列为 公。)=G(O)(I),a。)(6) = (2.67,3.113,3.253,3.40,3.5533

5、712)求模拟序列和原始序列的相关度x(0)= (X ,X0)(6) = (2.67313325,3.36356372)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 xl= (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A= (0, 0.006 , 0.001, 0.002 , 0.003)两级差 M=maxmax =0.006m=minmin =0计算关联系数取p = 0.50.0037(k) = I：)+o.oo3Tj(X) = 1(2) =

6、0.333(3) = 0.75 77(4) = 0.67/(5) = 0.5计算关联度1 5片一工= 0.6366 0.65 k=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公。)二(幺。),公。)(6) = (2.67,3.113,3.253340,3.553,3.712)原始序列为x(0)= (x(o)(l),.x(o)(6) = (2.67,3.13,3.25,3.36,3.56,3.72)由6的=卜叫女)一父(“得，残差序列为：e = (e ,e(6) = (0,0.017,0.003,0.04,0.007,0.008)所以，相对相对误差为:rel(k

7、) = - = (0,0.005,0.001,0.012,0.002,0.002) X (Z)平均相对误差为:1 6O = Nw/= 0.00376 A-=I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：(。)=(仆。),铲)(6)= (2.67,3.113,3.253,3.40,3.553,3.712) 原始序列：工=(X,%)=(267,313,3.25336,3.56,372)残差序列e = (e(l),e(6) = (0,0.017,0.003,0.04,0.007,0.008)% = -Xw() = 3.286 =采用VC编程完成均方差比值C

8、的解答程序：#include#includevoid main() int i;double x6=2.67, 3.13, 3.25, 3.36, 3.56, 3.72Hx 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=0.00,s,c=0.00,d,e=0.00,f,wyf 为 S2,s 为 sl,w 为均方差比值 Cfor(i=0;i6;i+)a+=(xi-3.28)*(xi-3.28);)s=sqrt(a6);printf(a=%f,s=%n,a,s);fbr(i=0;i6;i

9、+)(bi=xi-yi;printf(nb%d=%fn,i,bi);c+=bi;) d=c6;printf(,c=%f,d=%n,c,d);for(i=0;i6;i+)e+=(bi-d)*(bi-d);)f=sqrt(e6);w=f7s;printf(,f=%f,w=%fn,f,w);Sl = J-k(*)-fl =s = 08204V A=Is2=J7eW-e =/=0.0328V 6 k=,c = g = w = 0.0399(0.5 精度为 2 级，合格Sl小误差概率：P = Me -W 0.6745S= 46伙)-4 0.5533598= 1精度为 1 级好所以，模型的精度级别=MaX1的级别，C的级别, 因此模型的精度级别为1级，所以后验差检验通过。因此，三种检验方法检验都能通过，则可以用所建模型进行预测。当k=6时,得至IJaD= 23.58所以公。)= 23.58因此由预测模型得2005年的产品销售额为23.58亿元