1、建立GM (1, 1)模型對(duì)產(chǎn)品銷(xiāo)售額預(yù)測(cè)祁詩(shī)陽(yáng)馮曉凱申靜某大型企業(yè)1999年至2004年的產(chǎn)品銷(xiāo)售額如下表，試建立GM(1,1)預(yù)測(cè)模型，并預(yù)測(cè) 2005年的產(chǎn)品銷(xiāo)售額。年份199920002001200220032004銷(xiāo)售額 (億元)2.673.133.253.363.563.72有題目知X(0)= (X(0)(1),. X(0)(6) = (2.67,3.13,3.25,3.36,3.56,3.72)構(gòu)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)對(duì)X (1作緊鄰均值生成_ 1，、z (k) = _ (

2、z (k) + z (k -1)2k = 2,.6編程如下：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結(jié)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)于是構(gòu)造b矩陣和y矩陣如下：-4.235 1、(3.13)-7.425 13.2

3、5B =-10.73 1y =3.36-14.19 13.5617.83 L、3.72 /對(duì)參數(shù)&進(jìn)行最小二乘估計(jì)，采用matlab編程完成解答如下: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結(jié)果如下： a =-0.04396098154759662.92561659879905即 a =-0.044, u=2.96- =-66.55d則GM(1,1)白化方程為dxcr0.044x = 2.96 dt預(yù)測(cè)模型為：X(k +1) = 6

4、9.22e0.044*k - 66.551、關(guān)聯(lián)度檢驗(yàn)法：米用matlab編程得到模擬序列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結(jié)果如下：x =Columns 1 through 42.673.11367860537808 3.25373920141375 3.40010005288617Columns 5 through 63.553044560121343.71286887145915因此模擬序列為(0)= (0)(1),. (0)(

5、6) = (2.67,3.113,3.253,3.40,3.553,3.712)求模擬序列和原始序列的相關(guān)度 (0) = ( (0)(1),. (0)(6) = (2.67,3.13,3.25,3.36,3.56,3.72)初始化，即將該序列所有數(shù)據(jù)分別除以第一個(gè)數(shù)據(jù)。原始序列變?yōu)?x1=(1，1.172，1.217，1.258，1.333，1.393)模擬序列變?yōu)?x2=(1，1.166，1.218，1.273，1.331，1.390)序列差=(0，0.006, 0.001，0.002, 0.003)兩級(jí)差 M=maxmax A =0.006m=minmin A =0計(jì)算關(guān)聯(lián)系數(shù)取P = 0

6、.50.003門(mén)(k)= 一:A(k )| + 0.003n (1) = 1 n (2) = 0.3333) = 0.75 4) = 0.65) = 0.5計(jì)算關(guān)聯(lián)度Y = 1 寸 n (k) = 0.6366 0.65k=1因此此模型符合，預(yù)測(cè)出來(lái)的2005年的產(chǎn)品銷(xiāo)售額也可信。2、殘差檢驗(yàn)?zāi)M序列為 (0) = ( (0)(1),. (0)(6) = (2.67,3.113,3.253,3.40,3.553,3.712)原始序列為 (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，殘差序列為：e = (e(1)，e(6) = (0,0.017,0.003,0.04,0.007,0.008)所以，相對(duì)相對(duì)誤差為：rel (k) = M)= (0,0.005,0.001,0.012,0.002,0.002) x(0)(k)平均相對(duì)誤差為：。=1 寸 rel(k) = 0.0037k=1從上述可得到平均精度為99.63%，所以模型符合，預(yù)測(cè)結(jié)果可信。3、后驗(yàn)差檢驗(yàn)?zāi)M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)殘差序列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編程完成均方差比值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 為初始序列double y6=2.67, 3.113, 3.253, 3.4, 3.553, 3.712;/y 為模擬序列double b6;double a=0.

9、00,s,c=0.00,d,e=0.00,f,w;/f 為 S2,s 為 s1,w 為均方差比值 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精度為2級(jí)，合格1小誤差概率：0.5533598)= 1精度為1級(jí)好p = p|(k)-e 0.6745S=