数值分析实验报告6.docx

1、数值分析实验报告6实验名称:线性方程组迭代解法 1）实验目的：1. 熟悉Matlab编程。2. 学习线性方程组迭代解法的程序设计算法。2）实验题目：第一题：研究解线性方程组Ax=b迭代法收敛速度。A为20阶五对角距阵 要求：(1)选取不同的初始向量x0 及右端向量b，给定迭代误差要求，用雅可比迭代和高斯-赛德尔迭代法求解，观察得到的序列是否收敛？若收敛，记录迭代次数，分析计算结果并得出你的结论。(2)用SOR迭代法求解上述方程组，松弛系数取1 =1 for i=1:n m=0; %此步也可以用ifj=i条件判定一下。 for j=1:(i-1) m=m+a(i,j)*x(k,j); end f

2、or j=(i+1):n m=m+a(i,j)*x(k,j); end x(k+1,i)=(b(i)-m)/a(i,i); end l=0; %判定满足条件使循环停止迭代。 for i=1:n l=l+abs(x(k+1,i)-x(k,i); end if l=1 for i=1:n p=0;q=0; for j=1:(i-1) p=p+a(i,j)*x(k+1,j); end for j=(i+1):n q=q+a(i,j)*x(k,j); end x(k+1,i)=(b(i)-q-p)/a(i,i); end l=0; %判定满足条件使循环停止迭代。 for i=1:n l=l+abs(x

3、(k+1,i)-x(k,i); end if l=1 if w=2|w=1 请重新输入w的值，w在1与2之间; break end for i=1:n p=0;q=0; for j=1:(i-1) p=p+a(i,j)*x(k+1,j); end for j=i:n q=q+a(i,j)*x(k,j); end x(k+1,i)=x(k,i)+w*(b(i)-q-p)/a(i,i); end l=0; %判定满足条件使循环停止迭代。 for i=1:n l=l+abs(x(k+1,i)-x(k,i); end if le break end k=k+1;end%输出所有的x的值。 x(k+1,

4、:)k第二题：代码：雅克比迭代法实现的函数 function X = p211_1_JJ(n)Hn = GET_Hn(n);b = GET_b(n);temp = 0;X0 = zeros(1, n);X_old = zeros(1, n);X_new = zeros(1, n);disp(Now Jacobi method!);disp(Start with the vector that (0, 0, 0, .)T);for i = 1:n for k = 1:n X_old = X_new; temp = 0; for j = 1:n if(j = i) temp = temp + Hn

5、(i, j) * X_old(j); end end X_new(i) = (b(i) - temp) / Hn(i, i); endendX = X_new;endSOR迭代法实现的函数function X = p211_1_SOR(n, w)Hn = GET_Hn(n);b = GET_b(n);temp01 = 0;temp02 = 0;X0 = zeros(1, n);X_old = zeros(1, n);X_new = zeros(1, n);disp(Now Successive Over Relaxtion method!);disp(Start with the vector

6、 that (0, 0, 0, .)T);for i = 1:n for k = 1:n X_old = X_new; temp01 = 0; temp02 = 0; for j = 1:n if(j i) temp02 = temp02 + Hn(i, j) * X_old(j); end end end X_new(i) = w * (b(i) - temp01 - temp02) / Hn(i, i) + X_old(i);endX = X_new;End5）实验结果：第一题 雅克比（Jacobi）迭代法此时初值全取1； b=2.2 1.7 1.5 1.5 1.5 1.5 1.5 1.5

7、 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.7 2.2;yakebi(0.00001)ans = Columns 1 through 9 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 Columns 10 through 18 1.0000 1.0000 1.0000 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 Columns 19 through 20 0.9787 0.9793k = 12此时初值全取1； b=2.5 1.9 1.

8、5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.9 2.5;yakebi(0.00001)ans = Columns 1 through 12 1.0969 1.0707 1.0219 1.0103 1.0039 1.0016 1.0006 1.0003 1.0001 1.0001 1.0001 1.0001 Columns 13 through 20 1.0003 1.0006 1.0016 1.0039 1.0103 1.0219 1.0707 1.0969k = 14高斯赛得尔迭代法此时初值全取1； b=2.

9、2 1.7 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.7 2.2; gaoshisaideer(0.00001)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k =14此时初值全取

10、1； b=2.5 1.9 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.9 2.5;gaoshisaideer(0.00001)ans = Columns 1 through 12 1.0969 1.0707 1.0219 1.0103 1.0039 1.0016 1.0006 1.0003 1.0001 1.0001 1.0001 1.0001 Columns 13 through 20 1.0003 1.0006 1.0016 1.0039 1.0103 1.0219 1.0707 1.0969k = 1

11、4超松驰法 caosongci(0.00001,1.5)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 25 caosongci(0.00001,1.4)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970

12、0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 19 caosongci(0.00001,1.3)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.999

13、9 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 15 caosongci(0.00001,1.6)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 34 caosongci(0.00001,1.7)ans

14、= Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 47 caosongci(0.00001,1.9)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.00

15、00 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 150第二题：对于雅克比迭代法，通过执行以下代码 p211_1_JJ(6) p211_1_JJ(8) p211_1_JJ(10) 可以分别得到： Now Jacobi method!Start with the vector that (0, 0, 0, .)Tans = 2.4500 1.1036 0.6265 0.4060 0.2831 0.2071 Now Jacobi metho

16、d!Start with the vector that (0, 0, 0, .)Tans =2.7179 1.4101 0.8524 0.5809 0.4221 0.3198 0.2497 0.1995Now Jacobi method!Start with the vector that (0, 0, 0, .)Tans = Columns 1 through 9 2.9290 1.6662 1.0517 0.7423 0.5554 0.4315 0.3445 0.2807 0.2325 Column 10 0.1951 对于SOR迭代法，执行相对应代码n=6, =1,1.25,1.5的时

17、候 p211_1_SOR(6, 1)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans = 2.4500 1.1036 0.6265 0.4060 0.2831 0.2071 p211_1_SOR(6, 1.25)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans = 3.0625 0.2310 0.8704 0.3389 0.3141 0.2097 p211_1_SOR(6,

18、 1.5)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans = 3.6750 -1.1009 2.0106 -0.3994 0.7670 -0.0384与n=8, =1,1.25,1.5的时候 p211_1_SOR(8, 1)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans =2.7179 1.4101 0.8524 0.5809 0.4221 0.3198 0.2497

19、0.1995 p211_1_SOR(8, 1.25)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans =3.3973 0.4887 1.0898 0.5062 0.4501 0.3203 0.2573 0.2042 p211_1_SOR(8, 1.5)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans =4.0768 -0.9424 2.2923 -0.2753 0.9252

20、 0.0578 0.4071 0.1275与n=10, =1,1.25,1.5的时候 p211_1_SOR(10, 1)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans = Columns 1 through 9 2.9290 1.6662 1.0517 0.7423 0.5554 0.4315 0.3445 0.2807 0.2325 Column 10 0.1951 p211_1_SOR(10, 1.25)Now Successive Over Relaxtion method!

21、Start with the vector that (0, 0, 0, .)Tans = Columns 1 through 9 3.6612 0.7098 1.2835 0.6617 0.5807 0.4299 0.3506 0.2844 0.2363 Column 10 0.1984 p211_1_SOR(10, 1.5)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans = Columns 1 through 9 4.3935 -0.7958 2.5326 -0.1523 1.0720 0.1565 0.5050 0.2041 0.2819 Column 10 0.17666）实验结果分析与小结：1.在实习中，进行学习MATLAB，有助于将两者都更加的熟悉并加以充分利用。2.此次取的b对于雅克比迭代法、高斯赛得尔迭代法都是收敛的，对于相同的初值与右端向量明显可以看出高斯赛得尔迭代法比雅克比迭代法快，这与理论上的分析完全一致嘛。3.对于SOR迭代方法选择不同的松弛因子收敛次数大大不同，当松弛因子为1.1时，在同等条件下迭代最快.