线性方程组迭代解法

1、-/实验六：线性方程组迭代解法1) 实验目的? 熟悉Matlab编程；? 学习线性方程组迭代解法的程序设计算法2) 实验题目1.研究解线性方程组12Ax=b迭代法收敛速度。A为20阶五对角距阵14121412要求：x0及右端向量b,给定迭代误差要求，用雅可比迭代和高斯-赛记录迭代次数，分析计算结果并得出(1) 选取不同的初始向量德尔迭代法求解，观察得到的序列是否收敛？若收敛， 你的结论。(2) 用SOR迭代法求解上述方程组，松弛系数3取1< 3 <2勺不同值，在时停止迭代.记录迭代次数，分析计算结果并得出你的结论。2.给出线性方程组HnXb，其中系数矩阵 H n为希尔伯特矩阵:Hn

2、hjrh,i,j 2,n.假设x 1,1,1n,b HnX.若取 n6,8,10,分别用雅可比迭代法及SOR迭代(1,1.25,1.5)求解,比较计算结果。3) 实验原理与理论基础1.雅克比(Jacobi)迭代法算法设计： 输入矩阵a与右端向量b及初值x(1,i); 按公式计算得.(k1)丄 biaiin(k) aijXj(i 1,2, n)2.高斯一一赛得尔迭代法算法设计：1. 输入矩阵a与右端向量b及初值x(1,i).2.x(k 1)-aii1(k 1) aijXj1n(k) aijXj(i =1,2,n)3.超松驰法算法设计：输入矩阵a与右端向量b及初值x(1,i) o x(k 1)(1

3、 )x(k)aiibi1(k 1) aijXj1(k) aijXj1,04) 实验内容 第一题实验程序：1.雅克比迭代法：function =yakebi(e)%俞入矩阵a与右端向量bofor i=1:20a(i,i)=3;endfor i=3:20 for j=i-2 a(i,j)=-1/4; a(j,i)=-1/4; endendfor i=2:20for j=i-1a(i,j)=-1/2; a(j,i)=-1/2; end endb=2.2 1.7 1.5 1.5 1.5 1.5 k=1;n=len gth(a);for i=1:nx(1,i)=1;%数组中没有第endwhile k&g

4、t;=1for i=1:n1.51.50行。1.5 1.5 1.51.51.5 1.51.5 1.5 1.5 1.5 1.7 2.2;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);endfor j=(i+1):n m=m+a(i,j)*x(k,j);endx(k+1,i)=(b(i)-m)/a(i,i); end1=0;%判定满足条件使循环停止迭代。for i=1:nl=l+abs(x(k+1,i)-x(k,i); endif l<ebreakendk=k+1;end%俞出所有的X的值。x(k+1,:)k2. 高斯一赛德尔迭代

5、法: function =gaoshisaideer(e) for i=1:20a(i,i)=3;endfor i=3:20for j=i-2a(i,j)=-1/4;a(j,i)=-1/4;end end for i=2:20for j=i-1a(i,j)=-1/2;a(j,i)=-1/2;endendb=2.2 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; k=1;n=len gth(a);for i=1:nx(1,i)=0;%数组中没有第0行。end while k>=1for

6、 i=1:np=0;q=0;for j=1:(i-1)p=p+a(i,j)*x(k+1,j);endfor j=(i+1):n q=q+a(i,j)*x(k,j);endx(k+1,i)=(b(i)-q-p)/a(i,i); endl=0;辔U定满足条件使循环停止迭代。for i=1:nl=l+abs(x(k+1,i)-x(k,i);endif l<ebreakendk=k+1;end%俞出所有的X的值。x(k+1,:)k3.S0R迭代法程序：function =caos on gci(e,w) for i=1:20a(i,i)=3;endfor i=3:20 for j=i-2 a(i

7、,j)=-1/4; a(j,i)=-1/4; endendfor i=2:20 for j=i-1 a(i,j)=-1/2; a(j,i)=-1/2; endend b=2.2 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; k=1;n=len gth(a);for i=1:nx(1,i)=0;%数组中没有第0行。endwhile k>=1if w>=2|w<=1'请重新输入w的值，w在 1与2之间；breakendfor i=1:np=0;q=0;for j=1

8、:(i-1)p=p+a(i,j)*x(k+1,j);endfor j=i:n q=q+a(i,j)*x(k,j);endx(k+1,i)=x(k,i)+w*(b(i)-q-p)/a(i,i); endl=0;%判定满足条件使循环停止迭代。for i=1:nl=l+abs(x(k+1,i)-x(k,i);endif l<ebreakendk=k+1;end%俞出所有的X的值。x(k+1,:)k第二题实验程序：1.雅克比迭代法：fun ctio n X = p211_1_JJ( n)Hn = GET_ Hn(n);b = GET_b( n);temp = 0;X0 = zeros(1, n)

9、;X_old = zeros(1, n);X_new = zeros(1, n);dis p('Now Jacobi method!');dis p( 'Start with the vector that (0, 0, 0, .)T'); for i = 1:nfor k = 1:nX_old = X_n ew;temp = 0;for j = 1:nif(j = i)temp = temp + Hn (i, j) * X_old(j); endendX_n ew(i) = (b(i) - tem p) / Hn( i, i); endendX = X_new

10、;end2.S0R迭代法：fun ction X = p211_1_SOR( n, w)Hn = GET_ Hn(n);b = GET_b( n);temp 01 = 0;temp02 = 0;X0 = zeros(1, n);X_old = zeros(1, n);X_new = zeros(1, n);dis p('Now Successive Over Relaxti on method!'); dis p('Start with the vector that (0, 0, 0, .)T'); for i = 1:nfor k = 1:nX_old =

11、X_n ew; temp 01 = 0; temp02 = 0;forj = 1:nif(j < i)temp 01 = temp01 + Hn (i, j) * X_n ew(j); end if(j > i)temp02 = temp02 + Hn(i, j) * X_old(j);endendendX_new(i) = w * (b(i) - tem p01 - tem p0 2) / Hn (i, i) + X_old(i);endX = X_new; end5）实验结果 第一题实验结果:1.雅克比迭代法: 输入：>> b=2.2 1.7 1.5 1.5 1.5

12、 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; yakebi(0.00001)结果：ans =Colu mns 1 through 120.97930.97870.99410.99700.99890.99950.99980.99991.00001.00001.00001.0000Colu mns 13 through 200.99990.99980.97930.99950.99890.99700.99410.9787122.高斯一赛德尔迭代法: 此时初值全取1;输入：>> b=2.2 1.7 1.5 1.5

13、 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)结果：ans =Colu mns 1 through 120.97930.97870.99410.99700.99890.99950.99980.99991.0000 1.0000 1.0000 1.0000Colu mns 13 through 200.99990.99980.99950.9989k =0.99700.99410.97870.979314此时初值全取1;输入：>> b=2.

14、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)结果：ans =Colu mns 1 through 121.09691.07071.00031.00011.0001Colu mns 13 through 201.00031.00061.02191.00011.00161.0969143.SOR迭代法：>> caoso ngci(0.00001,1.1) ans =Colu mns 1 through 120.97930.9787

15、0.99410.99991.00001.00001.00001.01031.00011.00390.99701.00001.00391.01030.99890.99951.00161.02190.99981.00061.0707Colu mns 13 through 200.99990.99980.99950.99890.99700.99410.9787-/250.979311>> caoso ngci(0.00001,1.2) ans =Colu mns 1 through 120.97930.97870.99991.00001.0000Colu mns 13 through 2

16、00.99990.99980.99411.00000.99701.00000.99890.99950.99980.979312>> caoso ngci(0.00001,1.3) ans =Colu mns 1 through 120.97930.97870.99991.00001.0000Colu mns 13 through 200.99990.99980.979315>> caoso ngci(0.00001,1.4) ans =Colu mns 1 through 120.97930.97870.99991.00001.0000Colu mns 13 throu

17、gh 200.99990.99980.979319>> caoso ngci(0.00001,1.5) ans =Colu mns 1 through 120.97930.97870.99991.00001.0000Colu mns 13 through 200.99990.99980.9793k =>> caoso ngci(0.00001,1.6)0.99950.99890.99700.99410.97870.99411.00000.99701.00000.99890.99950.99980.99950.99890.99700.99410.97870.99411.0

18、0000.99701.00000.99890.99950.99980.99950.99890.99700.99410.97870.99411.00000.99701.00000.99890.99950.99980.99950.99890.99700.99410.9787-/ans =Colu mns 1 through 120.97930.97870.99410.99700.99890.99950.99980.99991.00001.00001.00001.0000Colu mns 13 through 200.99990.99980.99950.99890.99700.99410.97870

19、.979334>> caoso ngci(0.00001,1.7)ans =Colu mns 1 through 120.97930.97870.99410.99700.99890.99950.99980.99991.00001.00001.00001.0000Colu mns 13 through 200.99990.99980.99950.99890.99700.99410.97870.979347>> caoso ngci(0.00001,1.8)ans =Colu mns 1 through 120.97930.97870.99410.99700.99890.9

20、9950.99980.99991.00001.00001.00001.0000Colu mns 13 through 200.99990.99980.99950.99890.99700.99410.97870.979373>> caoso ngci(0.00001,1.9)ans =Colu mns 1 through 120.97930.97870.99410.99700.99890.99950.99980.99991.00001.00001.00001.0000Colu mns 13 through 200.99990.99980.99950.99890.99700.99410

21、.97870.9793150第二题实验结果:1.雅克比迭代法：>> p211_1_JJ(6)Now Jacobi method!Start with the vector that (0, 0, 0, .)Tans =2.45001.10360.62650.40600.28310.2071>> p211_1_JJ(8)Now Jacobi method!Start with the vector that (0, 0, 0, .)Tans =2.71791.41010.85240.58090.42210.31980.24970.1995>> p211_1_J

22、J(10)Now Jacobi method!Start with the vector that (0, 0, 0, .)Tans =Colu mns 1 through 92.92901.66621.05170.74230.55540.43150.34450.28070.2325Colu mn 100.19512.SOR迭代法：n=6,沪1,1.25,1.5 的时候>> p211_1_SOR(6, 1)Now Successive Over Relaxti on method!Start with the vector that (0, 0, 0, .)Tans =2.4500

23、1.1036 0.62650.4060 0.28310.2071>> p 211_1_SOR(6, 1.25)Now Successive Over Relaxti on method!Start with the vector that (0, 0, 0, .)Tans =3.06250.23100.87040.33890.31410.2097>> p211_1_SOR(6, 1.5)Now Successive Over Relaxti on method!Start with the vector that (0, 0, 0, .)Tans =3.6750 -1.

24、10092.0106 -0.39940.7670-0.0384与 n=8, 03=1,1.25,1.5 的时候>> p 211_1_SOR(8, 1)Now Successive Over Relaxti on method!Start with the vector that (0, 0, 0, .)Tans =2.71791.41010.85240.58090.42210.31980.24970.1995>> p 211_1_SOR(8, 1.25)Now Successive Over Relaxti on method!Start with the vector that (0, 0, 0, .)Tans =3.39730.48871.08980.50620.45010.32030.25730.2042 >> p 211_1_SOR(8, 1.5)Now Successive Over Relaxti on method!Start with the vector that (0, 0, 0, .)Tans =4.0768 -0.94242.2923 -0.27530.92520.05780.40710.1275与n=10,