Solve a square linear system usinggmreswith default settings, and then adjust the tolerance and number of iterations used in the solution process.

Create a random sparse matrixAwith 50% density and nonzeros on the main diagonal. Also create a random vectorbfor the right-hand side of $\mathrm{Ax}=\mathit{b}$。

RNGdefaultA = sprandn(400,400,0.5) + 12*speye(400); b = rand(400,1);

Solve $\mathrm{Ax}=\mathit{b}$使用gmres。The output display includes the value of the relative residual error $\frac{\Vert \mathit{b}-\mathrm{Ax}\Vert }{\Vert \mathit{b}\Vert }$。

x = gmres(A,b);

gmres stopped at iteration 10 without converging to the desired tolerance 1e-06 because the maximum number of iterations was reached. The iterate returned (number 10) has relative residual 0.35.

By defaultgmresuses 10 iterations and a tolerance of1e-6， 和the algorithm is unable to converge in those 10 iterations for this matrix. Since the residual is still large, it is a good indicator that more iterations (or a preconditioner matrix) are needed. You also can use a larger tolerance to make it easier for the algorithm to converge.

Solve the system again using a tolerance of1e-4和100 iterations.

tol = 1e-4;最大值= 100;x = gmres（a，b，[]，tol，maxit）;

GMRE在迭代100停止，而没有收敛到所需的公差0.0001，因为达到了最大迭代次数。迭代返回（数字100）具有相对残留的0.0045。

Even with a looser tolerance and more iterations the residual error does not improve enough for convergence. When an iterative algorithm stalls in this manner it is a good indication that a preconditioner matrix is needed. However,gmresalso has an input that controls the number of inner iterations. By specifying a value for the inner iterations,gmresdoes more work per outer iteration.

Solve the system again using arestartvalue of 100 and amaxitvalue of 20. Rather than doing 100 iterations once,gmresperforms 100 iterations between restarts and repeats this 20 times.

restart = 100; maxit = 20; x = gmres(A,b,restart,tol,maxit);

gmres(100) converged at outer iteration 2 (inner iteration 75) to a solution with relative residual 9.3e-05.

In this case specifying a large restart value forgmres使其能够收敛到允许的迭代次数中的解决方案。但是，大的重新启动值可以在Ais also large.

Examine the effect of using a preconditioner matrix with non-restartedgmres解决线性系统。

Load west0479, a real 479-by-479 nonsymmetric sparse matrix.

loadwest0479A = west0479;

Definebso that the true solution to $\mathrm{Ax}=\mathit{b}$is a vector of all ones.

b = sum（a，2）;

Set the tolerance and maximum number of iterations.

TOL = 1E-12;maxit = 20;

Usegmresto find a solution at the requested tolerance and number of iterations. Specify five outputs to return information about the solution process:

[x,fl0,rr0,it0,rv0] = gmres(A,b,[],tol,maxit); fl0

fl0 = 1

RR0

RR0= 0.7603

it0

it0 =1×21 20

fl0is 1 becausegmresdoes not converge to the requested tolerance1e-12在请求的20次迭代中。最好的近似解决方案gmres退货是最后一个（如it0(2) = 20）。MATLAB存储剩余历史rv0。

To aid with the slow convergence, you can specify a preconditioner matrix. SinceAis nonsymmetric, useiluto generate the preconditioner $\mathit{M}=\mathit{L}\mathit{U}$。Specify a drop tolerance to ignore nondiagonal entries with values smaller than1e-6。Solve the preconditioned system ${\mathit{M}}^{-1}\mathit{A}\mathit{x}={\mathit{M}}^{-1}\mathit{b}$by specifyingL和U作为输入gmres。Note that when you specify a preconditioner,gmrescalculates the residual norm of the preconditioned system for the outputsRR1和rv1。

[L,U] = ilu(A,struct('type','ilutp','droptol',1e-6)); [x1,fl1,rr1,it1,rv1] = gmres(A,b,[],tol,maxit,L,U); fl1

fl1 = 0

RR1

RR1= 1.3557e-13

it1

it1 =1×21 6

使用ilu预处理产生的相对残留少于规定的公差1e-12at the sixth iteration. The first residualrv1(1)isnorm(U\(L\b)), whereM = L*U。The last residualrv1(end)isnorm(U\(L\(b-A*x1)))。

You can follow the progress ofgmresby plotting the relative residuals at each iteration. Plot the residual history of each solution with a line for the specified tolerance.

semilogy(0:长度(rv0) 1, rv0 /规范(b),'-o'） 抓住上semilogy(0:length(rv1)-1,rv1/norm(U\(L\b)),'-o') yline(tol,'r--'）；legend('No preconditioner','ILU preconditioner','Tolerance','地点','East'）xlabel（'Iteration number'）ylabel（'Relative residual')

Using a preconditioner with restartedgmres。

Load west0479, a real 479-by-479 nonsymmetric sparse matrix.

loadwest0479A = west0479;

Definebso that the true solution to $\mathrm{Ax}=\mathit{b}$is a vector of all ones.

b = sum（a，2）;

构建一个不完整的LU预处理，具有下降的耐受性1e-6。

[L,U] = ilu(A,struct('type','ilutp','droptol',1e-6));

The benefit to using restartedgmresis to limit the amount of memory required to execute the method. Without restart,gmresrequiresmaxitvectors of storage to keep the basis of the Krylov subspace. Also,gmres必须支持hogonalize against all of the previous vectors at each step. Restarting limits the amount of workspace used and the amount of work done per outer iteration.

Execute转基因（3）,gmres(4)， 和gmres(5)使用the incomplete LU factors as preconditioners. Use a tolerance of1e-12和a maximum of 20 outer iterations.

TOL = 1E-12;maxit = 20;[x3，fl3，rr3，it3，rv3] = gmres（a，b，3，tol，maxit，l，u）;[x4，fl4，rr4，it4，rv4] = gmres（a，b，4，tol，maxit，l，u）;[x5，fl5，rr5，it5，rv5] = gmres（a，b，5，tol，maxit，l，u）;fl3

fl3= 0

fl4

fl4 = 0

fl5

fl5 = 0

fl3,fl4， 和fl5are all 0 because in each case restartedgmresdrives the relative residual to less than the prescribed tolerance of1e-12。

The following plot shows the convergence history of each restartedgmresmethod.转基因（3）converges at outer iteration 5, inner iteration 3 (it3 = [5, 3]) which would be the same as outer iteration 6, inner iteration 0, hence the marking of 6 on the final tick mark.

semilogy(1:1/3:6,rv3/norm(U\(L\b)),'-o'）；h1 = gca; h1.XTick = (1:6); title('gmres(N) for N = 3, 4, 5'）xlabel（“外迭代编号”）；ylabel（'Relative residual'）；hold上semilogy(1:1/4:3,rv4/norm(U\(L\b)),'-o'）；semilogy(1:1/5:2.8,rv5/norm(U\(L\b)),'-o'）；yline(tol,'r--'）；holdofflegend('gmres(3)','gmres(4)','gmres(5)','Tolerance') grid上

In general the larger the number of inner iterations, the more workgmresdoes per outer iteration and the faster it can converge.

Examine the effect of supplyinggmres最初猜测解决方案。

Create a tridiagonal sparse matrix. Use the sum of each row as the vector for the right-hand side of $\mathrm{Ax}=\mathit{b}$这样可以预期的解决方案 $\mathit{x}$is a vector of ones.

n = 900;e =一个（n，1）;a = spdiags（[E 2*e e]，-1：1，n，n）;b = sum（a，2）;

Usegmresto solve $\mathrm{Ax}=\mathit{b}$twice: one time with the default initial guess, and one time with a good initial guess of the solution. Use 200 iterations and the default tolerance for both solutions. Specify the initial guess in the second solution as a vector with all elements equal to0.99。

maxit = 200; x1 = gmres(A,b,[],[],maxit);

gmres converged at iteration 27 to a solution with relative residual 9.5e-07.

x0 = 0.99*e;x2 = gmres（a，b，[]，[]，maxit，[]，[]，x0）;

gmr聚合在迭代7到一个解决方案relative residual 6.7e-07.

In this case supplying an initial guess enablesgmresto converge more quickly.

x0 = zeros(size(A,2),1); tol = 1e-8; maxit = 100;fork = 1:4 [x,flag,relres] = gmres(A,b,[],tol,maxit,[],[],x0); X(:,k) = x; R(k) = relres; x0 = x;end

Solve a linear system by providinggmreswith a function handle that computesA*xin place of the coefficient matrixA。

威尔金森测试矩阵之一galleryis a 21-by-21 tridiagonal matrix. Preview the matrix.

A = gallery('wilk'，21）

A =21×21101 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 9 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 8 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 7 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 6 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 5 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 4 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 ⋮

The Wilkinson matrix has a special structure, so you can represent the operationA*x带有功能手柄。什么时候Amultiplies a vector, most of the elements in the resulting vector are zeros. The nonzero elements in the result correspond with the nonzero tridiagonal elements ofA。Moreover, only the main diagonal has nonzeros that are not equal to 1.

The expression $\mathrm{Ax}$becomes:

$\mathrm{Ax}=\left[\begin{array}{cccccccccc}10& 1& 0& \cdots & & \cdots & & \cdots & 0& 0\\ 1& 9& 1& 0& & & & & & 0\\ 0& 1& 8& 1& 0& & & & & ⋮\\ ⋮& 0& 1& 7& 1& 0& & & & \\ & & 0& 1& 6& 1& 0& & & ⋮\\ ⋮& & & 0& 1& 5& 1& 0& & \\ & & & & 0& 1& 4& 1& 0& ⋮\\ ⋮& & & & & 0& 1& 3& \ddots & 0\\ 0& & & & & & 0& \ddots & \ddots & 1\\ 0& 0& \cdots & & \cdots & & \cdots & 0& 1& 10\end{array}\right]\left[\begin{array}{c}{\mathit{x}}_1\\ {\mathit{x}}_2\\ {\mathit{x}}_3\\ {\mathit{x}}_4\\ {\mathit{x}}_5\\ ⋮\\ \\ ⋮\\ \\ {\mathit{x}}_{21}\end{array}\right]=\left[\begin{array}{c}10{\mathit{x}}_1+{\mathit{x}}_2\\ {\mathit{x}}_1+9{\mathit{x}}_2+{\mathit{x}}_3\\ {\mathit{x}}_2+8{\mathit{x}}_3+{\mathit{x}}_4\\ ⋮\\ {\mathit{x}}_{19}+9{\mathit{x}}_{20.}+{\mathit{x}}_{21}\\ {\mathit{x}}_{20.}+10{\mathit{x}}_{21}\end{array}\right]$。

The resulting vector can be written as the sum of three vectors:

$\mathrm{Ax}=\left[\begin{array}{c}0+10{\mathit{x}}_1+{\mathit{x}}_2\\ {\mathit{x}}_1+9{\mathit{x}}_2+{\mathit{x}}_3\\ {\mathit{x}}_2+8{\mathit{x}}_3+{\mathit{x}}_4\\ ⋮\\ {\mathit{x}}_{19}+9{\mathit{x}}_{20.}+{\mathit{x}}_{21}\\ {\mathit{x}}_{20.}+10{\mathit{x}}_{21}+0\end{array}\right]$= $\left[\begin{array}{c}0\\ {\mathit{x}}_1\\ ⋮\\ {\mathit{x}}_{20.}\end{array}\right]+\left[\begin{array}{c}10{\mathit{x}}_1\\ 9{\mathit{x}}_2\\ ⋮\\ 10{\mathit{x}}_{21}\end{array}\right]+\left[\begin{array}{c}{\mathit{x}}_2\\ ⋮\\ {\mathit{x}}_{21}\\ 0\end{array}\right]$。

In MATLAB®, write a function that creates these vectors and adds them together, thus giving the value ofA*x:

功能y = afun(x) y = [0; x(1:20)] +。。。[(10:-1:0)'; (1:10)'].*x +。。。[x(2:21); 0];end

(This function is saved as a local function at the end of the example.)

Now, solve the linear system $\mathrm{Ax}=\mathit{b}$by providinggmreswith the function handle that calculatesA*x。Use a tolerance of1e-12, 15 outer iterations, and 10 inner iterations before restart.

B =一个（21,1）;TOL = 1E-12;最大值= 15;重新启动= 10;x1 = gmres（@afun，b，ristart，tol，maxit）

gmres(10) converged at outer iteration 5 (inner iteration 10) to a solution with relative residual 5.3e-13.

x1 =21×10.0910 0.0899 0.0999 0.1109 0.1241 0.1443 0.1544 0。2383 0.1309 0.5000 ⋮

检查一下afun(x1)产生一个矢量。

afun(x1)

ans =21×11.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 ⋮

功能y = afun(x) y = [0; x(1:20)] +。。。[(10:-1:0)'; (1:10)'].*x +。。。[x(2:21); 0];end