The Convergence of block Gauss-Seidel method












1














The task:




To solve a system of equations with a nonsingular block matrix:
$$begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}f\gend{bmatrix}$$
where $A,Binmathbb{R}^{Ntimes N}$ and $A=A^{operatorname{T}}>0$ we use the following block Gauss-Seidel method:
$$Ax_{n+1}+By_n=f,\B^{operatorname{T}}x_{n+1}-y_{n+1}=g$$
Prove that if $left|left|B^{operatorname{T}}A^{-1}Bright|right|_2<1$, then the iteration converges to the solution of the system of equations for any initial $x_0,y_0$.




Solution attempt:



In the general case: An iterative method $Mmathcal{X}_{n+1}=mathcal{B}+Zmathcal{X}_n$, solving a system of equations $mathcal{A}mathcal{X}=mathcal{B}$, where $mathcal{A}=M-Z$, converges for any $mathcal{X_0}$ if $left|left|mathfrak{B}right|right|<1$, for $mathfrak{B}:=mathrm{I}-M^{-1}mathcal{A}$.



Here we have $mathcal{A}=begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix},mathcal{X}=begin{bmatrix}x\yend{bmatrix},mathcal{B}=begin{bmatrix}f\gend{bmatrix}$. Sorry for using fonts to distinguish tokens, but as you can see we have crazy conflicts here! Firstly let's try to find $M$:



$$begin{cases}Ax_{n+1}+By_n=f\B^{operatorname{T}}x_{n+1}-y_{n+1}=gend{cases}iffbegin{cases}Ax_{n+1}=f-By_n\B^{operatorname{T}}x_{n+1}-y_{n+1}=gend{cases}$$
In matrix form:
$$begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}begin{bmatrix}x_{n+1}\y_{n+1}end{bmatrix}=begin{bmatrix}f\gend{bmatrix}+begin{bmatrix}0&-B\0&0end{bmatrix}begin{bmatrix}x_n\y_nend{bmatrix}$$



Since clearly $begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}-begin{bmatrix}0&-B\0&0end{bmatrix}=begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=A$,



we can conclude that $begin{cases}M=begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}\Z=begin{bmatrix}0&-B\0&0end{bmatrix}end{cases}$.



Now let's try to find $M^{-1}$. We have $M^{-1}M=operatorname{I}$, so:
$$begin{bmatrix}M^{-1}_{1,1}&M^{-1}_{1,2}\M^{-1}_{2,1}&M^{-1}_{2,2}end{bmatrix}begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}$$



Filling in the blanks we easily get:
$$begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}$$
So $M^{-1}=begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}$.



Now let's compute $mathfrak{B}$:



$$mathfrak{B}=mathrm{I}-M^{-1}mathcal{A}=mathrm{I}-begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}-begin{bmatrix}mathrm{I}&A^{-1}B\0&mathrm{I}+B^{operatorname{T}}A^{-1}Bend{bmatrix}=begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}$$



Now don't really know what to do next. It would seem I have to somehow show that $left|left|B^{operatorname{T}}A^{-1}Bright|right|_2geqleft|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_2$, but I don't know how to do this.



I'm not sure if this is correct, but I think I remember that for any matrix $mathfrak{A}$, either all $p$-norms are $<1$, $=1$ or $>1$. If this holds, then we could perhpas do something like this: $$left|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_2<1iffleft|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_infty<1iffmaxleft(left|left|0right|right|_infty+left|left|A^{-1}Bright|right|_{infty},left|left|0right|right|_infty+left|left|B^{operatorname{T}}A^{-1}Bright|right|_{infty}right)<1iff\maxleft(left|left|A^{-1}Bright|right|,left|left|B^{operatorname{T}}A^{-1}Bright|right|right)<1$$



So if this reasoning is correct (I wouldn't bet a penny on that), the task would boil down to proving that $left|left|A^{-1}Bright|right|leqleft|left|B^{operatorname{T}}A^{-1}Bright|right|$, but I don't know how to prove it nor if it even holds.



And I haven't yet used the fact that $A$ is positive definite!



How to proceed?






linear-algebra matrices convergence numerical-methods







share|cite|improve this question





























    1














    The task:




    To solve a system of equations with a nonsingular block matrix:
    $$begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}f\gend{bmatrix}$$
    where $A,Binmathbb{R}^{Ntimes N}$ and $A=A^{operatorname{T}}>0$ we use the following block Gauss-Seidel method:
    $$Ax_{n+1}+By_n=f,\B^{operatorname{T}}x_{n+1}-y_{n+1}=g$$
    Prove that if $left|left|B^{operatorname{T}}A^{-1}Bright|right|_2<1$, then the iteration converges to the solution of the system of equations for any initial $x_0,y_0$.




    Solution attempt:



    In the general case: An iterative method $Mmathcal{X}_{n+1}=mathcal{B}+Zmathcal{X}_n$, solving a system of equations $mathcal{A}mathcal{X}=mathcal{B}$, where $mathcal{A}=M-Z$, converges for any $mathcal{X_0}$ if $left|left|mathfrak{B}right|right|<1$, for $mathfrak{B}:=mathrm{I}-M^{-1}mathcal{A}$.



    Here we have $mathcal{A}=begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix},mathcal{X}=begin{bmatrix}x\yend{bmatrix},mathcal{B}=begin{bmatrix}f\gend{bmatrix}$. Sorry for using fonts to distinguish tokens, but as you can see we have crazy conflicts here! Firstly let's try to find $M$:



    $$begin{cases}Ax_{n+1}+By_n=f\B^{operatorname{T}}x_{n+1}-y_{n+1}=gend{cases}iffbegin{cases}Ax_{n+1}=f-By_n\B^{operatorname{T}}x_{n+1}-y_{n+1}=gend{cases}$$
    In matrix form:
    $$begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}begin{bmatrix}x_{n+1}\y_{n+1}end{bmatrix}=begin{bmatrix}f\gend{bmatrix}+begin{bmatrix}0&-B\0&0end{bmatrix}begin{bmatrix}x_n\y_nend{bmatrix}$$



    Since clearly $begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}-begin{bmatrix}0&-B\0&0end{bmatrix}=begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=A$,



    we can conclude that $begin{cases}M=begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}\Z=begin{bmatrix}0&-B\0&0end{bmatrix}end{cases}$.



    Now let's try to find $M^{-1}$. We have $M^{-1}M=operatorname{I}$, so:
    $$begin{bmatrix}M^{-1}_{1,1}&M^{-1}_{1,2}\M^{-1}_{2,1}&M^{-1}_{2,2}end{bmatrix}begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}$$



    Filling in the blanks we easily get:
    $$begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}$$
    So $M^{-1}=begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}$.



    Now let's compute $mathfrak{B}$:



    $$mathfrak{B}=mathrm{I}-M^{-1}mathcal{A}=mathrm{I}-begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}-begin{bmatrix}mathrm{I}&A^{-1}B\0&mathrm{I}+B^{operatorname{T}}A^{-1}Bend{bmatrix}=begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}$$



    Now don't really know what to do next. It would seem I have to somehow show that $left|left|B^{operatorname{T}}A^{-1}Bright|right|_2geqleft|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_2$, but I don't know how to do this.



    I'm not sure if this is correct, but I think I remember that for any matrix $mathfrak{A}$, either all $p$-norms are $<1$, $=1$ or $>1$. If this holds, then we could perhpas do something like this: $$left|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_2<1iffleft|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_infty<1iffmaxleft(left|left|0right|right|_infty+left|left|A^{-1}Bright|right|_{infty},left|left|0right|right|_infty+left|left|B^{operatorname{T}}A^{-1}Bright|right|_{infty}right)<1iff\maxleft(left|left|A^{-1}Bright|right|,left|left|B^{operatorname{T}}A^{-1}Bright|right|right)<1$$



    So if this reasoning is correct (I wouldn't bet a penny on that), the task would boil down to proving that $left|left|A^{-1}Bright|right|leqleft|left|B^{operatorname{T}}A^{-1}Bright|right|$, but I don't know how to prove it nor if it even holds.



    And I haven't yet used the fact that $A$ is positive definite!



    How to proceed?






    linear-algebra matrices convergence numerical-methods







    share|cite|improve this question



























      1












      1








      1







      The task:




      To solve a system of equations with a nonsingular block matrix:
      $$begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}f\gend{bmatrix}$$
      where $A,Binmathbb{R}^{Ntimes N}$ and $A=A^{operatorname{T}}>0$ we use the following block Gauss-Seidel method:
      $$Ax_{n+1}+By_n=f,\B^{operatorname{T}}x_{n+1}-y_{n+1}=g$$
      Prove that if $left|left|B^{operatorname{T}}A^{-1}Bright|right|_2<1$, then the iteration converges to the solution of the system of equations for any initial $x_0,y_0$.




      Solution attempt:



      In the general case: An iterative method $Mmathcal{X}_{n+1}=mathcal{B}+Zmathcal{X}_n$, solving a system of equations $mathcal{A}mathcal{X}=mathcal{B}$, where $mathcal{A}=M-Z$, converges for any $mathcal{X_0}$ if $left|left|mathfrak{B}right|right|<1$, for $mathfrak{B}:=mathrm{I}-M^{-1}mathcal{A}$.



      Here we have $mathcal{A}=begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix},mathcal{X}=begin{bmatrix}x\yend{bmatrix},mathcal{B}=begin{bmatrix}f\gend{bmatrix}$. Sorry for using fonts to distinguish tokens, but as you can see we have crazy conflicts here! Firstly let's try to find $M$:



      $$begin{cases}Ax_{n+1}+By_n=f\B^{operatorname{T}}x_{n+1}-y_{n+1}=gend{cases}iffbegin{cases}Ax_{n+1}=f-By_n\B^{operatorname{T}}x_{n+1}-y_{n+1}=gend{cases}$$
      In matrix form:
      $$begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}begin{bmatrix}x_{n+1}\y_{n+1}end{bmatrix}=begin{bmatrix}f\gend{bmatrix}+begin{bmatrix}0&-B\0&0end{bmatrix}begin{bmatrix}x_n\y_nend{bmatrix}$$



      Since clearly $begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}-begin{bmatrix}0&-B\0&0end{bmatrix}=begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=A$,



      we can conclude that $begin{cases}M=begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}\Z=begin{bmatrix}0&-B\0&0end{bmatrix}end{cases}$.



      Now let's try to find $M^{-1}$. We have $M^{-1}M=operatorname{I}$, so:
      $$begin{bmatrix}M^{-1}_{1,1}&M^{-1}_{1,2}\M^{-1}_{2,1}&M^{-1}_{2,2}end{bmatrix}begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}$$



      Filling in the blanks we easily get:
      $$begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}$$
      So $M^{-1}=begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}$.



      Now let's compute $mathfrak{B}$:



      $$mathfrak{B}=mathrm{I}-M^{-1}mathcal{A}=mathrm{I}-begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}-begin{bmatrix}mathrm{I}&A^{-1}B\0&mathrm{I}+B^{operatorname{T}}A^{-1}Bend{bmatrix}=begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}$$



      Now don't really know what to do next. It would seem I have to somehow show that $left|left|B^{operatorname{T}}A^{-1}Bright|right|_2geqleft|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_2$, but I don't know how to do this.



      I'm not sure if this is correct, but I think I remember that for any matrix $mathfrak{A}$, either all $p$-norms are $<1$, $=1$ or $>1$. If this holds, then we could perhpas do something like this: $$left|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_2<1iffleft|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_infty<1iffmaxleft(left|left|0right|right|_infty+left|left|A^{-1}Bright|right|_{infty},left|left|0right|right|_infty+left|left|B^{operatorname{T}}A^{-1}Bright|right|_{infty}right)<1iff\maxleft(left|left|A^{-1}Bright|right|,left|left|B^{operatorname{T}}A^{-1}Bright|right|right)<1$$



      So if this reasoning is correct (I wouldn't bet a penny on that), the task would boil down to proving that $left|left|A^{-1}Bright|right|leqleft|left|B^{operatorname{T}}A^{-1}Bright|right|$, but I don't know how to prove it nor if it even holds.



      And I haven't yet used the fact that $A$ is positive definite!



      How to proceed?






      linear-algebra matrices convergence numerical-methods







      share|cite|improve this question















      The task:




      To solve a system of equations with a nonsingular block matrix:
      $$begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}begin{bmatrix}x\yend{bmatrix}=begin{bmatrix}f\gend{bmatrix}$$
      where $A,Binmathbb{R}^{Ntimes N}$ and $A=A^{operatorname{T}}>0$ we use the following block Gauss-Seidel method:
      $$Ax_{n+1}+By_n=f,\B^{operatorname{T}}x_{n+1}-y_{n+1}=g$$
      Prove that if $left|left|B^{operatorname{T}}A^{-1}Bright|right|_2<1$, then the iteration converges to the solution of the system of equations for any initial $x_0,y_0$.




      Solution attempt:



      In the general case: An iterative method $Mmathcal{X}_{n+1}=mathcal{B}+Zmathcal{X}_n$, solving a system of equations $mathcal{A}mathcal{X}=mathcal{B}$, where $mathcal{A}=M-Z$, converges for any $mathcal{X_0}$ if $left|left|mathfrak{B}right|right|<1$, for $mathfrak{B}:=mathrm{I}-M^{-1}mathcal{A}$.



      Here we have $mathcal{A}=begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix},mathcal{X}=begin{bmatrix}x\yend{bmatrix},mathcal{B}=begin{bmatrix}f\gend{bmatrix}$. Sorry for using fonts to distinguish tokens, but as you can see we have crazy conflicts here! Firstly let's try to find $M$:



      $$begin{cases}Ax_{n+1}+By_n=f\B^{operatorname{T}}x_{n+1}-y_{n+1}=gend{cases}iffbegin{cases}Ax_{n+1}=f-By_n\B^{operatorname{T}}x_{n+1}-y_{n+1}=gend{cases}$$
      In matrix form:
      $$begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}begin{bmatrix}x_{n+1}\y_{n+1}end{bmatrix}=begin{bmatrix}f\gend{bmatrix}+begin{bmatrix}0&-B\0&0end{bmatrix}begin{bmatrix}x_n\y_nend{bmatrix}$$



      Since clearly $begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}-begin{bmatrix}0&-B\0&0end{bmatrix}=begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=A$,



      we can conclude that $begin{cases}M=begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}\Z=begin{bmatrix}0&-B\0&0end{bmatrix}end{cases}$.



      Now let's try to find $M^{-1}$. We have $M^{-1}M=operatorname{I}$, so:
      $$begin{bmatrix}M^{-1}_{1,1}&M^{-1}_{1,2}\M^{-1}_{2,1}&M^{-1}_{2,2}end{bmatrix}begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}$$



      Filling in the blanks we easily get:
      $$begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}begin{bmatrix}A&0\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}$$
      So $M^{-1}=begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}$.



      Now let's compute $mathfrak{B}$:



      $$mathfrak{B}=mathrm{I}-M^{-1}mathcal{A}=mathrm{I}-begin{bmatrix}A^{-1}&0\B^{operatorname{T}}A^{-1}&-mathrm{I}end{bmatrix}begin{bmatrix}A&B\B^{operatorname{T}}&-mathrm{I}end{bmatrix}=begin{bmatrix}mathrm{I}&0\0&mathrm{I}end{bmatrix}-begin{bmatrix}mathrm{I}&A^{-1}B\0&mathrm{I}+B^{operatorname{T}}A^{-1}Bend{bmatrix}=begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}$$



      Now don't really know what to do next. It would seem I have to somehow show that $left|left|B^{operatorname{T}}A^{-1}Bright|right|_2geqleft|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_2$, but I don't know how to do this.



      I'm not sure if this is correct, but I think I remember that for any matrix $mathfrak{A}$, either all $p$-norms are $<1$, $=1$ or $>1$. If this holds, then we could perhpas do something like this: $$left|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_2<1iffleft|left|begin{bmatrix}0&-A^{-1}B\0&-B^{operatorname{T}}A^{-1}Bend{bmatrix}right|right|_infty<1iffmaxleft(left|left|0right|right|_infty+left|left|A^{-1}Bright|right|_{infty},left|left|0right|right|_infty+left|left|B^{operatorname{T}}A^{-1}Bright|right|_{infty}right)<1iff\maxleft(left|left|A^{-1}Bright|right|,left|left|B^{operatorname{T}}A^{-1}Bright|right|right)<1$$



      So if this reasoning is correct (I wouldn't bet a penny on that), the task would boil down to proving that $left|left|A^{-1}Bright|right|leqleft|left|B^{operatorname{T}}A^{-1}Bright|right|$, but I don't know how to prove it nor if it even holds.



      And I haven't yet used the fact that $A$ is positive definite!



      How to proceed?






      linear-algebra matrices convergence numerical-methods




      linear-algebra matrices convergence numerical-methods






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 18 at 19:23

























      asked Nov 18 at 14:58









      gaazkam

      437314




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          QoS: MAC-Priority for clients behind a repeater

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          1 1 I've setup a wireless network using a Linksys-Router (DD-WRT) as an AP and a TP-Link repeater for range extension. To manage bandwith for specific users, I've setup QoS-Rules on the AP (DD-WRT) using MAC-Priority: However the AP shows only the Repeater's MAC and the MACs of users inside the AP-Range. The MAC addresses of users behind the Repeater are not listed : The purple MAC is not listed in the AP-Clientlist. Its related to a device connected via the TP-Link repeater. Q: How can I setup MAC-Priority for repeater-clients? Do I simply add the purple MAC in AP-MAC priority list although it's not listed as a client? I want to setup specific priority instead of giving one priority for all clients behind the repeater by adding the repeater's MAC in the MAC priority list in the AP as ...
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