Solution with boundary condition
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Suppose we have $A in Bbb{R}^{n times 1}, B in Bbb{R}^{n times 1}, Q in Bbb{R}^{n times n}, h in (0,1) $ and $x in Bbb{R}^{n times 1}$.
How to I solve the following equation for $x$, where each element of $x$ must be in $(0,1)$:
$ A = B + (1-h)Qx $
I don't know how to incorporate the condition on $x$ into my equation.
linear-algebra
asked Dec 4 at 16:33
Quant Finance
498
add a comment |
up vote
0
down vote
favorite
Suppose we have $A in Bbb{R}^{n times 1}, B in Bbb{R}^{n times 1}, Q in Bbb{R}^{n times n}, h in (0,1) $ and $x in Bbb{R}^{n times 1}$.
How to I solve the following equation for $x$, where each element of $x$ must be in $(0,1)$:
$ A = B + (1-h)Qx $
I don't know how to incorporate the condition on $x$ into my equation.
linear-algebra
asked Dec 4 at 16:33
Quant Finance
498
1
If $Q$ is invertible, $x=(1-h)^{-1}Q^{-1}(A-B)$ is the unique solution, so you cannot really enforce $x_iin(0,1),forall i$.
– Federico
Dec 4 at 16:35
Is there a way to approximate it?
– Quant Finance
Dec 4 at 16:43
Yes, there are plenty of ways to compute it numerically. Search for algorithms to solve linear systems. My point is, if $Q$ happens to be invertible, then the solution is unique and you have no way to enforce the additional constraint on the $x_i$'s. Maybe you can tell us more about $A,B,Q,h$.
– Federico
Dec 4 at 16:47
It's an economic model where $Q$ is the adjacency matrix. I guess the problem is in my model setup. Thanks for your help!
– Quant Finance
Dec 4 at 16:50
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have $A in Bbb{R}^{n times 1}, B in Bbb{R}^{n times 1}, Q in Bbb{R}^{n times n}, h in (0,1) $ and $x in Bbb{R}^{n times 1}$.
How to I solve the following equation for $x$, where each element of $x$ must be in $(0,1)$:
$ A = B + (1-h)Qx $
I don't know how to incorporate the condition on $x$ into my equation.
linear-algebra
asked Dec 4 at 16:33
Quant Finance
498
Suppose we have $A in Bbb{R}^{n times 1}, B in Bbb{R}^{n times 1}, Q in Bbb{R}^{n times n}, h in (0,1) $ and $x in Bbb{R}^{n times 1}$.
How to I solve the following equation for $x$, where each element of $x$ must be in $(0,1)$:
$ A = B + (1-h)Qx $
I don't know how to incorporate the condition on $x$ into my equation.
linear-algebra
linear-algebra
asked Dec 4 at 16:33
Quant Finance
498
asked Dec 4 at 16:33
Quant Finance
498
asked Dec 4 at 16:33
Quant Finance
498
asked Dec 4 at 16:33
Quant Finance
498
asked Dec 4 at 16:33
Quant Finance
498
498
1
If $Q$ is invertible, $x=(1-h)^{-1}Q^{-1}(A-B)$ is the unique solution, so you cannot really enforce $x_iin(0,1),forall i$.
– Federico
Dec 4 at 16:35
Is there a way to approximate it?
– Quant Finance
Dec 4 at 16:43
Yes, there are plenty of ways to compute it numerically. Search for algorithms to solve linear systems. My point is, if $Q$ happens to be invertible, then the solution is unique and you have no way to enforce the additional constraint on the $x_i$'s. Maybe you can tell us more about $A,B,Q,h$.
– Federico
Dec 4 at 16:47
It's an economic model where $Q$ is the adjacency matrix. I guess the problem is in my model setup. Thanks for your help!
– Quant Finance
Dec 4 at 16:50
add a comment |
1
If $Q$ is invertible, $x=(1-h)^{-1}Q^{-1}(A-B)$ is the unique solution, so you cannot really enforce $x_iin(0,1),forall i$.
– Federico
Dec 4 at 16:35
Is there a way to approximate it?
– Quant Finance
Dec 4 at 16:43
Yes, there are plenty of ways to compute it numerically. Search for algorithms to solve linear systems. My point is, if $Q$ happens to be invertible, then the solution is unique and you have no way to enforce the additional constraint on the $x_i$'s. Maybe you can tell us more about $A,B,Q,h$.
– Federico
Dec 4 at 16:47
It's an economic model where $Q$ is the adjacency matrix. I guess the problem is in my model setup. Thanks for your help!
– Quant Finance
Dec 4 at 16:50
1
1
If $Q$ is invertible, $x=(1-h)^{-1}Q^{-1}(A-B)$ is the unique solution, so you cannot really enforce $x_iin(0,1),forall i$.
– Federico
Dec 4 at 16:35
If $Q$ is invertible, $x=(1-h)^{-1}Q^{-1}(A-B)$ is the unique solution, so you cannot really enforce $x_iin(0,1),forall i$.
– Federico
Dec 4 at 16:35
Is there a way to approximate it?
– Quant Finance
Dec 4 at 16:43
Is there a way to approximate it?
– Quant Finance
Dec 4 at 16:43
Yes, there are plenty of ways to compute it numerically. Search for algorithms to solve linear systems. My point is, if $Q$ happens to be invertible, then the solution is unique and you have no way to enforce the additional constraint on the $x_i$'s. Maybe you can tell us more about $A,B,Q,h$.
– Federico
Dec 4 at 16:47
Yes, there are plenty of ways to compute it numerically. Search for algorithms to solve linear systems. My point is, if $Q$ happens to be invertible, then the solution is unique and you have no way to enforce the additional constraint on the $x_i$'s. Maybe you can tell us more about $A,B,Q,h$.
– Federico
Dec 4 at 16:47
It's an economic model where $Q$ is the adjacency matrix. I guess the problem is in my model setup. Thanks for your help!
– Quant Finance
Dec 4 at 16:50
It's an economic model where $Q$ is the adjacency matrix. I guess the problem is in my model setup. Thanks for your help!
– Quant Finance
Dec 4 at 16:50
add a comment |
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1
If $Q$ is invertible, $x=(1-h)^{-1}Q^{-1}(A-B)$ is the unique solution, so you cannot really enforce $x_iin(0,1),forall i$.
– Federico
Dec 4 at 16:35
Is there a way to approximate it?
– Quant Finance
Dec 4 at 16:43
Yes, there are plenty of ways to compute it numerically. Search for algorithms to solve linear systems. My point is, if $Q$ happens to be invertible, then the solution is unique and you have no way to enforce the additional constraint on the $x_i$'s. Maybe you can tell us more about $A,B,Q,h$.
– Federico
Dec 4 at 16:47
It's an economic model where $Q$ is the adjacency matrix. I guess the problem is in my model setup. Thanks for your help!
– Quant Finance
Dec 4 at 16:50