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As per title: I have a convex payoff (finance) to approximate with available instruments traded.

I want to minimize $Ax-b$

s.t. $x>0$ and $Ax\ge b$

$A$ is a $n \times k$ matrix

$b$ is a $n \times 1$ vector

$x$ is a $k \times 1$ vector

$n > k $ always.

I cannot use least square error because in some parts $Ax$ would be lower than $b$ and this cannot happen.

I would use Matlab to solve for this, but I am struggling to find a solution (that is not the LSE...) and I do not know what is the best optimisation tool to use to solve it and how to input...

I am not expert about the topic, I was not able to find a similar question.

Thanks a lot

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    In what sense do you want to minimize the vector quantity $Ax-b$? Do you mean that you want to minimize the norm of $Ax-b$? Do you want to minimize the maximum positive entry in $Ax-b$? Or what? – 2017-02-13
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    The second one, minimizing the maximum positive entry in $Ax-b$. I want that $Ax$ to approximate $b$ from "upside" basically, thanks – 2017-02-13

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From chapter 6 of Boyd & Vandenberghe:

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    I did it, but I have still the problem that sometimes $Ax$ is lower than $b$, do you have any idea how to fix it? – 2017-02-15
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    Instead of the pair of constraints $-t 1_n \leq A x - b \leq t 1_n$, use only the constraint $A x - b \leq t 1_n$. – 2017-02-15
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    If you have the constraint $A x \geq b$, then how come you have that some entries of $A x - b$ are negative? – 2017-02-15
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    Because what I got using just the constraint ($Ax-b \leq t1_{n}$) as a result it is not exciting (the difference between $Ax$ and $b$ is too big), but I was able to get better results inputting no inequality constraint but just as equality constraint: $Ax+t_{1}1_{n}-t_{2}1_{n}$ with $t_{1}$ and $t_{2}$ more than 0. The problem is that if I add to this the inequality constraint $Ax-b \leq t1_{n}$, the system does not find any solution.... Do you know another way to solve such that I keep that constraint ($Ax-b \leq t1_{n}$) but I get a small difference between $Ax$ and $b$...? – 2017-02-20
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    You can minimize the 2-norm of $\rm A x - b$ with the constraint $\rm A x \geq b$. – 2017-02-20