# Maximum return portfolio using linear programming with quadratic constraints In the maximum return portfolio problem formulation above,

1. is $$A=\mu^\top \Sigma^{-1} \mu$$?
2. What is $$b$$ equal to, and
3. is the second constraint required? An inequality constraint for target portfolio volatility doesn't seem to have anything to do with maximizing portfolio return.

Source

In words, the formulation implies to maximize expected portfolio returns for a given maximum portfolio variance, $$\sigma$$ (Commonly noted as $$\sigma^2$$, though).

The linear restrictions $$Aw\leq b$$ imply the 'usual' conditions, i.e. $$w_i\geq 0$$, $$\sum_i w_i \leq 1$$ etc. Please note that your stated formulation does not include a restriction on 'full investment', i.e. you might come up with an optimal portfolio that is not fully invested. Hence, $$A$$ and $$b$$ encapsulate linear inequality conditions. Standard quadratic programming solvers let you specify equality and inequality constraints, though.

Edit: You finally construct the matrices by combining your investment constraints, each represented by a line in $$A$$ and a corresponding entry in $$b$$. Say you want to say $$\sum_i w_i\leq 1$$ then this is equivalent to $$w_1 + w_2 + \ldots + w_n \leq 1$$ which equals $$\mathbb{1}^Tw\leq 1$$. Hence, you add a row of ones to $$A$$ and a $$1$$ to $$b$$.

Example

Say you want to invest $$\geq0\%$$ in each asset, but no asset should be invested at more than, say, $$10\%$$. this would imply

$$Ax\leq b = \begin{pmatrix}-1&0&\ldots & 0\\ 0&-1&\ldots&0\\ \ldots &\ldots &\ldots &\ldots &\\ 0 & 0 & \ldots & -1\\ 1 & 1 & \ldots & 1\\ 1 & 0 & \ldots & 0 \\ 0 & 1 & \ldots & 0\\ \ldots&\ldots&\ldots&\ldots\\ 0 & 0 &\ldots & 1\end{pmatrix}x\leq\begin{pmatrix}0\\0\\ \ldots\\0\\1\\0.1\\0.1\\ \ldots\\ 0.1\end{pmatrix}$$

HTH?

• I agree that the authors forgot the 3rd constraint for full-investment, but at the same time you say $Aw\leq b$ includes this condition. These two statements conflict. Anyway, so $A\neq \mu^\top \Sigma^{-1} \mu$? then what is it really since how you've displayed it seems to be partitioned across the center ($-1$ diagonal matrix on top of a $1$ diagonal matrix) with a belt of 1's, as is $b$. "Encapsulate linear inequality conditions" gives no indication how to construct $A$ or $b$ – develarist Oct 30 '20 at 14:35
• I will update the answer later today. – Kermittfrog Oct 30 '20 at 14:42
• If i were to relax the non-negativity constraint so that short-selling is allowed, would I just remove the upper halves of $A$ and $b$? – develarist Nov 1 '20 at 1:24