All of the examples can be found in Jupyter notebook form here.

# Portfolio Optimization

In this problem, we will find the portfolio allocation that minimizes risk while achieving a given expected return $R_\text{target}$.

Suppose that we know the mean returns $\mu \in \mathbf{R}^n$ and the covariance $\Sigma \in \mathbf{R}^{n \times n}$ of the $n$ assets. We would like to find a portfolio allocation $w \in \mathbf{R}^n$, $\sum_i w_i = 1$, minimizing the *risk* of the portfolio, which we measure as the variance $w^T \Sigma w$ of the portfolio. The requirement that the portfolio allocation achieve the target expected return can be expressed as $w^T \mu >= R_\text{target}$. We suppose further that our portfolio allocation must comply with some lower and upper bounds on the allocation, $w_\text{lower} \leq w \leq w_\text{upper}$.

This problem can be written as

\[\begin{array}{ll} \text{minimize} & w^T \Sigma w \\ \text{subject to} & w^T \mu >= R_\text{target} \\ & \sum_i w_i = 1 \\ & w_\text{lower} \leq w \leq w_\text{upper} \end{array}\]

where $w \in \mathbf{R}^n$ is our optimization variable.

```
using Convex, SCS
# generate problem data
μ = [11.5; 9.5; 6]/100 #expected returns
Σ = [166 34 58; #covariance matrix
34 64 4;
58 4 100]/100^2
n = length(μ) #number of assets
R_target = 0.1
w_lower = 0
w_upper = 0.5;
```

If you want to try the optimization with more assets, uncomment and run the next cell. It creates a vector or average returns and a variance-covariance matrix that have scales similar to the numbers above.

using Random Random.seed!(123)

n = 15 #number of assets, CHANGE IT?

μ = (6 .+ (11.5-6)*rand(n))/100 #mean A = randn(n,n) Σ = (A * A' + diagm(0=>rand(n)))/500; #covariance matrix

```
w = Variable(n)
ret = dot(w,μ)
risk = quadform(w,Σ)
p = minimize( risk,
ret >= R_target,
sum(w) == 1,
w_lower <= w,
w <= w_upper )
solve!(p, () -> SCS.Optimizer()) #use SCS.Optimizer(verbose = false) to suppress printing
```

Optimal portfolio weights:

`evaluate(w)`

3-element Array{Float64,1}: 0.4090909113158389 0.5000000024210729 0.09090909092390967

`sum(evaluate(w))`

1.0000000046608215

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