The Quadratic Programming Solver -- Experimental

Example 12.2: Portfolio Optimization

Consider a portfolio optimization example. The two competing goals of investment are (1) long-term growth of capital and (2) low risk. A good portfolio grows steadily without wild fluctuations in value. The Markowitz model is an optimization model for balancing the return and risk of a portfolio. The decision variables are the amounts invested in each asset. The objective is to minimize the variance of the portfolio's total return, subject to the constraints that (1) the expected growth of the portfolio reaches at least some target level and (2) you do not invest more capital than you have.

Let x_1,  ... , x_n be the amount invested in each asset, \mathcal{b} be the amount of capital you have, \mathbf{r} be the random vector of asset returns over some period, and \mathbf{r} be the expected value of \mathbf{r}. Let g be the minimum growth you hope to obtain, and \mathcal{c} be the covariance matrix of \mathbf{r}. The objective function is {\rm var}( \sum   \limits_{i=1}^n x_i r_i ), which can be equivalently denoted as \mathbf{x}^{\rm t} \mathcal{c}\mathbf{x}.

Assume, for example, n = 4. Let \mathcal{b} = 10,000, g = 1000, \mathbf{r} = [0.05, -0.2, 0.15, 0.30], and

\mathcal{c} & = & [   0.08 & -0.05 & -0.05 & -0.05\   -0.05 & 0.16 & -0.02 & -0.02\   -0.05 & -0.02 & 0.35 & 0.06\   -0.05 & -0.02 & 0.06 & 0.35\ ]\

The QP formulation can be written as

{\min} & 0.08x_1^2 & - & 0.1 x_1 x_2 & - & 0.1 x_1 x_3 & - & 0.1 x_1 x_4 & + & \...   ... x_4 & \geq & 1000 \    & \multicolumn{7}r{x_1,\, x_2, \, x_3, \, x_4} & \ge & 0

Use the following SAS code to solve the problem:


    /* example 2: portfolio optimization */
    proc optmodel;
       /* let x1, x2, x3, x4 be the amount invested in each asset */
       var x{1..4} >= 0;
       
       num coeff{1..4, 1..4} = [0.08 -.05 -.05 -.05
                                -.05 0.16 -.02 -.02
                                -.05 -.02 0.35 0.06
                                -.05 -.02 0.06 0.35];
       num r{1..4}=[0.05 -.20 0.15 0.30];
       
       /* minimize the variance of the portfolio's total return */
       minimize f = sum{i in 1..4, j in 1..4}coeff[i,j]*x[i]*x[j];
       
       /* subject to the following constraints */
       con BUDGET: sum{i in 1..4}x[i] <= 10000;
       con GROWTH: sum{i in 1..4}r[i]*x[i] >= 1000;
       
       solve with qp;
       
       /* print the optimal solution */
       print x;
 

The summaries and the optimal solution are shown in Output 12.2.1.

Output 12.2.1: Portfolio Optimization
The OPTMODEL Procedure

Problem Summary
Objective Sense Minimization
Objective Function f
Objective Type Quadratic
   
Number of Variables 4
Bounded Above 0
Bounded Below 4
Bounded Below and Above 0
Free 0
Fixed 0
   
Number of Constraints 2
Linear LE (<=) 1
Linear EQ (=) 0
Linear GE (>=) 1
Linear Range 0



The OPTMODEL Procedure

Solution Summary
Solver QP
Objective Function f
Solution Status Optimal
Objective Value 2232313.8093
Iterations 7
   
Primal Infeasibility 0
Dual Infeasibility 0
Bound Infeasibility 0
Duality Gap 4.4734294E-7
Complementarity 0.9932470284

[1] x
1 3.4529E+03
2 7.0389E-04
3 1.0688E+03
4 2.2235E+03



Thus, the minimum variance portfolio that earns an expected return of at least 10% is x_1 = 3452, x_2 = 0, x_3 = 1068, x_4 = 2223. Asset 2 gets nothing because its expected return is -20% and its covariance with the other assets is not sufficiently negative for it to bring any diversification benefits. What if we drop the nonnegativity assumption?

Financially, that means you are allowed to short-sell - i.e., sell low-mean-return assets and use the proceeds to invest in high-mean-return assets. In other words, you put a negative portfolio weight in low-mean assets and "more than 100%" in high-mean assets.

To solve the portfolio optimization problem with the short-sale option, continue to submit the following SAS code:

       /* example 2: portfolio optimization with short-sale option */
       /* dropping nonnegativity assumption */
       for {i in 1..4} x[i].lb=-x[i].ub;
       
       solve with qp;
       
       /* print the optimal solution */
       print x;
    quit;
 

You can see in the optimal solution displayed in Output 12.2.2 that the decision variable x_2, denoting Asset 2, is equal to -1563.61, which means short sale of that asset.

Output 12.2.2: Portfolio Optimization with Short-Sale Option
The OPTMODEL Procedure

Solution Summary
Solver QP
Objective Function f
Solution Status Optimal
Objective Value 1907123.1559
Iterations 6
   
Primal Infeasibility 0
Dual Infeasibility 0
Bound Infeasibility 0
Duality Gap 6.1120033E-7
Complementarity 1.1690054718

[1] x
1 1684.35
2 -1563.61
3 682.51
4 1668.95



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