The Quadratic Programming Solver

Example 11.2 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 $x_1, \ldots , x_ n$ 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 $\mr{Var}\left( \sum \limits _{i=1}^ n x_ i R_ i \right)$, which can be equivalently denoted as $\mathbf{x}^\mr {T} \mathcal{C}\mathbf{x}$.

Assume, for example, n = 4. Let $\mathcal{B}$ = 10,000, G = 1,000, $\mathbf{r} = [0.05, -0.2, 0.15, 0.30]$, and

\[ \begin{array}{rcc} \mathcal{C} & = & \left[ \begin{array}{cccc} 0.08 & -0.05 & -0.05 & -0.05\\ -0.05 & 0.16 & -0.02 & -0.02\\ -0.05 & -0.02 & 0.35 & 0.06\\ -0.05 & -0.02 & 0.06 & 0.35\\ \end{array} \right]\\ \end{array} \]

The QP formulation can be written as:

\[ \begin{array}{rl} {\min } & 0.08x_1^2 - 0.1 x_1 x_2 - 0.1 x_1 x_3 - 0.1 x_1 x_4 + 0.16x_2^2 \\ & - 0.04 x_2 x_3 - 0.04 x_2 x_4 + 0.35 x_3^2 + 0.12 x_3 x_4 + 0.35 x_4^2 \\ \end{array} \]
\[ \begin{array}{rrcl} \mr{subject\ to} & & & \\ (\mr{budget}) & x_1 + x_2 + x_3 + x_4 & \leq & 10000 \\ (\mr{growth}) & 0.05 x_1 - 0.2 x_2 + 0.15 x_3 + 0.30 x_4 & \geq & 1000 \\ & x_1,\, x_2, \, x_3, \, x_4 & \ge & 0 \\ \end{array} \]

Use the following SAS statements 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 11.2.1.

Output 11.2.1: 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
   
Constraint Coefficients 8

Performance Information
Execution Mode Single-Machine
Number of Threads 4

Solution Summary
Solver QP
Algorithm Interior Point
Objective Function f
Solution Status Optimal
Objective Value 2232313.4432
   
Primal Infeasibility 0
Dual Infeasibility 8.038873E-14
Bound Infeasibility 0
Duality Gap 4.172004E-16
Complementarity 0
   
Iterations 7
Presolve Time 0.00
Solution Time 0.01

[1] x
1 3452.9
2 0.0
3 1068.8
4 2223.5 




            

            


            
Thus, the minimum variance portfolio that earns an expected return of at least 10% is $x_1$ = 3,452, $x_2$ = 0, $x_3$ = 1,068, $x_4 = 2,223$. 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 you drop the nonnegativity assumption? 
            

            
Financially, that means you are allowed to short-sell—that is, 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 statements: 


   /* 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 11.2.2 that the decision variable $x_2$, denoting Asset 2, is equal to $-$1,563.61, which means short sale of that asset. 
            

            
  

            


Output 11.2.2: Portfolio Optimization with Short-Sale Option

               

                  

                     
                     

                        
                        
                        
The OPTMODEL Procedure

                        
                        

                        
                        

                           
                           

                              
                              







Solution Summary
                                       
                                    




Solver
                                       
                                       QP
                                       
                                    



Algorithm
                                       
                                       Interior Point
                                       
                                    



Objective Function
                                       
                                       f
                                       
                                    



Solution Status
                                       
                                       Optimal
                                       
                                    



Objective Value
                                       
                                       1907122.2254
                                       
                                    



 
                                       
                                        
                                       
                                    



Primal Infeasibility
                                       
                                       0
                                       
                                    



Dual Infeasibility
                                       
                                       2.755587E-13
                                       
                                    



Bound Infeasibility
                                       
                                       0
                                       
                                    



Duality Gap
                                       
                                       2.441695E-16
                                       
                                    



Complementarity
                                       
                                       0
                                       
                                    



 
                                       
                                        
                                       
                                    



Iterations
                                       
                                       6
                                       
                                    



Presolve Time
                                       
                                       0.00
                                       
                                    



Solution Time
                                       
                                       0.02
                                       
                                    






                           
                        

                        



                     
                     
                  

                  

                     
                     

                        
                        
                        

                           
                           

                              
                              





[1]
                                       
                                       x
                                       
                                    




1
                                       
                                       1684.35
                                       
                                    



2
                                       
                                       -1563.61
                                       
                                    



3
                                       
                                       682.51
                                       
                                    



4
                                       
                                       1668.95