The Quadratic Programming Solver

Example 12.2 Portfolio Optimization

(View the complete code for this example.)

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 be the amount invested in each asset, be the amount of capital you have, be the random vector of asset returns over some period, and be the expected value of . Let G be the minimum growth you hope to obtain, and be the covariance matrix of . The objective function is , which can be equivalently denoted as .

Assume, for example, n = 4. Let = 10,000, G = 1,000, , and

The QP formulation can be written as:

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 12.2.1.

Output 12.2.1: Portfolio Optimization

The OPTMODEL Procedure

Problem Summary
Objective SenseMinimization
Objective Functionf
Objective TypeQuadratic
  
Number of Variables4
Bounded Above0
Bounded Below4
Bounded Below and Above0
Free0
Fixed0
  
Number of Constraints2
Linear LE (<=)1
Linear EQ (=)0
Linear GE (>=)1
Linear Range0
  
Constraint Coefficients8
  
Hessian Diagonal Elements4
Hessian Elements Below Diagonal6

Solution Summary
SolverQP
AlgorithmInterior Point
Objective Functionf
Solution StatusOptimal
Objective Value2232313.4432
  
Primal Infeasibility1.131114E-17
Dual Infeasibility2.799214E-13
Bound Infeasibility0
Duality Gap8.344009E-16
Complementarity0
  
Iterations7
Presolve Time0.00
Solution Time0.00

[1]x
13452.9
20.0
31068.8
42223.5


Thus, the minimum variance portfolio that earns an expected return of at least 10% is , , , . 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 12.2.2 that the decision variable , denoting Asset 2, is equal to 1,563.61, which means short sale of that asset.

Output 12.2.2: Portfolio Optimization with Short-Sale Option

The OPTMODEL Procedure

Solution Summary
SolverQP
AlgorithmInterior Point
Objective Functionf
Solution StatusOptimal
Objective Value1907122.2254
  
Primal Infeasibility4.997261E-14
Dual Infeasibility4.4944836E-8
Bound Infeasibility0
Duality Gap3.886201E-11
Complementarity0
  
Iterations5
Presolve Time0.00
Solution Time0.00

[1]x
11684.35
2-1563.61
3682.51
41668.95


Last updated: November 05, 2018