# Maximize Long-Term Investments Using Linear Programming: Solver-Based

This example shows how to use the `linprog`

solver in Optimization Toolbox® to solve an investment problem with deterministic returns over a fixed number of years `T`

. The problem is to allocate your money over available investments to maximize your final wealth. This example uses the solver-based approach.

### Problem Formulation

Suppose that you have an initial amount of money `Capital_0`

to invest over a time period of `T`

years in `N`

zero-coupon bonds. Each bond pays an interest rate that compounds each year, and pays the principal plus compounded interest at the end of a maturity period. The objective is to maximize the total amount of money after `T`

years.

You can include a constraint that no single investment is more than a certain fraction of your total capital.

This example shows the problem setup on a small case first, and then formulates the general case.

You can model this as a linear programming problem. Therefore, to optimize your wealth, formulate the problem for solution by the `linprog`

solver.

### Introductory Example

Start with a small example:

The starting amount to invest

`Capital_0`

is $1000.The time period

`T`

is 5 years.The number of bonds

`N`

is 4.To model uninvested money, have one option B0 available every year that has a maturity period of 1 year and a interest rate of 0%.

Bond 1, denoted by B1, can be purchased in year 1, has a maturity period of 4 years, and interest rate of 2%.

Bond 2, denoted by B2, can be purchased in year 5, has a maturity period of 1 year, and interest rate of 4%.

Bond 3, denoted by B3, can be purchased in year 2, has a maturity period of 4 years, and interest rate of 6%.

Bond 4, denoted by B4, can be purchased in year 2, has a maturity period of 3 years, and interest rate of 6%.

By splitting up the first option B0 into 5 bonds with maturity period of 1 year and interest rate of 0%, this problem can be equivalently modeled as having a total of 9 available bonds, such that for `k=1..9`

Entry

`k`

of vector`PurchaseYears`

represents the year that bond`k`

is available for purchase.Entry

`k`

of vector`Maturity`

represents the maturity period $${m}_{k}$$ of bond`k`

.Entry

`k`

of vector`InterestRates`

represents the interest rate $${\rho}_{k}$$ of bond`k`

.

Visualize this problem by horizontal bars that represent the available purchase times and durations for each bond.

% Time period in years T = 5; % Number of bonds N = 4; % Initial amount of money Capital_0 = 1000; % Total number of buying oportunities nPtotal = N+T; % Purchase times PurchaseYears = [1;2;3;4;5;1;5;2;2]; % Bond durations Maturity = [1;1;1;1;1;4;1;4;3]; % Interest rates InterestRates = [0;0;0;0;0;2;4;6;6]; plotInvestments(N,PurchaseYears,Maturity,InterestRates)

### Decision Variables

Represent your decision variables by a vector `x`

, where `x(k)`

is the dollar amount of investment in bond `k`

, for `k=1..9`

. Upon maturity, the payout for investment `x(k)`

is

$$x(k)(1+{\rho}_{k}/100{)}^{{m}_{k}}.$$

Define $${\beta}_{k}=1+{\rho}_{k}/100$$ and define $${r}_{k}$$ as the total return of bond `k`

:

$${r}_{k}=(1+{\rho}_{k}/100{)}^{{m}_{k}}={\beta}_{k}^{{m}_{k}}.$$

```
% Total returns
finalReturns = (1+InterestRates/100).^Maturity;
```

### Objective Function

The goal is to choose investments to maximize the amount of money collected at the end of year `T`

. From the plot, you see that investments are collected at various intermediate years and reinvested. At the end of year `T`

, the money returned from investments 5, 7, and 8 can be collected and represents your final wealth:

$$\underset{x}{\mathrm{max}}{x}_{5}{r}_{5}+{x}_{7}{r}_{7}+{x}_{8}{r}_{8}$$

To place this problem into the form `linprog`

solves, turn this maximization problem into a minimization problem using the negative of the coefficients of `x(j)`

:

$$\underset{x}{\mathrm{min}}{f}^{T}x$$

with

$$f=[0;0;0;0;-{r}_{5};0;-{r}_{7};-{r}_{8};0]$$

f = zeros(nPtotal,1); f([5,7,8]) = [-finalReturns(5),-finalReturns(7),-finalReturns(8)];

### Linear Constraints: Invest No More Than You Have

Every year, you have a certain amount of money available to purchase bonds. Starting with year 1, you can invest the initial capital in the purchase options $${x}_{1}$$ and $${x}_{6}$$, so:

$${x}_{1}+{x}_{6}={Capital}_{0}$$

Then for the following years, you collect the returns from maturing bonds, and reinvest them in new available bonds to obtain the system of equations:

$$\begin{array}{ll}{x}_{2}+{x}_{8}+{x}_{9}& ={r}_{1}{x}_{1}\\ {x}_{3}& ={r}_{2}{x}_{2}\\ {x}_{4}& ={r}_{3}{x}_{3}\\ {x}_{5}+{x}_{7}& ={r}_{4}{x}_{4}+{r}_{6}{x}_{6}+{r}_{9}{x}_{9}\end{array}$$

Write these equations in the form $$Aeqx=beq$$, where each row of the $$Aeq$$ matrix corresponds to the equality that needs to be satisfied that year:

$$Aeq=\left[\begin{array}{ccccccccc}1& 0& 0& 0& 0& 1& 0& 0& 0\\ -{r}_{1}& 1& 0& 0& 0& 0& 0& 1& 1\\ 0& -{r}_{2}& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& -{r}_{3}& 1& 0& 0& 0& 0& 0\\ 0& 0& 0& -{r}_{4}& 1& -{r}_{6}& 1& 0& -{r}_{9}\end{array}\right]$$

$$beq=\left[\begin{array}{c}{Capital}_{0}\\ 0\\ 0\\ 0\end{array}\right]$$

Aeq = spalloc(N+1,nPtotal,15); Aeq(1,[1,6]) = 1; Aeq(2,[1,2,8,9]) = [-1,1,1,1]; Aeq(3,[2,3]) = [-1,1]; Aeq(4,[3,4]) = [-1,1]; Aeq(5,[4:7,9]) = [-finalReturns(4),1,-finalReturns(6),1,-finalReturns(9)]; beq = zeros(T,1); beq(1) = Capital_0;

### Bound Constraints: No Borrowing

Because each amount invested must be positive, each entry in the solution vector $$x$$ must be positive. Include this constraint by setting a lower bound `lb`

on the solution vector $$x$$. There is no explicit upper bound on the solution vector. Thus, set the upper bound `ub`

to empty.

lb = zeros(size(f)); ub = [];

### Solve the Problem

Solve this problem with no constraints on the amount you can invest in a bond. The interior-point algorithm can be used to solve this type of linear programming problem.

options = optimoptions('linprog','Algorithm','interior-point'); [xsol,fval,exitflag] = linprog(f,[],[],Aeq,beq,lb,ub,options);

Solution found during presolve.

### Visualize the Solution

The exit flag is 1, indicating that the solver found a solution. The value `-fval`

, returned as the second `linprog`

output argument, corresponds to the final wealth. Plot your investments over time.

fprintf('After %d years, the return for the initial $%g is $%g \n',... T,Capital_0,-fval);

After 5 years, the return for the initial $1000 is $1262.48

plotInvestments(N,PurchaseYears,Maturity,InterestRates,xsol)

### Optimal Investment with Limited Holdings

To diversify your investments, you can choose to limit the amount invested in any one bond to a certain percentage `Pmax`

of the total capital that year (including the returns for bonds that are currently in their maturity period). You obtain the following system of inequalities:

$$\begin{array}{ll}{x}_{1}& \le Pmax\times {Capital}_{0}\\ {x}_{2}& \le Pmax\times ({\beta}_{1}{x}_{1}+{\beta}_{6}{x}_{6})\\ {x}_{3}& \le Pmax\times ({\beta}_{2}{x}_{2}+{\beta}_{6}^{2}{x}_{6}+{\beta}_{8}{x}_{8}+{\beta}_{9}{x}_{9})\\ {x}_{4}& \le Pmax\times ({\beta}_{3}{x}_{3}+{\beta}_{6}^{3}{x}_{6}+{\beta}_{8}^{2}{x}_{8}+{\beta}_{9}^{2}{x}_{9})\\ {x}_{5}& \le Pmax\times ({\beta}_{4}{x}_{4}+{\beta}_{6}^{4}{x}_{4}+{\beta}_{8}^{3}{x}_{8}+{\beta}_{9}^{3}{x}_{9})\\ {x}_{6}& \le Pmax\times {Capital}_{0}\\ {x}_{7}& \le Pmax\times ({\beta}_{4}{x}_{4}+{\beta}_{6}^{4}{x}_{4}+{\beta}_{8}^{3}{x}_{8}+{\beta}_{9}^{3}{x}_{9})\\ {x}_{8}& \le Pmax\times ({\beta}_{1}{x}_{1}+{\beta}_{6}{x}_{6})\\ {x}_{9}& \le Pmax\times ({\beta}_{1}{x}_{1}+{\beta}_{6}{x}_{6})\end{array}$$

Place these inequalities in the matrix form `Ax <= b`

.

To set up the system of inequalities, first generate a matrix `yearlyReturns`

that contains the return for the bond indexed by i at year j in row i and column j. Represent this system as a sparse matrix.

% Maximum percentage to invest in any bond Pmax = 0.6; % Build the return for each bond over the maturity period as a sparse % matrix cumMaturity = [0;cumsum(Maturity)]; xr = zeros(cumMaturity(end-1),1); yr = zeros(cumMaturity(end-1),1); cr = zeros(cumMaturity(end-1),1); for i = 1:nPtotal mi = Maturity(i); % maturity of bond i pi = PurchaseYears(i); % purchase year of bond i idx = cumMaturity(i)+1:cumMaturity(i+1); % index into xr, yr and cr xr(idx) = i; % bond index yr(idx) = pi+1:pi+mi; % maturing years cr(idx) = (1+InterestRates(i)/100).^(1:mi); % returns over the maturity period end yearlyReturns = sparse(xr,yr,cr,nPtotal,T+1); % Build the system of inequality constraints A = -Pmax*yearlyReturns(:,PurchaseYears)'+ speye(nPtotal); % Left-hand side b = zeros(nPtotal,1); b(PurchaseYears == 1) = Pmax*Capital_0;

Solve the problem by investing no more than 60% in any one asset. Plot the resulting purchases. Notice that your final wealth is less than the investment without this constraint.

[xsol,fval,exitflag] = linprog(f,A,b,Aeq,beq,lb,ub,options);

Minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the function tolerance, and constraints are satisfied to within the constraint tolerance.

fprintf('After %d years, the return for the initial $%g is $%g \n',... T,Capital_0,-fval);

After 5 years, the return for the initial $1000 is $1207.78

plotInvestments(N,PurchaseYears,Maturity,InterestRates,xsol)

### Model of Arbitrary Size

Create a model for a general version of the problem. Illustrate it using `T`

= 30 years and 400 randomly generated bonds with interest rates from 1 to 6%. This setup results in a linear programming problem with 430 decision variables. The system of equality constraints is represented by a sparse matrix `Aeq`

of dimension 30-by-430 and the system of inequality constraints is represented by a sparse matrix `A`

of dimension 430-by-430.

% for reproducibility rng default % Initial amount of money Capital_0 = 1000; % Time period in years T = 30; % Number of bonds N = 400; % Total number of buying oportunities nPtotal = N+T; % Generate random maturity durations Maturity = randi([1 T-1],nPtotal,1); % Bond 1 has a maturity period of 1 year Maturity(1:T) = 1; % Generate random yearly interest rate for each bond InterestRates = randi(6,nPtotal,1); % Bond 1 has an interest rate of 0 (not invested) InterestRates(1:T) = 0; % Compute the return at the end of the maturity period for each bond: finalReturns = (1+InterestRates/100).^Maturity; % Generate random purchase years for each option PurchaseYears = zeros(nPtotal,1); % Bond 1 is available for purchase every year PurchaseYears(1:T)=1:T; for i=1:N % Generate a random year for the bond to mature before the end of % the T year period PurchaseYears(i+T) = randi([1 T-Maturity(i+T)+1]); end % Compute the years where each bond reaches maturity SaleYears = PurchaseYears + Maturity; % Initialize f to 0 f = zeros(nPtotal,1); % Indices of the sale oportunities at the end of year T SalesTidx = SaleYears==T+1; % Expected return for the sale oportunities at the end of year T ReturnsT = finalReturns(SalesTidx); % Objective function f(SalesTidx) = -ReturnsT; % Generate the system of equality constraints. % For each purchase option, put a coefficient of 1 in the row corresponding % to the year for the purchase option and the column corresponding to the % index of the purchase oportunity xeq1 = PurchaseYears; yeq1 = (1:nPtotal)'; ceq1 = ones(nPtotal,1); % For each sale option, put -\rho_k, where \rho_k is the interest rate for the % associated bond that is being sold, in the row corresponding to the % year for the sale option and the column corresponding to the purchase % oportunity xeq2 = SaleYears(~SalesTidx); yeq2 = find(~SalesTidx); ceq2 = -finalReturns(~SalesTidx); % Generate the sparse equality matrix Aeq = sparse([xeq1; xeq2], [yeq1; yeq2], [ceq1; ceq2], T, nPtotal); % Generate the right-hand side beq = zeros(T,1); beq(1) = Capital_0; % Build the system of inequality constraints % Maximum percentage to invest in any bond Pmax = 0.4; % Build the returns for each bond over the maturity period cumMaturity = [0;cumsum(Maturity)]; xr = zeros(cumMaturity(end-1),1); yr = zeros(cumMaturity(end-1),1); cr = zeros(cumMaturity(end-1),1); for i = 1:nPtotal mi = Maturity(i); % maturity of bond i pi = PurchaseYears(i); % purchase year of bond i idx = cumMaturity(i)+1:cumMaturity(i+1); % index into xr, yr and cr xr(idx) = i; % bond index yr(idx) = pi+1:pi+mi; % maturing years cr(idx) = (1+InterestRates(i)/100).^(1:mi); % returns over the maturity period end yearlyReturns = sparse(xr,yr,cr,nPtotal,T+1); % Build the system of inequality constraints A = -Pmax*yearlyReturns(:,PurchaseYears)'+ speye(nPtotal); % Left-hand side b = zeros(nPtotal,1); b(PurchaseYears==1) = Pmax*Capital_0; % Add the lower-bound constraints to the problem. lb = zeros(nPtotal,1);

### Solution with No Holding Limit

First, solve the linear programming problem without inequality constraints using the interior-point algorithm.

% Solve the problem without inequality constraints options = optimoptions('linprog','Algorithm','interior-point'); tic [xsol,fval,exitflag] = linprog(f,[],[],Aeq,beq,lb,[],options);

Minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the function tolerance, and constraints are satisfied to within the constraint tolerance.

toc

Elapsed time is 0.043860 seconds.

fprintf('\nAfter %d years, the return for the initial $%g is $%g \n',... T,Capital_0,-fval);

After 30 years, the return for the initial $1000 is $5167.58

### Solution with Limited Holdings

Now, solve the problem with the inequality constraints.

% Solve the problem with inequality constraints options = optimoptions('linprog','Algorithm','interior-point'); tic [xsol,fval,exitflag] = linprog(f,A,b,Aeq,beq,lb,[],options);

Minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the function tolerance, and constraints are satisfied to within the constraint tolerance.

toc

Elapsed time is 1.055830 seconds.

fprintf('\nAfter %d years, the return for the initial $%g is $%g \n',... T,Capital_0,-fval);

After 30 years, the return for the initial $1000 is $5095.26

Even though the number of constraints increased by an order of 10, the time for the solver to find a solution increased by an order of 100. This performance discrepancy is partially caused by dense columns in the inequality system shown in matrix `A`

. These columns correspond to bonds with a long maturity period, as shown in the following graph.

% Number of nonzero elements per column nnzCol = sum(spones(A)); % Plot the maturity length vs. the number of nonzero elements for each bond figure; plot(Maturity,nnzCol,'o'); xlabel('Maturity period of bond k') ylabel('Number of nonzero in column k of A')

Dense columns in the constraints lead to dense blocks in the solver's internal matrices, yielding a loss of efficiency of its sparse methods. To speed up the solver, try the dual-simplex algorithm, which is less sensitive to column density.

% Solve the problem with inequality constraints using dual simplex options = optimoptions('linprog','Algorithm','dual-simplex'); tic [xsol,fval,exitflag] = linprog(f,A,b,Aeq,beq,lb,[],options);

Optimal solution found.

toc

Elapsed time is 0.242389 seconds.

fprintf('\nAfter %d years, the return for the initial $%g is $%g \n',... T,Capital_0,-fval);

After 30 years, the return for the initial $1000 is $5095.26

In this case, the dual-simplex algorithm took much less time to obtain the same solution.

### Qualitative Result Analysis

To get a feel for the solution found by `linprog`

, compare it to the amount `fmax`

that you would get if you could invest all of your starting money in one bond with a 6% interest rate (the maximum interest rate) over the full 30 year period. You can also compute the equivalent interest rate corresponding to your final wealth.

% Maximum amount fmax = Capital_0*(1+6/100)^T; % Ratio (in percent) rat = -fval/fmax*100; % Equivalent interest rate (in percent) rsol = ((-fval/Capital_0)^(1/T)-1)*100; fprintf(['The amount collected is %g%% of the maximum amount $%g '... 'that you would obtain from investing in one bond.\n'... 'Your final wealth corresponds to a %g%% interest rate over the %d year '... 'period.\n'], rat, fmax, rsol, T)

The amount collected is 88.7137% of the maximum amount $5743.49 that you would obtain from investing in one bond. Your final wealth corresponds to a 5.57771% interest rate over the 30 year period.

plotInvestments(N,PurchaseYears,Maturity,InterestRates,xsol,false)