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The general advice for least-squares problem setup is to formulate the problem in a way that allows `solve`

to recognize that the problem has a least-squares form. When you do that, `solve`

internally calls `lsqnonlin`

, which is efficient at solving least-squares problems. See Write Objective Function for Problem-Based Least Squares.

This example shows the efficiency of a least-squares solver by comparing the performance of `lsqnonlin`

with that of `fminunc`

on the same problem. Additionally, the example shows added benefits that you can obtain by explicitly recognizing and handling separately the linear parts of a problem.

Consider the following data:

```
Data = ...
[0.0000 5.8955
0.1000 3.5639
0.2000 2.5173
0.3000 1.9790
0.4000 1.8990
0.5000 1.3938
0.6000 1.1359
0.7000 1.0096
0.8000 1.0343
0.9000 0.8435
1.0000 0.6856
1.1000 0.6100
1.2000 0.5392
1.3000 0.3946
1.4000 0.3903
1.5000 0.5474
1.6000 0.3459
1.7000 0.1370
1.8000 0.2211
1.9000 0.1704
2.0000 0.2636];
```

Plot the data points.

t = Data(:,1); y = Data(:,2); plot(t,y,'ro') title('Data points')

The problem is to fit the function

` y = c(1)*exp(-lam(1)*t) + c(2)*exp(-lam(2)*t)`

to the data.

To begin, define optimization variables corresponding to the equation.

c = optimvar('c',2); lam = optimvar('lam',2);

Arbitrarily set the initial point `x0`

as follows: `c(1) = 1`

, `c(2) = 1`

, `lam(1) = 1`

, and `lam(2) = 0`

:

x0.c = [1,1]; x0.lam = [1,0];

Create a function that computes the value of the response at times `t`

when the parameters are `c`

and `lam`

.

diffun = c(1)*exp(-lam(1)*t) + c(2)*exp(-lam(2)*t);

Convert `diffun`

to an optimization expression that sums the squares of the differences between the function and the data `y`

.

diffexpr = sum((diffun - y).^2);

Create an optimization problem having `diffexpr`

as the objective function.

`ssqprob = optimproblem('Objective',diffexpr);`

Solve the problem using the default solver.

[sol,fval,exitflag,output] = solve(ssqprob,x0)

Solving problem using lsqnonlin. Local minimum possible. lsqnonlin stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.

`sol = `*struct with fields:*
c: [2x1 double]
lam: [2x1 double]

fval = 0.1477

exitflag = FunctionChangeBelowTolerance

`output = `*struct with fields:*
firstorderopt: 7.8852e-06
iterations: 6
funcCount: 35
cgiterations: 0
algorithm: 'trust-region-reflective'
stepsize: 0.0096
message: '...'
solver: 'lsqnonlin'

Plot the resulting curve based on the returned solution values `sol.c`

and `sol.lam`

.

resp = evaluate(diffun,sol); hold on plot(t,resp) hold off

The fit looks to be as good as possible.

`fminunc`

To solve the problem using the `fminunc`

solver, set the `'Solver'`

option to `'fminunc'`

when calling `solve`

.

[xunc,fvalunc,exitflagunc,outputunc] = solve(ssqprob,x0,'Solver',"fminunc")

Solving problem using fminunc. Local minimum found. Optimization completed because the size of the gradient is less than the value of the optimality tolerance.

`xunc = `*struct with fields:*
c: [2x1 double]
lam: [2x1 double]

fvalunc = 0.1477

exitflagunc = OptimalSolution

`outputunc = `*struct with fields:*
iterations: 30
funcCount: 185
stepsize: 0.0017
lssteplength: 1
firstorderopt: 2.9562e-05
algorithm: 'quasi-newton'
message: '...'
solver: "fminunc"

Notice that `fminunc`

found the same solution as `lsqcurvefit`

, but took many more function evaluations to do so. The parameters for `fminunc`

are in the opposite order as those for `lsqcurvefit`

; the larger `lam`

is `lam(2)`

, not `lam(1)`

. This is not surprising, the order of variables is arbitrary.

fprintf(['There were %d iterations using fminunc,' ... ' and %d using lsqcurvefit.\n'], ... outputunc.iterations,output.iterations)

There were 30 iterations using fminunc, and 6 using lsqcurvefit.

fprintf(['There were %d function evaluations using fminunc,' ... ' and %d using lsqcurvefit.'], ... outputunc.funcCount,output.funcCount)

There were 185 function evaluations using fminunc, and 35 using lsqcurvefit.

Notice that the fitting problem is linear in the parameters `c(1)`

and `c(2)`

. This means for any values of `lam(1)`

and `lam(2)`

, you can use the backslash operator to find the values of `c(1)`

and `c(2)`

that solve the least-squares problem.

Rework the problem as a two-dimensional problem, searching for the best values of `lam(1)`

and `lam(2)`

. The values of `c(1)`

and `c(2)`

are calculated at each step using the backslash operator as described above. To do so, use the `fitvector`

function, which performs the backslash operation to obtain `c(1)`

and `c(2)`

at each solver iteration.

`type fitvector`

function yEst = fitvector(lam,xdata,ydata) %FITVECTOR Used by DATDEMO to return value of fitting function. % yEst = FITVECTOR(lam,xdata) returns the value of the fitting function, y % (defined below), at the data points xdata with parameters set to lam. % yEst is returned as a N-by-1 column vector, where N is the number of % data points. % % FITVECTOR assumes the fitting function, y, takes the form % % y = c(1)*exp(-lam(1)*t) + ... + c(n)*exp(-lam(n)*t) % % with n linear parameters c, and n nonlinear parameters lam. % % To solve for the linear parameters c, we build a matrix A % where the j-th column of A is exp(-lam(j)*xdata) (xdata is a vector). % Then we solve A*c = ydata for the linear least-squares solution c, % where ydata is the observed values of y. A = zeros(length(xdata),length(lam)); % build A matrix for j = 1:length(lam) A(:,j) = exp(-lam(j)*xdata); end c = A\ydata; % solve A*c = y for linear parameters c yEst = A*c; % return the estimated response based on c

Solve the problem using `solve`

starting from a two-dimensional initial point `x02.lam`

= `[1,0]`

. To do so, first convert the `fitvector`

function to an optimization expression using `fcn2optimexpr`

. See Convert Nonlinear Function to Optimization Expression. To avoid a warning, give the output size of the resulting expression. Create a new optimization problem with objective as the sum of squared differences between the converted `fitvector`

function and the data `y`

.

x02.lam = x0.lam; F2 = fcn2optimexpr(@(x) fitvector(x,t,y),lam,'OutputSize',[length(t),1]); ssqprob2 = optimproblem('Objective',sum((F2 - y).^2)); [sol2,fval2,exitflag2,output2] = solve(ssqprob2,x02)

Solving problem using lsqnonlin. Local minimum possible. lsqnonlin stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.

`sol2 = `*struct with fields:*
lam: [2x1 double]

fval2 = 0.1477

exitflag2 = FunctionChangeBelowTolerance

`output2 = `*struct with fields:*
firstorderopt: 4.4071e-06
iterations: 10
funcCount: 33
cgiterations: 0
algorithm: 'trust-region-reflective'
stepsize: 0.0080
message: '...'
solver: 'lsqnonlin'

The efficiency of the two-dimensional solution is similar to that of the four-dimensional solution:

fprintf(['There were %d function evaluations using the 2-d ' ... 'formulation, and %d using the 4-d formulation.'], ... output2.funcCount,output.funcCount)

There were 33 function evaluations using the 2-d formulation, and 35 using the 4-d formulation.

Choosing a bad starting point for the original four-parameter problem leads to a local solution that is not global. Choosing a starting point with the same bad `lam(1)`

and `lam(2)`

values for the split two-parameter problem leads to the global solution. To show this, rerun the original problem with a start point that leads to a relatively bad local solution, and compare the resulting fit with the global solution.

x0bad.c = [5 1]; x0bad.lam = [1 0]; [solbad,fvalbad,exitflagbad,outputbad] = solve(ssqprob,x0bad)

Solving problem using lsqnonlin. Local minimum possible. lsqnonlin stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.

`solbad = `*struct with fields:*
c: [2x1 double]
lam: [2x1 double]

fvalbad = 2.2173

exitflagbad = FunctionChangeBelowTolerance

`outputbad = `*struct with fields:*
firstorderopt: 0.0057
iterations: 32
funcCount: 165
cgiterations: 0
algorithm: 'trust-region-reflective'
stepsize: 0.0021
message: '...'
solver: 'lsqnonlin'

respbad = evaluate(diffun,solbad); hold on plot(t,respbad,'g') legend('Data','Global fit','Bad local fit','Location','NE') hold off

fprintf(['The residual norm at the good ending point is %f,' ... ' and the residual norm at the bad ending point is %f.'], ... fval,fvalbad)

The residual norm at the good ending point is 0.147723, and the residual norm at the bad ending point is 2.217300.