Nonlinear Least-Squares, Problem-Based

This example shows how to perform nonlinear least-squares curve fitting using the Problem-Based Optimization Workflow.

Model

The model equation for this problem is

y(t)=A1exp(r1t)+A2exp(r2t),

where A1, A2, r1, and r2 are the unknown parameters, y is the response, and t is time. The problem requires data for times tdata and (noisy) response measurements ydata. The goal is to find the best A and r, meaning those values that minimize

ttdata(y(t)-ydata)2.

Sample Data

Typically, you have data for a problem. In this case, generate artificial noisy data for the problem. Use A = [1,2] and r = [-1,-3] as the underlying values, and use 200 random values from 0 to 3 as the time data. Plot the resulting data points.

rng default % For reproducibility
A = [1,2];
r = [-1,-3];
tdata = 3*rand(200,1);
tdata = sort(tdata); % Increasing times for easier plotting
noisedata = 0.05*randn(size(tdata)); % Artificial noise
ydata = A(1)*exp(r(1)*tdata) + A(2)*exp(r(2)*tdata) + noisedata;
plot(tdata,ydata,'r*')
xlabel 't'
ylabel 'Response'

The data appears to be noisy. Therefore, the solution probably will not match the original parameters A and r very well.

Problem-Based Approach

To find the best-fitting parameters A and r, first define optimization variables with those names.

A = optimvar('A',2);
r = optimvar('r',2);

Create an expression for the objective function, which is the sum of squares to minimize. Because the response has an exponential term, write an anonymous function for the response, and convert the response to an optimization expression using fcn2optimexpr. See Convert Nonlinear Function to Optimization Expression.

fun = @(A,r) A(1)*exp(r(1)*tdata) + A(2)*exp(r(2)*tdata);
response = fcn2optimexpr(fun,A,r);
obj = sum((response - ydata).^2);

Create an optimization problem with the objective function obj.

lsqproblem = optimproblem("Objective",obj);

For the problem-based approach, specify the initial point as a structure, with the variable names as the fields of the structure. Specify the initial A = [1/2,3/2] and the initial r = [-1/2,-3/2].

x0.A = [1/2,3/2];
x0.r = [-1/2,-3/2];

Review the problem formulation.

show(lsqproblem)
  OptimizationProblem : 

	Solve for:
       A, r

	minimize :
       sum((anonymousFunction1(A, r) - extraParams{1}).^2)

       where:

         anonymousFunction1 = @(A,r)A(1)*exp(r(1)*tdata)+A(2)*exp(r(2)*tdata);

         extraParams{1}:

         2.9278
         2.7513
         2.7272
         2.4199
         2.3172
         2.3961
         2.2522
         2.1974
         2.1666
         2.0944
         1.9566
         1.7989
         1.7984
         1.7540
         1.8318
         1.6745
         1.6874
         1.5526
         1.5229
         1.5680
         1.4784
         1.5280
         1.3727
         1.2968
         1.4012
         1.3602
         1.2714
         1.1773
         1.2119
         1.2033
         1.2037
         1.1729
         1.1829
         1.1602
         1.0448
         1.0320
         1.0397
         1.0334
         1.0233
         1.0275
         0.8173
         0.9373
         1.0202
         0.8896
         0.9791
         0.9128
         0.7763
         0.7669
         0.6579
         0.7135
         0.7978
         0.7164
         0.7071
         0.6429
         0.6676
         0.6782
         0.6802
         0.6328
         0.6301
         0.7406
         0.4908
         0.7126
         0.6136
         0.6269
         0.4668
         0.4963
         0.5007
         0.5226
         0.3764
         0.4824
         0.3930
         0.4390
         0.4665
         0.4490
         0.4841
         0.4539
         0.3698
         0.3974
         0.3356
         0.3045
         0.4131
         0.3561
         0.3506
         0.3960
         0.3625
         0.3446
         0.3778
         0.3565
         0.3187
         0.2677
         0.2664
         0.3572
         0.2129
         0.2919
         0.1764
         0.3210
         0.3016
         0.2572
         0.2514
         0.1301
         0.2825
         0.1372
         0.1243
         0.2421
         0.1888
         0.2547
         0.2559
         0.2632
         0.1801
         0.2309
         0.2134
         0.2495
         0.2332
         0.2512
         0.1875
         0.1861
         0.2397
         0.0803
         0.1579
         0.1196
         0.1541
         0.1978
         0.2034
         0.1095
         0.1332
         0.1567
         0.1345
         0.1635
         0.1661
         0.0991
         0.1366
         0.0387
         0.1922
         0.1031
         0.0714
         0.1178
         0.0568
         0.1255
         0.0957
         0.2313
         0.1710
        -0.0148
         0.1316
         0.0385
         0.0946
         0.1147
         0.1436
         0.0917
         0.1840
         0.0786
         0.1161
         0.1327
         0.1026
         0.1421
         0.1142
         0.0553
         0.0036
         0.1866
         0.0634
         0.0974
         0.1203
         0.0939
         0.0429
         0.0640
         0.0811
         0.1603
         0.0427
         0.1244
         0.0993
         0.0696
         0.0264
         0.0641
         0.0703
         0.0010
         0.0793
         0.0267
         0.0625
         0.0834
         0.0204
         0.0507
         0.0826
        -0.0272
         0.1161
         0.1832
         0.1100
         0.0453
         0.0826
         0.0079
         0.1531
         0.1052
         0.0965
         0.0132
         0.0729
         0.0287
         0.0410
         0.0280
         0.0049
         0.0102
         0.0442
        -0.0343

Problem-Based Solution

Solve the problem.

[sol,fval] = solve(lsqproblem,x0)
Solving problem using lsqnonlin.

Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.
sol = struct with fields:
    A: [2x1 double]
    r: [2x1 double]

fval = 0.4724

Plot the resulting solution and the original data.

figure
responsedata = evaluate(response,sol);
plot(tdata,ydata,'r*',tdata,responsedata,'b-')
legend('Original Data','Fitted Curve')
xlabel 't'
ylabel 'Response'
title("Fitted Response")

The plot shows that the fitted data matches the original noisy data fairly well.

See how closely the fitted parameters match the original parameters A = [1,2] and r = [-1,-3].

disp(sol.A)
    1.1615
    1.8629
disp(sol.r)
   -1.0882
   -3.2256

The fitted parameters are off by about 15% in A and 8% in r.

See Also

Related Topics