Matlab curve fitter error with exponential form with 2 terms

I am trying to curve fit the following data using the following exponetial form with two terms. I was expecting a concave curve along the points, but it is not producting that.

Please help. I am trying to curve fit the following points.

clc; 
clear;
close all; 

% Interior Square Footage
sqftL = [1500 1600 1750 1930 2250];
sqftUtilityL = [0 0.25 0.5 0.75 1];

Answer 1

[Matt J]

% Interior Square Footage
x = [1500 1600 1750 1930 2250];
y = [0 0.25 0.5 0.75 1];

theFit=fit(x' , y', 'exp2','Lower',[0 0 -inf -inf],'Upper',[inf 0 0 0 ])

theFit = 
     General model Exp2:
     theFit(x) = a*exp(b*x) + c*exp(d*x)
     Coefficients (with 95% confidence bounds):
       a =       1.279  (1.066, 1.492)
       b =           0  (fixed at bound)
       c =      -26.63  (-48.85, -4.412)
       d =   -0.002026  (-0.002675, -0.001378)

plot(theFit , x , y)

[Briana Canet]

What is the fit type for a linear curve following y = mx + b

[Torsten]

% Interior Square Footage
x = [1500 1600 1750 1930 2250];
y = [0 0.25 0.5 0.75 1];
theFit=fit(x' , y', 'poly1')

theFit = 
     Linear model Poly1:
     theFit(x) = p1*x + p2
     Coefficients (with 95% confidence bounds):
       p1 =    0.001301  (0.000765, 0.001837)
       p2 =      -1.849  (-2.827, -0.8712)

plot(theFit , x , y)

Answer 2

[Sulaymon Eshkabilov]

It does work very well if the initial conditions are given. Here is how it can be attained:

% Interior Square Footage
sqftL = [1500 1600 1750 1930 2250];
sqftUtilityL = [0 0.25 0.5 0.75 1];

FModel = @(b,x)(b(1)*exp(b(2)*x));
b0=[0.1, 0.001]; % Initial guess for b
opts = statset('Display','iter','TolFun',1e-10);
beta = nlinfit(sqftL,sqftUtilityL, FModel,b0, opts);

 
                                     Norm of         Norm of
   Iteration             SSE        Gradient           Step 
  -----------------------------------------------------------
           0        0.273074
           1        0.221476          286.87       0.0226301
           2        0.198314         224.135       0.0129788
           3        0.180237         186.212      0.00937518
           4        0.166374         158.151      0.00708214
           5        0.155587         135.318      0.00548545
           6        0.147093         116.609      0.00433898
           7        0.140339          101.15      0.00349367
           8        0.134923         88.2713      0.00285613
           9        0.130547         77.4559      0.00236579
          10        0.126988          68.305      0.00198214
          11        0.124079         60.5079      0.00167742
          12        0.121687          53.821      0.00143212
          13        0.118845         471.032      0.00770177
          14        0.114087         358.822      0.00496453
          15        0.109374         18.7422      0.00108585
          16         0.10936       0.0926197     7.33474e-05
          17         0.10936      0.00929679      2.0398e-05
          18         0.10936     0.000358592       3.812e-06
          19         0.10936     6.43899e-06     7.23391e-07
          20         0.10936     1.44957e-06     1.36534e-07
Iterations terminated: relative change in SSE less than OPTIONS.TolFun

x = linspace(min(sqftL), max(sqftL));
FModel_Vals = (beta(1)*exp(beta(2)*x));
plot(sqftL, sqftUtilityL, 'rd', 'MarkerSize',9, 'MarkerFaceColor','y')
hold on
plot(x, FModel_Vals, 'k', 'LineWidth',2)
legend('Data', 'Fit Model: a*exp(b*x)', 'Location', 'best')
xlabel('sqftL')
ylabel('sqftUtilityL & Fit Model')
grid on

[Torsten]

The fitted curve shouldn't be convex - the data are measurements of a utility curve.

Answer 3

[Sulaymon Eshkabilov]

It is better to fit with log fit.

% Interior Square Footage
sqftL = [1500 1600 1750 1930 2250];
sqftUtilityL = [0 0.25 0.5 0.75 1];

FModel = @(b,x)(b(1)*log(x)+b(2));
b0=[1, -1]; % Initial guess for b
opts = statset('Display','iter','TolFun',1e-10);
beta = nlinfit(sqftL,sqftUtilityL, FModel,b0, opts);

 
                                     Norm of         Norm of
   Iteration             SSE        Gradient           Step 
  -----------------------------------------------------------
           0         179.536
           1         0.35434         2.07117         3.24569
           2        0.254839        0.156622         2.12012
           3       0.0459699       0.0558453         7.45616
           4       0.0161053      0.00290256         3.87533
           5       0.0160244     1.58261e-05        0.211305
           6       0.0160244     8.67478e-09      0.00115783
           7       0.0160244     5.22904e-13     6.37516e-07
Iterations terminated: relative change in SSE less than OPTIONS.TolFun

x = linspace(min(sqftL), max(sqftL));
FModel_Vals = (beta(1)*log(x)+beta(2));
plot(sqftL, sqftUtilityL, 'rd', 'MarkerSize',9, 'MarkerFaceColor','y')
hold on
plot(x, FModel_Vals, 'k', 'LineWidth',2)
legend('Data', 'Fit Model: a*log(x)+b', 'Location', 'best')
xlabel('sqftL')
ylabel('sqftUtilityL & Fit Model')
grid on