lsqnonneg

Can I solve [x] in the relation [C]*[x]=[d] using x= lsqnonneg(C,d);? Vector dimensions are as follows C(5x2), d(5x1). If not please suggest me better way.

Thank you,

Answer 1

[Andrew Newell]

x = C\d;

EDIT: The least squares estimate for x is

xhat = (C'*C)\(C'*d);

You can estimate the error in the fit as follows:

n = length(d);
s2 = norm(x-xhat)^2/(n-2); %estimated variance of the data
covar = s2*inv(C'*C); %covariance matrix for the components of xhat
xerr = sqrt(diag(covar)) %standard errors for xhat components

[Abex]

Dear Andriw,

Thank you for your suggestion.

How can i use matrix division as the C and d posses different size? I tried it using x=inv(C)*d but C is not a square matrix and has no inverse.

Would you please have a say on this?

[Titus Edelhofer]

The backslash is solving the linear system of equations you wrote, even for overdetermined system (e.g. C has more rows then columns). It does this in the least squares sense (because there (usually) is no exact solution).

[Andrew Newell]

MATLAB does not take the inverse of C to solve this equation. In effect, you have two equations for the components of x, and it solves them using Gaussian elimination.

[Titus Edelhofer]

If the matrix C is 5x2 you have 5 equations for 2 unknowns, that is an overdetermined system ...

[Abex]

Dear Andrew,

Thank you for elaborating, it is very helpful.I am interested in posetive value of x and that is why I used lsqnonneg(C,d). Why is posetive value of C\d and lsqnonneg(C,d) show different result for same C and d?

thank you

[Andrew Newell]

Can you give an example?

[Abex]

Dear Andrew,thank you

Here is my full data

C =

0.6065 0.6065

0.6065 0.6065

0.6065 0.6065

0.6065 0.6065

0.6065 0.6065

0.6065 0.6065

0.6065 0.6065

0.6065 0.6065

0.6065 0.6065

0.6064 0.6064

0.6002 0.6003

0.5648 0.5649

0.5705 0.5705

0.5783 0.5783

0.5786 0.5786

0.5860 0.5859

0.5925 0.5925

0.5962 0.5961

0.5962 0.5961

0.5967 0.5965

0.6015 0.6014

0.6000 0.5998

0.5994 0.5993

0.6016 0.6015

0.6034 0.6034

0.6046 0.6045

0.6031 0.6030

0.6049 0.6049

0.6036 0.6035

0.6050 0.6049

0.6047 0.6046

0.6044 0.6044

0.6041 0.6040

0.6043 0.6042

0.6043 0.6043

0.6044 0.6043

0.6047 0.6047

0.6046 0.6045

0.6047 0.6046

0.6041 0.6040

0.6029 0.6028

0.6027 0.6026

0.6032 0.6031

0.6040 0.6039

0.6024 0.6022

0.6015 0.6013

0.6024 0.6022

0.6038 0.6037

0.6040 0.6039

0.6049 0.6048

0.6044 0.6043

0.6043 0.6042

0.6028 0.6026

0.6026 0.6024

0.6047 0.6046

0.6043 0.6042

0.6053 0.6052

0.6042 0.6041

0.6046 0.6045

0.6034 0.6033

0.6054 0.6053

0.6055 0.6054

0.6038 0.6037

0.6045 0.6043

0.6052 0.6051

0.6059 0.6058

0.6056 0.6056

0.6053 0.6052

0.6054 0.6053

0.6058 0.6057

0.6062 0.6062

0.6063 0.6062

0.6064 0.6064

0.6064 0.6064

0.6065 0.6065

d =

0.2500

0.2600

0.2700

0.2800

0.2900

0.3000

0.3100

0.3200

0.3300

0.3400

0.3500

0.3600

0.3700

0.3800

0.3900

0.4000

0.4100

0.4200

0.4300

0.4400

0.4500

0.4600

0.4700

0.4800

0.4900

0.5000

0.5100

0.5200

0.5300

0.5400

0.5500

0.5600

0.5700

0.5800

0.5900

0.6000

0.6100

0.6200

0.6300

0.6400

0.6500

0.6600

0.6700

0.6800

0.6900

0.7000

0.7100

0.7200

0.7300

0.7400

0.7500

0.7600

0.7700

0.7800

0.7900

0.8000

0.8100

0.8200

0.8300

0.8400

0.8500

0.8600

0.8700

0.8800

0.8900

0.9000

0.9100

0.9200

0.9300

0.9400

0.9500

0.9600

0.9700

0.9800

0.9900

>> C\d

ans =

1.0e+003 *

1.7095

-1.7087

>> lsqnonneg(C,d)

ans =

1.0313

0

My expectation for the value of x was 0 to 1.

[Andrew Newell]

Your matrix C is very ill-conditioned because the two rows are so nearly equal. The estimated errors for xhat, using the equations I have added above, are

1.0e+05 *

3.4387

3.4391

i.e., much larger than the components themselves. In other words, it is futile to try to fit such a data set.

Answer 2

[Titus Edelhofer]

Hi,

if you take a closer look at your data you observe:

1. the two columns of C are very similar, therefore it's no surprise that using lsqnonneg you get one positive value and a zero.

2. If you do the following:

x = C(:,1)./d

you get values 2.4, 2.3, 2.2, 2.1, ..., 0.62, 0.61. The quadratic mean will be about the value you get from lsqnonneg and (from the data) it's not clear why the value should be less then one.

Titus

[Abex]

the value of d is normalised mass which varies from 0 to 1 and C is obtained from other algorithm