Korsmeyer - Peppas "kinetic model" it is in fact power equation or power function. You can use graphical method (after transformation to linear form) or better nonlinear regression method. Remember, however, that it is increasing function, so only aproximation of part of kinetic curve is possible. There is an asymptote (limit) in case of compounds release. Regards,

What specifically do you not understand from literature references that show examples of kinetic analysis of data using Korsmeyer-Peppas models? Alternatively or in addition, how is your question different from these?

https://www.researchgate.net/post/How_we_can_predict_the_n_value_in_Korsmeyer_peppas_release_model2

https://www.researchgate.net/post/How_can_I_determine_the_n_exponent_of_korsmeyer_peppas

https://www.researchgate.net/post/What_is_an_acceptable_R-square_value_for_dissolution_curve_fitting_drug_release_modelling

The process of taking the log of equation and fitting the result to a straight line is a common approach, but it distorts the presumed error distribution in a way that significantly overemphasizes the low-value results at the expense of the high-value results. A better and simpler approach, that weights each measurement equally, is to use the original equation (without taking the log) as a predictor. In Excel or some other suitable spreadsheet, compute the difference between these predictions and the observations with the value of n recorded at the top of the sheet. Sum the squares of these differences and use Solver to minimize this sum by adjusting the value of the exponent.

From a statistical standpoint, the important things are:

1. That you are comparing your actual measurements with the predictions (not the log or any other nonlinear transformation of the measurements)

2. That differences in the measurements and your predictions are appropriately treated (sum of squares assumes your measurements have errors that are normally distributed, as suggested by the Central Limit Theorem, and that the standard deviation of this normal distribution is the same at all measurement conditions - this is probably the best assumption unless you have compelling evidence to the contrary).

3. That your model is capable of describing all of the non-random variation in the data. If this is only true for the first portion of the measurements, as some suggest is the case for this model, then only sum the portion of the measurements for which it is true. That is, if the model is no longer valid after, say, 60% conversion, do not include data beyond 60% conversion in the sum of squares.

Larry

There is no problem here, with a change in the statistical distribution of new variables after the logarithmization of the power equation. The problem is not meeting the boundary condition in infinity. Of course, there would be a problem, but the case of statistical analysis of the regression equation, where it is necessary to assume a normal distribution. Regards,

With respect, I disagree with Dr. Grzesik . The standard formulas that compute the slope and intercept based on fitting data to a straight line assume errors that are additive and normally distributed. If the errors in the original equation are additive and normally distributed, the errors in the logarithm of the equation cannot be additive and normally distributed. The Central Limit Theorem gives you good reason to believe that the errors in your measurements should be additive and normally distributed. If you measure y, then you should fit the data to the original function. If you measure log(y), then you should fit the data to the log of the function. The fact that you get two different answers if you do it both ways shows that you are making fundamentally different assumptions about the error distributions when you take the log of the equation than when you do not. The differences, in many cases, can be quite large.

Larry, I do not know what you disagree with and what you agree with. I hope you agree that K-P equation is a power equation (y = kxn), that it does not satisfy the boundary condition in infinity (release of compounds), that k and n parameters in this equation are correlated numerically. In addition, I have always been againts such a transformation of non-linear equations, so that after introducing new variables (with different statistical distribution than normal), one can obtain a linear equation. This approach can lead to various errors and misunderstandings. I wrote about it many times in this forum.

Referring again to K-P equation (in linear or non linear form), the use of it to the description of individual kinetic curves of compounds release, it is in fact a replacement of several points with two numbers with unspecified meaning. That is the point. Regards,

If I might interject here. The fundamental stand that L Baxter (and I) take is as follows: Using an approach that linearizes a non-linear function in order to perform a linear-regression fit to obtain the fitting parameters from the function is lazy by today's standards if not invalid based on fundamentals. The statement that y = kxn is singular at infinity is a different issue, and I believe we all agree with it.

@L Baxter -- I hope I have not mis-stated your stand.

Using a fitting program that tries thousands of different equation forms and selects the one that happens to fit the error in your data is also "lazy". If your data doesn't fit a reasonable model form you have to examine the quality of the data and the correctness any assumptions you have made. You should also check if you experiment is reproducible.

Dr. Grzesik, I may have misunderstood your post. In any case. Dr. Weimer summarized my view succinctly. There a many examples in my work where linearizing an equation leads to poor parameter estimates. Indeed, there is a whole host of statistical issues that were and largely still are common practice that can be done much better now, and often with less effort. I have formulated a graduate statistics course around them. This is probably not the forum for this discussion but I would be happy to learn from others what issues we can improve.

Larry

There are many reasons that the value of parameters determined on the basis of a given equation differs from its linear equivalent. A few of them pointed Rick. I will add only one. It is a numerical correlation between the parameters of equations, especially visible in the case of exponential equation and power equation.

It seems that several other problems have been raised above. I think it's worth discussing about them. Regards

A few words to Sunil Kumar. In your case n = 0.456. It is a slope of the linear function. There are those who interpret the value of this parameter from the point of view of the mass transport mechanism. I do not belong to them, because for me it is only a number with no physical meaning. But here the reviewers opinion is important, not mine. Regards,

Rick Manner Larry L. Baxter I have a question about n determination, we should use the data (mi/mt) up to 0.6, right? what if I got just two points < 0.6! then how could I get n value?

Thank you very much in advance.

I am intensfied by your question. we determined the exponent from drug releases or diffusion releases from the polymeric matrix tablets based on the linear slope of the pilot which is equivalent to the slope of the curve+ 0.456. in another case you simply calculate it through KINET DS software

I think there is no answer given for the question. I am also expecting the answer for it. As of my knowledge there is a formula for it, it can use to find the n value. But I am also unable to calculate the n value by using this formula. F = (Mt /M ) = Km tn.

Can any one solve it.

(I have found the equation as follow:

y = 0.456x + 1.579 R² = 0.914

how can determine the n exponent to determine the mechanism of release ?)

Take the log of both sides. It becomes

ln y=0.456lnx + ln 1.579. Compare it with this (Mt/M) = Kmtn ;taking log of both sides

In (Mt/M) = n In t + In Km

Comparing the parameters

ln (Mt/M) = In y

n = 0.456

In t = In x

In Km = In 1.579