mws_gen_reg_txt_nonlinear - Chapter 06.04 Nonlinear Models...

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06.04.1 Chapter 06.04 Nonlinear Models for Regression After reading this chapter, you should be able to 1. derive constants of nonlinear regression models, 2. use in examples, the derived formula for the constants of the nonlinear regression model, and 3. linearize (transform) data to find constants of some nonlinear regression models. From fundamental theories, we may know the relationship between two variables. An example in chemical engineering is the Clausius-Clapeyron equation that relates vapor pressure P of a vapor to its absolute temperature, T . ( ) T B A P + = log (1) where A and B are the unknown parameters to be determined. The above equation is not linear in the unknown parameters. Any model that is not linear in the unknown parameters is described as a nonlinear regression model. Nonlinear models using least squares The development of the least squares estimation for nonlinear models does not generally yield equations that are linear and hence easy to solve. An example of a nonlinear regression model is the exponential model. Exponential model Given ( ) 1 1 ,y x , ( ) 2 2 ,y x , . . . ( ) n n ,y x , best fit bx ae y = to the data. The variables a and b are the constants of the exponential model. The residual at each data point i x is i bx i i ae y E = (2) The sum of the square of the residuals is = = n i i r E S 1 2 ( ) = = n i bx i i ae y 1 2 (3)
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Chapter 06.04 To find the constants a and b of the exponential model, we minimize r S by differentiating with respect to a and b and equating the resulting equations to zero. ( )( ) 0 2 1 = = = i i bx n i bx i r e ae y a S ( )( ) 0 2 1 = = = i i bx i n i bx i r e ax ae y b S (4a,b) or 0 1 2 1 = + = = n i bx n i bx i i i e a e y 0 1 2 1 = = = n i bx i n i bx i i i i e x a e x y (5a,b) Equations (5a) and (5b) are nonlinear in a and b and thus not in a closed form to be solved as was the case for linear regression. In general, iterative methods (such as Gauss-Newton iteration method, method of steepest descent, Marquardt's method, direct search, etc) must be used to find values of a and b . However, in this case, from Equation (5a), a can be written explicitly in terms of b as = = = n i bx n i bx i i i e e y a 1 2 1 (6) Substituting Equation (6) in (5b) gives 0 1 2 1 2 1 1 = = = = = n i bx i n i bx bx n i i bx i n i i i i i i e x e e y e x y (7) This equation is still a nonlinear equation in b and can be solved best by numerical methods such as the bisection method or the secant method. Example 1 Many patients get concerned when a test involves injection of a radioactive material. For example for scanning a gallbladder, a few drops of Technetium-99m isotope is used. Half of the technetium-99m would be gone in about 6 hours. It, however, takes about 24 hours for the radiation levels to reach what we are exposed to in day-to-day activities. Below is given the relative intensity of radiation as a function of time. Table 1
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This note was uploaded on 06/12/2011 for the course EML 3041 taught by Professor Kaw,a during the Spring '08 term at University of South Florida.

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mws_gen_reg_txt_nonlinear - Chapter 06.04 Nonlinear Models...

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