160 likes | 304 Views
Some Examples. Example: daily auto accidents in Saskatchewan to 1984 to 1992. Data collected: Date Number of Accidents Factors we want to consider: Trend Yearly Cyclical Effect Day of the week effect Holiday effects. Trend. This will be modeled by a Linear function : Y = b 0 + b 1 X
E N D
Example: daily auto accidents in Saskatchewan to 1984 to 1992 Data collected: • Date • Number of Accidents Factors we want to consider: • Trend • Yearly Cyclical Effect • Day of the week effect • Holiday effects
Trend This will be modeled by a Linear function : Y = b0 +b1X (more generally a polynomial) Y = b0 +b1X +b2X2+ b3X3+ …. Yearly Cyclical Trend This will be modeled by a Trig Polynomial – Sin and Cos functions with differing frequencies(periods) : Y = d1 sin(2pf1X)+ g1 cos(2pf2X)+ d1 sin(2pf2X) + g2 cos(2pf2X) + …
Day of the week effect: This will be modeled using “dummy”variables : a1D1 + a2D2 + a3D3 + a4D4 + a5D5 + a6D6 Di = (1 if day of week = i, 0 otherwise) Holiday Effects Also will be modeled using “dummy”variables :
Independent variables X = day,D1,D2,D3,D4,D5,D6,S1,S2,S3,S4,S5, S6,C1,C2,C3,C4,C5,C6,NYE,HW,V1,V2,cd,T1, T2. Si=sin(0.017202423838959*i*day). Ci=cos(0.017202423838959*i*day). Dependent variable Y = daily accident frequency
Independent variables ANALYSIS OF VARIANCE SUM OF SQUARES DF MEAN SQUARE F RATIO REGRESSION 976292.38 18 54238.46 114.60 RESIDUAL 1547102.1 3269 473.2646 VARIABLES IN EQUATION FOR PACC . VARIABLES NOT IN EQUATION STD. ERROR STD REG F . PARTIAL F VARIABLE COEFFICIENT OF COEFF COEFF TOLERANCE TO REMOVE LEVEL. VARIABLE CORR. TOLERANCE TO ENTER LEVEL (Y-INTERCEPT 60.48909 ) . day 1 0.11107E-02 0.4017E-03 0.038 0.99005 7.64 1 . IACC 7 0.49837 0.78647 1079.91 0 D1 9 4.99945 1.4272 0.063 0.57785 12.27 1 . Dths 8 0.04788 0.93491 7.51 0 D2 10 9.86107 1.4200 0.124 0.58367 48.22 1 . S3 17 -0.02761 0.99511 2.49 1 D3 11 9.43565 1.4195 0.119 0.58311 44.19 1 . S5 19 -0.01625 0.99348 0.86 1 D4 12 13.84377 1.4195 0.175 0.58304 95.11 1 . S6 20 -0.00489 0.99539 0.08 1 D5 13 28.69194 1.4185 0.363 0.58284 409.11 1 . C6 26 -0.02856 0.98788 2.67 1 D6 14 21.63193 1.4202 0.273 0.58352 232.00 1 . V1 29 -0.01331 0.96168 0.58 1 S1 15 -7.89293 0.5413 -0.201 0.98285 212.65 1 . V2 30 -0.02555 0.96088 2.13 1 S2 16 -3.41996 0.5385 -0.087 0.99306 40.34 1 . cd 31 0.00555 0.97172 0.10 1 S4 18 -3.56763 0.5386 -0.091 0.99276 43.88 1 . T1 32 0.00000 0.00000 0.00 1 C1 21 15.40978 0.5384 0.393 0.99279 819.12 1 . C2 22 7.53336 0.5397 0.192 0.98816 194.85 1 . C3 23 -3.67034 0.5399 -0.094 0.98722 46.21 1 . C4 24 -1.40299 0.5392 -0.036 0.98999 6.77 1 . C5 25 -1.36866 0.5393 -0.035 0.98955 6.44 1 . NYE 27 32.46759 7.3664 0.061 0.97171 19.43 1 . HW 28 35.95494 7.3516 0.068 0.97565 23.92 1 . T2 33 -18.38942 7.4039 -0.035 0.96191 6.17 1 . ***** F LEVELS( 4.000, 3.900) OR TOLERANCE INSUFFICIENT FOR FURTHER STEPPING
Example 2 • Data on the reaction rate of the catalytic isomerization of л-pentane to isopentane versus the partial pressures of hydrogen, л-pentane, and isopentane are reproduced in the Table on the following slide
Isomerization is a chemical process in which a complex chemical is converted into more simple units, called isomers: catalytic isomerization employs catalysts to speed the reaction. The reaction rate depends on various factors, such as partial pressures of the products and the concentration of the catalyst.
The differential reaction rate was expressed as grams of isopentane produced per gram of catalyst per hour (hr-1), and the instantaneous partial pressure of a component was calculated as the mole fraction of the component times the total pressure, in pounds per square inch absolute (psia).
A common form of model for the reaction rate is the Hougen-Watson model: Fit this model