Part1: Full factorial designs for 2-level factors Prepared by: Paul Funkenbusch, Department of Mechanical Engineering, University of Rochester
Terminology Experiment Model relationship Full factorial experiments Interactions and sparsity of effects DOE mini-course, part 1, Paul Funkenbusch, 2015 2
Example Terms (measure the volume of a balloon as a function of temperature and pressure) Temperature, Factors variables whose influence you want to study. Pressure Levels specific values given 50 C, 100 C to a factor during experiments 1Pa, 2Pa (initially limit ourselves to 2-levels) Set T = 50 C, P = 1Pa Treatment condition one running of the experiment and measure volume Response result measured measured volume for a treatment condition DOE mini-course, part 1, Paul Funkenbusch, 2015 3
Leve vel Use X1, X2, etc. to designate factors. Factor -1 +1 Use -1, +1 to designate levels 50 100 X1. Temperature ( C) X2. Pressure (Pa) 1 2 X1 at level -1 means T = 50 C y response y = volume of balloon TC TC X1 X1 X2 X2 y Use a table to show factor levels and response (a) for each 1 -1 -1 y 1 treatment condition. 2 +1 -1 y 2 For example, during TC2, 3 -1 +1 y 3 set T = 100 C and P = 1Pa, 4 +1 +1 y4 measure the balloon volume = y 2 etc. DOE mini-course, part 1, Paul Funkenbusch, 2015 4
“Effects” calculated from Leve vel responses (y 1, y 2 ) Factor -1 +1 50 100 X1. Temperature ( C) m* TC TC X1 X1 y ◦ =(y 1 + y 2 )/2 1 -1 y 1 ◦ = overal all l average age 2 +1 y 2 D X1 ◦ = (avg. at +1) – (avg. at -1) y ◦ = m +1 – m -1 = y 2 – y 1 ◦ = “effect of X1” -1 1 X1 X1 DOE mini-course, part 1, Paul Funkenbusch, 2015 5
TC TC X1 X1 y Can fit a straight line 1 -1 y 1 y pred = a o + a 1 X 1 2 +1 y 2 a o ◦ intercept y ◦ related to m* ◦ (= m* for X1on ±1 scale) m* D a 1 ◦ slope ◦ related to D X1 -1 1 X1 X1 ◦ (= D X1 /2 for X1on ±1 scale) DOE mini-course, part 1, Paul Funkenbusch, 2015 6
DOF or DF TC TC X1 X1 y ◦ Counter for information 1 -1 y 1 2 +1 y 2 Experimental data ◦ 2 data points (y1 & y2) ◦ = 2 DOF y Analysis ◦ 2 results (m*, D X1 a o , a 1 ) ◦ = 2 DOF -1 1 X1 X1 Information is conserved ◦ 2 DOF 2 DOF DOE mini-course, part 1, Paul Funkenbusch, 2015 7
Leve vel One At a Time experiment Factor -1 +1 ◦ “OAT” 50 100 X1. Temperature ( C) ◦ what most people are taught X2. Pressure (Pa) 1 2 ◦ pick a baseline (X1=-1, X2=-1) ◦ Change one factor at a time TC TC X1 X1 X2 X2 y Results 1 -1 -1 y 1 ◦ baseline = y 1 2 +1 -1 y 2 ◦ D X1 = y 2 – y 1 3 -1 +1 y 3 ◦ D x2 = y 3 – y 1 DOE mini-course, part 1, Paul Funkenbusch, 2015 8
Leve vel Choice of baseline is Factor -1 +1 critical (next slide) 50 100 X1. Temp ( C) X2. Pressure (Pa) 1 2 Inefficient use of data ◦ baseline = y 1 ◦ D X1 = y 2 – y 1 TC TC X1 X1 X2 X2 y ◦ D x2 = y 3 – y 1 1 -1 -1 y 1 ◦ Only use part of data to 2 +1 -1 y 2 calculate each effect 3 -1 +1 y 3 DOE mini-course, part 1, Paul Funkenbusch, 2015 9
Leve vel Study plant growth as a Factor -1 +1 function of watering and X1. watering never daily sunlight X2. daily sunlight none 10 hrs Results ◦ baseline = y 1 = 0 TC TC X1 X1 X2 X2 growth th ◦ D X1 = y 2 – y 1 = 0 1 -1 -1 0 ◦ D x2 = y 3 – y 1 = 0 2 +1 -1 0 3 -1 +1 0 Conclusion ◦ Watering and sunlight do not affect plant growth DOE mini-course, part 1, Paul Funkenbusch, 2015 10
Test all combinations Leve vel Factor -1 +1 Results 50 100 X1. Temp ( C) ◦ m* = (y 1 +y 2 +y 3 +y 4 )/4 X2. Pressure (Pa) 1 2 ◦ D X1 = (y 3 +y 4 )/2 - (y 1 +y 2 )/2 ◦ D x2 = (y 2 +y 4 )/2 - (y 1 +y 3 )/2 TC TC X1 X1 X2 X2 y Effect of X1 1 -1 -1 y 1 ◦ Averaged over both X2 levels ◦ No baseline 2 -1 +1 y 2 3 +1 -1 y 3 All results all effects 4 +1 +1 y 4 ◦ Use all data in each calculation ◦ Efficient use of data DOE mini-course, part 1, Paul Funkenbusch, 2015 11
TC TC X1 X1 X2 X2 X1*X2 *X2 y Four degrees of freedom 1 -1 -1 +1 y 1 Model can include an extra (interaction) term 2 -1 +1 -1 y 2 3 +1 -1 -1 y 3 y pred = a o +a 1 X 1 +a 2 X 2 +a 12 12 X 1 X 2 4 +1 +1 +1 y 4 a 12 D 12 ◦ = (avg. for X 1 X 2 =+1) – (avg. for X 1 X 2 =-1) ◦ = (y 1 +y 4 )/2 - (y 2 +y 3 )/2 = “effect of X1*X2 interaction” Can show interaction as a column to help calculations ◦ (Note: This simple algebra works only for 2-level factors, with levels set to ±1, but the underlying ideas also apply to other designs.) DOE mini-course, part 1, Paul Funkenbusch, 2015 12
Data on the removal Leve vel rate of osteotomy drills Factor -1 +1 is collected as a function of the load X1. applied load (kg) 2 3 applied and the number of previous X2. previous cuts 0 20 cuts made. (new) Find the effects of the two factors and the TC TC X1 X1 X2 X2 X1*X2 *X2 Remov oval al rate interaction. Which (mm3/s) /s) is/are most important? 1 -1 -1 +1 3 Build a simple model to 2 -1 +1 -1 2 predict the removal rate. 3 +1 -1 -1 5 4 +1 +1 +1 2 DOE mini-course, part 1, Paul Funkenbusch, 2015 13
TC TC X1 X1 X2 X2 X1*X2 *X2 Remov oval al rate (mm3/s) /s) 1 -1 -1 +1 3 2 -1 +1 -1 2 3 +1 -1 -1 5 4 +1 +1 +1 2 Effects ◦ m* = ? ◦ D X1 = ? ◦ D x2 =? ◦ D x1x2 = ? DOE mini-course, part 1, Paul Funkenbusch, 2015 14
TC TC X1 X1 X2 X2 X1*X2 *X2 Remov oval al rate (mm3/s) /s) Number of previous cuts (i.e. X2) is the 1 -1 -1 +1 3 more important factor 2 -1 +1 -1 2 (given the range of 3 +1 -1 -1 5 values tested). 4 +1 +1 +1 2 Effects ◦ m* = (y 1 +y 2 +y 3 +y 4 )/4 =(3+2+5+2)/4 = 3 ◦ D X1 = (y 3 +y 4 )/2 - (y 1 +y 2 )/2 = (5+2)/2 – (3+2)/2 = 1 ◦ D x2 = (y 2 +y 4 )/2 - (y 1 +y 3 )/2 = (2+2)/2 – (3+5)/2 = -2 ◦ D x1x2 = (y 1 +y 4 )/2 - (y 2 +y 3 )/2 = (3+2)/2 – (2+5)/2 = -1 DOE mini-course, part 1, Paul Funkenbusch, 2015 15
Use the normalized (±1) scale for the levels. Effects y ◦ m* = 3 ◦ D X1 = 1 y m* ◦ D x2 = -2 D a 0 2 ◦ D x1x2 = -1 0 -1 1 -1 1 X1 X1 X1 X1 Then the midpoint of the design corresponds to the intercept (0) for both factors a 0 = m* = 3.0 And the slope (a) for each effect is given by D /2 ◦ a 1 = D X1 /2 = 0.5; a 2 = D X2 /2 = -1.0; a 12 = D X12 /2 = -0.5 y pred = 3.0 + (0.5)X 1 - (1.0)X 2 - (0.5)X 1 X 2 (y in units of mm 3 /s) DOE mini-course, part 1, Paul Funkenbusch, 2015 16
L = load (kg) Leve vel C = # of cuts Factor -1 +1 X1. applied load (kg) 2 3 X1 = -5 + 2L X2. previous cuts 0 20 X2 = -1 + 0.1C y pred = 3.0 + (0.5)X 1 - (1.0)X 2 - (0.5)X 1 X 2 (y in mm 3 /s, for L in kg) DOE mini-course, part 1, Paul Funkenbusch, 2015 17
4 data points, 4 constants ◦ will hit all of the original data exactly New trials under same conditions? ◦ How good was the original data? (error) Other conditions (interpolation, extrapolation)? ◦ How good is the linear fit to the real dependence? DOE mini-course, part 1, Paul Funkenbusch, 2015 18
require more TC to include all combinations ◦ # of TC = 2 n , where n= # of factors model includes more terms # of interaction terms increases rapidly includes “higher - order” interactions ◦ “order” = number of factors in the interaction term ◦ X 1 X 2 second order; X 1 X 2 X 3 third order, etc. DOE mini-course, part 1, Paul Funkenbusch, 2015 19
TC TC X1 X1 X2 X2 X3 X3 y 2 3 = 8 1 -1 -1 -1 y 1 8 responses (8 y’s) 2 -1 -1 +1 y 2 3 -1 +1 -1 y 3 m* , D x1 , D x2 , D x3 , 4 -1 +1 +1 y 4 D x1x2 , D x1x3 , D x2x3 , 5 +1 -1 -1 y 5 D x1x2x3 6 +1 -1 +1 y 6 7 +1 +1 -1 y 7 8 +1 +1 +1 y 8 y pred = a o +a 1 X 1 +a 2 X 2 + a 2 X 3 8 DOF + a 12 X 1 X 2 +a 13 X 1 X 3 +a 23 X 2 X 3 ◦ m* 1 DOF +a 123 X 1 X 2 X 3 ◦ factors 3 DOF ◦ 2-factor inter. 3 DOF ◦ 3-factor inter. 1 DOF DOE mini-course, part 1, Paul Funkenbusch, 2015 20
4 factors at 2 levels 5 factors at 2 levels 2 4 = 16 2 5 = 32 16 DOF 32 DOF ◦ m* 1 DOF ◦ m* 1 DOF ◦ factors 4 DOF ◦ factors 5 DOF ◦ 2-factor inter. 6 DOF ◦ 2-factor inter. 10 DOF ◦ 3-factor inter. 4 DOF ◦ 3-factor inter. 10 DOF ◦ 4-factor inter. 1 DOF ◦ 4-factor inter. 5 DOF ◦ 5-factor inter. 1 DOF More factors much more effort to measure interactions DOE mini-course, part 1, Paul Funkenbusch, 2015 21
System is usually dominated by the effects of factors and lower-order interactions. Usually don’t really need 3 -factor, 4-factor, etc. interactions ~ 0. But have just seen, that we may spend a lot of effort (i.e. lots of DOF) on them. Can we make better use of these DOF? ◦ will discuss in Part II. DOE mini-course, part 1, Paul Funkenbusch, 2015 22
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