Difference between revisions of "Worksheets/Week6"

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Case study: Achieve a stability value of 50 days or more, for a new product. We had a full factorial set of experiments in 3 factors:
Case study: Achieve a stability value of 50 days or more, for a new product. We had a full factorial set of experiments in 3 factors:
* B:
* C: mixer type:  -1 = R mixer; +1 = W mixer


{| class="wikitable"
{| class="wikitable"
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# The response: stability [units=days]
# The response: stability [units=days]
y = ...
y = ...
model_stability_poshalf
 
model_stability_poshalf = lm(y ~ A*B*C)
model_stability_poshalf = lm(y ~ A*B*C)
summary(model_stability_poshalf)
summary(model_stability_poshalf)
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Data from a bioreactor experiment is available, were we were investigating four factors:
Data from a bioreactor experiment is available, were we were investigating four factors:
* '''A''' = feed rate: 5 g/min or 8 g/min
* '''B''' = initial inoculate amount: 300 g or 400 g
* '''C''' = feed substrate concentration: 40 g/L or 60 g/L
* '''D''' = dissolved oxygen set-point: 4 mg/L or 5 mg/L


The 16 experiments from a full factorial, 24, were randomly run, and the yields, y, the outcome variable were given in standard order: <tt>[60, 59, 63, 61, 69, 61, 94, 93, 56, 63, 70, 65, 44, 45, 78, 77]</tt>
{| class="wikitable"
! Factor name
! Description
! Low value
! High value
! Type of factor
|-
| '''A'''
| Feed rate
| 5 g/min
| 8 g/min
| Numeric factor
|-
| '''B'''
| Initial inoculant amount
| 300 g
| 400 g
| Numeric
|-
| '''C'''
| Feed substrate concentration
| 40 g/L
| 60 g/L
| Numeric
|-
| '''D'''
| Dissolved oxygen set-point
| 4 mg/L
| 5 mg/L
| Numeric
|}
 
See the [https://yint.org/w6 link full case study], and to help you, there is a [https://yint.org/w6table table you can fill out].
 


<html><div data-datacamp-exercise data-lang="r" data-height="auto">
The 16 experiments from a full factorial, 2^4 = 16, were randomly run, and the yields, y, the outcome variable were given in standard order: <tt>[60, 59, 63, 61, 69, 61, 94, 93, 56, 63, 70, 65, 44, 45, 78, 77]</tt>
 
<html><div data-datacamp-exercise data-lang="r" data-height="800px">
     <code data-type="sample-code">
     <code data-type="sample-code">


base <- c(-1, +1)
base = c(-1, +1)
design <- expand.grid(A=base, B=base, C=base)
design = expand.grid(A=base, B=base, C=base)
A <- design$A
A = design$A
B <- design$B
B = design$B
C <- design$C
C = design$C
D <- A * B * C
D = A * B * C


# Confirm that you can find these 8 runs yourself:
# These are correct, but confirm that you can find these 8 runs yourself:
y <- c(60, 63, 70, 61, 44, 61, 94, 77)
y = c(60, 63, 70, 61, 44, 61, 94, 77)


model.bio <- lm(y ~ A*B*C*D)
model_bio = lm(y ~ A*B*C*D)
summary(model.bio)
summary(model_bio)


# Uncomment this line if you run the code in RStudio
# Uncomment this line if you run the code in RStudio
Line 102: Line 129:
source('https://yint.org/paretoPlot.R')
source('https://yint.org/paretoPlot.R')
source('https://yint.org/contourPlot.R')
source('https://yint.org/contourPlot.R')
paretoPlot(model.bio)
paretoPlot(model_bio)


     </code>
     </code>
</div></html>
</div></html>

Latest revision as of 09:40, 17 October 2019

Part 1

Case study: Achieve a stability value of 50 days or more, for a new product. We had a full factorial set of experiments in 3 factors:

Factor name Description Low value High value Type of factor
A Enzyme strength 20% 30% Numeric factor
B Feed concentration 75% 85% Numeric
C Mixer type R W Categorical

We will show what we loose out if we pretend we only did half the experiments. In other words, we actually have 8 experiments, but we will see what happens if we only use 4 of them.


# This is the half-fraction, when C = A*B A = c(-1, +1, -1, +1) B = c(-1, -1, +1, +1) C = A * B # The response: stability [units=days] y = ... model_stability_poshalf = lm(y ~ A*B*C) summary(model_stability_poshalf) # Uncomment this line if you run the code in RStudio #library(pid) # Comment these 2 lines if you run this code in RStudio source('https://yint.org/paretoPlot.R') source('https://yint.org/contourPlot.R') paretoPlot(model_stability_poshalf) # This is the other half-fraction, when C = -A*B A = c(-1, +1, -1, +1) B = c(-1, -1, +1, +1) C = -1 * A * B # The response: stability [units=days] y = ... model_stability_neghalf = lm(y ~ A*B*C) summary(model_stability_neghalf)

Part 2

Data from a bioreactor experiment is available, were we were investigating four factors:

Factor name Description Low value High value Type of factor
A Feed rate 5 g/min 8 g/min Numeric factor
B Initial inoculant amount 300 g 400 g Numeric
C Feed substrate concentration 40 g/L 60 g/L Numeric
D Dissolved oxygen set-point 4 mg/L 5 mg/L Numeric

See the link full case study, and to help you, there is a table you can fill out.


The 16 experiments from a full factorial, 2^4 = 16, were randomly run, and the yields, y, the outcome variable were given in standard order: [60, 59, 63, 61, 69, 61, 94, 93, 56, 63, 70, 65, 44, 45, 78, 77]

base = c(-1, +1) design = expand.grid(A=base, B=base, C=base) A = design$A B = design$B C = design$C D = A * B * C # These are correct, but confirm that you can find these 8 runs yourself: y = c(60, 63, 70, 61, 44, 61, 94, 77) model_bio = lm(y ~ A*B*C*D) summary(model_bio) # Uncomment this line if you run the code in RStudio #library(pid) # Comment these 2 lines if you run this code in RStudio source('https://yint.org/paretoPlot.R') source('https://yint.org/contourPlot.R') paretoPlot(model_bio)