Homework 7
Victor Torres
2024-10-31
Instructions
In Kuhn and Johnson do problems 6.2 and 6.3. There are only two but they consist of many parts. Please submit a link to your Rpubs and submit the .rmd file as well.
library(tidyverse)
library(caret)
library(AppliedPredictiveModeling)
library(corrplot)Problems
6.2
Developing a model to predict permeability (see Sect. 1.4) could save significant resources for a pharmaceutical company, while at the same time more rapidly identifying molecules that have a sufficient permeability to become a drug
Sections
a
Start R and use these commands to load the data
data("permeability")The matrix fingerprints contains the 1,107 binary molecular predictors for the 165 compounds, while permeability contains permeability response.
b
The fingerprint predictors indicate the presence or absence of substructures of a molecule and are often sparse meaning that relatively few of the molecules contain each substructure. Filter out the predictors that have low frequencies using the nearZeroVar function from the caret package. How many predictors are left for modeling?
dim(fingerprints)## [1] 165 1107fingerprints <- fingerprints[, -nearZeroVar(fingerprints)]
dim(fingerprints)## [1] 165 388Fingerprints matrix has 1107 predictors, after applying the -nearZeroVar function I got 388 predictors.
c
Split the data into a training and a test set, pre-process the data, and tune a PLS model. How many latent variables are optimal and what is the corresponding resampled estimate of R2?
set.seed(624)
# index for training
index <- createDataPartition(permeability, p = .8, list = FALSE)
# train
train_perm <- permeability[index, ]
train_fp <- fingerprints[index, ]
# test
test_perm <- permeability[-index, ]
test_fp <- fingerprints [-index, ]
# 10-fold cross-validation to make reasonable estimates
ctrl <- trainControl(method = "cv", number = 10)
plsTune <- train(train_fp, train_perm, method = "pls", metric = "Rsquared",
tuneLength = 20, trControl = ctrl, preProc = c("center", "scale"))
plot(plsTune) 
plsTune## Partial Least Squares
##
## 133 samples
## 388 predictors
##
## Pre-processing: centered (388), scaled (388)
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 118, 119, 120, 120, 121, 120, ...
## Resampling results across tuning parameters:
##
## ncomp RMSE Rsquared MAE
## 1 13.25656 0.2953172 10.202424
## 2 11.90191 0.4662745 8.710959
## 3 12.05579 0.4624570 9.271134
## 4 12.03297 0.4793759 9.314195
## 5 12.06645 0.4870447 8.986610
## 6 11.82964 0.5012972 8.788270
## 7 11.91363 0.5011672 9.153386
## 8 11.79990 0.4960881 9.119966
## 9 11.81946 0.4959475 9.301787
## 10 11.85288 0.4924486 9.176136
## 11 11.79654 0.5025443 9.128199
## 12 11.62869 0.5131115 8.965070
## 13 11.78348 0.5080595 8.920097
## 14 12.01377 0.4935108 9.101865
## 15 12.09297 0.4862359 9.131109
## 16 12.10087 0.4953868 9.053161
## 17 12.38093 0.4847366 9.218277
## 18 12.59348 0.4768971 9.402569
## 19 12.61895 0.4807222 9.338592
## 20 12.77045 0.4682401 9.549745
##
## Rsquared was used to select the optimal model using the largest value.
## The final value used for the model was ncomp = 12.We can see in the table above that the optimal tuning has 12 components. In the graph we can see that the highest R2 value is 12.5
d
Predict the response for the test set. What is the test set estimate of R2?
fp_predict <- predict(plsTune, test_fp)
postResample(fp_predict, test_perm)## RMSE Rsquared MAE
## 11.4895371 0.4741832 9.3113125The test set estimate of R2 is 0.4741832.
e
Try building other models discussed in this chapter. Do any have better predictive performance?
Models
ENET
set.seed(247)
# grid of penalties
enetGrid <- expand.grid(.lambda = c(0, 0.01, .1), .fraction = seq(.05, 1, length = 20))
# tuning penalized regression model
enetTune <- train(train_fp, train_perm, method = "enet",
tuneGrid = enetGrid, trControl = ctrl, preProc = c("center", "scale"))## Warning: model fit failed for Fold02: lambda=0.00, fraction=1 Error in if (zmin < gamhat) { : missing value where TRUE/FALSE needed## Warning in nominalTrainWorkflow(x = x, y = y, wts = weights, info = trainInfo,
## : There were missing values in resampled performance measures.plot(enetTune)
enet_predict <- predict(enetTune, test_fp)
postResample(enet_predict, test_perm)## RMSE Rsquared MAE
## 11.2682652 0.4652716 9.2858416The R2 is 0.5307087 and the RMSE is lower using a penalized Elastic Net regression model.
LARS
set.seed(624)
larsTune <- train(train_fp, train_perm, method = "lars", metric = "Rsquared",
tuneLength = 20, trControl = ctrl, preProc = c("center", "scale"))
plot(larsTune)
lars_predict <- predict(larsTune, test_fp)
postResample(lars_predict, test_perm)## RMSE Rsquared MAE
## 11.6906258 0.4342625 9.6425013The Least Angle Regression is slightly worse than the PLS method as the R2 is lower and the RMSE is higher
f
Would you recommend any of your models to replace the permeability laboratory experiment?
I would recommend the Elastic Net regression model as it produced better statistics. It had a higher R2 and lower RMSE and MAE.
6.3
A chemical manufacturing process for a pharmaceutical product was discussed in Sect. 1.4. In this problem, the objective is to understand the relationship between biological measurements of the raw materials (predictors), measurements of the manufacturing process (predictors), and the response of product yield. Biological predictors cannot be changed but can be used to assess the quality of the raw material before processing. On the other hand, manufacturing process predictors can be changed in the manufacturing process. Improving product yield by 1% will boost revenue by approximately one hundred thousand dollars per batch:
Sections
a
Start R and use these commands to load the data
data(ChemicalManufacturingProcess)The matrix processPredictors contains the 57 predictors (12 describing the input biological material and 45 describing the process predictors) for the 176 manufacturing runs. yield contains the percent yield for each run.
b
A small percentage of cells in the predictor set contain missing values. Use an imputation function to fill in these missing values (e.g., see Sect. 3.8)
sum(is.na(ChemicalManufacturingProcess))## [1] 106miss <- preProcess(ChemicalManufacturingProcess, method = "bagImpute")
Chemical <- predict(miss, ChemicalManufacturingProcess)
sum(is.na(Chemical))## [1] 0There were 106 missing values in ChemicalManufacturingProcess. Bagged trees were used to impute the data. Bagged trees are made using all the other variables.
c
Split the data into a training and a test set, pre-process the data, and tune a model of your choice from this chapter. What is the optimal value of the performance metric?
# filtering low frequencies
Chemical <- Chemical[, -nearZeroVar(Chemical)]set.seed(279)
# index for training
index <- createDataPartition(Chemical$Yield, p = .8, list = FALSE)
# train
train_chem <- Chemical[index, ]
# test
test_chem <- Chemical[-index, ]Models
PLS
set.seed(624)
plsTune <- train(Yield ~ ., Chemical , method = "pls",
tuneLength = 20, trControl = ctrl, preProc = c("center", "scale"))
plot(plsTune) 
plsTune## Partial Least Squares
##
## 176 samples
## 56 predictor
##
## Pre-processing: centered (56), scaled (56)
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 160, 157, 158, 159, 158, 159, ...
## Resampling results across tuning parameters:
##
## ncomp RMSE Rsquared MAE
## 1 1.436891 0.4568805 1.147590
## 2 1.872742 0.4711564 1.185897
## 3 1.292614 0.5633698 1.020010
## 4 1.480526 0.5381868 1.085319
## 5 1.707358 0.5156812 1.131007
## 6 1.821904 0.4903840 1.156300
## 7 2.006142 0.4802835 1.211850
## 8 2.092370 0.4622998 1.253598
## 9 2.220647 0.4485854 1.290999
## 10 2.322021 0.4410360 1.315081
## 11 2.446697 0.4264842 1.352393
## 12 2.475260 0.4206118 1.367989
## 13 2.464162 0.4197124 1.377783
## 14 2.418661 0.4227280 1.364288
## 15 2.375812 0.4242080 1.350175
## 16 2.368363 0.4267259 1.337946
## 17 2.386174 0.4254577 1.339398
## 18 2.376321 0.4271412 1.334877
## 19 2.400843 0.4255697 1.348122
## 20 2.422785 0.4231162 1.359970
##
## RMSE was used to select the optimal model using the smallest value.
## The final value used for the model was ncomp = 3.Optimal tuning has 3 components with R2 of 0.5623262.
ENET
set.seed(624)
enetTune <- train(Yield ~ ., Chemical , method = "enet",
tuneGrid = enetGrid, trControl = ctrl, preProc = c("center", "scale"))
plot(enetTune)
enetTune## Elasticnet
##
## 176 samples
## 56 predictor
##
## Pre-processing: centered (56), scaled (56)
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 160, 157, 158, 159, 158, 159, ...
## Resampling results across tuning parameters:
##
## lambda fraction RMSE Rsquared MAE
## 0.00 0.05 1.263954 0.6225860 1.0294423
## 0.00 0.10 1.171467 0.6195893 0.9388695
## 0.00 0.15 1.285779 0.6043223 0.9537550
## 0.00 0.20 1.455526 0.5547255 1.0089814
## 0.00 0.25 1.743737 0.4948981 1.1067214
## 0.00 0.30 1.939587 0.4680604 1.1795594
## 0.00 0.35 1.908244 0.4619370 1.1924860
## 0.00 0.40 1.850254 0.4606806 1.1877837
## 0.00 0.45 1.794856 0.4594057 1.1771407
## 0.00 0.50 2.101187 0.4481459 1.2573220
## 0.00 0.55 2.407435 0.4725106 1.3304398
## 0.00 0.60 2.744013 0.5217990 1.4001566
## 0.00 0.65 3.225543 0.4863491 1.5339631
## 0.00 0.70 3.771484 0.4747361 1.6693508
## 0.00 0.75 4.094773 0.4681231 1.7557563
## 0.00 0.80 4.135408 0.4589587 1.7764653
## 0.00 0.85 4.225706 0.4419317 1.8076852
## 0.00 0.90 4.318549 0.4257147 1.8358006
## 0.00 0.95 4.415063 0.4120784 1.8636370
## 0.00 1.00 4.501021 0.4019904 1.8870890
## 0.01 0.05 1.536442 0.5973476 1.2453271
## 0.01 0.10 1.309336 0.6258493 1.0618555
## 0.01 0.15 1.201950 0.6195671 0.9778981
## 0.01 0.20 1.176854 0.6175939 0.9523896
## 0.01 0.25 1.161852 0.6236987 0.9319051
## 0.01 0.30 1.175811 0.6198597 0.9301295
## 0.01 0.35 1.246551 0.6032199 0.9540424
## 0.01 0.40 1.296291 0.5952774 0.9647967
## 0.01 0.45 1.362599 0.5626257 0.9909640
## 0.01 0.50 1.460207 0.5489330 1.0197566
## 0.01 0.55 1.600805 0.5172511 1.0731396
## 0.01 0.60 1.793769 0.4868596 1.1322410
## 0.01 0.65 1.885961 0.4712944 1.1672050
## 0.01 0.70 1.953361 0.4609687 1.1927301
## 0.01 0.75 2.016199 0.4534714 1.2156241
## 0.01 0.80 2.076369 0.4480579 1.2356802
## 0.01 0.85 2.143715 0.4433120 1.2568727
## 0.01 0.90 2.179751 0.4401770 1.2700016
## 0.01 0.95 2.129338 0.4398117 1.2610389
## 0.01 1.00 2.043851 0.4421279 1.2426724
## 0.10 0.05 1.649315 0.5363459 1.3354459
## 0.10 0.10 1.487591 0.6095911 1.2076167
## 0.10 0.15 1.354079 0.6236486 1.0996757
## 0.10 0.20 1.258077 0.6219261 1.0220867
## 0.10 0.25 1.200809 0.6201101 0.9778990
## 0.10 0.30 1.181463 0.6179041 0.9615112
## 0.10 0.35 1.171335 0.6193380 0.9468400
## 0.10 0.40 1.165095 0.6229576 0.9424421
## 0.10 0.45 1.167185 0.6235463 0.9390511
## 0.10 0.50 1.212130 0.6118562 0.9527442
## 0.10 0.55 1.310136 0.6005375 0.9787766
## 0.10 0.60 1.383384 0.5963431 0.9968869
## 0.10 0.65 1.442303 0.5805640 1.0192090
## 0.10 0.70 1.511715 0.5573960 1.0504530
## 0.10 0.75 1.598893 0.5315716 1.0836263
## 0.10 0.80 1.707558 0.5079105 1.1181461
## 0.10 0.85 1.772034 0.4960139 1.1387951
## 0.10 0.90 1.812766 0.4895472 1.1528253
## 0.10 0.95 1.846664 0.4841682 1.1650245
## 0.10 1.00 1.873437 0.4803403 1.1741747
##
## RMSE was used to select the optimal model using the smallest value.
## The final values used for the model were fraction = 0.25 and lambda = 0.01.The optimal model has a fraction of 0.1 and λ of 0. The R2 is 0.6253182.
Lasso Model
lm_model <- lm(Yield ~ ., Chemical)
summary(lm_model)##
## Call:
## lm(formula = Yield ~ ., data = Chemical)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2.17844 -0.53656 -0.02842 0.50526 2.00415
##
## Coefficients: (1 not defined because of singularities)
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 4.234e+00 8.608e+01 0.049 0.96085
## BiologicalMaterial01 2.483e-01 3.342e-01 0.743 0.45900
## BiologicalMaterial02 -1.120e-01 1.281e-01 -0.874 0.38375
## BiologicalMaterial03 1.636e-01 2.354e-01 0.695 0.48843
## BiologicalMaterial04 -1.044e-01 5.235e-01 -0.199 0.84233
## BiologicalMaterial05 1.513e-01 1.061e-01 1.426 0.15641
## BiologicalMaterial06 3.336e-03 3.014e-01 0.011 0.99119
## BiologicalMaterial08 3.808e-01 6.358e-01 0.599 0.55034
## BiologicalMaterial09 -8.180e-01 1.370e+00 -0.597 0.55162
## BiologicalMaterial10 7.954e-02 1.367e+00 0.058 0.95370
## BiologicalMaterial11 -8.954e-02 8.230e-02 -1.088 0.27874
## BiologicalMaterial12 3.493e-01 6.346e-01 0.551 0.58300
## ManufacturingProcess01 6.695e-02 9.596e-02 0.698 0.48672
## ManufacturingProcess02 1.343e-02 4.311e-02 0.311 0.75601
## ManufacturingProcess03 -3.377e+00 5.103e+00 -0.662 0.50934
## ManufacturingProcess04 6.282e-02 2.940e-02 2.137 0.03464 *
## ManufacturingProcess05 7.326e-04 3.859e-03 0.190 0.84974
## ManufacturingProcess06 3.261e-02 4.341e-02 0.751 0.45401
## ManufacturingProcess07 -1.810e-01 2.126e-01 -0.851 0.39623
## ManufacturingProcess08 -6.282e-02 2.522e-01 -0.249 0.80374
## ManufacturingProcess09 2.614e-01 1.812e-01 1.443 0.15176
## ManufacturingProcess10 -1.166e-01 5.742e-01 -0.203 0.83950
## ManufacturingProcess11 1.942e-01 7.132e-01 0.272 0.78590
## ManufacturingProcess12 3.761e-05 1.013e-04 0.371 0.71120
## ManufacturingProcess13 -2.670e-01 3.843e-01 -0.695 0.48859
## ManufacturingProcess14 3.058e-04 1.115e-02 0.027 0.97816
## ManufacturingProcess15 1.972e-03 8.903e-03 0.222 0.82506
## ManufacturingProcess16 -4.937e-05 3.190e-04 -0.155 0.87728
## ManufacturingProcess17 -1.402e-01 3.011e-01 -0.466 0.64240
## ManufacturingProcess18 4.245e-03 4.450e-03 0.954 0.34211
## ManufacturingProcess19 -2.233e-03 7.301e-03 -0.306 0.76021
## ManufacturingProcess20 -4.517e-03 4.721e-03 -0.957 0.34062
## ManufacturingProcess21 NA NA NA NA
## ManufacturingProcess22 -1.666e-02 4.209e-02 -0.396 0.69299
## ManufacturingProcess23 -4.181e-02 8.289e-02 -0.504 0.61495
## ManufacturingProcess24 -1.931e-02 2.340e-02 -0.825 0.41100
## ManufacturingProcess25 -6.493e-03 1.365e-02 -0.476 0.63506
## ManufacturingProcess26 6.101e-03 1.041e-02 0.586 0.55909
## ManufacturingProcess27 -7.061e-03 7.781e-03 -0.907 0.36601
## ManufacturingProcess28 -7.882e-02 3.094e-02 -2.547 0.01212 *
## ManufacturingProcess29 1.393e+00 8.961e-01 1.555 0.12261
## ManufacturingProcess30 -3.693e-01 6.233e-01 -0.592 0.55463
## ManufacturingProcess31 4.783e-02 1.203e-01 0.398 0.69168
## ManufacturingProcess32 3.333e-01 6.833e-02 4.877 3.34e-06 ***
## ManufacturingProcess33 -4.068e-01 1.286e-01 -3.164 0.00197 **
## ManufacturingProcess34 -1.496e+00 2.753e+00 -0.543 0.58792
## ManufacturingProcess35 -1.879e-02 1.765e-02 -1.064 0.28926
## ManufacturingProcess36 2.833e+02 3.132e+02 0.904 0.36765
## ManufacturingProcess37 -6.935e-01 2.889e-01 -2.401 0.01789 *
## ManufacturingProcess38 -1.900e-01 2.417e-01 -0.786 0.43333
## ManufacturingProcess39 7.077e-02 1.307e-01 0.542 0.58907
## ManufacturingProcess40 4.605e-01 6.545e+00 0.070 0.94403
## ManufacturingProcess41 2.549e-01 4.736e+00 0.054 0.95716
## ManufacturingProcess42 4.372e-02 2.102e-01 0.208 0.83557
## ManufacturingProcess43 2.268e-01 1.182e-01 1.919 0.05741 .
## ManufacturingProcess44 -4.385e-01 1.186e+00 -0.370 0.71222
## ManufacturingProcess45 9.547e-01 5.444e-01 1.754 0.08204 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.039 on 120 degrees of freedom
## Multiple R-squared: 0.7826, Adjusted R-squared: 0.683
## F-statistic: 7.854 on 55 and 120 DF, p-value: < 2.2e-16The ordinary linear regression model has a Multiple R2 of 0.7813 and an Adjusted R2 of 0.6811.
RIDGE
set.seed(624)
## Define the candidate set of values
ridgeGrid <- data.frame(.lambda = seq(0, .1, length = 15))
ridgeTune <- train(Yield ~ ., Chemical , method = "ridge",
tuneGrid = ridgeGrid, trControl = ctrl, preProc = c("center", "scale"))
plot(ridgeTune)
ridgeTune## Ridge Regression
##
## 176 samples
## 56 predictor
##
## Pre-processing: centered (56), scaled (56)
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 160, 157, 158, 159, 158, 159, ...
## Resampling results across tuning parameters:
##
## lambda RMSE Rsquared MAE
## 0.000000000 4.501021 0.4019904 1.887089
## 0.007142857 1.951471 0.4493837 1.224896
## 0.014285714 2.093121 0.4429601 1.247997
## 0.021428571 2.098017 0.4475006 1.242244
## 0.028571429 2.077066 0.4519785 1.232565
## 0.035714286 2.050861 0.4560167 1.222873
## 0.042857143 2.024673 0.4596597 1.213933
## 0.050000000 1.999988 0.4629748 1.206382
## 0.057142857 1.977156 0.4660169 1.200121
## 0.064285714 1.956157 0.4688278 1.194388
## 0.071428571 1.936855 0.4714397 1.189526
## 0.078571429 1.919085 0.4738781 1.185097
## 0.085714286 1.902686 0.4761635 1.180984
## 0.092857143 1.887513 0.4783129 1.177421
## 0.100000000 1.873437 0.4803403 1.174175
##
## RMSE was used to select the optimal model using the smallest value.
## The final value used for the model was lambda = 0.1.The optimal model has λ of 0.1 and R2 of 0.4785667.
LARS
set.seed(624)
larsTune <- train(Yield ~ ., Chemical , method = "lars", metric = "Rsquared",
tuneLength = 20, trControl = ctrl, preProc = c("center", "scale"))
plot(larsTune)
larsTune## Least Angle Regression
##
## 176 samples
## 56 predictor
##
## Pre-processing: centered (56), scaled (56)
## Resampling: Cross-Validated (10 fold)
## Summary of sample sizes: 160, 157, 158, 159, 158, 159, ...
## Resampling results across tuning parameters:
##
## fraction RMSE Rsquared MAE
## 0.05 1.268059 0.6252093 1.037325
## 0.10 1.157751 0.6229682 0.936117
## 0.15 1.163964 0.6233226 0.922333
## 0.20 1.425304 0.5567018 1.005712
## 0.25 1.717206 0.4959408 1.101890
## 0.30 1.936760 0.4676172 1.176116
## 0.35 1.922115 0.4611422 1.188942
## 0.40 1.875680 0.4585226 1.187174
## 0.45 1.838067 0.4577825 1.183074
## 0.50 1.731858 0.4637976 1.161242
## 0.55 1.508554 0.4924491 1.104212
## 0.60 1.286634 0.5676612 1.027343
## 0.65 1.637094 0.4985405 1.138608
## 0.70 2.069278 0.4782574 1.247884
## 0.75 2.478866 0.4698927 1.355965
## 0.80 2.854003 0.4595411 1.460073
## 0.85 3.231161 0.4425963 1.561937
## 0.90 3.619450 0.4262587 1.662857
## 0.95 4.056912 0.4124876 1.774808
## 1.00 4.501021 0.4019904 1.887089
##
## Rsquared was used to select the optimal model using the largest value.
## The final value used for the model was fraction = 0.05.The optimal model has a fraction of 0.05 and R2 of 0.6256310.
d
Predict the response for the test set.What is the value of the performance metric and how does this compare with the resampled performance metric on the training set?
lars_predict <- predict(larsTune, test_chem[ ,-1])
postResample(lars_predict, test_chem[ ,1])## RMSE Rsquared MAE
## 1.192691 0.662927 1.012612Ordinary linear model had the highest R2 , but it comes with consequences. Therefore, the lars method was chosen as it had the highest R2. The R2 is 0.7170636, which is higher than the training set.
e
Which predictors are most important in the model you have trained? Do either the biological or process predictors dominate the list?
varImp(larsTune)## loess r-squared variable importance
##
## only 20 most important variables shown (out of 56)
##
## Overall
## ManufacturingProcess32 100.00
## ManufacturingProcess13 90.02
## BiologicalMaterial06 84.56
## ManufacturingProcess36 76.03
## ManufacturingProcess17 74.88
## BiologicalMaterial03 73.53
## ManufacturingProcess09 70.37
## BiologicalMaterial12 67.98
## BiologicalMaterial02 65.33
## ManufacturingProcess31 60.38
## ManufacturingProcess06 58.03
## ManufacturingProcess33 49.39
## BiologicalMaterial11 48.11
## BiologicalMaterial04 47.13
## ManufacturingProcess11 42.47
## BiologicalMaterial08 41.88
## BiologicalMaterial01 39.14
## ManufacturingProcess12 33.02
## ManufacturingProcess30 32.91
## BiologicalMaterial09 32.41The 5 most important variables used in the modeling are ManufacturingProcess32, ManufacturingProcess13, BiologicalMaterial06, ManufacturingProcess36, and ManufacturingProcess17. Process predictors dominate the list. The ratio of process to biological predictors is 11:9.
f
Explore the relationships between each of the top predictors and the response. How could this information be helpful in improving yield in future runs of the manufacturing process?
top10 <- varImp(larsTune)$importance %>%
arrange(-Overall) %>%
head(10)
Chemical %>%
select(c("Yield", row.names(top10))) %>%
cor() %>%
corrplot()
According to the correlation plot, ManufacturingProcess32 has the highest positive correlation with Yield. Three of the top ten variables are negatively correlated with Yield. This information can be helfup in the future runs of the manufacturing process, as these were the predictors that affected the yield. If they want to maximize or improve their yield, they way want to improve their measurements of the manufacturing process and biological measurements of the raw materials.