Business Analytics Lab Worksheet 07
CME Group Foundation Business Analytics Lab
Iain Buehler
November 1st, 2017
About
In this lab, we will focus on linear programming and non-linear regression modeling. Linear regression modeling, as discussed in the previous lab, works with simple and multiple linear regression models. Sometimes the relationships are not best represented by linear models, and a non-linear regression modeling is required. The general concept remains the same; minimizing the error between the observed/actual values and the values predicted by the model. Linear programming on the other hands seek to find the optimal solution to a problem with multivariables and multiconstraints described by linear relationships.
In this lab, we will perform a non-linear regression modeling on the cost of servers, and setup a linear programming model to solve the marketing use case discussed in class.
Setup
Remember to always set your working directory to the source file location. Go to ‘Session’, scroll down to ‘Set Working Directory’, and click ‘To Source File Location’. Read carefully the below and follow the instructions to complete the tasks and answer any questions. Submit your work to RPubs as detailed in previous notes.
Note
For your assignment you may be using different data sets than what is included here. Always read carefully the instructions on Sakai. Tasks/questions to be completed/answered are highlighted in larger bolded fonts and numbered according to their particular placement in the task section.
Task 1: Non-Linear Regression Modeling
First, we must read the file ‘ServersCost.csv’ into R, and extract the two columns of interest.
mydata <- read.csv("data/serverscost.csv", header=TRUE)
head(mydata)## servers cost
## 1 1 27654
## 2 2 24789
## 3 3 21890
## 4 4 21633
## 5 5 15843
## 6 6 12567servers = mydata$servers
servers## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20cost = mydata$cost
cost## [1] 27654 24789 21890 21633 15843 12567 8943 6789 4533 4678 5321
## [12] 5765 5432 9995 13522 17563 22732 22643 24621 28111We start by creating a simple linear regression model. Next, we plot the points to visually inspect the data and unravel any potential relationships.
linear_model = lm(cost ~ servers)
summary(linear_model)##
## Call:
## lm(formula = cost ~ servers)
##
## Residuals:
## Min 1Q Median 3Q Max
## -10646.2 -8646.2 -544.7 7066.0 12858.8
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 14747.2 4035.5 3.654 0.00181 **
## servers 48.0 336.9 0.142 0.88828
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 8687 on 18 degrees of freedom
## Multiple R-squared: 0.001127, Adjusted R-squared: -0.05437
## F-statistic: 0.0203 on 1 and 18 DF, p-value: 0.8883# add linear line based on regression model
plot(servers,cost, pch=16) # the pch option is to accentuate the points
abline(linear_model, col="blue", lwd=2)
The blue line here represents the model based predicted data, and the black dots are the actual data points. Clearly the line model is far from being a good predictor. Next we will use a nonlinear quadratic model to see how the model can be improved.
A linear model is of the form y ~ x0 + x1 +….xn where y is the dependent variable and the xn are the independent variables. For the non-linear quadratic regression model the equation takes the form y ~ x + x^2
# First it is best to define a new variable which is the squared value of servers
servers2 = servers^2
# The model formula is based on the form y = x + x^2
quad_model = lm(cost ~ servers + servers2)
summary(quad_model)##
## Call:
## lm(formula = cost ~ servers + servers2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2897.8 -1553.4 -513.2 1152.4 4752.7
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 35417.77 1742.64 20.32 2.30e-13 ***
## servers -5589.43 382.19 -14.62 4.62e-11 ***
## servers2 268.45 17.68 15.19 2.55e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2342 on 17 degrees of freedom
## Multiple R-squared: 0.9314, Adjusted R-squared: 0.9233
## F-statistic: 115.4 on 2 and 17 DF, p-value: 1.282e-10From the summary, we see the R-squared value has increased dramatically from the linear model, indicating a big improvement. That is not the only value we must check though. Let’s inspect visually how the model based data compare to the actual data.
First, we must calculate the predicted values. Then, we can plot the points.
# compute the predicted values based on the quad model
predicted2 = predict(quad_model,data=mydata)
# Plot cost versus servers based on actual values
plot(servers,cost, pch=16)
# The par (parameter setting) command is needed to overlay the predicted model based values without the labels and annotations
par(new=TRUE, xaxt="n", yaxt="n", ann=FALSE)
# Use the red color for the quadratic model
plot(predicted2, col="red", pch=16) 
It is easy to observe now that the predicted values are more in line with the observed values.
A common misconception is that the higher order the non-linear model is the better it is. Lets try a cubic model and see how it performs. This model is of the form y = x + x^2 + x^3.
1A) Create a cubic non-linear regression model, and display the summary statistics.
#cubic model
#First define the additional new variable. the cubic of servers, needed for your model
servers3=servers^3 #=serverscubed
# The model formula is of the form x + x^2 + x^3. For consistency best to call your new model cubic_model.
cubicmodel=lm(cost~servers+servers2+servers3)
cubicmodel##
## Call:
## lm(formula = cost ~ servers + servers2 + servers3)
##
## Coefficients:
## (Intercept) servers servers2 servers3
## 36133.696 -5954.738 310.895 -1.347summary(cubicmodel)##
## Call:
## lm(formula = cost ~ servers + servers2 + servers3)
##
## Residuals:
## Min 1Q Median 3Q Max
## -2871.0 -1435.1 -473.6 1271.8 4600.3
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 36133.696 2625.976 13.760 2.77e-10 ***
## servers -5954.738 1056.596 -5.636 3.72e-05 ***
## servers2 310.895 115.431 2.693 0.016 *
## servers3 -1.347 3.619 -0.372 0.715
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 2404 on 16 degrees of freedom
## Multiple R-squared: 0.932, Adjusted R-squared: 0.9193
## F-statistic: 73.11 on 3 and 16 DF, p-value: 1.478e-09We can now visually inspect the goodness of the quadratic model and the cubic model.
1B) Create a plot of cost of versus servers based on actual data. Overlay the predicted values based on the cubic model, and on the quadratic model. Follow the sequence of commands described below.
# compute the predicted values based on the cubic model. For consistency best to call predicted3.
#predicted3
predicted3=predict(cubicmodel,data=mydata)
predicted3## 1 2 3 4 5 6 7
## 30488.507 25457.022 21031.159 17202.831 13963.954 11306.443 9222.212
## 8 9 10 11 12 13 14
## 7703.177 6741.253 6328.355 6456.398 7117.297 8302.966 10005.322
## 15 16 17 18 19 20
## 12216.278 14927.751 18131.654 21819.904 25984.414 30617.101# Plot cost versus servers based on actual values
plot(servers, cost, pch=16)
plot## function (x, y, ...)
## UseMethod("plot")
## <bytecode: 0x0000000013ad8b08>
## <environment: namespace:graphics># The par (parameter setting) command is needed to overlay the predicted model based values without the labels and annotations
par(new=TRUE, xaxt="n", yaxt="n", ann=FALSE)
# Use the green color for the cubic model
plot(predicted3, col="green", pch=16)
# The par (parameter setting) command is needed to overlay the predicted model based values without the labels and annotations
par(new=TRUE, xaxt="n", yaxt="n", pch=16)
# Use the red color for the quadratic model
#quadratic model = predicted2
plot(predicted2, col="red", pch=16)
From the graphs it should be hard to tell which of the two models, the quadratic or the cubic, is a better fit and predictor. A better way to quantify which model is best is to look at the Adjusted-R2.
1C) List here the R2 and Adjusted R2 for all three models linear, quandratic, and cubic. Identify which model is a better predictor and why
linear # line 48
R^2 = 0.001127 Adjusted R^2 = -0.05437
quadratic # line 69
R^2 = .9314 Adjusted R^2 = .9233
cubic # line 101
R^2 = .932 Adjusted R^2 = .9193
The best predictor model is the quadratic model. This is because the quadratic model has the highest Adjusted R^2 model, and when you examine the curve, the values are very close together and don’t deviate much from the plot.
Task 2: Linear Programming & Optimization
For this task, we need to install an optimization package in R.
# Require will load the package only if not installed
# Dependencies = TRUE makes sure that dependencies are install
if(!require("lpSolveAPI",quietly = TRUE))
install.packages("lpSolveAPI",dependencies = TRUE, repos = "https://cloud.r-project.org")Solving Marketing Model
First create the linear programming model object in R. This is the starting point. The object will eventually contain all definitions and results.
# We start with `0` constraint and `2` decision variables. The object name `lpmark` is discretionary.
lpmark <- make.lp(0, 2) Next we need to define the type of optimization, set the objective function, and add the constraints to our model object.
# Define type of optimization as maximum and dump the screen output into `dump`
dump = lp.control(lpmark, sense="max")
# Set the objective function
set.objfn(lpmark, c(275.691, 48.341))
# add constraints
add.constraint(lpmark, c(1, 1), "<=", 350000)2A) Add the remaining missing five constraints to the model above. Follow the guidelines as described in the lectures.
#binary outcomes 1,0/0,1/2,-1/1,0/0,1 #all equal to 1
# see values in slide 5 in Powerpoint 'BSAD343F17-08'
add.constraint(lpmark, c(1,0), ">=", 15000)
add.constraint(lpmark, c(0,1), ">=", 75000)
add.constraint(lpmark, c(2,-1), "=", 0)
add.constraint(lpmark, c(1,0), ">=", 0)
add.constraint(lpmark, c(0,1), ">=", 0)Finally we can explore and solve the model using the lpSolveAPI package.
# View the problem formulation in tabular/matrix form
lpmark## Model name:
## C1 C2
## Maximize 275.691 48.341
## R1 1 1 <= 350000
## R2 1 0 >= 15000
## R3 0 1 >= 75000
## R4 2 -1 = 0
## R5 1 0 >= 0
## R6 0 1 >= 0
## Kind Std Std
## Type Real Real
## Upper Inf Inf
## Lower 0 0# Solve
solve(lpmark) ## [1] 0# Display the objective function optimum value
get.objective(lpmark)## [1] 43443517# Display the decision variables optimum values
get.variables(lpmark) ## [1] 116666.7 233333.32B) Write down and clearly mark the optimum values for sales, radio, and tv ads. Show how the optimum decision values satisfy all constraints.
Slides in Powerpoint ‘BSAD343F17-08’/ ‘BSAD343F17-07’
Sales
Optimum sales value or Z = $43,443,517 (when at maximum).
Radio
= $11,666.70 budget for Radio Ads
TV Ads
= $233,333.30 budget for TV Ads