Business Analytics Lab Worksheet 06

CME Group Foundation Business Analytics Lab

---

Notebook Instructions

---

About

• In this lab, we will focus on linear and non-linear programming.

• Linear programming, as discussed in the previous lab, works with simple and multiple linear regression techniques; sometimes the variables have completely direct or completely non-direct relationships and these techniques can model them.

• Sometimes, however, the variables do not predict each other in a linear way. For example, looking at the stock market vs. time, we know that generally the market was booming before the crash, then the market crashed and the great depression hit, and slowly the market started to rise again.

• This pattern is not linear, and in fact a non-linear programming technique can be used to model it and predict the value of the market based on the year.

• In this lab, we will explore topics like optimization, solve a marketing model, and perform linear and non-linear regression on the cost of servers.

Load Packages in R/RStudio

We are going to use tidyverse a collection of R packages designed for data science.

• Info: https://www.tidyverse.org/

# Here we are checking if the package is installed
if(!require("tidyverse")){
  
  # If the package is not in the system then it will be install
  install.packages("tidyverse", dependencies = TRUE)
  
  # Here we are loading the package
  library("tidyverse")
}

Loading required package: tidyverse
-- Attaching packages --------------------------------------- tidyverse 1.2.1 --
v ggplot2 2.2.1     v purrr   0.2.4
v tibble  1.4.2     v dplyr   0.7.4
v tidyr   0.7.2     v stringr 1.2.0
v readr   1.1.1     v forcats 0.2.0
-- Conflicts ------------------------------------------ tidyverse_conflicts() --
x dplyr::filter() masks stats::filter()
x dplyr::lag()    masks stats::lag()

# Here we are checking if the package is installed
if(!require("plotly")){
  
  # If the package is not in the system then it will be install
  install.packages("plotly", dependencies = TRUE)
  
  # Here we are loading the package
  library("plotly")
}

Loading required package: plotly

Attaching package: <U+393C><U+3E31>plotly<U+393C><U+3E32>

The following object is masked from <U+393C><U+3E31>package:ggplot2<U+393C><U+3E32>:

    last_plot

The following object is masked from <U+393C><U+3E31>package:stats<U+393C><U+3E32>:

    filter

The following object is masked from <U+393C><U+3E31>package:graphics<U+393C><U+3E32>:

    layout

---

Task 1: Linear Programming - Solving Marketing Model

---

1A) Create the model object in R.

lprec <- make.lp(0, 2) 

Set the constrains and objective function for the model.

• Set for maximum

lp.control(lprec, sense="max")  

$anti.degen
[1] "fixedvars" "stalling" 

$basis.crash
[1] "none"

$bb.depthlimit
[1] -50

$bb.floorfirst
[1] "automatic"

$bb.rule
[1] "pseudononint" "greedy"       "dynamic"      "rcostfixing" 

$break.at.first
[1] FALSE

$break.at.value
[1] 1e+30

$epsilon
      epsb       epsd      epsel     epsint epsperturb   epspivot 
     1e-10      1e-09      1e-12      1e-07      1e-05      2e-07 

$improve
[1] "dualfeas" "thetagap"

$infinite
[1] 1e+30

$maxpivot
[1] 250

$mip.gap
absolute relative 
   1e-11    1e-11 

$negrange
[1] -1e+06

$obj.in.basis
[1] TRUE

$pivoting
[1] "devex"    "adaptive"

$presolve
[1] "none"

$scalelimit
[1] 5

$scaling
[1] "geometric"   "equilibrate" "integers"   

$sense
[1] "maximize"

$simplextype
[1] "dual"   "primal"

$timeout
[1] 0

$verbose
[1] "neutral"

set.objfn(lprec, c(275.691, 48.341))

1B) Add constrains

add.constraint(lprec, c(1, 1), "<=", 350000)
add.constraint(lprec, c(1, 0), ">=", 15000)
add.constraint(lprec, c(0, 1), ">=", 75000)
add.constraint(lprec, c(2, -1), "=", 0)

View the problem formulation in tabular/matrix form to confirm that the model was created correctly.

lprec

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
Kind          Std      Std            
Type         Real     Real            
Upper         Inf      Inf            
Lower           0        0            

1C) Solve the optimization problem

# solve 
solve(lprec) 

[1] 0

Display the objective function optimum value

get.objective(lprec)

[1] 43443517

Display the variables optimum values

get.variables(lprec) 

[1] 116666.7 233333.3

---

Task 2: Regression Analysis - Linear Regression

---

• A linear model is of the form y = x0 + x1 + …+ x_n

2A) Read the csv file into R Studio and display the dataset.

• Name your dataset ‘mydata’ so it easy to work with.

• Commands: read_csv() head()

mydata <- read.csv("data/ServersCost.csv")
head(mydata)

Extract the assigned features (columns) to perform some analytics.

servers <- mydata$servers
cost <- mydata$cost

2B) Create a correlation table for your to compare the correlations between all variables. What can you tell about the correlation between the variables.

Corr <- cor(mydata)
Corr

           servers       cost
servers 1.00000000 0.03356606
cost    0.03356606 1.00000000

The correlation between these two variables is direct, positive, but very weak.

2C) Create a plot for the dependent (y) and independent (x) variables. Note any patterns or relation between the two variables describe the trend line.

• The blue line here represents the linear model we created and the black dots are the data points.

Commands: p <- qplot( x = INDEPENDENT, y = DEPENDENT, data = mydata) + geom_point()

p <- qplot( x = servers, y = cost, data = mydata) +geom_point()
p

Commmand: p + geom_smooth(method = “lm”)

Add a trend line plot using the a linear model

p + geom_smooth(method = "lm")

2D) Create a linear regression model by identifying the dependent variable (y) and independent variable (x_n)

• Commands: linear_model <- lm( DEPENDENT ~ INDEPENDENT )

linear_model <- lm( cost ~ servers, data =mydata ) 
predict (linear_model, data = mydata)

       1        2        3        4        5        6        7        8 
14795.19 14843.19 14891.19 14939.19 14987.19 15035.19 15083.19 15131.20 
       9       10       11       12       13       14       15       16 
15179.20 15227.20 15275.20 15323.20 15371.20 15419.21 15467.21 15515.21 
      17       18       19       20 
15563.21 15611.21 15659.21 15707.21 

Use the regression model to create a report. Note the R-Squared and Adjusted R-Squared values, determine if this is a good or bad fit for your data?

• Commands: summary( linear_model )

 summary( linear_model )

Call:
lm(formula = cost ~ servers, data = mydata)

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

R-squared is .0112 and the Adjusted R-squared is -0.05437, which indicates that this relationship is not very strong.

Task 3: Regression Analysis - Non-linear Regression

---

• We use a transformation and use a nonlinear quadratic model to see how the model fits to the data.

• A quadratic model transforms the predictor by squaring it and adding to the model.

• Quadratic Model: y = x + x^2

3A) Create a non-linear quadratic regression model by identifying the dependent variable (y) and independent variables (x). Transforms the independent variable by squaring it and adding to the model.

• The Quadratic model formula is: y = x + x^2
• Commands: quad_model <- lm(y ~ x + x_squared)
• Commands: To squared a variable use (^) such as x^2

setwd("C:\\Users\\hp\\Documents\\Spring 2018\\BSAD 343H\\Labs\\Lab 7\\07-notebook-lab")
# y = x + x^2
servers = mydata$servers
servers2 = mydata$servers^2
quad_model <- lm(cost ~ servers + servers2, data = mydata)

Use the quadratic model to create a report. Note the R-Squared and Adjusted R-Squared values, determine if this is a good or bad fit for your data?

• Commands: summary( quad_model )

summary (quad_model)

Call:
lm(formula = cost ~ servers + servers2, data = mydata)

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-10

3B) Compute the predicted values based on the quadratic model.

Commands: predicted_2 <- predict( quad_model, data = mydata )

servers2 = servers^2
quad_model = lm(cost ~ servers + servers2 ) 
predicted2 = predict(quad_model,data=mydata)

Create a plot using the quadratic model predicted values in color red. Noted the shape, looking at the plot is this a good or bad fit for your data?

Commands: qplot( x = DEPENDENT, y = INDEPENDENT/PREDICTED, colour = “red” )

qplot( x = servers, y = predicted2, colour = "red" )

3C) Create a non-linear cubic regression model by identifying the dependent variable (y) and independent variables (x). Transforms the independent variable by squaring it to second (x^2) and third )x^3) degrees and adding them to the model.

• The Cubic model formula is: y = x + x^2 + x^3
• Commands: cubic_model <- lm(y ~ x + x_squared + x_cubic)
• Commands: To squared a variable use (^) such as x^2, x^3

servers <- mydata$servers
servers2 <- mydata$servers^2
servers3 <- mydata$servers^3
cubic_model <- lm(cost ~ servers + servers2 + servers3)

Use the cubic model to create a report. Note the R-Squared and Adjusted R-Squared values, determine if this is a good or bad fit for your data?

• Commands: summary( cubic_model )

summary( cubic_model )

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-09

Multiple R-squared is .932 and adjusted R-squared is .9193, which indicates the model and variables have a strong relationship and they are a good fit for the data.

3D) Compute the predicted values based on the cubic model.

Commands: predicted3 <- predict( cubic_model, data = mydata )

predicted3 <- predict( cubic_model, data = mydata )

Create a plot using the cubic model predicted values in color green. Noted the shape, looking at the plot is this a good or bad fit for your data? Is this model better than the previous?

Commands: qplot( x = DEPENDENT, y = INDEPENDENT/PREDICTED, colour = “red” )

qplot( x = servers, y = predicted3, colour = "red" )

This model appears to have data points that are better suited to the model, because they follow a quandratic formula with little deviations. Furthermore, the values R-Squared and adjusted R-squared values are higher. ### 3E) Overlay the all models on top of the data. Which model seems to fit the best in your opinion? Justify your answer.

variables: LINEAR_MODEL , PREDICTED_QUADRATIC, PREDICTED_CUBIC

# Black = Actual Data
plot(servers, cost, pch = 16) 
# Blue = Linear Line based on Linear Regression Model
abline(linear_model,  col = "blue", lwd = 2)  

# Red = Quadratic Model based on Quadratric Regression found above
# Needed to overlay new points without the labels and annotations
par(new = TRUE, xaxt = "n", yaxt = "n", ann = FALSE) 
plot(predicted2, col = "red", pch = 16) 

# Green = Cubic Model based on Cubic Regression found above
# Overlay new points without the labels and annotations 
par(new = TRUE, xaxt = "n", yaxt = "n", ann = FALSE) 
plot(predicted3, col = "green", pch = 16)

Model 3 appears to be the best fit for the data, as there is little variation between the actual and plotted data.