Business Analytics Lab Worksheet 06

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/

## Loading required package: lpSolveAPI

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)

##   servers  cost
## 1       1 27654
## 2       2 24789
## 3       3 21890
## 4       4 21633
## 5       5 15843
## 6       6 12567

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 servers and cost is positive, however it is not strong.

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()

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

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

Add a trend line plot using the a linear model

#plot + geom_smooth(method = "lm")

These did not work

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 ) 
linear_model

## 
## Call:
## lm(formula = cost ~ servers, data = mydata)
## 
## Coefficients:
## (Intercept)      servers  
##       14747           48

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

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

servers = mydata$servers
servers_squared = mydata$servers^2
quad_model = lm(cost ~ servers + servers_squared)
quad_model

## 
## Call:
## lm(formula = cost ~ servers + servers_squared)
## 
## Coefficients:
##     (Intercept)          servers  servers_squared  
##         35417.8          -5589.4            268.4

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 + servers_squared)
## 
## 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 ***
## servers_squared   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")

qplot seems to not be working

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_cubed = mydata$servers^3
cubic_model = lm(cost ~ servers + servers_squared + servers_cubed)
cubic_model

## 
## Call:
## lm(formula = cost ~ servers + servers_squared + servers_cubed)
## 
## Coefficients:
##     (Intercept)          servers  servers_squared    servers_cubed  
##       36133.696        -5954.738          310.895           -1.347

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 + servers_squared + servers_cubed)
## 
## 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 ***
## servers_squared   310.895    115.431   2.693    0.016 *  
## servers_cubed      -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

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?