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

## Loading required package: tidyverse

## ── Attaching packages ─────────────────────────────────────── tidyverse 1.2.1 ──

## ✔ ggplot2 2.2.1     ✔ purrr   0.2.4
## ✔ tibble  1.4.2     ✔ dplyr   0.7.4
## ✔ tidyr   0.8.0     ✔ stringr 1.2.0
## ✔ readr   1.1.1     ✔ forcats 0.2.0

## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag()    masks stats::lag()

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() ##Might need to load tidyverse? package

mydata <- read.csv('data/ServersCost.csv')

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.

There is a weak, positive correlation between costs and servers

cor(mydata)

##            servers       cost
## servers 1.00000000 0.03356606
## cost    0.03356606 1.00000000

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.

Initially, additionall servers descrease the cost, but after a certain amount of servers (10), additional servers increase the cost. The shape of the trend line is not linear.

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 )

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?

This is not a good fit for the data. The R-squared value is 0.001127 and the adjusted R-squared value is -0.05437, which shows that linear model is not a good fit.

Commands: summary( linear_model )

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

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

#y = x + x^2
servers2 <- servers^2
quad_model <- lm(cost ~ servers + servers2)

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?

The R-Squared Value is 0.9314 and the adjusted R-Squared value is 0.9233, which is much better than the linear model. The quadtratic model is a good fit for the data, as the data follows a non-linear pattern.

Commands: summary( quad_model )

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-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)

predicted2

##         1         2         3         4         5         6         7 
## 30096.790 25312.706 21065.520 17355.233 14181.844 11545.354  9445.762 
##         8         9        10        11        12        13        14 
##  7883.068  6857.273  6368.376  6416.377  7001.277  8123.076  9781.772 
##        15        16        17        18        19        20 
## 11977.367 14709.861 17979.252 21785.543 26128.731 31008.818

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?

The plot is a good fit for the data. It follows the same pattern (after a certain number of servers-10- the cost starts to increase rather than decrease) and it the same shape (U)

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

##y = x + x^2 + x^3
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?

The R-Sqaured value is 0.932 and the Adjusted R-Squared value is 0.9193, which is a good fit for the data, but almost the same as the previous quadtratic model (R-Sqaured: 0.9314, Adjusted R-squared: 0.9233).

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

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

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

predicted3 <- predict( cubic_model, 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

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?

The plot is a good for the data, but it is very similar to the previous model and only slightly better.