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.

## Loading required package: lpSolveAPI
library("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()

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

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(file = "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$servers
head(servers)
## [1] 1 2 3 4 5 6
cost <- mydata$cost
head(cost)
## [1] 27654 24789 21890 21633 15843 12567

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

cor(mydata)
##            servers       cost
## servers 1.00000000 0.03356606
## cost    0.03356606 1.00000000

The correlation between the two variables, cost and servers, is very weak since it is around .034. ### 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

The relationship between cost and servers decreases at a steep rate when the number of servers is 0 to 9. Then it levels out from 9 to around 13 and then increases at a steep rate. 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?

  • 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

This model is a bad fit for the data because the r-squared is .001127 and the adjusted r-squared in -.05437.

Task 3: Regression Analysis - Non-linear Regression

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
sqservers <- servers^2
quad_model <- lm(cost ~ servers + sqservers)
quad_model
## 
## Call:
## lm(formula = cost ~ servers + sqservers)
## 
## Coefficients:
## (Intercept)      servers    sqservers  
##     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 + sqservers)
## 
## 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 ***
## sqservers     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

This is a great fit for the data because the r-squared is .9314 and the adjusted r-squared is .9233. These numbers are a lot higher than that of the linear model meaning this model is a much better fit.

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

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

sqservers = servers^2
quad_model = lm(cost ~ servers + sqservers ) 
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?

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

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

This graph looks very similar to the real data and it has the same negative slope to around 10 then increases afterwards.

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
cubeservers <- servers^3
cubic_model <- lm(cost ~ servers + sqservers + cubeservers)
cubic_model
## 
## Call:
## lm(formula = cost ~ servers + sqservers + cubeservers)
## 
## Coefficients:
## (Intercept)      servers    sqservers  cubeservers  
##   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 + sqservers + cubeservers)
## 
## 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 ***
## sqservers     310.895    115.431   2.693    0.016 *  
## cubeservers    -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

This is a great model for the data as well because r-squared is .932 and adjusted r-squared is .9193 which are both very high. The r-square is higher than the second model but the adjusted r-squared is a bit lower.

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 data looks very similar to the second model because they both had a decreasing slope to around 10 then increased from then on.

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)

The green dots which are the cubic model seem to be slightly closer to the black dots which are the real data points than the red dots of the quadratic model. Therefore, the cubic model is the best fit for this data set. The linear model does not predict any of the data properly.