About

This project will explore the concepts of simple linear regression, multiple linear regression and finding the optimum solution with possible business limitations.

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 html, pdf or word output to Sakai, If you want you can also publish your work into rpubs and share the link on Sakai. If you worked as a team you can have 1 submission with all the names of the team on it.

Task 1: Simple Linear Regression (Total 2.5 points)

First, read in the marketing data that was used in the previous lab. Make sure the file is read in correctly. (0.25 points)

#Read data correctly
marketing = read.csv(file = "marketing.csv")
head(marketing)
##   case_number sales radio paper  tv pos
## 1           1 11125    65    89 250 1.3
## 2           2 16121    73    55 260 1.6
## 3           3 16440    74    58 270 1.7
## 4           4 16876    75    82 270 1.3
## 5           5 13965    69    75 255 1.5
## 6           6 14999    70    71 255 2.1

Next, apply the cor() function to the data to understand the correlations between variables. This is a great way to compare the correlations between all variables.(0.25 points)

#Correlation matrix of all columns in the data
corr = cor(marketing)
corr
##             case_number      sales       radio       paper          tv
## case_number   1.0000000  0.2402344  0.23586825 -0.36838393  0.22282482
## sales         0.2402344  1.0000000  0.97713807 -0.28306828  0.95797025
## radio         0.2358682  0.9771381  1.00000000 -0.23835848  0.96609579
## paper        -0.3683839 -0.2830683 -0.23835848  1.00000000 -0.24587896
## tv            0.2228248  0.9579703  0.96609579 -0.24587896  1.00000000
## pos           0.0539763  0.0126486  0.06040209 -0.09006241 -0.03602314
##                     pos
## case_number  0.05397630
## sales        0.01264860
## radio        0.06040209
## paper       -0.09006241
## tv          -0.03602314
## pos          1.00000000
#Correlation matrix of all columns except the first column. This is convenient since case_number is only an indicator for the month and should be excluded from the calculations.
corr = cor( marketing[ c(2:6) ] )
corr
##            sales       radio       paper          tv         pos
## sales  1.0000000  0.97713807 -0.28306828  0.95797025  0.01264860
## radio  0.9771381  1.00000000 -0.23835848  0.96609579  0.06040209
## paper -0.2830683 -0.23835848  1.00000000 -0.24587896 -0.09006241
## tv     0.9579703  0.96609579 -0.24587896  1.00000000 -0.03602314
## pos    0.0126486  0.06040209 -0.09006241 -0.03602314  1.00000000
Q 1A) Why the value of “1.0” along the diagonal? (0.25 points)

A 1A) The value of 1 is along the diagonal because the values correlated with each other to produce the value of 1 are the exact same variable.

Q 1B) Evaluate the correlation between the variables (0.25 points)

A 1B) Sales is strongly correlated with tv and radio; it is also negatively correlated with paper. Paper has a negative correlation with radio and tv. POS does not have much correlation with the other variables.

Now, try to find a simple linear regression model between sales and marketing expences for radio commercials.(0.25 points)

#Extract variables
pos  = marketing$pos
paper = marketing$paper
tv = marketing$tv
sales = marketing$sales
radio = marketing$radio

#Simple Linear Regression 
reg <- lm(sales ~ radio)

#Summary of Model
summary(reg)
## 
## Call:
## lm(formula = sales ~ radio)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1732.85  -198.88    62.64   415.26   637.70 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -9741.92    1362.94  -7.148 1.17e-06 ***
## radio         347.69      17.83  19.499 1.49e-13 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 571.6 on 18 degrees of freedom
## Multiple R-squared:  0.9548, Adjusted R-squared:  0.9523 
## F-statistic: 380.2 on 1 and 18 DF,  p-value: 1.492e-13

Q 1C) What is the values for the intercept and the slope for radio? Write down the equation for the linear regression model and your interpretation for the intercept and slope.(0.5 points)

A 1C) The slope for radio is 347.69 and the y-intercept is -9471.92.

Y= -9741.92 + 347.69x This equation means that for every “x” radio sold, sales increase by about $347.69. The y-intercept means that at 0 radios, the amount of money (sales) is -$9741.92.

Given this equation we can predict the value of sales for any given value of radio ads. Use your equation to calculate the predicted sales value for 75 (investing $75,000 in radio ads).(0.25 points)

### Sales_predicted  ~ Radio expences
sales_predicted = -9741.92 + 347.69 * (75)
sales_predicted
## [1] 16334.83

A high R-Squared value indicates that the model is a good fit, but not perfect. For the case of Sales versus Radio we will overlay the trend line representing the regression equation over the original plot. This will show how far the calculated results are from the actual value. The difference between the actual sales (circles) and the fitted sales (solid line) is captured in the residual error calculations.

#Plot Radio and Sales 
plot(radio,sales)

scatter.smooth(radio,sales)

#Add a trend line plot using the linear model we created above
abline(reg, col="blue",lwd=2)

Now you can derive the linear regression model for Sales versus TV (0.25 points)

#Simple Linear Regression 
regtv <- lm(sales ~ tv)

#Summary of Model
summary(regtv)
## 
## Call:
## lm(formula = sales ~ tv)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1921.87  -412.24     7.02   581.59  1081.61 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -42229.21    4164.12  -10.14 7.19e-09 ***
## tv             221.10      15.61   14.17 3.34e-11 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 771.3 on 18 degrees of freedom
## Multiple R-squared:  0.9177, Adjusted R-squared:  0.9131 
## F-statistic: 200.7 on 1 and 18 DF,  p-value: 3.336e-11
Q 1D) Write down the equation for the linear regression model.(0.25 points)

A 1D) y= -42229.21 + 221.10x

Task 2: Multiple Linear Regression (Total 1.5 points)

Create a multiple linear regression predicting sales using both independent variables radio and tv. (0.25 points)

#Multiple Linear Regression Model
mlr1 <-lm(sales ~ radio + tv)

#Summary of Multiple Linear Regression Model
summary(mlr1)
## 
## Call:
## lm(formula = sales ~ radio + tv)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1729.58  -205.97    56.95   335.15   759.26 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -17150.46    6965.59  -2.462 0.024791 *  
## radio          275.69      68.73   4.011 0.000905 ***
## tv              48.34      44.58   1.084 0.293351    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 568.9 on 17 degrees of freedom
## Multiple R-squared:  0.9577, Adjusted R-squared:  0.9527 
## F-statistic: 192.6 on 2 and 17 DF,  p-value: 2.098e-12
Q 2A) Write down the equation for the linear regression model and your interpretation for the intercept and the slopes. (0.5 points)

A 2A) y= -17,150.46 + 275.69(Radio) + 48.34(TV)

For every radio ad, the sales will increase by $275.69 and for every TV ad, the sales will increase by $48.34. At 0 radio ads and 0 TV ads, the predicted sales would be -$17,150.46.

The predicted sales can again be calculated given the coefficients of the regression model.

Calculate the predicted sales for TV = 270 and Radio = 75 (0.25 points)

# sales_predicted = radio + tv
sales_predicted = coef(mlr1)[1] + coef(mlr1)[2]*(75) + coef(mlr1)[3]*(270)
sales_predicted
## (Intercept) 
##     16578.3

Create a multiple linear regression model for each of the following, and display the summary statistics (0.25 points)

#Multiple regression model for sales predicted by radio, tv, and pos
mlr2 <-lm(sales ~ radio + tv + pos)


#Summary of Multiple Linear Regression Model
summary(mlr2)
## 
## Call:
## lm(formula = sales ~ radio + tv + pos)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1748.20  -187.42   -61.14   352.07   734.20 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)   
## (Intercept) -15491.23    7697.08  -2.013  0.06130 . 
## radio          291.36      75.48   3.860  0.00139 **
## tv              38.26      48.90   0.782  0.44538   
## pos           -107.62     191.25  -0.563  0.58142   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 580.7 on 16 degrees of freedom
## Multiple R-squared:  0.9585, Adjusted R-squared:  0.9508 
## F-statistic: 123.3 on 3 and 16 DF,  p-value: 2.859e-11
#Multiple regression model for sales predicted by radio, tv, pos, and paper
mlr3 <-lm(sales ~ radio + tv + pos + paper)


#Summary of Multiple Linear Regression Model
summary(mlr3)
## 
## Call:
## lm(formula = sales ~ radio + tv + pos + paper)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1558.13  -239.35     7.25   387.02   728.02 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)   
## (Intercept) -13801.015   7865.017  -1.755  0.09970 . 
## radio          294.224     75.442   3.900  0.00142 **
## tv              33.369     49.080   0.680  0.50693   
## pos           -128.875    192.156  -0.671  0.51262   
## paper           -9.159      8.991  -1.019  0.32449   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 580 on 15 degrees of freedom
## Multiple R-squared:  0.9612, Adjusted R-squared:  0.9509 
## F-statistic: 92.96 on 4 and 15 DF,  p-value: 2.13e-10
Q 2B) Which of the three multiple linear regression models is best in predicting sales. Explain why. (0.25 points)

A 2B) The first multiple linear regression is the best in predicting sales. This is because the models’ Adjusted R-squared is the highest of all of the 3 models at .9527, this is because the adjusted R-squared as close to 1 as possible means the strongest correlation to sales.

###Task 3: Linear Programming & Optimization (total 4 points) For this task, we need to install an optimization package in R.

if(!require("lpSolveAPI",quietly = TRUE))
  install.packages("lpSolveAPI",dependencies = TRUE, repos = "https://cloud.r-project.org")
#install special package required for the solver
#install.packages("lpSolveAPI", repos = "https://cran.us.r-project.org") 
#install.packages("lpSolveAPI", repos = "https://cloud.r-project.org")
# load package library
library("lpSolveAPI")

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. (0.25 points)

# Start creating the linear programming model
lpmark <- make.lp(0,2)
?make.lp

Next, define the type of optimization, set the objective function, and add the constraints to our model object.(0.5 points)

# Define type of optimization,and dump the screen output into a variable `dump`
dump = lp.control(lpmark, sense="max")

# Set the objective function
set.objfn(lpmark, c(275.69, 48.34))

# add constraints
add.constraint(lpmark, c(1, 1), "<=", 350000)
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 solve the model. (0.5 points)

# View the problem formulation in tabular/matrix form
lpmark
## Model name: 
##               C1      C2            
## Maximize  275.69   48.34            
## 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] 43443167
# Display the decision variables optimum values
  get.variables(lpmark)
## [1] 116666.7 233333.3
# Display the constraints
 get.constr.value(lpmark)
## [1] 350000  15000  75000      0      0      0

Q 3A) Write down and clearly mark the optimum values for sales, radio, and tv ads. Show how the optimum values satisfy all constraints (0.5 points)

A 3A) Optimal Sales: $43443167

This satisfies all constraints because the Sales must be greater than $350,000, the sum of the optimal radio and TV Ads values, in order for a profit, in which the number for sales definitely exceeds this number.

Optimal Radio: $116666.70 This satisfies all constraints because the Radio must be greater than of equal to $15,000, which the answer exceeds.

Optimal TV Ads: $233333.3 This satisfies all constraints because the TV Ads must be greater than or equal to $75,000, which the answer exceeds.

Display the sensitivity of the coefficiants(0.25 points)

# display sensitivity to coefficients of objective function. 
get.sensitivity.obj(lpmark)
## $objfrom
## [1]  -96.680 -137.845
## 
## $objtill
## [1] 1e+30 1e+30
Q 3B) Explain in coincise manner what the sensitivity results represent in reference to the marketing model.(1 points)

A 3B) The -96.680 and -137.845 represent the lower boundary limits. The values of X1 and X2 can’t drop below these constraints, or it will change the optimum solution.

Display the sensitivity to right hand side constraints (0.25 points)

# display sensitivity to right hand side constraints. 
# There will be a total of m+n values where m is the number of contraints and n is the number of decision variables
get.sensitivity.rhs(lpmark) 
## $duals
## [1] 124.12333   0.00000   0.00000  75.78333   0.00000   0.00000   0.00000
## [8]   0.00000
## 
## $dualsfrom
## [1]  1.125e+05 -1.000e+30 -1.000e+30 -3.050e+05 -1.000e+30 -1.000e+30 -1.000e+30
## [8] -1.000e+30
## 
## $dualstill
## [1] 1.00e+30 1.00e+30 1.00e+30 4.75e+05 1.00e+30 1.00e+30 1.00e+30 1.00e+30
Q 3C) For this exercise we are only interested in the first part of the output labeled duals. Explain in coincise manner what the two non-zero sensitivity results represent. Distinguish the binding/non-binding constraints, the surplus/slack, and marginal values.(1 points)

A 3C) The 2 non-zero sensitivity results are binding constraints, meaning that any change in values will change the optimal value. If you change teh value by 1, the result will be 124.12333 and 75.78333. Every other constraint that hsa a value of ‘0’ means that they are non-binding, or unbinding, constraints.