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	# Set Working Directory
	setwd("F:/IIITB - Upgrad/Course 3 - Predictive Analysis/Case Study")


	# Install and Load the required packages
	install.packages("MASS")
	install.packages("car")
	install.packages("e1071")
	install.packages("ROCR")
	install.packages("caret")
	install.packages("Hmisc")
	install.packages("plyr")
	install.packages("ggplot2")
	install.packages("moments")
	install.packages("arules")
	install.packages("ROCR")
	install.packages("class")
	library(ROCR)
	library(arules)
	library(MASS)
	library(car)
	library(caret)
	library(ROCR)
	library(e1071)
	library(Hmisc)
	library(plyr)
	library(ggplot2)
	library(arules)
	library(class)

	#Checkpoint 1
	# Load the given files.
	churn_data <- read.csv("churn_data.csv", stringsAsFactors = FALSE)
	customer_data <-
	read.csv("customer_data.csv", stringsAsFactors = FALSE)
	internet_data <-
	read.csv("internet_data.csv", stringsAsFactors = FALSE)
	str(churn_data)
	str(customer_data)
	str(internet_data)

	# Collate the 3 files in a single file.
	merge1 <- merge(churn_data, customer_data, by = 'customerID')
	churn <- merge(merge1, internet_data, by = 'customerID')

	# Understand the structure of the collated file.
	str(churn)

	#Checkpoint 2 - EDA
	# Make bar charts to find interesting relationships between variables.
	#Distribution of monthly charges along with churn
	ggplot(churn, aes(x = churn$MonthlyCharges)) + geom_histogram() + aes(fill = churn$Churn)
	#Distribution of tenure along with churn
	ggplot(churn, aes(x = churn$tenure)) + geom_histogram() + aes(fill = churn$Churn)
	#Gender wise churn
	ggplot(churn, aes(x = churn$Churn, fill = churn$gender)) + geom_bar()
	#internet service wise churn
	ggplot(churn, aes(x = churn$InternetService, fill = churn$Churn)) + geom_bar()
	#payment method wise churn
	ggplot(churn, aes(x = churn$PaymentMethod, fill = churn$Churn)) + geom_bar()
	#Techsupport wise churn
	ggplot(churn, aes(x = churn$TechSupport, fill = churn$Churn)) + geom_bar()

	#Checkpoint 3 - Data Preparation
	# Make Box plots for numeric variables to look for outliers.
	boxplot(churn$tenure)
	boxplot.stats(churn$tenure) #No Outlier

	boxplot(churn$MonthlyCharges)
	boxplot.stats(churn$MonthlyCharges) #No Outlier

	boxplot(churn$TotalCharges)
	boxplot.stats(churn$TotalCharges) #No Outlier

	# Perform De-Duplication if required
	which(duplicated(churn) == 'TRUE') #No Duplicates

	# Impute the missing values, and perform the outlier treatment.
	sapply(churn, function(x)
	sum(is.na(x)))
	churn$TotalCharges[which(is.na(churn$TotalCharges) == 'TRUE')] <-
	mean(churn$TotalCharges, na.rm = TRUE)



	#CHECKPOINT 4: Modeling

	#Model 1: Logistics Regression


	#Creating object for modeling
	churn_df = cbind.data.frame(churn$tenure, churn$MonthlyCharges, churn$TotalCharges)
	names(churn_df)[names(churn_df) == 'churn$tenure'] <- 'tenure'
	names(churn_df)[names(churn_df) == 'churn$MonthlyCharges'] <-
	'MonthlyCharges'
	names(churn_df)[names(churn_df) == 'churn$TotalCharges'] <-
	'TotalCharges'

	# Bring the variables in the correct format
	dummy_PhoneService = as.data.frame(model.matrix( ~ PhoneService - 1, data = churn))
	churn_df = cbind(churn_df, dummy_PhoneService[, -1])
	names(churn_df)[names(churn_df) == 'dummy_PhoneService[, -1]'] <-
	'dummy_PhoneService'

	dummy_PaperlessBilling = as.data.frame(model.matrix( ~ PaperlessBilling -
	1, data = churn))
	churn_df = cbind(churn_df, dummy_PaperlessBilling[, -1])
	names(churn_df)[names(churn_df) == 'dummy_PaperlessBilling[, -1]'] <-
	'dummy_PaperlessBilling'

	dummy_Churn = as.data.frame(model.matrix( ~ Churn - 1, data = churn))
	churn_df = cbind(churn_df, dummy_Churn[, -1])
	names(churn_df)[names(churn_df) == 'dummy_Churn[, -1]'] <-
	'dummy_Churn'

	dummy_gender = as.data.frame(model.matrix( ~ gender - 1, data = churn))
	churn_df = cbind(churn_df, dummy_gender[, -1])
	names(churn_df)[names(churn_df) == 'dummy_gender[, -1]'] <-
	'dummy_gender'

	#Senior Citizen data is already in 0-1 format, changing it to numeric
	churn_df = cbind(churn_df, as.numeric(churn$SeniorCitizen))
	names(churn_df)[names(churn_df) == 'as.numeric(churn$SeniorCitizen)'] <-
	'dummy_SeniorCitizen'

	dummy_Partner = as.data.frame(model.matrix( ~ Partner - 1, data = churn))
	churn_df = cbind(churn_df, dummy_Partner[, -1])
	names(churn_df)[names(churn_df) == 'dummy_Partner[, -1]'] <-
	'dummy_Partner'

	dummy_Dependents = as.data.frame(model.matrix( ~ Dependents - 1, data = churn))
	churn_df = cbind(churn_df, dummy_Dependents[, -1])
	names(churn_df)[names(churn_df) == 'dummy_Dependents[, -1]'] <-
	'dummy_Dependents'

	dummy_contract = as.data.frame(model.matrix( ~ Contract - 1, data = churn))
	churn_df = cbind(churn_df, dummy_contract[, -3])

	dummy_PaymentMethod = as.data.frame(model.matrix( ~ PaymentMethod - 1, data = churn))
	churn_df = cbind(churn_df, dummy_PaymentMethod[, -4])

	dummy_InternetService = as.data.frame(model.matrix( ~ InternetService -
	1, data = churn))
	churn_df = cbind(churn_df, dummy_InternetService[, -3])

	dummy_OnlineSecurity = as.data.frame(model.matrix( ~ OnlineSecurity - 1, data = churn))
	churn_df = cbind(churn_df, dummy_OnlineSecurity[, -3])

	dummy_OnlineBackup = as.data.frame(model.matrix( ~ OnlineBackup - 1, data = churn))
	churn_df = cbind(churn_df, dummy_OnlineBackup[, -3])

	dummy_multiplelines = as.data.frame(model.matrix( ~ MultipleLines - 1, data = churn))
	churn_df = cbind(churn_df, dummy_multiplelines[, -3])

	dummy_DeviceProtection = as.data.frame(model.matrix( ~ DeviceProtection -
	1, data = churn))
	churn_df = cbind(churn_df, dummy_DeviceProtection[, -3])

	dummy_TechSupport = as.data.frame(model.matrix( ~ TechSupport - 1, data = churn))
	churn_df = cbind(churn_df, dummy_TechSupport[, -3])

	dummy_StreamingTV = as.data.frame(model.matrix( ~ StreamingTV - 1, data = churn))
	churn_df = cbind(churn_df, dummy_StreamingTV[, -3])

	dummy_StreamingMovies = as.data.frame(model.matrix( ~ StreamingMovies -
	1, data = churn))
	churn_df = cbind(churn_df, dummy_StreamingMovies[, -3])

	str(churn_df)

	#Split data set
	set.seed(100)
	train.indices = sample(1:nrow(churn_df), 0.7 * nrow(churn_df))
	train.data = churn_df[train.indices,]
	test.data = churn_df[-train.indices,]


	# Initial Model with all variables
	initial_model = glm(dummy_Churn ~ ., data = train.data, family = "binomial")
	summary(initial_model)

	# Stepwise selection
	step_model = step(initial_model, direction = "both")
	summary(step_model)
	vif(step_model)

	dummy_Churn ~ tenure + MonthlyCharges + TotalCharges + dummy_PhoneService +
	dummy_PaperlessBilling + dummy_SeniorCitizen + `ContractMonth-to-month` + `ContractOne year` +
	`PaymentMethodElectronic check` + InternetServiceDSL + OnlineSecurityNo +
	OnlineBackupNo + TechSupportNo

	#Remove Tenure (High VIF)
	model1 = glm(
	dummy_Churn ~ MonthlyCharges + TotalCharges + dummy_PhoneService +
	dummy_PaperlessBilling + dummy_SeniorCitizen + `ContractMonth-to-month` + `ContractOne year` +
	`PaymentMethodElectronic check` + InternetServiceDSL + OnlineSecurityNo +
	OnlineBackupNo + TechSupportNo,
	data = train.data,
	family = "binomial"
	)
	summary(model1)
	vif(model1)

	#Remove ContractMonth-to-month (High VIF)
	model2 = glm(
	dummy_Churn ~ MonthlyCharges + TotalCharges + dummy_PhoneService +
	dummy_PaperlessBilling + dummy_SeniorCitizen + `ContractOne year` +
	`PaymentMethodElectronic check` + InternetServiceDSL + OnlineSecurityNo +
	OnlineBackupNo + TechSupportNo,
	data = train.data,
	family = "binomial"
	)
	summary(model2)
	vif(model2)

	#Remove OnlineBackupNo (Insignificant)
	model3 = glm(
	dummy_Churn ~ MonthlyCharges + TotalCharges + dummy_PhoneService +
	dummy_PaperlessBilling + dummy_SeniorCitizen + `ContractOne year` +
	`PaymentMethodElectronic check` + InternetServiceDSL + OnlineSecurityNo +
	TechSupportNo,
	data = train.data,
	family = "binomial"
	)
	summary(model3)
	vif(model3)

	#Checking correlation between MonthlyCharges & TotalCharges
	cor(train.data$MonthlyCharges, train.data$TotalCharges)

	#Since strong correlation, removing MonthlyCharges
	model4 = glm(
	dummy_Churn ~ TotalCharges + dummy_PhoneService +
	dummy_PaperlessBilling + dummy_SeniorCitizen + `ContractOne year` +
	`PaymentMethodElectronic check` + InternetServiceDSL + OnlineSecurityNo +
	TechSupportNo,
	data = train.data,
	family = "binomial"
	)
	summary(model4)
	vif(model4)

	#Removing dummy_PhoneService (low significance)
	model5 = glm(
	dummy_Churn ~ TotalCharges + dummy_PaperlessBilling + dummy_SeniorCitizen + `ContractOne year` +
	`PaymentMethodElectronic check` + InternetServiceDSL + OnlineSecurityNo +
	TechSupportNo,
	data = train.data,
	family = "binomial"
	)
	summary(model5)
	vif(model5)

	#Removing dummy_SeniorCitizen (low significance)
	model6 = glm(
	dummy_Churn ~ TotalCharges + dummy_PaperlessBilling + `ContractOne year` +
	`PaymentMethodElectronic check` + InternetServiceDSL + OnlineSecurityNo +
	TechSupportNo,
	data = train.data,
	family = "binomial"
	)
	summary(model6)
	vif(model6)

	best_model_log = model6
	summary(best_model_log)

	## C-statistic

	train.data$predicted_prob = predict(best_model_log, type = "response")
	rcorr.cens(train.data$predicted_prob, train.data$dummy_Churn)

	test.data$predicted_prob = predict(best_model_log, newdata = test.data, type = "response")
	rcorr.cens(test.data$predicted_prob, test.data$dummy_Churn)

	#KS-statistic

	#Train Data
	model_score <-
	prediction(train.data$predicted_prob, train.data$dummy_Churn)
	model_perf <- performance(model_score, "tpr", "fpr")
	ks_table <-
	attr(model_perf, "y.values")[[1]] - (attr(model_perf, "x.values")[[1]])
	ks = max(ks_table)
	ks
	which(ks_table == ks)

	#Test Data
	model_score_test <-
	prediction(test.data$predicted_prob, test.data$dummy_Churn)
	model_perf_test <- performance(model_score_test, "tpr", "fpr")
	ks_table_test <-
	attr(model_perf_test, "y.values")[[1]] - (attr(model_perf_test, "x.values")[[1]])
	ks_test = max(ks_table_test)
	ks_test
	which(ks_table_test == ks_test)


	# Selecting threshold value

	# ROC curve
	plot(model_perf, col = "red", lab = c(10, 10, 10))
	plot(model_perf_test, col = "red", lab = c(10, 10, 10))

	#confusion matrix 1 (Threshold Value = 0.5)
	confusionMatrix(as.numeric(train.data$predicted_prob > 0.5),
	train.data$dummy_Churn,
	positive = "1")
	confusionMatrix(as.numeric(test.data$predicted_prob > 0.5),
	test.data$dummy_Churn,
	positive = "1")

	#confusion matrix 2 (Threshold Value = 0.3)
	confusionMatrix(as.numeric(train.data$predicted_prob > 0.3),
	train.data$dummy_Churn,
	positive = "1")
	confusionMatrix(as.numeric(test.data$predicted_prob > 0.3),
	test.data$dummy_Churn,
	positive = "1")

	#confusion matrix 3 (Threshold Value = 0.7)
	confusionMatrix(as.numeric(train.data$predicted_prob > 0.7),
	train.data$dummy_Churn,
	positive = "1")
	confusionMatrix(as.numeric(test.data$predicted_prob > 0.7),
	test.data$dummy_Churn,
	positive = "1")


	# Model2

	# K-NN Model:

	#Creating "churn_knn" dataframe from "churn_df" and bringing original churn data

	churn_knn = churn_df[, -6]
	churn_knn = cbind(churn_knn, as.factor(churn$Churn))
	names(churn_knn)[names(churn_knn) == 'as.factor(churn$Churn)'] <-
	'Churn'


	# Bring the data in the correct format to implement K-NN model.
	churn_knn$MonthlyCharges <- scale(churn_knn$MonthlyCharges)
	churn_knn$TotalCharges <- scale(churn_knn$TotalCharges)

	# Implement the K-NN model for optimal K.
	set.seed(2)
	s1 = sample(1:nrow(churn_knn), 0.7 * nrow(churn_knn))
	churn_knn_train = churn_knn[s1, ]
	churn_knn_test = churn_knn[-s1, ]


	cl <- churn_knn_train[, 31]
	#Removing Class label "dummy_churn" from the train and test data set
	churn_knn_train1 <- churn_knn_train[, -31]
	churn_knn_test1 <- churn_knn_test[, -31]

	#Using the train() command to find the best K.
	model <- train(
	Churn ~ .,
	data = churn_knn_train,
	method = 'knn',
	tuneGrid = expand.grid(.k = 1:50),
	metric = 'Accuracy',
	trControl = trainControl(
	method = 'repeatedcv',
	number = 10,
	repeats = 10
	)
	)

	#Generating the plot of the model
	model
	plot(model) # Looking at graph we can see that optimum K = 19 (Post which not much difference in accuracy)

	#Creating the model with optimum cost
	impknn2 <-
	knn(churn_knn_train1,
	churn_knn_test1,
	cl,
	k = 19,
	prob = TRUE)
	table(impknn2, churn_knn_test[, 31])
	confusionMatrix(impknn2, churn_knn_test[, 31], positive = "Yes")

	#KNN Model Solution
	# Accuracy = 79.32%
	# Sensitivity = 50.98%
	# Specificity = 89.51%

	##########################################################################

	# Model 3
	# Naive Bayes Model:

	#Using the original data set
	str(churn)

	# Bring the data in the correct format to implement Naive Bayes algorithm.
	churn$customerID <- factor(churn$customerID)
	churn$PhoneService <- factor(churn$PhoneService)
	churn$Contract <- factor(churn$Contract)
	churn$PaperlessBilling <- factor(churn$PaperlessBilling)
	churn$PaymentMethod <- factor(churn$PaymentMethod)
	churn$Churn <- factor(churn$Churn)
	churn$gender <- factor(churn$gender)
	churn$Partner <- factor(churn$Partner)
	churn$Dependents <- factor(churn$Dependents)
	churn$MultipleLines <- factor(churn$MultipleLines)
	churn$InternetService <- factor(churn$InternetService)
	churn$OnlineSecurity <- factor(churn$OnlineSecurity)
	churn$OnlineBackup <- factor(churn$DeviceProtection)
	churn$TechSupport <- factor(churn$TechSupport)
	churn$StreamingTV <- factor(churn$StreamingTV)
	churn$StreamingMovies <- factor(churn$StreamingMovies)
	churn$DeviceProtection <- factor(churn$DeviceProtection)

	set.seed(2)
	s = sample(1:nrow(churn), 0.7 * nrow(churn))
	churn_NB_train = churn[s, ]
	churn_NB_test = churn[-s, ]


	churn_NB_test1 <- churn_NB_test[, -9]

	# Implement the Naive Bayes algorithm.
	model <- naiveBayes(churn_NB_train$Churn ~ . , data = churn_NB_train)


	pred <- predict(model, churn_NB_test1)
	table(pred, churn_NB_test$Churn)

	confusionMatrix(pred, churn_NB_test$Churn)

	#Naive Bayes Solution
	#??? Accuracy=71.46%
	#??? Sensitivity = 68.79%
	#??? Specificit y=78.89%
	#??? ROC Curve

	###############################################################################

	# Model 4

	# SVM:

	# Bring the data in the correct format to implement the SVM algorithm.

	# Taking data frame from Logistic Regression model
	churn_svm = churn_df[, -6]
	churn_svm = cbind(churn_svm, as.factor(churn$Churn))
	names(churn_svm)[names(churn_svm) == 'as.factor(churn$Churn)'] <-
	'Churn'
	# Also taking the train and test data from Logistic regression model (minus predicted probibility)
	set.seed(2)
	svm.set = sample(1:nrow(churn_svm), 0.7 * nrow(churn_svm))
	churn_svm_train = churn_knn[svm.set, ]
	churn_svm_test = churn_knn[-svm.set, ]

	# Implement the SVM algorithm using the optimal cost.

	# model 0 with cost = 0.1
	#model.svm.0 = svm(churn_svm_train$Churn~., data = churn_svm_train, kernel = "linear", cost = 0.1, scale = F)

	#plot(model.svm.0,churn_svm_train)
	#summary(model.svm.0)



	# finding the optimal value of cost using cross-validation using the tune function

	tune.svm = tune(
	svm,
	Churn ~ .,
	data = churn_svm_train,
	kernel = "linear",
	ranges = list(cost = c(0.001, 0.01, 0.1, 1, 10, 100))
	)
	summary(tune.svm)


	bestmodel_svm <- tune.svm$best.model
	bestmodel_svm #Best Performance is for Cost = 0.01
	summary(bestmodel_svm)


	# predicting test classes using the best model and analyzing the table


	svm.predict = predict(bestmodel_svm, churn_svm_test)


	table(svm.predict, churn_svm_test$Churn)
	confusionMatrix(svm.predict, churn_svm_test$Churn)


	# Accuracy: 78.99
	# Sensitivity: 88.22
	# Specificity: 53.31

	# Plotting the SVM
	plot(svm.predict, churn_svm_test$Churn)

Smitan Pradhan

Data Science Portfolio

Melbourne, Australia
Custom Social Profile Link

Predicting churn in Telecom Industry - Advanced ML

less than 1 minute read

In the telecom industry, customers are able to choose from multiple service providers and actively switch from one operator to another. In this highly competitive market, the telecommunications industry experiences an average of 15-25% annual churn rate. Given the fact that it costs 5-10 times more to acquire a new customer than to retain an existing one, customer retention has now become even more important than customer acquisition.

For many incumbent operators, retaining high profitable customers is the number one business goal.

Background: To reduce customer churn, telecom companies need to predict which customers are at high risk of churn. We have been hired by a telecom industry giant to look at customer level data and identify customers at high risk of churn and identify the main indicators of churn.

Problem Statement: We need to build a predictive model using advanced Machine Learning algorithms in order to predict the customers at high risk of churn along with the key indicators of churn.

Link to the project code

This case study has been completed with the help of my team mate Koushal Deshpande. Thanks Koushal for your help and your key insights!

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THE HISTORY OF THE COMPANY CREATION

1993 how the construction company remstroy was created the year 1993 was a period when a lot of construction companies, which had been working successfully during the soviet times and had rich staff capacity, were forced to cease their activity for various reasons. a lot of capable specialists either had to look for another job or change their field. but there were also those who were willing to realise their potential in the field of construction in accordance with the received degree and the experience they had accumulated. thus, in 1993 in elektrostal (moscow oblast) a group of specialists and people sharing each other’s ideas, who had enormous educational background and the highest degree in architecture, organized and registered ooo firm erg which began its rapid development and successful work, offering its service both on the construction market and other areas. 2000 industrial construction is the main area seven years of successful work have shown that combining different types of activities in the same company is not always convenient. and in the year 2000 the founders of ooo firm erg decided to create and register a monoprofile construction company ooo remstroy construction company. industrial construction was chosen as the priority area. it was in this area that the directors of ooo sk remstroy began their working life and grew as specialists. in order to achieve the set goal, they selected a mobile team of professionals in the field of industrial construction, which allows us to cope with the tasks assigned to ooo sk remstroy throughout russia and the near abroad. 2010 manufacturing of metal structures we possess modern equipment that allows us to carry out the entire cycle of works on the manufacture of metal structures of any complexity without assistance. designing – production – installation of metal structures. a staff of professionals and well-coordinated interaction of the departments let us carry out the work as soon as possible and in accordance with all customer’s requirements.” extract from the list of members of self-regulatory organizations, construction.

LICENSE OF MINISTRY OF EMERGENCY SITUATIONS

Certificates, system of managing quality.

SYSTEM OF ECOLOGIAL MANAGEMENT

SYSTEM OF OCCUPATIONAL SAFETY AND HEALTH MANAGEMENT

LETTERS OF RECOMMENDATION

THE GEOGRAPHY OF CONSTRUCTION SITES

YOU CAN FIND MORE INFORMATION ON THE CONSTRUCTION SITES OF OOO REMSTROY ON THE PAGE OF THE SITE

OUR CLIENTS

http://remstroi.pro/yandex-promyshlennoe-stroitelstvo

Geographic coordinates of Elektrostal, Moscow Oblast, Russia

Coordinates of elektrostal in decimal degrees, coordinates of elektrostal in degrees and decimal minutes, utm coordinates of elektrostal, geographic coordinate systems.

WGS 84 coordinate reference system is the latest revision of the World Geodetic System, which is used in mapping and navigation, including GPS satellite navigation system (the Global Positioning System).

Geographic coordinates (latitude and longitude) define a position on the Earth’s surface. Coordinates are angular units. The canonical form of latitude and longitude representation uses degrees (°), minutes (′), and seconds (″). GPS systems widely use coordinates in degrees and decimal minutes, or in decimal degrees.

Latitude varies from −90° to 90°. The latitude of the Equator is 0°; the latitude of the South Pole is −90°; the latitude of the North Pole is 90°. Positive latitude values correspond to the geographic locations north of the Equator (abbrev. N). Negative latitude values correspond to the geographic locations south of the Equator (abbrev. S).

Longitude is counted from the prime meridian ( IERS Reference Meridian for WGS 84) and varies from −180° to 180°. Positive longitude values correspond to the geographic locations east of the prime meridian (abbrev. E). Negative longitude values correspond to the geographic locations west of the prime meridian (abbrev. W).

UTM or Universal Transverse Mercator coordinate system divides the Earth’s surface into 60 longitudinal zones. The coordinates of a location within each zone are defined as a planar coordinate pair related to the intersection of the equator and the zone’s central meridian, and measured in meters.

Elevation above sea level is a measure of a geographic location’s height. We are using the global digital elevation model GTOPO30 .

Elektrostal , Moscow Oblast, Russia

Zhukovsky International Airport

Zhukovsky International Airport, formerly known as Ramenskoye Airport or Zhukovsky Airfield - international airport, located in Moscow Oblast, Russia 36 km southeast of central Moscow, in the town of Zhukovsky, a few kilometers southeast of the old Bykovo Airport. After its reconstruction in 2014–2016, Zhukovsky International Airport was officially opened on 30 May 2016. The declared capacity of the new airport was 4 million passengers per year.

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Telecom churn case study where, Based on customer behavior (such as the monthly bill, internet usage, etc.) to predict whether a particular customer will switch to another telecom provider or not (i.e. churn or not). Visualised data and provided some insights on the same. The case study has the following files:
GitHub
In the telecom industry, customers are able to choose from multiple service providers and actively switch from one operator to another. In this highly competitive market, the telecommunications industry experiences an average of 15-25% annual churn rate.
Telecom Churn Prediction Group Case Study
To reduce customer churn, telecom companies need to predict which customers are at high risk of churn. In this project, you will analyze customer-level data of a leading telecom firm, build predictive models to identify customers at high risk of churn. In this competition, your goal is to build a machine learning model that is able to predict ...
Telecom Churn Case Study · GitHub
Telecom Churn Case Study. # Load the given files. # Collate the 3 files in a single file. # Understand the structure of the collated file. # Make bar charts to find interesting relationships between variables. # Make Box plots for numeric variables to look for outliers. # Impute the missing values, and perform the outlier treatment.
Case Study: Predict Customer Churn Using Machine Learning
We're told by our colleagues at the hypothetical company that customer churn is at 50% within 3 months. That means that within 3 months of a set of customers that sign up for the paid product, by the end of 3 months half of them will have cancelled. This is an urgent problem we need to help fix with machine learning!
Predicting churn in Telecom Industry
In this highly competitive market, the telecommunications industry experiences an average of 15-25% annual churn rate. Given the fact that it costs 5-10 times more to acquire a new customer than to retain an existing one, customer retention has now become even more important than customer acquisition. For many incumbent operators, retaining ...
PDF Telecom Customer Churn Analysis
churn currently ranges from 5% to 32% per year[11]. Usually if a customer leaves the company, he or she will not come back within one year. Thus an over 20% of the churn rate would be a heavy loss for the company. However, with telecom companies become more digitized, more insights have been generated.
Telecom Churn Case Study
If the issue persists, it's likely a problem on our side. Unexpected token < in JSON at position 4. keyboard_arrow_up. content_copy. SyntaxError: Unexpected token < in JSON at position 4. Refresh. Identify customers at high risk of churn and the main indicators of churn.
GitHub
To reduce customer churn, telecom companies need to predict which customers are at high risk of churn. In this project, we analysed customer-level data of a leading telecom firm, build predictive models to identify customers at high risk of churn and identify the main indicators of churn. You can also access the Live web-app here.
Elektrostal, Moscow Oblast, Russia
Elektrostal Geography. Geographic Information regarding City of Elektrostal. Elektrostal Geographical coordinates. Latitude: 55.8, Longitude: 38.45. 55° 48′ 0″ North, 38° 27′ 0″ East. Elektrostal Area. 4,951 hectares. 49.51 km² (19.12 sq mi) Elektrostal Altitude.
OOO Remstroy Construction Company
2000. Seven years of successful work have shown that combining different types of activities in the same company is not always convenient. And in the year 2000 the founders of OOO Firm ERG decided to create and register a monoprofile construction company OOO Remstroy Construction Company. Industrial construction was chosen as the priority area.
GitHub
Contribute to ashishc3090/Telecom-Churn-Group-Case-Study development by creating an account on GitHub.
Geographic coordinates of Elektrostal, Moscow Oblast, Russia
Geographic coordinates of Elektrostal, Moscow Oblast, Russia in WGS 84 coordinate system which is a standard in cartography, geodesy, and navigation, including Global Positioning System (GPS). Latitude of Elektrostal, longitude of Elektrostal, elevation above sea level of Elektrostal.
Case Study: Churn Prediction
Case Study: Churn Prediction 4 minute read Business Problem Overview. In the telecom industry, customers are able to choose from multiple service providers and actively switch from one operator to another. In this highly competitive market, the telecommunications industry experiences an average of 15-25% annual churn rate.
PDF Telecom-Churn-Case-Study/Upgrad+hackathon.pdf at main
Contribute to rmythili2/Telecom-Churn-Case-Study development by creating an account on GitHub.
Zhukovsky International Airport
Zhukovsky International Airport, formerly known as Ramenskoye Airport or Zhukovsky Airfield - international airport, located in Moscow Oblast, Russia 36 km southeast of central Moscow, in the town of Zhukovsky, a few kilometers southeast of the old Bykovo Airport. After its reconstruction in 2014-2016, Zhukovsky International Airport was officially opened on 30 May 2016.
GitHub
Contribute to Vishal9628712304/Telecom-Churn-Group-Case-Study development by creating an account on GitHub.
GitHub
Telecom Churn Case Study Problem Statement Business Problem Overview. In the telecom industry, customers are able to choose from multiple service providers and actively switch from one operator to another. In this highly competitive market, the telecommunications industry experiences an average of 15-25% annual churn rate.
Telecome-Churn/Telecom_Churn_Case_Study.ipynb at main
Contribute to indu1911/Telecome-Churn development by creating an account on GitHub.
Telecom Churn
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Releases · Sid205/telecom-churn-case-study · GitHub
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Pull requests · Sid205/telecom-churn-case-study · GitHub
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