Basic Time Series Forecasting Models
##Project Objective
Look at three examples using basic time series forecasting methods.Question 1: Naive Approach, Part a-d:
a.Most recent value as forecast/Mean Absolute Error?
b.Average all of the data as forecast/Mean Squared Error?
c.Which method is better/Mean Absolute Percentage Error?
d.What is the forecast for week 7?
Step 0: Import the data
weeks <- 1:6
values <- c(17, 13, 15, 11, 17, 14)
ts_data <- data.frame(Week = weeks, Value = values)Step 1: Most Recent Value as Forecast & Mean Absolute Error
#Most Recent Value as Forecast
forecasts_a <- values[-length(values)] #exclude the last value
actual_a <- values[-1] #exclude the first value
mse_a <- mean((actual_a - forecasts_a)^2)
mse_a #16.2## [1] 16.2forecast_week_7_a <- tail(values, 1)
forecast_week_7_a #14## [1] 14#Mean Absolute Error
absolute_errors <- abs(forecasts_a - actual_a)
mae <- mean(absolute_errors)
mae## [1] 3.8###Step 2: Average all of the data as forecast & Mean Squared Error
#Average of all the data as forecast
cumulative_averages <- cumsum(values[-length(values)]) / (1:(length(values) - 1))
cumulative_averages## [1] 17.0 15.0 15.0 14.0 14.6forecasts_b <- cumulative_averages
actual_b <- values[-1] #exclude the first value
mse_b <- mean((actual_b - forecasts_b)^2)
mse_b #8.272## [1] 8.272forecast_week_7_b <- tail(values, 1) #average of all weeks as forecast for week 7
forecast_week_7_b #14## [1] 14#Mean squared error
naive_forecast <- ts_data$Value[length(ts_data$Value)]
mse <- mean((ts_data$Value - naive_forecast)^2)
mse## [1] 4.833333###Step 3: Which method is better & Mean Absolute Percentage Error
#Which method is better
better_method <- ifelse(mse_a < mse_b, "Most Recent Value", "Average of All Data")
list(
MSE_most_recent_value = mse_a,
forecast_week_7_a_most_recent = forecast_week_7_a,
MSE_average_of_all_data = mse_b,
forecast_week_7_average = forecast_week_7_b,
Better_Method = better_method
) #Result: Average of all data## $MSE_most_recent_value
## [1] 16.2
##
## $forecast_week_7_a_most_recent
## [1] 14
##
## $MSE_average_of_all_data
## [1] 8.272
##
## $forecast_week_7_average
## [1] 14
##
## $Better_Method
## [1] "Average of All Data"#Mean Absolute Percentage Error
mae_a <- mean(abs(actual_a - forecasts_a))
mae_a## [1] 3.8Question 2: Part a & b
a. Construct a time series plot. What type of pattern exists in the data?
b. Compare the three-month moving average approach wth the exponential smoothing forecast using alpha = 0.2. Which provides more accurate forecasts based on MSE?
Step 1: Install & load required libraries
# Set CRAN mirror
options(repos = "https://cloud.r-project.org")
#Install packages
install.packages("dplyr") ##
## The downloaded binary packages are in
## /var/folders/lm/7q9jv5rx5gz0mnz2f3p9v59c0000gn/T//RtmpusjeyX/downloaded_packagesinstall.packages("zoo")##
## The downloaded binary packages are in
## /var/folders/lm/7q9jv5rx5gz0mnz2f3p9v59c0000gn/T//RtmpusjeyX/downloaded_packages#Load libraries
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, unionlibrary(zoo)##
## Attaching package: 'zoo'## The following objects are masked from 'package:base':
##
## as.Date, as.Date.numericStep 2: Import the data
df <- data.frame(month=c(1,2,3,4,5,6,7,8,9,10,11,12),
data=c(240, 352, 230, 260, 280, 322, 220, 310, 240, 310, 240, 230))###Step 3: Construct a time series plot
#Part A. Time Series Plot
plot(df$month, df$data, type = "o", col = "red", xlab = "Month", ylab = "Building Contracts",
main = "Alabama Building Contracts for a 12-month period")
#Interpretation: The time-series plot exhibits a horizontal pattern###Step 4: Compare the three-month moving average approach wth the exponential smoothing forecast using alpha = 0.2.
#Part B. Three-Month Moving Average
df$avg_contract <- c(NA, NA, NA,
(df$data[1] + df$data[2] + df$data[3]) / 3,
(df$data[2] + df$data[3] + df$data[4]) / 3,
(df$data[3] + df$data[4] + df$data[5]) / 3,
(df$data[4] + df$data[5] + df$data[6]) / 3,
(df$data[5] + df$data[6] + df$data[7]) / 3,
(df$data[6] + df$data[7] + df$data[8]) / 3,
(df$data[7] + df$data[8] + df$data[9]) / 3,
(df$data[8] + df$data[9] + df$data[10]) / 3,
(df$data[9] + df$data[10] + df$data[11]) / 3)
#Calculate the squared errors (only for the months where moving average is available)
df <- df %>%
mutate(
squared_error = ifelse(is.na(avg_contract), NA, (data - avg_contract)^2)
)
#Compute the MSE (excluding the initial months with NA)
mse <- mean(df$squared_error, na.rm = TRUE)
mse #2040.44## [1] 2040.444#Exponential Smoothing
alpha = 0.2
exp_smooth <- rep(NA, length(df$data))
exp_smooth[1] <- df$data[1] #Starting point
for(i in 2:length(df$data)) {
exp_smooth[i] <- alpha * df$data[i-1] + (1 - alpha) * exp_smooth[i-1]
}
mse_exp_smooth <- mean((df$data[2:12] - exp_smooth[2:12])^2)
mse_exp_smooth #2593.76## [1] 2593.762###Step 5: Compare MSE
#Compare
better_method <- ifelse(mse < mse_exp_smooth, "Three-Month Moving Average", "Exponential Smoothing")
list(
MSE_moving_average = mse,
MSE_exponential_smoothing = mse_exp_smooth,
Better_Method = better_method
) #Result: Better Method - Three-Month Moving Average## $MSE_moving_average
## [1] 2040.444
##
## $MSE_exponential_smoothing
## [1] 2593.762
##
## $Better_Method
## [1] "Three-Month Moving Average"Question 3: Part a-c:
##a. Construct a time series plot. What type of pattern exists in the data? ##b. Develop the linear trend equation for this time series ##c. Using the linear trend equation from b, forecast the average interest rate for period 25
Step 1: Install & load required libraries
library(readxl)
library(ggplot2)Step 2: Import the data
df1 <- read_excel("/Users/ethanma/Desktop/R markdown 2/Mortgage.xlsx")Step 3: Time Series Plot
#Part A. Time Series Plot
ggplot(df1, aes(x= Year, y = Interest_Rate)) +
geom_line() +
geom_point() +
xlab("Year") +
ylab("Interest_Rate") +
ggtitle("Time Series Plot of Interest Rates for a 30-Year Fixed-Rate Mortgage Over a 20-Year Period")
#Interpretation: We observe a decreasing pattern or trend in the time series plot from 2000 to 2021,
#followed by an increasing pattern or trend from 2021 to 2023###Step 4: Linear Trend Equation
#Part B. Develop a Linear Trend Equation
model <- lm(Interest_Rate ~ Period, data = df1)
summary(model)##
## Call:
## lm(formula = Interest_Rate ~ Period, data = df1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.3622 -0.7212 -0.2823 0.5015 3.1847
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.69541 0.43776 15.295 3.32e-13 ***
## Period -0.12890 0.03064 -4.207 0.000364 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.039 on 22 degrees of freedom
## Multiple R-squared: 0.4459, Adjusted R-squared: 0.4207
## F-statistic: 17.7 on 1 and 22 DF, p-value: 0.0003637#Result - linear trend equation: Revenue = 6.7 + (-0.13)*Period OR y = 6.7 + (-0.13)t#Step 5: Forecast the average interest rate for period 25
#Part C. Forecasat for period 25
forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25 #3.47% average interest rate for period 25## 1
## 3.472942