** Class Exercise 16: Basic Time Series Forecasting Methods**
Project Objective
To use naive method, moving average and exponential smoothing to calculate MSE
and forecast accuracy.
Example 1: Naive Apporach
Time Series data
months <- 1:6
values <- c(17, 13, 15, 11, 17, 14)
Part A. 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 #result: 16.2
## [1] 16.2
forecast_month_7_a <- tail(values, 1)
forecast_month_7_a # 14
## [1] 14
Calculate the mean absolute error and mean absolute percentage
error
# Calculate Mean Absolute Error (MAE)
mae <- mean(abs(actual_a - forecasts_a))
mae #3.8
## [1] 3.8
# Calculate Mean Absolute Percentage Error (MAPE)
mape <- mean(abs((actual_a - forecasts_a) / actual_a)) * 100
mape #27.44
## [1] 27.43778
Part B. Average all of 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 14.6
forecasts_b <- cumulative_averages
actual_b <- values[-1] #exclude the first value
mse_b <- mean((actual_b - forecasts_b)^2)
mse_b #8.27
## [1] 8.272
forecast_month_7_b <- mean(values) #average of all months as forecast for month 7
forecast_month_7_b # 14.5
## [1] 14.5
Part C. 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_month_7_most_recent = forecast_month_7_a,
MSE_average_of_all_data = mse_b,
forecast_month_7_average = forecast_month_7_b,
Better_Method = better_method
) #Result: Better method is the average of all data
## $MSE_most_recent_value
## [1] 16.2
##
## $forecast_month_7_most_recent
## [1] 14
##
## $MSE_average_of_all_data
## [1] 8.272
##
## $forecast_month_7_average
## [1] 14.5
##
## $Better_Method
## [1] "Average of All Data"
Example 2: Three-month moving average and exponential smoothing
forecast
Install some packages
options(repos = c(CRAN = "https://cran.rstudio.com"))
# Install packages
install.packages("dplyr")
##
## The downloaded binary packages are in
## /var/folders/5f/2vqvsqc91055s6nc4m0ldk2r0000gn/T//RtmpOFEZFQ/downloaded_packages
install.packages("zoo")
##
## The downloaded binary packages are in
## /var/folders/5f/2vqvsqc91055s6nc4m0ldk2r0000gn/T//RtmpOFEZFQ/downloaded_packages
# Load libraries
library(dplyr)
##
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
##
## filter, lag
## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union
library(zoo)
##
## Attaching package: 'zoo'
## The following objects are masked from 'package:base':
##
## as.Date, as.Date.numeric
Import the data
df <- data.frame(month=c(1,2,3,4,5,6,7,8,9,10,11,12),
data=c(240, 352, 230, 260, 280, 322, 220, 310, 240, 310, 240, 230))
Part A. Time series plot
plot(df$month, df$data, type = "o", col = "blue", xlab = "Month", ylab = "Values",
main = "The values of Alabama building contracts")
Interpretation: The time-series plot exhibits a seasonal pattern, with peak values in month 2, 6, 8 and 10 and it seems to repeat monthly.
Part B. Three-month moving average
df$avg_value <- 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$data[11]) / 3)
Calculate the squared errors( only for the months where moving
average is available)
df <- df %>%
mutate(
squared_error = ifelse(is.na(avg_value), NA, (data - avg_value)^2)
)
Compute MSE (excluding the initial months with NA)
mse <- mean(df$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
Compare the three-month moving average and exponential
smoothing
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"
Example 3: The average interest rate (%) for a 30-year fixed-rate
mortgage over a 20-year period
Load the packages and data
library(readxl)
library(ggplot2)
# Load the data
df1 <- read_excel("Mortgage.xlsx")
Part A. Time series plot
ggplot(df1, aes(x = Period, y = Interest_Rate))+
geom_line() +
geom_point() +
xlab("Period") +
ylab("Interest_Rate (%") +
ggtitle("Time Series Plot of Mortgage")
Interpretation: We observe a decreasing trend in the first half of time series plot with a sharp increase towards the latest period. It could be a cyclical pattern.
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
Interpretation: linear trend equation: Revenue = 6.70 - 0.13*Period
Part C. Forecast for period 25
forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25 # 3.47
## 1
## 3.472942
Interpretation: Using the linear trend equation from question 3B, we can
forecast the average interest rate for period 25 is 3.47.