Basic Time Series Forecasting Methods

Project Objective

To forecast data using different time series methods.

Example 1: Naive Approach

Time series data

weeks <- 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 
## [1] 16.2
Result: Mean Squared Error A = 16.20
forecast_week_7_a <- tail(values, 1)
forecast_week_7_a
## [1] 14
Result: Forecast A for week 7 is 14.

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 
## [1] 8.272
Result: Mean Squared Error B = 8.27.
forecast_week_7_b <- mean(values) #average of all weeks as forecast for week 7
forecast_week_7_b 
## [1] 14.5
Result: Forecast B for week 7 is 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_week_7_most_recent = forecast_week_7_a,
  MSE_average_of_all_fata = mse_b,
  forecast_week_7_average = forecast_week_7_b,
  Better_Method = better_method
) 
## $MSE_most_recent_value
## [1] 16.2
## 
## $forecast_week_7_most_recent
## [1] 14
## 
## $MSE_average_of_all_fata
## [1] 8.272
## 
## $forecast_week_7_average
## [1] 14.5
## 
## $Better_Method
## [1] "Average of All Data"
Result: The better method is the Average of all data.

Part D. Finding Mean Absolute Error

mae <- mean(abs(actual_a - forecasts_a))
mae 
## [1] 3.8
Result: The mean absolute error is 3.80.

### Part E. Finding Mean Absolute Percentage Error


```r
mape <- mean(abs((actual_a - forecasts_a) / actual_a)) * 100
mape
## [1] 27.43778
Result: The mean absolute percentage error is 27.44%.

Example 2

Install and load packages

#install.packages("dplyr")
#install.packages("zoo")

library("dplyr")
## Warning: package 'dplyr' was built under R version 4.3.3
## 
## 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")
## Warning: package 'zoo' was built under R version 4.3.3
## 
## Attaching package: 'zoo'
## The following objects are masked from 'package:base':
## 
##     as.Date, as.Date.numeric

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 of Alabama Business Contracts",
     main = "Value of Alabama Building Contracts For A 12-month Period")

Interpretation: The time-series plot exhibits a seasonal pattern.

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 (exclusing the initial months with NA)

mse <- mean(df$squared_error, na.rm = TRUE)
mse
## [1] 2040.444
Result: The mean squared error for the three-month moving average is 2040.44.

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 
## [1] 2593.762
Result: The MSE for exponential smoothing is 2593.76.

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
) 
## $MSE_moving_average
## [1] 2040.444
## 
## $MSE_exponential_smoothing
## [1] 2593.762
## 
## $Better_Method
## [1] "Three-month moving average"
Result: The better method is the three-month moving average.

Example 3

Load packages

library("readxl")
library("ggplot2")
## Warning: package 'ggplot2' was built under R version 4.3.3

Load the data

df1 <- read_excel("C:/Users/Admin/OneDrive/Desktop/school/rsconnect/documents/Mortgage.xlsx")

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 FreddieMac Average Interest Rate")

Interpretation: We observe a cyclical pattern in the time series plot.

Part B. Develop a linear trend equation

model <- lm(Interest_Rate ~ Period, data <- df1)
summary(model)
## 
## Call:
## lm(formula = Interest_Rate ~ Period, data = 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: Interest Rate = 6.70 - 0.13*Period OR y = 6.70 - 0.13t

Part C. Forecast for period 25 (i.e., 2024)

forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25
##        1 
## 3.472942
Result: The forecast for period 25 is 3.47.