Project Objective
Learn how to use various forecasting techniques and forecast accuracy equations to solve various time series forecasting issues in R programming.
Question 1: Using the naive method (most recent value) as the
forecast for the next week, compute the following measures of forecast
accuracy.
Part A: Solve for mean absolute error (MAE)
Step 1: Load time series data
week <- 1:10 # This is the independent variable ~ time
values <- c(17,13,15,11,17,14) # Dependent variable
Step 2: Calculate Actual and forcasted values
forecast <- values[-length(values)] #Excludes the last value
actual <- values[-1] # Excludes the first sale
Step 3: Calculate Mean absolute error
mae <- mean(abs(actual - forecast))
mae # Mean absolute error is 3.8
## [1] 3.8
Part B: Solve for Mean Square Error (MSE)
mse <- mean((actual - forecast)^2)
mse # Mean square error is 16.2
## [1] 16.2
Part C: Solve for Mean absolute (MAPE) percentage error
mape <- mean(abs((actual - forecast) / actual)*100)
mape # Mean absolute percentage error is 27.44%
## [1] 27.43778
Part D: What is the forecast for week 7?
forecast_week7 <- tail(values, 1)
forecast_week7 # value = 14
## [1] 14
# interpretation: the value projected at week 7 in this time series data using the naive approach is 14
Question 2: The values of Alabama building contracts (in $ millions)
for a 12-month period is as follows: 240, 352, 230, 260, 280, 322, 220,
310, 240, 310, 240, 230. Use this data to answer question 2A and 2B
below.
Part A: Construct a time series plot. What type of pattern exists in
the data?
Step 1: Install and load packages
# install.packages("dplyr")
# install.packages("zoo")
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)
## Warning: package 'zoo' was built under R version 4.4.2
##
## Attaching package: 'zoo'
## The following objects are masked from 'package:base':
##
## as.Date, as.Date.numeric
Step 2: Import the data
df <- data.frame(month=c(1,2,3,4,5,6,7,8,9,10,11,12), values=c(240,352,230,260,280,322,220,310,240,310,240,230))
Step 3: Descrpitive statistics
summary(df)
## month values
## Min. : 1.00 Min. :220.0
## 1st Qu.: 3.75 1st Qu.:237.5
## Median : 6.50 Median :250.0
## Mean : 6.50 Mean :269.5
## 3rd Qu.: 9.25 3rd Qu.:310.0
## Max. :12.00 Max. :352.0
Interpretation: the average value of a contract over a 12 month period is 269.5 million
Step 4: Create time series plot
plot(df$month, df$values, type = "o", col = "pink", xlab = "Month", ylab = "Values (In Millions)", main = "Alabama Building Contracts Plot")
Interpretation: the time series plot exhibits a horizontal pattern as it is steady on the mean of 269.5
Part B: Compare the three-month moving average approach with the
exponential smoothing forecast using alpha = 0.2. Which provides more
accurate forecasts based on MSE?
Step 1: Manually calculate the Thee-Month Moving Average
df$avg_value3 <- c(NA, NA, NA,
(df$values[1] + df$values[2] + df$values[3]) / 3,
(df$values[2] + df$values[3] + df$values[4]) / 3,
(df$values[3] + df$values[4] + df$values[5]) / 3,
(df$values[4] + df$values[5] + df$values[6]) / 3,
(df$values[5] + df$values[6] + df$values[7]) / 3,
(df$values[6] + df$values[7] + df$values[8]) / 3,
(df$values[7] + df$values[8] + df$values[9]) / 3,
(df$values[8] + df$values[9] + df$values[10]) / 3,
(df$values[9] + df$values[10] + df$values[11]) / 3
)
Step 2: Calculate the Squared errors (only for months where moving
average is behavioral)
df <- df %>%
mutate(
squared_error = ifelse(is.na(avg_value3), NA, (values - avg_value3)^2)
)
Step 3: Compute MSE (exluding the initial months with NA)
mse <- mean(df$squared_error, na.rm = TRUE)
mse
## [1] 2040.444
Step 4: Calculate MSE using Exponential Smoothing
alpha <- 0.2
exp_smooth <- rep(NA, length(df$values))
exp_smooth[1] <- df$values[1] #starting point
for(i in 2: length(df$values)) {
exp_smooth[i] <- alpha * df$values[i-1] + (1 - alpha) * exp_smooth[i-1]
}
mse_exp_smooth <- mean((df$values[2:12] - exp_smooth[2:12])^2)
mse_exp_smooth
## [1] 2593.762
Step 5: Comparison of both methods
better_method <- ifelse(mse < mse_exp_smooth, "Three-Month Moving Average", "Exponential
Smoothing")
Step 6: Results of Comparison
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"
Interpretatiopn: As the code displays, the three-month moving average is the more accurate forecast based on MSE. The primary objective of this model is achieved by the three-month moving average, which has a smaller mean squared error (MSE): The three-month moving average performs better than the exponential smoothing forecast in terms of minimizing error to produce a more accurate forecast.
Question 3: The following data shows the average interest rate (%)
for a 30-year fixed-rate mortgage over a 20-year period (FreddieMac
website).
Part A: Construct a time series plot. What type of pattern exists in
the data?
Step 1: Install and load libraries
#install.packages("ggplot2")
#install.packages("readxl")
library(readxl)
library(ggplot2)
Step 2: Load the data
df <- read_excel("Mortgage.xlsx")
Step 3: Descriptive statistics
summary(df)
## Year Period Interest_Rate
## Min. :2000-01-01 00:00:00 Min. : 1.00 Min. :2.958
## 1st Qu.:2005-10-01 18:00:00 1st Qu.: 6.75 1st Qu.:3.966
## Median :2011-07-02 12:00:00 Median :12.50 Median :4.863
## Mean :2011-07-02 18:00:00 Mean :12.50 Mean :5.084
## 3rd Qu.:2017-04-02 06:00:00 3rd Qu.:18.25 3rd Qu.:6.105
## Max. :2023-01-01 00:00:00 Max. :24.00 Max. :8.053
Interpretation: On average the interest rate across the 25 year period is 5.08%
Step 4: Construct a time series plot
ggplot(df, aes(x = Period, y = Interest_Rate)) +
geom_line() +
geom_point() +
xlab("Period (In Years)") +
ylab("Interest rate") +
ggtitle("Time Series Plot of Interest Rates")
Interpretation: We can observe an decreasing pattern or trend in the time series plot
Part B: Develop the linear trend equation for this time series.
Step 1: Develop linear trend equation for this time series
model <- lm(Interest_Rate ~ Period, data = df)
summary(model)
##
## Call:
## lm(formula = Interest_Rate ~ Period, data = df)
##
## 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 ~ estimated linear trend equation: Interest Rates = 6.70 - 0.13*period. The R-squared is 0.45 (Moderately fits the data) so the overall model is significant as p-value < 0.05
Step 2: Find the MSE & MAPE values : Step Calculate the fitted
values from the model
df$predicted_IntRates <- predict(model)
Step 3: Calculate the residuals
df$residuals <- df$Interest_Rate - df$predicted_IntRates
Step 4: Calculate the Mean squared error (MSE)
mse <- mean(df$residuals^2)
cat("Mean Squarred Error (MSE):", mse, "\n")
## Mean Squarred Error (MSE): 0.989475
Step 5: Calculate Mean Absolute Percentage Error (MAPE)
df$percentage_error <- abs(df$residuals / df$Interest_Rate) * 100
mape <- mean(df$percentage_error)
cat("Mean Absolute Percentage Error (MAPE):", mape, "%\n")
## Mean Absolute Percentage Error (MAPE): 15.79088 %
Part C: Using the linear trend equation from question 3B, forecast
the average interest rate for period 25 (i.e., 2024).
forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25
## 1
## 3.472942
Interpretation: The forecasted interest rates in year 25 is 3.47%