Class Exercise Objective
Figuring out the forecast according to the measurements provided.
Data - Identification of Values
months <- 1:6
values <- c(17, 13, 15, 11, 17, 14)
Step 1: Computation of the most recent value as the Forecast
forecasts_a <- values[-length(values)] # Exclude last value
actual_a <- values[-1] # Exclude 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 #Result = 14
## [1] 14
# Interpretations: The MSE (Mean Square Error) for the most recent value is 16.2 and the most recent forecast for week 7 is 14.
Calculating MAE and MAPE
# MAE =
mae <- mean(abs(actual_a - forecasts_a))
mae # Result = 3.8
## [1] 3.8
# MAPE =
mape <- mean(abs((actual_a - forecasts_a) / actual_a)) * 100
mape # Result = 27.44
## [1] 27.43778
Step 2: 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] # Excludes first value
mse_b <- mean((actual_b - forecasts_b)^2)
mse_b # Result = 8.27
## [1] 8.272
forecast_month_7_b <- mean(values) # Avg of all the months as forecast for month 7
forecast_month_7_b # Result = 14.5
## [1] 14.5
# Interpretation: The MSE for the average of the data is 8.27 and the average forecast of week 7 is 14.5.
Step 3: 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
)
## $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
##
## $Better_Method
## [1] "Average of All Data"
# Interpretation: Using the code above, we can interpret this as the average of all the data is the better method.
EXAMPLE 2: The Three-Month Moving Average and Exponential Smoothing
Forecast
Install Packages
options(repos = c(CRAN = "https://cran.rstudio.com"))
install.packages("dplyr")
##
## The downloaded binary packages are in
## /var/folders/03/0n7qr7yj4lngn931pnys1fxc0000gn/T//RtmpON16xl/downloaded_packages
install.packages("zoo")
##
## The downloaded binary packages are in
## /var/folders/03/0n7qr7yj4lngn931pnys1fxc0000gn/T//RtmpON16xl/downloaded_packages
library(dplyr)
##
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
##
## filter, lag
## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union
library(zoo)
##
## Attaching package: 'zoo'
## The following objects are masked from 'package:base':
##
## as.Date, as.Date.numeric
Import Data
df <- data.frame(month=c(1,2,3,4,5,6,7,8,9,10,11,12),
data=c(240, 352, 230, 260, 280, 322, 220, 310, 240, 310, 240, 230))
Step 1: Time Series Plot
plot(df$month, df$data, type = "o", col = "blue", xlab = "Month", ylab = "Values",
main = "Values of the building contracts")
#Interpretation: The plot shows to have a seasonal pattern and has certain peaks in certain months such as 2, 6, 8, 10.
Step 2: The Three-Month Moving Average
df$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$data[11]) / 3)
Find the Squared Error for the available months of the moving
average
df <- df %>%
mutate(
squared_error = ifelse(is.na(avg_value), NA, (data - avg_value)^2)
)
Compute the Mean Squared Error
mse <- mean(df$squared_error, na.rm = TRUE)
mse # Result = 2040.444
## [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 # Result = 2593.76
## [1] 2593.762
Comparison between the two; Three-Month Moving Avg. vs the
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 = 3 Month Moving Average
## $MSE_moving_average
## [1] 2040.444
##
## $MSE_exponential_smoothing
## [1] 2593.762
##
## $Better_Method
## [1] "Three-month moving average"
EXAMPLE 3: Average interest rate for a 30 year fixed rate mortgage
going over a 20 year period
library(readxl)
library(ggplot2)
# Load Data
df1 <- read_excel("/Users/salimh/Desktop/r markdown/Mortgage.xlsx")
Step 1: 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 The Fixed Mortgage Rate")
#Interpretation: The trend is decreasing in the first half, while heading towards the end, it has a sudden increase. The movements depicted are shown to be a cyclical period.
Step 2: Developing 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: The Linear Trend Equation is: Revenue = 6.70 - 0.13*Period
Step 3: The Forecast for Period 25
forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25 # Result = 3.47
## 1
## 3.472942
# Interpretation: As we use the linear trend equation in Step 2, we can forecast that the average interest rate will be 3.47% for period 25.