Class Exercise 16: Chapter 17 Forecasting

Data Set 1

Time series data

weeks <- 1:6 
values <- c(17, 13, 15, 11, 17, 14)

Most recent value as forecast

forecasts_a <- values[-length(values)]
actual_a <- values[-1]
mse_a <- mean((actual_a - forecasts_a)^2)
mse_a

## [1] 16.2

forecasts_week_7_a <- tail(values, 1)
forecasts_week_7_a

## [1] 14

Result: mse_a = 16.2, forecasts_week_7_a = 14

Average all of the data as forecast

cumulative_avergaes <- cumsum(values[-length(values)])/ (1:(length(values) - 1))
cumulative_avergaes

## [1] 17.0 15.0 15.0 14.0 14.6

forecasts_b <- cumulative_avergaes
actual_b <- values[-1]
mse_b <- mean((actual_b - forecasts_b)^2)
mse_b

## [1] 8.272

forecasts_week_7_b <- tail(values, 1)
forecasts_week_7_b

## [1] 14

mae <-mean(abs(actual_a - forecasts_a))
mae

## [1] 3.8

mape <- mean(abs((actual_a - forecasts_a) / actual_a)) * 100
mape

## [1] 27.43778

Results: mse_b = 8.272, forecasts_week_7_b = 14, mae = 3.8, mape = 27.44

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,
  forecasts_week_7_most_recent = forecasts_week_7_a,
  MSE_average_of_all_data = mse_b,
  forecasts_week_7_average = forecasts_week_7_b,
  Better_Method = better_method
)

## $MSE_most_recent_value
## [1] 16.2
## 
## $forecasts_week_7_most_recent
## [1] 14
## 
## $MSE_average_of_all_data
## [1] 8.272
## 
## $forecasts_week_7_average
## [1] 14
## 
## $Better_Method
## [1] "Average of all Data"

Results:
$Better_Method = "Average of all Data"
Interpretation: The better method is the average of all data

Data Set 2

Install and load required libraries

#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)

## 
## Attaching package: 'zoo'

## The following objects are masked from 'package:base':
## 
##     as.Date, as.Date.numeric

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))

Time Series Plot

plot(df$month, df$data, type = "o", col = "blue", xlab = "Month", ylab = "Value of building contracts",
     main = "Monthly Value of Alabamy Building Contract in $")

Interpretation: This time series plot exhibits a seasonal pattern

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

df <- df %>%
  mutate(
    squared_error = ifelse(is.na(avg_value), NA, (data - avg_value) ^2)
  )

Compute MSE (exclude NA months)

mse <- mean((df$squared_error), na.rm = TRUE)
mse

## [1] 2040.444

Result: 2040.44

#Exponential smoothing

alpha = 0.2 
exp_smooth <- rep(NA, length(df$data))
exp_smooth[1] <- df$data[1] 
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] 2626350

Result: 2626350

Choosing Better Method

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] 2626350
## 
## $Better_Method
## [1] "Three-month Moving Average"

Interpretation: The better method is the Three-Month moving average

Data Set 3

#Install packages

# install.packages("ggplot2")
# install.packages("readxl")

library(readxl)
library(ggplot2)

Load the Data

df1 <- read_excel("Mortgage.xlsx")

Time plot series

ggplot(df1, aes(x = Year, y = Interest_Rate)) + 
  geom_line() + 
  geom_point() + 
  xlab("Year") + 
  ylab("Interest Rate") + 
  ggtitle("Time Series Plot of Mortgage Revenue")

Interpretation: We observe a trend patter in the time series plot

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: 6.70 - 0.13*Period

Forecast for period 25

forecast_period_25 <- predict(model, newdata = data.frame(Period = 25))
forecast_period_25

##        1 
## 3.472942

# Result: $3.47 interest rate