MBA678 - Predictive Analytics
Helen Smith
March 5, 2018
knitr::opts_chunk$set(echo = TRUE)Assignment 5, Problem 1
Data set books contains the daily sales of paperback and hardcover books at the same store. The task is to forecast the next four days’ sales for paperback and hardcover books (data set books).
# Load libraries and set environment options
#library(dplyr)
#library(tidyr)
library(knitr)
#library(readxl)
library(forecast)
library(ggplot2)
#library(zoo)
library(fpp)
# Use this option to supress scientific notation in printing values
options(scipen = 10, digits = 2)str(books)## Time-Series [1:30, 1:2] from 1 to 30: 199 172 111 209 161 119 195 195 131 183 ...
## - attr(*, "dimnames")=List of 2
## ..$ : NULL
## ..$ : chr [1:2] "Paperback" "Hardcover"summary(books)## Paperback Hardcover
## Min. :111 Min. :128
## 1st Qu.:167 1st Qu.:170
## Median :189 Median :200
## Mean :186 Mean :199
## 3rd Qu.:207 3rd Qu.:222
## Max. :247 Max. :283head(books)## Time Series:
## Start = 1
## End = 6
## Frequency = 1
## Paperback Hardcover
## 1 199 139
## 2 172 128
## 3 111 172
## 4 209 139
## 5 161 191
## 6 119 168tail(books)## Time Series:
## Start = 25
## End = 30
## Frequency = 1
## Paperback Hardcover
## 25 190 214
## 26 182 200
## 27 222 201
## 28 217 283
## 29 188 220
## 30 247 259a. Plot the series and discuss the main features of the data.
# Take a quick look at each time series
autoplot(books) + xlab("day") + ylab("count")
Based on an initial view of the plot there seems to be an upward trend to both the paperback and hardcover sales volume for the month. Since we only have one month’s worth of data it is hard to see if there might be monthly, quarterly, or annual seasonality. There could be some weekly seasonality. A decomposition may help identify if any exists.
# Create a seasonal (weekly) time series and see if a decomposition will give any more clues to the data
books_weeklyts <- ts(books, frequency=7)
print("Decomposition of Paperback data")## [1] "Decomposition of Paperback data"plot(decompose(books_weeklyts[,1]))
print("Decomposition of Hardcover data")## [1] "Decomposition of Hardcover data"plot(decompose(books_weeklyts[,2]))
Decomposition shows a definite upward trend for both paperback and hardcover sales for the month. There could be some weekly seasonality in each, but for this exercise we will assume none.
b. Use simple exponential smoothing with the ses function (setting initial=“simple”) and explore different values of alpha for the paperback series. Record the within-sample SSE for the one-step forecasts. Plot SSE against alpha and find which value of alpha works best. What is the effect of alpha on the forecasts?
We will start with the SES function for the paperback data
# Create vectors for setting values for alpha and SSEs, for paperback
alphas<-numeric()
SSEs <- numeric()
for(i in seq(0, 1, 0.05)) {
alphas <-c(alphas, i)
s <-ses(pback, initial="simple", alpha=i, h=4)
SSEs<-c(SSEs, s$model$SSE)
}
# Create data frames for paperback with alphas and SSEs, then show values and plot
pbackDF<-data.frame(alphas, SSEs)
pbackDF## alphas SSEs
## 1 0.00 41270
## 2 0.05 39245
## 3 0.10 37785
## 4 0.15 36738
## 5 0.20 36329
## 6 0.25 36438
## 7 0.30 36931
## 8 0.35 37716
## 9 0.40 38738
## 10 0.45 39967
## 11 0.50 41384
## 12 0.55 42977
## 13 0.60 44743
## 14 0.65 46675
## 15 0.70 48774
## 16 0.75 51035
## 17 0.80 53456
## 18 0.85 56035
## 19 0.90 58769
## 20 0.95 61655
## 21 1.00 64690plot(pbackDF, main="Alpha Values for Paperback Data")
Looking at the output from the SSE function, and the plot of SSE, it appears that the best value for alpha is around 0.20.
The smaller the value of alpha, the more weight that is placed on more recent data.
c. Now let ses select the optimal value of alpha. Use this value to generate forecasts for the next four days. Compare your results with 2.
# let SES select the optimal value for alpha for paperback using "simple" for the initial value, then generate a forecast for the next 4 days.
optpbackSimple <- ses(pback, initial="simple", h=4)
summary(optpbackSimple)##
## Forecast method: Simple exponential smoothing
##
## Model Information:
## Simple exponential smoothing
##
## Call:
## ses(y = pback, h = 4, initial = "simple")
##
## Smoothing parameters:
## alpha = 0.2125
##
## Initial states:
## l = 199
##
## sigma: 35
## Error measures:
## ME RMSE MAE MPE MAPE MASE ACF1
## Training set 1.7 35 29 -2.8 17 0.72 -0.13
##
## Forecasts:
## Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
## 31 210 166 255 142 278
## 32 210 165 256 140 280
## 33 210 164 257 139 281
## 34 210 163 258 137 283The SES function chose an alpha of 0.2125, very close to the estimate of 0.20. And the forecast for the next four days is 210 paperback books per day. SES does not take the upward trend into consideration for the forecast.
d. Repeat but with initial=“optimal”. How much difference does an optimal initial level make?
# Now, run the forecast for paperback using "optimal" for the initial value, then generate a forecast for the next 4 days.
optpbackOptimal <- ses(pback, initial="optimal", h=4)
summary(optpbackOptimal)##
## Forecast method: Simple exponential smoothing
##
## Model Information:
## Simple exponential smoothing
##
## Call:
## ses(y = pback, h = 4, initial = "optimal")
##
## Smoothing parameters:
## alpha = 0.1685
##
## Initial states:
## l = 170.8257
##
## sigma: 34
##
## AIC AICc BIC
## 319 320 323
##
## Error measures:
## ME RMSE MAE MPE MAPE MASE ACF1
## Training set 7.2 34 28 0.47 16 0.7 -0.21
##
## Forecasts:
## Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
## 31 207 164 250 141 273
## 32 207 163 251 140 274
## 33 207 163 251 139 275
## 34 207 162 252 138 276Using the SES function with the initial parameter set to “optimal”, SES chose an alpha of 0.1685. This is still relatively close to the earlier estimate of 0.20. And the forecast for the next four days is 207 paperback books per day. This lower forecast is the result of additional weighting on the older data values in the series with the smaller alpha value.
e. Repeat steps (b)-(d) with the hardcover series.
First, use SSE to generate values of different alphas.
# Create vectors for setting values for alpha and SSEs, for hardcover
alphas<-numeric()
SSEs <- numeric()
for(i in seq(0, 1, 0.05)) {
alphas <-c(alphas, i)
s <-ses(hcover, initial="simple", alpha=i, h=4)
SSEs<-c(SSEs, s$model$SSE)
}
# Create data frames for hardcover with alphas and SSEs, then show values and plot
hcoverDF<-data.frame(alphas, SSEs)
hcoverDF## alphas SSEs
## 1 0.00 154503
## 2 0.05 70483
## 3 0.10 45715
## 4 0.15 36814
## 5 0.20 33148
## 6 0.25 31554
## 7 0.30 30910
## 8 0.35 30758
## 9 0.40 30895
## 10 0.45 31224
## 11 0.50 31703
## 12 0.55 32314
## 13 0.60 33060
## 14 0.65 33948
## 15 0.70 34994
## 16 0.75 36217
## 17 0.80 37642
## 18 0.85 39295
## 19 0.90 41210
## 20 0.95 43423
## 21 1.00 45982plot(hcoverDF, main="values for alpha for hardcover data")
For Hardcover book sales, the best alpha looks to be around 0.35.
Now, run the SES function with initial set to “simple”.
# let SES select the optimal value for alpha for hardcover using "simple" for the initial value, then generate a forecast for the next 4 days.
opthcoverSimple <- ses(hcover, initial="simple", h=4)
summary(opthcoverSimple)##
## Forecast method: Simple exponential smoothing
##
## Model Information:
## Simple exponential smoothing
##
## Call:
## ses(y = hcover, h = 4, initial = "simple")
##
## Smoothing parameters:
## alpha = 0.3473
##
## Initial states:
## l = 139
##
## sigma: 32
## Error measures:
## ME RMSE MAE MPE MAPE MASE ACF1
## Training set 9.7 32 26 3.1 13 0.79 -0.16
##
## Forecasts:
## Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
## 31 240 199 281 178 303
## 32 240 197 284 174 307
## 33 240 195 286 170 310
## 34 240 192 288 167 314As expected, the alpha came in very close to 0.35, at 0.3473. The simple forecast is for 240 hardcover books sold each day for the next four days.
Now, run SES with initial set to “optimal”.
# Now, run the forecast for hardcover using "optimal" for the initial value, then generate a forecast for the next 4 days.
opthcoverOptimal <- ses(hcover, initial="optimal", h=4)
summary(opthcoverOptimal)##
## Forecast method: Simple exponential smoothing
##
## Model Information:
## Simple exponential smoothing
##
## Call:
## ses(y = hcover, h = 4, initial = "optimal")
##
## Smoothing parameters:
## alpha = 0.3283
##
## Initial states:
## l = 149.2836
##
## sigma: 32
##
## AIC AICc BIC
## 316 317 320
##
## Error measures:
## ME RMSE MAE MPE MAPE MASE ACF1
## Training set 9.2 32 27 2.6 13 0.8 -0.14
##
## Forecasts:
## Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
## 31 240 199 280 177 302
## 32 240 196 283 174 305
## 33 240 194 285 171 309
## 34 240 192 287 168 312Using an initial setting of “optimal” generates an alpha of 0.3283, but the same 240 books per day for the next four days. The change in alpha was not significant enough to change the forecast, MAPE, or RMSE.
Assignment 5, Problem 2
Apply Holt’s linear method to the paperback and hardback series and compute four-day forecasts in each case.
a. Compare the SSE measures of Holt’s method for the two series to those of simple exponential smoothing in the previous question. Discuss the merits of the two forecasting methods for these data sets.
Paperback
# paperback Simple
pbackHS <- holt(pback, initial="simple", h=4)
summary(pbackHS)##
## Forecast method: Holt's method
##
## Model Information:
## Holt's method
##
## Call:
## holt(y = pback, h = 4, initial = "simple")
##
## Smoothing parameters:
## alpha = 0.2984
## beta = 0.4984
##
## Initial states:
## l = 199
## b = -27
##
## sigma: 40
## Error measures:
## ME RMSE MAE MPE MAPE MASE ACF1
## Training set 7.8 40 34 1.6 18 0.85 -0.11
##
## Forecasts:
## Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
## 31 222 171 273 145 300
## 32 230 165 294 131 329
## 33 237 145 330 96 378
## 34 245 116 375 47 443Using Holt’s method with initial set to simple, the forecast for the next four days is significantly higher than the SES forecast. The Holt method also reflects the upward trend in the four day forecast.
# paperback optimal
pbackH <- holt(pback, initial="optimal", h=4)
summary(pbackH)##
## Forecast method: Holt's method
##
## Model Information:
## Holt's method
##
## Call:
## holt(y = pback, h = 4, initial = "optimal")
##
## Smoothing parameters:
## alpha = 0.0001
## beta = 0.0001
##
## Initial states:
## l = 174.6003
## b = 1.0606
##
## sigma: 32
##
## AIC AICc BIC
## 319 322 326
##
## Error measures:
## ME RMSE MAE MPE MAPE MASE ACF1
## Training set -4.5 32 26 -6 16 0.67 -0.14
##
## Forecasts:
## Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
## 31 207 166 248 145 269
## 32 208 168 249 146 270
## 33 209 169 250 147 271
## 34 210 170 251 148 272With Initial set to optimal, the forecast is closer to the forecast for both the SES simple and optimal forecasts.
Hardcover
# hardcover Smiple
hcoverHS <- holt(hcover, initial="simple", h=4)
summary(hcoverHS)##
## Forecast method: Holt's method
##
## Model Information:
## Holt's method
##
## Call:
## holt(y = hcover, h = 4, initial = "simple")
##
## Smoothing parameters:
## alpha = 0.439
## beta = 0.1574
##
## Initial states:
## l = 139
## b = -11
##
## sigma: 35
## Error measures:
## ME RMSE MAE MPE MAPE MASE ACF1
## Training set 7.2 35 28 2.4 14 0.84 -0.077
##
## Forecasts:
## Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
## 31 251 206 296 182 319
## 32 255 202 307 175 335
## 33 259 196 321 163 354
## 34 263 188 337 149 377Using Holt’s method with initial set to simple and optimal both generated forecasts for the next four days higher than the SES model.
Again, the Holt method reflected the upward trend in the four day forecast.
# hardcover Optimal
hcoverH <- holt(hcover, initial="optimal", h=4)
summary(hcoverH)##
## Forecast method: Holt's method
##
## Model Information:
## Holt's method
##
## Call:
## holt(y = hcover, h = 4, initial = "optimal")
##
## Smoothing parameters:
## alpha = 0.0001
## beta = 0.0001
##
## Initial states:
## l = 154.6543
## b = 2.9961
##
## sigma: 27
##
## AIC AICc BIC
## 311 313 318
##
## Error measures:
## ME RMSE MAE MPE MAPE MASE ACF1
## Training set -2.2 27 23 -3.4 12 0.7 -0.027
##
## Forecasts:
## Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
## 31 247 212 283 194 301
## 32 250 215 286 197 304
## 33 253 218 288 200 307
## 34 256 221 291 203 310With Initial set to optimal, the forecast is closer to the forecast for both the SES simple and optimal forecasts.
b. Compare the forecasts for the two series using both methods. Which do you think is best?
Compare the SSE measures of Holt’s method for paperbacks and hardcovers with those of the SES models
Paperback
# Paperback plot of all forecasts
autoplot(pback) + xlab("day") + ylab("Sales") + autolayer(optpbackSimple, PI=FALSE, series="SES Simple") + autolayer(optpbackOptimal, PI=FALSE, series="SES optimal") + autolayer(pbackH, series="Holt Optimal", PI=FALSE) + autolayer(pbackHS, series="Holt Simple", PI=FALSE)
Comparing the simple and optimal forecasts using both SES and Holt’s methods, I would recommend the Holt optimal forecast. This forecast reflects the upward trend that the SES forecasts don’t, and it puts adequate weight on the earlier data.
Running the accuracy function confirms that Holt’s optimal forecast has the lowest MAPE and RMSE.
# Run Accuracy check for forecasts
print("Paperback - SES Simple")## [1] "Paperback - SES Simple"accuracy(optpbackSimple$fitted, pback)## ME RMSE MAE MPE MAPE ACF1 Theil's U
## Test set 1.7 35 29 -2.8 17 -0.13 0.68print("Paperback - SES Optimal")## [1] "Paperback - SES Optimal"accuracy(optpbackOptimal$fitted, pback)## ME RMSE MAE MPE MAPE ACF1 Theil's U
## Test set 7.2 34 28 0.47 16 -0.21 0.67print("Paperback - Holt Simple")## [1] "Paperback - Holt Simple"accuracy(pbackHS$fitted, pback)## ME RMSE MAE MPE MAPE ACF1 Theil's U
## Test set 7.8 40 34 1.6 18 -0.11 0.88print("Paperback - Holt Optimal")## [1] "Paperback - Holt Optimal"accuracy(pbackH$fitted, pback)## ME RMSE MAE MPE MAPE ACF1 Theil's U
## Test set -4.5 32 26 -6 16 -0.14 0.61Hardcover
# Hardcover plot of all forecasts
autoplot(hcover) + xlab("day") + autolayer(opthcoverSimple, PI=FALSE, series="SES Simple") + autolayer(opthcoverOptimal, PI=FALSE, series="SES optimal") + autolayer(hcoverH, series="Holt Optimal", PI=FALSE) + autolayer(hcoverHS, series="Holt Simple", PI=FALSE)
Comparing the simple and optimal forecasts using both SES and Holt’s methods for hardcover, I would also recommend the Holt optimal forecast, for the same reason. NOTE, the SES optimal and Simple both generated the same forecast.
Running the accuracy function confirms that Holt’s optimal forecast has the lowest MAPE and RMSE.
# Run Accuracy check for forecasts
print("Hardcover - SES Simple")## [1] "Hardcover - SES Simple"accuracy(opthcoverSimple$fitted, hcover)## ME RMSE MAE MPE MAPE ACF1 Theil's U
## Test set 9.7 32 26 3.1 13 -0.16 0.81print("Hardcover - SES Optimale")## [1] "Hardcover - SES Optimale"accuracy(opthcoverOptimal$fitted, hcover)## ME RMSE MAE MPE MAPE ACF1 Theil's U
## Test set 9.2 32 27 2.6 13 -0.14 0.81print("Hardcover - Holt Simple")## [1] "Hardcover - Holt Simple"accuracy(hcoverHS$fitted, hcover)## ME RMSE MAE MPE MAPE ACF1 Theil's U
## Test set 7.2 35 28 2.4 14 -0.077 0.92print("Hardcover - Holt Optimal")## [1] "Hardcover - Holt Optimal"accuracy(hcoverH$fitted, hcover)## ME RMSE MAE MPE MAPE ACF1 Theil's U
## Test set -2.2 27 23 -3.4 12 -0.027 0.66c. Calculate a 95% prediction interval for the first forecast for each series using both methods, assuming normal errors. Compare your forecasts with those produced by R.
Paperback - Holt’s Optimal forecast with Predication Intervals
# Calculate 95% Prediction interval, and plot
autoplot(forecast(pbackH, h=4), ylab="Paperback Books Sold", xlab="Day")
Calculating the 95th percent interval for the first day’s forecast manually would give:
ForecastDay31= ValueDay30 +/- 1.96 x sigma, where sigma = 32 & ValueDay30=247 —> ForecastDay31= 184 to 310
From the output of the forecast summary the 95th% interval was 145 to 269
Hardcover - Holt’s Optimal forecast with Predication Intervals
# Calculate 95% Prediction interval, and plot
autoplot(forecast(hcoverH, h=4), ylab="Hardcover Books Sold", xlab="Day")