Time Series

Time Series Assignment

The attached data shows monthly demand of two different types of consumable items in a certain store from January 2002 to September 2017. The ultimate objective of this exercise is to predict sales for the period October 2017 to December 2018. I. Read the data as time series objects in R. Plot the data. What are the major features you notice in the series? How do the two series differ?

Before a formal extraction of time series components is done, can you check for seasonal changes in the data for the two series separately? Particularly whether there are more variability in a season compared to the others, whether seasonal variations are changing across years etc. Compare the behaviour of the two series.
Decompose each series to extract trend and seasonality, if there are any. Which seasonality is more appropriate – additive or multiplicative? Explain the seasonal indices. In which month(s) do you see higher sales and which month(s) you see lower sales? Any difference in the nature of demand of the two items?
Can you extract the residuals for the two decomposition exercises and check if they form a stationary series? Do a formal test for stationary writing down the null and alternative hypothesis. What is your conclusion in each case?
Before the final forecast is undertaken one would like to compare a few models. Use the last 21 months as hold-out sample fit a suitable exponential smoothing model to the rest of the data and calculate MAPE. What are the values of a, ß and ??? What role do they play in the modeling? For the same hold-out period compare forecast by decomposition and compute MAPE. Which model gives smaller MAPE? Give a comparison for the two demands.
Use the ‘best’ model obtained from above to forecast demand for the period Oct 2017 to December 2018 for both items. Provide forecasted values as well as their upper and lower confidence limits. If you are the store manager what decisions would you make after looking at the demand of the two items over years?

Approach:

In the given dataset we have demand data for Item A and B for the period January 2002 to July 2017 (as against Sep 2017 mentioned in the problem statement). The data is continuous monthly data for the whole period without any breaks. This qualifies for a time series analysis on the demand for Item A & B, subject to other assumptions being valid.

The plan is to do the following:
1. Convert to time series for the 2 types of data
2. Perform Exploratory analysis of data by visual check of trend, seasonality
3. Time series decomposition into trend, seasonality and residuals
4. Check assumptions for stationary time series. In case of non-stationary, convert to stationary
5. Create various models of Forecasting the time series value
6. Interpretation of result

Solution:

Read the dataset

library(readxl)
  
#Remove the first record from the excel file (skip = 1)  
demand <- read_excel("C:/Users/ranvkumar/Desktop/TimeSeries/Demand.xlsx", skip = 1)
    
str(demand)

## Classes 'tbl_df', 'tbl' and 'data.frame':    187 obs. of  4 variables:
##  $ Year  : num  2002 2002 2002 2002 2002 ...
##  $ Month : num  1 2 3 4 5 6 7 8 9 10 ...
##  $ Item A: num  1954 2302 3054 2414 2226 ...
##  $ Item B: num  2585 3368 3210 3111 3756 ...

Rename the columns as per the series

names(demand)[3] <- c("IteamA")
names(demand)[4] <- c("IteamB")
  
str(demand)

## Classes 'tbl_df', 'tbl' and 'data.frame':    187 obs. of  4 variables:
##  $ Year  : num  2002 2002 2002 2002 2002 ...
##  $ Month : num  1 2 3 4 5 6 7 8 9 10 ...
##  $ IteamA: num  1954 2302 3054 2414 2226 ...
##  $ IteamB: num  2585 3368 3210 3111 3756 ...

Data summary

summary(demand[,3:4])

##      IteamA         IteamB    
##  Min.   :1954   Min.   :1153  
##  1st Qu.:2748   1st Qu.:2362  
##  Median :3134   Median :2876  
##  Mean   :3263   Mean   :2962  
##  3rd Qu.:3741   3rd Qu.:3468  
##  Max.   :5725   Max.   :5618

Convert the data into a time series dataset

#Plottig Time Series for Item A for Monthly data from year 2012 Jan to 2017 July
dem_ItA <- ts(demand[,3], start=c(2002,1), end=c(2017,7), frequency=12) 
plot(dem_ItA)

#Plottig Time Series for Item B for Monthly data from year 2012 Jan to 2017 July 
dem_ItB <- ts(demand[,4], start=c(2002,1), end=c(2017,7), frequency=12) 
plot(dem_ItB)

#Plotting the Time Series across Item A and Item B 
ts.plot(dem_ItA, dem_ItB, gpars = list(col = c("black", "red")),xlab="year", ylab="demand") 
legend("topleft", colnames(demand[3:4]), col=1:ncol(demand), lty=1.9, cex=.45)

From the above plots, we can see Item A has an increasing demand, whereas Item B has fall in demand. Also, there is some seasonality and trend in demands. Both Item A and B doesn’t seem to have cyclic in nature. Item A variation increases with time whereas Item B variation decreases.

Decomposition of time series dataset

A time series decomposition is procedure which transform a time series into multiple different series. The original time series is often computed (decompose) into 3 sub-time series:

Seasonal: patterns that repeat with fixed period of time.
Trend: the underlying trend of the metrics.
Random: (also call “noise”, “Irregular” or “Remainder”) Is the residuals of the time series after allocation into the seasonal and trends time series.

Other than above three component there is Cyclic component which occurs after long period of time.

Additive or multiplicative decomposition? To get a successful decomposition, it is important to choose between the additive or multiplicative model. To choose the right model we need to look at the time series.

The additive model is useful when the seasonal variation is relatively constant over time.
The multiplicative model is useful when the seasonal variation increases over time.

#Iteam A
monthplot(dem_ItA)

boxplot (dem_ItA ~cycle(dem_ItA))

#Iteam B
monthplot(dem_ItB)

boxplot (dem_ItA ~cycle(dem_ItB))

The seasonal variation looked to be about the same magnitude across time, so an additive decomposition might give good results.

Decomposing the time series using STL

STL is a very versatile and robust method for decomposing time series. STL is an acronym for “Seasonal and Trend decomposition using Loess”. It does an additive decomposition and the four graphs are the original data, seasonal component, trend component and the remainder.

#Item A 
ItA_Sea <- stl(dem_ItA[,1], s.window="p") #constant seasonality 
plot(ItA_Sea)

#Item B 
ItB_Sea <- stl(dem_ItB[,1], s.window="p") #constant seasonality 
plot(ItB_Sea)

From the above decomposed details, we can see that there is continous incease in demand for Item A, but on contraty similar drop pattern observed for Item B.

Decompose the time series and plot the deseasoned series
If the focus is on figuring out whether the general trend of demand is up, we deseasonalize, and possibly forget about the seasonal component. However, if you need to forecast the demand in next mnoth, then you need take into account both the secular trend and seasonality.

Item A

series_names <- c('Deseasoned', 'Actual') 
Deseason_ItA <- (ItA_Sea$time.series[,2]+ItA_Sea$time.series[,3]) 
ts.plot(dem_ItA, Deseason_ItA, col=c("red", "blue"), main="ItemA Demand vs Deseasoned Demand")

Above plot show demand in Red and de-seasoned demand in Blue, we can see that there is increasing trend of demand. The residual component is also part of analysis.

Item B

series_names <- c('Deseasoned', 'Actual') 
Deseason_ItB <- (ItB_Sea$time.series[,2]+ItB_Sea$time.series[,3]) 
ts.plot(dem_ItB, Deseason_ItB, col=c("red", "blue"), main="ItemB Demand vs Deseasoned Demand")

The above plot show demand in Red and de-seasoned demand in Blue, we can see that there is decreasing trend of demand.

Divide data into test and train

DataATrain <- window(dem_ItA, start=c(2002,1), end=c(2015,12), frequency=12) 
DataATest <- window(dem_ItA, start=c(2016,1), frequency=12) 

DataBTrain <- window(dem_ItB, start=c(2002,1), end=c(2015,12), frequency=12) 
DataBTest <- window(dem_ItB, start=c(2016,1), frequency=12)

Convert into seasonal, trend and irregular component using STL

ItmATrn <- stl(DataATrain[,1], s.window="p") 
ItmBTrn <- stl(DataBTrain[,1], s.window="p")

Model building

Random walk with drift model - Forecast on train data

It assumes that, at each point in time, the series merely takes a random step away from its last recorded position, with steps whose mean value is zero.

library(forecast)
fcst.ItA.stl <- forecast(ItmATrn, method="rwdrift", h=19) 
fcst.ItB.stl <- forecast(ItmBTrn, method="rwdrift", h=19) 

VecA<- cbind(DataATest,fcst.ItA.stl$mean) 
VecB<- cbind(DataBTest,fcst.ItB.stl$mean)

Item A

#par(mfrow=c(1,1), mar=c(2, 2, 2, 2), mgp=c(3, 1, 0), las=0) 
ts.plot(VecA, col=c("blue", "red"),xlab="year", ylab="demand", main="Quarterly Demand A: Actual vs Forecast")

Mean absolute percentage error (MAPE)
Calculates the mean absolute percentage error (Deviation) function for the forecast and the eventual outcomes.

MAPEA <- mean(abs(VecA[,1]-VecA[,2])/VecA[,1]) 
MAPEA

## [1] 0.1408798

Box-Ljung Test
To check is resuidual are independent
H0: Residuals are independent
Ha: Residuals are not independent

Box.test(fcst.ItA.stl$residuals, lag=10, type="Ljung-Box")

## 
##  Box-Ljung test
## 
## data:  fcst.ItA.stl$residuals
## X-squared = 72.292, df = 10, p-value = 1.597e-11

Conclusion: Reject Ha: Residuals are not independent

Item B

ts.plot(VecB, col=c("blue", "red"),xlab="year", ylab="demand", main="Quarterly Demand B: Actual vs Forecast")

Mean absolute percentage error (MAPE)
Calculates the mean absolute percentage error (Deviation) function for the forecast and the eventual outcomes.

MAPEB <- mean(abs(VecB[,1]-VecB[,2])/VecB[,1]) 
MAPEB

## [1] 0.1082608

Box-Ljung Test
To check is resuidual are independent
H0: Residuals are independent
Ha: Residuals are not independent

Box.test(fcst.ItB.stl$residuals, lag=30, type="Ljung-Box")

## 
##  Box-Ljung test
## 
## data:  fcst.ItB.stl$residuals
## X-squared = 123.22, df = 30, p-value = 2.931e-13

Conclusion: Reject Ha: Residuals are not independent

From the above MAPE results we can see the 14 % and 10.8% less accuracy in model.

Holt Winter method

The Holt-Winters seasonal method comprises the forecast equation and three smoothing equations — one for the level, one for trend, and one for the seasonal component, with smoothing parameters alpha(a), beta(ß) and gama(??).

Item A

hwA <- HoltWinters(as.ts(DataATrain),seasonal="additive") 
hwA

## Holt-Winters exponential smoothing with trend and additive seasonal component.
## 
## Call:
## HoltWinters(x = as.ts(DataATrain), seasonal = "additive")
## 
## Smoothing parameters:
##  alpha: 0.1241357
##  beta : 0.03174654
##  gamma: 0.3636975
## 
## Coefficients:
##             [,1]
## a    3753.348040
## b       7.663395
## s1  -1250.098605
## s2   -438.592232
## s3   -224.017731
## s4   -407.395313
## s5   -507.668223
## s6   -667.267246
## s7     63.659702
## s8    197.909330
## s9   -301.525945
## s10    25.272325
## s11   712.529546
## s12  1545.291998

plot(hwA)

hwAForecast <- forecast(hwA, h=19) 
VecA1 <- cbind(DataATest,hwAForecast) 

par(mfrow=c(1,1), mar=c(2, 2, 2, 2), mgp=c(3, 1, 0), las=0) 

ts.plot(VecA1[,1],VecA1[,2], col=c("blue","red"),xlab="year", ylab="demand", main="Demand A: Actual vs Forecast")

Box.test(hwAForecast$residuals, lag=20, type="Ljung-Box")

## 
##  Box-Ljung test
## 
## data:  hwAForecast$residuals
## X-squared = 14.227, df = 20, p-value = 0.8188

Conclusion: Reject H0: Residuals are independent

#install.packages("MLmetrics")
library(MLmetrics)

## Warning: package 'MLmetrics' was built under R version 3.5.1

## 
## Attaching package: 'MLmetrics'

## The following object is masked from 'package:base':
## 
##     Recall

MAPE(VecA1[,1],VecA1[,2])

## [1] 0.1160528

Item B

hwB <- HoltWinters(as.ts(DataBTrain),seasonal="additive") 
hwB

## Holt-Winters exponential smoothing with trend and additive seasonal component.
## 
## Call:
## HoltWinters(x = as.ts(DataBTrain), seasonal = "additive")
## 
## Smoothing parameters:
##  alpha: 0.0166627
##  beta : 0.4878834
##  gamma: 0.5000132
## 
## Coefficients:
##            [,1]
## a    2297.12724
## b     -15.29024
## s1  -1222.01821
## s2  -1012.34884
## s3   -442.56913
## s4   -307.95973
## s5     79.56065
## s6    258.33260
## s7    697.64492
## s8    241.68337
## s9   -246.12729
## s10  -465.09216
## s11   120.77708
## s12   412.50043

plot(hwB)

hwBForecast <- forecast(hwB, h=19) 
VecB1 <- cbind(DataBTest,hwBForecast) 

par(mfrow=c(1,1), mar=c(2, 2, 2, 2), mgp=c(3, 1, 0), las=0) 

ts.plot(VecB1[,1],VecB1[,2], col=c("blue","red"),xlab="year", ylab="demand", main="Demand B: Actual vs Forecast")

Box.test(hwBForecast$residuals, lag=20, type="Ljung-Box")

## 
##  Box-Ljung test
## 
## data:  hwBForecast$residuals
## X-squared = 13.101, df = 20, p-value = 0.873

Conclusion: Do not reject H0: Residuals are independent

MAPE(VecB1[,1],VecB1[,2])

## [1] 0.1867152

MAPE is 11.6 % and 18.6 % for item A and Item B resp.

ARIMA Model

Check for stationary time series

Dickey-Fuller test
Statistical tests make strong assumptions about your data. They can only be used to inform the degree to which a null hypothesis can be accepted or rejected. The result must be interpreted for a given problem to be meaningful. Nevertheless, they can provide a quick check and confirmatory evidence that your time series is stationary or non-stationary.

Null Hypothesis (H0): If accepted, it suggests the time series has a unit root, meaning it is non-stationary. It has some time dependent structure.
Alternate Hypothesis (H1): The null hypothesis is rejected; it suggests the time series does not have a unit root, meaning it is stationary. It does not have time-dependent structure.

p-value > 0.05: Accept the null hypothesis (H0), the data has a unit root and is non-stationary.
p-value <= 0.05: Reject the null hypothesis (H0), the data does not have a unit root and is stationary.

Item A

library(tseries) 
adf.test(dem_ItA)

## Warning in adf.test(dem_ItA): p-value smaller than printed p-value

## 
##  Augmented Dickey-Fuller Test
## 
## data:  dem_ItA
## Dickey-Fuller = -7.8632, Lag order = 5, p-value = 0.01
## alternative hypothesis: stationary

diff_dem_ItA <- diff(dem_ItA) 
plot(diff_dem_ItA)

adf.test(diff(dem_ItA))

## Warning in adf.test(diff(dem_ItA)): p-value smaller than printed p-value

## 
##  Augmented Dickey-Fuller Test
## 
## data:  diff(dem_ItA)
## Dickey-Fuller = -8.0907, Lag order = 5, p-value = 0.01
## alternative hypothesis: stationary

Item B

library(tseries) 
adf.test(dem_ItB)

## Warning in adf.test(dem_ItB): p-value smaller than printed p-value

## 
##  Augmented Dickey-Fuller Test
## 
## data:  dem_ItB
## Dickey-Fuller = -12.967, Lag order = 5, p-value = 0.01
## alternative hypothesis: stationary

diff_dem_ItB <- diff(dem_ItB) 
plot(diff_dem_ItB)

adf.test(diff(dem_ItB))

## Warning in adf.test(diff(dem_ItB)): p-value smaller than printed p-value

## 
##  Augmented Dickey-Fuller Test
## 
## data:  diff(dem_ItB)
## Dickey-Fuller = -9.8701, Lag order = 5, p-value = 0.01
## alternative hypothesis: stationary

From the above ADF test the Null Hypothesis is rejected.
The time series of differences (above) does appear to be stationary in mean and variance, as the level of the series stays roughly constant over time, and the variance of the series appears roughly constant over time.

ACF and PACF (performing to check the stationary data and autocorrelation)

The function Acf computes an estimate of the autocorrelation function of a (possibly multivariate) time series. Function Pacf computes an estimate of the partial autocorrelation function of a (possibly multivariate) time series.

ACF
> Autocorrelation of different orders gives inside information regarding time series. It will help determine order p of the series
> Significant autocorrelations imply observations of long past influences current observation.
> lag.max: maximum lag at which to calculate the acf. Default is 10\(*\)log10(N/m)
>>where N is the number of observations and m the number of series.
>>Will be automatically limited to one less than the number of observations in the series

PACF
> Partial autocorrelation adjusts for the intervening periods

acf(dem_ItA,lag=15)

acf(diff_dem_ItA, lag=15)

acf(dem_ItB,lag=15)

acf(diff_dem_ItB, lag=15)

Checking with lag 50

acf(dem_ItA,lag=50)

acf(diff_dem_ItA, lag=50)

pacf(dem_ItA)

pacf(diff_dem_ItA)

acf(dem_ItB,lag=50)

acf(diff_dem_ItB, lag=50)

pacf(dem_ItB)

pacf(diff_dem_ItB)

ARIMA model
ARMA models are commonly used in time series modeling. In ARMA model, AR stands for auto-regression and MA stands for moving average.
The above ACF and PACF we have found out that the positive and negative values mean (that is because of data is stationary); they are not cuts for AR(2) series and no gradually decrease in the value of PACF, no significance of MA(2).

Item A

ItA.arima.fit.train <- auto.arima(DataATrain, seasonal=TRUE) 
ItA.arima.fit.train

## Series: DataATrain 
## ARIMA(0,0,0)(0,1,1)[12] with drift 
## 
## Coefficients:
##          sma1   drift
##       -0.6581  3.9132
## s.e.   0.0798  0.9188
## 
## sigma^2 estimated as 116022:  log likelihood=-1133.35
## AIC=2272.71   AICc=2272.86   BIC=2281.86

plot(ItA.arima.fit.train$residuals)

plot(ItA.arima.fit.train$x,col="blue")
lines(ItA.arima.fit.train$fitted,col="red",main="Demand A: Actual vs Forecast")

MAPE(ItA.arima.fit.train$fitted,ItA.arima.fit.train$x)

## [1] 0.0733376

The MAPE percentage error is now reduced to 7.3% for ARIMA model

acf(ItA.arima.fit.train$residuals)

pacf(ItA.arima.fit.train$residuals)

Box-Ljung Test
To check is resuidual are independent
H0: Residuals are independent
Ha: Residuals are not independent

Box.test(ItA.arima.fit.train$residuals, lag = 10, type = c("Ljung-Box"), fitdf = 0)

## 
##  Box-Ljung test
## 
## data:  ItA.arima.fit.train$residuals
## X-squared = 16.716, df = 10, p-value = 0.0809

Conclusion: Reject H0: Residuals are independent

Forecasting on hold out dataset

ArimafcastA <- forecast(ItA.arima.fit.train, h=19) 
VecA2 <- cbind(DataATest,ArimafcastA) 

par(mfrow=c(1,1), mar=c(2, 2, 2, 2), mgp=c(3, 1, 0), las=0) 

ts.plot(VecA2[,1],VecA2[,2], col=c("blue","red"),xlab="year", ylab="demand", main="Demand A: Actual vs Forecast")

From the plot and data, we can see the forecasted value follows almost the same as actual value, there are point of interaction at Jan 2016, May 2016, Dec 2016, Jan 2017.

Item B

ItB.arima.fit.train <- auto.arima(dem_ItB, seasonal=TRUE) 
ItB.arima.fit.train

## Series: dem_ItB 
## ARIMA(0,0,0)(2,1,1)[12] with drift 
## 
## Coefficients:
##         sar1    sar2     sma1    drift
##       0.1763  0.0339  -0.7149  -9.3358
## s.e.  0.2083  0.1433   0.1880   0.8680
## 
## sigma^2 estimated as 102973:  log likelihood=-1258.89
## AIC=2527.78   AICc=2528.13   BIC=2543.6

plot(ItB.arima.fit.train$residuals)

plot(ItB.arima.fit.train$x,col="blue") 
lines(ItB.arima.fit.train$fitted,col="red", main="Demand B: Actual vs Forecast")

MAPE(ItB.arima.fit.train$fitted,ItB.arima.fit.train$x)

## [1] 0.07654621

The MAPE percentage error is now reduced to 9% for ARIMA model

acf(ItB.arima.fit.train$residuals)

pacf(ItB.arima.fit.train$residuals)

Box-Ljung Test
To check is resuidual are independent
H0: Residuals are independent
Ha: Residuals are not independent

Box.test(ItB.arima.fit.train$residuals, lag = 10, type = c("Ljung-Box"), fitdf = 0)

## 
##  Box-Ljung test
## 
## data:  ItB.arima.fit.train$residuals
## X-squared = 16.285, df = 10, p-value = 0.09177

Conclusion: Reject H0: Residuals are independent

Forecasting on hold out dataset

ArimafcastB <- forecast(ItB.arima.fit.train, h=19) 
VecB2 <- cbind(DataBTest,ArimafcastB) 

par(mfrow=c(1,1), mar=c(2, 2, 2, 2), mgp=c(3, 1, 0), las=0) 

ts.plot(VecB2[,1],VecB2[,2], col=c("blue","red"),xlab="year", ylab="demand", main="Demand B: Actual vs Forecast")

From the plot and data, we can see the forecasted value doesn’t exactly follows the actual value, but there are point of interaction at Mar 2016, Apr 2016, May 2016 Nov 2016, Mar 2017.

Conclusion

For Time Series Forecasting problem, we observed the trend and seasonality in the data.
We have observed that the Item A has increasing trend, but for Item B the trend is declining.
Also, we observed for both item there are few months with high variation in seasonality; and for Item A there are few outliers.

As the seasonality was not following the trend pattern we have used the “Additive” seasonality. We have performed the three models
1. Random Walk with Drift,
2. Holt Winters and
3. ARIMA model.

Below are MAPE and Box-Ljung test observations for Models.

Random Walk with Drift
Item A# 0.1408798 (14%), p-value < 2.2e-16
Item B# 0.1082608 (10.8%), p-value = 2.931e-13

Holt Winters
Item A# 0.1160528 (11.6%), p-value = 0.8188
Item B# 0.1867152 (18.6%), p-value = 0.873

ARIMA
Item A# 0.0733376 (7%), p-value = 0.0809
Item B# 0.07654621 (7%), p-value = 0.09177

From the MAPE values observed the ARIMA model provided the lowest values and we selected the model for the Forecasting.

Forecasting using ARIMA Model

Iteam A

ItA.arima.fit <- auto.arima(dem_ItA, seasonal=TRUE) 
fcastA <- forecast(ItA.arima.fit, h=19) 

plot(fcastA)

fcastA

##          Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
## Aug 2017       4320.211 3879.991 4760.431 3646.953 4993.469
## Sep 2017       4169.513 3725.551 4613.476 3490.531 4848.495
## Oct 2017       4428.791 3981.385 4876.197 3744.542 5113.040
## Nov 2017       5102.669 4652.091 5553.246 4413.570 5791.767
## Dec 2017       5879.220 5425.721 6332.719 5185.653 6572.787
## Jan 2018       2819.535 2363.343 3275.727 2121.849 3517.221
## Feb 2018       3990.984 3532.307 4449.660 3289.498 4692.469
## Mar 2018       4181.449 3720.480 4642.419 3476.458 4886.441
## Apr 2018       4081.089 3618.003 4544.174 3372.860 4789.317
## May 2018       3888.336 3423.296 4353.376 3177.118 4599.554
## Jun 2018       4029.525 3562.679 4496.370 3315.545 4743.504
## Jul 2018       4390.292 3921.777 4858.807 3673.760 5106.823
## Aug 2018       4407.590 3900.778 4914.402 3632.487 5182.693
## Sep 2018       4257.019 3747.019 4767.019 3477.041 5036.997
## Oct 2018       4516.419 4003.480 5029.358 3731.946 5300.892
## Nov 2018       5190.414 4674.763 5706.065 4401.794 5979.034
## Dec 2018       5967.079 5448.925 6485.232 5174.632 6759.526
## Jan 2019       2907.502 2387.039 3427.966 2111.522 3703.483
## Feb 2019       4079.056 3556.458 4601.654 3279.812 4878.301

Item B

ItB.arima.fit <- auto.arima(dem_ItB, seasonal=TRUE) 
fcastB <- forecast(ItB.arima.fit, h=19) 

plot(fcastB)

fcastB

##          Point Forecast     Lo 80    Hi 80     Lo 95    Hi 95
## Aug 2017      2356.3605 1945.1156 2767.605 1727.4157 2985.305
## Sep 2017      2082.9473 1671.7024 2494.192 1454.0025 2711.892
## Oct 2017      1784.7949 1373.5500 2196.040 1155.8501 2413.740
## Nov 2017      2436.4019 2025.1570 2847.647 1807.4571 3065.347
## Dec 2017      2429.8611 2018.6162 2841.106 1800.9163 3058.806
## Jan 2018       965.2270  553.9834 1376.471  336.2842 1594.170
## Feb 2018      1278.2300  866.9865 1689.474  649.2873 1907.173
## Mar 2018      1693.3730 1282.1294 2104.617 1064.4302 2322.316
## Apr 2018      2088.9240 1677.6805 2500.168 1459.9813 2717.867
## May 2018      2342.6726 1931.4290 2753.916 1713.7298 2971.615
## Jun 2018      2587.5703 2176.3268 2998.814 1958.6276 3216.513
## Jul 2018      2903.9519 2492.7084 3315.196 2275.0092 3532.895
## Aug 2018      2267.6121 1814.7093 2720.515 1574.9570 2960.267
## Sep 2018      1945.3845 1492.4816 2398.287 1252.7293 2638.040
## Oct 2018      1663.8271 1210.9242 2116.730  971.1719 2356.482
## Nov 2018      2293.2590 1840.3561 2746.162 1600.6038 2985.914
## Dec 2018      2325.0672 1872.1643 2777.970 1632.4120 3017.722
## Jan 2019       843.6106  390.7094 1296.512  150.9580 1536.263
## Feb 2019      1150.9134  698.0123 1603.815  458.2608 1843.566

Time Series

Ranvir

July 5, 2018

Time Series Assignment

Approach:

Solution:

Read the dataset

Rename the columns as per the series

Data summary

Convert the data into a time series dataset

Decomposition of time series dataset

Decomposing the time series using STL

Divide data into test and train

Convert into seasonal, trend and irregular component using STL

Model building

Random walk with drift model - Forecast on train data

Item A

Item B

Holt Winter method

ARIMA Model

Check for stationary time series

ACF and PACF (performing to check the stationary data and autocorrelation)

Conclusion

Forecasting using ARIMA Model