Solved – Forecasting Function R (Holt-Winters HW approach)

forecastingrtime series

I was trying to forecast using the "hw" method in R.
I have data which follows:

[...] And more periods

The period here represent weeks and each of the data in periods are quantities on a given day. So we can see a pattern that the first 3 days usually shows the largest quantities.
I decided to forecast with HW method.

So I did the following in R:

library(forecast)
x <- data$weeklydata
x <- ts(x, frequency = 7)
plot(hw(x, 6), byt = "1")

However, the graph I get ranges from -200 to 300, with my forecast being showed to near 0 to negative quantity. Am making a fundamental mistake here with my concepts, or is there something wrong with my R coding?

Best Answer

The following code produces the plot shown below (with a warning message saying that "byt" is not a graphical parameter). The range of forecasts are noticeably different from -200 to 300. From here, my only suggestion is that maybe there is an error elsewhere in your code.

I understand that you've got more data points, so the forecasts here will be different for the full dataset. With this in mind, your plot should still look something similar to what's shown below. If it differs significantly, check for errors elsewhere in your code.

# Load forecast package
library(forecast)
# Store data as a numeric vector
x <- c(568,485,360,523,514,370,332,841,685,719,647,615,389,367,731,721,819,662,581,436,394,865,805,952,759,677,429,424,598,868,888,849,707,458,426,950,806,826,804,730,541,439,1070,770,989,863,737,525,461,1041,863,989,833,783,506,496)
# Convert data to a 'ts' object
x <- ts(x, frequency = 7)
# Plot forecasts from Holt-Winters' additive method
plot(hw(x, 6), byt = "1")

enter image description here

This answer probably should have been a comment, but I decided to make it an answer because it contains code and a plot.

Related Solutions

Solved – Forecasting beyond one season using Holt-Winters’ exponential smoothing

I am not very familiar with Holt-Winters, however I have this excellent book by @Rob Hyndman. The package forecast (which is based on the book) of statistical package R gives the following result on your data:

> hw<-read.table("~/R/stackoverflow/hw.txt")
> tt<-ts(hw[,3],start=c(1999,1),freq=12)

> aa<-forecast(tt)
> plot(aa)
> summary(aa)

Forecast method: ETS(M,N,A)

Model Information:
ETS(M,N,A) 

Call:
 ets(y = object) 

  Smoothing parameters:
    alpha = 0.1701 
    gamma = 1e-04 

  Initial states:
    l = 870.4847 
    s = -278.0815 -143.6584 151.959 -135.595 514.2527 236.9216
           -32.7679 128.8337 115.0829 47.5922 -234.4105 -370.1288

  sigma:  0.1122

     AIC     AICc      BIC 
1892.756 1896.346 1933.115 

In-sample error measures:
         ME        RMSE         MAE         MPE        MAPE        MASE 
 18.1543007 121.8594668  70.7086492   0.8480306   7.0006920   0.2893504

Here is the graph of the forecast together with the confidence intervals: enter image description here

Note that the function forecast picks automatically the best exponential smoothing model from 30 models which are classified by the type of trend model, seasonal part model and the additivity or multiplicity of error.

The best model found in your data is with multiplicative error, no trend and additive seasonality, which is less complicated model than you are trying to fit. The way function forecast works is however that the more complicated model was considered and rejected in favor the final model.

If you provide the exact formulas it would be possible to fit the precise model to see whether the problem you described is really property of the model.

Solved – Using Holt-Winters for forecasting in Python

I think the R forecast package you mentioned is a better fit for this problem than just using Holt-Winters. The two functions you are interested in are ets() and auto.arima(). ets() will fit an exponential smoothing model, including Holt-Winters and several other methods. It will choose parameters (alpha, beta, and gama) for a variety of models and then return the one with the lowest AIC (or BIC if you prefer). auto.arima() works similarly.

However, as IrishStat pointed out, these kinds of models may not be appropriate for your analysis. In that case, try calculating some covariates, such as dummy variables for weekends, holidays, and their interactions. Once you've specified covariates that make sense, use auto.arima() to find a ARMAX model, and then forecast() to make predictions. You will probably end up with something much better than a simple Holt-Winters model in python with default parameters.

You should also note that both ets() and auto.arima can fit seasonal models, but you need to format your data as a seasonal time series. Let me know if you need any help with that.

You can read more about the forecast package here.

Best Answer

Related Solutions

Solved – Forecasting beyond one season using Holt-Winters’ exponential smoothing

Solved – Using Holt-Winters for forecasting in Python

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