A solar-powered calculator\'s sales history for the first four months of 2015 ar

Question

A solar-powered calculator's sales history for the first four months of 2015 are given below: a) What is the three month moving average forecast for demand in May? b) What is the three month moving average forecast for July? c) If the demand forecast for January was 25000, determine the one-step ahead forecasts for February through May using exponential smoothing with alpha = 0.1. Repeat with alpha = 0.25. Compare the MAD, MSE, and MAPE for these two forecasts for months February through April. What smoothing constant gives the more accurate forecast? Identify the manufacturing/service processes (flow, batch, job etc.) that would best suit the following firms: a. Barr Tax Attorneys b. Mazda Automotive c. Build a bear The number of cars using a toll bridge on every day for 2 weeks is recorded as follows. The county transportation department would like to determine the level, trend, and seasonal factors for usage of the bridge. Find these for the county. If the county records usage for Monday of week 3 as 18000 cars, determine the one-step ahead forecast for usage on Tuesday of Week 3 (all smoothing constants = 0.1)

Explanation / Answer

1) I am using R software to solve the question

First a time series object needs to be created as follows:

sales <- c(23300,72300,30300,15500)
tsales <- ts(sales,start = c(2015,1),end=c(2015,4),frequency = 12)

forecast package can be used for forecasting a simple moving average model.

#Loading the forecast package

library(forecast)
forecast(ma(tsales,3), h = 3)

It gives the below result:

Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
Apr 2015 41858.33 39502.23 44214.44 38254.98 45461.68
May 2015 41858.33 39500.19 44216.48 38251.86 45464.81
Jun 2015 41858.33 39498.14 44218.52 38248.73 45467.93

So three month moving average forecast for demand in May is 41,858.33

b) As seen above three month moving average forecast for demand in July is also 41,858.33

c) For making forecasts through simple exponential smoothing in R, we can use “HoltWinters()” function as below:

#Using alpha=0.1

fit1 <- HoltWinters(tsales, alpha = 0.1, beta=FALSE, gamma = FALSE)
forecast1 <- forecast(fit1,1)

Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
May 2015 28358.3 -12829.12 69545.72 -34632.42 91349.02

##Using alpha=0.25

fit2 <- HoltWinters(tsales, alpha = 0.25, beta=FALSE, gamma = FALSE)
forecast2 <- forecast(fit2,1)

Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
May 2015 30270.31 -15216.37 75757 -39295.57 99836.19

Accuracy measures can be found using accuracy() function in R

accuracy(forecast1)

ME RMSE MAE MPE MAPE MASE ACF1
Training set 11194.33 28529.13 20719 -8.290384 53.15908 NaN -0.05464071

accuracy(forecast2)

ME RMSE MAE MPE MAPE MASE ACF1
Training set 7027.083 29820.07 24506.25 -27.72308 71.33765 NaN -0.07289299

So we can see that smoothing constant of 0.1 is giving better foreacsts as it has lower RMSE.

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