PPP - Homework 2

Nama : I Made Wiratha Nungrat NIM : 13209057 Homework 2 Problem on Moving Averages Problems 16 and 17 are based on the following data. Observation of the demand for a certain part stocked at a part supply depot during the calendar year 1999 were: Month January February March April May June July August September October November December

Demand 89 57 144 221 177 280 223 286 212 275 188 312

Problem 16. Determine the one-step-ahead forecasts for the demand for January 2000 using 3-, 6-, and 12-month moving averages. Answer: Using Moving Average equation: ( ) ∑

We get: Quarter

Year

1 2 3 4 5 6 7 8 9 10 11 12 1

Month Demand MA(3) MA(6) MA(12) 1999 January 89 1999 February 57 1999 March 144 1999 April 221 97 1999 May 177 141 1999 June 280 181 1999 July 223 226 161 1999 August 286 227 184 1999 September 212 263 222 1999 October 275 240 233 1999 November 188 258 242 1999 December 312 225 244 2000 January 258 249 205

The one-step-ahead forecasts for January 2000 is: -

MA(3) = 258 MA(6) = 249 MA(12) = 205

Problem 17. Using a four-month moving average, determine one-step-ahead forecasts for July through December 1999 Answer: Quarter 1 2 3 4 5 6 7 8 9 10 11 12

Year

Month Demand MA(4) Error 1999 January 89 1999 February 57 1999 March 144 1999 April 221 1999 May 177 1999 June 280 1999 July 223 206 1999 August 286 225 1999 September 212 242 1999 October 275 250 1999 November 188 249 1999 December 312 240

-17 -61 30 -25 61 -72

Problem for Exponential Smoothing Problem 22. Handy, Inc., produces a solar-powered electronic calculator that has experienced the following monthly sales history for the first four months of the year, in thousands of units: Month January February March April

Sales History 23,3 72,3 30,3 15,5

a. If the forecast for January was 25, determine the one-step-ahead forecasts for February through May using exponential smoothing with a smoothing constant of α = 0,15 Answer: Given α = 0,15, assumption for forecast for January was 25, New forecast = α(Current observation of Demand) + (1 - α)(Last forecast) In equation: ( ) ( Using Ms. Excel then we got:

)(

)

(

)(

)

Quarter

Month 1 January 2 February 3 March 4 April 5 May

Sales History Forecast 23,3 25 72,3 24,7 30,3 31,8 15,5 31,6 29,2

b. Repeat the calculation in part (a) for the value of α = 0,40. What difference in the forecasts do you observe? Answer: Using the same method and procedure for problem 22.a, with Ms. Excel we have: Quarter

Month 1 January 2 February 3 March 4 April 5 May

Sales History Forecast 23,3 25 72,3 24,3 30,3 43,5 15,5 38,2 29,1

The difference between α = 0,15 and 0,40 could be seen from the graph below:

Figure 1. Exponential smoothing for different values of alpha In order to appreciate the effect of the smoothing constant, we have to graphed it. For α = 0,15 the predicted value of demand result in a relatively smooth pattern, whereas α = 0,40, the predicted value exhibits significantly greater variation c. Compute the MSEs for the forecast you obtained in parts (a) and (b) for February through April. Which value of α gave more accurate forecasts, based on the MSE? Answer: Comparing the result from (a) and (b), then we got the number in the table below:

Month Sales History α = 0,15 α = 0,40 January 23,3 25 25 February 72,3 24,7 24,3 March 30,3 31,8 43,5 April 15,5 31,6 38,2

Using MSEs or Mean Square Error, given by the following formula:

( )∑ The result from Ms. Excel Month Sales History α = 0,15 |Error| α = 0,40 |Error| January 23,3 25 1,7 25 1,7 February 72,3 24,7 47,6 24,3 48 March 30,3 31,8 1,5 43,5 13,2 April 15,5 31,6 16,1 38,2 22,7

MSE for February through april: MSE α = 0,15 = 842,4067 MSE α = 0,40 = 997,8433 Based on MSE, forecast with alpha = 0,15 gave more accurate than 0,40. Problem 30. For the data in problem 28; use the result of the regression equation to estimate the slope and intercept of the series at the end of June. Use these numbers as the initial values of slope and intercept required in Holt’s method. Assume that α = 0,15, β = 0,10 for all calculations. a. Suppose that the actual number of visitors using the park in July was 2.150 and the number in august was 2.660. Use Holt’s method to update the estimates of the slope and intercept based on the observations. Answer: The data from problem 28. Month January February March April May June July August

Number of patrons 133 183 285 640 1.876 2.550 2.150 2.660

We use this data as a baseline in order to estimate the regression parameter. Using the regression equation: ( )

Where (

∑ (

)(

) )

∑ (

)

Then

Therefore,

The regression result from problem 28 for intercept a and slope b then used for the initial value of slope and intercept required in Holt’s method. Assume that α = 0,15 and β = 0,10 for all calculation and utilizing Double Exponential Smoothing method: ( (

)( )

) (

)

With α = 0,15 and β = 0,1, Then we obtain: Month Number of patrons Si Gi January 133 -240,87857 507,141 February 183 253,772821 505,892 March 285 688,464913 498,772 April 640 1105,15122 490,563 May 1.876 1637,75731 494,768 June 2.550 2195,14612 501,03 July 2.150 2614,24943 492,837 August 2.660 3040,02349 486,131

The intercept and slope for July and August based on Holt’s method is:

b. What the one-step-ahead and two-step-ahead forecasts that Holt’s method gives for the number of park visitor in September and October? Answer: One-step-ahead forecast made in august for September is: Two-step-ahead forecast made in august for October is: ( ) c. What is the forecast made at the end of July for the number of park attendees in December? Answer: ( )

Problem 33. Sales of the walking shorts at Hugo’s Department Store in downtown Rolla appear to exhibit a seasonal pattern. The proprietor of the store, Wally Hugo, has kept carefull records of the sales of several of his popular items, including walking shorts. During the past 2 years the monthly sales of the shorts have been Answer: 1. First we compute the average all of data which is 58,598 2. Next, divide each value by the mean value. Doing so gives the following: Month January February March April May June July August September October November December

Year 1

Year 2 0,204 0,305 0,611 0,899 1,34 2,273 1,9 1,527 1,119 0,763 0,39 0,356

0,271 0,237 0,78 0,814 1,493 2,714 2,205 1,408 0,882 0,831 0,237 0,441

3. Finally, we average factors corresponding to the same month of the season and determine the forecast. The result is:

Month January February March April May June July August September October November December

Year Seasonal 1 Factor Forecast 0,238 14,032 0,271 15,978 0,696 41,035 0,857 50,527 1,417 83,544 2,494 147,042 2,053 121,041 1,468 86,551 1,001 59,017 0,797 46,99 0,314 18,513 0,399 23,524

The monthly seasonal factor can be see clearly by this chart below: