Week 1 Assignment - Speech or Presentation Example

The following gives the number of pints of type A blood used at Woodlawn Hospital in the past 6 weeks: Week Of Pints Used August 31 360 September 7389September 14410September 21381September 28368October 5374a) Forecast the demand for the week of October 12 using a 3-week moving average.[381+368+374]/3 = 374.33 pintsb) Use a 3-week weighted moving average, with weights of .1, .3, and .6, using .6 for the most recent week. Forecast demand for the week of October 12.381*0.138.1368*0.3110.4374*0.6224.

4Forecast (October 12).372.9c) Compute the forecast for the week of October 12 using exponential smoothing with a forecast for August 31 of 360 andα = .2.Week Of Pints Used ForecastForecasting errorError*0.20ForecastAugust 3136036000360September 7389360295.8365.8September 14410365.844.28.84374.64September 21381374.646.361.272375.912September 28368375.912-7.912-1.5824374.3296October 5374374.32296-0.3296-0.06592374.2636The Carbondale Hospital is considering the purchase of a new ambulance. The decision will rest partly on the anticipated mileage to be driven next year.

The miles driven during the past 5 years are as follows:Year Mileage 13,00024,00033,40043,80053,700*Note: means the problem may be solved with POM for Windows and/or Excel OM.a. Forecast the mileage for next year using a 2-year moving average. [3,700+3,800]/2 = 3,750 ml.b. Find the MAD based on the 2-year moving average forecast in part (a).(Hint: You will have only 3 years of matched data.) Year Mileage Two-year Moving AverageError/error/13,00024,00033,4003,500-10010043,8003,70010010053,7003,600100100Totals100100Mfile:///D:/Downloads/878980_t2_202013_20econ11026_20_20assessment_20question_20.

pdfAD = 300/3 = 100c. Use a weighted 2-year moving average with weights of .4 and .6 to forecast next year’s mileage. (The weight of .6 is for the most recent year.) What MAD results from using this approach to forecasting? (Hint: You will have only 3 years of matched data.) Year Mileage ForecastError/error/13,00024,00033,4003,600-20020043,8003,64016016053,7003,6406060420Forecasting for year 6 = 3,740MAD = 140[420/3]d. Compute the forecast for year 6 using exponential smoothing, an initial forecast for year 1 of 3,000 miles, and α = .5.Year Mileage ForecastForecast ErrorError*0.

50New Forecast13,0003,000003,00024,0003,0001,0005003,50033,4003,600-100-503,45043,8003,6403501753,62553,7003,64075383,663Total1,325Therefore, forecast = 3,663 miles.4.9 Dell uses the CR5 chip in some of its laptop computers. The prices for the chip during the past 12 months were as follows:Month Price per Chip Month Price per Chip January$1.80July1.80February1.67August1.83March1.70September1.70April1.85October1.65May1.90November1.70June1.87December1.75a) Use a 2-month moving average on all the data and plot the averages and the prices.

MonthPrice per Chip ($)2-month moving average January1.8 February1.67 March1.71.735April1.851.685May1.91.775June1.871.875July1.81.885August1.831.835September1.71.815October1.651.765November1.71.675December1.751.675b) Use a 3-month moving average and add the 3-month plot to the graph created in part (a).MonthPrice per Chip ($)3-month moving average January1.8 February1.67 March1.7 April1.851.72May1.91.74June1.871.82July1.81.87August1.831.86September1.71.83October1.651.78November1.71.73December1.751.68December + 1 Month 1.70c) Which is better (using the mean absolute deviation): the 2-month average or the 3-month average?

MonthPrice per Chip ($)2-month moving average ErrorAbsoluteJanuary1.8   February1.67   March1.71.735-0.0350.03April1.851.6850.1650.17May1.91.7750.1250.13June1.871.875-0.0050.00July1.81.885-0.0850.09August1.831.835-0.0050.00September1.71.815-0.1150.12October1.651.765-0.1150.12November1.71.6750.0250.03December1.751.6750.0750.08MAD0.08MonthPrice per Chip ($)3-month moving average ErrorAbsoluteJanuary1.8   February1.67   March1.7 April1.851.720.130.13May1.91.740.160.16June1.871.820.050.05July1.81.87-0.070.07August1.831.86-0.030.

03September1.71.83-0.130.13October1.651.78-0.130.13November1.71.73-0.030.03December1.751.680.070.07MAD0.09The 2-month average is better because it has a lower MAD, hence more accurate.d) Compute the forecasts for each month using exponential smoothing, with an initial forecast for January of $1.80. Use α = .1, then α = .3, and finally α = .5. Using MAD, which α is the best?MonthPrice per Chip ($)Forecast using exponential smoothing ( alpha = 0.1)ErrorAbsoluteJanuary1.81.80.000.000February1.671.8-0.130.130March1.71.79-0.090.087April1.851.780.070.072May1.91.790.110.115June1.871.800.070.073July1.81.800.000.004August1.831.800.030.026September1.71.81-0.110.106October1.651.80-0.150.146November1.71.78-0.080.081December1.751.77-0.020.023MAD0.

072MonthPrice per Chip ($)Forecast using exponential smoothing ( alpha = 0.3)ErrorAbsoluteJanuary1.81.80.000.000February1.671.8-0.130.130March1.71.76-0.060.061April1.851.740.110.107May1.91.770.130.125June1.871.810.060.058July1.81.83-0.030.030August1.831.820.010.009September1.71.82-0.120.124October1.651.79-0.140.136November1.71.75-0.050.046December1.751.730.020.018   MAD0.070MonthPrice per Chip ($)Forecast using exponential smoothing ( alpha = 0.5)ErrorAbsoluteJanuary1.81.80.000.000February1.671.8-0.130.130March1.71.74-0.030.035April1.851.720.130.133May1.91.780.120.116June1.871.840.030.028July1.81.86-0.060.056August1.831.830.000.002September1.71.83-0.130.129October1.651.76-0.110.114November1.71.71-0.010.007December1.751.700.050.046MAD0.

066 The Forecast using exponential smoothing using alpha = 0.5 is better because it has the lowest MAD (Abraham & Leddolter, 2005).4.11 a) Use exponential smoothing with a smoothing constant of 0.3 to forecast the registrations at the seminar given in Problem 4.10. To begin the procedure, assume that the forecast for year 1 was 5,000 people signing up. (Abraham & Leddolter, 2005).YearRegistrations (000)Forecast registrations (‘000) using exponential smoothing ( alpha = 0.3)145264.7345.09454.765104.83686.38776.87896.919127.5410148.87111510.41b) What is the MAD?

YearRegistrations (000)Forecast registrations (‘000) using exponential smoothing ( alpha = 0.3)ErrorAbsolute145-1.001.00264.71.301.30345.09-1.091.09454.760.240.245104.835.175.17686.381.621.62776.870.130.13896.912.092.099127.544.464.4610148.875.135.13111510.414.594.59MAD2.44ReferenceAbraham, B., & Leddolter, J. (2005). Statistical Methods for Forecasting. New York: Wiley.