Forecasting model

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Please see the attached file all the details. Calculations and solutions are in blue and different colors. Hope this helps you. For any further clarification, feel free to contact.

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Problem 2 a)
The number of cans of soft drinks sold in a machine each week is recorded below from left to right, with oldest data to the left of the table.
323 299 259 287 302

Read instructions carefully:
Develop two forecasting models which are described below, and use MAD to compare them:
• Model 1: Three period weighted moving average (weights being 5, 2, 1 respectively for one period ago, 2 periods ago and 3 periods ago).
• Model 2: Single exponential smoothing, with smoothing constant = 0.35. Use 323 as Forecast for time period 2 to start your calculations.
Model 1: Calculation of weights is as follows
For period t-1 = 5/(5+2+1) = 0.625
For period t-2 = 2/(5+2+1) = 0.25
For period t-3 = 1/(5+2+1) = 0.125
For model 1, forecast for periods 5 and 6 can be done as prior three weeks data is available. However, this will result in only two values of MAD for comparison. Otherwise, it is simply assumed that the demand in three weeks prior to week 1 is 323. Following are the calculations with these assumptions. Generally, this is not required as forecasting is done for more number of periods and there are sufficient values for comparison.
MAD for model 1 = (0+24+49+10+20.5)/5 = 20.7

Model 2: The forecast for period t = Forecast for period t-1 + smoothing constant*(Actual demand in t-1 - forecast for t-1). The assumption about demand in week 0 as in case of model 1 holds for this model also.
MAD for model 2 = (0+24+55.6+8.14+9.7)/5 = 19.488

Fill out the table below to report your forecasts and accuracy for each model.

t
Y Model 1
Develop Forecasts for
Weighted Moving Average Absolute
Error
Model 1 Model 2
Develop Forecasts for Exponential Smoothing Absolute
Error
Model 2
1
323 0.625*323 + 0.25*323 + 0.125*323
= 323 |323-323|
= 0 323 + 0.35*(323-323)
= 323 |323-323|
= 0
2 299 0.625*323 + 0.25*323 + 0.125*323
= 323 |299-323|
= 24 323 + 0.35*(323-323)
= 323 |299-323|
= 24
3 259 0.625*299 + 0.25*323 + 0.125*323
= 308 |259-308|
= 49 323 + 0.35*(299-323)
= 314.6 |259-314.6|
= 55.6
4 287 0.625*259 + 0.25*299 + 0.125*323
= 277 |287-277|
= 10 314.6 + 0.35*(259-314.6)
= 295.14 |287-295.14|
= 8.14
5 302 0.625*287 + 0.25*259 + 0.125*299
= 281.5 |302-281.5|
= 20.5 295.14 + 0.35*(287-295.14)
= 292.3 |302-292.3|
= 9.7

6 ? 292.3 + 0.35*(302-292.3)
= 295.7

a. Based on MAD as the forecast accuracy, which of the two models should be used to forecast this time series? _____ Model 1 _____ Model 2
As MAD for exponential smoothing method is less than the moving average method, former is chose.

b. Use the model you selected to forecast number of cans sold in the following week: ___295.7________

Problem 2 b)
In the space provided, draw the network for this project. (The network is as shown below)
Use the 3 time estimates provided for each activity (in weeks) to estimate the expected time and variance for the activities. Then work the network forward and backward to arrive at the expected project completion time.
Expected time for each activity = (Optimistic time + 4*most likely time+ pessimistic time)/6
Standard deviation of time for each activity = (Pessimistic time - Optimistic time)/6
Variance = (Standard deviation)^2
To calculate completion time, expected times are used for calculations.
Earliest start time = earliest start time of preceding activity + expected completion time for preceding activity
Earliest start time for the activity having two preceding activity = max {earliest start time of preceding activity i + expected completion time for activity i}
Earliest finish time = earliest start time + ...