FDM Assignment 1 (Babar Ali) PDF

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Assignment#1

Course Title

:

Forecasting and Decision Making

Course Code

:

EMG 605

Class

:

MS-EM 3

Teacher’s Name

:

MUHAMMAD SAJID

Student Name

:

Babar Ali

Enrollment No.

:

01-246192-003

Marks

:

________________

Department of Software Engineering Bahria University Islamabad Campus (BUIC)

Forecasting and Decision Making (MS-EM)

1.What are some of the problems and drawbacks of the moving average forecasting model? Answer: Moving averages can be used for measuring the trend of any series. This method is applicable to linear as well as non-linear trends. Drawback & Disadvantages: 1) The trend obtained by moving averages generally is neither a straight line nor a standard curve. For this reason, the trend cannot be extended for forecasting future values. Trend values are not availab le for some periods at the start and some values at the end of the time series. This method is not applicable to short time series. 2) The disadvantages of the moving average forecasting model are that the averages always stay within past levels, and the moving averages do not consider seasonal variations 2.A seasonal index may be less than one, equal to one, or greater than one. Explain what each of these values would mean. Answer: A seasonal index is a measure of how a particular season through some cycle compares with the average season of that cycle. a) If a seasonal index equals 1, that season is just an average season. b) If the index is less than 1, that season tends to be lower than average. c) If the index is greater than 1, that season tends to be higher than average. 3.Explain what would happen if the smoothing constant in an exponential smoothing model was equal to zero. Explain what would happen if the smoothing constant was equal to one. Answer: As we know that 𝑭𝒕+𝟏 = 𝑭𝒕 + 𝛂(𝒀𝒕 − 𝑭𝒕 ) So, If the smoothing constant equals 0, then 𝑭𝒕+𝟏 = 𝑭𝒕 + 𝟎 (𝒀𝒕 − 𝑭𝒕 ) = 𝑭𝒕 This means that the forecast never changes, If the smoothing constant equals 0. And, If the smoothing constant equals 1, then 𝑭𝒕+𝟏 = 𝑭𝒕 + 𝟏 (𝒀𝒕 − 𝑭𝒕 ) = 𝒀𝒕 This means that the forecast is always equal to the actual value in the prior period if the smoothing constant equals 1

Department of Software Engineering Bahria University Islamabad Campus

4.Data collected on the yearly demand for 50-pound bags of fertilizer at Wallace Garden Supply are shown in the following table. Develop a 3-year moving average to forecast sales. Then estimate demand again with a weighted moving average in which sales in the most recent year are given a weight of 2 and sales in the other 2 years are each given a weight of 1. Which method do you think is best?

Answer:

MAD for 3-year average = 20.36/8 = 2.55 MAD for weighted 3-year average = 18.55/8 = 2.32 So, The weighted moving average appears to be slightly more accurate in its annual forecasts Department of Software Engineering Bahria University Islamabad Campus

5. Develop a trend line for the demand for fertilizer. Answer: As we know that, A trend line is simply a linear regression equation in which the independent variable (X) is the time period. And,  = 𝒃𝟎 + 𝒃𝟏𝑿 𝒀 And.  =predicted value 𝒀

b0 = intercept

b1 = slope of the line

X = time period

And

Σ [(𝑥𝑖 − 𝑥 )(𝑦𝑖 − 𝑦)] 𝑏1 = Σ(𝑥𝑖 − 𝑥 )2 Also

𝑏0 = 𝑦 − 𝑏1 𝑥 𝒙 =𝟔

𝒚  = 𝟖. 𝟓

So the calculations are: YEARS(𝒙𝒊 ) 1 2 3 4 5 6 7 8 9 10 11 𝒏 = 𝟏𝟏

Demand(𝒚 𝒊) 4 6 4 5 10 8 7 9 12 14 15 𝚺𝒚𝒊 = 𝟗𝟒

( 𝒙𝒊 − 𝒙 )

(𝒚 𝒊 − 𝒚 )

( 𝒙𝒊 − 𝒙 )( 𝒚 𝒊 − 𝒚 )

( 𝒙𝒊 − 𝒙 )𝟐

-5 -4 -3 -2 -1 0 1 2 3 4 5

-4.5 -2.5 -4.5 -3.5 1.5 -0.5 -1.5 0.5 3.5 5.5 6.5

22.5 10 13.5 7 -1.5 0 -1.5 1 10.5 22 32.5 𝚺(𝒙𝒊 − 𝒙 )(𝒚 𝒊 − 𝒚 )= 116

25 16 9 4 1 0 1 4 9 16 25 𝚺(𝒙𝒊 − 𝒙 )𝟐 = 110

So

𝑏1 =

116

=1.05

110

And 𝑏0 = 8.5 − 1.05(6) = 2.22

Department of Software Engineering Bahria University Islamabad Campus

So for the above values of fertilizers  𝒀 = 2.22 + 1.05𝑋

So the Trend line will be:

Department of Software Engineering Bahria University Islamabad Campus

6.In the above two problems, three different forecasts were developed for the demand for fertilizer. These three forecasts are a 3-year moving average, a weighted moving average, and a trend line. Which one would you use? Explain your answer. Answer:

Using the forecasts in the previous problem i obtained the absolute deviations given in the table below.

And, MAD (3-year moving average) = 20.36/8 = 2.55 MAD (3-year weighted moving average) = 18.55/8 = 2.32 MAD (trend line) = 15.24/12 = 1.39

The trend line is best because the MAD for that method is lowest.

Department of Software Engineering Bahria University Islamabad Campus

7.Use exponential smoothing with a smoothing constant of 0.3 to forecast the demand for fertilizer given in above table. Assume that last period’s forecast for year 1 is 5,000 bags to begin the procedure. Answer: 𝜶 = 𝟎. 𝟑

𝑭𝒐𝒓 𝒕𝒉𝒆 𝒇𝒊𝒓𝒔𝒕 𝒚𝒆𝒂𝒓 = 𝟓, 𝟎𝟎𝟎 𝒐𝒓 𝟓(𝟏𝟎𝟎𝟎𝒔 𝒐𝒇 𝒃𝒂𝒈𝒔)

As we know that New forecast for year 2 = last period’s forecast + 𝛼(last period’s actual demand −last period’s forecast) 𝑭𝒕+𝟏 = 𝑭𝒕 + 𝛂(𝒀𝒕 − 𝑭𝒕 ) So, New forecast for year 2 = 5,000 + (0.3) (4,000 – 5,000) = 5,000 + (0.3) (– 1,000) = 5,000 – 300 = 4,700 or 4.7(1000s of Bags)

The Calculations are (1000s of bags) YEARS DEMAND 1 2 3 4 5 6 7 8 9 10 11

4 6 4 5 10 8 7 9 12 14 15

FORCAST

DEVIATION

|DEVIATION|

----4.7 = 5 + (0.3)(4− 5) 5.09 = 4.7 + (0.3)(6 – 4.7) 4.76 = 5.09 + (0.3)(4 – 5.09) 4.83 = 4.76 + (0.3)(5 – 4.76) 6.38 = 4.83 + (0.3)(10 – 4.83) 6.86 = 6.38 + (0.3)(8 – 6.38) 6.90 = 6.86 + (0.3)(7 – 6.86) 7.53 = 6.90 + (0.3)(9 – 6.90) 8.87 = 7.53 + (0.3)(12 – 7.53) 10.41 = 8.87 + (0.3)(14 – 8.87)

---1.3 -1.09 0.24 5.17 1.62 0.14 2.1 4.47 5.13 4.59

---1.3 1.09 0.24 5.17 1.62 0.14 2.1 4.47 5.13 4.59

As, 𝑀𝐴𝐷 = Σ

|𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛| 𝑛

So, 𝑀𝐴𝐷 =

1.3 + 1.09 + 0.24 + 5.17 + 1.62 + 0.14 + 2.1 + 4. 47 + 5.13 + 4.59 10

𝑀𝐴𝐷 =

25.85 10

𝐌𝐀𝐃 = 𝟐. 𝟓𝟖

Department of Software Engineering Bahria University Islamabad Campus...