LE ANH KHOA – BABAIU14124 QUANTITATIVE METHODS FOR BUSINESS – ASSIGNMENT: FORECASTING 5–15: 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? DEMAND FOR FERTILIZER (1,000S OF BAGS) 4 6 4 5 10 8 7 9 12 14 15
YEAR 1 2 3 4 5 6 7 8 9 10 11
Solution: *We use the 3–year moving average to forecast sales. We have the table as following: YEAR 1 2 3 4 5 6 7 8 9 10 11
FERTILIZER DEMAND (1,000S OF BAGS) 4 6 4 5 10 8 7 9 12 14 15
3–YEAR MOVING AVERAGE
ERROR
– – – (4 + 6 + 4)/3 = 4.67 (6 + 4 + 5)/3 = 5.00 (4 + 5 + 10)/3 = 6.33 (5 + 10 + 8)/3 = 7.67 (10 + 8 + 7)/3 = 8.33 (8 + 7 + 9)/3 = 8.00 (7 + 9 + 12)/3 = 9.33 (12 + 14 + 15)/3 = 13.67
– – – 0.33 5.00 1.67 –0.67 0.67 4.00 4.67 3.33 MAD = 2.5425
*We use the weighted moving average to forecast sales. We have the table as following: YEA R 1 2
DEMAND FOR FERTILIZER (1,000S OF BAGS) 4 6
3–YEAR WEIGHTED MOVING AVERAGE
ERROR
– –
– –
LE ANH KHOA – BABAIU14124 3
4
4
5
5
10
6
8
7
7
8
9
9
12
10
14
11
15
– [(2 X 4) + (1 X 6) + (1 X 4)]/4 = 4.50 [(2 X 5) + (1 X 4) + (1 X 6)]/4 = 5.00 [(2 X 10) + (1 X 5) + (1 X 4)]/4 = 7.25 [(2 X 8) + (1 X 10) + (1 X 5)]/4 = 7.75 [(2 X 7) + (1 X 8) + (1 X 10)]/4 = 8.00 [(2 X 9) + (1 X 7) + (1 X 8)]/4 = 8.25 [(2 X 12) + (1 X 9) + (1 X 7)]/4 = 10.00 [(2 X 14) + (1 X 12) + (1 X 9)]/4 = 12.25
– 0.50 5.00 0.75 –0.75 1.00 3.75 4.00 2.75 MAD = 2.3125
Thus, since the MAD of the weighted moving average is smaller than of the 3–year moving average, the better method to forecast the sales will be the weighted moving average.
LE ANH KHOA – BABAIU14124 5–16: Develop a trend line for the demand for fertilizer in Problem 5–15, using any computer software. Solution: Let X be the time period (1, 2, …) and Y be the demand of fertilizer. Using the function “Data Analysis – Regression” of Excel, we have the following tables: Regression Statistics 0.906914487 0.822493888 0.802770986 1.712697677 11
Multiple R R Square Adjusted R Square Standard Error Observations ANOVA Regression Residual Total
Intercept YEAR (X)
df 1 9 10
SS 122.3272727 26.4 148.7272727
Coefficients 2.218181818 1.054545455
MS 122.3272727 2.933333333
Standard Error 1.107549848 0.163299316
F 41.70247934
Significance F 0.000117077
t–Stat 2.002782829 6.457745686
P–value 0.076210093 0.000117077
So, based on the tables, we can see that the trend line for the demand for fertilizer will be: Ŷ = b0 + b1X = 2.22 + 1.05X where X = time period (1, 2, …) and Ŷ = predicted demand
LE ANH KHOA – BABAIU14124 5–17: In Problems 5–15 and 5–16, 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. Solution: Based on the trend line ( Ŷ YEAR (X) 1 2 3 4 5 6 7 8 9 10 11
= 2.22 + 1.05X), we have the following table:
ACTUAL DEMAND 4 6 4 5 10 8 7 9 12 14 15
PREDICTED DEMAND (Y) 3.27 4.32 5.38 6.44 7.49 8.55 9.60 10.65 11.71 12.76 13.82
ERROR 0.73 1.68 –1.38 –1.44 2.51 –0.55 –2.60 –1.65 0.29 1.24 1.18 MAD = 1.39
Thus, since the MAD of the trend line is lowest, the trend line is the best method in order to forecast the demand of fertilizer.
LE ANH KHOA – BABAIU14124 5–18: Use exponential smoothing with a smoothing constant of 0.3 to forecast the demand for fertilizer given in Problem 5–15. Assume that last period’s forecast for year 1 is 5,000 bags to begin the procedure. Would you prefer to use the exponential smoothing model or the weighted average model developed in Problem 5–15? Explain your answer. Solution: We use the exponential smoothing with a smoothing constant of 0.3 to forecast the demand for fertilizer. We have the table as following (with the new assumption): YEAR
DEMAND
1
4,000
2
6,000
3
4,000
4
5,000
5
10,000
6
8,000
7
7,000
8
9,000
9
12,000
10
14,000
11
15,000
FORECASTED DEMAND 5,000 5,000 + (0.3)*(4,000 – 5,000) = 4,700 4,700 + (0.3)*(6,000 – 4,700) = 5,090 5,090 + (0.3)*(4,000 – 5,090) = 4,763 4,763 + (0.3)*(5,000 – 4,763) = 4,834 4,834 + (0.3)*(10,000 – 4,834) = 6,384 6,384 + (0.3)*(8,000 – 6,384) = 6,869 6,869 + (0.3)*(7,000 – 6,869) = 6,908 6,908 + (0.3)*(9,000 – 6,908) = 7,536 7,536 + (0.3)*(12,000 – 7,536) = 8,857 8,875 + (0.3)*(14,000 – 8,875) = 10,412
ERROR –1,000 1,300 –1,090 237 5,166 1,616 131 2,092 4,464 5,125 4,588 MAD (in thousands) = 2,437
Thus, since the MAD of the weighted moving average is smaller than of the exponential smoothing, the better method to forecast the sales will be the weighted moving average.
LE ANH KHOA – BABAIU14124 5–25: Sales of industrial vacuum cleaners at R. Lowenthal Supply Co. over the past 13 months are as follows: SALES ($1,000s) 11 14 16 10 15 17 11
MONTH January February March April May June July
SALES ($1,000s) 14 17 12 14 16 11
MONTH August September October November December January
(a) Using a moving average with three periods, determine the demand for vacuum cleaners for next February. (b) Using a weighted moving average with three periods, determine the demand for vacuum cleaners for February. Use 3, 2, and 1 for the weights of the most recent, second most recent, and third most recent periods, respectively. For example, if you were forecasting the demand for February, November would have a weight of 1, December would have a weight of 2, and January would have a weight of 3. (c) Evaluate the accuracy of each of these methods. (d) What other factors might R. Lowenthal consider in forecasting sales? Solution: (a), (b) Perio d 1 2 3
Mont h Jan. Feb. Mar.
Deman d 11 14 16
4
Apr.
10
5
May
15
6
June
17
7
July
11
8
Aug.
14
9
Sept.
17
10
Oct.
12
11
Nov.
14
12
Dec.
16
3–period Moving Average – – – (11 + 14 + 16)/3 = 13.67 (14 + 16 + 10)/3 = 13.33 (16 + 10 + 15)/3 = 13.67 (10 + 15 + 17)/3 = 14.00 (15 + 17 + 11)/3 = 14.33 (17 + 11 + 14)/3 = 14.00 (11 + 14 + 17)/3 = 14.00 (14 + 17 + 12)/3 = 14.33 (17 + 12 + 14)/3 = 14.33
Error – – – –3.67 1.67 3.33 –3.00 –0.33 3.00 –2.00 –0.33 1.67
3–period Weighted Moving Average – – – [(3 X 16) + (2 X 14) + (1 X 11)]/6 = 14.50 [(3 X 10) + (2 X 16) + (1 X 14)]/6 = 12.67 [(3 X 15) + (2 X 10) + (1 X 16)]/6 = 13.50 [(3 X 17) + (2 X 15) + (1 X 10)]/6 = 15.17 [(3 X 11) + (2 X 17) + (1 X 15)]/6 = 13.67 [(3 X 14) + (2 X 11) + (1 X 17)]/6 = 13.50 [(3 X 17) + (2 X 14) + (1 X 11)]/6 = 15.00 [(3 X 12) + (2 X 17) + (1 X 14)]/6 = 14.00 [(3 X 14) + (2 X 12) + (1 X 17)]/6 = 13.83
Error – – – –4.50 2.33 3.50 –4.17 0.33 3.50 –3.00 0 2.17
LE ANH KHOA – BABAIU14124
13
Jan.
11
14
Feb.
–
(12 + 14 + 16)/3 = 14.00 (14 + 16 + 11)/3 = 13.67
–3.00 – MAD = 2.20
[(3 X 16) + (2 X 14) + (1 X 12)]/6 = 14.67 [(3 X 11) + (2 X 16) + (1 X 14)]/6 = 13.17
–3.67 – MAD = 2.72
(c) As can be seen from the table, the MAD for the moving average is 2.20, and the MAD for the weighted moving average is 2.72. Thus, based on this analysis, the moving average appears to be more accurate; and, the forecast for February is 13.67 (about 14). (d) There are many other factors to consider, including seasonality and any underlying causal variables such as advertising budget.
LE ANH KHOA – BABAIU14124 5–27: Emergency calls to Winter Park, Florida’s 911 system, for the past 24 weeks are as follows: WEEK 1 2 3 4 5 6 7 8
CALLS 50 35 25 40 45 35 20 30
WEEK 9 10 11 12 13 14 15 16
CALLS 35 20 15 40 55 35 25 55
WEEK 17 18 19 20 21 22 23 24
CALLS 55 40 35 60 75 50 40 65
(a) Compute the exponentially smoothed forecast of calls for each week. Assume an initial forecast of 50 calls in the first week and use α =0.1 . What is the forecast for the 25th week? (b) Reforecast each period using α =0.6 (c) Actual calls during the 25th week were 85. Which smoothing constant provides a superior forecast? Solution: (a), (b), (c) We have the following table:
1
Actua l Value 50
2
35
3
25
4
40
5
45
6
35
7
20
8
30
9
35
10
20
11
15
12
40
13
55
14
35
Week
Smoothed Value ( α =0.1 ) 50.00 50.00 + (0.1)*(50 – 50.00) = 50.00 50.00 + (0.1)*(35 – 50.00) = 48.50 48.50 + (0.1)*(25 – 48.50) = 46.15 46.16 + (0.1)*(40 – 46.15) = 45.53 45.53 + (0.1)*(45 – 45.53) = 45.47 45.47 + (0.1)*(35 – 45.47) = 44.42 44.42 + (0.1)*(20 – 44.42) = 41.98 41.98 + (0.1)*(30 – 41.98) = 40.78 40.78 + (0.1)*(35 – 40.78) = 40.20 40.20 + (0.1)*(20 – 40.20) = 38.18 38.18 + (0.1)*(15 – 38.18) = 35.86 35.86 + (0.1)*(40 – 35.86) = 36.27 36.27 + (0.1)*(55 – 36.27)
Forecas t Error 0 –15.00 –23.50 –6.15 –0.53 –10.47 –24.42 –11.98 –5.78 –20.20 –20.18 4.14 18.73 –3.14
Smoothed Value ( α =0.6 ) 50.00 50.00 + (0.6)*(50 – 50.00) = 50 50.00 + (0.6)*(35 – 50.00) = 41.00 41.00 + (0.6)*(25 – 41.00) = 31.40 31.40 + (0.6)*(40 – 31.40) = 36.56 36.56 + (0.6)*(45 – 36.56) = 41.62 41.62 + (0.6)*(35 – 41.62) = 37.65 37.65 + (0.6)*(20 – 37.65) = 27.06 27.06 + (0.6)*(30 – 27.06) = 28.82 28.82 + (0.6)*(35 – 28.82) = 32.53 32.53 + (0.6)*(20 – 32.53) = 25.01 25.01 + (0.6)*(15 – 25.01) = 19.00 19.00 + (0.6)*(40 – 19.00) = 31.60 31.60 + (0.6)*(55 – 31.60)
Forecas t Error 0 –15.00 –16.00 8.60 8.44 –6.62 –17.65 2.94 6.18 –12.53 –10.01 21.00 23.40 –10.64
LE ANH KHOA – BABAIU14124
15
25
16
55
17
55
18
40
19
35
20
60
21
75
22
50
23
40
24
65
25
85
= 38.14 38.14 + (0.1)*(35 – 38.14) = 37.83 37.83 + (0.1)*(25 – 37.83) = 36.55 36.55 + (0.1)*(55 – 36.55) = 38.40 38.40 + (0.1)*(55 – 38.40) = 40.06 40.06 + (0.1)*(40 – 40.06) = 40.05 40.05 + (0.1)*(35 – 40.05) = 39.55 39.55 + (0.1)*(60 – 39.55) = 41.60 41.60 + (0.1)*(75 – 41.60) = 44.94 44.94 + (0.1)*(50 – 44.94) = 45.45 45.45 + (0.1)*(40 – 45.45) = 44.91 44.91 + (0.1)*(65 – 44.91) = 46.92
–12.83 18.45 16.60 –0.06 –5.50 20.45 33.40 5.06 –5.45 20.09 38.08 MAD = 13.61
Thus, since the MAD of should be use to forecast is
α=0.1
is smaller than of
α =0.1 .
= 45.64 45.64 + (0.6)*(35 – 45.64) = 39.26 39.26 + (0.6)*(25 – 39.26) = 30.70 30.70 + (0.6)*(55 – 30.70) = 45.28 45.28 + (0.6)*(55 – 45.28) = 51.11 51.11 + (0.6)*(40 – 51.11) = 44.44 44.44 + (0.6)*(35 – 44.44) = 38.78 38.78 + (0.6)*(60 – 38.78) = 51.51 51.51 + (0.6)*(75 – 51.51) = 65.60 65.60 + (0.6)*(50 – 65.60) = 56.24 56.24 + (0.6)*(40 – 56.24) = 46.50 46.50 + (0.6)*(65 – 46.50) = 57.60
–14.26 24.30 9.72 –11.11 –9.44 21.22 23.49 –15.60 –16.24 18.50 27.40 MAD = 14.01
α =0.6 , the better smoothing constant
