For the following list of data (number of sold cars in every year ) , apply “exponential smoothing with trend” for all the years below. (i.e find the Forecast including trend) : case1 use : α= 0.01 and β = 0.3 case2 use: for α= 0.1 and β = 0.05 Calculate MAD for both cases. Decide which case values are better and why? Assume trend is zero for the 2012 years. Year Number of sold cars 2012 110 2013 123 2014 160 2015 133 2016 151 2017 175 2018 189 2019 211 2020 255 Based on least squared regression method find the projections for the years 2021, 2022, and 2023. Plot the number of sold cars versus years (points graph) and show the trend line based on your projections.
For the following list of data (number of sold cars in every year ) , apply “exponential smoothing with trend” for all the years below. (i.e find the Forecast including trend) : case1 use : α= 0.01 and β = 0.3 case2 use: for α= 0.1 and β = 0.05 Calculate MAD for both cases. Decide which case values are better and why? Assume trend is zero for the 2012 years. Year Number of sold cars 2012 110 2013 123 2014 160 2015 133 2016 151 2017 175 2018 189 2019 211 2020 255 Based on least squared regression method find the projections for the years 2021, 2022, and 2023. Plot the number of sold cars versus years (points graph) and show the trend line based on your projections.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Please check the documents uploaded, which have my answers to the below questions. Please check if my answers are correct, especially the graph for the 2nd question, as I am unsure which one is correct.
(USE ONLY EXCEL SHEET FOR DATA, FORMULAS, and the GRAPHS)
- For the following list of data (number of sold cars in every year ) , apply “exponential smoothing with trend” for all the years below. (i.e find the Forecast including trend) :
- case1 use : α= 0.01 and β = 0.3
- case2 use: for α= 0.1 and β = 0.05
- Calculate MAD for both cases.
- Decide which case values are better and why?
Assume trend is zero for the 2012 years.
|
Year |
Number of sold cars |
|
2012 |
110 |
|
2013 |
123 |
|
2014 |
160 |
|
2015 |
133 |
|
2016 |
151 |
|
2017 |
175 |
|
2018 |
189 |
|
2019 |
211 |
|
2020 |
255 |
- Based on least squared regression method find the projections for the years 2021, 2022, and 2023. Plot the number of sold cars versus years (points graph) and show the trend line based on your projections.
Transcribed Image Text:Year
2012
2013
2014
2015
2016
2017
2018
2019
2020
Year
2012
2013
2014
2015
2016
2017
2018
2019
2020
Answer question 1 - 1st Case:
Period
Number
(x)
1
2
3
4
5
6
7
8
9
Period
Number
(x)
Number of
sold cars
Answer question 1 - 2nd Case:
1
21
3
4
5
6
7
8
9
O
110
110
0
123
110.13
0.039
160
110.6287
0.17691
133
110.852413 0.1909509
151
111.2538889 0.254108391
175
111.89135 0.369114207
112.6624365 0.489705895
189
211 113.6458121 0.637806817
255
115.059354 0.870527336
Smoothed Value Trend value Forecast including trend
Number of
sold cars
110
123
160
133
151
175
189
211
255
Here are my answers to be verified:
110
0
111.3
116.17
117.853 0.3741375
0.065
0.30525
121.1677 0.521165625
126.55093 0.764268844
132.795837 1.038300752
140.6162533 1.377406529
152.054628 1.880454936
110
110.169
110.80561
111.0433639
111.5079973
112.2604642
113.1521424
114.2836189
115.9298813
MAD
Smoothed Value Trend value Forecast including trend
110
111.365
116.47525
118.2271375
121.6888656
127.3151988
133.8341378
141.9936598
153.9350829
MAD
Absolute error
0
12.831
49.19439
21.9566361
39.49200274
62.73953581
75.84785762
96.71638107
139.0701187
55.31643578
Absolute error
11.635
43.52475
14.7728625
29.31113438
47.68480116
55.16586225
69.00634017
101.0649171
41.35174084
Value
Value
300
250
200
150
100
50
0
300
250
200
150
100
50
0
0
0
2
Number of sold cars
(a=0.01 & B=0.3)
y = 15.733x+88.778
4
6
Number of Period (x)
Number of sold cars
(α=0.1 & 3=0.05)
y = 15.733x+88.778
Number of Period (x)
8
Since the MAD value for case 2 (41.3) is lower than the MAD value for case 1 (55.3), this suggests that
the forecasting method used in case 2 (where a=0.1 and B=0.05) is more accurate than the forecasting
method used in case 1 (where a=0.01 and B=0.3).
8
10
10
Transcribed Image Text:Year
Answer 2nd Question: Which graph is the correct one?
2012
2013
2014
2015
2016
2017
2018
2019
2020
2021
2022
2023
Period
Number
(x)
1
2
3
4
5
6
7
8
9
10
11
12
Number of sold
cars
110
123
160
133
151
175
189
211
255
246.508
262.281
278.054
VALUE
VALUE
VALUE
300
250
200
150
100
50
0
300
250
200
150
100
50
0
300
250
200
150
100
50
0
0
0
1
0
1
Number of sold cars
----- Linear (Projected Number of sold Cars)
2
1
2
3 4
Number of sold cars.
2
3 4
5 6 7 8
NUMBER OF PERIOD (X)
3
Number of sold cars
----- Linear (Number of sold cars)
------ Projected Number of sold Cars
5 6 7 8
NUMBER OF PERIOD (X)
9 10 11
R² = 0.6713
----- Projected Number of sold Cars
4 5 6 7 8 9 10
NUMBER OF PERIOD (X)
12
9 10 11 12 13
13
11 12 13
Projected Number of sold Cars
----- Linear (Projected Number of sold Cars)
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