Obtain exponential smoothing for α = 0.1 and α = 0.2 . Identify the preferred smoothing constant using MSE measure of forecast accuracy.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 8, Problem 8P

(a)

To determine

Obtain exponential smoothing for α=0.1 and α=0.2.

Identify the preferred smoothing constant using MSE measure of forecast accuracy.

(a)

Expert Solution

Explanation of Solution

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	16.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	2.56
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	29.59
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.01
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	4.38
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	4.48
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.01
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	15.32
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	2.32
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	13.18
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	13.94
			Total	101.78

MSE=∑|er|2n−k=101.7811=9.253

Thus, the mean squared error is 9.253.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	16.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.44
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	24.60
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.07
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	7.98
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	3.03
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.37
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	12.34
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.66
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	18.94
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	12.38
			Total	98.80

MSE=∑|er|2n−k=98.8011=8.982

Thus, the mean squared error is 8.982.

MSE when α=0.2 is less than MSE when α=0.1. Thus, α=0.2 is preferred.

(b)

To determine

Check whether the results are same when MAE is used as measure of accuracy.

(b)

Expert Solution

Answer to Problem 8P

No, the results are not same.

Explanation of Solution

The MSE for four-week moving average is obtained as given below:

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Absolute Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	4.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	1.60
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	5.44
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.10
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	2.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	2.12
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.10
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	3.91
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	1.52
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	3.63
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	3.73
			Total	28.25

MAE=∑|er|n−k=28.258=2.568

Thus, the mean absolute error is 2.568.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	4.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.20
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	4.96
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.03
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	2.83
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	1.74
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.61
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	3.51
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.81
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	4.35
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	3.52
			Total	28.56

MAE=∑|er|n−k=28.5611=2.596

Thus, the mean absolute error is 2.596.

MAE when α=0.1 is less than MAE when α=0.2. Thus, α=0.1 is preferred.

Hence, the results are not same when MAE is used as measure of accuracy.

(c)

To determine

Obtain the results when MAPE is used as measure of accuracy.

(c)

Expert Solution

Answer to Problem 8P

MAPE when α=0.1 is 12.95.

MAPE when α=0.2 is 13.40.

Explanation of Solution

Exponential smoothing for α=0.1:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	19.05
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	8.42
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	23.65
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.58
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	13.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	10.58
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.53
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	17.79
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	7.61
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	24.20
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	16.97
			Total	142.46

MAPE=|100×Forecast errorTime series value|n−k=142.4611=12.95

Thus, the value of MAPE is 12.95.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	19.05
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	6.32
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	21.57
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	5.73
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	17.66
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	8.70
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	3.38
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	15.97
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	4.05
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	29.01
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	15.99
			Total	147.43

MAPE=|100×Forecast errorTime series value|n−k=147.4311=13.40

Thus, the value of MAPE is 13.40.

MAPE when α=0.1 is less than MAPE when α=0.2. Thus, α=0.1 is preferred.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

For each of the time series, construct a line chart of the data and identify the characteristics of the time series (that is, random, stationary, trend, seasonal, or cyclical). Year Month Rate (%)2009 Mar 8.72009 Apr 9.02009 May 9.42009 Jun 9.52009 Jul 9.52009 Aug 9.62009 Sep 9.82009 Oct 10.02009 Nov 9.92009 Dec 9.92010 Jan 9.82010 Feb 9.82010 Mar 9.92010 Apr 9.92010 May 9.62010 Jun 9.42010 Jul 9.52010 Aug 9.52010 Sep 9.52010 Oct 9.52010 Nov 9.82010 Dec 9.32011 Jan 9.12011 Feb 9.02011 Mar 8.92011 Apr 9.02011 May 9.02011 Jun 9.12011 Jul 9.02011 Aug 9.02011 Sep 9.02011 Oct 8.92011 Nov 8.62011 Dec 8.52012 Jan 8.32012 Feb 8.32012 Mar 8.22012 Apr 8.12012 May 8.22012 Jun 8.22012 Jul 8.22012 Aug 8.12012 Sep 7.82012 Oct…

For each of the time series, construct a line chart of the data and identify the characteristics of the time series (that is, random, stationary, trend, seasonal, or cyclical). Date IBM9/7/2010 $125.959/8/2010 $126.089/9/2010 $126.369/10/2010 $127.999/13/2010 $129.619/14/2010 $128.859/15/2010 $129.439/16/2010 $129.679/17/2010 $130.199/20/2010 $131.79 a. Construct a line chart of the closing stock prices data. Choose the correct chart below.

For each of the time series, construct a line chart of the data and identify the characteristics of the time series (that is, random, stationary, trend, seasonal, or cyclical) Date IBM9/7/2010 $125.959/8/2010 $126.089/9/2010 $126.369/10/2010 $127.999/13/2010 $129.619/14/2010 $128.859/15/2010 $129.439/16/2010 $129.679/17/2010 $130.199/20/2010 $131.79

Answer 2

Question

Chapter 8, Problem 8P

(a)

To determine

Obtain exponential smoothing for α=0.1 and α=0.2.

Identify the preferred smoothing constant using MSE measure of forecast accuracy.

(a)

Expert Solution

Explanation of Solution

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	16.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	2.56
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	29.59
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.01
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	4.38
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	4.48
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.01
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	15.32
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	2.32
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	13.18
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	13.94
			Total	101.78

MSE=∑|er|2n−k=101.7811=9.253

Thus, the mean squared error is 9.253.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	16.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.44
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	24.60
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.07
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	7.98
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	3.03
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.37
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	12.34
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.66
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	18.94
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	12.38
			Total	98.80

MSE=∑|er|2n−k=98.8011=8.982

Thus, the mean squared error is 8.982.

MSE when α=0.2 is less than MSE when α=0.1. Thus, α=0.2 is preferred.

(b)

To determine

Check whether the results are same when MAE is used as measure of accuracy.

(b)

Expert Solution

Answer to Problem 8P

No, the results are not same.

Explanation of Solution

The MSE for four-week moving average is obtained as given below:

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Absolute Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	4.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	1.60
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	5.44
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.10
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	2.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	2.12
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.10
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	3.91
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	1.52
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	3.63
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	3.73
			Total	28.25

MAE=∑|er|n−k=28.258=2.568

Thus, the mean absolute error is 2.568.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	4.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.20
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	4.96
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.03
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	2.83
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	1.74
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.61
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	3.51
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.81
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	4.35
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	3.52
			Total	28.56

MAE=∑|er|n−k=28.5611=2.596

Thus, the mean absolute error is 2.596.

MAE when α=0.1 is less than MAE when α=0.2. Thus, α=0.1 is preferred.

Hence, the results are not same when MAE is used as measure of accuracy.

(c)

To determine

Obtain the results when MAPE is used as measure of accuracy.

(c)

Expert Solution

Answer to Problem 8P

MAPE when α=0.1 is 12.95.

MAPE when α=0.2 is 13.40.

Explanation of Solution

Exponential smoothing for α=0.1:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	19.05
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	8.42
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	23.65
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.58
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	13.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	10.58
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.53
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	17.79
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	7.61
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	24.20
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	16.97
			Total	142.46

MAPE=|100×Forecast errorTime series value|n−k=142.4611=12.95

Thus, the value of MAPE is 12.95.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	19.05
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	6.32
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	21.57
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	5.73
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	17.66
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	8.70
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	3.38
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	15.97
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	4.05
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	29.01
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	15.99
			Total	147.43

MAPE=|100×Forecast errorTime series value|n−k=147.4311=13.40

Thus, the value of MAPE is 13.40.

MAPE when α=0.1 is less than MAPE when α=0.2. Thus, α=0.1 is preferred.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 8, Problem 8P

(a)

To determine

Obtain exponential smoothing for α=0.1 and α=0.2.

Identify the preferred smoothing constant using MSE measure of forecast accuracy.

(a)

Expert Solution

Explanation of Solution

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	16.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	2.56
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	29.59
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.01
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	4.38
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	4.48
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.01
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	15.32
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	2.32
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	13.18
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	13.94
			Total	101.78

MSE=∑|er|2n−k=101.7811=9.253

Thus, the mean squared error is 9.253.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	16.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.44
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	24.60
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.07
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	7.98
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	3.03
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.37
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	12.34
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.66
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	18.94
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	12.38
			Total	98.80

MSE=∑|er|2n−k=98.8011=8.982

Thus, the mean squared error is 8.982.

MSE when α=0.2 is less than MSE when α=0.1. Thus, α=0.2 is preferred.

(b)

To determine

Check whether the results are same when MAE is used as measure of accuracy.

(b)

Expert Solution

Answer to Problem 8P

No, the results are not same.

Explanation of Solution

The MSE for four-week moving average is obtained as given below:

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Absolute Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	4.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	1.60
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	5.44
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.10
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	2.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	2.12
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.10
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	3.91
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	1.52
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	3.63
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	3.73
			Total	28.25

MAE=∑|er|n−k=28.258=2.568

Thus, the mean absolute error is 2.568.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	4.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.20
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	4.96
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.03
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	2.83
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	1.74
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.61
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	3.51
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.81
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	4.35
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	3.52
			Total	28.56

MAE=∑|er|n−k=28.5611=2.596

Thus, the mean absolute error is 2.596.

MAE when α=0.1 is less than MAE when α=0.2. Thus, α=0.1 is preferred.

Hence, the results are not same when MAE is used as measure of accuracy.

(c)

To determine

Obtain the results when MAPE is used as measure of accuracy.

(c)

Expert Solution

Answer to Problem 8P

MAPE when α=0.1 is 12.95.

MAPE when α=0.2 is 13.40.

Explanation of Solution

Exponential smoothing for α=0.1:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	19.05
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	8.42
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	23.65
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.58
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	13.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	10.58
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.53
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	17.79
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	7.61
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	24.20
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	16.97
			Total	142.46

MAPE=|100×Forecast errorTime series value|n−k=142.4611=12.95

Thus, the value of MAPE is 12.95.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	19.05
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	6.32
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	21.57
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	5.73
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	17.66
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	8.70
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	3.38
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	15.97
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	4.05
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	29.01
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	15.99
			Total	147.43

MAPE=|100×Forecast errorTime series value|n−k=147.4311=13.40

Thus, the value of MAPE is 13.40.

MAPE when α=0.1 is less than MAPE when α=0.2. Thus, α=0.1 is preferred.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 8, Problem 8P

(a)

To determine

Obtain exponential smoothing for α=0.1 and α=0.2.

Identify the preferred smoothing constant using MSE measure of forecast accuracy.

(a)

Expert Solution

Explanation of Solution

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	16.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	2.56
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	29.59
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.01
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	4.38
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	4.48
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.01
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	15.32
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	2.32
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	13.18
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	13.94
			Total	101.78

MSE=∑|er|2n−k=101.7811=9.253

Thus, the mean squared error is 9.253.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	16.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.44
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	24.60
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.07
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	7.98
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	3.03
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.37
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	12.34
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.66
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	18.94
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	12.38
			Total	98.80

MSE=∑|er|2n−k=98.8011=8.982

Thus, the mean squared error is 8.982.

MSE when α=0.2 is less than MSE when α=0.1. Thus, α=0.2 is preferred.

(b)

To determine

Check whether the results are same when MAE is used as measure of accuracy.

(b)

Expert Solution

Answer to Problem 8P

No, the results are not same.

Explanation of Solution

The MSE for four-week moving average is obtained as given below:

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Absolute Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	4.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	1.60
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	5.44
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.10
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	2.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	2.12
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.10
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	3.91
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	1.52
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	3.63
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	3.73
			Total	28.25

MAE=∑|er|n−k=28.258=2.568

Thus, the mean absolute error is 2.568.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	4.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.20
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	4.96
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.03
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	2.83
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	1.74
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.61
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	3.51
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.81
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	4.35
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	3.52
			Total	28.56

MAE=∑|er|n−k=28.5611=2.596

Thus, the mean absolute error is 2.596.

MAE when α=0.1 is less than MAE when α=0.2. Thus, α=0.1 is preferred.

Hence, the results are not same when MAE is used as measure of accuracy.

(c)

To determine

Obtain the results when MAPE is used as measure of accuracy.

(c)

Expert Solution

Answer to Problem 8P

MAPE when α=0.1 is 12.95.

MAPE when α=0.2 is 13.40.

Explanation of Solution

Exponential smoothing for α=0.1:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	19.05
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	8.42
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	23.65
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.58
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	13.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	10.58
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.53
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	17.79
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	7.61
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	24.20
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	16.97
			Total	142.46

MAPE=|100×Forecast errorTime series value|n−k=142.4611=12.95

Thus, the value of MAPE is 12.95.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	19.05
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	6.32
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	21.57
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	5.73
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	17.66
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	8.70
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	3.38
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	15.97
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	4.05
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	29.01
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	15.99
			Total	147.43

MAPE=|100×Forecast errorTime series value|n−k=147.4311=13.40

Thus, the value of MAPE is 13.40.

MAPE when α=0.1 is less than MAPE when α=0.2. Thus, α=0.1 is preferred.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 8, Problem 8P

(a)

To determine

Obtain exponential smoothing for α=0.1 and α=0.2.

Identify the preferred smoothing constant using MSE measure of forecast accuracy.

(a)

Expert Solution

Explanation of Solution

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	16.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	2.56
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	29.59
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.01
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	4.38
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	4.48
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.01
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	15.32
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	2.32
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	13.18
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	13.94
			Total	101.78

MSE=∑|er|2n−k=101.7811=9.253

Thus, the mean squared error is 9.253.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	16.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.44
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	24.60
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.07
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	7.98
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	3.03
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.37
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	12.34
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.66
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	18.94
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	12.38
			Total	98.80

MSE=∑|er|2n−k=98.8011=8.982

Thus, the mean squared error is 8.982.

MSE when α=0.2 is less than MSE when α=0.1. Thus, α=0.2 is preferred.

(b)

To determine

Check whether the results are same when MAE is used as measure of accuracy.

(b)

Expert Solution

Answer to Problem 8P

No, the results are not same.

Explanation of Solution

The MSE for four-week moving average is obtained as given below:

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Absolute Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	4.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	1.60
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	5.44
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.10
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	2.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	2.12
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.10
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	3.91
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	1.52
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	3.63
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	3.73
			Total	28.25

MAE=∑|er|n−k=28.258=2.568

Thus, the mean absolute error is 2.568.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	4.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.20
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	4.96
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.03
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	2.83
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	1.74
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.61
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	3.51
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.81
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	4.35
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	3.52
			Total	28.56

MAE=∑|er|n−k=28.5611=2.596

Thus, the mean absolute error is 2.596.

MAE when α=0.1 is less than MAE when α=0.2. Thus, α=0.1 is preferred.

Hence, the results are not same when MAE is used as measure of accuracy.

(c)

To determine

Obtain the results when MAPE is used as measure of accuracy.

(c)

Expert Solution

Answer to Problem 8P

MAPE when α=0.1 is 12.95.

MAPE when α=0.2 is 13.40.

Explanation of Solution

Exponential smoothing for α=0.1:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	19.05
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	8.42
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	23.65
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.58
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	13.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	10.58
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.53
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	17.79
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	7.61
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	24.20
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	16.97
			Total	142.46

MAPE=|100×Forecast errorTime series value|n−k=142.4611=12.95

Thus, the value of MAPE is 12.95.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	19.05
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	6.32
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	21.57
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	5.73
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	17.66
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	8.70
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	3.38
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	15.97
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	4.05
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	29.01
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	15.99
			Total	147.43

MAPE=|100×Forecast errorTime series value|n−k=147.4311=13.40

Thus, the value of MAPE is 13.40.

MAPE when α=0.1 is less than MAPE when α=0.2. Thus, α=0.1 is preferred.

Answer 6

Chapter 8, Problem 8P

(a)

To determine

Obtain exponential smoothing for α=0.1 and α=0.2.

Identify the preferred smoothing constant using MSE measure of forecast accuracy.

(a)

Expert Solution

Explanation of Solution

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	16.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	2.56
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	29.59
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.01
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	4.38
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	4.48
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.01
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	15.32
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	2.32
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	13.18
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	13.94
			Total	101.78

MSE=∑|er|2n−k=101.7811=9.253

Thus, the mean squared error is 9.253.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	16.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.44
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	24.60
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.07
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	7.98
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	3.03
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.37
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	12.34
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.66
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	18.94
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	12.38
			Total	98.80

MSE=∑|er|2n−k=98.8011=8.982

Thus, the mean squared error is 8.982.

MSE when α=0.2 is less than MSE when α=0.1. Thus, α=0.2 is preferred.

Answer 7

Chapter 8, Problem 8P

(a)

To determine

Obtain exponential smoothing for α=0.1 and α=0.2.

Identify the preferred smoothing constant using MSE measure of forecast accuracy.

Answer 8

(a)

Expert Solution

Explanation of Solution

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	16.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	2.56
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	29.59
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.01
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	4.38
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	4.48
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.01
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	15.32
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	2.32
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	13.18
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	13.94
			Total	101.78

MSE=∑|er|2n−k=101.7811=9.253

Thus, the mean squared error is 9.253.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	16.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.44
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	24.60
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.07
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	7.98
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	3.03
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.37
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	12.34
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.66
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	18.94
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	12.38
			Total	98.80

MSE=∑|er|2n−k=98.8011=8.982

Thus, the mean squared error is 8.982.

MSE when α=0.2 is less than MSE when α=0.1. Thus, α=0.2 is preferred.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

Check whether the results are same when MAE is used as measure of accuracy.

(b)

Expert Solution

Answer to Problem 8P

No, the results are not same.

Explanation of Solution

The MSE for four-week moving average is obtained as given below:

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Absolute Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	4.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	1.60
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	5.44
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.10
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	2.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	2.12
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.10
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	3.91
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	1.52
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	3.63
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	3.73
			Total	28.25

MAE=∑|er|n−k=28.258=2.568

Thus, the mean absolute error is 2.568.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	4.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.20
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	4.96
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.03
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	2.83
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	1.74
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.61
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	3.51
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.81
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	4.35
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	3.52
			Total	28.56

MAE=∑|er|n−k=28.5611=2.596

Thus, the mean absolute error is 2.596.

MAE when α=0.1 is less than MAE when α=0.2. Thus, α=0.1 is preferred.

Hence, the results are not same when MAE is used as measure of accuracy.

Answer 13

(b)

To determine

Check whether the results are same when MAE is used as measure of accuracy.

Answer 14

(b)

Expert Solution

Answer to Problem 8P

No, the results are not same.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

The MSE for four-week moving average is obtained as given below:

Exponential smoothing for α=0.1:

Exponential smoothing is obtained using the formula given below:

y^t+1=αyt+(1−α)y^t

Week	Time Series Value	Forecast	Forecast Error	Absolute Forecast Error
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	4.00
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	1.60
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	5.44
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.10
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	2.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	2.12
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.10
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	3.91
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	1.52
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	3.63
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	3.73
			Total	28.25

MAE=∑|er|n−k=28.258=2.568

Thus, the mean absolute error is 2.568.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	Squared Forecast Error
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	4.00
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	1.20
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	4.96
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	1.03
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	2.83
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	1.74
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	0.61
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	3.51
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	0.81
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	4.35
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	3.52
			Total	28.56

MAE=∑|er|n−k=28.5611=2.596

Thus, the mean absolute error is 2.596.

MAE when α=0.1 is less than MAE when α=0.2. Thus, α=0.1 is preferred.

Hence, the results are not same when MAE is used as measure of accuracy.

Answer 19

(c)

To determine

Obtain the results when MAPE is used as measure of accuracy.

(c)

Expert Solution

Answer to Problem 8P

MAPE when α=0.1 is 12.95.

MAPE when α=0.2 is 13.40.

Explanation of Solution

Exponential smoothing for α=0.1:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	19.05
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	8.42
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	23.65
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.58
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	13.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	10.58
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.53
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	17.79
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	7.61
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	24.20
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	16.97
			Total	142.46

MAPE=|100×Forecast errorTime series value|n−k=142.4611=12.95

Thus, the value of MAPE is 12.95.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	19.05
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	6.32
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	21.57
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	5.73
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	17.66
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	8.70
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	3.38
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	15.97
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	4.05
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	29.01
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	15.99
			Total	147.43

MAPE=|100×Forecast errorTime series value|n−k=147.4311=13.40

Thus, the value of MAPE is 13.40.

MAPE when α=0.1 is less than MAPE when α=0.2. Thus, α=0.1 is preferred.

Answer 20

(c)

To determine

Obtain the results when MAPE is used as measure of accuracy.

Answer 21

(c)

Expert Solution

Answer to Problem 8P

MAPE when α=0.1 is 12.95.

MAPE when α=0.2 is 13.40.

Answer 22

(c)

Expert Solution

Answer 23

(c)

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

Exponential smoothing for α=0.1:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.1(17)+(1−0.1)(17)=17	4.00	19.05
3	19	0.1(21)+(1−0.1)(17)=17.40	1.60	8.42
4	23	0.1(19)+(1−0.1)(17.4)=17.56	5.44	23.65
5	18	0.1(23)+(1−0.1)(17.56)=18.10	−0.10	0.58
6	16	0.1(18)+(1−0.1)(18.10)=18.09	−2.09	13.09
7	20	0.1(16)+(1−0.1)(18.09)=17.88	2.12	10.58
8	18	0.1(20)+(1−0.1)(17.88)=18.10	−0.10	0.53
9	22	0.1(18)+(1−0.1)(18.1)=18.09	3.91	17.79
10	20	0.1(22)+(1−0.1)(18.09)=18.48	1.52	7.61
11	15	0.1(20)+(1−0.1)(18.48)=18.63	−3.63	24.20
12	22	0.1(15)+(1−0.1)(18.63)=18.27	3.73	16.97
			Total	142.46

MAPE=|100×Forecast errorTime series value|n−k=142.4611=12.95

Thus, the value of MAPE is 12.95.

Exponential smoothing for α=0.2:

Week	Time Series Value	Forecast	Forecast Error	\|100×Forecast errorTime series value\|
1	17
2	21	0.2(17)+(1−0.2)(17)=17	4.00	19.05
3	19	0.2(21)+(1−0.2)(17)=17.8	1.20	6.32
4	23	0.2(19)+(1−0.2)(17.8)=18.04	4.96	21.57
5	18	0.2(23)+(1−0.2)(18.04)=19.03	−1.03	5.73
6	16	0.2(18)+(1−0.2)(19.03)=18.83	−2.83	17.66
7	20	0.2(16)+(1−0.2)(18.83)=18.26	1.74	8.70
8	18	0.2(20)+(1−0.2)(18.26)=18.61	−0.61	3.38
9	22	0.2(18)+(1−0.2)(18.61)=18.49	3.51	15.97
10	20	0.2(22)+(1−0.2)(18.49)=19.19	0.81	4.05
11	15	0.2(20)+(1−0.2)(19.19)=19.35	−4.35	29.01
12	22	0.2(15)+(1−0.2)(19.35)=18.48	3.52	15.99
			Total	147.43