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

Let x be a random variable that represents the percentage of successful free throws a professional basketball player makes in a season. Let y be a random variable that represents the percentage of successful field goals a professional basketball player makes in a season. A random sample of n =6 professional basketball players gave the following information. x 63 79 70 80 84 87 y 46 49 45 55 57 58 Find Se. Round your answer to three decimal places.

1. Define probability: 2. Define statistics:

Results of tossing a coin four times: H, H, H, H How many times is the Coin expected to come up heads? How did you determine this number? Calculate the % deviation. Can these results be used to conclude that a coin is not fair? Why or why not?

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