**Chapter 8: Forecasting**

- Sales for a product for the past three months have been 200, 350, and 287. Use a three-month moving average to calculate a forecast for the fourth month. If the actual demand for month 4 turns out to be 300, calculate the forecast for month 5.

**Answer:**

*F*_{4} = (*A*_{1} + *A*_{2} + *A*_{3})/3 = (200 + 350 + 287)/3 = 279.0

*F*_{5} = (*A*_{2} + *A*_{3} + *A*_{4})/3 = (350 + 287 + 300)/3 = 312.33

- Lauren’s Beauty Boutique has experienced the following weekly sales:

Week | Sales |

1 | 432 |

2 | 396 |

3 | 415 |

4 | 458 |

5 | 460 |

Forecast sales for week 6 using the naïve method, a simple average, and a three-period moving average.

**Answer:**

Naïve Method: F_{6} = A_{5} = 460

Simple Average: *F*_{6} = (*A*_{1 }+ *A*_{2 }+ *A* _{3}+ *A*_{4 }+ *A*_{5})/5 = (432 + 396 + 415 + 458 +460)/5

= 432.2

3-Period Moving Average: F_{6} = (A_{3 }+ A_{4 }+ A_{5})/3 = (415 + 458 + 460)/3 = 444.3

- Hospitality Hotels forecasts monthly labor needs.
- Given the following monthly labor figures, make a forecast for June using a three-period moving average and a five-period moving average.

Month | Actual Values |

January | 32 |

February | 41 |

March | 38 |

April | 39 |

May | 43 |

- What would be the forecast for June using the naïve method?
- If the actual labor figure for June turns out to be 41, what would be the forecast for July using each of these models?
- Compare the accuracy of these models using the mean absolute deviation (
*MAD*). - Compare the accuracy of these models using the mean squared error (
*MSE*).

**Answer:**

- 3-Period Moving Average:
*F*_{June}= (*A*_{March}+*A*_{April}+*A*_{May})/3 = (38 + 39 + 43)/3 = 40

5-Period Moving Average: *F*_{June} = (*A*_{January} + *A*_{February} + *A*_{March} + *A*_{April }+ *A*_{May})/5

= (32 + 41 + 38 + 39 + 43)/5 = 38.6

- Naïve: F
_{June }= A_{May}= 43

- 3-Period Moving Average:
*F*_{July}= (*A*_{April}+*A*_{May}+*A*_{June})/3 = (39 + 43 + 41)/3 = 41

5-Period Moving Average: *F*_{July} = (*A*_{February} + *A*_{March} + *A*_{April} + *A*_{May} + *A*_{June})/5

= (41 + 38 + 39 + 43 + 41)/5 = 40.4

Naïve: *F*_{July} = *A*_{June }= 41

Month | Actual | 3-period moving average | Absolute Error | 5-period moving average | Absolute Error | Naïve | Absolute Error |

January | 32 | ||||||

February | 41 | 32 | 9 | ||||

March | 38 | 41 | 3 | ||||

April | 39 | 37 | 2 | 38 | 1 | ||

May | 43 | 39.33 | 3.67 | 39 | 4 | ||

June | 41 | 40 | 1 | 38.6 | 2.4 | 43 | 2 |

*MAD* (3-period moving average) = = (2 + 3.67 + 1)/3 = 2.22

*MAD* (5-period moving average) = = 2.4/1 = 2.4

*MAD* (Naïve) = = (9 + 3 + 1 + 4 + 2)/5 = 3.8

The 3-period moving average provides the best historical fit using the *MAD* criterion and would be better to use.

Month | Actual | 3-period moving average | Squared Error | 5-period moving average | Squared Error | Naïve | Squared Error |

January | 32 | ||||||

February | 41 | 32 | 81 | ||||

March | 38 | 41 | 9 | ||||

April | 39 | 37 | 2 | 38 | 1 | ||

May | 43 | 39.33 | 13.47 | 39 | 16 | ||

June | 41 | 40 | 1 | 38.6 | 5.76 | 43 | 4 |

*MSE* (3-period moving average) = = (2 + 13.47 + 1) = 6.15

*MSE* (5-period moving average) = = 5.76/1 = 5.76

*MSE* (Naïve) = = (81 + 9 + 1 + 16 + 4)/5 = 22.20

The 5-period moving average provides the best historical fit using the *MSE* criterion, but it only measures one error term.

- The following data are monthly sales of jeans at a local department store. The buyer would like to forecast sales of jeans for the next month, July.
- Forecast sales of jeans for March through June using the naïve method, a two-period moving average, and exponential smoothing with an α = 0.2. (Hint: Use naïve to start the exponential smoothing process.)
- Compare the forecasts using
*MAD*and decide which is best. - Using your method of choice, make a forecast for the month of July.

Month | Sales |

January | 45 |

February | 30 |

March | 40 |

April | 50 |

May | 55 |

June | 47 |

**Answer:**

Month | Sales | Naïve Forecast | Absolute Error | 2-period moving average | Absolute Error | Exponential smoothing | Absolute Error |

January | 45 | ||||||

February | 30 | ||||||

March | 40 | 30 | 10 | 37.5 | 2.5 | 30 | 10 |

April | 50 | 40 | 10 | 35 | 15 | 32 | 18 |

May | 55 | 50 | 5 | 45 | 10 | 35.6 | 19.4 |

June | 47 | 55 | 8 | 52.5 | 5.5 | 39.48 | 7.52 |

*MAD*(Naïve) = = (10 + 10 + 5 + 8)/4 = 8.25

*MAD* (2-period moving average) = = (2.5 + 15 + 10 + 5.5)/4

= 8.25

*MAD* (exponential smoothing) =

= (10 + 18 + 19.4 + 7.52)/4 = 13.73

The 2-period moving average and the naïve approach provide the best historical fit using the *MAD* criterion and would be better to use.

- Naïve:
*F*_{July}=*A*_{June}= 47

2-period moving average: *F*_{July} = (55 + 47)/2 = 51

- The manager of a small health clinic would like to use exponential smoothing to forecast demand for laboratory services in the facility. However, she is not sure whether to use a high or low value of α. To make her decision, she would like to compare the forecast accuracy of a high and low α on historical data. She has decided to use α = 0.7 for the high value and α = 0.1 for the low value. Given the following historical data, which do you think would be better to use?

Week | Demand (lab requirements) |

1 | 330 |

2 | 350 |

3 | 320 |

4 | 370 |

5 | 368 |

6 | 343 |

**Answer:**

Week | Demand (lab requirements) | Exponential Smoothing (α = 0.1) | Absolute Error | Exponential Smoothing (α = 0.7) | Absolute Error |

1 | 330 | ||||

2 | 350 | 330 | 20 | 330 | 20 |

3 | 320 | 332 | 12 | 344 | 24 |

4 | 370 | 330.8 | 39.2 | 327.2 | 42.8 |

5 | 368 | 334.72 | 33.28 | 357.16 | 10.84 |

6 | 343 | 338.05 | 4.952 | 364.748 | 21.748 |

*MAD *(exponential smoothing α = 0.1) =

= (20 + 12 + 39.2 + 33.28 + 4.952) = 21.89

*MAD* (exponential smoothing α = 0.7) =

= (20 + 24 + 42.8 + 10.84 + 21.748) = 23.88

Using α = 0.1 provides a better historical fit based on the *MAD* criterion.

- The manager of the health clinic in Problem 5 would also like to use exponential smoothing to forecast demand for emergency services in the facility. As in Problem 5, she is not sure whether to use a high or low value of α. To make her decision, she would like to compare the forecast accuracy of a high and low α on historical data. Again, she has decided to use α = 0.7 for the high value and α = 0.1 for the low value.
- Given the following historical data, which value of α do you think would be better to use?
- Is your answer the same as in Problem 5? Why or why not?

Week | Demand (in patients serviced) |

1 | 430 |

2 | 289 |

3 | 367 |

4 | 470 |

5 | 468 |

6 | 365 |

**Answer:**

Week | Demand (in patients serviced) | Exponential Smoothing (α = 0.1) | Absolute Error | Exponential Smoothing (α = 0.7) | Absolute Error |

1 | 430 | ||||

2 | 289 | 430 | 141 | 430 | 141 |

3 | 367 | 415.9 | 48.9 | 331.3 | 35.7 |

4 | 470 | 411.01 | 58.99 | 356.29 | 113.71 |

5 | 468 | 416.909 | 51.091 | 435.887 | 32.113 |

6 | 365 | 422.02 | 57.0181 | 458.3661 | 93.3661 |

*MAD* (exponential smoothing α = 0.1) =

= (141 + 48.99 + 58.99 + 51.091 + 57.0181)/5 = 71.40

*MAD* (exponential smoothing α = 0.7) =

= (141 + 35.7 + 113.71 + 32.113 + 93.3661) = 83.18

Using α = 0.1 provides a better historical fit based on the *MAD* criterion.

- The answer is the same since an α value of 0.1 provides a better historical fit. Both data sets do not exhibit a trend and the variation in the data appears to be random.

- The following historical data have been collected representing sales of a product. Compare forecasts using a three-period moving average, exponential smoothing with α = 0.2, and linear regression. Using
*MAD*and*MSE*, which forecasting model is best? Are your results the same using the two error measures?

Week | Demand |

1 | 20 |

2 | 31 |

3 | 36 |

4 | 38 |

5 | 42 |

6 | 40 |

**Answer:**

Week | Demand | 3-period moving average | Absolute Error | Squared Error | Exponential smoothing | Absolute Error | Squared Error |

1 | 20 | ||||||

2 | 31 | 20 | 11 | 121 | |||

3 | 36 | 22.2 | 13.8 | 190.44 | |||

4 | 38 | 29 | 9 | 81 | 24.96 | 13.04 | 170.04 |

5 | 42 | 35 | 7 | 49 | 27.568 | 14.432 | 208.28 |

6 | 40 | 38.67 | 1.33 | 1.78 | 30.454 | 9.545 | 91.118 |

*MAD* (3-period moving average) = (9 + 7 + 1.33)/3 = 5.77

*MSE* (3-period moving average) = (81 + 49 + 1.78)/3 = 43.92

*MAD* (exponential smoothing) = (11 + 13.8 + 13.04 + 14.432 + 9.545) = 12.36

*MSE* (exponential smoothing) = (121 + 190.44 + 170.04 + 208.28 + 91.118) = 156.17

Regression Model:

Time (X) | Demand (Y) | X^{2} | XY | Regression Line | Absolute Error | Squared Error | |

1 | 20 | 1 | 20 | 24.857 | 4.857 | 23.591 | |

2 | 31 | 4 | 62 | 28.714 | 2.286 | 5.224 | |

3 | 36 | 9 | 108 | 32.571 | 3.429 | 11.755 | |

4 | 38 | 16 | 152 | 36.428 | 1.572 | 2.469 | |

5 | 42 | 25 | 210 | 40.285 | 1.715 | 2.938 | |

6 | 40 | 36 | 240 | 44.142 | 4.142 | 17.163 | |

Total | 21 | 207 | 91 | 792 |

= 21/6 = 3.5

= 207/6 = 34.5

= 3.857

= 21

*MAD* = (4.857 + 2.286 + 3.429 + 1.572 + 1.715 + 4.142)/6 = 3.00

*MSE* = (23.591 + 5.224 + 11.755 + 2.469 + 2.938 + 17.163) = 10.52

The linear regression model provides the best historical fit using the *MAD* and the *MSE* criteria.

- A manufacturer of printed circuit boards uses exponential smoothing with trend to forecast monthly demand of its product. At the end of December, the company wishes to forecast sales for January. The estimate of trend through November has been 200 additional boards sold per month. Average sales have been around 1000 units per month. The demand for December was 1100 units. The company uses α = 0.20 and β = 0.10. Make a forecast including trend for the month of January.

**Answer:**

*A*_{Dec} = 1100 units/month

*S*_{Nov} = 1000 units/month

*T*_{Nov} = 200 units/month

α = 0.20

β = 0.10

**Step 1:** Smoothing the level of the series

*S*_{Dec} = α*A*_{Dec} + (1 – α) (*S*_{Nov} + *T*_{Nov}) = 0.20(1100) + 0.80(1000 + 200) = 1180 units

**Step 2:** Smoothing the trend

*T*_{Dec} = β (*S*_{Dec} – *S*_{Nov}) + (1 – β) *T*_{Nov} = 0.10(1180 – 1000) + 0.90(200) = 198 units

**Step 3:** Forecast including trend

*FIT* = *S*_{Dec} + *T*_{Dec} = 1180 + 198 = 1378 units

- Demand at Nature Trails Ski Resort has a seasonal pattern. Demand is highest during the winter, as this is the peak ski season. However, there is some ski demand in the spring and even fall months. The summer months can also be busy as visitors often come for summer vacation to go hiking on the mountain trails. The owner of Nature Trails would like to make a forecast for each season of the next year. Total annual demand has been estimated at 4000 visitors. Given the last two years of historical data, what is the forecast for each season of the next year?

| Visitors | |

Season | Year 1 | Year 2 |

Fall | 200 | 230 |

Winter | 1400 | 1600 |

Spring | 520 | 580 |

Summer | 720 | 831 |

**Answer:**

**Step 1:** Average demand for each season:

Year 1: 2840/4 = 710

Year 2: 3241/4 = 810.25

**Step 2: **Seasonal index for each season:

Season | Year 1 | Year 2 |

Fall | 200/710 = 0.282 | 230/810.25 = 0.284 |

Winter | 1400/710 = 1.972 | 1600/810.25 = 1.975 |

Spring | 520/710 = 0.732 | 580/810.25 = 0.716 |

Summer | 720/710 = 1.014 | 831/810.25 = 1.026 |

**Step 3: **Average seasonal index for each season:

Fall | 0.283 |

Winter | 1.973 |

Spring | 0.724 |

Summer | 1.020 |

**Step 4: **Average demand per season = 4000/4 = 1000

**Step 5: **Multiply next year’s average seasonal demand by each seasonal index.

Season | Forecast |

Fall | 283 |

Winter | 1973 |

Spring | 724 |

Summer | 1020 |

- Rosa’s Italian restaurant wants to develop forecasts of daily demand for the next week. The restaurant is closed on Mondays and experiences a seasonal pattern for the other six days of the week. Mario, the manager, has collected information on the number of customers served each day for the past two weeks. If Mario expects total demand for next week to be around 350, what is the forecast for each day of next week?

| Number of Customers | |

Day | Week 1 | Week 2 |

Tuesday | 52 | 48 |

Wednesday | 36 | 32 |

Thursday | 35 | 30 |

Friday | 89 | 97 |

Saturday | 98 | 99 |

Sunday | 65 | 69 |

**Answer:**

**Step 1: **Average demand for each week:

Week 1: 375/4 = 62.5

Week 2: 375/4 = 62.5

**Step 2:** Seasonal index for each week:

Day | Week 1 | Week 2 |

Tuesday | 0.832 | 0.768 |

Wednesday | 0.576 | 0.512 |

Thursday | 0.56 | 0.48 |

Friday | 1.424 | 1.552 |

Saturday | 1.568 | 1.584 |

Sunday | 1.04 | 1.104 |

**Step 3:** Average daily index for each season:

Day | Average Daily Index |

Tuesday | 0.8 |

Wednesday | 0.544 |

Thursday | 0.52 |

Friday | 1.488 |

Saturday | 1.576 |

Sunday | 1.072 |

**Step 4: **Average demand per day = 350/6 = 58.33

**Step 5: **Multiply average demand per day by each average daily index

Day | Forecast |

Tuesday | 46.664 |

Wednesday | 31.73152 |

Thursday | 30.3316 |

Friday | 86.79504 |

Saturday | 91.92808 |

Sunday | 62.52976 |

- The president of a company was interested in determining whether there is a correlation between sales made by different sales teams and hours spent on employee training. These figures are shown.

Sales (in thousands) | Training Hours |

25 | 10 |

40 | 12 |

36 | 12 |

50 | 15 |

11 | 6 |

- Compute the correlation coefficient for the data. What is your interpretation of this value?
- Using the data, what would you expect sales to be if training was increased to 18 hours?

**Answer:**

- ∑
*XY*= 1978; ∑*X*^{2}= 649; ∑*Y*^{2}= 6142; ∑*X*= 55; ∑*Y*= 162

Correlation coefficient *r* = = 0.9887

The correlation coefficient is 0.9887. This high correlation indicated that there is a strong positive linear association between sales and training hours.

- = 55/5 = 11

= 162/5 = 32.4

= 4.455

= –16.55

= 63.59

Sales (in thousands) = 63.59

- The number of students enrolled at Spring Valley Elementary has been steadily increasing over the past five years. The school board would like to forecast enrollment for years 6 and 7 in order to better plan capacity. Use a linear trend line to forecast enrollment for years 6 and 7.

Year | Enrollment |

1 | 220 |

2 | 245 |

3 | 256 |

4 | 289 |

5 | 310 |

**Answer:**

= 15/5 = 3

= 1320/5 = 264

= 22.4

= 196.8

Year 6 forecast: 196.8 + 22.4 (6) = 331.2

Year 7 forecast: 196.8 + 22.4 (7) = 353.6

- Happy Lodge Ski Resorts tries to forecast monthly attendance. The management has noticed a direct relationship between the average monthly temperature and attendance.
- Given five months of average monthly temperatures and corresponding monthly attendance, compute a linear regression equation of the relationship between the two.

If next month’s average temperature is forecast to be 45 degrees, use your linear regression equation to develop a forecast.

Month | Average Temperature | Resort Attendance (in thousands) |

1 | 24 | 43 |

2 | 41 | 31 |

3 | 32 | 39 |

4 | 30 | 38 |

5 | 38 | 35 |

- Compute a correlation coefficient for the data and determine the strength of the linear relationship between average temperature and attendance. How good a predictor is temperature for attendance?

**Answer:**

- ∑
*XY*= 6021; ∑*X*^{2}= 5625; ∑*Y*^{2}= 7000; ∑*X*= 165; ∑*Y*= 186

= 165/5 = 33

= 186/5 = 37.2

= –0.65

= 58.65

Resort attendance forecast when the average temperature is 45 degrees:

= 58.65 + (–0.65) (45) = 29.4 thousand attendees

- Correlation coefficient
*r*= = – 0.97

The correlation coefficient is – 0.97. Since this value is very close to negative 1, it indicates that the average temperature is a strong predictor of resort attendance. Note that since the sign is negative, it indicates that an inverse or negative relationship exists between the two variables.

- Small Wonder, an amusement park, experiences seasonal attendance. It has collected two years of quarterly attendance data and made a forecast of annual attendance for the coming year. Compute the seasonal indexes for the four quarters and generate quarterly forecasts for the coming year, assuming annual attendance for the coming year to be 1525.

| Park Attendance (in thousands) | |

Quarter | Year 1 | Year 2 |

Fall | 352 | 391 |

Winter | 156 | 212 |

Spring | 489 | 518 |

Summer | 314 | 352 |

**Answer:**

**Step 1:**

Average demand for each quarter for year 1 = (352+156+489+314)/4 = 327.75

Average demand for each quarter for year 2 = (391+212+518+352)/4 = 368.25

**Step 2:**

Compute a seasonal index for every season of every year:

Quarter | Year 1 | Year 2 |

Fall | 352/327.75 = 1.07 | 391/368.25 = 1.06 |

Winter | 156/327.75 = 0.48 | 212/368.25 = 0.58 |

Spring | 489/327.75 = 1.49 | 518/368.25 = 1.41 |

Summer | 314/327.75 = 0.96 | 352/368.25 = 0.95 |

**Step 3: **Calculate the average seasonal index for each season:

Quarter | Average Seasonal Index |

Fall | (1.07+1.06)/2 = 1.065 |

Winter | (0.48+0.58)/2 = 0.53 |

Spring | (1.49+1.41)/2 = 1.45 |

Summer | (0.96+0.95)/2 = 0.955 |

**Step 4: **Calculate the average demand per season for next year = 1525/4 = 381.25

**Step 5: **Multiply next year’s average seasonal demand by each seasonal index:

Quarter | Forecast |

Fall | (381.25)(1.065) = 406.03 |

Winter | (381.25)(0.53) = 202.06 |

Spring | (381.25)(1.45) = 552.81 |

Summer | (381.25)(0.955) = 364.09 |

- Burger Lover Restaurant forecasts weekly sales of cheeseburgers. Based on historical observations over the past five weeks, make a forecast for the next period using the following methods: simple average, three-period moving average, and exponential smoothing with α = 0.3, given a forecast of 328 cheeseburgers for the first week.

Week | Cheeseburger Sales |

1 | 354 |

2 | 345 |

3 | 367 |

4 | 322 |

5 | 356 |

If actual sales for week 6 turn out to be 368, compare the three forecasts using *MAD*. Which method performed best?

**Answer:**

Period 6 Forecasts:

Simple average: *F*_{6} = 348.8

3-Period moving average: *F*_{6} = 348.33

Exponential smoothing: Using the fifth period forecast of 328, *F*_{6} = 336.40

Looking at the error only in period 6,

*MAD* (simple average) = |368 – 348.8| = 19.2

*MAD* (3-Period moving average) = |368 – 348.33| = 19.67

*MAD* (exponential smoothing) = |368 – 336.40| = 31.60

The simple average provides the best historical fit using the *MAD* criterion.

- A company uses exponential smoothing with trend to forecast monthly sales of its product, which show a trend pattern. At the end of week 5, the company wants to forecast sales for week 6. The trend through week 4 has been 20 additional cases sold per week. Average sales have been 85 cases per week. The demand for week 5 was 90 cases. The company uses α = 0.20 and β = 0.10. Make a forecast including trend for week 6.

**Answer:**

Given: T_{4} = 20, A_{5} = 90, S_{4} = 85

α = 0.20

β = 0.10

**Step 1:** Smoothing the level of the series:

*S*_{5} = α*A*_{5} + (1 – α) (*S*_{4} + *T*_{4}) = 0.20(90) + 0.80(85 + 20) = 102

**Step 2:** Smoothing the trend:

*T*_{5} = β (*S*_{5} – *S*_{4}) + (1 – β) *T*_{4} = 0.10(102 – 85) + 0.90(20) = 19.7

**Step 3:** Forecast including trend:

*FIT*_{6} = *S*_{5} + *T*_{5} = 102 + 19.7 = 121.7

- The number of patients coming to the Healthy Start maternity clinic has been increasing steadily over the past eight months. Given the following data, use a linear trend line to forecast attendance for months 9 and 10.

Month | Clinic Attendance (in thousands) |

1 | 3.4 |

2 | 3.9 |

3 | 4.5 |

4 | 5.0 |

5 | 5.8 |

6 | 5.9 |

7 | 6.5 |

8 | 6.7 |

**Answer:**

= 36/8 = 4.5

= 41.7/8 = 5.21

= 0.489

= 3.011

*F*_{9} = 3.011 + 0.489(9) = 7.412 attendees (in thousands)

*F*_{10 }= 3.011 + 0.489(10) = 7.901 attendees (in thousands)

- Given the following data, use exponential smoothing with α = 0.2 and α = 0.5 to generate forecasts for periods 2 through 6. Use
*MAD*and*MSE*to decide which of the two models produced a better forecast.

Period | Actual | Forecast |

1 | 15 | 17 |

2 | 18 | |

3 | 14 | |

4 | 16 | |

5 | 13 | |

6 | 16 |

**Answer:**

Period | Actual | Forecast (α=0.2) | Absolute Error | Squared Error | Forecast (α=0.5) | Absolute Error | Squared Error |

1 | 15 | 17 | 2 | 4 | 17 | 2 | 4 |

2 | 18 | 16.6 | 1.4 | 1.96 | 16 | 2 | 4 |

3 | 14 | 16.88 | 2.88 | 8.294 | 17 | 3 | 9 |

4 | 16 | 16.304 | 0.304 | 0.092 | 15.5 | 0.5 | 0.25 |

5 | 13 | 16.243 | 3.243 | 10.518 | 15.75 | 2.75 | 7.563 |

6 | 16 | 15.595 | 0.405 | 0.164 | 14.375 | 1.625 | 2.641 |

*MAD* (α = 0.2) = (2 + 1.4 + 2.88 + 0.304 + 3.243 + 0.405)/6 = 1.705

*MAD *(α = 0.5) = (2 + 2 + 3 + 0.5 + 2.75 + 1.625) = 1.979

Exponential smoothing using α = 0.2 yields lower *MAD*.

*MSE* (α = 0.2) = 4 + 1.96 + 8.294 + 0.092 + 10.518 + 0.164 = 4.172

*MSE* (α = 0.5) = 4 + 4 + 9 + 0.25 + 7.563 + 2.641) = 4.576

Exponential smoothing using α = 0.2 yields lower *MSE*.

- Pumpkin Pies Galore is trying to forecast sales of pies for the month of December. Demand for pies in September, October, and November has been 230, 304, and 415, respectively. Edith, the company’s owner, uses a three-period weighted moving average to forecast sales. Based on her experience, she chooses to weight September as 0.1, October as 0.3, and November as 0.6.
- What would Edith’s forecast for December be?
- What would her forecast be using the naïve method?
- If actual sales for December turned out to be 420 pies, which method was better (use
*MAD*)?

**Answer:**

- Forecast using a weighted moving average = 230(0.1) + 304(0.3) + 415(0.6) = 363.2

- Forecast using the naïve approach = 415

*MAD*(weighted moving average) = 420 – 363.2 = 56.8

*MAD* (naïve method) = 420 – 415 = 5

The naïve approach is better.

- A company has used three different methods to forecast sales for the past five months. Use
*MAD*and*MSE*to evaluate the performance of the three methods. - Which forecasting method performed best? Do
*MAD*and*MSE*give the same results?

Period | Actual | Method A | Method B | Method C |

1 | 10 | 10 | 9 | 8 |

2 | 8 | 11 | 10 | 11 |

3 | 12 | 12 | 8 | 10 |

4 | 11 | 13 | 12 | 11 |

5 | 12 | 14 | 11 | 12 |

- Which of these is actually the naïve method?

**Answer:**

Period | Actual | Method A | Absolute Error | Squared Error |

1 | 10 | 10 | 0 | 0 |

2 | 8 | 11 | 3 | 9 |

3 | 12 | 12 | 0 | 0 |

4 | 11 | 13 | 2 | 4 |

5 | 12 | 14 | 2 | 4 |

*MAD*_{A} = (0 + 3 + 0 + 2 + 2)/5 = 1.4

*MSE*_{A} = (0 + 9 + 0 + 4 + 4)/5 = 3.4

Period | Actual | Method B | Absolute Error | Squared Error |

1 | 10 | 9 | 1 | 1 |

2 | 8 | 10 | 2 | 4 |

3 | 12 | 8 | 4 | 16 |

4 | 11 | 12 | 1 | 1 |

5 | 12 | 11 | 1 | 1 |

*MAD*_{B} = (1 + 2 + 4 + 1 + 1)/5 = 1.8

*MSE*_{B} = (1 + 4 + 16 + 1 + 1)/5 = 4.6

Period | Actual | Method C | Absolute Error | Squared Error |

1 | 10 | 8 | 2 | 4 |

2 | 8 | 11 | 3 | 9 |

3 | 12 | 10 | 2 | 4 |

4 | 11 | 11 | 0 | 0 |

5 | 12 | 12 | 0 | 0 |

*MAD*_{C} = (2 + 3 + 2 + 0 + 0)/5 = 1.4

*MSE*_{C} = (4 + 9 + 4 + 0 + 0)/5 = 3.4

Both Method A and Method C provide the best historical fit using the *MAD* and *MSE* criteria.

- Forecast method B is the naïve method.

- Two different forecasting models were used to forecast sales of a popular soda on a college campus. Actual demand and the two sets of forecasts are shown. Use
*MAD*to explain which method provided a better forecast.

Period | Actual Demand | Forecast 1 | Forecast 2 |

1 | 90 | 78 | 87 |

2 | 87 | 85 | 88 |

3 | 92 | 84 | 90 |

4 | 95 | 92 | 97 |

5 | 98 | 100 | 102 |

6 | 98 | 102 | 101 |

**Answer:**

Period | Actual Demand | Forecast 1 | Absolute Error | Forecast 2 | Absolute Error |

1 | 90 | 78 | 12 | 87 | 3 |

2 | 87 | 85 | 2 | 88 | 1 |

3 | 92 | 84 | 8 | 90 | 2 |

4 | 95 | 92 | 3 | 97 | 2 |

5 | 98 | 100 | 2 | 102 | 4 |

6 | 98 | 102 | 4 | 101 | 3 |

*MAD* (Forecast 1) = (12 + 2 + 8 + 3 + 2 + 4)/6 = 5.17

*MAD* (Forecast 2) = (3 + 1 + 2 + 2 + 4 + 3)/6 = 2.50

Forecast 2 provides a better historical fit using the *MAD* criterion.

- A producer of picture frames uses a tracking signal with limits of ±4 to decide whether a forecast should be reviewed. Given historical information for the past four weeks, compute the tracking signal and decide whether the forecast should be reviewed. The
*MAD*for this item was computed as 2.

Weeks | Actual Sales | Forecast | Deviation | Cumulative Deviation | Tracking Signal |

6 | 3 | ||||

1 | 12 | 11 | |||

2 | 14 | 13 | |||

3 | 14 | 14 | |||

4 | 16 | 14 |

**Answer:**

The tracking signal is the sum of the forecast errors (the cumulative deviation) divided by the *MAD*. In this problem *MAD* has been computed as 2. You need to compute the cumulative deviation for each period and divide by the *MAD*.

Weeks | Actual Sales | Forecast | Deviation | Cumulative Deviation | Tracking Signal |

6 | 3 | ||||

1 | 12 | 11 | 1 | 7 | 3.5 |

2 | 14 | 13 | 1 | 8 | 4 |

3 | 14 | 14 | 0 | 8 | 4 |

4 | 16 | 14 | 2 | 10 | 5 |

The forecast should be reviewed in week 4 because the tracking signal has exceeded +4.

- Mop and Broom Manufacturing has tracked the number of units sold of their most popular mop over the past 24 months. This is shown.

Month | Sales | Month | Sales | Month | Sales |

1 | 239 | 9 | 310 | 17 | 369 |

2 | 248 | 10 | 335 | 18 | 378 |

3 | 256 | 11 | 348 | 19 | 367 |

4 | 260 | 12 | 353 | 20 | 383 |

5 | 271 | 13 | 355 | 21 | 394 |

6 | 280 | 14 | 368 | 22 | 393 |

7 | 295 | 15 | 379 | 23 | 405 |

8 | 305 | 16 | 358 | 24 | 412 |

- Develop a linear trend line for the data.
- Compute a correlation coefficient for the data and evaluate the strength of the linear relationship.
- Using the linear trend line equation, develop a forecast for the next period, month 25.

**Answer:**

- = 300/24 = 12.5

= 8061/24 = 335.875

= 7.424

= 243.07

Sales = 243.07 + 7.424 (month)

- Correlation coefficient
*r*= = 0.97

It indicates a strong, positive linear relationship.

- Forecast for month 25 = 243.07 + (7.424)25 = 428.68 or 429 units

- Given the sales data from Problem 23, generate forecasts for months 7–24 using a six-period and a three-period moving average. Use MAD to compare the forecasts. Which forecast is more stable? Which is more responsive and why?

**Answer:**

Month (X) | Sales (Y) | 6-period moving average | Absolute error | 3-period moving average | Absolute Error |

1 | 239 | ||||

2 | 248 | ||||

3 | 256 | ||||

4 | 260 | 247.67 | 12.33 | ||

5 | 271 | 254.67 | 16.33 | ||

6 | 280 | 262.33 | 17.67 | ||

7 | 295 | 259.00 | 36.00 | 270.33 | 24.67 |

8 | 305 | 268.33 | 36.67 | 282.00 | 23.00 |

9 | 310 | 277.83 | 32.17 | 293.33 | 16.67 |

10 | 335 | 286.83 | 48.17 | 303.33 | 31.67 |

11 | 348 | 299.33 | 48.67 | 316.67 | 31.33 |

12 | 353 | 312.17 | 40.83 | 331.00 | 22.00 |

13 | 355 | 324.33 | 30.67 | 345.33 | 9.67 |

14 | 368 | 334.33 | 33.67 | 352.00 | 16.00 |

15 | 379 | 344.83 | 34.17 | 358.67 | 20.33 |

16 | 358 | 356.33 | 1.67 | 367.33 | 9.33 |

17 | 369 | 360.17 | 8.83 | 368.33 | 0.67 |

18 | 378 | 363.67 | 14.33 | 368.67 | 9.33 |

19 | 367 | 367.83 | 0.83 | 368.33 | 1.33 |

20 | 383 | 369.83 | 13.17 | 371.33 | 11.67 |

21 | 394 | 372.33 | 21.67 | 376.00 | 18.00 |

22 | 393 | 374.83 | 18.17 | 381.33 | 11.67 |

23 | 405 | 380.67 | 24.33 | 390.00 | 15.00 |

24 | 412 | 386.67 | 25.33 | 397.33 | 14.67 |

*MAD* (6-period moving average) = 469.33/18 = 26.074

*MAD* (3-period moving average) = 333.33/21 = 15.873

The very nature of averaging makes the 3-period moving average is more responsive. The 6-period moving average has more weight on older (and lower) sales data. The lower *MAD* confirms the 3-period moving average was better in forecasting the increasing sales in this data set.

- The following data were collected on the study of the relationship between a company’s retail sales and advertising dollars:

Retails Sales ($) | Advertising ($) |

29,789 | 16,893 |

35,434 | 18,398 |

38,732 | 20,376 |

43,585 | 22,982 |

46,821 | 25,732 |

49,283 | 27,281 |

52,271 | 32,182 |

55,289 | 35,298 |

57,298 | 36,281 |

58,293 | 38,178 |

- Obtain a linear regression line for the data.
- Compute a correlation coefficient and determine the strength of the linear relationship.
- Using the linear regression equation, develop a forecast of retail sales for advertising dollars of $40,000.

**Answer:**

- = 466795/10 = 46679.5

= 273601/10 = 27360.1

= 0.784

= –9232.57

Value of sales = –9232.57 + 0.784 (advertising dollars)

- Correlation coefficient
*r*= = 0.98

It indicates a strong, positive linear relationship.

- Forecast of retail sales = –9232.57 + 0.784 (40000) = $22123

All papers are written by ENL (US, UK, AUSTRALIA) writers with vast experience in the field. We perform a quality assessment on all orders before submitting them.

Do you have an urgent order? We have more than enough writers who will ensure that your order is delivered on time.

We provide plagiarism reports for all our custom written papers. All papers are written from scratch.

Contact us anytime, any day, via any means if you need any help. You can use the Live Chat, email, or our provided phone number anytime.

We will not disclose the nature of our services or any information you provide to a third party.

Get your money back if your paper is not delivered on time or if your instructions are not followed.