**⭳**82 Download

**📄**16 Pages / 3948 Words

**Forecasting examples**

**Forecasting example 1996 UG exam**

The demand for a product in each of the last five months is shown below.

Month 1 2 3 4 5

Demand ('00s) 13 17 19 23 24

- Use a two month moving average to generate a forecast for demand in month 6.
- Apply exponential smoothing with a smoothing constant of 0.9 to generate a forecast for demand for demand in month 6.
- Which of these two forecasts do you prefer and why?

**Solution**

The two month moving average for months two to five is given by:

m_{2} = (13 + 17)/2 = 15.0

m_{3} = (17 + 19)/2 = 18.0

m_{4} = (19 + 23)/2 = 21.0

m_{5} = (23 + 24)/2 = 23.5

The forecast for month six is just the moving average for the month before that i.e. the moving average for month 5= m_{5} = 2350.

Applying exponential smoothing with a smoothing constant of 0.9 we get:

M_{1} = Y_{1} = 13

M_{2} = 0.9Y_{2} + 0.1M_{1} = 0.9(17) + 0.1(13) = 16.60

M_{3} = 0.9Y_{3} + 0.1M_{2} = 0.9(19) + 0.1(16.60) = 18.76

M_{4} = 0.9Y_{4} + 0.1M_{3} = 0.9(23) + 0.1(18.76) = 22.58

M_{5} = 0.9Y_{5} + 0.1M_{4} = 0.9(24) + 0.1(22.58) = 23.86

As before the forecast for month six is just the average for month 5= M_{5} = 2386

To compare the two forecasts we calculate the mean squared deviation (MSD). If we do this we find that for the moving average

- MSD = [(15 - 19)² + (18 - 23)²+ (21 - 24)²]/3 = 16.67

and for the exponentially smoothed average with a smoothing constant of 0.9

- MSD = [(13 - 17)² + (16.60 - 19)² +(18.76 - 23)²+ (22.58 - 24)²]/4 = 10.44

Overall then we see that exponential smoothing appears to give the best one month ahead forecasts as it has a lower MSD. Hence we prefer the forecast of 2386 that has been produced by exponential smoothing.

**Forecasting example 1994 UG exam**

The table below shows the demand for a new aftershave in a shop for each of the last 7 months.

Month 1 2 3 4 5 6 7

Demand 23 29 33 40 41 43 49

- Calculate a two month moving average for months two to seven. What would be your forecast for the demand in month eight?
- Apply exponential smoothing with a smoothing constant of 0.1 to derive a forecast for the demand in month eight.
- Which of the two forecasts for month eight do you prefer and why?
- The shop keeper believes that customers are switching to this new aftershave from other brands. Discuss how you might model this switching behaviour and indicate the data that you would require to confirm whether this switching is occurring or not.

**Solution**

The two month moving average for months two to seven is given by:

m_{2} = (23 + 29)/2 = 26.0

m_{3} = (29 + 33)/2 = 31.0

m_{4} = (33 + 40)/2 = 36.5

m_{5} = (40 + 41)/2 = 40.5

m_{6} = (41 + 43)/2 = 42.0

m_{7} = (43 + 49)/2 = 46.0

The forecast for month eight is just the moving average for the month before that i.e. the moving average for month 7 = m_{7} = 46.

Applying exponential smoothing with a smoothing constant of 0.1 we get:

M_{1} = Y_{1} = 23

M_{2} = 0.1Y_{2} + 0.9M_{1} = 0.1(29) + 0.9(23) = 23.60

M_{3} = 0.1Y_{3} + 0.9M_{2} = 0.1(33) + 0.9(23.60) = 24.54

M_{4} = 0.1Y_{4} + 0.9M_{3} = 0.1(40) + 0.9(24.54) = 26.09

M_{5} = 0.1Y_{5} + 0.9M_{4} = 0.1(41) + 0.9(26.09) = 27.58

M_{6} = 0.1Y_{6} + 0.9M_{5} = 0.1(43) + 0.9(27.58) = 29.12

M_{7} = 0.1Y_{7} + 0.9M_{6} = 0.1(49) + 0.9(29.12) = 31.11

As before the forecast for month eight is just the average for month 7 = M_{7} = 31.11 = 31 (as we cannot have fractional demand).

To compare the two forecasts we calculate the mean squared deviation (MSD). If we do this we find that for the moving average

- MSD = [(26.0 - 33)² + ... + (42.0 - 49)²]/5 = 41.1

and for the exponentially smoothed average with a smoothing constant of 0.1

- MSD = [(23 - 29)² + ... + (29.12 - 49)²]/6 = 203.15

Overall then we see that the two month moving average appears to give the best one month ahead forecasts as it has a lower MSD. Hence we prefer the forecast of 46 that has been produced by the two month moving average.

To examine switching we would need to use a Markov process model, where states = brands and we would need initial state information and customer switching probabilities (from surveys). We would need to run the model on historical data to see if we have a fit between the model and historical behaviour.

**Forecasting example 1992 UG exam**

The table below shows the demand for a particular brand of razor in a shop for each of the last nine months.

Month 1 2 3 4 5 6 7 8 9

Demand 10 12 13 17 15 19 20 21 20

- Calculate a three month moving average for months three to nine. What would be your forecast for the demand in month ten?
- Apply exponential smoothing with a smoothing constant of 0.3 to derive a forecast for the demand in month ten.
- Which of the two forecasts for month ten do you prefer and why?

**Solution**

The three month moving average for months 3 to 9 is given by:

m_{3} = (10 + 12 + 13)/3 = 11.67

m_{4} = (12 + 13 + 17)/3 = 14.00

m_{5} = (13 + 17 + 15)/3 = 15.00

m_{6} = (17 + 15 + 19)/3 = 17.00

m_{7} = (15 + 19 + 20)/3 = 18.00

m_{8} = (19 + 20 + 21)/3 = 20.00

m_{9} = (20 + 21 + 20)/3 = 20.33

The forecast for month 10 is just the moving average for the month before that i.e. the moving average for month 9 = m_{9} = 20.33.

Hence (as we cannot have fractional demand) the forecast for month 10 is 20.

Applying exponential smoothing with a smoothing constant of 0.3 we get:

M_{1} = Y_{1} = 10

M_{2} = 0.3Y_{2} + 0.7M_{1} = 0.3(12) + 0.7(10) = 10.60

M_{3} = 0.3Y_{3} + 0.7M_{2} = 0.3(13) + 0.7(10.60) = 11.32

M_{4} = 0.3Y_{4} + 0.7M_{3} = 0.3(17) + 0.7(11.32) = 13.02

M_{5} = 0.3Y_{5} + 0.7M_{4} = 0.3(15) + 0.7(13.02) = 13.61

M_{6} = 0.3Y_{6} + 0.7M_{5} = 0.3(19) + 0.7(13.61) = 15.23

M_{7} = 0.3Y_{7} + 0.7M_{6} = 0.3(20) + 0.7(15.23) = 16.66

M_{8} = 0.3Y_{8} + 0.7M_{7} = 0.3(21) + 0.7(16.66) = 17.96

M_{9} = 0.3Y_{9} + 0.7M_{8} = 0.3(20) + 0.7(17.96) = 18.57

As before the forecast for month 10 is just the average for month 9 = M_{9} = 18.57 = 19 (as we cannot have fractional demand).

To compare the two forecasts we calculate the mean squared deviation (MSD). If we do this we find that for the moving average

- MSD = [(11.67 - 17)² + ... + (20.00 - 20)²]/6 = 10.57

and for the exponentially smoothed average with a smoothing constant of 0.3

- MSD = [(10 - 12)² + ... + (17.96 - 20)²]/8 = 15.08

Overall then we see that the three month moving average appears to give the best one month ahead forecasts as it has a lower MSD. Hence we prefer the forecast of 20 that has been produced by the three month moving average.

**Forecasting example 1991 UG exam**

The table below shows the demand for a particular brand of fax machine in a department store in each of the last twelve months.

Month 1 2 3 4 5 6 7 8 9 10 11 12

Demand 12 15 19 23 27 30 32 33 37 41 49 58

- Calculate the four month moving average for months 4 to 12. What would be your forecast for the demand in month 13?
- Apply exponential smoothing with a smoothing constant of 0.2 to derive a forecast for the demand in month 13.
- Which of the two forecasts for month 13 do you prefer and why?
- What other factors, not considered in the above calculations, might influence demand for the fax machine in month 13?

**Solution**

The four month moving average for months 4 to 12 is given by:

m_{4} = (23 + 19 + 15 + 12)/4 = 17.25

m_{5} = (27 + 23 + 19 + 15)/4 = 21

m_{6} = (30 + 27 + 23 + 19)/4 = 24.75

m_{7} = (32 + 30 + 27 + 23)/4 = 28

m_{8} = (33 + 32 + 30 + 27)/4 = 30.5

m_{9} = (37 + 33 + 32 + 30)/4 = 33

m_{10} = (41 + 37 + 33 + 32)/4 = 35.75

m_{11} = (49 + 41 + 37 + 33)/4 = 40

m_{12} = (58 + 49 + 41 + 37)/4 = 46.25

The forecast for month 13 is just the moving average for the month before that i.e. the moving average for month 12 = m_{12} = 46.25.

Hence (as we cannot have fractional demand) the forecast for month 13 is 46.

Applying exponential smoothing with a smoothing constant of 0.2 we get:

M_{1} = Y_{1} = 12

M_{2} = 0.2Y_{2} + 0.8M_{1} = 0.2(15) + 0.8(12) = 12.600

M_{3} = 0.2Y_{3} + 0.8M_{2} = 0.2(19) + 0.8(12.600) = 13.880

M_{4} = 0.2Y_{4} + 0.8M_{3} = 0.2(23) + 0.8(13.880) = 15.704

M_{5} = 0.2Y_{5} + 0.8M_{4} = 0.2(27) + 0.8(15.704) = 17.963

M_{6} = 0.2Y_{6} + 0.8M_{5} = 0.2(30) + 0.8(17.963) = 20.370

M_{7} = 0.2Y_{7} + 0.8M_{6} = 0.2(32) + 0.8(20.370) = 22.696

M_{8} = 0.2Y_{8} + 0.8M_{7} = 0.2(33) + 0.8(22.696) = 24.757

M_{9} = 0.2Y_{9} + 0.8M_{8} = 0.2(37) + 0.8(24.757) = 27.206

M_{10} = 0.2Y_{10} + 0.8M_{9} = 0.2(41) + 0.8(27.206) = 29.965

M_{11} = 0.2Y_{11} + 0.8M_{10} = 0.2(49) + 0.8(29.965) = 33.772

M_{12} = 0.2Y_{12} + 0.8M_{11} = 0.2(58) + 0.8(33.772) = 38.618

As before the forecast for month 13 is just the average for month 12 = M_{12} = 38.618 = 39 (as we cannot have fractional demand).

- MSD = [(17.25 - 27)² + ... + (40 - 58)²]/8 = 107.43

and for the exponentially smoothed average with a smoothing constant of 0.2

- MSD = [(12 - 15)² + ... + (33.772 - 58)²]/11 = 176.05

Overall then we see that the four month moving average appears to give the best one month ahead forecasts as it has a lower MSD. Hence we prefer the forecast of 46 that has been produced by the four month moving average.

Other factors:

- seasonal demand
- advertising
- price changes, both this brand and other brands
- general economic situation
- new technology

**Forecasting example 1989 UG exam**

The table below shows the demand for a particular brand of microwave oven in a department store in each of the last twelve months.

Month 1 2 3 4 5 6 7 8 9 10 11 12

Demand 27 31 29 30 32 34 36 35 37 39 40 42

- Calculate a six month moving average for each month. What would be your forecast for the demand in month 13?
- Apply exponential smoothing with a smoothing constant of 0.7 to derive a forecast for the demand in month 13.
- Which of the two forecasts for month 13 do you prefer and why?

**Solution**

Now we cannot calculate a six month moving average until we have at least 6 observations - i.e. we can only calculate such an average from month 6 onward. Hence we have:

m_{6} = (34 + 32 + 30 + 29 + 31 + 27)/6 = 30.50

m_{7} = (36 + 34 + 32 + 30 + 29 + 31)/6 = 32.00

m_{8} = (35 + 36 + 34 + 32 + 30 + 29)/6 = 32.67

m_{9} = (37 + 35 + 36 + 34 + 32 + 30)/6 = 34.00

m_{10} = (39 + 37 + 35 + 36 + 34 + 32)/6 = 35.50

m_{11} = (40 + 39 + 37 + 35 + 36 + 34)/6 = 36.83

m_{12} = (42 + 40 + 39 + 37 + 35 + 36)/6 = 38.17

The forecast for month 13 is just the moving average for the month before that i.e. the moving average for month 12 = m_{12} = 38.17.

Hence (as we cannot have fractional demand) the forecast for month 13 is 38.

Applying exponential smoothing with a smoothing constant of 0.7 we get:

M_{1} = Y_{1} = 27

M_{2} = 0.7Y_{2} + 0.3M_{1} = 0.7(31) + 0.3(27) = 29.80

M_{3} = 0.7Y_{3} + 0.3M_{2} = 0.7(29) + 0.3(29.80) = 29.24

M_{4} = 0.7Y_{4} + 0.3M_{3} = 0.7(30) + 0.3(29.24) = 29.77

M_{5} = 0.7Y_{5} + 0.3M_{4} = 0.7(32) + 0.3(29.77) = 31.33

M_{6} = 0.7Y_{6} + 0.3M_{5} = 0.7(34) + 0.3(31.33) = 33.20

M_{7} = 0.7Y_{7} + 0.3M_{6} = 0.7(36) + 0.3(33.20) = 35.16

M_{8} = 0.7Y_{8} + 0.3M_{7} = 0.7(35) + 0.3(35.16) = 35.05

M_{9} = 0.7Y_{9} + 0.3M_{8} = 0.7(37) + 0.3(35.05) = 36.42

M_{10} = 0.7Y_{10} + 0.3M_{9} = 0.7(39) + 0.3(36.42) = 38.23

M_{11} = 0.7Y_{11} + 0.3M_{10} = 0.7(40) + 0.3(38.23) = 39.47

M_{12} = 0.7Y_{12} + 0.3M_{11} = 0.7(42) + 0.3(39.47) = 41.24

As before the forecast for month 13 is just the average for month 12 = M_{12} = 41.24 = 41 (as we cannot have fractional demand).

- MSD = [(30.50 - 36)² + ... + (36.83 - 42)²]/6 = 21.66

and for the exponentially smoothed average with a smoothing constant of 0.7

- MSD = [(27 - 31)² + ... + (39.47 - 42)²]/11 = 5.25

a smoothing constant of 0.7 appears to give the best one month ahead forecasts as it has a lower MSD. Hence we prefer the forecast of 41 that has been produced by exponential smoothing with a smoothing constant of 0.7.

**Forecasting example 1987 UG exam**

The table below shows the temperature (degrees C), at 11 p.m., over the last ten days:

Day 1 2 3 4 5 6 7 8 9 10

Temperature 1.5 2.3 3.7 3.0 1.4 -1.3 -2.4 -3.7 -0.5 1.3

- Calculate a three day moving average for each day.
- What would be your forecast for the temperature at 11 p.m. on day 11?
- Apply exponential smoothing with a smoothing constant of 0.8 to derive a forecast for the temperature at 11 p.m. on day 11.
- Which of the two forecasts for the temperature at 11 p.m. on day 11 do you prefer and why?

**Solution**

Now we cannot calculate a 3 day moving average until we have at least 3 observations i.e. we can only calculate such an average from month 3 onward. Hence we have:

m_{3} = (1.5 + 2.3 + 3.7)/3 = 2.50

m_{4} = (2.3 + 3.7 + 3.0)/3 = 3.00

m_{5} = (3.7 + 3.0 + 1.4)/3 = 2.70

m_{6} = (3.0 + 1.4 - 1.3)/3 = 1.03

m_{7} = (1.4 - 1.3 - 2.4)/3 = -0.77

m_{8} = (-1.3 - 2.4 - 3.7)/3 = -2.47

m_{9} = (-2.4 - 3.7 - 0.5)/3 = -2.20

m_{10} = (-3.7 - 0.5 + 1.3)/3 = -0.97

Hence the forecast for the temperature at 11 p.m. on day 11 is just m_{10} = -0.97.

Applying exponential smoothing with a smoothing constant of 0.8 we get:

M_{1} = Y_{1} = 1.5

M_{2} = 0.8Y_{2} + 0.2M_{1} = 0.8(2.3) + 0.2(1.5) = 2.14

M_{3} = 0.8Y_{3} + 0.2M_{2} = 0.8(3.7) + 0.2(2.14) = 3.39

M_{4} = 0.8Y_{4} + 0.2M_{3} = 0.8(3.0) + 0.2(3.39) = 3.08

M_{5} = 0.8Y_{5} + 0.2M_{4} = 0.8(1.4) + 0.2(3.08) = 1.74

M_{6} = 0.8Y_{6} + 0.2M_{5} = 0.8(-1.3) + 0.2(1.74) = -0.69

M_{7} = 0.8Y_{7} + 0.2M_{6} = 0.8(-2.4) + 0.2(-0.69) = -2.06

M_{8} = 0.8Y_{8} + 0.2M_{7} = 0.8(-3.7) + 0.2(-2.06) = -3.37

M_{9} = 0.8Y_{9} + 0.2M_{8} = 0.8(-0.5) + 0.2(-3.37) = -1.07

M_{10} = 0.8Y_{10} + 0.2M_{9} = 0.8(1.3) + 0.2(-1.07) = 0.83

Hence the forecast for the temperature at 11 p.m. on day 11 is just M_{10} = 0.83.

To compare the two forecasts we calculate the mean squared deviation (MSD). If we do this we find that for the moving average MSD=7.90 and for the exponentially smoothed average with a smoothing constant of 0.8 MSD=3.86.

Hence overall prefer the exponentially smoothed forecast as that seems to give the best one day ahead forecasts as it has a smaller MSD.

**Forecasting example 1985 UG exam**

The table below shows the sales of a toy robot over the last 11 months.

Month 1 2 3 4 5 6 7 8 9 10 11

Sales 3651 4015 3874 3501 3307 3105 2986 3100 3209 3450 3507

- Calculate a four month moving average for each month. What would be your forecast for the sales in month 12?
- Apply exponential smoothing with a smoothing constant of 0.9 to derive a forecast for the sales in month 12.
- Which of the two forecasts for month 12 do you prefer and why?

**Solution**

Now we cannot calculate a 4 month moving average until we have at least 4 observations - i.e. we can only calculate such an average from month 4 onward. Hence we have:

m_{4} = (3651 + 4015 + 3874 + 3501)/4 = 3760.25

m_{5} = (4015 + 3874 + 3501 + 3307)/4 = 3674.25

m_{6} = (3874 + 3501 + 3307 + 3105)/4 = 3446.75

m_{7} = (3501 + 3307 + 3105 + 2986)/4 = 3224.75

m_{8} = (3307 + 3105 + 2986 + 3100)/4 = 3124.50

m_{9} = (3105 + 2986 + 3100 + 3209)/4 = 3100.00

m_{10} = (2986 + 3100 + 3209 + 3450)/4 = 3186.25

m_{11} = (3100 + 3209 + 3450 + 3507)/4 = 3316.50

The forecast for month 12 is just the moving average for the month before that i.e. the moving average for month 11 = m_{11} = 3316.50

Applying exponential smoothing with a smoothing constant of 0.9 we get:

M_{1} = Y_{1} = 3651

M_{2} = 0.9Y_{2} + 0.1M_{1} = 0.9(4015) + 0.1(3651) = 3978.60

M_{3} = 0.9Y_{3} + 0.1M_{2} = 0.9(3874) + 0.1(3978.60) = 3884.46

M_{4} = 0.9Y_{4} + 0.1M_{3} = 0.9(3501) + 0.1(3884.46) = 3539.35

M_{5} = 0.9Y_{5} + 0.1M_{4} = 0.9(3307) + 0.1(3539.35) = 3330.24

M_{6} = 0.9Y_{6} + 0.1M_{5} = 0.9(3105) + 0.1(3330.24) = 3127.52

M_{7} = 0.9Y_{7} + 0.1M_{6} = 0.9(2986) + 0.1(3127.52) = 3000.15

M_{8} = 0.9Y_{8} + 0.1M_{7} = 0.9(3100) + 0.1(3000.15) = 3090.02

M_{9} = 0.9Y_{9} + 0.1M_{8} = 0.9(3209) + 0.1(3090.02) = 3197.10

M10= 0.9Y10 + 0.1M_{9} = 0.9(3450) + 0.1(3197.10) = 3424.71

M_{11} = 0.9Y_{11} + 0.1M10 = 0.9(3507) + 0.1(3424.71) = 3498.77

As before the forecast for month 12 is just the average for month 11 = M_{11} = 3498.77.

- MSD= [(3760.25 - 3307)² + ... + (3186.25 - 3507)²]/7 = 141407.9

and for the exponentially smoothed average with a smoothing constant of 0.9

- MSD= [(3651 - 4015)² + ... + (3424.71 - 3507)²]/10 = 51008.3

Overall then we see that exponential smoothing with a smoothing constant of 0.9 appears to give the best one month ahead forecasts as it has a lower MSD. Hence we prefer the forecast of 3498.77 that has been produced by exponential smoothing with a smoothing constant of 0.9.

**Forecasting example**

The table below shows the movement of the price of a commodity over 12 months.

Month 1 2 3 4 5 6 7 8 9 10 11 12

Price 25 30 32 33 32 31 30 29 28 28 29 31

- Calculate a 6 month moving average for each month. What is the forecast for month 13?
- Apply exponential smoothing with smoothing constants of 0.7 and 0.8 to derive forecasts for month 13.
- Which of the two forecasts based on exponential smoothing for month 13 do you prefer and why?

**Solution**

Now we cannot calculate a 6 month moving average until we have at least 6 observations - i.e. we can only calculate such an average from month 6 onward. Hence we have:

m_{6} = (25 + 30 + 32 + 33 + 32 + 31)/6 = 30.50

m_{7} = (30 + 32 + 33 + 32 + 31 + 30)/6 = 31.33

m_{8} = (32 + 33 + 32 + 31 + 30 + 29)/6 = 31.17

m_{9} = (33 + 32 + 31 + 30 + 29 + 28)/6 = 30.50

m_{10} = (32 + 31 + 30 + 29 + 28 + 28)/6 = 29.67

m_{11} = (31 + 30 + 29 + 28 + 28 + 29)/6 = 29.17

m_{12} = (30 + 29 + 28 + 28 + 29 + 31)/6 = 29.17

The forecast for month 13 is just the moving average for the month before that i.e. the moving average for month 12 = m_{12} = 29.17.

Applying exponential smoothing with a smoothing constant of 0.7 we get:

M_{1} = Y_{1} = 25

M_{2} = 0.7Y_{2} + 0.3M_{1} = 0.7(30) + 0.3(25) = 28.50

M_{3} = 0.7Y_{3} + 0.3M_{2} = 0.7(32) + 0.3(28.50) = 30.95

M_{4} = 0.7Y_{4} + 0.3M_{3} = 0.7(33) + 0.3(30.95) = 32.39

M_{5} = 0.7Y_{5} + 0.3M_{4} = 0.7(32) + 0.3(32.39) = 32.12

M_{6} = 0.7Y_{6} + 0.3M_{5} = 0.7(31) + 0.3(32.12) = 31.34

M_{7} = 0.7Y_{7} + 0.3M_{6} = 0.7(30) + 0.3(31.34) = 30.40

M_{8} = 0.7Y_{8} + 0.3M_{7} = 0.7(29) + 0.3(30.40) = 29.42

M_{9} = 0.7Y_{9} + 0.3M_{8} = 0.7(28) + 0.3(29.42) = 28.43

M_{10} = 0.7Y_{10} + 0.3M_{9} = 0.7(28) + 0.3(28.43) = 28.13

M_{11} = 0.7Y_{11} + 0.3M_{10} = 0.7(29) + 0.3(28.13) = 28.74

M_{12} = 0.7Y_{12} + 0.3M_{11} = 0.7(31) + 0.3(28.74) = 30.32

As before the forecast for month 13 is just the average for month 12 = M_{12} = 30.32.

Applying exponential smoothing with a smoothing constant of 0.8 we get:

M_{1} = Y_{1} = 25

M_{2} = 0.8Y_{2} + 0.2M_{1} = 0.8(30) + 0.2(25) = 29.00

M_{3} = 0.8Y_{3} + 0.2M_{2} = 0.8(32) + 0.2(29.00) = 31.40

M_{4} = 0.8Y_{4} + 0.2M_{3} = 0.8(33) + 0.2(31.40) = 32.68

M_{5} = 0.8Y_{5} + 0.2M_{4} = 0.8(32) + 0.2(32.68) = 32.14

M_{6} = 0.8Y_{6} + 0.2M_{5} = 0.8(31) + 0.2(32.14) = 31.23

M_{7} = 0.8Y_{7} + 0.2M_{6} = 0.8(30) + 0.2(31.23) = 30.25

M_{8} = 0.8Y_{8} + 0.2M_{7} = 0.8(29) + 0.2(30.25) = 29.25

M_{9} = 0.8Y_{9} + 0.2M_{8} = 0.8(28) + 0.2(29.25) = 28.25

M_{10} = 0.8Y_{10} + 0.2M_{9} = 0.8(28) + 0.2(28.25) = 28.05

M_{11} = 0.8Y_{11} + 0.2M_{10} = 0.8(29) + 0.2(28.05) = 28.81

M_{12} = 0.8Y_{12} + 0.2M_{11} = 0.8(31) + 0.2(28.81) = 30.56

As before the forecast for month 13 is just the average for month 12 = M_{12} = 30.56.

To decide which of the two forecasts based on exponential smoothing we prefer we calculate the MSD for the two exponentially smoothed averages. The resulting figures represent the historical accuracy of the two forecasting procedures with respect to one month ahead forecasts. Knowing this accuracy tells us which of the two exponentially smoothed forecasts for month 13 we prefer.

Performing the calculations we find that for exponential smoothing with a smoothing constant of 0.7 MSD=4.97 whilst for exponential smoothing with a smoothing constant of 0.8 MSD=4.43.

Overall then we see that exponential smoothing with a smoothing constant of 0.8 appears to give the best one month ahead forecasts over the last 11 months. Hence we prefer the forecast of 30.56 for month 13 that has been produced by exponential smoothing with a smoothing constant of 0.8.

## This problem has been solved.

## Cite This work.

To export a reference to this article please select a referencing stye below.

Urgent Homework (2022) . Retrive from https://www.urgenthomework.com/sample-homework/forecasting-examples-with-solution

"." Urgent Homework ,2022, https://www.urgenthomework.com/sample-homework/forecasting-examples-with-solution

Urgent Homework (2022) . Available from: https://www.urgenthomework.com/sample-homework/forecasting-examples-with-solution

[Accessed 29/11/2022].

Urgent Homework . ''(Urgent Homework ,2022) https://www.urgenthomework.com/sample-homework/forecasting-examples-with-solution accessed 29/11/2022.

## Buy Forecasting Examples With Solution Answers Online

Talk to our expert to get the help with Forecasting Examples With Solution to complete your assessment on time and boost your grades now

The main aim/motive of the management assignment help services is to get connect with a greater number of students, and effectively help, and support them in getting completing their assignments the students also get find this a wonderful opportunity where they could effectively learn more about their topics, as the experts also have the best team members with them in which all the members effectively support each other to get complete their diploma assignments. They complete the assessments of the students in an appropriate manner and deliver them back to the students before the due date of the assignment so that the students could timely submit this, and can score higher marks.Â The experts of the assignment help services at urgenthomework.com are so much skilled, capable, talented, and experienced in their field of programming homework help writing assignments, so, for this, they can effectively write the best economics assignment help services.

## Follow Us