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Forecasting

Chapter 8

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What is a Forecast?

Forecast

A prediction of future events used for planning purposes.

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Demand Patterns

A time series is the repeated observations of demand for a service or product in their order of occurrence
There are five basic time series patterns

Horizontal
Trend
Seasonal
Cyclical
Random

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Demand Patterns

Quantity

Time

(a) Horizontal: Data cluster about a horizontal line

Figure 8.1

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Demand Patterns

Quantity

Time

(b) Trend: Data consistently increase or decrease

Figure 8.1

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Demand Patterns

Quantity

| | | | | | | | | | | |

J F M A M J J A S O N D

Months

(c) Seasonal: Data consistently show peaks and valleys

Year 1

Year 2

Figure 8.1

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Demand Patterns

Quantity

| | | | | |

1 2 3 4 5 6

Years

(d) Cyclical: Data reveal gradual increases and decreases over extended periods

Figure 8.1

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Demand Management Options

Demand Management

The process of changing demand patterns using one or more demand options

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Demand Management Options

Complementary Products
Promotional Pricing
Prescheduled Appointments
Reservations
Revenue Management
Backlogs
Backorders and Stockouts

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Key Decisions on Making Forecasts

Deciding What to Forecast

Level of aggregation
Units of measurement

Choosing the Type of Forecasting Technique

Judgment methods
Causal methods
Time-series analysis
Trend projection using regression

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Forecast Error

For any forecasting method, it is important to measure the accuracy of its forecasts.

Forecast error is simply the difference found by subtracting the forecast from actual demand for a given period, or

where

E_t = forecast error for period t

D_t = actual demand in period t

F_t = forecast for period t

E_t = D_t – F_t

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Measures of Forecast Error

(Σ|E_t|/ D_t)(100)

n

MAPE =

CFE = ΣE_t

ΣE_t²

n

MSE =

Σ|E_t|

n

MAD =

Cumulative sum of forecast errors (Bias)

Average forecast error

Mean Squared Error

Mean Absolute Percent Error

Mean Absolute Deviation

Standard deviation

CFE

n

Ē=

σ =

Σ(E_t– Ē)²

n– 1

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Example 8.1

The following table shows the actual sales of upholstered chairs for a furniture manufacturer and the forecasts made for each of the last eight months.

Calculate CFE, MSE, σ, MAD, and MAPE for this product.

Month t	Demand D_t	Forecast F_t	Error E_t		Error² E_t²		Absolute Error \|E_t\|		Absolute % Error (\|E_t\|/D_t)(100)
1	200	225		–25
2	240	220		20
3	300	285		15
4	270	290		–20
5	230	250		–20		400		20		8.7
6	260	240		20		400		20		7.7
7	210	250		–40		1,600		40		19.0
8	275	240		35		1,225		35		12.7
		Total		–15		5,275		195		81.3%

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Example 8.1

The following table shows the actual sales of upholstered chairs for a furniture manufacturer and the forecasts made for each of the last eight months.

Calculate CFE, MSE, σ, MAD, and MAPE for this product.

625	25	12.5%
400	20	8.3
225	15	5.0
400	20	7.4

Month t	Demand D_t	Forecast F_t	Error E_t		Error² E_t²		Absolute Error \|E_t\|		Absolute % Error (\|E_t\|/D_t)(100)
1	200	225		–25
2	240	220		20
3	300	285		15
4	270	290		–20
5	230	250		–20		400		20		8.7
6	260	240		20		400		20		7.7
7	210	250		–40		1,600		40		19.0
8	275	240		35		1,225		35		12.7
		Total		–15		5,275		195		81.3%

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Example 8.1

Using the formulas for the measures, we get:

CFE =

–15

Average forecast error (mean bias):

Mean squared error:

Cumulative forecast error (mean bias)

MSE =

ΣE_t²

n

5,275

8

=

659.4

=

CFE

n

–1.875

= =

15

8

Ē =

8-15

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Example 8.1

Standard deviation:

Mean absolute deviation:

Mean absolute percent error:

Σ[E_t – (–1.875)]²

n – 1

σ =

Σ|E_t|

n

MAD =

(Σ|E_t|/ D_t)(100)

n

MAPE =

= 27.4

= = 24.4

195

8

= = 10.2%

81.3%

8

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Example 8.1

A CFE of –15 indicates that the forecast has a slight bias to overestimate demand.
The MSE, σ, and MAD statistics provide measures of forecast error variability.
A MAD of 24.4 means that the average forecast error was 24.4 units in absolute value.
The value of σ, 27.4, indicates that the sample distribution of forecast errors has a standard deviation of 27.4 units.
A MAPE of 10.2 percent implies that, on average, the forecast error was about 10 percent of actual demand.

These measures become more reliable as the number of periods of data increases.

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Judgment Methods

Other methods (casual, time-series, and trend projection using regression) require an adequate history file, which might not be available.
Judgmental forecasts use contextual knowledge gained through experience.

Salesforce estimates
Executive opinion
Market research
Delphi method

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Causal Methods: Linear Regression

A dependent variable is related to one or more independent variables by a linear equation
The independent variables are assumed to “cause” the results observed in the past
Simple linear regression model is a straight line

Y = a + bX

where

Y = dependent variable

X = independent variable

a = Y-intercept of the line

b = slope of the line

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Linear Regression

Dependent variable

Independent variable

X

Y

Estimate of

Y from

regression

equation

Regression

equation:

Y = a + bX

Actual

value

of Y

Value of X used

to estimate Y

Deviation,

or error

Figure 8.3

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Linear Regression

The sample correlation coefficient, r

Measures the direction and strength of the relationship between the independent variable and the dependent variable.
The value of r can range from –1.00 ≤ r ≤ 1.00

The sample coefficient of determination, r²

Measures the amount of variation in the dependent variable about its mean that is explained by the regression line
The values of r² range from 0.00 ≤ r² ≤ 1.00

The standard error of the estimate, s_yx

Measures how closely the data on the dependent variable cluster around the regression line

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Example 8.2

The supply chain manager seeks a better way to forecast the demand for door hinges and believes that the demand is related to advertising expenditures. The following are sales and advertising data for the past 5 months:

Month	Sales (thousands of units)	Advertising (thousands of $)
1	264	2.5
2	116	1.3
3	165	1.4
4	101	1.0
5	209	2.0

The company will spend $1,750 next month on advertising for the product. Use linear regression to develop an equation and a forecast for this product.

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Example 8.2

We used POM for Windows to determine the best values of a, b, the correlation coefficient, the coefficient of determination, and the standard error of the estimate

The regression equation is

Y = –8.135 + 109.229X

a = –8.135

b = 109.229X

r = 0.980

r² = 0.960

s_yx = 15.603

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Example 8.2

The r of 0.98 suggests an unusually strong positive relationship between sales and advertising expenditures. The coefficient of determination, r², implies that 96 percent of the variation in sales is explained by advertising expenditures.

| |

1.0 2.0

Advertising ($000)

250 –

200 –

150 –

100 –

50 –

0 –

Sales (000 units)

Brass Door Hinge

X

Data

Forecasts

Figure 8.4

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Example 8.2

Forecast for month 6:

Y = –8.135 + 109.229X

Y = –8.135 + 109.229(1.75)

Y = 183.016 or 183,016 units

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Time Series Methods

Naïve forecast

The forecast for the next period equals the demand for the current period (Forecast = D_t)

Horizontal Patterns: Estimating the average

Simple moving average
Weighted moving average
Exponential smoothing

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Simple Moving Averages

Specifically, the forecast for period t + 1 can be calculated at the end of period t (after the actual demand for period t is known) as

F_t₊₁ = =

Sum of last n demands

n

D_t + D_t_-1 + D_t_-2 + … + D_t_-_n₊₁

n

where

D_t = actual demand in period t

n = total number of periods in the average

F_t₊₁ = forecast for period t + 1

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Example 8.3

a. Compute a three-week moving average forecast for the arrival of medical clinic patients in week 4. The numbers of arrivals for the past three weeks were as follows:

Week	Patient Arrivals
1	400
2	380
3	411

b. If the actual number of patient arrivals in week 4 is 415, what is the forecast error for week 4?

c. What is the forecast for week 5?

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Example 8.3

a. The moving average forecast at the end of week 3 is:

Week	Patient Arrivals
1	400
2	380
3	411

b. The forecast error for week 4 is

F₄ =

= 397.0

411 + 380 + 400

3

E₄ = D₄ – F₄

= 415 – 397 = 18

c. The forecast for week 5 requires the actual arrivals from weeks 2 through 4, the three most recent weeks of data

F₅ =

= 402.0

415 + 411 + 380

3

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Application 8.1

Estimating with Simple Moving Average using the following customer-arrival data:

Month	Customer arrival
1	800
2	740
3	810
4	790

Use a three-month moving average to forecast customer arrivals for month 5

F₅ =

= 780

D₄ + D₃ + D₂

3

790 + 810 + 740

3

=

Forecast for month 5 is 780 customer arrivals

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Application 8.1

If the actual number of arrivals in month 5 is 805, what is the forecast for month 6?

F₆ =

= 801.667

D₅ + D₄ + D₃

3

805 + 790 + 810

3

=

Forecast for month 6 is 802 customer arrivals

Month	Customer arrival
1	800
2	740
3	810
4	790

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Application 8.1

Forecast error is simply the difference found by subtracting the forecast from actual demand for a given period, or

Given the three-month moving average forecast for month 5, and the number of patients that actually arrived (805), what is the forecast error?

Forecast error for month 5 is 25

E_t = D_t – F_t

E₅ =

805 – 780

= 25

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Weighted Moving Averages

In the weighted moving average method, each historical demand in the average can have its own weight, provided that the sum of the weights equals 1.0.

The average is obtained by multiplying the weight of each period by the actual demand for that period, and then adding the products together

F_t₊₁ = W₁D₁ + W₂D₂ + … + W_nD_t_-_n₊₁

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Application 8.2

Using the customer arrival data in Application 14.1, let

W₁ = 0.50, W₂ = 0.30, and W₃ = 0.20. Use the weighted moving average method to forecast arrivals for month 5.

= 0.50(790) + 0.30(810) + 0.20(740)

F₅ = W₁D₄ + W₂D₃ + W₃D₂

= 786

Forecast for month 5 is 786 customer arrivals.

Given the number of customers that actually arrived (805),

what is the forecast error?

Forecast error for month 5 is 19.

E₅ =

805 – 786

= 19

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Application 8.2

If the actual number of arrivals in month 5 is 805, compute the forecast for month 6:

= 0.50(805) + 0.30(790) + 0.20(810)

F₆ = W₁D₅ + W₂D₄ + W₃D₃

= 801.5

Forecast for month 6 is 802 customer arrivals.

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Exponential Smoothing

A sophisticated weighted moving average that calculates the average of a time series by implicitly giving recent demands more weight than earlier demands
Requires only three items of data

The last period’s forecast
The demand for this period
A smoothing parameter, alpha (α), where 0 ≤ α ≤ 1.0

The equation for the forecast is

F_t₊₁ = α(Demand this period) + (1 – α)(Forecast calculated last period)

= αD_t + (1 – α)F_t

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Exponential Smoothing

The emphasis given to the most recent demand levels can be adjusted by changing the smoothing parameter.
Larger α values emphasize recent levels of demand and result in forecasts more responsive to changes in the underlying average.
Smaller α values treat past demand more uniformly and result in more stable forecasts.

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Example 8.4

Reconsider the patient arrival data in Example 14.3. It is now the end of week 3 so the actual arrivals is known to be 411 patients. Using α = 0.10, calculate the exponential smoothing forecast for week 4.

b. What was the forecast error for week 4 if the actual demand turned out to be 415?

c. What is the forecast for week 5?

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Example 8.4

a. To obtain the forecast for week 4, using exponential smoothing with and the initial forecast of 390*, we calculate the average at the end of week 3 as:

F₄ =

Thus, the forecast for week 4 would be 392 patients.

0.10(411) + 0.90(390) = 392.1

* Here the initial forecast of 390 is the average of the first two weeks of demand. POM for Windows and OM Explorer, on the other hand, simply use the actual demand for the first week as the default setting for the initial forecast for period 1, and do not begin tracking forecast errors until the second period.

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Example 8.4

b. The forecast error for week 4 is

c. The new forecast for week 5 would be

E₄ =

F₅ =

or 394 patients.

415 – 392 = 23

0.10(415) + 0.90(392.1) = 394.4

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Application 8.3

Suppose that there were 790 arrivals in month 4 (D_t), whereas the forecast (F_t) was for 783 arrivals. Use exponential smoothing with α = 0.20 to compute the forecast for month 5.

F_t₊₁ = F_t + α(D_t – F_t)

783 + 0.20(790 – 783)

= 784.4

Forecast for month 5 is 784 customer arrivals

Given the number of patients that actually arrived (805), what is the forecast error?

E₅ =

Forecast error for month 5 is 21

805 – 784 = 21

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Application 8.3

Given the actual number of arrivals in month 5, what is the forecast for month 6?

F_t₊₁ = F_t + α(D_t – F_t)

= 784.4 + 0.20(805 – 784.4)

= 788.52

Forecast for month 6 is 789 customer arrivals

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Trend Patterns: Using Regression

A trend in a time series is a systematic increase or decrease in the average of the series over time
The forecast can be improved by calculating an estimate of the trend
Trend Projection with Regression accounts for the trend with simple regression analysis.

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Example 8.5

Medanalysis, Inc., provides medical laboratory services
Managers are interested in forecasting the number of blood analysis requests per week
There has been a national increase in requests for standard blood tests.
The arrivals over the next 16 weeks are given in Table 8.1.
What is the forecasted demand for the next three periods?

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Example 8.5

Week	Arrivals	Week	Arrivals
1	28	9	61
2	27	10	39
3	44	11	55
4	37	12	54
5	35	13	52
6	53	14	60
7	38	15	60
8	57	16	75

Table 8.1

Arrivals at Medanalysis, Inc.

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Example 8.5

Figure 8.6(a)

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Example 8.5

Figure 8.6(b)

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Application 8.4

Use OM Explorer to project the following weekly demand data using trend projection with regression.
What is the forecasted demand for periods 11-14?

Week	Demand	Week	Demand
1 2 3 4 5	24 34 29 27 39	6 7 8 9 10	42 39 56 45 43

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Application 8.4

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Application 8.4

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Seasonal Patterns: Using Seasonal Factors

Multiplicative seasonal method

A method whereby seasonal factors are multiplied by an estimate of average demand to arrive at a seasonal forecast.

Additive seasonal method

A method in which seasonal forecasts are generated by adding a constant to the estimate of average demand per season.

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Multiplicative Seasonal Method

For each year, calculate the average demand for each season by dividing annual demand by the number of seasons per year.
For each year, divide the actual demand for each season by the average demand per season, resulting in a seasonal factor for each season.
Calculate the average seasonal factor for each season using the results from Step 2.
Calculate each season’s forecast for next year.

Multiplicative seasonal method

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Example 8.6

The manager of the Stanley Steemer carpet cleaning company needs a quarterly forecast of the number of customers expected next year. The carpet cleaning business is seasonal, with a peak in the third quarter and a trough in the first quarter. Following are the quarterly demand data from the past 4 years:

The manager wants to forecast customer demand for each quarter of year 5, based on an estimate of total year 5 demand of 2,600 customers.

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Example 8.6

	YEAR 1		YEAR 2

Q	Demand	Seasonal Factor (1)	Demand	Seasonal Factor (2)
1	45	45/250 = 0.18	70	70/300 = 0.23
2	335	335/250 = 1.34	370	370/300 = 1.23
3	520	520/250 = 2.08	590	590/300 = 1.97
4	100	100/250 = 0.40	170	170/300 = 0.57
Total	1,000		1,200
Average	1,000/4 = 250		1,200/4 = 300

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Example 8.6

	YEAR 3		YEAR 4

Q	Demand	Seasonal Factor (3)	Demand	Seasonal Factor (4)
1	100	100/450 = 0.22	100	100/550 = 0.18
2	585	585/450 = 1.30	725	725/550 = 1.32
3	830	830/450 = 1.84	1160	1160/550 = 2.11
4	285	285/450 = 0.63	215	215/550 = 0.39
Total	1,800		2,200
Average	1,800/4 = 450		2,200/4 = 550

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Example 8.6

Quarterly Forecasts

Quarter	Forecast
1	650 x 0.2043 = 132.795
2	650 x 1.2979 = 843.635
3	650 x 2.001 = 1,300.06
4	650 x 0.4977 = 323.505

Quarter	Average Seasonal Factor
1	0.2043
2	1.2979
3	2.0001
4	0.4977

Average Seasonal Factor

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Example 8.6

Figure 8.7

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Application 8.5

Suppose the multiplicative seasonal method is being used to forecast customer demand. The actual demand and seasonal indices are shown below.

	Year 1				Year 2				Average Index
Quarter	Demand		Index		Demand		Index		Average Index
1		100		0.40		192		0.64	0.52
2		400		1.60		408		1.36	1.48
3		300		1.20		384		1.28	1.24
4		200		0.80		216		0.72	0.76
Average		250				300

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Application 8.5

1320 units ÷ 4 quarters = 330 units

Quarter	Average Index
1	0.52
2	1.48
3	1.24
4	0.76

If the projected demand for Year 3 is 1320 units, what is the forecast for each quarter of that year?

Forecast for Quarter 1 =

Forecast for Quarter 2 =

Forecast for Quarter 3 =

Forecast for Quarter 4 =

0.52(330) ≈ 172 units

1.48(330) ≈ 488 units

1.24(330) ≈ 409 units

0.76(330) ≈ 251 units

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Criteria for Selecting �Time-Series Method

Criteria:

Minimizing bias (CFE)
Minimizing MAPE, MAD, or MSE
Maximizing r²for trend projections using regression
Using a holdout sample analysis
Using a tracking signal
Meeting managerial expectations of changes in the components of demand.
Minimizing the forecast errors in recent periods.

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Choosing a Time-Series Method

Using Statistical Criteria:

For more stable demand patterns, use lower α values or larger n values to emphasize historical experience.
For more dynamic demand patters, use higher α values or smaller n values.

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Choosing a Time-Series Method

Holdout sample

Actual demands from the more recent time periods in the time series that are set aside to test different models developed from the earlier time periods.

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Tracking Signals

A measure that indicates whether a method of forecasting is accurately predicting actual changes in demand.

Tracking signal =

CFE

MAD

Each period, the CFE and MAD are updated to reflect current error, and the tracking signal is compared to some predetermined limits.

CFE

MAD_t

or

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Tracking Signals

The MAD can be calculated as the simple average of all absolute errors or as a weighted average determined by the exponential smoothing method

MAD_t = α|E_t| + (1 – α)MAD_t_-1

If forecast errors are normally distributed with a mean of 0, the relationship between σ and MAD is simple

σ = ( π /2)(MAD) ≅ 1.25(MAD)

MAD = 0.7978σ ≅ 0.8σ where π = 3.1416

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Tracking Signals

+2.0 –

+1.5 –

+1.0 –

+0.5 –

0 –

–0.5 –

–1.0 –

–1.5 –

| | | | |

0 5 10 15 20 25

Observation number

Tracking signal

Out of control

Control limit

Figure 8.8

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Forecasting as a Process

Finalize

and communicate

6

Review by Operating Committee

5

Revise forecasts

4

Consensus meetings and collaboration

3

Prepare initial forecasts

2

Adjust history file

1

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Using Multiple Forecasting Methods

Combination forecasts
Judgmental adjustments
Focus forecasting

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Forecasting Principles

SOME PRINCIPLES FOR THE FORECASTING PROCESS
Better processes yield better forecasts
Demand forecasting is being done in virtually every company, either formally or informally. The challenge is to do it well—better than the competition
Better forecasts result in better customer service and lower costs, as well as better relationships with suppliers and customers
The forecast can and must make sense based on the big picture, economic outlook, market share, and so on
The best way to improve forecast accuracy is to focus on reducing forecast error
Bias is the worst kind of forecast error; strive for zero bias
Whenever possible, forecast at more aggregate levels. Forecast in detail only where necessary
Far more can be gained by people collaborating and communicating well than by using the most advanced forecasting technique or model

Table 8.2

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Adding Collaboration to the Process

CPFR

Collaborative planning, forecasting, and replenishment

A process for supply chain integration that allows a supplier and its customers to collaborate on making the forecast by using the Internet.

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Solved Problem 1

Chicken Palace periodically offers carryout five-piece chicken dinners at special prices. Let Y be the number of dinners sold and X be the price. Based on the historical observations and calculations in the following table, determine the regression equation, correlation coefficient, and coefficient of determination. How many dinners can Chicken Palace expect to sell at $3.00 each?

Observation	Price (X)		Dinners Sold (Y)
1		$2.70		760
2		$3.50		510
3		$2.00		980
4		$4.20		250
5		$3.10		320
6		$4.05		480
Total		$19.55		3,300
Average		$ 3.26		550

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Solved Problem 1

We use the computer to calculate the best values of a, b, the correlation coefficient, and the coefficient of determination

r ² = 0.71

r = –0.84

b = –277.63

a = 1,454.60

The regression line is

Y = a + bX =

1,454.60 – 277.63X

For an estimated sales price of $3.00 per dinner

Y = a + bX =

1,454.60 – 277.63(3.00)

= 621.71 or 622 dinners

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Solved Problem 2

The Polish General’s Pizza Parlor is a small restaurant catering to patrons with a taste for European pizza. One of its specialties is Polish Prize pizza. The manager must forecast weekly demand for these special pizzas so that he can order pizza shells weekly. Recently, demand has been as follows:

Week	Pizzas	Week	Pizzas
June 2	50	June 23	56
June 9	65	June 30	55
June 16	52	July 7	60

a. Forecast the demand for pizza for June 23 to July 14 by using the simple moving average method with n = 3 then using the weighted moving average method with and weights of 0.50, 0.30, and 0.20, with 0.50.

b. Calculate the MAD for each method.

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Solved Problem 2

a. The simple moving average method and the weighted moving average method give the following results:

Current Week	Simple Moving Average Forecast for Next Week	Weighted Moving Average Forecast �for Next Week
June 16
June 23
June 30
July 7

= 55.7 or 56

52 + 65 + 50

3

[(0.5 × 52) + (0.3 × 65) + (0.2 × 50)] = 55.5 or 56

= 57.7 or 58

56 + 52 + 65

3

= 54.3 or 54

55 + 56 + 52

3

[(0.5 × 56) + (0.3 × 52) + (0.2 × 65)] = 56.6 or 57

[(0.5 × 55) + (0.3 × 56) + (0.2 × 52)] = 54.7 or 55

= 57.0 or 57

60 + 55 + 56

3

[(0.5 × 60) + (0.3 × 55) + (0.2 × 56)] = 57.7 or 58

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Solved Problem 2

b. The mean absolute deviation is calculated as follows:

		Simple Moving Average		Weighted Moving Average
Week	Actual Demand	Forecast for This Week	Absolute Errors \|E_t\|	Forecast for This Week	Absolute Errors \|E_t\|
June 23	56	56		56
June 30	55	58		57
July 7	60	54		55

|56 – 56| = 0

|55 – 58| = 3

|60 – 54| = 6

MAD = = 3

0 + 3 + 6

3

MAD = = 2.3

0 + 2 + 2

3

|56 – 56| = 0

|55 – 57| = 2

|60 – 55| = 5

For this limited set of data, the weighted moving average method resulted in a slightly lower mean absolute deviation. However, final conclusions can be made only after analyzing much more data.

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Solved Problem 3

The monthly demand for units manufactured by the Acme Rocket Company has been as follows:

Month	Units	Month	Units
May	100	September	105
June	80	October	110
July	110	November	125
August	115	December	120

a. Use the exponential smoothing method to forecast June to January. The initial forecast for May was 105 units; α = 0.2.

b. Calculate the absolute percentage error for each month from June through December and the MAD and MAPE of forecast error as of the end of December.

c. Calculate the tracking signal as of the end of December. What can you say about the performance of your forecasting method?

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Solved Problem 3

a.

Current Month, t	Calculating Forecast for Next Month F_t₊₁ = αD_t + (1 – α)F_t	Forecast for Month t + 1
May		June
June		July
July		August
August		September
September		October
October		November
November		December
December		January

0.2(100) + 0.8(105)

= 104.0 or 104

0.2(80) + 0.8(104.0)

0.2(110) + 0.8(99.2)

= 99.2 or 99

= 101.4 or 101

0.2(115) + 0.8(101.4)

0.2(105) + 0.8(104.1)

0.2(110) + 0.8(104.3)

0.2(125) + 0.8(105.4)

0.2(120) + 0.8(109.3)

= 104.1 or 104

= 104.3 or 104

= 105.4 or 105

= 109.3 or 109

= 111.4 or 111

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Solved Problem 3

b.

–24

24

30.0%

11

		11

		10.0

Month, t	Actual Demand, D_t		Forecast, F_t		Error, E_t = D_t – F_t	Absolute Error, \|E_t\|	Absolute Percent Error, (\|E_t\|/D_t)(100)
June		80		104
July		110		99
August		115		101
September		105		104
October		110		104
November		125		105
December		120		109
Total		765

14	14	12.0
1	1	1.0
6	6	5.5
20	20	16.0
11	11	9.2
39	87	83.7%

Σ|E_t|

n

MAD =

(Σ|E_t|/D_t)(100)

n

MAPE =

= = 11.96%

83.7%

7

= = 12.4

87

7

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Solved Problem 3

c. As of the end of December, the cumulative sum of forecast errors (CFE) is 39. Using the mean absolute deviation calculated in part (b), we calculate the tracking signal:

The probability that a tracking signal value of 3.14 could be generated completely by chance is small. Consequently, we should revise our approach. The long string of forecasts lower than actual demand suggests use of a trend method.

Tracking signal =

CFE

MAD

= = 3.14

39

12.4

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Solved Problem 4

The Northville Post Office experiences a seasonal pattern of daily mail volume every week. The following data for two representative weeks are expressed in thousands of pieces of mail:

Day	Week 1		Week 2
Sunday		5		8
Monday		20		15
Tuesday		30		32
Wednesday		35		30
Thursday		49		45
Friday		70		70
Saturday		15		10
Total		224		210

a. Calculate a seasonal factor for each day of the week.

b. If the postmaster estimates 230,000 pieces of mail to be sorted next week, forecast the volume for each day.

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Solved Problem 4

	Week 1			Week 2
Day	Mail Volume		Seasonal Factor (1)	Mail Volume		Seasonal Factor (2)	Average Seasonal Factor [(1) + (2)]/2
Sunday		5			8
Monday		20			15
Tuesday		30			32
Wednesday		35			30
Thursday		49			45
Friday		70			70
Saturday		15			10
Total		224			210
Average	224/7 = 32			210/7 = 30

5/32 = 0.15625

20/32 = 0.62500

30/32 = 0.93750

8/30 = 0.26667

15/30 = 0.50000

32/30 = 1.06667

0.21146

0.56250

1.00209

35/32 = 1.09375

49/32 = 1.53125

70/32 = 2.18750

15/32 = 0.46875

30/30 = 1.00000

45/30 = 1.50000

70/30 = 2.33333

10/30 = 0.33333

1.04688

1.51563

2.26042

0.40104

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Solved Problem 4

b. The average daily mail volume is expected to be 230,000/7 = 32,857 pieces of mail. Using the average seasonal factors calculated in part (a), we obtain the following forecasts:

6,948

18,482

32,926

0.21146(32,857) =

0.56250(32,857) =

1.00209(32,857) =

34,397

49,799

74,271

13,177

230,000

1.04688(32,857) =

1.51563(32,857) =

2.26042(32,857) =

0.40104(32,857) =

Day	Calculations	Forecast
Sunday
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
	Total

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