CH.8Forecasting Learning objectives:

After completing this chapter, you should be able to:1.Explain the importance of forecasting in

organization.

2.Describe the three major approaches to forecasting.

3.Measure the accuracy of a forecast over time.

SummaryDecision maker rely on forecasting to reduce uncertainty about the future, for this reason forecasts play a vital role in planning.

Decision makers have a wide variety of forecasting models to choose from, these can be classified as qualitative, projection of historical pattern, and explanatory models.

GlossaryTime series: A time- ordered sequence of observation.

Trend: A long- term movement in time series data, often linear, but not necessarily.

CH.8

Forecasting

If there is considerable uncertainty , it is much more difficult to formulate plans that will produce the desired results than when there is little or no uncertainty involved.

That is

Forecasts provide decision makers with an improved picture of probable future events.

So, forecasts are important because they, can help to reduce some of the uncertainty?

In fact, forecasts are vital inputs for almost all planning processes.

Forecasts are used planning the system

Product and service design, process design, capacity planning, and equipment investment decision.

As well as for planning the use of system

Advertising production and inventory planning and scheduling, purchasing, planning the size of work force, budgeting and cost estimation.

Good forecasts are more likely to result using this process:

1 .Determine the purpose of the forecast.

2 .Determine the time horizon.

3 .Select an appropriate technique.

4 .Identify the necessary date.

5 .Make the forecast.

6 .Monitor forecast errors .

For forecasting we use

Qualitative forecasts Quantitative forecasts

We will use this approach

Quantitative models

Time series models:

1 .Naïve

2 .Moving averages.

3 .Exponential smoothing.

4 .Trend.

Associative models

1 .Simple regression.

Let’s go to example to see how we could apply each type

Forecasts that use

Time series

DATA

*Time series data are historical values of a variable that have been intervals.

Daily demand, weekly sales, quarterly revenues, annual demand.

Forecasting techniques that use time series data typically involve the assumption that past experience reflects probable future experience.

So, what are the observed patterns in historical data ?

These are:

Trends, seasonal variation, cyclical variations, and random variations.

Look at figures:D

eman

d

Dem

and

0 1 2 3month

Year10 2 3

Random variation around an average average

Seasonal variation D

eman

d

Dem

and

0 1 2 3year

4

Year10 2 3 4

Cyclical variation

Trend with random

Techniques for averaging

1 .Naïve forecasts.

2 .Moving averages.

3 .Exponential smoothing.

Naïve forecasts

It’s is simple , for any period the previous periods actual value.

How?

If demand last week was 50 units the forecasts for the upcoming week is 50 units.

If the demand in the upcoming week turns to be 54 units, the forecast for the following week would be 54 units.

Moving averages

Uses a number of the most recent actual data values in generating forecast.

The moving average forecast can be computed using the following equation:

MAn = Ai

M

n

i= 1

n

Where:

i = the ‘age’ of the data I = 1,2,3..………,

n = the number of periods in the moving average.

Ai = actual value with age i.

Go to example

Example 1.

Compute a three-period moving average forecast given demand for shopping product for the last 5 years:

Period Age Demand

1

2

3

4

5

5

4

3

2

1

40

44

36

42

40

MA3 = 36 + 42 + 40 = 38.33

3

If actual demand in period 6 turns out to be 41, the moving average forecast for period 7 would be:

MA3 = 42 + 40 + 41 = 41.00

3

If we want to make a weight for each data used in computing moving average?

Go to the example

Example 1.

Weighted MA.

WMA = Ai (w) MM w

Weight determined base number of period MA.

Period Weight Demand

1

2

3

4

5

3

2

1

40

44

36

42

40Weight

If we use:

WMA3 = 36(1) + 42(2) + 40(3) = 40.006

If actual demand in period 6 turns to be 40,the weighted MA forecast for period

7 would be:

WMA3 = 42(1) + 40(2) + 40(3) = 40.336For 7

Note :- as we move forward we change the weighted ( it’s arrangement ).

Exponential smoothing model

Is an averaging technique that reduces difficulties associated the previous model.

The exponentially smoothed forecast can be computed using this equation:

Ft = F + X ( A – F )

Where:

Ft = the forecast for period (t).

F t-1 = the forecast for period ( t-1 ).

X = smoothing constraint.

A t-1 = actual demand or sales for period ( t-1 ).

Go to understanding smoothing constant, and how to use it?

t-1 t-1 t-1

The smoothing constant , represent a percentage of the forecast error.

Each new forecast is equal to the previous forecast plus a percentage of the previous error.

For example: suppose the previous forecast was 100 units, actual demand was 90 units, and = .10 . The new forecast would be computed as follows

Ft = 100 + .10 ( 90- 100 ) = 99

Then, if the next actual demand turns out to be 102, the next forecast would be:

Ft = 99 + .10 ( 102- 99 ) =99.3

For forecasting which a percentage to use low or high?

The sensitivity of forecast adjustment to error is determined by the smoothing constant, alpha.

The closer its value ( ) is to zero , the slower the forecast will be to adjust to forecast errors ( the greater the smoothing ) conversely , the closer the value of

is to 1.00 the greater the sensitivity and the less the smoothing.

Selecting a smoothing constant is, basically, a matter of judgment and trial and error.

If you want to be sure about that use .05 and .50

Go to the example

Period A Ft Error A Ft Error

. =05 . =50

1

2

3

4

5

6

7

8

9

10

11

12

42

40

43

40

41

39

46

44

45

38

40

- - - - -

Ft = F - ( A t-1 – F t-1 ) t-1 Error = A – Ft

For accuracy use mean absolute deviation ( MAD ) equation:

MAD = e = ( A – Ft )n

MM

n

value with MAD less is the best for forecasting?

Suppose you have these info.

Alternative MAD

1

2

3

26.3

18.2

37.0

Best accuracy

Complete the previous example to see that works .

Techniques for trend

Trend is a persistent upward or downward movement in a time series. The trend may be linear or nonlinear.

We will focus on linear trends.

There are two important techniques that can be used to develop forecasts when trend present.

1 .Trend equation.

2 .Trend- adjusted exponential smoothing.

Go to examples

Trend equation : a linear trend equation:

Yt = a + bt

Where:

t = a specified number of time periods from t=0

Yt = the forecast for period t.

a = value of Yt at t=0.

b = slope of the line .

The coefficient of the line, a and b , can be computed from historical data using these two equation:

b = n Ty - t y MMM

n t - ( t) MM 22

a = Y - b t M M

n

Where n the number of periods

Note-:

Consider the trend equation Yt = 45 + 5t.

The value of Yt when t = 0 is 45 , and the slope of the line is 5 , which means that the value of Yt will increase by 5 units for each one unit increase in t. If t = 10, the forecast, Yt, is 45 +5(10) = 95 units.

Example: Month Actual demand Mar Apr May Jun Jul Auq Sept Oct Nov

112 125 120 133 136 146 140 155 152

1 .Plot the data to determine if linear trend equation is appropriate .

2 .Obtain a trend equation .

3 .Forecast demand for the next two months.

Solution:

1 .The data seem to show an upward, roughly linear trend .

Period

Dem

and

1 2 3 4 5 6 7 8 9

110120130

140

150

160

MonthMar Apr May Jun Jul Auq

Nov

Sept Oct

112 125 120 133 136 146 140 155 152

y t ty 1 2 3 4 5 6 7 8 9

112 250 360 532 660 876 980

1240 1360

t = 45M M y =1218 Ty M =6398

From the table

n = 9 , t = 45 , t = 285M M

2

b = 9 ( 6398 ) – 45 ( 1219 ) = 2727 = t 5.05

5409) 285 – (45 ) 45(

a = 1219 – 5,05 ( 45 ) = 110.19

9

Yt = 110.19 + 5.05 t

3 .The next two months, t= 10 , t = 1.

Y dec = 110.19 + 5.05 ( 10 ) = 160.69.

Y jan = 110.19 + 5.05 ( 11 ) = 165.74.

Trend- adjusted exponential smoothing

Note: we apply this technique when time series includes trend.

The equation for trend adjusted forecast is:

TAFt+1 = St + Tt

Where

St = smoothed error

Tt = current trend estimate.

And

St = TAF t + ( A t – TAF t )

Tt = T t-1 + ( TAF t – TAF t-1 – T t-1 ) 2

Suppose a manager estimates a trend of + 10 units, based on observed changes in the first four data, and uses a starting forecast of 250, and =.5 and = .4 if the next actual value is 255, TAF for the following period would be computed in this manager:

St = 250 + .5 ( 255 – 250 ) = 252.5

Tt = 10 + .4 (0)*

*Since this is the initial forecast, no previous error is available, so the value zero is used.

TAF = St + Tt = 252.5 + 10 = 262.5

If the next actual value is 265 the next TAF would be

St = 262.5 + 05 ( 265 – 262.5 ) = 263.75

Tt = 10+ .4 ( 262.5 – 250 – 10 ) = 11.00

TAF = 263.75 + 11.00 = 274.75

We can use another equation for TAF

2

FITL = F t + T t

Where:

FIT t forecast including trend

F t = new forecast simple smoothing.

T t = trend for period

For F t = F t-1 + ( A t-1 – F t-1 )

For T t = ( 1 – B ) T t-1 + B ( F t – F t-1 )

To compute FIT t follow these steps:

1 .Find F t.

2 .Compute T t

3 .Find FIT t = 1 + 2.

Go to example

Example

You have the following info:Month demand

1 2 3 4 5 6 7 8 9

12 172019 24 26 31 32 36 . =2 B = .4

F 2 = 11 + .2 ( 12 – 11 ) = 11.2

T 2 = ( 1 - .4) 0 + .4 ( 11.2 – 11.0 ) = .08

FIT 2 = 11.2 + .08 = 11.28

And so on for the other periods.

Assume the initial forecast for month first was 11 unit

M d1 2 3 4 5 6 7 8 9

12 172019 24 26 31 32 36

Ft Tt FITt11

11.212.56 13.89 14.91 16.75 18.58 21.07 23.25

0 .08 .51 .92 .96

1.30 1.52 1.91

2.02

-11.28 12.87 14.81 15.87 18.03 20.10 22.98 25.27

Techniques for seasonality

Seasonal variation in time series data are regularly upward or downward movements.

Weather variation and it’s impact on travel sports product.

To compute seasonal relative, use this model:

Centered moving average, which based on

Identical to a moving average forecast as are the resulting values. However the values are not projected as in a forecast, but, instead, they are positioned in the middle of the periods to compute average.

Go to example

Period Demand Three-period centered average

1

2

3

40

46

42

Average = 40 + 46 + 42

3 =42.67

It would be positioned of period 2, the average representative of the series at that point. The ratio of demand at period 2 to this centered average at period 2 is an estimate of the seasonal relative at that point.

Because the ratio is 46 = 1.08 the series is about 8% above average at this point.

What if both trend and seasonality appear in a series?

42.67

Trend + seasonality

Suppose a furniture manufacturer wants to predict quarterly demand for a certain seat for period 15 and 16, which happen to be the second and third quarters of a particular year. The series consists of both trend and seasonality. The trend portion of demand can be projected the equation Yt= 124 + 7.5t . Quarter

relatives are:

Q1 = 1.20 Q2 = 1.10 Q3 = .75 Q4 = .95

Use the info. To predict demand for period 15 and 16.

Solution:

The trend values at t = 15 and t = 16

Y15 = 124 + 7.5 ( 15 ) = 235.5

Y 16 = 124 + 7.5 ( 16 ) = 244.0

Period 15 235.5 ( 1.10 ) = 259.05

Period 16 244.0 ( .75 ) = 183.00

Quarter relative

Forecasts that use explanatory model

These model incorporate one or more variables that are related to the variable of interest and, therefore, they can be used to predict future values of that variable.

Are called regression models

We are going to use the simple linear regression.

*For multiple regression go to the computer?

Simple linear regression

Consists two variable

The variable to be forecasted is dependent variable?

The other value used to explain or predict the dependent variable is called independent variable?

To make forecast , the two variables must be expressed in a linear equation of the form:

Y = a + bx

Where:

Y = the dependent variable.

X = the independent variable.

B = slope of the line.

b = n x y - x y M M M

N x - ( x ) 2M M

2A = y - b x = y – bx

n

M M

Note:

In order to successfully use this approach , it is necessary to be able to:

1 .Identify an appropriate independent variable.

2 .Obtain a sample of at least 10 observations.

3 .Develop an equation.

4 .Identify any restriction on prediction.

5 .Measure accuracy in a given forecast.

Go to the example ??

Conceder, for example , 10 paired observations about sales and ads .

)millions of( $

observation Ads expenditure, x Sales , y 123456789

10

1.11.4

20.9

1.4

1.62

1.71.20.8

1010

107

78

111196

You can plot the pairs to see if a linear relationship is appropriate??,

It appear / linear relationship

Solution:

Calculations for regression coefficients

observation Ads expenditure, x Sales , y 123456789

10

1.11.4

20.9

1.4

1.62

1.71.20.8

1010

107

78

111196

x y x y2 2

14.020.0

16.06.3

7.711.2

22.018.710.8

4.8

1.694.00

2.560.81

1.211.96

4.002.891.440.64

100100

10049

4964

1211218136

14.1 89 131.5 21.47 821

b = 10 ( 131.5 ) – 14.1 ( 89 ) = 60.1 = + 3.78

10 ) 21.47 – ( 14.1 ) 14.1( 15.89

a = 89 – 3.78 ( 14.1 ) = 3.57

10

y = 3.57 + 3.87

Before we go in prediction process ,we must measure the strength of the relationship between the two variable . For that, we use correlation coefficient ?

r = n x y - x y

n ( x ) - ( x ) X n ( y ) - ( y ) 2 2 2 2M M M M

M M M

r = 10 ( 131.5 ) - ( 14.1 ) ( 89 )

10) 21.47 ) - ( 14.1 ( X 10 ( 821 ) - ( 89 )

. + =887

The same as the sign of b.

Relationship is good to use for prediction .

Square of the correlation coefficient, r , can be used as a measure of goodness of fit : the closer the value is to 1.00 the better the fit, for Ads data r = 0.787 this means that .79 of the sum of the squared deviation of Y value around the mean of

the Y . We say that the independent variable explain .79 of the variation.

There are a number of important questions that must be answered if a regression equation is to be used for forecasting??

2 2

1 .Is the relationship significant?

2 .How accurate will a particular forecast probably be?

These questions and other depend on the extend to which the points scatter around the regression line.

This is measured by standard error of estimate, se ,

Se =

n – 2 / because we have two coefficient a and b

y - a y - b x y

n – 2

M M M 2

For our example

Se = 821 – 3.57) 89 – (3.78 ) 131.5(

10 – 2 . = 880

Note : if the value of r is less than , say .40 the computed relationship will not be strong enough to yield acceptable forecast accuracy.

To test for the existence of relationship ( than the slope is non zero )

Compute the ration

T test = b

2

S b

Where

b = slope of regression line.

S b = Se 1

X - ( x )n

2 2M M

For our example

S b = .880 121.47 ) – 14.1 ( 2

10

. =880 .629 . =698

T test = 3.78 = 5.42

.698

Compare this value with value of t .025 for n – 2 / from the table of T test

n 10 11 12 13 14 15 16 17 18 19 20 21 22

t 2.31 2.26 2.23 2.20 2.18 2.16 2.15 2.13 2.12 2.11 2.10 2.09 2.08

If the test is greater than the table value, as it is in our case ( 5.42 ), it can be concluded that the relationship is significant.

Not due to the chance ??

How accurate will a given predication be?

Here we should use the regression equation to compute expected value of the dependent variable for a selected or given value of X the construct a 95%

confidence interval around that value.

Suppose expected value of X is 1.3?

Y + t S reg

Where Y = a + b x

t = from the T test table

S reg = 1 + 1 +

n )x g – x (

x - ( x )n

M M 2

22

X g = value of x to be used

X = 1.3 , X = 14.1 = 1.41 10

S reg = .880 1 + 1 +

10 )1.3 – 1.41 (

21.47) - 14.1(10

2

2 . =926

For our case:

X g = 1.3 , 95% confidence interval

For the predicted value of y is

Y = a + b x = 3.57 + 3.78 (1.3) = 8.48

95% interval = 8.48 + 2.31 ( .926 )

= 8.48 + 2.14

or 6.34 to 10.62

for test table.