lecture 18 forecasting (continued) books introduction to materials management, sixth edition, j. r....

Lecture 18

Forecasting (Continued)

Books• Introduction to Materials Management, Sixth Edition, J. R. Tony Arnold, P.E., CFPIM, CIRM, Fleming

College, Emeritus, Stephen N. Chapman, Ph.D., CFPIM, North Carolina State University, Lloyd M. Clive, P.E., CFPIM, Fleming College

• Operations Management for Competitive Advantage, 11th Edition, by Chase, Jacobs, and Aquilano, 2005, N.Y.: McGraw-Hill/Irwin.

• Operations Management, 11/E, Jay Heizer, Texas Lutheran University, Barry Render, Graduate School of Business, Rollins College, Prentice Hall

Objectives

When you complete this chapter you should be able to :

Compute three measures of forecast accuracy

Develop seasonal indexes Conduct a regression and correlation

analysis Use a tracking signal

Common Measures of Error

Mean Absolute Deviation (MAD)

MAD =∑ |Actual - Forecast|

n

Mean Squared Error (MSE)

MSE =∑ (Forecast Errors)2

n

Common Measures of Error

Mean Absolute Percent Error (MAPE)

MAPE =∑100|Actuali - Forecasti|/Actuali

n

n

i = 1

Comparison of Forecast Error

Rounded Absolute Rounded AbsoluteActual Forecast Deviation Forecast Deviation

Tonnage with for with forQuarter Unloaded a = .10 a = .10 a = .50 a = .50

1 180 175 5.00 175 5.002 168 175.5 7.50 177.50 9.503 159 174.75 15.75 172.75 13.754 175 173.18 1.82 165.88 9.125 190 173.36 16.64 170.44 19.566 205 175.02 29.98 180.22 24.787 180 178.02 1.98 192.61 12.618 182 178.22 3.78 186.30 4.30

82.45 98.62




1 180 175 5.00 175 5.002 168 175.5 7.50 177.50 9.503 159 174.75 15.75 172.75 13.754 175 173.18 1.82 165.88 9.125 190 173.36 16.64 170.44 19.566 205 175.02 29.98 180.22 24.787 180 178.02 1.98 192.61 12.618 182 178.22 3.78 186.30 4.30

82.45 98.62

MAD =∑ |deviations|

n

= 82.45/8 = 10.31For a = .10

= 98.62/8 = 12.33For a = .50




1 180 175 5.00 175 5.002 168 175.5 7.50 177.50 9.503 159 174.75 15.75 172.75 13.754 175 173.18 1.82 165.88 9.125 190 173.36 16.64 170.44 19.566 205 175.02 29.98 180.22 24.787 180 178.02 1.98 192.61 12.618 182 178.22 3.78 186.30 4.30

82.45 98.62MAD 10.31 12.33

= 1,526.54/8 = 190.82For a = .10

= 1,561.91/8 = 195.24For a = .50

MSE =∑ (forecast errors)2

n




1 180 175 5.00 175 5.002 168 175.5 7.50 177.50 9.503 159 174.75 15.75 172.75 13.754 175 173.18 1.82 165.88 9.125 190 173.36 16.64 170.44 19.566 205 175.02 29.98 180.22 24.787 180 178.02 1.98 192.61 12.618 182 178.22 3.78 186.30 4.30

82.45 98.62MAD 10.31 12.33MSE 190.82 195.24

= 44.75/8 = 5.59%For a = .10

= 54.05/8 = 6.76%For a = .50

MAPE =∑100|deviationi|/actuali

n

n

i = 1




1 180 175 5.00 175 5.002 168 175.5 7.50 177.50 9.503 159 174.75 15.75 172.75 13.754 175 173.18 1.82 165.88 9.125 190 173.36 16.64 170.44 19.566 205 175.02 29.98 180.22 24.787 180 178.02 1.98 192.61 12.618 182 178.22 3.78 186.30 4.30

82.45 98.62MAD 10.31 12.33MSE 190.82 195.24MAPE 5.59% 6.76%

Exponential Smoothing with Trend Adjustment

When a trend is present, exponential smoothing must be modified

Forecast including (FITt) = trend

Exponentially Exponentiallysmoothed (Ft) + (Tt) smoothedforecast trend

Exponential Smoothing with Trend Adjustment

Ft = a(At - 1) + (1 - a)(Ft - 1 + Tt - 1)

Tt = b(Ft - Ft - 1) + (1 - b)Tt - 1

Step 1: Compute Ft

Step 2: Compute Tt

Step 3: Calculate the forecast FITt = Ft + Tt

Exponential Smoothing with Trend Adjustment Example

ForecastActual Smoothed Smoothed Including

Month(t) Demand (At) Forecast, Ft Trend, Tt Trend, FITt

1 12 11 2 13.002 173 204 195 246 217 318 289 36

10




1 12 11 2 13.002 173 204 195 246 217 318 289 36

10

F2 = aA1 + (1 - a)(F1 + T1)

F2 = (.2)(12) + (1 - .2)(11 + 2)

= 2.4 + 10.4 = 12.8 units

Step 1: Forecast for Month 2




1 12 11 2 13.002 17 12.803 204 195 246 217 318 289 36

10

T2 = b(F2 - F1) + (1 - b)T1

T2 = (.4)(12.8 - 11) + (1 - .4)(2)

= .72 + 1.2 = 1.92 units

Step 2: Trend for Month 2




1 12 11 2 13.002 17 12.80 1.923 204 195 246 217 318 289 36

10

FIT2 = F2 + T1

FIT2 = 12.8 + 1.92

= 14.72 units

Step 3: Calculate FIT for Month 2




1 12 11 2 13.002 17 12.80 1.92 14.723 204 195 246 217 318 289 36

10

15.18 2.10 17.2817.82 2.32 20.1419.91 2.23 22.1422.51 2.38 24.8924.11 2.07 26.1827.14 2.45 29.5929.28 2.32 31.6032.48 2.68 35.16


| | | | | | | | |1 2 3 4 5 6 7 8 9

Time (month)

Prod

uct d

eman

d

35 –

30 –

25 –

20 –

15 –

10 –

5 –

0 –

Actual demand (At)

Forecast including trend (FITt)with a = .2 and b = .4

Trend Projections

Fitting a trend line to historical data points to project into the medium to long-range

Linear trends can be found using the least squares technique

y = a + bx^

where y = computed value of the variable to be predicted (dependent variable)a = y-axis interceptb = slope of the regression linex = the independent variable

^

Least Squares Method

Time period

Valu

es o

f Dep

ende

nt V

aria

ble

Deviation1

(error)

Deviation5

Deviation7

Deviation2

Deviation6

Deviation4

Deviation3

Actual observation (y value)

Trend line, y = a + bx^


Time period

Valu

es o

f Dep

ende

nt V

aria

ble

Deviation1

Deviation5

Deviation7

Deviation2

Deviation6

Deviation4

Deviation3

Actual observation (y value)

Trend line, y = a + bx^

Least squares method minimizes the sum of the squared errors

(deviations)


Equations to calculate the regression variables

b = Sxy - nxySx2 - nx2

y = a + bx^

a = y - bx

Least Squares Example

b = = = 10.54∑xy - nxy∑x2 - nx2

3,063 - (7)(4)(98.86)140 - (7)(42)

a = y - bx = 98.86 - 10.54(4) = 56.70

Time Electrical Power Year Period (x) Demand x2 xy

2001 1 74 1 742002 2 79 4 1582003 3 80 9 2402004 4 90 16 3602005 5 105 25 5252005 6 142 36 8522007 7 122 49 854

∑x = 28 ∑y = 692 ∑x2 = 140 ∑xy = 3,063x = 4 y = 98.86


b = = = 10.54Sxy - nxySx2 - nx2

3,063 - (7)(4)(98.86)140 - (7)(42)

a = y - bx = 98.86 - 10.54(4) = 56.70

Time Electrical Power Year Period (x) Demand x2 xy

1999 1 74 1 742000 2 79 4 1582001 3 80 9 2402002 4 90 16 3602003 5 105 25 5252004 6 142 36 8522005 7 122 49 854

Sx = 28 Sy = 692 Sx2 = 140 Sxy = 3,063x = 4 y = 98.86

The trend line is

y = 56.70 + 10.54x^


| | | | | | | | |2001 2002 2003 2004 2005 2006 2007 2008 2009

160 –150 –140 –130 –120 –110 –100 –

90 –80 –70 –60 –50 –

Year

Pow

er d

eman

d

Trend line,y = 56.70 + 10.54x^

Seasonal Variations In Data

The multiplicative seasonal model can adjust trend data for seasonal variations in demand

Seasonal Variations In Data

1. Find average historical demand for each season 2. Compute the average demand over all seasons 3. Compute a seasonal index for each season 4. Estimate next year’s total demand5. Divide this estimate of total demand by the

number of seasons, then multiply it by the seasonal index for that season

Steps in the process:

Seasonal Index Example

Jan 80 85 105 90 94Feb 70 85 85 80 94Mar 80 93 82 85 94Apr 90 95 115 100 94May 113 125 131 123 94Jun 110 115 120 115 94Jul 100 102 113 105 94Aug 88 102 110 100 94Sept 85 90 95 90 94Oct 77 78 85 80 94Nov 75 72 83 80 94Dec 82 78 80 80 94

Demand Average Average Seasonal Month 2005 2006 2007 2005-2007 Monthly Index


Jan 80 85 105 90 94Feb 70 85 85 80 94Mar 80 93 82 85 94Apr 90 95 115 100 94May 113 125 131 123 94Jun 110 115 120 115 94Jul 100 102 113 105 94Aug 88 102 110 100 94Sept 85 90 95 90 94Oct 77 78 85 80 94Nov 75 72 83 80 94Dec 82 78 80 80 94


0.957

Seasonal index = average 2005-2007 monthly demand

average monthly demand

= 90/94 = .957


Jan 80 85 105 90 94 0.957Feb 70 85 85 80 94 0.851Mar 80 93 82 85 94 0.904Apr 90 95 115 100 94 1.064May 113 125 131 123 94 1.309Jun 110 115 120 115 94 1.223Jul 100 102 113 105 94 1.117Aug 88 102 110 100 94 1.064Sept 85 90 95 90 94 0.957Oct 77 78 85 80 94 0.851Nov 75 72 83 80 94 0.851Dec 82 78 80 80 94 0.851



Jan 80 85 105 90 94 0.957Feb 70 85 85 80 94 0.851Mar 80 93 82 85 94 0.904Apr 90 95 115 100 94 1.064May 113 125 131 123 94 1.309Jun 110 115 120 115 94 1.223Jul 100 102 113 105 94 1.117Aug 88 102 110 100 94 1.064Sept 85 90 95 90 94 0.957Oct 77 78 85 80 94 0.851Nov 75 72 83 80 94 0.851Dec 82 78 80 80 94 0.851


Expected annual demand = 1,200

Jan x .957 = 961,200

12

Feb x .851 = 851,200

12

Forecast for 2008


140 –

130 –

120 –

110 –

100 –

90 –

80 –

70 –| | | | | | | | | | | |J F M A M J J A S O N D

Time

Dem

and

2008 Forecast2007 Demand 2006 Demand2005 Demand

San Diego Hospital

10,200 –

10,000 –

9,800 –

9,600 –

9,400 –

9,200 –

9,000 –| | | | | | | | | | | |

Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec67 68 69 70 71 72 73 74 75 76 77 78

Month

Inpa

tient

Day

s

9530

9551

9573

9594

9616

9637

9659

9680

9702

9724

9745

9766

Trend Data

San Diego Hospital

1.06 –

1.04 –

1.02 –

1.00 –

0.98 –

0.96 –

0.94 –

0.92 – | | | | | | | | | | | |Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec67 68 69 70 71 72 73 74 75 76 77 78

Month

Inde

x fo

r Inp

atie

nt D

ays 1.04

1.021.01

0.99

1.031.04

1.00

0.98

0.97

0.99

0.970.96

Seasonal Indices

San Diego Hospital

10,200 –

10,000 –

9,800 –

9,600 –

9,400 –

9,200 –

9,000 –| | | | | | | | | | | |

Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec67 68 69 70 71 72 73 74 75 76 77 78

Month

Inpa

tient

Day

s

9911

9265

9764

9520

9691

9411

9949

9724

9542

9355

10068

9572

Combined Trend and Seasonal Forecast

Associative Forecasting

Used when changes in one or more independent variables can be used to predict the changes in the

dependent variable

Most common technique is linear regression analysis

We apply this technique just as we did in the time series example

Associative Forecasting

Forecasting an outcome based on predictor variables using the least squares technique

y = a + bx^

where y = computed value of the variable to be predicted (dependent variable)a = y-axis interceptb = slope of the regression linex = the independent variable though to predict the value of the dependent variable

^

Associative Forecasting Example

Sales Local Payroll($ millions), y ($ billions), x

2.0 13.0 32.5 42.0 22.0 13.5 7

4.0 –

3.0 –

2.0 –

1.0 –

| | | | | | |0 1 2 3 4 5 6 7

Sale

s

Area payroll


Sales, y Payroll, x x2 xy

2.0 1 1 2.03.0 3 9 9.02.5 4 16 10.02.0 2 4 4.02.0 1 1 2.03.5 7 49 24.5

∑y = 15.0 ∑x = 18 ∑x2 = 80 ∑xy = 51.5

x = ∑x/6 = 18/6 = 3

y = ∑y/6 = 15/6 = 2.5

b = = = .25∑xy - nxy∑x2 - nx2

51.5 - (6)(3)(2.5)80 - (6)(32)

a = y - bx = 2.5 - (.25)(3) = 1.75


4.0 –

3.0 –

2.0 –

1.0 –

| | | | | | |0 1 2 3 4 5 6 7

Sale

s

Area payroll

y = 1.75 + .25x^ Sales = 1.75 + .25(payroll)

If payroll next year is estimated to be $6 billion, then:

Sales = 1.75 + .25(6)Sales = $3,250,000

3.25

Standard Error of the Estimate

A forecast is just a point estimate of a future value

This point is actually the mean of a probability distribution

4.0 –

3.0 –

2.0 –

1.0 –

| | | | | | |0 1 2 3 4 5 6 7

Sale

s

Area payroll

3.25


where y = y-value of each data pointyc = computed value of the dependent variable, from the regression equationn = number of data points

Sy,x =∑(y - yc)2

n - 2


Computationally, this equation is considerably easier to use

We use the standard error to set up prediction intervals around the point

estimate

Sy,x =∑y2 - a∑y - b∑xy

n - 2


4.0 –

3.0 –

2.0 –

1.0 –

| | | | | | |0 1 2 3 4 5 6 7

Sale

s

Area payroll

3.25

Sy,x = =∑y2 - a∑y - b∑xyn - 2

39.5 - 1.75(15) - .25(51.5)6 - 2

Sy,x = .306

The standard error of the estimate is $306,000 in sales

How strong is the linear relationship between the variables?

Correlation does not necessarily imply causality! Coefficient of correlation, r, measures degree of

association Values range from -1 to +1

Correlation

Correlation Coefficient

r = nSxy - SxSy

[nSx2 - (Sx)2][nSy2 - (Sy)2]

Correlation Coefficient

r = nSxy - SxSy

[nSx2 - (Sx)2][nSy2 - (Sy)2]

y

x(a) Perfect positive correlation: r = +1

y

x(b) Positive correlation: 0 < r < 1

y

x(c) No correlation: r = 0

y

x(d) Perfect negative correlation: r = -1

Coefficient of Determination, r2, measures the percent of change in y predicted by the change in x Values range from 0 to 1 Easy to interpret

Correlation

For the Nodel Construction example:r = .901r2 = .81

Multiple Regression Analysis

If more than one independent variable is to be used in the model, linear regression can be extended to

multiple regression to accommodate several independent variables

y = a + b1x1 + b2x2 …^

Computationally, this is quite complex and generally done on the computer

Multiple Regression Analysis

y = 1.80 + .30x1 - 5.0x2^

In the Nodel example, including interest rates in the model gives the new equation:

An improved correlation coefficient of r = .96 means this model does a better job of predicting the change in construction sales

Sales = 1.80 + .30(6) - 5.0(.12) = 3.00Sales = $3,000,000

Measures how well the forecast is predicting actual values

Ratio of running sum of forecast errors (RSFE) to mean absolute deviation (MAD) Good tracking signal has low values If forecasts are continually high or low, the forecast

has a bias error

Monitoring and Controlling Forecasts

Tracking Signal

Monitoring and Controlling Forecasts

Tracking signal

RSFEMAD=

Tracking signal =

∑(Actual demand in period i -

Forecast demand in period i)

(∑|Actual - Forecast|/n)

Tracking Signal

Tracking signal

+

0 MADs

–

Upper control limit

Lower control limit

Time

Signal exceeding limit

Acceptable range

Tracking Signal Example

CumulativeAbsolute Absolute

Actual Forecast Forecast ForecastQtr Demand Demand Error RSFE Error Error MAD

1 90 100 -10 -10 10 10 10.02 95 100 -5 -15 5 15 7.53 115 100 +15 0 15 30 10.04 100 110 -10 -10 10 40 10.05 125 110 +15 +5 15 55 11.06 140 110 +30 +35 30 85 14.2

CumulativeAbsolute Absolute

Actual Forecast Forecast ForecastQtr Demand Demand Error RSFE Error Error MAD

1 90 100 -10 -10 10 10 10.02 95 100 -5 -15 5 15 7.53 115 100 +15 0 15 30 10.04 100 110 -10 -10 10 40 10.05 125 110 +15 +5 15 55 11.06 140 110 +30 +35 30 85 14.2

Tracking Signal Example

TrackingSignal

(RSFE/MAD)

-10/10 = -1-15/7.5 = -2

0/10 = 0-10/10 = -1

+5/11 = +0.5+35/14.2 = +2.5

The variation of the tracking signal between -2.0 and +2.5 is within acceptable limits

Adaptive Forecasting

It’s possible to use the computer to continually monitor forecast error and adjust the values of the a and b coefficients used in exponential smoothing to continually minimize forecast errorThis technique is called adaptive smoothing

Focus Forecasting

Developed at American Hardware Supply, focus forecasting is based on two principles:

1. Sophisticated forecasting models are not always better than simple ones

2. There is no single technique that should be used for all products or services

This approach uses historical data to test multiple forecasting models for individual itemsThe forecasting model with the lowest error is then used to forecast the next demand

Forecasting in the Service Sector

Presents unusual challenges Special need for short term records Needs differ greatly as function of industry

and product Holidays and other calendar events Unusual events

Fast Food Restaurant Forecast

20% –

15% –

10% –

5% –

11-12 1-2 3-4 5-6 7-8 9-1012-1 2-3 4-5 6-7 8-9 10-11

(Lunchtime) (Dinnertime)Hour of day

Perc

enta

ge o

f sal

es

FedEx Call Center Forecast

12% –

10% –

8% –

6% –

4% –

2% –

0% –

Hour of dayA.M. P.M.

2 4 6 8 10 12 2 4 6 8 10 12

Measuring Forecast Errors

• Mean Absolute Deviation (MAD) The sum of the absolute value of the individual forecast errors divided by the number of periods

Bias

• Bias exists when the cumulative actual demand varies from the cumulative forecast.

Seasonal Index

Seasonal Index = period average demandavg. demand for all periods

End of Lecture 18

lecture 18 forecasting (continued) books introduction to materials management, sixth edition, j. r....

Documents