Forecasting The Process of Predicting the Future presented by Your Local Engineering Management...

ForecastingThe Process of Predicting the Future

presented by

Your Local Engineering Management Office (LEMO)

My concern is the futuresince I plan to spend the restof my life there.

"Those who have knowledge, don't predict. Those who predict, don't have knowledge. " --Lao Tzu, 6th Century BC Chinese Poet Wise words from a long time ago.

Applications• sales• strategic planning• financial investments• inventory levels• production levels• work force sizing• energy requirements• economic planning

– unemployment– housing starts– inflation rates

Typical Forecasts

• Product Sales• Replacement part demands• Lead-times• Machine break rates• Expenditures• Market share• Unit costs• Labor rates

• Forecasts are used in manufacturing engineering in:– Inventory models– Machine loading– Production planning models– MRP systems – Manufacturing simulations

• A good forecast model is– accurate– computationally efficient– robust (to changes in patterns)

Laws of Forecasting• First Law: Forecasts are always wrong!• Second Law: Forecasts always change!• Third Law: The further into the future, the less

reliable the forecast!• Fourth Law: A good forecast includes a measure of

error!• Fifth Law: Aggregate forecasts are more accurate!• Sixth Law: Forecasts should not replace known

values!

Accuracy of forecasts depends on

• Accuracy of data

• sample size

• stability of the random process– variability– stationary vs non-stationary process

• length of forecasting period

• method used

• model selected

Forecasting Methods• Qualitative (subjective)

– historical analogy– market research

• customer surveys

– Expert opinion– Delphi technique– sales force composites

• Quantitative Models– regression analysis (causal models)– time-series models

• moving averages• exponential smoothing• Box-Jenkins• auto-regressive

Time Horizon

• Short Term– Sales, shift schedules, material and part requirements,

equipment failures– Days and weeks

• Intermediate– Product sales, labor requirements, resources– Weeks and months

• Long-term– Capacity requirements, long-term sales patterns, growth

trends, resource and labor costs– Months and years

Time SeriesTime Series: random variable indexed on timee.g. Dt = demands during month t

D1, D2 , …, Dt,… form a time series

Basic Premise: Can predict the future from thepast - the underlying process will continue as it has in the (recent) past.

Forecast:

where ai is the weight placed on the ith observation

1

n

t i t ii

F a D

Elements of Time Series Data• Trend (Gt)

– Constant (stationary) (Gt = b)

– linear (constant) (Gt = bt)

– quadratic (accelerated) (Gt = bt2)

– exponential (growth) (Gt = bt)

• Seasonal (St)

• Cyclical (Ct)

• Randomness (Noise) (et)

– no recognizable pattern

Trends

time

population

Seasonal Snow blower sales

spring summer fall winter spring summer fall winter

trend present

Cycles Unemployment Rate

1990 1991 1992 1993 1994 1995 1996

Cycles• long swings away from trend due to factors other than

seasonality– generally occurs over a number of years

• difficult to model– not as stable – rarely repeats at fixed intervals– amplitude varies– need several years of data (complete cycles) to distinguish from

trends

• causes of cycles include– psychological forces (fashions, music, food)– population demographics (college enrollments)– institutional (public policy, business practices, tax policies)– replacement cycles (technology changes, obsolescence)– education

The Road Ahead

• Stationary (constant) process– moving averages– exponential smoothing

• Trend only process– linear regression– Holt’s method (double exponential smoothing)

• Seasonal process– seasonal factors (stationary process)– Winter’s method (trend process)

but first a detour…

Evaluating ForecastsForecast Error

et = Ft - Dt

1

1

2

1

1

1

1

1100

n

t tt

n

tt

n

tt

nt

t t

E e

MAD en

MSE en

eMAPE x

n D

errors

Bad forecast

Wrong again

(measures bias)

Forecast Error Example

Time period (i)

Forecast (F) Actual (D) ei = F - A MSE MAD MAPE

1 120 122 -2.0 4 2 0.0163932 156 145 11.0 121 11 0.0758623 147 137 10.0 100 10 0.0729934 132 125 7.0 49 7 0.0565 122 119 3.0 9 3 0.025216 184 178 6.0 36 6 0.0337087 171 165 6.0 36 6 0.0363648 168 177 -9.0 81 9 0.0508479 145 153 -8.0 64 8 0.052288

10 136 148 -12.0 144 12 0.08108111 145 135 10.0 100 10 0.07407412 179 164 15.0 225 15 0.091463

sum 37.0 969.0 99.0 0.7average 3.1 80.8 8.3 5.6%

e = F - D

Forecast Model

Modelhistorical data forecast

error

Observedvalue

Multiplicative Model

Ft = Trend x seasonal x cyclical x irregular= aGt St Ct et

Additive Model

Ft = Trend + seasonal + cyclical + irregular = a + Gt + St + Ct + et

Notation

Given D1 , D2 , … , Dt demands have been observed and assuming an additive linear model:

Ft,t+ = forecast made at time tfor period t+ let Ft = Ft-1,t

,0

t t n t nn

F a D

D1, D2 , … , Dt are observed values of demand duringperiods 1, 2, …, t.

At time t, we have observed Dt , Dt-1 , …

2. Average forecast:

1

1

1

t

ii

t

DF

t

3. Moving averages:

Ft = (Dt-1+ Dt-2 + … + Dt-N ) / N and Ft-1,t+ = Ft

1. Last data point (LDP) Forecast: Ft = Dt-1

1

11

(1/ ) (1/ )

(1/ )

t t

t i t i t Ni t N i t N

t t t N

F N D N D D D

F N D D

Potential Forecasts

Time Series - Moving Averagesno trends/no cycles/no seasonal effects

Model: Dt = + t

Underlying constant of the process

E[t] = 0 and Var[t] = 2

Let’s hear it for the moving

average model!

Moving Averages Lag behind Trends

Period Demand MA(3) MA(6)1 22 43 64 8 45 10 66 12 87 14 10 78 16 12 99 18 14 11

10 20 16 1311 22 18 1512 24 20 17

(Simple) Exponential Smoothing

1 1

1 1 1 1 1

(1 ) ; 0 1t t t

t t t t t

F D F

F F D F

Why Ft is just the weighted sum of the current observation

and the previous estimate.

previousforecast

error frompreviousforecast

More (Simple) Exponential Smoothing

1 1

1 2 2

21 2 2

(1 )

(1 ) (1 )

(1 ) (1 )

t t t

t t t

t t t

F D F

D D F

D D F

10

(1 )it t i

i

F D

continuing:

0

(1 ) 11 (1 )

it

i

F

Note that the weights sum to one:

Moving Averages versus Exponential Smoothing

Average age of data:

moving average = (1/N) (1 + 2 + 3 + … + N) = (1/N) N (N+1)/2 = (N + 1)/2

exp smooth = 1

1

(1 ) 1/i

i

i

equating ages:1 1

22

N

N

Exampleif N = 10, then = .18182 if = .1, then N = 19

A Little Math Trick

1

1

1

1 1 1

20

(1 ) 1/

(1 )(1 ) (1 )

1 1 1(1 ) 1 1 1

1 (1 )

i

i

ii i

i i i

i

i

i

d di

d d

d d d

d d d

More Moving Averages versus Exponential Smoothing

age of data1 2 3 4 5 6

exponential smoothing with = .3

moving average with n = 6

weightplaced onith value

.3( ) .3 if i e

Considerations in the selection of the smoothing constant

If is small – response to change will be slowIf is large – response to change will be fast Normally .1 < < .3 Average age of the data:

Set: = 2/(n+1) to correspond to n-period moving average Minimize forecast error (MAD, MSE, RMSE, etc.)

1

1

11

. . .1, 10

.3, 3.33

i

k

i

e g A

A

Remember to go to Excel!

Moving Averages versus Exponential Smoothing – A Comparison

Similarities

• Assume stationary process (with adjustment of shifts in the mean)

• Single parameter model (N and α)

• Lags behind trend data

• Similar levels of accuracy

Differences

• Smoothing uses all past data, MA uses only the last N values

• Need to save N data points for MA

• MA weighs each observation by N while smoothing weights the Nth observation by (1-)N-1

Trend Based Methods

The Journey Continues…

"Wall Street indices predicted nine out of the last five recessions ! " --Paul A. Samuelson in Newsweek, Science and Stocks, 19 Sep. 1966.

Trend Data

x

xx

x xx

Model: Dt = + Bt + t

E[t] = 0 and Var[t] = 2

time

demands

Regression Analysis

yt = A + Bt + et with t = 1, 2, …, n

Ft+k = a +b (t + k)

where a & b are Least Square estimates

Least-Squares Estimates

1 1

2 2 2

1

( 1) / 2

( 1)

2

( 1)(2 1) ( 1)

6 4

where /

xy

xx

n n

xy i ii i

xx

n

ii

Sb

S

a D b n

n nS n iD D

n n n n nS

D D n

and Di is the demand fortime period i, i = 1,2, …, n

The Necessary ExampleQuarter Index Engine

Demands (failures)

1/2007 1 20

2/2007 2 25

3/2007 3 22

4/2007 4 28

1/2008 5 30

2/2008 6 32

3/2008 7 33

4/2008 8 31

GE's F110 engine family provides the most reliable power for the F-16C/D fighter aircraft. With a reputation for stall-free operation, the F110 continues to be the choice for F-16 operators and has been selected for twin-engine F-15 application.

Go to Excel…

A Second ExampleQuarter

Machine Failures

1 1982 2503 3214 4565 534

forecast6 615.27 7038 790.8

Machine Failuresy = 87.8x + 88.4

R2 = 0.9755

0

100

200

300

400

500

600

0 1 2 3 4 5 6

quarter

e.g. F6 = 88.4 + 87.8 (6) = 615.2

Trend Data - continued Double Exponential Smoothing (Holt’s method)

1 1

1 1

,

(1 )( )

( ) (1 )

0 1 ; 0 1

t t t t

t t t t

t t t t

S D S G

G S S G

F S G

Model: Dt = + Bt + t

Value of theseries (intercept)

Value of thetrend (slope)

A Holt’s Method Example

Set α = = .1; S0 = 200; G0 = 10

S1 =(.1)(200) + (.9)(200+10)=209.0

G1 = (.1)(209 - 200) + (.9)(10) = 9.9

S2 =(.1)(250) + (.9)(209+9.9)= 222.0

G2 = (.1)(222 - 209) + (.9)(9.9) = 10.2

S3 =(.1)(175) + (.9)(222+10.2)=226.5

G3 = (.1)(226.5 - 222) + (.9)(10.2) = 9.6

F3,4 = 226.5 + 9.6 = 236.1 and F3,5 = 226.5 +(2) 9.6 = 245.7

Dt

200250175186225285305190

Forecasting with Seasonal Effects

Experience the ups and downs of the four seasons

"I always avoid prophesying beforehandbecause it is much better to prophesy after the event has already taken place. "

--Winston Churchill

Seasonal Data with no trend

Model: Dt = ck + t , 1 k N

where = average (annual) demandck = seasonal factor for period k

t = random component

N = number of periods in a season

The road ahead is no longer straight

Seasonal Factors for a Series with no Trend

1. Compute the sample mean of the data

2. Divide each observation by the sample mean

3. Average the factors for like periods within each season.

4. Result are N seasonal factors

I can do that.

average for periodSeasonal Factor

overall average

Seasonal Indices – a real nice example

week 1 week 2 week 3 week 4Mon 16.2 17.3 14.6 16.1Tues 12.2 11.5 13.1 11.8Weds 14.2 15 13 12.9Thurs 17.3 17.6 16.9 16.6Fri 22.5 23.5 21.9 24.3

week 1 week 2 week 3 week 4Mon 0.986 1.053 0.889 0.980Tues 0.743 0.700 0.798 0.718Weds 0.865 0.913 0.791 0.785Thurs 1.053 1.072 1.029 1.011Fri 1.370 1.431 1.333 1.479

Divide by the Mean = 16.425

averageMon 0.977Tues 0.740Weds 0.839Thurs 1.041Fri 1.403

Example 2.6Cars on a toll bridgeData is in 1,000

forecasts16.0512.1513.78

17.123.05

x 16.425

Another example

Qtr 2001 2002 2003 avg index

1 124.5 157.4 144.1 142.0 0.691

2 181.0 192.3 178.4 183.9 0.896

3 287.1 281.8 251.5 273.4 1.332

4 240.1 217.1 208.6 221.9 1.081

205.3 1

Data are quarterly sales in 1,000 of gallons and are normally distributed with a mean of 200 and a std. dev. of 20

De-seasonalizing Our ExampleQtr Raw De-seasonnal

2001 1 124.5 180.0

2 181.0 202.1

3 287.1 215.5

4 240.1 222.1

2002 1 157.4 227.6

2 192.3 214.7

3 281.8 211.6

4 217.1 200.8

2003 1 144.1 208.4

2 178.4 199.1

3 251.5 188.8

4 208.6 193.0

Qtr index

1 0.691

2 0.896

3 1.332

4 1.081

for example:124.5/.691 = 180.0181.0/.896 = 202.1

Applying a MA Forecasting ModelMonth Sales 1-mo 2-mo 3-mo avg 4-mo 5-mo 6-mo avg 7-mo avg

2001/1 180

2 202.1 180.03 215.5 202.1 191.14 222.1 215.5 208.8 199.2

2002/1 227.6 222.1 218.8 213.2 204.92 214.7 227.6 224.9 221.7 216.8 209.53 211.6 214.7 221.2 221.5 220.0 216.4 210.34 200.8 211.6 213.2 218.0 219.0 218.3 215.6 210.5

2003/1 208.4 200.8 206.2 209.0 213.7 215.4 215.4 213.52 199.1 208.4 204.6 206.9 208.9 212.6 214.2 214.43 188.8 199.1 203.8 202.8 205.0 206.9 210.4 212.04 193 188.8 194.0 198.8 199.3 201.7 203.9 207.3

forecast 193.0 190.9 193.6 197.3 198.0 200.3 202.3MAD = 9.62 10.22 11.06 11.11 10.70 11.77 13.52RMSE = 10.9 12.1 12.7 13.0 11.9 13.4 14.8

Qtr sales 1-qtr 2-qtr 3-qtr 4-qtr 5-qtr 6-qtr 7-qtr

The Forecast for the next yearforecast 1-mo 2-mo 3-mo avg 4-mo 5-mo 6-mo avg 7-mo avg

2004 index 193.0 190.9 193.6 197.3 198.0 200.3 202.31 0.691 133.4 131.9 133.8 136.4 136.8 138.4 139.82 0.896 172.9 171.0 173.5 176.8 177.4 179.5 181.33 1.332 257.1 254.3 257.9 262.8 263.8 266.8 269.54 1.081 208.6 206.4 209.3 213.3 214.1 216.5 218.7

1-qtr 2-qtr 3-qtr 4-qtr 5-qtr 6-qtr 7-qtr

Seasonal Data with Trend

Model: Dt = ( + Gt)ct + t

where = the base constant at t = 0G = slope of trend componentct = seasonal factor for period tt = random component

The road ahead not only curves but also climbs.

Another better way when trend is present…

Could you review with us the 8 easy steps to applying the moving average method?

The 8 easy steps…

1. Obtain moving averages where N = length of season

2. Average and center adjacent values

3. Divide results of #2. into Dt

4. For each season, compute the “average” (i.e. the mean or median)

5. Adjust so sum is N by multiply each average by N / Total

6. Deseasonalize series by dividing each Dt by corresponding seasonal index

7. Forecast deseasonalized series using appropriate model

8. Apply corresponding seasonal indices to “reseasonalize” the series

He is right. There are 8 steps.

Example of the first 6 easy steps…

year Period actuals - Dt #1- MA#2 -

Centered #3 - Index #4 Avg qtr #5 Adjust#6 - De-season

Y1 Qtr 1 1 10 0.531 0.532 18.8Y1 Qtr 2 2 20 1.123 1.124 17.8Y1 Qtr 3 3 26 18.25 18.50 1.405 1.372 1.374 18.9Y1 Qtr 4 4 17 18.75 19.13 0.889 0.969 0.970 17.5Y2 Qtr 1 5 12 19.50 20.00 0.600 22.6Y2 Qtr 2 6 23 20.50 21.25 1.082 20.5Y2 Qtr 3 7 30 22.00 22.00 1.364 21.8Y2 Qtr 4 8 23 22.00 22.50 1.022 23.7Y3 Qtr 1 9 12 23.00 23.25 0.516 22.6Y3 Qtr 2 10 27 23.50 23.63 1.143 24.0Y3 Qtr 3 11 32 23.75 23.75 1.347 23.3Y3 Qtr 4 12 24 23.75 24.13 0.995 24.7Y4 Qtr 1 13 12 24.50 25.13 0.478 22.6Y4 Qtr 2 14 30 25.75 26.25 1.143 26.7Y4 Qtr 3 15 37 26.75 Totals 3.995 4.000 26.9Y4 Qtr 4 16 28 28.9

Steps 1and 2 of the first 6 easy steps…




More Steps 1 and 2 of the first 6 easy steps…




Step 3 of the first 6 easy steps…




26 / 18.50 = 1.405

17 / 19.13 = .889





(.600 + .516 + .478)/3 = .531

Another Step 4 of the first 6 easy steps…




(1.082+ 1.143 + 1.143)/3 = 1.123





(4/ 3.995) .531 = .5321 1 1 1

if ; thenn n

i i

nx T x n

T

More Step 5 of the first 6 easy steps…




(4/ 3.995) 1.123 = 1.124





12 / .532 = 22.6

More Step 6 of the first 6 easy steps…




23 / 1.124 = 20.5

Forecast period indices

#7 - Deseason Forecast

Y5 Qtr 1 17 0.532 28.0Y5 Qtr 2 18 1.124 28.7Y5 Qtr 3 19 1.374 29.3Y5 Qtr 4 20 0.970 29.9

Now easy step 7 …Forecast

year Period#6 - De-season

Y1 Qtr 1 1 18.8Y1 Qtr 2 2 17.8Y1 Qtr 3 3 18.9Y1 Qtr 4 4 17.5Y2 Qtr 1 5 22.6Y2 Qtr 2 6 20.5Y2 Qtr 3 7 21.8Y2 Qtr 4 8 23.7Y3 Qtr 1 9 22.6Y3 Qtr 2 10 24.0Y3 Qtr 3 11 23.3Y3 Qtr 4 12 24.7Y4 Qtr 1 13 22.6Y4 Qtr 2 14 26.7Y4 Qtr 3 15 26.9Y4 Qtr 4 16 28.9

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

0 5 10 15 20

y = 0.6402x + 17.137 R2 = 0.8444

7. Forecast using deseasonalized values

.6402 (18) + 17.137 = 28.7

Forecast period indices

#7 - Deseason Forecast

#8 - season adjusted

Y5 Qtr 1 17 0.532 28.0 14.9Y5 Qtr 2 18 1.124 28.7 32.2Y5 Qtr 3 19 1.374 29.3 40.3Y5 Qtr 4 20 0.970 29.9 29.0

Finally easy step 8…

y = 0.6402x + 17.137

8. Seasonalize forecast

.532 x 28.0 = 14.9

1.124 x 28.7 = 32.2

Now Begins Winter’s Model

Let Dt = demand in period tN = the number of periods (length of season)St = estimate of deseasonalized series in period tGt = estimate of trend term in period tct = estimate of seasonal component for period t

Yikes, this model has it all!

1 1

1 1

(1 ) ( )

( ) (1 )

(1 )

0 1 ; 0 1; 0 1

tt t t

t N

t t t t

tt t N

t

DS S G

c

G S S G

Dc c

S

The smoothing equations:

, ( )t t t t t NF S G c

Series

Trend

Seasonal

Forecast

Step 1: Calculate the average of each of the seasons:V1 = Davg1; V2 = Davg2 …, Vm = Davgm

Step 2: Set G0 = (Vm – V1 ) / [(m-1)N] (initial slope estimate)

Step 3: Calculate S0 = Davgm + G0 (N-1)/2 (value of series at t = 0)

a.

Step 4: Calculate the seasonal factors:

0

; 2 1 0[( 1) / 2 ]

tt

i

DC N t

V N j G

Vi = average of season i, j = period of season

Could you review with us the 6 easy steps to applying Winter’s Model?

Step 4 Explained

a.

Step 4: Calculate the seasonal factors:

0

; 2 1 0[( 1) / 2 ]

tt

i

DC N t

V N j G

Vi = average of season i, j = period of season

0[2.5 ]t iY V j G

mean of theith year

initial slopeestimate

for N = 4(quarters)

j=1,2,3,4

series+ trend

Step 4 b. Average the seasonal factors

2 1 1 01 0, ...,

2 2n N N

N

c c c cc c

Step 4 c. Normalize the seasonal factors

1

0

for 1 0jj N

ii

cc N N j

c

Step 5: Forecast for period t+

Ft+ = (St + Gt) ct+

Step 6: Next period, update model parameters with new data point.

1 1

1 1

(1 ) ( )

( ) (1 )

(1 )

0 1 ; 0 1; 0 1

tt t t

t N

t t t t

tt t N

t

DS S G

c

G S S G

Dc c

S

year quarterSales in 100's

Step 1 average j

Step 4a Initial Ct

Step 4b Avg each qtr

Step 4c normalize

Year 1 1 10 1.0 0.5904 0.5888 0.5932 20 2 1.1228 1.1010 1.1083 26 3 1.3913 1.3717 1.3814 17 18.25 4.0 0.8690 0.9115 0.918

Year 2 1 12 1 0.58722 23 2 1.0792 3.9730 4.003 30 3.0 1.35214 22 21.75 4 0.9539

Step 2. Initial Gt= 0.875Step 3. Initial ST= 23.0625

Example 2.8

0[( 1) / 2 ]t

ti

DC

V N j G

1

0

jj N

ii

cc N

c

G0 = (Vm – V1 ) / [(m-1)N]

S0 = Davg2 + G0 (N-1)/2

Sales in 100's alpha=.2

beta=.1

gamma=.1

before yr3 observed

Dt Ft St Gt Ct forecastyr 3 Qtr 1 16 14.19 24.58 0.940 0.595 14.190yr 3 Qtr 2 33 28.29 26.41 1.029 1.131 27.504yr 3 Qtr 3 34 37.90 26.91 0.976 1.350 35.476yr 3 Qtr 4 26 25.59 28.01 0.9883 0.915 24.376

More of Example 2.8

1 1

1 1

(1 ) ( )

( ) (1 )

(1 )

tt t t

t N

t t t t

tt t N

t

DS S G

c

G S S G

Dc c

S

Ft+ 1 = (St + Gt) c ; t = 0 Ft+ 1 = (St + Gt) ct+ ; t = 0, = 1,2,3,4

Year 4 forecast

Sales in 100's alpha=.2 beta=.1 gamma=.1 Normalize

Dt Ft St Gt Ct yr 3

yr 3 Qtr 1 16 14.19 24.58 0.940 0.595 0.5964yr 3 Qtr 2 33 28.29 26.41 1.029 1.131 1.1334yr 3 Qtr 3 34 37.90 26.91 0.976 1.350 1.3533yr 3 Qtr 4 26 25.59 28.015 0.9883 0.915 0.9170Yr 4 Qtr 1 1 17.30 3.990 4.000Yr 4 Qtr 2 2 33.99Yr 4 Qtr 3 3 41.92Yr 4 Qtr 4 4 29.31

Ft+ 1 = [28.015 + .9883] .5964 = 17.30Ft+ 2 = [28.015 + 2(.9883)] 1.1334 = 33.99Ft+ 2 = [28.015 + 3(.9883)] 1.3533 = 41.92Ft+ 2 = [28.015 + 4(.9883)] .9170 = 29.31

Bonus Topic

Casual Regression

Regression Model (Causal)

Model: Dt = a + bxt + et

where Dt = (demand) forecast for period txt = independent variable for period tet = random noisea,b slope, intercept to be estimated using least-squares

Two requirements are necessary to use this model:1. A causal relationship exists between x and D2. The value for x can be determined prior to

period t (a time lag exists)

Least-Squares Formulae

n n

tttt=1 t=1

n22

t

t=1

- xx D Db =

x xn

a = D - bx

Why not show the class how it

works in Excel?

Example – casual model

Month

Housing starts 3 mo. prior (in 1000's)

Sales of drywall (in 1000's)

House Starts Sales1 1.9 1022 2.1 1133 3.1 1634 4.2 2155 3.7 1926 5.5 2837 4.1 2128 2.9 1509 3.5 180

10 2.2 12011 1.4 8012 1.5 82

The Regression ModelDrywall Sales

y = 49.491x + 8.7823

R2 = 0.9993

0

50

100

150

200

250

300

0 1 2 3 4 5 6

Housing Starts

A Multiple Regression Example

0 1 2 1 2 2 2 3t t t t tY b b t b x b x b x e

Yt = drywall sales in month txt-1 = housing starts in month t-1xt-2 = housing starts in month t-2xt-3 = housing starts in month t-3

linear trend

Demand Sales

• Demands for items that are not in stock will result in backorders or lost sales

• Lost sales data is usually not available• Assume true monthly demand is normal

with a mean of 100 and a standard deviation of 30– If 110 items are stock each month– Then Pr{Demands > 110} .37

What about lost Sales?

Can we have some homework

problems?

Chapter 2: 12, 13 ,16-22, 24, 28- 30, 33 – 36.

"The future isn't what it used to be !" -- anonymous

Forecasting The Process of Predicting the Future presented by Your Local Engineering Management...

Documents

Transcript of Forecasting The Process of Predicting the Future presented by Your Local Engineering Management...