ie5403 facilities design and planning

IE5403 Facilities Design and Planning

Instructor: Assistant Prof. Dr.

Rıfat Gürcan Özdemir

http://web.iku.edu.tr/~rgozdemir/IE551/index(IE551).htm

Course topics

Chapter 1: Forecasting methods Chapter 2: Capacity planning Chapter 3: Facility location Chapter 4: Plant layout Chapter 5: Material handling and storage

systems

Grading

Participation 5% Quizzes 15% (4 quizzes) Assignment 15% (every week) Midterm 130% (chapters 1 and 2) Final 35% (all chapters)

3

4

IE5403 - Chapter 1

Forecasting methods

5

Forecasting

Forecasting is the process of analyzing the past data of a time – dependent variable & predicting its future values by the help of a qualitative or quantitative method

6

Why is forecasting important?

Proper forecasting

Better use of capacity

Reduced inventory costs

Lower overall personel costs

Increased customer satisfaction

Poor forecasting Decreased profitability

Collapse of the firm

7

Planning horizondemand

timenow

past demand

planning horizon

actual demand?

actual demand?

Forcast demand

8

Designing a forcasting systemForecast need

Data avilable?

Quantitative?

Analyze data

YES

Causal factors?

YES

Causal approach

YES

Collect data?

NO

YES

Qualitative approach

NO

Time series

NO

NO

9

Regression Methods

Model

dependent variable

independent variable

Unknown parameters

Random error componenttt tbax

xt

t

Simple linear model

xt = a + b t

10

Estimating a and b parameters

xt

t

xt = a + b t

e1

e2

e5

e3

e4

T = 50

Such that sum squares of the errors (SSE) is minimized

forecast erroret = ( xt – xt )

11

Least squares normal equations

Least squares normal equations

T

tt

E tbaxa

SS

1

0)ˆˆ(2ˆ

T

tt

T

tttE tbaxxxSS

1

2

1

2 )ˆˆ()ˆ(

T

tt

E ttbaxb

SS

1

0)ˆˆ(2ˆ

T

tt

T

t

xtbTa11

ˆˆ

T

tt

T

t

T

t

xttbta11

2

1

ˆˆ

12

(xt – xt)2unexplained deviation =

(xt – xt)2 = explained deviation

xt

Coefficient of determination (r2)

xt

xt

t

(xt – xt)2 = total deviation

2

22

)(

)ˆ(

total

explained

tt

tt

xx

xxr , 0 r2 1

13

Coefficient of corelation (r)

r = coeff. of determination = r2

Sign of r ,(– / +), shows the direction,of the relationship between xt and t

( )or

r shows the strength of relationship between xt and t

]][[2222

2

tt

tt

xxTttT

xttxTrr

– 1 r 1

14

Example – 3.1 It is assumed that the monthly furniture sales in a city is directly proportional to the establishment of new housing in that month

a) Determine regression parameters, a and b

b) Determine and interpret r and r2

c) Estimate the furniture sales, when expected establishment of new housing

is 250

15

Example – 3.1(continued)

Month

New Housing

/month

Furniture

Sales /month

($1000)

Jan. 100 461

Feb. 110 473

March 96 450

April 114 472

May 120 481

June 160 538

Month

New Housing

/month

Furniture

Sales /month

($1000)

July 150 540

Aug. 124 517

Sep. 93 449

Oct. 88 452

Nov. 104 454

Dec. 116 495

16

Example – 3.1(solution to a)

T = 12

xt = 5782

txt = 670,215

t = 1375

t2 = 162,853

t xt

100 461

110 473

96 450

114 472

120 481

160 538

150 540

124 517

93 449

88 452

104 454

116 495

t2

10,000

12,100

9216

12,996

14,400

25,600

22,500

15,376

8649

7744

10,816

13,456

txt

46,100

52,030

43,200

53,808

57,720

86,080

81,000

64,108

41,757

39,776

47,216

57,420

a T + b t = xt

a t + b t2 = txt

17


12 a + 1375 b = 5782

a 1375 + b 162,853 = 670,215

1375 x

– 12 x

b (1375 2 – 12 x 162,853)= (1375 x 5782 – 12 x 670,215)

b(1375 2 – 12 x 162,853)

(1375 x 5782 – 12 x 670,215)= = 1.45

18


b = 1.4512 a + 1375 b = 5782

a 1375 + b 162,853 = 670,215

a12

(5782 – 1375 x 1.45)= = 315.5

xt = a + b t

xt = 315.5 + 1.45 t

19

Example – 3.1(solution to b)

t xt

100 461

110 473

96 450

114 472

120 481

160 538

150 540

124 517

93 449

88 452

104 454

116 495

T 12xt =

xt 5782= = 482

xt = 315.5 + 1.45 (100) = 461

461

475

455

481

490

548

533

495

450

443

466

484

xt

20


t xt

100 461 461 -21 -21

110 473 475 -7 -9

96 450 455 -27 -32

114 472 481 -1 -10

120 481 490 8 -1

160 538 548 66 56

150 540 533 51 58

124 517 495 13 35

93 449 450 -32 -33

88 452 443 -39 -30

104 454 466 -16 -28

116 495 484 2 13

xt xt xtxt xt

Explained deviation

xtxt

xt

t

Total deviation

xtxt

xt

t

21


t xt

100 461 461 -21 -21 441 441

110 473 475 -7 -9 49 81

96 450 455 -27 -32 729 1.024

114 472 481 -1 -10 1 100

120 481 490 8 -1 64 1

160 538 548 66 56 4.356 3.136

150 540 533 51 58 2.601 3.364

124 517 495 13 35 169 1.225

93 449 450 -32 -33 1.024 1.089

88 452 443 -39 -30 1.521 900

104 454 466 -16 -28 256 784

116 495 484 2 13 4 169

xt xt xtxt xt xt xt( )2xt xt( )2

22


r2 =

xt xt( )2

xt xt( )2=

11.215

12.314= 0.91

Coefficient of determination:

91% of the deviation in the furniture sales can be explained by the

establishment of new housing in the city

23


= = 0.95r r2 = 0.91

Coefficient of corelation:

a very strong (+) relationship (highly corelated)

24


]][[ 2222

2

tt

tt

xxTttT

xttxTrr

xt2 = 2.798.274

r =12 x 670,215 – 1375 x 5782

[12 x 162,853 – (1375)2 ][12 x 2,798,274 – (5782)2]=0.95

r2 = (0.95)2 = 0.91

25

Example – 3.1(solution to c)

xt = a + b t

xt = 315.5 + 1.45 t t = 250

xt = 315.5 + 1.45 (250) = 678

xt = $ 678,000 x $1000

26

Components of a time series1. Trend ( a continious long term directional

movement, indicating growth or decline, in the data)

2. Seasonal variation ( a decrease or increase in the data during certain time intervals, due to calendar or climatic changes. May contain yearly, monthly or weekly cycles)

3. Cyclical variation (a temporary upturn or downturn that seems to follow no observable pattern. Usually results from changes in economic conditions such as inflation, stagnation)

4. Random effects (occasional and unpredictable effects due to chance and unusual occurances. They are the residual after the trend, seasonali and cyclical variations are removed)

27

Components of a time series

0

xt

t1 2 3 4 5 6 7 8

a2

a1

seasonal variation

trend slope

random effect

Year 1 Year 2

28

Simple Moving Average

Model

xt = a + t

xt

Constant process

t

xt = a

a

Forecast error

Simple Moving Average Forecast is average of N previous observations

or actuals Xt :

Note that the N past observations are equally weighted.

Issues with moving average forecasts: All N past observations treated equally; Observations older than N are not included at all; Requires that N past observations be retained.

T

NTttt

NTTTT

XN

X

XXTXN

X

11

111

1ˆ

)(1ˆ

1ˆ

TX

Simple Moving Average Include N most recent observations Weight equally Ignore older observations

weight

today

TT-1T-2...T+1-N

1/N

31

Parameter N for Moving Average

If the process is relatively stable choose a large N

If the process is changing choose a small N

32

Example 3.2Week Demand

1 6502 6783 7204 7855 8596 9207 8508 7589 892

10 92011 78912 844

What are the 3-week and 6-week Moving Average Forecasts for demand of periods 11, 12 and 13?

Weighted Moving Average Include N most recent observations Weight decreases linearly when age

of demand increases

34

Weighted Moving Average

WMT =

t=T-N+1

T

wt xt

t=T-N+1

T

wt

wt = weight value for xt

The value of wtis higher

for more recent data

35

Example 3.3

Month Sales

Jan. 10

Feb. 12

March 13

April ?

May ?

a) Use 3-month weighted moving average with the following weight valuesto predict the demand of april

b) Assume demand of april is realized as 16, what is the demand of may?

wT = 3

wT-1 = 2

wT-2 = 1

36

Exponential Smoothing Method

A moving average technique which places weights on past observations exponentially

ST = xT + ST-1

Smoothed value Smoothing constant

Realized demand at period T

Exponential Smoothing Include all past observations Weight recent observations much more heavily

than very old observations:

weight

today

Decreasing weight given to older observations



weight

today


10



weight

today


10

)1(



weight

today


10

2)1(

)1(



weight

today


10

3

2

)1(

)1(

)1(

Exponential Smoothing

211

22

11

)1()1(ˆ

)1()1(ˆ

tttT

tttT

XaXXX

XXXX


TTT XaaXX ˆ)1(ˆ1

211

22

11

)1()1(ˆ

)1()1(ˆ

tttT

tttT

XaXXX

XXXX

1)1( TTT SaaXS

44

The meaning of smoothing equation

ST = xT + ST-1

xTST = + ST-1ST-1

ST = xT+ ST-1ST-1( )

ST = xT+New forecast forfuture periods

ST-1 xT=Old forecast forthe most recent period

eT xT= xT –Forecast error


Thus, new forecast is weighted sum of old forecast and actual demand

Notes:Only 2 values ( and ) are required,

compared with N for moving averageParameter a determined empirically (whatever

works best)Rule of thumb: < 0.5Typically, = 0.2 or = 0.3 work well

TX̂TX

46

Choice of

Small Slower response

Large Quicker response

Equivelance between and N

=2

N+ 1 =2

N–

47

Example 3.4

Week Demand

1 820

2 775

3 680

Given the weekly demand data, what are the exponential smoothing forecasts for periods 3 and 4 using = 0.1 and = 0.6 ?

Assume that S1 = x1 = 820

48

Example 3.4 (solution for = 0.1) S1 = x1 = 820

S2 = x2 + S1

=x2 820

t xt St1 820

2 775

3 680

4

xt

S2 = 0.1(775) + 0.9(820) = 815.5

=x3 815.5

820

815.5 820

815.5801.95

801.95

49

Example 3.4 (solution for = 0.6) S1 = x1 = 820

S2 = x2 + S1

=x2 820

t xt St1 820

2 775

3 680

4

xt

S2 = 0.6(775) + 0.4(820) = 793.0

=x3 793.0

820

793.0 820

793.0725.2

725.2

50

Winters’ Method for Seasonal Variation

Model

xt = a + tb t( ) ct +

Constant parameter

Trend parameter

Seasonal factor for period t

Random error component

t

xt

51

Initial values of a cb, and

t=1

L

a0 =xt

L

b0 = 0

ct =xt

a0

t=1

Lct = L

ct values are valid

ct =Lct

t=1

Lct

ct values arenormalized :

YES

NO

for one year available demand data

52

Smoothing equations

11ˆˆ)1(

ˆˆ

TT

LT

TT ba

c

xa

]ˆ)[1(ˆˆˆ11 TTTT baab

0 < < 1

0 < < 1

0 < < 1LT

T

TT c

a

xc

ˆ)1(

ˆˆ

53

Forecast Equation

aT bT c(T+-L)x(T+) =( )+

= the smallest integer ≥L

54

Example 3.6

Month(2005) 1 2 3 4 5 6 7 8 9 10 11 12

Demand 4 2 5 8 11 13 18 15 9 6 5 4

a) Forecast the demand of Jan.’06 using Winters method with = 0.2, = 0.1, = 0.5

b) Forecast the demand of Feb.’06 when Jan.’06 realizes as 5 using Winters method with = 0.2, = 0.1, = 0.5

c) Forecast the demand of Mar.’06 and Mar.’07 when Feb.’06 realizes as 4 using Winters method with = 0.2, = 0.1, = 0.5

55

Example 3.6 (solution to a)

b0 = 0

t=1

L

a0 =xt

L=

100

12= 8.3

c1x(T+) = ( )+1a0 b0

c1 =x1

a0

=4

8.3= 0.48

t

1 0.48

2 0.24

3 0.60

4 0.96

5 1.32

6 1.56

7 2.16

8 1.80

9 1.08

10 0.72

11 0.60

12 0.48

12

ct

0.48x(12+1)= ( )+8.3 0 = 4

56

Example 3.6 (solution to b)

001213

1313

ˆˆ)1(ˆ

ˆ bac

xa

x13 = 5JAN.’06 realizes as:

]ˆ)[1(ˆˆˆ11 TTTT baab

LTT

TT c

a

xc

ˆ)1(

ˆˆ

= 0.2 50.48 + 0.8 8.3+ 0 = 8.72

= 0.1 8.72 – 8.3 + 0.9 0 = 0.043

= 0.5 58.72 + 0.5 0.48 = 0.53

c2x(T+) = ( )+1a13 b13

0.24x(13+1) =( )+8.72 0.043 = 2.11 x

Updated seasonal values should be normalized!

57

58

Example 3.6 (solution to c)FEB.’06 realizes as:

x14 = 4

0.5 410.34 + 0.5 0.24 = 0.31=c14

= 0.2 40.24 + 0.8 8.72 + 0.043 = 10.34a14

= 0.1 10.34 – 8.72 + 0.9 0.043 = 0.2b14

0.60x(14+1) =( )+10.34 0.2 = 6.31 x

0.60x(14+13) =( )+10.34 0.2 = 7.7613 x

Forecast of Mar.’06

Forecast of Mar.’07

59

Forecast accuracyForecast accuracy shows the performance of the model for complying with the demand process, and is measured by using forecast error

Forecast error is the difference between the actual demand and the forecast

Forecast Actual demand

xtet = xt –

Forecast error

60

Error measuresLooking at the error for an isolated period does not provide much useful information

Rather we will look at errors over the history of the forecasting system. There are several methods for this process, although each has different meaning

1. Cumulative (sum) error, Et

2. Mean error, ME3. Mean square error, MSE4. Mean absolute deviation, MAD5. Mean absolute percentage error, MAPE

61

Cumulative (sum) error, Et

Et should be close to zero if the forecast is behavingproperly. That is, sometimes it overestimates and sometimes it underestimates, but in the long run these should cancel out

T

ET =t=1

et

62

Mean error, ME

ME should be interpreted same as sum error, Et , that is, it shows whether the model is biased toward certain direction or not

n

ME = t=1

etn1

A forecast consistently larger than actual is called biased highA forecast consistently lower than actual is called biased low

63

Example 3.9

t xt

1 92 123 154 185 216 247 278 309 33

Validate the moving average (N=3) if it is suited to the given past data using ME and Et , and say if it is bias

64

Example 3.9 (solution)t xt MA(t-1) et

1 9

2 12

3 15

4 18 12 6

5 21 15 6

6 24 18 6

7 27 21 6

8 30 24 6

9 33 27 6

E9 = 6 + 6 +... + 6 = 36.00

ME = 1/n (et)= E9 / 6 = 6.00

BIASED LOW !

65

Mean square error, MSEn

MSE = t=1

etn1 2

MSE is mainly used to counteract the inefficiency in error measuring as negative errors (– et) cancel out the positive error terms (+ et)

By squaring the error terms, the “penalty” is increasedfor large errors. Thus a single large error greatly increases MSE

66

Mean absolute deviation, MAD

n

MAD = t=1

etn1

MAD is another error measure for solving the neutralizing problem

MAD measures the dispersion of the errors, and if MAD is small the forecast should be close to actual demand

67

Mean absolute percentage error, MAPEn

MAPE = t=1

PEtn1

PEt =xtxt –

xt

(100)

MAPE is mainly used to counteract the inefficiency in error measuring as the previously defined mesaures depend on the magnitude of the numbers being forecast

If the numbers are large the error tends to be large. Itmay more meaningful to look at error relative to the magnitude of the forecasts, which is done by MAPE

68

Example 3.11t xt

1 102 123 154 455 1306 1807 1708 1209 125

10 10011 12512 135

Compare a 3-period moving average model and a 6-period moving average model using given past data and show which suits better with respect to MSE, MAPE.

69

Example 3.11 (solution)t xt MA[3]t-1 et

1 102 123 154 45 12.335 130 24.006 180 63.337 170 118.33 51.678 120 160.00 -40.009 125 156.67 -31.67

10 100 138.33 -38.3311 125 115.00 10.0012 135 116.67 18.33

t=76

12MSE= et

1 21196.30=

70

Example 3.11 (solution)t xt MA[6]t-1 et

1 102 123 154 455 1306 1807 170 65.33 104.678 120 92.00 28.009 125 110.00 15.00

10 100 128.33 -28.3311 125 137.50 -12.5012 135 136.67 -1.67

t=76

12MSE= et

1 22154.32=

71

Example 3.11 (solution)

t xt MA[3]t-1 et Pet

7 170 118.33 51.67 30.398 120 160.00 -40.00 33.339 125 156.67 -31.67 25.33

10 100 138.33 -38.33 38.3311 125 115.00 10.00 8.0012 135 116.67 18.33 13.58

MA[6]t-1 et Pet

65.33 104.67 61.5792.00 28.00 23.33

110.00 15.00 12.00128.33 -28.33 28.33137.50 -12.50 10.00136.67 -1.67 1.23

MA[3] model results MA[6] model results

MAPE Pet12

= t=76

124.83= MAPE Pet

12=

t=761

22.74=

72

Example 3.11 (solution)

Error

measures

Forecast models

MA[3] MA[6]

MSE 2154.32

MAPE 24.83

1196.30

22.74

ie5403 facilities design and planning

Documents

establishment of new

monthly furniture sales

forecasting methodschapter

a1375 x

time seriestrend

b parametersxtt0

time dependent variable

certain time intervals