Lecturer Notes of Unit -3

5/24/2018 Lecturer Notes of Unit -3

1/59

Unit -3

Dr. G.Rajendra

Production and Inventory control activities:

A few years ago, a small test equipment manufacturer in Bombay received a corporation

directive to improve their business operations.

With the help of a consultant, they decided to discard their manual production control system and

undertake a five-phase program to gain better control of their costs. Heres what happened.

The material requirements, estimated costs and inventory records were computerizedwithin four months, they began to check actual inventories against the computerized data

base and analyze any variances.

The general ledger and financial data were integrated into the system a month later. Ten months after that, the payroll and labor distribution information was transferred from

their bank to the system;

It was automatically interfaced to the job costing system. Finally, the order entryinformation and invoicing was incorporated.

The manufacturing control system took about two years to implement and saved the firm

Rs. 152000/- in the first year of operations.

In the second chapter we had discussed on the operations strategy, which is embodied in the long

range operations/production plan.

While all elements of operations management are important, I view forecasting as one of the key

elements in the operations structure. In this chapter, helps us to recognize the models and when

to use for our needs.


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3/59

The plan specifies positioning, strategy, product process and technology plans, strategic

allocation of resources and facility planning shown in fig 1.

Information feedback

Fig:1 Strategic perspective

I

M

O

P

S

P

O

C : P

D

P

C

M

Q

E

S

R


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Forecast of a product is an estimate of its future demand. However, it is not a prediction. A

forecast is however based upon scientific analysis of past data, if available and by other

techniques.

Once these are in place, the fundamental structure of the operation function is established

Before, resources can be planned but, it is critical to estimate or forecast long-range and short-

range demand for products and services.

These forecasts guide the strategic allocation of resources. Based on the expected levels of

demand, decisions are made concerning product, process and service designs, facility capacity,

location and layout, operations technologies and allocation of operations resources.

Other issues involving the strategic allocation of resources include managing quality, planning

service operations and managing projects.

Forecasting meals on airline flightsProviding in-flight meals to the airline passengers is big business.

Few companiess which have business are listed below

Northwest airlines and continentals food budget per year: $ 300 million dollars

Delta serves about 135,000 meals per day

American airlines spends around $800 million each ear on food with each meals cost is $8.20

With this huge expense, the airlines are interested in accurately forecasting the number of meals

that will be needed in each flight.

Factors that make airline meal forecasting:

Passengers purchasing tickets just before a flight Cancelled flights Passengers no-shows Complicate maters Some passengers decide not to have meals, Children can request a kid meals Some passengers request special-diet meals, First class passengers receive different meals than economy class passengers and may

have two or more choices of meals.

Some flights may have 60% full and while others may be 100% If an airline orders too many meals for a flight, extra meals must be thrown away,

although some items such as boxes cereal might be given to charity. If it does not order enough meals, then hungry passengers may be upset and may not fly

on that airline in the future.


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Shortages of meals statics:

Last year: 1% shortageContinental had average meal shortage : 0.6 %

Excess meal : 3.5%

At Home base: 5%To satisfies the customers of first class passengers the airline orders 125percent instead of 100percent.Accurate demand forecasting is critical to providing good customer service in a cost-efficientmanner.Forecasting enables his company to respond more quickly and accurately to market changes.How does forecasting relate to the management processes of planning, organizing andcontrolling? These processes are not independent processes. They interrelate and overlap.If operations have been properly planned and organized, control is easier and smoother. Thesewere forecasting comes in. cost can be reduced and accurately goods and services can beestimated and this in turn improves operating efficiently increases. The figure below explain therelationships of P O C and the forecasting plan.

Fig: 2 operations and production management activities

Operations managers need long range forecasts to make strategic decisions about products,

processes and facilities. They also need short-range forecasts to assist them in making decisions

about operations issues that span for few days or weeks. The following table 1 shows

summarizes some of the reasons why operations managers must develop forecasts.


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In the table 2 shows examples of things that are commonly forecasted.

T 1 S R W F I E O M

1. .I

. S POM

.

2. . D . P

. I

. O

.

3. . D . T

, , , . O

.

F H T S E T S

T B F

L Y N P D

O P D

F C G, , U

T

C D

F S,

M R M P U

D C H, ,,

G,

U

T

W W,

P U, , I U,

S W S U

LS W,

M C U, , ,

P

P

C D

I U,


7/59

Forecasting is an integral part of business planning. The inputs are processed throughforecasting models or method to develop demand estimates.

Theses demand estimates are not the sales forecasts; rather, they are the starting point formanagement teams to develop sales forecasts.

The sales forecasts become inputs to both business strategy and production resourceforecasts.

Fig: 3 Forecasting as an Integral Part of Business Planning

F E/

F


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WHAT IS A FORECAST?

F . I .

F .

W . F , , . T ,

.

WHY DO FIRMS FORECAST?

F , , .

F I . F .

A .

T , , .

G .

F . B 1. I 2. I M 3. B 4. I

COST OF FORECASTING

A , .

T . O

, , , ,

.


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OUR APPROACH TO FORECASTING

T

, .

1. J2. T 3. E 4. R

FORECASTING VARIABLES

F

1. T 2. T 3. D 4. M

W

T .

M , , .

F .O

P:

T .

M ;1. P 2. S 3. C 4. S .


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T :

T ;

1. C2. U

E: S

.

F .

A:

F .

I , . W

( ), ,

, .

O , . B ,

.

T :

E; ( )

. H

.

M .

F .

R .

T ,

.


11/59

E , , .

E , ,,

.

Forecasting and operations subsystems:

In the production units number for televisions in a plant, the number of patients fed in ahospital, the number of books circulated in a library, or the number of lots of common stock soldin a brokerage house the resource forecasts are used to plan and control operation subsystems,as shown in figure 4.

F 4: /

I

D

C

P

I

L

C

S

A

O

P

()

P

P

E

C

O

I

D


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()

M

.

T : D .

A

C M I ()

T . O ,

,

.

T

.

C /

. I ,

, E: , M D D A, .

W ,

.

M ,

.

F .

J

.

M , ,

.

A , .

150

140

130

80

5 10 15 20

F: 5.1 S D


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150

140

130

80

5 10 15 20

F: 5.2 D .

150

140

130

80

5 10 15 20

F: 5.3 S D.

150

140

130

80

5 10 15 20

F: 5.4 S D .

L N

H N

T

T , NOISE.

NOISE

LOW NOISE: M .

HIGH NOISE: M .

P

(U)


14/59

I , . T

.

F :

A , .

I , .

We know forecasts are of two kinds

Long range short range

The long range is making forecast on capacity, location and layout. The short range makesforecast on the individual items. The following figure3 shows different types of planningdecisions depend on different types of information, which in turn depend on what are called theforecasting time horizons, of the future times to which the forecasting points.

Let us differentiate between Forecast and Prediction:

Forecast Prediction

A forecast is an estimate of a future eventachieved by systematically combining andcasting forward in a predetermined way dataabout the past.

A predication is an estimate of a future eventachieved through subjective considerationother than just past data; this subjectiveconsideration need not occur in anypredetermined way.

Type of RepresentativeDecision Information

Needs

Short SpecificPlanning ItemDecisions Demands

Aggregate

Demands

Long run StrategiesPlanning andDecisions Facilities

Present Five Years Hence

Fig. 3 Forecasting time Horizon


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EX: for Forecasting

A TV manufacturer, for example can use past data to forecast the number of picture screensrequired for next weeks TV assembly schedule.A fast food restaurant can use past data to forecast the number of hamburger buns required for

this weekends operations.

EX: for Prediction:

Suppose the manufacturer offers a new TV model or the restaurant decides to offer a new item.Since, no past data exist to estimate first year sales of the new product, prediction, notforecasting is required.For predicting good subjective estimates can be based on the mangers skill, experience andjudgment; but, forecasting requires statistical and management science techniques.

TIME HORIZON

F . S 1 ( 03 );

.

M R1 3 ; L 5 ; M

.

A 35 , .

P .

P .

T .

DATABASE: QUANTITATIVE AND QUALITATIVE

M . I .

S , . I , ( )

. S

.

E . T .


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I . T

,

T

T /

.

T

M D

D Q

H M

N G .

( )

S A

E W

R D

E P (GNP)

1. D T: A

.

T :

A , ; . E .

T , .

O , . T .

A, ,

.


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T D . T ; , , ,

.

T .

2. H I .

3. N :

.

T . S

, .

A . E

.

A ,

.

A .

T

.

W , ,

, .

M .

T .

S : (SA)

. T :

S

SA=

N

D

= , =

D=

E: , 50 , 60,

,, 40 . T :

D1+ D2+D3

SA=

3

50+60+40


18/59

=

3

= 50

A 50

.

A (MA)

, . O

, . W 3

20 .

A :

S

MA=

C

D

MA = D =

N =

T .

T ; ,

.I .

I .

I . H,

() X

MA =

F

E= 1D1 + 2D2 +3D3 + ..+ D

W: D = I

= D

=


19/59

FORECASTING METHODOLOGY

T .

A , .

W , . C

.

S , .

O , , . T L , ,

. T .

J , , .

T .

T . E , ,

. T ,

.

R .

BJ .

TIME SERIES METHODS

A . T

.

C ;1. T (T)2. C (C)3. S (S)4. R (R) I , (Y)

: Y=TCSR


20/59

T , .

C .

S . R

.

FORECASTING PROCEDURE:1. P ( , ..)2. D 3. D 4.

P

5. M 6. M :

) C (C)) A ( R)

A C .

A , .

MA = X

N

W X .

C 3 .


21/59

year Shipments(tons)

Three-YearMovingTotal

Three-YearMovingAverage

1977 2

1978 3 11 3 = 3.7

1979 6 19 6.3

1980 10 24 8.0

1981 8 25 8.3

1982 7 27 9.0

1983 12 33 11.0

1984 14 40 13.3

1985 14 46 15.3

1986 18 51 17.0

1987 19

N . T 3.7

J 1, 1978.

E: F 11

0.5, 0.3 0.2 10, 9, 8. T 66, 67 70 8, 9,

10.

S: E 11= 66 0.2+67 0.3 +70 0.5 = 67.7.

W E:

1. the sales pattern of a manufacturing firm is given below.compute the 3yearly movingtrend and find out the sales forecast for the year 1993.

Year 1985 1986 1987 1988 1989 1990 1991 1992

sales 8 8.5 9 10 9.5 11 11.5 12


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SOLUTION: the 3 yearly moving average trend is computed in the following table.

year Sales(Rs.

Lakhs)

3 yearly moving

total

3 yearly moving

average

1985 8

1986 8.5

1987 9 25.5 8.5

1988 10 27.5 9.2

1989 9.5 28.5 9.5

1990 11 30.5 10.2

1991 11.5 32.0 10.7

1992 12 34.5 11.5

The forecast for 1993 in 11.5 which is the average of last 3 years.

2.T 20

. F

.

D M A

(4 )

M A

(2 )1 121

2 125

3 124 123

4 118 125

5 134 122 121

6 127 125 126

7 124 126 130

8 141 126 126

9 133 131 133

10 135 131 137

11 141 133 134

12 139 138 138

13 144 137 140

14 152 140 141

15 142 144 148

16 149 144 147


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F2: .

17 145 146 145

18 140 147 147

19 132 144 143

20 130 141 136


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3.A . P (

) .

) C 5 52.) C 3

3,2,1.

M A

43 105

44 106

45 110

46 110

47 114

48 121

49 130

50 128

51 13752

S:

X

) MA=

114+121+130+128+137

=

5

= 126

)

()(X)

MA=

W X =

3 X 137 = 411

2 X 128 = 256

1 X 130 = 130

6 797797

MA= = 133

6


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4. The ABC Floral shop sold the following number of geraniums during the last 2 weeks.

Day Demand Day Demand

1 200 8 150

2 134 9 182

3 157 10 197

4 165 11 136

5 177 12 163

6 125 13 157

7 146 14 169

i. Determine the forecast for the number of Geraniums demanded on the 15thday usingthree period moving average as well as five period moving average.

ii. Depict graphically the difference between forecast and the actual demandSOLUTION:

i. 3 period moving average.Day Demand 3 day moving

total

3 day moving

average

Round off to

nearest figure

1 200

2 134

3 157 491 163.7 164

4 165 456 152.0 152

5 177 499 166.3 166

6 125 467 155.7 156

7 146 448 149.3 149

8 150 421 140.3 140

9 182 478 159.3 159


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10 197 529 176.3 176

11 136 515 171.6 172

12 163 496 165.3 165

13 157 456 152.0 152

14 169 489 163.0 163

Forecast for the 15th

day is 163.0

ii. 5 period moving averageDay Demand 5 day moving

total

5 day moving

average

Round off to

nearest figure

1 200

2 134

3 157

4 165

5 177 833 166.6 167

6 125 758 151.6 152

7 146 770 154.0 154

8 150 763 153.0 153

9 182 780 156.0 156

10 197 800 160.0 160

11 136 811 162.0 162

12 163 828 166.0 166

13 157 835 167.0 167

14 169 822 165.0 165


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ii.The forecast value is the average of past sales. The forecast values (trend values) of 3period and 5 period moving averages and the actual demand can be shown on the graph by

plotting the trend values and actual demand on y-axis and No. of days on x-axis.

1. L

2. T T ;) T ) T ) T X Y

L E

Y = + X

XY = X + X(1)

W X =0, , :

Y=

XY = X(2)

C ,

X=0 .


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E;

U . S

, 1992

X

Y

Y

S

()

XY X

1977 5 2 10 25

1978 4 3 12 16

1979 3 6 18 9

1980 2 10 20 4

1981 1 8 8 1

1982 0 7 0 0

1983 1 12 12 1

1984 2 14 28 4

1985 3 14 42 9

1986 4 18 72 16

1987 5 19 95 25

0 113 181 110

R 2,

113

= = = 10.3

11

XY 181

= = = 1.6X 110

T Y= + X

Y= 10.3=1.6X (1982=0, X= , Y=)

F 1992: B 1992 10 ,

Y=10.3+1.6(10)=26.3 .


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T . T ;

.

2.The sales of a product during the last five years is tabulated below

Year: 1973 1974 1975 1976 1977

Sales: 4 8 6 10 4

Using least square method forecaster,

SOLUTION:

Year x y x xy

1973 -2 4 4 -8

1974 -1 8 1 -8

1975 0 6 0 0

1976 +1 10 1 +10

1977 +2 4 4 +8

x=0 y=32 x=10 xy=2

The linear equation is y=a+bx

Equation of regression line is given by

y=6.4+0.2x

Regression and correlation methods

Regression and correlation techniques are means of describing the association between two or more

such variables. They make no claim to establishing cause and effect but, instead merely quantify the

statistical dependence or extent to which the two or more variable are related.

Regression means dependence and involves estimating the value of a dependent variable, Y from

an independent variable X.

In simple regression only one independent variable is used, whereas in multiple regression two or

more independent variables are involved.

Simple Linear regression model

Yc = a+bX where Y=dependent variable, X=independent variable.

Multiple linear regression equation

Yc=a+bX+cX2+dX

3.


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T

. T

. T

,

:

XY XY

=

X2X

=Y X

W X = ()/ Y = (Y)/

1. T

. T .

C

(X)

P

(Y)

15 6

9 4

40 6

20 6

25 3

25 9

15 1035 16

F

) G .

) U .

) C () ()

) D

30.

S:

) G


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A

.

)

C

(X)

P

(Y)

X Y X2 Y2

15 6 90 225 36

9 4 36 81 16

40 6 640 1600 256

20 6 120 400 36

25 3 325 625 169

25 9 225 625 81

15 10 150 225 100

35 16 560 225 256

184 80 2146 5006 950

N=8

Y=+X 80 = 8+184 (1)

XY = X + X2 2146 = 184+5006 (2)

M (1) (23):* 1840 =184 4232 (3)

A (2) (3) 360= 774 (4)

T = 306/774 =0.395

S (1) 80 = 8 + 184 (0.395)

8=80 72.7

A= 7.3 / 8 = 0.91


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E Y=0.91+ 0.395X

W X = Y=

) A,

X = ()/ = 184/8=23

Y = (Y)/ = 80/8 =10

XY XY 2,1468(23)(10)

= = = 0.395

X2X 5006 8(23)(23)

=Y X = 100.395 (23) =0.91

) X=30,Y =0.91 +0.395 (30)

= 12.76 = 30 .

X= X=

= Y=

= Y=+X

= =

= =

=

=

()

= ()

Y=+X

2=

[() ][()]


33/59

1. A D C I. P . T . W

. ,

3 . S

10 :

( )

( )

1 1000 6 2000

2 1300 7 2200

3 1800 8 2600

4 2000 9 29005 2000 10 3200

W

(I M E

.)

(

)

()

1 1000 1 1 1000

2 1300 2 4 26003 1800 3 9 5400

4 2000 4 16 8000

5 2000 5 25 10000

6 2000 6 36 12000

7 2200 7 49 15400

8 2200 8 64 20800

9 2600 9 81 26100

10 3200 10 100 32000

T =21000 =55 =385 =133,300

Solution:1. L :

= = (385) (21,000) (55) (133,300)

() 10(385) (55)

= 8,085000 7,331,500 = 753,500 =913.333

3,850 3,025 825

1. S L R A: A T S


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= = (10) (133,300) (55) (21,000)

() 825

= 1,333,0001,155,000 = 178000 = 215.758

825 825

2. N , :

Y=+X =913.333+215.758X

3. I , 11, 12, 13, , X:

Y11 = 913.333 + 215.758(11) = 3,286.7 3,290

Y12 = 913.333 + 215.758(12) = 3,502.4 3,500

Y13 =913.333 + 215.758(13) = 3,718.2 3,720

T . N ; .

2. J W, E C,

. H

? J M C,

, . M C,

: . D .

. . .

D .

Solution:. D :

1. M , , G .

2. S .3. T :

( )

( )

1 Q1 8 150

Q2 10 170

Q3 15 190

Q4 9 170

2 Q1 12 180

Q2 13 190

Q3 12 200

Q4 16 220


35/59

4. M . T = 5. (I

.)

() ()

1 8 150 22500 1200 64

2 10 170 28900 1700 100

3 15 190 36100 2850 225

4 9 170 28900 1530 81

5 12 180 32400 2160 144

6 13 190 36100 2470 169

7 12 200 40000 2400 144

8 16 220 48400 3520 256

9 9 81 26100T =95 =1470 =273,300 =17,830 =1,183

5. U 3.45 :

= = (273,300) (95) (1470) (17,830)

() 8(273,300) (1470)

= 25,963,500 26,210,100 = 753,500 = 9.671

2,186,400 2,160,900 25,500

= = (8) (17,830) (1470) (95)

() 25,500

= 142,640139,650 = 2,990 = 0.1173

25,500 25,500

6. T Y= 9.671 +0.1173X.


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. F :

1. M

. T 260,290,300

270.

2. N, M (

) Y9.671+0.1173X:

Y1=9.671 +0.1173(260) Y2=9.671 +0.1173(290)

= 9.671 + 30.498 = 9.671 + 34.017

= 20.827 = 24.346

Y3=9.671 +0.1173(300) Y4=9.671 +0.1173(290)

= 9.671 + 35.190 = 9.671 + 31.671

= 25.519 = 22.000

T ( ) :

20.827 +24.346 +25.5819 + 22.000 =$92.7

N .

. E := = 2900

[() ][()] [25,550][8(1,183) (95)

= 2,900 = 2,900 = 2,900

[25,550][9,464 9,025] (25,500)(439) 11,194,500

= 2990 =.894

3,345.8

= 0.799

T 80 %( = 0.799)

.


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3. The table below gives a sales record of a firm. Determine the regression line for the firmand find the forecast of sales in the month of Jan for next year.

Month Sales(in

units)/(Demand)

Jan 90

Feb 111Mar 99

April 89

May 87

June 84

July 104

Aug 102

Sept 95

Oct 114

Nov 103

Dec 113

SOLUTION:

Regression equation is,

y=a+ bx

Where

Month x y x xy

Jan 0 90 0 0

Feb 1 111 1 111

Mar 2 99 4 198

April 3 89 9 267

May 4 87 16 348

June 5 84 25 420

July 6 104 36 624

Aug 7 102 49 714

Sept 8 95 64 760Oct 9 114 81 1026

Nov 10 103 100 1030

Dec 11 113 121 1243

x=66 y=1191 x=506 xy=6741


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Using above values, the constants a and b are calculated as follows

x=66, y=1191, x=506, xy=6741

The equation of the regression line is given by

y=a+ bx

++++

( )

+*

.

2nd

method: the above problem can be worked in a simplified manner by taking the deviation

from the middle year, such that x will be equal to zero, and the values of a and b can begiven by

;

Month x y x xy

Jan -6 90 36 -540

Feb -5 111 25 -555

Mar -4 99 16 -396

April -3 89 9 -267May -2 87 4 -174

June -1 84 1 -84

July +1 104 1 +104

Aug +2 102 4 +204

Sept +3 95 9 +285

Oct +4 114 16 +456

Nov +5 103 25 +515

W;

X =

Y= sales in the respective month

n= number of observations

W;

X =

Y= sales in the respective month

n= number of observations


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Dec +6 113 36 +678

x=0 y=1191 x=182 xy=226

The required equation is y=a +bx

=99.25

1.2417

Put x=+7(since we are taking the deviation from the middle year, such that x=0)

To find forecast for Jan

Forecast for next Jan is 108 units.

4. The sales of a product during the last five years is tabulated belowYear: 1973 1974 1975 1976 1977

Sales: 4 8 6 10 4

Using linear forecaster, calculatei. Sales in the years 1978 and 1979ii. Standard error of estimate and give its significance.

SOLUTION:

Year x y x xy

1973 -2 4 4 -8

1974 -1 8 1 -8

1975 0 6 0 0

1976 +1 10 1 +10

1977 +2 4 4 +8

x=0 y=32 x=10 xy=2

The linear equation is y=a+bx


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Equation of regression line is given by

y=6.4+0.2x

i. Estimated sales for,Year 1978

Put x=3 in the regression equation, y=6.4+0.2*3=7

Year 1979

Put x=4 in the regression equation, y=6.4+0.2*4=7.2

Forecast values y1978=7; y1979=7.2

ii. To determine standard error estimate standard deviationWhere y=Demand values (sales)

y'=calculated values from regression equation

Year y y' (y'-y) (y'-y)

1973 4 6 +2 4

1974 8 6.2 +1.8 3.241975 6 6.4 +0.4 0.16

1976 10 6.6 +3.4 11.56

1977 4 6.8 +2.8 7.84

26.80

(y'-y)=26.80

=2.315

Its significance:the standard error estimate simply shows that 95% of the data are expected

to fall within 2 limits of the regression line.2 =2*2.315=4.63

2

5. the sales of machine tools in the last 8 years is lakhs of Rs are 5.0,4.5,10,9.0,11.0,18.5,17.5 an22.0.Find the linear regression line and calculate the sales for the 10

thyear.

Solution:

Regression line is given by

y=a+bx

;

Year y x x xy

1 5.0 -7 49 -35.0

2 4.5 -5 25 -22.5

3 10.0 -3 9 -30.0

4 9.0 -1 1 -9.0


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12.187

y=12.19+13x

Estimated sales in the 10thyear

Put x=11 in the equation

y10=12.19+1.3*11=26.49

6. A manufacturer of childrens cycle believes that the demand for the cycles is correlatedto the birth of babies in the area during the previous year. The following data shows the

relationship.

Compute the probable sales in the ninth year, given the number of births in the previous

year as 1, 66,000

5 11.0 +1 1 +11.0

6 18.5 +3 9 +55.5

7 17.5 +5 25 +87.5

8 22.0 +7 49 +154

y=97.5 x=0 x=168 xy=211.5

Year No. of births in the previous year Cycles sold during the year

1 40000 3000

2 48000 3200

3 66000 3700

4 78000 4000

5 92000 5200

6 1,05,000 7900

7 1,25,000 9000

8 1,40,000 10000


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SOLUTION: y=a+bx

Year y*1000 x*1000 xy x

1 3 40 120 1600

2 3.2 48 153.6 2304

3 3.7 66 244.2 4356

4 4 78 312 6084

5 5.2 92 478.4 8464

6 7.9 105 829.5 11025

7 9 125 1125.0 15625

8 10 140 1400.0 19600

y=46 x=694 xy=4662.7 x=69078

y=-0.82+0.0757x

Forecast for 9th

year is given x=1,66,000

y9= [-0.82+0.0757*1,66,000] =11744 cycles.

7. A manufacturer of tyres believes that a relationship exists between the automobiles soldin the year and the sales of the tyres two years later. The data for the past 10 years are

given below.

Sales ofautomobiles(in

lakhs)

4.0 4.5 4.2 5.5 5.8 5.5 6.2 7.2 6.7 7.9

Sales of tyres

2 years later(in

lakhs)

8.0 7.9 8.1 8.4 8.1 8.6 9.1 8.9 9.1 9.6


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Establish a linear regression fit to forecast the sales of tyres on the basis of the sales of

automobiles 2 years earlier. Also find what will be the sales of tyres given that the sale of

automobiles 2 years earlier was 6.1 lakhs.

Solution: Required equation is,

y=a+bx

y=16.23-1.16x

Given sales of automobiles 2 years earlier=6.1 lakhs

Forecast for tyres

Put x=6.1 in the regression equation

y=16.23+116*61=9.154

=9.2

Forecast sales of tyres=9.2 lakhs

Sales of

automobiles in

lakhs

x

Sales of tyres 2

years later.

y

x xy

4 8 16 32

4.5 7.9 20.25 35.55

4.2 8.1 17.64 34.02

5.5 8.4 30.25 46.2

5.8 8.1 33.64 46.98

5.5 8.6 30.25 47.3

6.2 9.1 38.44 56.42

7.2 8.9 51.84 64.08

6.7 9.1 60.97 60.97

7.9 9.6 75.84 75.84

x=57.5 y=95.8 x=375.12 xy=499.36


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( )

E .

E

.

S .

W .

S :

T

.

F= F1 + (D1 F1)

W

F=

F1=

=

D1 = I ,

, .

T , .

I 0 1 F .

T 0.005 0.30 S

.

T .

P 1. A =0.1 .T F 1 500 , 45

.

) F F 8) A F 8 505 .

F F 15.C M 15,

516, 488, 467, 554 10

S;

() F = F1 = (D1 F1)= 500=0.1(450500) = 495

() A , ;W D

D1O

F1F

D1 F1C

(D1 F1)N (F)

F= F1= (D1 F1)

F.1 450 500 50 5 495

8 505 495 10 1 496


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15 516 496 20 2 498

22 488 498 10 1 497

M.1 467 497 30 3 494

8 554 494 60 6 500

15 510 500 10 1 501

I , . I ,

.

Smoothing coefficient selection:Smoothing coefficient: A numerical parameter that determines the weighting of old demands inexponential smoothing.To begin forecasting, some reasonable estimate for an old beginning forecast is necessary.Likewise, a smoothing coefficient, , must be selected.

A high smoothing coefficient could be more appropriate for new products or items for which theunderlying demand is shifting about (dynamic or unstable).A of 0.7, 0.8, 0.9 might be best for these conditions, i.e. if unstable conditions are known toexist.If demand is very stable and believed to be representative of the future, the forecaster wants toselect a low value to smooth out any sudden noise that might have occurred.Under the stable conditions, an appropriate value may be 0.1, 0.2, and 0.3. When demand isslightly unstable smoothing coefficients of 0.4, 0.5 or 0.6 might provide the most accurateforecasts.Selecting forecasting parameters and comparing models:The procedure for selecting forecasting parameters is given in the first four steps that follow; the

fifth step is used for comparing and selecting models.1. Partition the avialble data into two subjects, one for fitting parameters (the test set) and

the other for forecasting.2. Select an error measure to evaluate forecast accuracy of the parameters to be tried. MAD

and /or bias are useful error measures.3. Select a range of values. Using one of the values apply the forecasting model to the

test set of data, recording the resulting ofrecast errors. Then, selecting a new values in theselected range have been tested.

4. Select the value that resulted in the lowest forecast error when applied to the test set.Your model is now fitted to the demand data.

5. Forecast using the balance of the data with the exponential (or moving average) modelthat you have fitted to the test set. Use the results to compare alternative models that havepreviously been fitted to representative demand data.

Forecast Error:When we evaluate different forecasting methods, it is necessary to measure the effectiveness.Forecast error is the numeric difference of forecasted demand and actual demand.Mean Absolute Deviation (MAD):


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A forecast error measure that is the average forecast error without regard to direction;calculated as the sum of the absolute value of forecast error for all periods divided by thetotal number of periods evaluated.

Sum of the absolute value of forecast error for all periodsMAD = -----------------------------------------------------------------------

number of periods|forecast error|= -----------------------

n

|forecast demand actual demand|= ------------------------------------------------

W

T

, (). I ,

. W , (SMAD)

.

1.25SMAD

S MAD MAD .

;

.

S =

=

( )

=

,

2

=

+1


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How to monitor and control a forecasting model.

Forecasts can be monitored and controlled by setting upper and lower limits on how much theperformance characteristics of a model can deteriorate before we change the parameters of themodel. One common way that we can track the performance of forecasting models is to use whatis called a tracking signal:

Algebraic sum of errors over n periodsTracking signal =----------------------------------------------------Mean absolute deviation over n periods(Actual Demand forecast demand)

=--------------------------------------------------MAD

(Actual demand Forecast demand)=-------------------------------------------------|Actual demand Forecast demand|------------------------------------------------

n

The tracking signal measures the cumulative forecast error over n period in terms of MAD.

Problem 1.Forecast for 9thyear using the following data by exponential technique.

Year 1 2 3 4 5 6 7 8

Demand(Rs

in lakhs)

90 100 107 113 123 136 144 155

Smoothing constant =0.5 and initial forecast F=85.

SOLUTION:

Year Demand

Dt-1

Forecast

Ft-1

New forecast

Ft= Ft-1+ (Dt-1- Ft-1)

1 90 85 85.75

2 100 85.75 87.89

3 107 87.89 90.76

4 113 90.76 94.1

5 123 94.1 98.44

6 136 98.44 104.07


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7 144 104.07 110.06

8 155 110.06 116.8

The forecast value 116.8 made in the 8thyear, is the forecast for the next year (i.e., 9th year)

Rs. In lakhs.

Problem 2. monthly sales of a product in thousands of rupees for the past 2 years are shown

below:

Month Jan Feb Mar Apr May June July Aug Sep Oct Nov Dec

2

years

ago

253 236 245 246 260 251 249 242 234 244 246 251

1 year

ago250 252 248 241 247 244 244 249 511 238 249 252

a) Fit a line to the data and determine a forecast of the next month.b) Select an initial forecast from part a and use =0.2 to determine the forecast for the next

Jan by exponential smoothing.

c) Compare the forecast from (a) and (b), which one would you select.SOLUTION:

a) Required equation y=a+bxMonth Average

y

x xy x y' Error

E=(y'-y)

Jan 251.5 -11 -2766.5 121 245.8 -5.7

Feb 244.0 -9 -2196.0 81 247.9 +3.9

W;

Ft= current period forecast

Ft-1= last period forecast

Dt-1= last period demand

= smoothing constant


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Mar 246.5 -7 -1725.5 49 250.1 +3.6

Apr 243.5 -5 -1217.5 25 252.2 +8.7

May 253.5 -3 -760.5 9 254.5 +0.9

June 247.5 -1 -247.5 1 256.5 +9.0

July 246.5 +1 +246.5 1 258.7 +12.2

Aug 245.5 +3 +736.5 9 260.8 +15.3

Sep 372.5 +5 +1862.5 25 262.9 -109.6

Oct 241.0 +7 +1687.0 49 265.1 +24.1

Nov 247.5 +9 +2227.5 81 267.3 +19.8

dec 251.5 +11 +2766.5 121 269.4 +17.9

y=3091 x=0 xy=613 x=572 |(y'-

y)|=230.7

257.6

Regression y is given by

y=257.6+1.072x

Forecast for the next month (i.e.,jan)

Put x=13 in the above equation

+*

The forecast values for all the months are calculated using regression equation and

tabulated.

The mean absolute deviation is calculated by taking the average value of absolute error.

=19.23

MAD=19.23

b) Initial Forecast = 271.5Smoothing constant = = 0.2


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Month Demand Dt-1 Forecast Ft-1 New Forecast

Ft= Ft-1+ (Dt-1- Ft-1)

Error

E = (Ft-1- Dt-1 )

Jan 251.5 271.5 260.3 +56.0

Feb 244.0 260.3 257.04 +16.3

Mar 246.5 257.04 254.9 +10.54

Apr 243.5 254.9 252.6 +11.4

May 253.5 252.6 252.8 -0.9

June 247.5 252.8 251.7 +5.3

July 246.5 251.7 250.7 +5.2

Aug 245.5 250.7 249.7 +5.2

Sep 372.5 249.7 274.3 -122.8

Oct 241.0 274.3 267.6 +33.3

Nov 247.5 267.6 263.6 +20.1

Dec 251.5 263.6 261.2 +12.1

|(Ft-1- Dt-1 )| = 299.14

= 24.92

MAD = 24.92

c) The deviation (MAD = 19.23) in the I method is less. Hence the least square method ispreferred.


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Problem 3.The pesticide manufacture has experienced the following monthly demand for an

environmentally sound pesticide.

Month Actual demand(in tones)

Feb 62

Mar 84April 77

May 95

June 100

Using first order exponential smoothing technique forecast the demand for the month of July.

Choosing = 0.3, compare graphically the forecast with the actual demand from the months

march through June, assuming that the forecast for the February was 60.

Solution :

Month Actual demand Initial forecast New forecast (Ft)

Feb 62 60 60.6

Mar 84 60.6 67.62

April 77 67.62 70.43

May 95 70.43 77.8

June 100 77.8 84.5

Specimen Calculation

F1= Ft-1+ (Dt-1- Ft-1)=60+0.3(62-60)

=60+0.6

=60.6 = Forecast for the month of July = FJuly =84.5


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Problem 4:Number of daily calls for repair has been recorded as follows:

Day 1 2 3 4 5 6

Calls 132 170 95 110 120 135

Prepare exponentially smoothed forecasts for =0.1 and F1=130. Compute the errors of

Bias and absolute Deviation. Forecast for 7thday.

Solution:

Given =0.1, F1=130

Day CallsDt-1

Initial forecastFt-1

New forecastFt

Error = (Ft-1- Dt-1)

1 132 130 130.2 -2

2 170 130.2 134.18 -39.8

3 95 134.18 130.26 39.18

4 110 130.26 128.23 20.265 120 128.23 127.41 8.23

6 135 127.41 128.17 -7.59

7 128.17

New forecast of 6thday is the forecast for 7thday.

The forecast value for 7th

day =129 calls

(i) Mean Absolute Deviation|Error| 117.06

=------------- = ------------ = 19.51n 6

Error 18.28(ii) Bias = ---------- = ----------- = 3.05

n 6

Problem 5: sales of plywood in rupees of a particular size have been tabulated below.

Year 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989

Sales

in Rs.

x 105

15 16 12 22 16 21 30 12 31 40

(i) What is the expected sale in 1990 by method of least squares.(ii)Select an initial forecast from (i) and use = 0.1 to determine the forecast for 1990 by

exponential smoothing.

(iii)Compare the forecast (i) and (ii) which one would you select.


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Solution:

(i) By method of least squaresYear Sales y in

Rs.x 105X XY X Forecast from regression

equations y

Error (Y1-Y)

1980 15 -9 -135 81 11.69 -3.31

1981 16 -7 -112 49 13.87 -2.13

1982 12 -5 -60 25 16.05 4.05

1983 22 -3 -66 09 18.23 -3.77

1984 16 -1 -16 01 20.41 -4.41

1985 21 1 21 01 25.59 1.59

1986 30 3 90 09 24.77 -5.23

1987 12 5 60 25 26.95 14.95

1988 31 7 217 49 29.13 -1.87

1989 40 9 360 81 31.31 -8.69

y=215 X=0

XY=359

X

=330

|Y1-Y|=50

Y=+

215

= = = 21.5

10

XY 359

= = = 1.09

X 330

Y=+

= 215+1.09X

Forecast for the year 1990 Put X=10


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Y =215 + 1.09 x10 =32.4

Y1990 = 32.4 rs. x105 y1990 =32.4 x 10

5

| Error| 50

MAD = ------------ = -------- = 5

n 10

(ii)Exponential smoothing Technique= 0.1

Initial forecast = 32.4

Month SalesDt-1

Initial Forecast Ft-1 New ForecastFt= Ft-1+ (Dt-1- Ft-1)

ErrorE = (Ft-1-Dt-1 )

1980 15 32.4 30.66 17.4

1981 16 30.66 29.2 14.66

1982 12 29.2 27.48 17.2

1983 22 27.48 26.93 5.48

1984 16 26.93 25.84 10.93

1985 21 25.84 25.36 4.87

1986 30 25.36 25.82 -4.64

1987 12 25.82 24.44 13.82

1988 31 24.44 25.09 -6.56

1989 40 25.09 26.58 -14.91

|Error|= 110.44

Forecast for the year 1980 is 26.58 x105

|Error | 110.44

MAD = ------------ = ------------- = 11.044

n 10

(iii) The Absolute deviation is less in case of method of least squares. Hence, it can be

preferred.


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Seasonal indexes:

A seasonal index (SI) is a ratio that relates a recurring seasonal variation to the correspondingtrend value at that given time.Several methods of computing are available, but the most widely used is a ratio-to-moving-average method.

The procedure is to tabulate the data in monthly terms and compute 12-month moving-averagevalues over a period of several years.Seasonalized forecast= seasonal index(trend forecast)

Ysz=(SI)Yc

Problem 1.The production manager of a natural gas pipeline company has projected trendvalues for next august, September, and October of 2.1, 2.2 and 2.3 million cubic meters,respectively. Seasonal indexes for the three months have been found to be, 0.80, 1.05, and 1.20,respectively. What actual seasonalized (adjusted) production should the manager plan for?Solution:

Ysz =SI (Yc)

For August: = (0.80) (2.1) = 1.68 million cubic metersFor September: = (1.05) (2.2) = 2.31 million cubic metersFor October: = (1.20) (2.3) = 2.76 million cubic meters

After seasonal adjustments have been made, similar adjustments can be made for cyclicalor irregular effects if data are available.

2. Wayne Conners, the plant manager of Aroma drip coffee Inc., is trying to plan cash,personnel and materials and supplies requirements for each quarter of next year. Thequarterly sales data for the past three years seem to reflect fairly the seasonal outputpattern that should be expected in the future. If Wayne could estimate quarterly sales fornext year, the cash, personnel, and materials and supplies needs could be determined.

Solution:1. We compute the seasonal indexes.

Year Quarterly Sales (thousands of Units) Annual total

Q1 Q2 Q3 Q4

8 520 730 820 530 2600

9 590 810 900 600 2900

10 650 900 1000 650 3200

Totals 1760 2440 2720 1780 8700

Quarteraverage

586 2/3 813 2/3 906 2/3 593 1/3 725*

Seasonalindex (S.I)

**0.809 1.122 1.251 0.818

*Overall quarter average= 8700/12 =725

**S.I = quarter average / overall quarter average.2. Next, we deseasonalize the data by dividing each quarterly value by its S.I (seasonal

index) for instance, 520 0.809=642.8, 7301.122=650.6 and so on


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YearDeseasonalized Adjusted quarterly Data

Q1 Q2 Q3 Q4

8 642.8 650.6 655.5 647.9

9 729.3 721.9 719.4 733.5

10 803.5 802.1 799.4 794.6

3. Now, we perform a regression analysis on the deseasonalized data (12 quarters) andforecast for the next 4 quarters:

Time Period X Y Y X XY

Year 8, Q1 1 642.8 413,191.84 1 642.8

Year 8, Q2 2 650.6 423,280.36 4 1301.2

Year 8, Q3 3 655.5 429,680.25 9 1966.5

Year 8, Q4 4 647.9 419,774.41 16 2591.6

Year 9, Q1 5 729.3 531,878.49 25 3646.5

Year 9, Q2 6 721.9 521,139.61 36 4331.4

Year 9, Q3 7 719.4 517,536.36 49 5035.8Year 9, Q4 8 733.5 538,022.25 64 5868.0

Year 10, Q1 9 803.5 645,612.25 81 7231.5

Year 10, Q2 10 802.1 643,364.41 100 8021.0

Year 10, Q3 11 799.4 639,040.36 121 8793.4

Year 10, Q4 12 794.6 631,389.16 144 9535.2

totals x=78 y=8700.5 y =6,353,909.75 x =650 xy=58,964.9

4. Now, to find the value of a, b and Y= 650(8700.5) 78(58,964.9) / 12(650) (78)2

= 615.421

= 12(58,964.9) -78(8700.5) / 12(650) (78)2

= 16.865

Y = a + b x = 615.421 +16.865X

5. Now, we substitute the values 13,14, and 16 the next four values for x into theregression equation. These are deseasonalized forecasts, in thousands of units, for the next

four quarters.Y13=615.421+16.865(13)=834.666 Y14=615.421+16.865(14)=851.531

Y15= 615.421+16.865(15) =868.531 Y16 =615.421+16.865(16)=885.261


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6. Now, we use the seasonal indexes (SI) to seasonalize the forecasts:Quarter S.I. Deseasonalized

forecastsSeasonalized forecasts (S.I.XDeseasonalized forecasts)(thousands of Units)

Q1 0.809 834.666 675Q2 1.122 851.531 955

Q3 1.251 868.396 1086

Q4 0.818 885.261 724

Life cycle effects upon forecasting methodology:

Introduction

Data No data available : rely on qualitative method

time Need long horizon

methods Judgment, Delphi and historical analogy were useful, market surveys importantGrowth

Data Some data available for analysis

time Still need long horizon; trends and cause-effect relationships important

methods Market surveys and historical comparison still useful. Regression and computersimulation models justified. Tracking product history now important

Maturity

Data Considerable data available on demand, inventory levels et.

time More uses of short-term forecasts; still need long-term projections, but, trendschange only gradually.

methods Statistical and quantitative methods more useful. Time series help for trend,seasonal. Regression and correlation use associations and leading indicators.Exponential smoothing very useful. Econometric methods feasible.

Decline

Data Abundant data (but not necessarily on decline).

time Shorter horizon.

methods Continue use of maturity methods as applicable. Judgment, historical analogies,and market surveys may signal changes.

N

T


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How to select a forecasting method:

Several factors should be considered in the selection of a forecasting method

Cost

Accuracy

Data available

Time span

Nature of product and services

Impulse response and noise dampening

Some reasons for ineffective forecasting:

1. Failure of the organizations to involve a broad cross section of people in forecasting.Individual effort is important, but the need to involve everyone who has pertinent

information and who will need to implement the forecast is also important.

2. Failure to recognize that forecasting is integral to business planning3. Failure to recognize that forecasting will always be wrong. Estimates of future demand

are bound to be subject to error, and the magnitude of error tends to be greater for

forecasts that cover very long spans of time. When operations mangers have unrealistic

expectations of forecasts , the fact that the forecasts were not on the nose is often used as

an excuse for poor performance in operations

4. Failure to forecast the right things. Organizations may forecast the demand for rawmaterials that go into finished products. The demands for raw materials need not be

forecast because theses demands can be computed form the forecasts for the finished

products. Forecasting too many things can be overload the forecasting system and cause

it to be too expensive and time consuming.

5. Failure to select an appropriate forecasting method.6. Failure to track the performance of the forecasting models so that the forecast accuracy

can be improved. The forecasting models can be modified as needed to control the

performance of the forecasts.

Sources of forecasting data

Auto sales Consumer confidence index Consumer price index Durable goods Employment Factory orders Gross domestic product Housing starts Index of leading economic indicators Industrial production


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Merchandise trade Personal income and consumption Producer price index Purchasing price indies

Retail sales

****************THE END**********

Lecturer Notes of Unit -3

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