UNIT II 9 Hours

Layout Planning: Layout Types: Process Layout, Product Layout, Fixed Position
Layout Planning, line balancing, computerized layout planning- overview. CRAFT
Algorithm, ROC algorithm
Forecasting: Principles & methods, moving average, double moving average,
Regression methods, exponential smoothing, double exponential smoothing,
Forecasting error analysis.
PLANT LAYOUT
PROCESS LAYOUT
PRODUCT LAYOUT
GROUP TECHNOLOGY
Advantages. Group technology layout can increase the items given in List A and it can decrease
the 1tems given in List B.

List A
|. Component standardization and rationalization
Reliability of estimates
AR
-o o
Effective machine operation
Producuvity
Costing accuracy
Customer service

Order potential

List B
I. Planning effort
2 . Paper work
3 . Setting time
4. Down time
3 . Work 1n progress
6. Work movement
.
. Dverall production times
8. Finished part stock
9 . Owverall cost

Limitations. This type of layout may not be feasible for all situations. If the product mix is
completely dissimilar, then we may not have meaningful cell formation.
This type of layout is the least important for today’s manufacturing industries. In this type of layout the major
component remain in a fixed location, other materials, parts, tools, machinery man power and other supporting
equipment’s are brought to this location. The major component or body of the product remain in a fixed position
because it is too heavy or too big and as such it is economical and convenient to bring the necessary tools and
equipment’s to work place along with the man power. This type of layout is used in the manufacture of
boilers, hydraulic and steam turbines and ships etc.
Advantages Offered by Fixed Position Layout:
(i) Material movement is reduced
(ii) Capital investment is minimized.
(iii) The task is usually done by gang of operators, hence continuity of operations is ensured
(iv) Production centers are independent of each other. Hence, effective planning and loading can
be made. Thus total production cost will be reduced.
(v) It offers greater flexibility and allows change in product design, product mix and production
volume.
Limitations of Fixed Position Layout:
(i) Highly skilled man power is required.
(ii) Movement of machines equipment’s to production centre may be time consuming.
(iii) Complicated fixtures may be required for positioning of jobs and tools. This may increase
the cost of production.
LAYOUT DESIGN PROCEDURES

Manual methods. Under this category, there are some conventional methods like, travel chart
and Systematic Layout Planning (SLP).

Computerized methods. Under this method, again the layout design procedures can be classified
into constructive type algorithms and improvement type algorithms.
Construction tvpe algorithms

e Automated Layout Design Program (ALDEP)

(Computerized Relationship Layout Planning (CORELAP)
Improvement tvpe algorithms
o Computerized Relative Allocation of Facilittes Techmique (CRAFT).
 Systematic Layout Design Procedure
An organmezed approach w lavoul planning has been developed by Muther and has recerved considerable
publicity due to the success derived from its application in selving a large variety of layout problems.
This approach is referred o as systematic layout planning or simply SLP. This procedure is shown
in Fig. 6.4, From the figure, it 15 clear that once the appropriate information 1s gathered, a flow
analysis can be combined with an activity analysis 10 develop the relanonship diagram. The space-
relaticnship diagram 15 construcied by combining space considerations with the relationship diagram.
Based on the space-relationship diagram, modifying considerations and practical limitations, a number
of alternative layouts are designed and evaluated. In companson with the steps in the design process,
SLFP begins after the problem is [ormulated.

Inpul data and aclivilies

BREEAREE,
1. Flow o 2. Activil
materials 5""'--._ rulal:i-unshgz-s

" ._.,_--'"'___f i
3. Relation ship E
diagrasm <

4. Spacs . 5. EJ:IEEJH
reuirements il svailabds

B. Space
relabonship
cliBgram

. - -
T, Modilying —m — B Frachcal E
consideralions — — hmilalions '
S - iy
L!
9. Develop
|yt
altamatives
____________o e esanaian Salechon

10. Evaluation
 Computerized Helative Allocation of Facilities Technigue (CRAFT)

CEAFT algorithm was originally developed by Armour and Buffa. CEAFT 15 more widely used than
ALDEF and COEELAT. It 15 an improvement algorithm. It stams with an initial lavout and improves
the layout by interchanging the departments pairwise so that the transportation cost is minimized.

The algorithm continues until no further interchanges are possible o reduce the transportation cost.
The result given by CRAFT 15 not optuimum in terms of mimmum ¢ost of transportabion. But the
result will be good and close w opuimum in majority of applications. Hence, CRAFT 15 mainly a
heuristic algonthm. Unformunately, plant lavout problem comes under combinatorial category. 5o,
usage of efficient heuristic like CRAFT 1s inevitable for such prohlem.

CRAFT requirements
1. Imitial layout
2. Flow data
3, Cost per unit distance
4, Total pumber of depanmenis
3, Fixed departments
Number of such departments
Locaton of those departments
f. Area of departments.
 The steps of CRAFT algorithm are summarized below.
Step [, Input: 1. Number of departments
2. Mumber of interchangeable departments
J. Imitial layout
4, Cost matnx
3. Flow matrix
6. Area of depariments
Step 2. Compute centroids of departments in the present layout.

Step 3. Form distance matrix wsing the centrodds,

Step 4. Given data on (low, distance and cost, compute the wial handling cost of the present
layout,
Step 5. Find all the possible pamrwise interchanges of departments based on common border
or equal area cntenon. For each possibility, interchange the comesponding centronds and compulte
approximate cosis,
Step 6. Find the pair of departments corresponding to the minimum handling cost from among
all the possible pairs of interchanges.
Step 7. Is the cost in the previous step less than the total cost of the present layout? If ves,
go w Step 8. If not, go w Step 11,

Step 8 Interchange the selected pair of departments. Call this as the NEW LAYOUT. Compute
centroids, distance matrix and total cost.

Step %. Is the cost of new layout less than the cost of the present layout?

If yes, go to Step 10, If not, go o Step 11,

Step [0, The new layout 15 here after considered as the PRESENT LAYOUT. Its data on
centronds, layoul matnix and the wial cost 1% retamned. Go w Step 5.
Step 11, Primt the present layout as the FINAL LAYOUT,
Step 12, Stop.
APPLICATION OF CRAFT
 In thas table, cell values represent the flow in terms of number of rps In & given penod of
tme from Depanment § o Department j

Area of Departments

[department I 2 A 4 b

Are)
{5q. units) 16 16 24 16 5

Mep 2. Centroids of all the departments in the initial layout are calculaled and presented
below, Here, the left side of the layout 15 assumed as ¥ axis and the botom side of the layoul is
pssumed as the X axis.
(X,. ¥)= 2.6
(X Fai= 2,2
(X3 ¥a) = 7, 2
(X, ¥ = 8. 6
Xy, ¥l =5, 6
afep 3, The dislance between any two depariments 15 represented by rechlmear distance between
the centroads of the two departments.
d;=IX;
- X} + ¥, - ¥}
where (X, ¥} and (X, ¥} ore the centrowds of the Departments @ and j, respectively.
The distance matrix 15 shown below

Dristance Marrix [d;]

From/To I 2 3 4 5
| - R Q9 b 3
2 Rl 5 11 1
3 i 5 i1
4 b 1) > - A
3 3 7 i 3 -

Sep 4. Total cost of handling for the present lavout is calculated.

2 13
Towal cost = T E fiy = dy = o
dall =l

where
fii= Flow Trom Deparimem¢ 1o the Department §.
dyy = Distance from Department ¢ 1© the Depanment j.
¢y = CostfUnit distance of travel/tnp.
 Total Cost Matnx [TC,1

FromsTo | 2 3 R 3
I - 21] I8 14 L]
. U 1] ) [
3 |& i = 0 all
- |& (] 3 - 1
3 0 (] 12 0 -

Total cost = 205

Step 5. Consider various departmental interchanges for improvement. Departmental interchanges

that are possible are given below.
* Departments having common border.
¢« Departments having equal area.
Parrwise interchanges are considered, If there are & departments, theoretically, WCs pairwise
mmierchanges are possible. For the present problem ot this stage, eight imerchanges are possible,

Fair ol Departmeants Remark

1 and 2 Interchange based on common border

| and 3 Mol possible
1 and 4 Interchange based on agual area
1 and 5 Interchange based an common border
£ and 3 Inferchange based on common border
2 and 4 Interchange based on agual area
2and 5 Mol possible
3 and 4 Interchange based an common border
4 and & Inferchange based on comman border
4 and 5 Interchange based on common border

Fig. 6.6 N, pairwise imterchanges.

For the purpose of cosl calculation, an interchange between two depariments would mean thar
their present centroids are interchanged,
For each interchange, the pssocipted distance matnx 1= calculated. Then, the total cost of
handling 15 calcalated.

Interchange Detween T and 2, Centrmds after interchange:

(X, ¥p=212
(X5 ¥opm 2,6
Xy ¥oh =7, 2
(X Y50 =8, 6
{xjl- I'..:IJ' = :i'l 'E'
 Diistance Matrix

FromiT o | 2 3 5

| - | 3 | (] 1
2 il L 6 1
1 3 ‘| —~ 3 6
4 1{) A 3 - 3
5 7 3 fy 3 -

Total cost = 205

Interchange Detween T and 4. Centroids after interchange:
{Ep i'r|} = fl. 'fi

{x]r rj} = .EI- 2

(X5, =T, 2

'[-!fq- Fq]' =216

{.’I.';, I'rj} | :|'| 'El

Distance MMairix

FromuT o | 2 3 R 5

| - | 1] 3 6 3
2 1) 5 4 7
1 3 3 —~ 0 6
4 b - L - 3
5 3 7 [y 3

Total cost = 193

Imterchange between 1T and 5, Centronds after interchange:

I:Ji’|. ]".|:| = 5, E:l

(X, Y;i= 8 6
(X, Yol = 2.6

Distance MMatrix

FromuTo | 2 3 | 5

] I b h J
s T — 5 [ -
3 fy 5 - 5 g
Rl 3 | () 3 - £
5 3 4 ] I

Total cost = 208

Imerchange between 2 and 3. Centroids after interchange:
(X, Fil = 2.6

(&3, Fal = 7, 2

(X, Fil =B 6O

Distance Matrix

FromuTo | 2 3 | 5

[ Ly d Ly 1
2 B - 3 3 0
3 4 5 1) T
| 4 5 1M - 3
) 3 5 7 3

Total cost = 197

Interchange between 2 and 4. Centroids after interchange:
'[.:I|:I||. I'r|_|' = 2. 'El

(X., =8, 6

Xy, K30 =T, 2

{.’l’., I'rqb | I. I

{xjl- rj} = 5‘| B

Distance Mairix

FromuTo | 2 & R 5

| - £ Ll | 3
2 Iy A 10 3
4 L 5 - 3 0
4 - 100 3 - 7
5 3 3 fy 7 -

Total cost = 20

Interchange between 3 and 4, Centroids after interchange:

(X, Fil=2.6
(Xs Vol = 2,2
(X, Vi1 =86
(Xy. Yol = 7.2
(Xe, Fs1 =356
 Diistance Motrix

From'T o | 2 3 4 2
| - g fy g 3
) 4 I 3 7
3 f 14] —~ > i
4 14 5 5 E 6
5 3 7 3 6 -

Towal cost = |78

Interchange between 3 and 5. Centroids afier interchange:
{.’I.',, I'r|f i 1. 'El

(X Yopm 2 2
(X3, Fip=3, 06
{-'f-l- Fq} =8 6
{xfil- I'..:I} = -|r| z

Distance Mairix

FromiT o | 2 3 5

| B g 3 6 il
7 4 T |} 5
1 5 1 —~ 5 0
4 b 11l 3 E 5
5 + 5 fy 3 -

Toral cost = 183

Inmterchange between 4 and 5. Centronds alter interchange:
[JI[,.. ]".|:| _ E.. fi

Distance Mairix

From'T o | . K 4 5
I - - f 3 6
2z | - 5 7 [1]
1 i § £ 5
dq 3 Y, fy = 3
5 b 11l 5 3 -

Total cost = 163

 FPairwiga Interchange Ceoal

1 and 2 2056
1 and 4 193
1 and 5 208
2 and 3 187
2 and 4 201
d and 4 178
3 and 5§ 183
4 and 5 163

Fig. 6.7 Summary ol approximate total cost due o pairwise mterchange of centrgids.

Step6. The mterchange which promises mimimum hamdling cost 15 selected for actual interchange
in the layouat.
The imerchange between 4 and 3 resulis into minimum cost of 163,

Step 7. This cost 15 compared with the cost of the present layout, Only i the cost due o
proposed interchange 15 less than the present layout cost, the mmterchange 1= & be actually made,
Otherwise go o Step 1.
In the problem considered, the approximate handling cost of 1623 is less than the present lavour
cost of 205. Hence, go 1w Step 8.
Step & Interchange can be made between 4 and 5. The lavout afier making actual interchange
between Departments 4 and 3 15 shown in Fig, 6.8 This s called Mew Lavout.

] 4 2

1 4 = q

2 3 4

F. =]

Fig. 6.8 New lavoul

The centronds are

(X, ¥} = 2,6

(X, ¥y} = 7.2

(X Fal =6, 6
(Xs, F5) =90
 Dhistance Maimnx

FromuTo 2 3

pE
A g

e e
etk

S
e |
3
5 =

e S

T
i 3
| ] [

Lad
—
Total cost = 181

Step 9. The cost of the New Layoul 13 compared with the cost of the Present Layoul.
If the cost of the Mew Lavout 15 less than the cost of the Present Loyowt, then treat the New
Layout as the Present Layout. Then go to Step 3. Otherwise 2o 1o Step 11.
In the problem, the new layout cost is |81 which is less than the present layout cost of 203.
Hence, trean the new layoul as the present layoul and go 10 Swep 5.
Step 5. Applving the rule of common border or equal area, all the possible puirwise miterchanges
are considered.
For each possibility, the centroids are interchanged and the resulting distance matrix and the
il cost of handling are computed. The results are given in Fig. 6.9,

Intarchangs Fair Interchange Rule Taotal Gosi

1 and 2 Comman border 181

1 and 3 - =
1 and 4 Common border | G5
1 and 5 - -
2 and 3 Comman border 208
2 and 4 Equal area 181
2 and 5 - -
3 and 4 Common border 166
3 and 5 Common border 1B3
4 and 5 Hecenlly inter- -
changed. So
maad not ba
considarad

Fig. 6.9 Towal cost due 1o pairwise interchange.

Step & In the previous step, 1686 15 the minimum total cost. Corresponding pair 1s 3
and 4.

Step 7. Thiz minimuam cost of 166 1= less than the present layout cost of 181, Hence go o
Step 8,
Step 8. Interchange the selected pair of Departments 3 and 4. Call this as the New Layout. This
i3 shown in Fig. &.10.
 1 5 =
K

2 o
. 4

4 i
Fig. 6,10 MNew layout,

The centrosds are

(X, ¥)=26
(X, F)=22
(X, Fy)mb, 5
(X, ¥ )=75 15
(X, ¥ =9 6
wate, The formuls 1© compute centrond for wrregolar shaped depariment 15 given below;

J_fl. = X-coordinate of the centroid of the ith porion of the depariment.

¥ = Fcoordinate of the centroid of the ith porton of the departiment.
a; = Area of the ith portion.
r = Mumber of regular shaped portions in the departmeni.
If the shape of a particular department is irregular, then that is divided into several regular
shaped portions, Then the following formulas can be applhied 10 compute the X- and Y-coordinates
of the centrond of the given depariment,

f =u|E] +a,
K, +---+ak +--+a
XA
gl +H:.+"'+'fl__+"'+fl.

o_afitaf+otaf +osalf
a4, +dd o da
This concept 15 demonstraied vwsing Department 4 1m0 Fig. 6.11.
 In Fig. &.11, the Department 4 15 divided into two regular shaped departments, P, and P,

= axb4+mnxl
¥ = =T.5
! R+8
. 5 . B
= =1.5
! B+8
(X,.¥1=(7515
Dietancg MMatrix

Fromd'T o I 3 3 4 9

I -~ - 3 | £} 7
2 - - () i | ]
3 5 fi - 5 4
il 1) i 5 f
9 T 11 4 o _

Towal cost = 187

Step ¥ The cost of the New Layout 15 compared with the cost of the Present Layout.

Since, the cost of the Mew Layout (187) 18 not less than the cost of the Present Layouat {181}, go o
Step 11,

Step 10, Print the present layout as the final layout,

- S 2

E d = i

c & &

4 b
Fig. 6.12 Final layoul.

Total cost = |81

 Rank Order Clustering Algorithm (ROC)

This algorithm was developed by LR, King (19800, This algorithm considers the following data.

« Number of components.

= LOomponenl sequences,

This algorithm 15 used to obtain machine-component cells in group technology.

Based on the component sequences, o machine-component mcidence matrix 15 developed. The

rows of the machine-component incidence matrix represent the machines which are required to
process the components. The columns of the matrix represent the component numbers.
Let ws treat thas madrix as [C ] This matrix containg U=1 entries, If the sequence of component
J has the machine number ¢, then IHH respective matnx ¢lement -I':'Lr 15 assumed s 1; otherwiss, it s
assumed as 1.
Each row and each ¢olumn of the matrix are considered as binary words, For example, in a
row, 11 we have numbers (1 0 10 1), then the decimal equivalent 15 computed as Tollows:
Row decimal equivalent = 1 2 2" 0= 27 4 1 2 2 4+ 0w 2 & 1 x 27
=16+0+44+04+1=21
Simularly, i a column has the following entries from wop 10 bottom, the decimal egquivalent 15
compuied as explaned below:
Column enines = (] 1 0 | 0

Decimal equivalent e M s 12 +0 2+ 1 w2 4020

6+ E+0+2+0=2A
The row (column) with the largest decimal equivalent 15 considered to have the highest rank
| among the rows. Similarly, the column with the largest decimal eguivalent 1s considered o have
the highest rank among the columns, The procedure 1 obtain final machine component incidence
matrx 15 summarized below,
Steps in ROC. The steps in ROC algorithm are given below.
Step O Input: Total number of components and component sequences
step 1. Form the machine-component incsience matnx wsing the component sequences.
Step 2. Compute binary eguivalent of ¢ach row.
Step 3. Rearrange the rows of the matnx in rankwise (high 0 low rom wp 0 bottom).

Step 4, Compute binary equivalent of each ¢column and check whether the columns of the
matrnx are arranged in rankwise (high o kow from left 1o nght)? If not, go o Step 3; otherwise, 2o
o Step 7.

Step 5. Rearrange the columns of the matnx rankwise and compute the hinary equivalent of
ench row,

Step &, Check whether the rows of the matrix are arranged rankwise? If not go w Step 3;
otherwize, go o Step 7.

Ktep 7. Print the final machine-component incidence matrix.,

Application of AOC algonthm

Ktep 0, Number of components = 10
 Component sequences arg given below,

Component
MNumber Sequence
| | -2-3-4-2-5-6-3-4-1
2 13 - T =14 -15-14
- 15
3 B =12 =8 =11
4 P f-5-4-8
a 14 - 15 -142
£ O-10D-8-11-12
T T =14 = 15
£ | c3-h-2-5_-4_-8_4
4 T -13-14
| (3 9 -8 =10

Setep 1. The inital machine-component incidence matnx is shown below.

Compoenent ()

| 2 3 = 3 & 7 # 0 | ()

1 | [ 0 ( { (} 0 | [ 0
2 | [ 0 ( 0 (b 0 I [ 0
3 | 1 0 ] { ( o I 1 i
= | 1 0 | 0 0 0 | 1 {
3 I 1 0 1 { () 0 I [ 0
b | [l 0 1 0 (b 0 I [l 0
7 i ] 0 (] 'L'I Ch ] (b ] i
Machine i 5 | 1 | 1 0 | 0 I 1 |
9 | [ 0 ( 0 I 0 () [ |
1 | [l 0 (l 0 1 0 {0 [ |
11 0 1 | ( 0 1 0 0 [ 0
12 {0 1 | (] 0 | 0 (b 1 {0
13 0 | 0 (] 0 ( o () l {
14 | 1 0 ( } (b ] 0 1 0
15 0 ] 0 ( | (b ] 0 [ 0

Step 2. Compute binary eguivalent of each row.

Eow No. | 2 o4 5 6 7 B 9 1 11 12 13 14 15
Binary
Egquivalent 5316 316 380 5RO 380 580 266 725 17 17 144 144 258 I08 296

Rank Mo. 6 7 2 & 4 5 10 1l 14 15 12 13 11 5 4

Step 3. Rearrange the rows of the matnix rankwise (high to low from top 1o bottom). The
rowwise rearranged machine-component incidence matrix as follows:
 Component {J)

L]

o T e T e B e R eT e

sg eg e

e e
—

—_——— ==

— 0
I
A

0o

= aor o R = e
=
SR =

oo
-
=

B = =
(3

- = =2
Machine | Id
15

= B3 = = —
L
)

= = =

= = =

eR
E—
m—
I

Step 4. The binary eguivalent of each column s computed and the ranks are summarized in
the following maltrix.,

Component {f)

Ll

Lo I e
R

o B
k- -E-E-N-

e T o B

el =
eB e
s U

e
s

==
B

o
o o oo

e B e e
Y P

e I

o [ s
v [

= =
B
e I e |
i

= e e A =3 =
Machine | 14
L

II.“.

= =
15
g -

¥
L3 -

= = = e
=
—_

= =

= =

= O

fl.l—
Il
Fank 10

ainee the arrangement of the columns 15 not rankwise, po o Step 3,

Step 5. The columns of the matrix are rearranged rankwise as shown below. Then the ranks
of the rows are also summarized in the same table.
 Component { )

9 Rank

s Y e
P — -

s~

= = e ==

o
e B B e B e IR
-

D e 23

Lh e e P
e
e Y .
e

Y e

s
o

o I
—

e
Machine ¢ 14 0 |2
|5 |3
|4
| 3

= e

-e —

OO =
s Y
| 1

b Y e eB [

eB [
e
|2

R
0
- [ o Ys
e

e
b Y o
= =

=0
e
1}

—
o

o
Srep &, Smmee the mows of the matnx 1n the previous step are not amanged mnkwise, go o
Step 3
Step 3. The rows of matnix in Step 5 are rearranged rankwise as shown below,

Component | )
DD

p—
e

p—
e L

- e

e B
. Y .Y

eB e eY

e Y
e o i
el

e——
G

el

T
—
Fod sn O

R o
e

sO

eR
—

3 =
G o
hMachine & I
-

—_ D D B i3

12
o Y o

OO i
L
i
—

— = =

— = e
10
o B B
i
s[ o

L4
-

=
T

15

15
Rank |

Step 4. The columns of the matrix in the previous iteration are given rankwise. Hence, go o
Step 7,
Step 7. The final machine-component incidence matnix is presented in the following matrix,
 Final Machine-Component Incidence Motrix,

Component { )

| 5 4 N L 10 2 T 5 ':?
! I ] | | | | { ] [ [
3 | I | (] {) i} {l ] [ ()
i | ] | L} 0 i 0 () [} 0
5 | 1 | [l { () 0 i 1 i
fi I ] | (l { (h { i, [l {
] | I { (] () ] {l (b [ ()
2 | 1 0 L] 0 i { () 1 0
1 () [l { 1 | (} { () [l i
Machine¢ 12 i [l {0 | | i { i Ll b
0 i, 1 { ] {) | {l i, 1 ()
I0 i 1 { I 0 I 0 i [ {
Id i, 1 { (] { (} 1 I 1 |
15 i} [l {) (l { (} ] | 1 )
7 ] 1 { (l { ] ] I [ |
13 [ 1 {) (] {] [} ] [ [ |

Kank | 2 4 4 % & 7 8 L | ()

There are three machine-component cells. In the above matrix, there are some exceptional
elements, 2. machine ¥ 18 required in cell | and cell 2. If we get machine-component cells formation
withoul any overlapping on machine regquirement, then the formaton 15 said 1w have perfect block
chagonal form.
Nare, UOmne can assumie a final machine-component incidence matnx with perfect block diagonal
form. Then it may be disturbed suitably o obtain an initial machine-component form for RO
algorithm and again one will get the machine-component cell formaton with perfect block diagonal
form on application of ROC algorithm. This is left a5 an exercise,
Based on the above matrix, the machine-component groupimgs arg given in the following table.

Table 6.2 Machine-Component Groupings

Croup No, Machine Cluster Component Family

| 1.2, 5 4, 5 6and B |, & and4
8.9, 10, 11 and 12 3,6 and 10
T. 13 14 and 15 2,5, 7 and Y
 LINE BALANCING

Machine Capacity Machine Capacity Machine Capacity

20 Pcs/ Hour 10 Pcs/ Hour 5 Pcs/ Hour

20 °
10Pce —-15 Pcs
Pcs/Hr 10 -
5 -
 Machine Capacity
5 Pcs/ Hour

Machine Capacity
5 Pcs/ Hour
Machine Capacity
10 Pcs/ Hour

Machine Capacity
5 Pcs/ Hour
Machine Capacity
Machine Capacity 10 Pes/ Hour
20 Pcs/ Hour
Machine Capacity
5 Pes/ Hour

20-
Pcs/ Hr
10-
- s
 » “Line Balancing” in a layout means arrangement of machine
capacity to secure relatively uniform flow at capacity
operation.
» It can also be said as "a layout which has equal operating
times at the successive operations in the process as a
whole”.

Line balancing operates under two circumstances:

Precedence Constraint: Products cannot progress to other

station if it doesn’'t complete necessary task at that station, It
should not across other station because certain part needs to
be performed before other activities.
Cycle time Restriction: Cycle time is maximum time for
products spend in every workstation. Different workstation has
different cycle time.
OBJECTIVE OF LINE BALANCING

Manage the workloads among assemblers

Y
PO
Decide number of workstation
Decrease production cost.
W

Assigning task to each work station in such a way that there

e

s little idle time

BALANCED LINE UNBALANCED LINE

» Promotes one piece flow » High work load in some stages

> Avoids excessive work load In
» Maximizes wastes (over-
some stages
processing , waiting, rework,
» Minimizes wastes (over- transportation, motion)
processing, inventory, waiting,
rework, transportation, motion) High variation in output
..

» Reduces vanation Restrict one piece flow

» Increased Efficiency
Maximizes Idle time
..

» Minimizes Idle time

Poor efficiency
IMPORTANT TERMINOLOGIES

» TASK PRECEDENCE: The sequence In which Tasks are performed.

» CYCLE TIME: The time expressed in minutes between two simultaneous
products coming off the end of a production line.
» PRODUCTIVE TIME PER HOUR: The average number of minutes a
workstation is working in an hour.
» WORKSTATION: A physical area where a worker with tools/ one or more
machines. or an unattended machines like a robot performs a particular
set of task in a production line.
» WORK CENTER: A small group of identical workstations, where each
workstation performs the same set of task.
» NUMBER OF WORKSTATIONS WORKING: The amount of work done at a
work center expressed in number of workstations.
» MINIMUM NUMBER OF WORKSTATIONS: The least number of
workstation that provides the required production,
» ACTUAL NUMBER OF WORKSTATIONS: This is the total number of
workstations required on the entire production line. It is calculated as the
next higher integer of the number of workstations working.
» UTILIZATION: The percentage of time a production line is working

LINE BALANCING PROCESS

1. Draw and label a precedence diagram.

2. Calculate the desired cycle tUme required for the line.
3. Calculate the theoretical minimum number of
workstations.
4. Group elements Into workstations, recognizing cycle
time and precedence constraints,
5. Calculate the efficiency of the line.
6. Stop If theoretical minimum number of workstations on
an acceptable efficiency level reached. If not, go back
to step 4,
 STEPS IN SOLVING LINE BALANCING

Drawing Precedence Diaaram: Precegence diagram needs to be

drawn to demonstrate a relationship between warkstations.
Certain process oegins when Previcus process was aone

Determining Cycle Time: Cycle time is longest time allowed at each

station. This can be expressed by this formula

Available time
Cycle time =
Desired output

This means the products needs to leave the workstations before it

reaches its cycle time
Assigning
tasks to workstation: The tasks distributions should be
taken after completing a time cycle. It's good to allocate tasks to
workstation in the order of longest task times

Y Task Time
Number of work Stations =
Desired Actual Time

» Calculating an Efficiency Line: This is done to find effectiveness of

the line. The formula is given by:

Sum of task times

Line Efficiency =
Number of workstanon
X Desired cycle time
LIMITATION OF LINE BALANCING

belts paced the

» Froduction lines were designed <o that conveyaor
gemen t wasn 't
speed of the employees work. This arran
appreciated by the empioyees
Oul of balance.
» Inevitable changes lead to production Tineés being
» Rebalancing causes disruptions 1o production
7.3 RANK POSITIONAL WEIGHT METHOD
The ranked positional weight method was developed by Helgeson and Brinmie. This method assigns
those jobs first whose followers have the largest total time. The positional weight of a work element
15 115 own processing time plus the processing tumes of all the following work elements. In RPW,
as stated earlier, the work element with the highest positional weight 15 selected and assigned to the
current station. The generalized algorithm stated earher can be used for rank positional weight

method as such without any modification by using the maximum rank positional weight as the
selection criterion in Step 11 for selection of a work element if we have a tie in List A.
The application of the whole procedure to a problem 1s summarized in the form of a table.

Example 7.1. Consider the assembly network shown in Fig. 7.2 which shows the precedence
relationships 1n assembling a product.

4
5 et
Fig. 7.2 Assembly network of a product. (The nodes from 1-10 represent the real tasks of the
product, while the number by the side of each node represents 1ts duration.)

The number by the side of each node represents the processing time in minutes. The required
production volume in 8-hour shift 15 24 completed assemblies. Design an assembly line using RPW
method.
Solution.
Production time
Cycle time (CT) =
Production volume
~ 8x60
= 20 minutes.
24
 iy|

Compute the positonal weight of all the work elements in the network as shown in Fig. 7.3
The same details are also shown in Table 7.1.
A work element can be included in the List A if 1t can satisfy the following two conditions:
1. All its immediate predecessors are already assigned to some stations.
2. Its processing time 15 less than or equal to the UACT.
The calculations are summarized in Table 7.2.

Balancing efficiency = [1 _ SUACT ] x 100

CT = 5N

=[1 — w 100 =93,33%

20x3
 Fig. 7.3 Precedence diagram with positional weights.

Table 7.1 Details of Work Elements

Work Element Positional Immediate

Number Time Weight Predecessors
56
=

16
b

3l
e L

FUNs Y s

31
21
Wh

18
=0 Ch

13
S

15 =
00

el s a

17
= WD

9
e

Table 7.2 Details of Stations

 B ERLFER, ol Ll
T R T R S LR
R[

Station Work Element Unassigned

Number List A Selected Cycle Time (UACT) Remarks
| — - 20 —~
1 1 15 -
2,3, 4 4 10 Break tie between 3 and 4 randomly
2,39 3 3 -
2.6 6 0 Change over to next station
2 - - 20) -
2.5 09 5 14 —
2. 8,09 9 6 -
2, 8 2 3 -
- 3 Change over to next station
3 - - 20 -
7, 8 5 14 —
7 7 10 —
10 10 1 E
A forecast 15 an estimate of an event which will happen n future. The event may be demand of a
product, rainfall at a particular place. populanon of a country, or growth of a technology. The
forecast value 15 not a deterministic quantity. Since, i1t 15 only an estumate hased on the past data
related to a particular event, proper care must be given in estimating it.
In any industrial enterprise, forecasting 1s the first level decision actuvity. That 1s the
demand of a particular product must be available before taking up any other decision problems like,
materials planning, scheduling, type of production system (Mass or batch production) to be
implemented, etc.
S0, forecasting provides a basis for coordination of plans for activities in various parts of a
company. All the functional managers in any organization will base their decisions on the forecast
value. So, 1t 15 a vital information for the organization. Due to these reasons, proper care should be
exercised while esumating forecast values.,
In business, forecasts may be classified into technology forecasts, economic forecasts and
demand forecasts.
Technology forecast. Technology 1s a combination of hardware and software. Hardware 15 any
physical product while software i1s the know-how, technique or procedure. Technology forecast
deals with certain characteristics such as level of technical performance, rate of technological
advances.
Technological forecast 1s a prediction of the future characteristics of useful machines, products,
process, procedures or techniques. Based on the importance of this activity, Government of India has
established a “Technology Information Forecasting and Assessment Council (TIFAC)”, under the
Ministry of Science and Technology to promote action oriented studies and forecasting in selected
dreds,

Economic forecasts. Government agencies and other organizations involve in collecting data
and prediction of estimate on the general business environment. These will be useful to government
agencies in predicting future tax revenues, level of business growth, level of employment, level of
inflation, etc. Also, these will be useful to business circles to plan their future actuvities based on the
level of business growth.

Demand forecast. The demand forecast gives the expected level of demand for goods or services.
This 15 the basic input for business planning and control. Hence, the decisions for all the functions
of any corporate house are mmfluenced by the demand forecast.
The factors affecting forecast are given below:
Business cycle
e Random variation
Customer’s plan
e Product’s life cycle
e Competition’s efforts and prices
« Customer’s confidence and attitude
e Quality
Credit policy
¢ Design of goods or services
¢ Reputation for service
e Sales effort
o Advertising
Forecasting in different functional areas of management such as Marketing, Production, Finance and
Personnel play a crucial role for planming ahead. The wvarious types of forecast i each of the
aforementioned areas are as follows:
Marketing
* Demand forecasting of products
e Forecast of market share
» Forecasting trend in prices
Production

Forecast of
e Materials requirements
Trends in material and labour costs
« Mamntenance requirements
e Plant capacity
Finance

Forecast of
o Cash flows
Rates ol expenses
e Revenues
Personnel
Forecast of
« Number of workers in each category
« Labour turn over
¢ Absenteeism
Forecasting 15 based on the pattern of events in the past. A pattern may solely exist as a function of
time. Such a pattern can be identified directly from historical data. Another pattern consists of
relationship between two or more variables. 50 it 15 important 0 understand the most common
demand patterns.

Historical pattern (Stationary pattern). This exists when there is no trend in data and when
the mean value does not change over time.

VARV VYA

Fig. 4.1 Historical pattern (Stationary pattern).

Products with stable sales, number of defective items from a stable production process are
some of the examples.
Seasonal demand pattern. This demand pattern exists when the series fluctuates according to
some seasonal factor. The season may be months, quarters, weeks, etc. sale of refrigerators, sale of
soft drink, sale of wool items, elc.

Forecast T
variable
\/

Time
Fig. 4.2 Seasonal demand pattern.
Cyclical pattern. In this type of pattern, the length of a single cycle is longer than a year. The cycle
does not repeat at constant intervals of time.
The best examples are the prices of some metals, gross national product, etc.
A

Forecast L=
varnable #

Time

Fig. 4.3 Cyclical pattern.

Trend pattern. This type of pattern exists when there is an increase or decrease in the value of
the variable over ume. The examples are sales of many products, stock prices, business and economic
indicators.

Forecasl
varibale

Time
Fig. 4.4 Trend pattern.
4.4 FORECASTING MODELS
The forecasting techniques can be classified into gualitative technigues and quantitative techniques.
These are presented helow.
Qualitative techniques use subjective approaches. These are useful where no data 1s available
and are useful for new products. Quantitative techniques are based on historical data. These are more
accurate and computers can be used to speed up the process.

Quantitative forecasting techniques

« Simple moving average
e Single exponential smoothing
* Double moving average
* Double exponential smoothing
e Simple regression
¢ Semi-average method
Multiple regression
¢ Box Jenkins

Qualitative forecasting techniques

¢« Delphi type method
o Market surveys
4.4.2 Measures of Forecast Accuracy
Demand forecast influences most of the decisions in all the functions. Hence, 1t must be estimated
with the highest level of precision. Some common measures are inevitable to measure the accuracy
of a forecasting techmigque. This measure may be an aggregate error (deviation) of the forecast values
from the actual demands. The different types of errors which are generally computed are as
presented below.

I. Mean Absolute Deviation (MAD)

2. Mean Square Error (MSE)

3. Mean Forecast Error (MFE)

4. Mean Absolute Percent Error (MAPE)

The formula for error i1s given below.

£, = 'D.' - F:

where
Dy, = Demand for the period &
F, = Forecast demand for the period f, and
e, = Forecast error for the period .

Mean Absolute Deviation (MAD). 1t is the mean of absolute deviations of forecast demands
from actual demand values. The MAD i1s sometimes called as the mean absolute error (MAE).

SID
—F |
MAD ==
1
where
D, = Actual demand for the period 1
F, = Forecast demand for the period ¢
n = Number of time period used.
Mean Square Error (MSE). Mean square error is the mean of the squares of the deviations of
the forecast demands from the actual demand values. Usually the effects on operations of small
errors are not serious. These errors may be smoothed out by inventory or overtime work. It will be
difficult to have smoothed values for forecast even if there are few large errors. Consequently, a
method of measuring errors that penalizes large errors more than small errors 15 sometime desired.
The mean square error (MSE) provides this type of measure of forecast error.

(D - F;}I
MSE — r=|

where
D, = Actual demand for the period 1
F, = Forecast demand for the period ¢
n = Total number of errors used.

Mean Forecast Error (MFE). Mean forecast error (MFE) is the mean of the deviations of the
forecast demands from the actual demands,

where
D, = Actual demand for the period ¢
F, = Forecast demand for the period «
n = Number of years (time period) used.
Mean Absolute Percentage Error (MAPE). Mean absolute percentage error (MAPE) 15 the
mean of the percent deviations of the forecast demands from the actual demands.
1 = 1D —F|
MAPE =— X » 100
=l
I

Suppose that a forecast of 165 units had been made for the demand in every period for the data given
in Table 4.1. The calculations of these errors are as shown in the last two columns of Table 4.1.

Table 4.1 Calculation of YVarious Errors

Absolute Squared Percentage Absolute

Demand Forecast Deviation Dewviation Error Error %t Error
D -F D = F |
T D, F. D,-F, |D,-F,| (D,-F) 2 ’
D, — = 100 '
1D |
— » 1(0)

] 150) 165 —15 15 225 — 10,000 10,000

2 16 1635 -3 3 25 — 3125 3.125
3 165 165 0 0 0 0.000 0L OG0
4 175 165 +110) 10) 130 5.710 5.710
5 |8 163 +15 15 225 5. 330 B. 330

k) + 5 45 575 27.165

MAD = E =4
5

375
MSE=——=113
3

MAPE = 2?;55 =3.433%

MFE=+?5=+I
4.4.3 Simple Moving Average Method
A simple moving average 15 a method of computing the average of a specified number of the most
recent data values 1n a series.
The formula w compute the simple moving average (SMA) 15 as lollows.
l
M, = {D:—{n—l:l + Dr—[n—’-'_] + -+ DJ'—E + D.l—] + Dr]
I

where
M, = Simple moving average at the end of period r (It 18 to be used as a forecast for period
I+ 1).
D, = Actual demand in period 1.
H MNumhber of neriods included 1n each averave.

Table 4.2 Three Months Moving Averages

Time Demand Moving Forecast Error

(Month) () for Month (¢) Average M(t) F, €,

l 93
2 100
3 87 94.00
4 123 [003.33 94.00 29.00
3 90 100,00 103,00 ~13.00
6 96 103.00 100.00 — 4.00
7 [ &7.00 103.00 —28.00
8 78 83.00 87.00 ~ 9.00
9 106 86.33 83.00 23.00
10 104 96.00 56.33 17.67
11 89 99.67 96.00 =7.00
12 83 92.00 99.67 —-16.67

Mean Forecast Error (MFE) = —().889

Mean Absolute Deviation (MAD) = 16.37
The values printed under column M(r) are obtained as:
Moving Average Period (n) = 3
M (3) = (95 + 100 + 87)/3 = 04
M (d) = (100 + 87 + 123)/3 = 103.33

The forecast for period 13 would be:

Fla= M, =91
Comments:
1. In most cases, this method 15 applied 1o forecast for only one period into the future.
2. The forecaster must wait until demand entries are available for making the first forecast.
3. This method 1s applicable only for horizontal demand data patterns.
4 . This model requires to store the first n observed values, which consumes considerable
amount of computer storage space 1f the demands of several products are to be forecasted.
5. Greater smoothing effect could be obtained by including more observations in the moving
average.
4.4.4 Weighted Moving Average Method
Equal weights were assigned to all periods in the computation of the simple moving average. The
welghted moving average assigns more weight to some demand values (usually the more recent
ones) than to others, Table 4.3 shows the computation for a three months weighted moving average
with a weight of (0.5 assigned to the most recent demand value, a weight of (.30 assigned to the next
most recent value and a weight of (.20 assigned to the oldest of the demand value included in the
dVEIdZC.

Table 4.3 Three Months Weighted Moving Averages

Time Demand Moving Forecast

(Month) (1) D, Average M(1) F,
1 120
2 130
3 110 118
1 1 40) |29 I1s
3 110 119 129
6 130 126 119

Weight MA, = 02x120+0.3x130 +0.5x110 _ 118

| 0.2+03+05

Weight MA, = =
IW,
=]

where i = 1, 2, 3 1f we use three periods moving averages. [ = 3 corresponds to the most recent time
period and § = 1 corresponds to the oldest time period.
W, = Weight for the time period 1.
In the example, W, = (0.2, W, = 0.3 and W; = (.5
4.4.5 Double Moving Average Method
Maodel:
Let the moving average period = n
First single moving average = M(n)

(D, +D, +--+D,)

B n

Subsequent single moving average for any period = M\|(1)

[DWr) — Dt —n)])
H

First double moving average Ms(2n - 1)

ML +Mi(2)+--+M(2n-1)
fn

Subsequent double moving average M,(1)

N M (t)—M
(1t —n)
M,(t—1)
"

Forecast for any future period £ from period #(F,

7)) = 2M (1) — M) + (2Z0(n = 1)) [M (1) — M.(1)]
This 15 demonstrated 1n Table 4.4.
 Table 4.4 Double Moving Averages
Time Demand
(Month)(r) INi) M (1) M,(1) F(r) Error

1 60
2 70
3 B 71.66
4 60 71.66
3 B 77.66 73.66
6 68 72.00 73,77 B5.60 ~17.66
7 106 B7.33 79.00 68.44 37.56
B 75 83.00 80.77 104.00 —20.00
49 86 89.00 ®H.44 B7.44 - 1.44
10) 124 05.00 ®9.00 04.11 29.89
11 122 110.66 08.22 107.00 15.00)
12 87 111.00 1()5.55 135.55 —48.55
Explanation of solution.
n= 23
M(3) = (60 + 70 + 85)/3
71.66
M(4) M(3) + (Di4) — D(1))/3
71.66 + (60 - 60)/3
71.66
(Other single (simple) moving averages are calculated, in the similar way.
First double moving average = M.(2n - 1) = M,(5)
M,(5) = (71.66 + 71.66 + 77.66)/3 = 73.66
M3(6) = My(5) + |M,(6) — M(3)1/3
= 73.60 + (72 = 71.66)/3 = 73.77
New forecast for period say 10, made as of period 9 is
FI9 + 1) =2 % M{9) - My9) + {2 % 113 - 1) * [M(9) - My9)]}
2 x 89 - Bodd + (1 * 203 = 1)) (89 - BH.44)
178 — B6.44 + 2.56
Fr10) 9412
Comments:
1. This method 1s applicable to wrend data patterns.
2. The forecaster must have Zn data points to find a forecast and thus substantial data storage
15 needed.
3. This method can be used to forecast for any number of periods into the future.
4.4.6 Simple (Single) Exponential Smoothing Method
Another form of weighted moving average 15 the exponential smoothed average. This method keeps
a running average of demand and adjusts 1t for each period in proportion to the difference between
the latest actual demand figure and the latest value of the average.
FJ- = F.I'—l + (X r.:fl_] - Fl'—l::l
where
F, = Smoothed average forecast for period t.
F, ) = Previous perniod forecast.
¢ = Smoothing constant, weight given to previous data () < @ < 1).
D,, = Previous period demand.

If 15 equal to 1, then the latest forecast would be egual to the previous period actual demand
value. The preferred range for « is from 0.1 to 0.3.
The apphication of this technique 15 demonstrated using an example,
Example 4.1. A firm uses simple exponential smoothing with @ = (0.2 to forecast demand. The
forecast for the first week of Januvary was 400 units, whereas actual demand turned out to be 450
units.

(a) Forecast the demand for the second week of January.

(b) Assume that the actual demand during the second week of January turned out to be 46()
units. Forecast the demand up to February third week, assuming the subsequent demands
as 465, 434, 420, 498, and 462 units.

Solution.
(a) The forecast for the second week of January 1s computed as shown below.

Fo=F_+aD_,-F_
Fo =400 4 0.2 (450 = 400)
= 410 units,
(b) The workings for the remaining weeks are shown in tabular form.

Table 4.5 Smoothed Average Forecasts

Demand Old Fore- Forecast Error Correction New Forecast (F))

Week D, cast (F,.
;) D, -F,, D, -F.,) F,+alD_,-F]

Jan. Week 1 450) 400 + 50 10 410

Week 2 46() 410 50 10 420)
Week 3 4635 420) 45 9 429
Week 4 434 429 5 ] 430
Feb, Week 1 42(0) 43() - 10 - 2 428
Week 2 498 428 70 14 442
Week 3 462 442 20) -+ 446

In Table 4.5 ninal forecast was available. If no previous forecast value 15 known, the “old
forecast’ starting point may be estimated or taken to be an average of the values of some preceding
periods.
 4.4.7 Adjusted Exponential Smoothing Method
The simple exponential smoothing forecast 1s a smoothed average positioned on the current period.
It 15 taken as a next period forecast. In reality, trend exists in demand pattern of many businesses,
Hence, due recognition should be given to make correction in the demand forecast for trend also.
Adjusted exponential smoothed forecast model actually projects the next period forecast by
adding a trend component to the current period smoothed forecast, F.

Fr+] = F.r + Tr
where
Fo=ab,
(+ (1 —a)ilF,_ +1, )
Tr=fi|:F:—Fr—]]+“—m T,

where o and [ are smoothing constants. The trend adjustment T, utilizes a second smoothing
co-efficient, 3.

Example 4.2. Compute the adjusted exponential forecast for the first week of March for a firm with
the following data. Assume the forecast for the first week of January (F;;) as 60 and corresponding
initial trend (Ty) as 0. Let ¢ = 0.1, and B = 0.2.
 Month

Jan. Feb.

Week I 2 3 + I 2 3
Demand 650 600 550 650 625 675 700 710

Solution.
First week of Jan.
F,=oD, _+(1-a)(F,_+T,_
0.1 (650) + 0.9 (600 + 0) = 605

I, =BF,-F_)+(-/T,_,
= (.2 (605 — 600) + 0.8 ()
= 1.00
Fooy =F,+T,=605+1
=606
The adjusted exponential smoothed forecast for the second week of January is 606,
Second week of Jan.
F, = 0.1 (600) + 0.9 (605 + 1) = 605.4
T, = 0.2 (6054 - 603) + 0.8 x 1.0 = (.88
F., =06054 + (.88 = 606.28

The remainder of the calculations are given in Table 4.6. The trend adjusted exponential
smoothed forecast for the first week of March 1s 644.04, 1.e. which 15 approximately 644 units.
 Table 4.6 Adjusted Exponential Smooth Forecast for January and February

Previous Actual Smoothed Smoothed Next Period

Average Demand Average Trend Projection
Week Fe D, F, I, Fiy
Jan., Week 1 600,00 650 605.00 1,000 606.00
Week 2 605.00 GO0 605.40 0.880 606.38
Week 3 605.40 550 600.65 ~(.246 600,40
Week 4 GO0.635 650 6035.36 0.742 606.10
Feb., Week 1 6035.36 625 607.99 1.120 609,11
Week 2 6(7.99 675 615.70 2.440 61%8.14
Week 3 615.70 700 (26.33 4080 630.4]
Week 4 626.33 710 T38.37 5.670 644.04

Comments:
l. Smoothing methods are well suited for short or immediate term predictions of a large
number of items.
2. Suitable for stationary data patterns.

3. Eliminates the need for storing the historical values of the variable,
4. A starting smoothed value and values for the smoothing constants (¢, ) are needed.
5. There 15 no good rule for determining the approximate values of the weighits.
4.4.8 Linear Regression
Regression means dependence and involves estimating the value of a dependent variable ¥, from an
independent variable X. In simple regression, only one independent variable 15 used, whereas n
multiple regression two or more independent variables are involved. The simple regression takes the
following form.
Y=a+ bX
where
Y = Dependent variable
X = Independent variable
a = Intercept
b = Slope (trend)

These are represented in the following graph.

T

Y o X
Fig. 4.5 Graph showing simple regression.
 Consider the model for the simple linear n_:g,rtflflinn which 15 shown below.
Y=a+ bX
The formulas to compute the constants of the model are given below.
~ EZXY —nXY
b
- IX - nX?
a=Y —bX
where
X =ZX/n and ¥ = ZV/n
and n 15 the number of pairs of observations made.
 Exampie 4.3. A firm believes that 1ts annual profit depends on 1ts expenditures for research. The
information for the preceding six years is given below. Estimate the profit when the expenditure
15 6 units.

Expenditure for Annual Profit

Year Research (X) (¥)

] 2 20
2 3 25
3 5 34
4 4 3
3 11 4()
) 5 31
7 6 ?

Solution.

Year X ¥ XY X*
] 2 20) 4() 4
2 3 25 TS L
3 5 34 170 25
4 4 30 120 16
5 11 440} () 121
] 3 31 155 25
Total (&) 30 180 1000 200

— 30 — 180
I = — 1" — T—

6 6
=5 =30

_ZXY - nXY 1000 -6 x5 x 30

b
COEXP —pX? 200—6x5x5
_ 1000 -900 _,
200 — 150

a=Y —bX
=3 -2=x5=20
The model 1s:
Y=a+bX=20+2X
The profit when expenditure is 6 units:
Example 4.4. Alpha company has the following sales pattern. Compute the sales forecast for the
yvear 11).

Year ] 2 3 4 5 6 7 8 .
Sales
(in Lakhs) W) ! I1 23 29 34 4() 45 56

Solution.

Year Sales (F)

(X) in Lakhs x=X-35 x¥ x*
| 6 - 4 - 24 16
2 ) -3 - 24 9
3 11 -2 — 22 =
d 23 - 1 - 23 1
3 29 0 () 0
6 34 1 34 1
7 4() 2 B0 4
! 435 3 135 G
0 36 - 224 16
Total (X) 252 0 380 i)

F=2-0
O

— 252
¥ = > =28
 - Ex¥ —nmxY
b
R -l
380
-9 x0x28
=6.33
60 =9 %0 x0

a=Y
- bX
=28 - 633 X0 =18
Y =a+ bx
= 28 + 6.33x
=28 +633(X-3)
Y =28
+ 6.33 (X
- 5)
The forecast of sales for the year 10 15 computed by substituting X = 1(} in the above equation.

Y =28+ 633 (10 - 5)

= 28 + 6.33 x5 = 39.65 = 60 Unmts (Approx.)
Note. When £ X = (), then the formulae for & and a are reduced as given below.
XY
b=
X
a=Y
ASSIGNMENT -2
- The Super Snow paint shop has recorded the demand for a particular colour during the past
6 weeks as shown below.

Week Demand in Litre

| 1st Week May 19
2nd Week May 17
ird Week May 22
d4th Week May 21

[ st Week June 20
2nd Week June 33

(a) Calculate a 3-week moving average for the data to forecast demand for the next week.
(b) Calculate a weighted average forecast for the data, using a weight of (0.6 for the most
recent data and weights of 0.3 and 0.1 for successive older data.
- Compute the adjusted exponential forecast for the first week of June for a firm with the
following data. Assume the forecast for the first week of April (FO) as 700 and corresponding
initial trend (T0) as 0. Let @ = 0.15, and f = (.25.

Month
April May
Week ] 2 3 4 | 2 3 4

Demand 675 625 575 675 630 630 725 740

The following table represents sales data for LITRES of milk (in hundreds) sold by a milk
bhooth,

Month 1]3 4 3 fi]? 8 [ 9 [10 12]

Sales 96 106 | 92 114 108 98 | 99 115 | 106 | 91 102 99|
(a) Use single exponential smoothing to forecast demand, with a of 0.20 and an imunal
forecast of 100.00.
(b) Use trend adjusted exponential smoothing to forecast sales through period 13 for the
following data. Use an « of 0.30, a § of 0.50, an initial base of 29.0 and a trend of 1.0.
The sales particular of a company for 13 vears of operation 15 furnished below.

Y ear \I 2 3 4 3 6 \? a8 |Y 10 \11 \12 13

Lumber Sales ‘ 06 116 | 119 127 146 145 \ 153 158 | 160 165 | 177 | 190 205
(a) Fit a simple regression for the above data.
(b) Forecast the sales for the 14th year of operation.
A firm believes that 1ts annual profit depends on 1ts expenditures for research. The information
for the preceding six years is given below. Estimate the profit when the expenditure 1s 8 units.

Expenditure for | Annual Profit

Year Y)
Research (X)
" =
' 2004 .
2003 4
6 2
2006
5 -
2007
) 45
2008
7 26
2009
,
2010 Q

Beta company has the following sales pattern during 2002 to 2010, Compute the sales forecast
for 2011.

Year | 2002 | 2003 | 2004 | 2005 | 2006 | 2007 | 2008 | 2009 | 2010
inLakhs) | 8 | 10 [ 13 | 25 | 35 | 39 | 45 | 50 | 60
Sales

The details of sales turnover of a Cement Company for the period 2004-2010 are given In the
following table. Compute the estimated sales for the year 2011.

[ Year [ 2004 [2005 [ 2006 [ 2007 [ 2008 [ 2009 [ 2010

Sales
(1n Crores)
30 | 40 | 55 | 68 | 95 | 90 | 120
What are the types of layout? Explain them with examples.
Discuss the ments and dements of Process Layout and Product Layout.
Discuss the advantages of group technology layout.

Consider the following initial layout with unit cost matrix. Use the CRAFT pairwise interchange

| Production and Operalions Management

B i

4 A B

¢ C D
Initial Layout

Ta A B C L
From
A ~ l &
B 2 - B |
C 0 1) - 2
D 5 2 2 —

Flow Matrix
 Consider the following initial layout with unit cost matnx. Use the CRAFT pairwise interchange
techmgue 10 oblain the desirable layouwt

(3 i\

E A &

Initial Layout

To .! B L [
From

A — | | A
B 2 - 2 |
i () 2 - 2
N | 4 3 -

Flow Matrnix

. Consider the followang mitial layout with umit cost matrix. Use the CRAFT pairwise interchange
lechmque o obtain the desirable lavoul.

4 i B

Initial Layout

To A B L
From

A - ] il
B 2 - |
- 6 2 —

Flow Maitrix
Consider the following machine-component incidence matrix with 100 machines and 5
components. Obtain the fnal machine-component cells using Rank Order Clustering Algorithm.
Component
1 2 3 4 3
1 ] L 1 (]
2 ] () (] | (]
k. ] () { | {
Machine S| 1 0 0 0 0
5 ] o0 1 (]
y (] I (] () ]
T 0 I { () {
3 i 1 1 () ]
9 0 0 ] () ]
1 () ] () ]
Machine-Component Incidence Matrix

Consider the following machine-component incidence matrix with 7 machines and 3

components. Obtain the final machine-component cells using Rank Order Clustering Algonithm.
Component
1 i 3 | 3
| 0 | 0 | {
2 ] ( (] () ]
3 { I ] | {
Machine 1 1 () { () ]
3 () ] 1 (]
& (] ( (] () ]
7 { I ] | {
Machine-Component Incidence Mairix
 Consider the following assembly network relationships of a product. The number of shifts per
day 1% one and the number of working hours per shift 1 B, The company aims o produce 44
units of the product per shatt,
Group the activities into work stations using Rank Positonal Weight Method and compute
Balancing Efficiency.

| Production and Opsrafions Managemens

Operation Immediate Diuration

Mumber Preceding Tasks (i)
] - R
2 | h
3 | y
- | L]
3 | T
f 17 4
[ 2 N
! 4, 5 £
a .6 B
I 7.5 4 o
Il 3 4
12 . 3
13 &, 10, 11 6
14 12, 13 A
Consider the following assembly network relationships ofa product. The number of shifts per
day i3 three and the number of working hours per shift 15 8. The company aims w produce
60 units of the product per day,
Crogp the activities into work siatons wsing Rank Posinonal Weight Method and compute
Balancing Efficiency.

Line Balancing | 237

{peration Immediate Duration

M unber Preceding Tasks (Alim
] — k.
I ] 12
3 ] 15
4 2 10
% 2 14
& 2 i
T 3 11
. 3 16
0 3 20
10 4, 5 R
1] b, 7 13
12 - 17
13 10, 11 £
14 11, 12 3
13 15, 14 12
Explain the significance of mass production in the present day context,
Explain the concept of Cycle Time with a suitable example.
Discuss the steps of RPW method lor line balancing.