07 Line Balancing

Arif Rahman – The Production Systems 1
Slide 7
Manual
Assembly Line
Arif Rahman, ST MT

Arif Rahman – The Production Systems
Most consumer products are assembled on
manual assembly lines
Factors favoring the use of assembly lines:
¤ High or medium demand for product
¤ Identical or similar products
¤ Total work content can be divided into work elements
¤ It is technologically impossible or economically
infeasible to automate the assembly operations
Manual Assembly Lines
2

Specialization of labor
¤ Learning curve
Interchangeable parts
¤ Components made to close tolerances
¤ Assemblies do not need fitting of mating components
Work flow principle
¤ Products are brought to the workers
Line pacing
¤ Workers must complete their tasks within the cycle
time of the line
Why Assembly Lines are so Productive
3

Typical Products Made on Assembly Lines
Automobiles
Cooking ranges
Dishwashers
Dryers
Furniture
Lamps
Luggage
Microwave ovens
Personal computers
Power tools
Refrigerators
Telephones
Toasters
Trucks
Video DVD players
Washing machines
4

A production line consisting of a sequence of
workstations where assembly tasks are performed by
human workers as the product moves along the line
¤ Work systems consisting of multiple workers organized to
produce a single product or a limited range of products
• Each product consists of multiple components joined together by
various assembly work elements
 Total work content - the sum of all work elements required to assemble
one product unit on the line
¤ Assembly workers perform tasks at workstations located along
the line-of-flow of the product
• Usually a powered conveyor is used
• Some of the workstations may be equipped with portable powered
tools
Manual Assembly Line Defined
5

Manual Assembly Line
6
Configuration of a manual assembly line
with n manually operated workstations

The production rate of an assembly line is
determined by its slowest station.
Assembly workstation: A designated
location along the work flow path at which
one or more work elements are performed
by one or more workers
7

Two assembly
operators working on
an engine assembly
line (photo courtesy of
Ford Motor Company)
8

Products are assembled as they move
along the line
¤ At each station a portion of the total work
content is performed on each unit
Base parts are launched onto the
beginning of the line at regular intervals
(cycle time)
¤ Workers add components to progressively
build the product
9

Typical operations performed at manual assembly stations
Adhesive application
Sealant application
Arc welding
Spotwelding
Electrical connections
Componentinsertion
Pressfitting
Riveting
Snap fitting
Soldering
Stitching/stapling
Threadedfasteners
Assembly Workstation
10
A designated location along the work flow path at which one
or more work elements are performed by one or more workers

There may be more than one worker per
station.
Utility workers: are not assigned to specific
workstations.
They are responsible for
¤ helping workers who fall behind,
¤ relieving for workers for personal breaks,
¤ maintenance and repair
Manning level
11

Average manning level:
where :
M = average manning level of the line,
n = number of workstations
w = number of workers on the line
wu = number of utility workers assigned to the system,,
wi = number of workers assigned specifically to
station i for i=1,…,n
Manning level
12
n
ww
n
w
M
n
i
iu ∑=
+
== 1

Two basic categories:
¤ Manual
¤ Mechanized
Work Transport Systems
13

Work units are moved between stations by
the workers without the aid of a powered
conveyor
¤ Types:
• Work units moved in batches
• Work units moved one at a time
¤ Problems:
• Starving of stations
• Blocking of stations
• No pacing
Manual Work Transport Systems
14

To reduce starving,
¤ use buffers
To prevent blocking,
¤ provide space between upstream and
downstream stations.
But both solutions can result in higher
WIP,
¤ which is economically undesirable.
Work Transport Systems-Manual Methods
15

Work units are moved by powered
conveyor or other mechanized apparatus
¤ Categories:
• Work units attached to conveyor
• Work units are removable from conveyor
¤ Problems
• Starving of stations
• Incomplete units
Mechanized Work Transport Systems
16

Continuous transport
¤ Conveyor moves at constant speed
Synchronous transport/intermittent transport
¤ Work units are moved simultaneously with stop-
and-go (intermittent) motion to next stations
Asynchronous transport
¤ Work units are moved independently between
workstations
¤ Queues of work units can form in front of each
station
Types of Mechanized Work Transport
17

Continuous Transport
18
Conveyor moves at constant velocity vc

Synchronous Transport
19
All work units are moved simultaneously to their respective
next workstations with quick, discontinuous motion

Asynchronous Transport
20
Work units move independently, not simultaneously. A work
unit departs a given station when the worker releases it.
Small queues of parts can form at each station.

Continuous transport
¤ Overhead trolley conveyor
¤ Belt conveyor
¤ Roller conveyor
¤ Drag chain conveyor
Synchronous transport
¤ Walking beam transport equipment
¤ Rotary indexing mechanisms
Asynchronous transport
¤ Power-and-free overhead conveyor
¤ Cart-on-track conveyor
¤ Automated guided vehicle systems
¤ Monorail systems
¤ Chain-driven carousel systems
Material Handling Equipment for Mechanized Work Transport
21

A manual assembly line operates at a certain
cycle time to achieve the required production rate
of the line – On average, each worker must
complete his/her assigned task within this cycle
time
Pacing provides a discipline for the assembly line
workers that more or less guarantees a certain
production rate for the line
Several levels of pacing:
1. Rigid pacing
2. Pacing with margin
3. No pacing
Line Pacing
22

Each worker is allowed only a certain fixed
time each cycle to complete the assigned
task
¤ Allowed time is set equal to the cycle time less
repositioning time
¤ Synchronous work transport system provides
rigid pacing
Undesirable aspects of rigid pacing:
¤ Incompatible with inherent human variability
¤ Emotionally and physically stressful to worker
¤ Incomplete work units if task not completed
Rigid Pacing
23

Worker is allowed to complete the task within
a specified time range, the upper limit of
which is greater than the cycle time
On average, the worker’s average task time
must balance with the cycle time of the line
How to achieve pacing with margin:
¤ Allow queues of work units between stations
¤ Provide for tolerance time to be longer than cycle time
¤ Allow worker to move beyond station boundaries
Pacing with Margin
24

No time limit within which task must be
completed
Each assembly worker works at his/her
own pace
No pacing can occur when:
¤ Manual transport of work units is used
¤ Work units can be removed from the conveyor
to perform the task
¤ An asynchronous conveyor is used
No Pacing
25

Single model assembly line (SMAL)
¤ Every work unit is the same
Batch model assembly line (BMAL)
¤ Hard product variety
¤ Products must be made in batches
Mixed model assembly line (MMAL)
¤ Soft product variety
¤ Models can be assembled simultaneously
without batching
Coping with Product Variety
26

Advantages of mixed model lines over
batch models lines:
¤ No lost production time between models
¤ High inventories typical of batch production
are avoided
¤ Production rates of different models can be
adjusted as product demand changes
MMAL vs. BMAL
27

Difficulties with mixed model line
compared to batch model line
¤ Line balancing problem more complex due to
differences in work elements among models
¤ Scheduling the sequence of the different models is a
problem
¤ Logistics is a problem - getting the right parts to each
workstation for the model currently there
¤ Cannot accommodate as wide model variations as
BMAL
MMAL vs. BMAL
28

Problems on high production assembly lines
¤ Complain about the monotony of the repetitive tasks
that must be performed and the unrelenting pace that
must be maintained when a moving conveyor is used
¤ Poor quality workmanship
¤ Sabotage of the line equipment
Alternative assembly systems
¤ Single station manual assembly cells
¤ Assembly cells based on worker teams
• Single station manual assembly cell with multiple workers
• Moving the product through multiple workstations, but having
the same worker team follow the product
¤ Automated assembly systems
Alternative Assembly Systems
29

The key to successful DFA
¤ Design the product with as few parts as possible
¤ Design the remaining parts so they are easy to
assemble
General principles of DFA
¤ Use the fewest number of parts possible to
reduce the amount of assembly required
¤ Reduce the number of threaded fasteners
required
¤ Standardize fasteners
¤ Reduce parts orientation difficulties
¤ Avoid parts that tangle
Design for Assembly (DFA)
30

The formulas and the algorithms in this
section are developed for single model
lines, but they can be extended to batch
and mixed models.
The assembly line must be designed to
achieve a production rate sufficient to
satisfy the demand.
Demand rate → production rate→ cycle
time
Analysis of Single Model Lines
31

Annual demand Da must be reduced to an hourly
production rate Rp
where
Da = annual demand (pieces)
Rp = hourly production rate (pieces/hour)
Sw = number of shifts/week (shift/week)
Hsh = number of hours/shift (hour/shift)
The constant 52 converts years to weeks, and vise
versa
Analysis of Single Model Lines
32
shw
a
p
HS
D
R
52
=

Now our aim is to convert production rate,
Rp, to cycle time, Tc.
One should take into account that some
production time will be lost due to
¤ equipment failures
¤ power outages,
¤ material unavailability,
¤ quality problems,
¤ labor problems.
Determining Cycle Time
33

Line efficiency (uptime proportion): only a
certain proportion of the shift time will be
available.
Production rate, Rp, is converted to a cycle
time, Tc, accounting for line efficiency, E.
34
p
c
R
E
T
60
=

Rc = cycle rate for the line (cycles/hour)
Tc = cycle time of the line (minutes/cycle)
Rp = required production rate (pieces/hour)
E = line efficiency
Rc must be greater than Rp because E is less than 100%
35
c
p
R
R
E =

EAT
T
ET
WL
TRWL
AT
WL
w
c
wc
wcp
60
60
=
=
=
=
Number of Stations Required
36

Work content time (Twc): The total time of all work
elements that must be performed to produce one unit of
the work unit.
The theoretical minimum number of stations that will be
required to on the line to produce one unit of the work
unit, w*:
where
Twc = work content time (min)
Tc = cycle time (min/station)
If we assume one worker per station then this gives the minimum
number of workers
Number of Stations Required
37
c
wc
T
T
w ≥= intmin*

Repositioning losses
¤ Some time will be lost at each station every cycle
for repositioning the worker or the work unit; thus,
the workers will not have the entire Tc each cycle
Line balancing problem (imperfect
balancing):
¤ It is not possible to divide the work content time
evenly among workers, and some workers will
have an amount of work that is less than Tc
Task time variability
Quality problems
Theoretical Minimum Not Possible
38

Repositioning losses occur on a production
line because some time is required each
cycle to reposition the worker, the work unit,
or both
¤ On a continous transport line, time is required for
the worker to walk from the unit just completed to
the the upstream unit entering the station
¤ In conveyor systems, time is required to remove
work units from the conveyor and position it at the
station for worker to perform his task.
Repositioning Losses
39

Repositioning time, (Tr) = time available
each cycle for the worker to position
Service time, (Ts) = time available each
cycle for the worker to work on the product
Service time, Ts = Max{Tsi} ≤ Tc – Tr
where Tsi= service time for station i, i=1,2,..,n
Repositioning efficiency
Repositioning Losses
40
c
rc
c
s
r
T
TT
T
T
E
−
==

Cycle Time (Tc) on an Assembly Line
41
Components of cycle time at several stations on a
manual assembly line.
At the bottleneck station, there is no idle time.
Tsi = service time, Tr = repositioning time

Given:
¤ Total work content consists of many distinct work
elements
¤ The sequence in which the elements can be
performed is restricted
¤ The line must operate at a specified cycle time
Problem:
¤ To assign the individual work elements to
workstations so that all workers have an equal
amount of work to perform
Line Balancing Problem
42

Element times are constant values
¤ But in fact they are variable
Work element times are additive
¤ The time to perform two/more work elements
in sequence is the sum of the individual
element times
¤ Additivity assumption can be violated (due to
motion economies)
Assumptions About Work Element Times
43

Total work content time Twc
¤ where Tek = work element time for element k
Work elements are assigned to station i that add up to the
service time for that station
The station service times must add up to the total work
content time
Work Element Times
44
∑=
=
en
k
ekwc TT
1
∑∈
=
ik
eksi TT
∑=
=
n
i
siwc TT
1

Different work elements require different times.
When elements are grouped into logical tasks
and assigned to workers, the station service
times, Tsi, are likely not to be equal.
Simply because of the variation among work
element times, some workers will be assigned
more work.
Thus, variations among work elements make it
difficult to obtain equal service times for all
stations.
Constraints of Line Balancing Problem
45

Precedence Constraints
46
Restrictions on the order in which work elements can be performed
Precedence Diagram

Largest Candidate Rule
¤ Assignment of work elements to stations based on amount
of time each work element requires
Kilbridge and Wester Method
¤ Assignment of work elements to stations based on position
in the precedence diagram
¤ Elements at front of diagram are assigned first
Ranked Positional Weights
¤ Combines the two preceding approaches by calculating an
RPW for each element
COMSOAL Random Sequence Generation
¤ Sequences are generated by selecting at random from the
set of available tasks and placing that task next in
sequence
Line Balancing Algorithms
47

Example:
48
Grommet : sealant like ring
No Work Element Description Tek (min) Predecessors Successors
1 Place frame in workholder and clamp 0.2 - 3,4
2 Assemble plug, grommet to power cord 0.4 - 4,5
3 Assemble brackets to frame 0.7 1 6,7,8
4 Wire power cord to motor 0.1 1,2 8
5 Wire power cord to switch 0.3 2 10
6 Assemble mechanism plate to bracket 0.11 3 9
7 Assemble blade to bracket 0.32 3 9
8 Assemble motor to brackets 0.6 3,4 9,10
9 Align blade and attach to motor 0.27 6,7,8 11
10 Assemble switch to motor bracket 0.38 5,8 11
11 Attach cover,lnspect, and test 0.5 9,10 12
12 Place in tote pan for packing 0.12 11 -

Example:
49

Given: The previous precedence diagram and the
standard times. Annual demand=100,000 units/year. The
line will operate 50 wk/yr, 5 shifts/wk, 7.5 hr/shift. Uptime
efficiency=96%. Repositioning time lost=0.08 min.
Determine
a) total work content time,
b) required hourly production rate to achieve the annual
demand,
c) cycle time,
d) theoretical minimum number of workers required on the
line,
e) service time to which the line must be balanced.
Example: A problem for line balancing
50

The total work content time is the sum of
the work element times given in the table
Twc = 4.0 min
The hourly production rate
Example: Solution
51
units/hr33.53
)5.7)(5(50
000,100
==pR
∑=
=
en
k
ekwc TT
1
shw
a
p
HS
D
R
50
=

The corresponding cycle time with an
uptime efficiency of 96%
The minimum number of workers:
w*
= (Min Int) ≥ 4.0 /1.08=3.7
= 4 workers
The available service time
Ts = 1.08 - 0.08 = 1.00 min
Example: Solution
52
Production
System
29/10/17
min08.1
33.53
)96.0(60
==cT
p
c
R
E
T
60
=
c
wc
T
T
w =*
rcs TTT −=

It is almost imposible to obtain a perfect line balance
Line balance efficiency, Eb:
Perfect line: Eb = 1
Balance delay, d:
Perfect line: d = 0
Note that Eb + d = 1 (they are complement of each
other)
Measures of Balance Efficiency
53
s
wc
b
wT
T
E =
s
wcs
wT
TwT
d
−
=

Factors that reduce the productivity of a manual
line
¤ Line efficiency (Availability), E,
¤ Repositioning efficiency (repositioning), Er,
¤ Balance efficiency (balancing), Eb,
Overall Labor efficiency on the assembly line =
Overall Efficiency
54
c
rc
c
s
r
T
TT
T
T
E
−
==
s
wc
b
wT
T
E =
p
c
R
E
T
60
=
rb EEE ⋅⋅=EfficiencyLabor

The actual number of workers on the
assembly line is given by:
where :
w = number of workers required
Rp = hourly production rate (pieces/hour)
Twc = work content time per product, (min/piece)
Worker Requirements
55
c
rc
c
s
r
T
TT
T
T
E
−
==
s
wc
b
wT
T
E =
sb
wc
cbr
wc
br
wcp
TE
T
TEE
T
EEE
TR
w ==≥=
60
intmin
p
c
R
E
T
60
=

Number of stations:
Total length of the assembly line
where
L = length of the assembly line (m)
Lsi= length of station i (m)
Continously moving conveyors - Workstation considerations
56
∑=
=
n
i
isLL
1
M
w
n =

Constant speed conveyor: (if the base parts
remain fixed during their assembly)
¤ Feed rate
where : fp=feed rate on the line, (products/min)
¤ Center-to-center spacing between base parts
where :
sp = center-to-center spacing between base parts (m/part)
vc = velocity of the conveyor (m/min)
Continously moving conveyors - Workstation considerations
57
c
p
T
f
1
=
cc
p
c
p Tv
f
v
s ==

Defined as the time a work unit spends inside the
boundaries of the workstation
Provides a way to allow for product-to-product
variations in task times at a station
where
Tt = tolerance time (min)
Ls = station length (m)
vc = conveyor speed, (m/min)
Continously moving conveyors - Tolerance Time
58
c
s
t
v
L
T =

The time a work unit spends on the assembly
line.
where
ET = total elapsed time, (min)
n = number of workstations
Tt = tolerance time, (min)
L = length of the assembly line, (m)
vc = conveyor speed, (m/min)
Continously moving conveyors -Total Elapsed Time
5929/10/17
t
c
nT
v
L
ET ==

To distribute the total work content on the assembly
line as evenly as possible among the workers
Minimize (wTs – Twc)
or
Minimize
Subject to:
(1)
(2) all precedence requirements are obeyed
Line Balancing Objective
60
( )∑=
−
w
i
sis TT
1
s
ik
ek TT ≤∑∈

0. List all work elements in descending order based on their Tek
values; then, proceed three-step procedure:
1. Start at the top of the list and selecting the first element that
satisfies precedence requirements and does not cause the
total sum of Tek to exceed the allowable Ts value.
¤ When an element is assigned, start back at the top of the list
and repeat selection process
1. When no more elements can be assigned to the current
station, proceed to next station
2. Repeat steps 1 and 2 until all elements have been assigned
to as many stations as needed
61

Example: Largest Candidate Rule
62

63
Work Elements Arranged According to Tek Value for the
Largest Candidate Rule. (Ts = 1 min)
No Tek (min) Predecessors Successors
3 0.7 1 6,7,8
8 0.6 3,4 9,1
11 0.5 9,1 12
2 0.4 - 4,5
10 0.38 5,8 11
7 0.32 3 9
5 0.3 2 10
9 0.27 6,7,8 11
1 0.2 - 3,4
12 0.12 11 -
6 0.11 3 9
4 0.1 1,2 8

Iteration 1:
3, 8, and 11 require preceding
elements
Element 2 is selected.
Alternatives are 5 and 1.
It assigns element 5.
ΣTek = 0.4 + 0.3 = 0.7 min
Alternative is 1
It assigns element 1
ΣTek = 0.7 + 0.2 = 0.9 min
Alternatives are 3 and 4
3 can not be assigned, since ΣTek
> Ts
ΣTek = 0.9 + 0.1 = 1.0 min 64
3 0.7 1 6,7,8
8 0.6 3,4 9,1
11 0.5 9,1 12
2 0.4 - 4,5
10 0.38 5,8 11
7 0.32 3 9
5 0.3 2 10
9 0.27 6,7,8 11
1 0.2 - 3,4
12 0.12 11 -
6 0.11 3 9
4 0.1 1,2 8

Iteration 2:
Alternatives are 8, 7 and 6.
8 and 7 can not be assigned, since
ΣTek > Ts
ΣTek = 0.7 + 0.11 = 0.81 min
ΣTek > Ts
65
3 0.7 1 6,7,8
8 0.6 3,4 9,1
11 0.5 9,1 12
2 0.4 - 4,5
10 0.38 5,8 11
7 0.32 3 9
5 0.3 2 10
9 0.27 6,7,8 11
1 0.2 - 3,4
12 0.12 11 -
6 0.11 3 9
4 0.1 1,2 8

Iteration 3:
ΣTek = 0.6 + 0.38 = 0.98 min
Alternative is 7.
> Ts
66
3 0.7 1 6,7,8
8 0.6 3,4 9,1
11 0.5 9,1 12
2 0.4 - 4,5
10 0.38 5,8 11
7 0.32 3 9
5 0.3 2 10
9 0.27 6,7,8 11
1 0.2 - 3,4
12 0.12 11 -
6 0.11 3 9
4 0.1 1,2 8

Iteration 4:
11 requires preceding elements
Alternative is 9.
ΣTek = 0.32 + 0.27 = 0.59 min
Alternative is 11.
> Ts
67
3 0.7 1 6,7,8
8 0.6 3,4 9,1
11 0.5 9,1 12
2 0.4 - 4,5
10 0.38 5,8 11
7 0.32 3 9
5 0.3 2 10
9 0.27 6,7,8 11
1 0.2 - 3,4
12 0.12 11 -
6 0.11 3 9
4 0.1 1,2 8

Iteration 5:
Alternative is 12.
ΣTek = 0.5 + 0.12 = 0.62 min
All elements have been assigned.
68
3 0.7 1 6,7,8
8 0.6 3,4 9,1
11 0.5 9,1 12
2 0.4 - 4,5
10 0.38 5,8 11
7 0.32 3 9
5 0.3 2 10
9 0.27 6,7,8 11
1 0.2 - 3,4
12 0.12 11 -
6 0.11 3 9
4 0.1 1,2 8

Final Solution:
Workstation 1 : 2-5-1-4 = 1.0 min
Workstation 2 : 3-6 = 0.81 min
69
3 0.7 1 6,7,8
8 0.6 3,4 9,1
11 0.5 9,1 12
2 0.4 - 4,5
10 0.38 5,8 11
7 0.32 3 9
5 0.3 2 10
9 0.27 6,7,8 11
1 0.2 - 3,4
12 0.12 11 -
6 0.11 3 9
4 0.1 1,2 8

Example: Solution
70
Work Elements Arrangement . Ts = 1 min.
Workstatio
n
No Tek (min) ΣTek
1 1 0.2 1.0
2 0.4
4 0.1
5 0.3
2 3 0.7 0.81
6 0.11
3 8 0.6 0.98
10 0.38
4 7 0.32 0.59
9 0.27
5 11 0.5 0.62
12 0.12

Example: Solution
71

Solution for Largest Candidate Rule
72
Physical layout of workstations and assignment of
elements to stations using the largest candidate rule

Selects work elements for assignment to stations
according to their position in the precedence
diagram
Work elements in the precedence diagram are
arranged into columns
The elements can be organized into a list
according to their columns, with the elements in
the first column listed first
If a given element can be located in more than
one column, then list all of the columns for that
element
Proceed with same steps 1, 2, and 3 as in the
largest candidate rule
73

Example: Kilbridge and Wester Method
74

75
Kilbridge and Wester Method. (Ts = 1 min)
No Tek (min) Region Predecessors Successors
2 0.4 I - 4,5
1 0.2 I - 3,4
3 0.7 II 1 6,7,8
4 0.1 II 1,2 8
5 0.3 II / III 2 10
8 0.6 III 3,4 9,1
7 0.32 III 3 9
6 0.11 III 3 9
10 0.38 III / IV 5,8 11
9 0.27 IV 6,7,8 11
11 0.5 V 9,1 12
12 0.12 VI 11 -

Iteration 1:
ΣTek = 0.4 + 0.2 = 0.6 min
Alternatives are 3, 4 and 5
> Ts
ΣTek = 0.6 + 0.1 = 0.7 min
> Ts
ΣTek = 0.7 + 0.3 = 1.0 min 76
2 0.4 I - 4,5
1 0.2 I - 3,4
3 0.7 II 1 6,7,8
4 0.1 II 1,2 8
5 0.3 II / III 2 10
8 0.6 III 3,4 9,1
7 0.32 III 3 9
6 0.11 III 3 9
10 0.38 III / IV 5,8 11
9 0.27 IV 6,7,8 11
11 0.5 V 9,1 12
12 0.12 VI 11 -

Iteration 2:
Alternatives are 8, 7 and 6.
ΣTek > Ts
ΣTek = 0.7 + 0.11 = 0.81 min
ΣTek > Ts
77
2 0.4 I - 4,5
1 0.2 I - 3,4
3 0.7 II 1 6,7,8
4 0.1 II 1,2 8
5 0.3 II / III 2 10
8 0.6 III 3,4 9,1
7 0.32 III 3 9
6 0.11 III 3 9
10 0.38 III / IV 5,8 11
9 0.27 IV 6,7,8 11
11 0.5 V 9,1 12
12 0.12 VI 11 -

Iteration 3:
ΣTek = 0.6 + 0.32 = 0.92 min
ΣTek > Ts
78
2 0.4 I - 4,5
1 0.2 I - 3,4
3 0.7 II 1 6,7,8
4 0.1 II 1,2 8
5 0.3 II / III 2 10
8 0.6 III 3,4 9,1
7 0.32 III 3 9
6 0.11 III 3 9
10 0.38 III / IV 5,8 11
9 0.27 IV 6,7,8 11
11 0.5 V 9,1 12
12 0.12 VI 11 -

Iteration 4:
Alternatives is 9.
ΣTek = 0.38 + 0.27 = 0.65 min
Alternative is 11
11 can not be assigned, since ΣTek >
Ts
79
2 0.4 I - 4,5
1 0.2 I - 3,4
3 0.7 II 1 6,7,8
4 0.1 II 1,2 8
5 0.3 II / III 2 10
8 0.6 III 3,4 9,1
7 0.32 III 3 9
6 0.11 III 3 9
10 0.38 III / IV 5,8 11
9 0.27 IV 6,7,8 11
11 0.5 V 9,1 12
12 0.12 VI 11 -

Iteration 5:
Alternatives is 12.
ΣTek = 0.5 + 0.12 = 0.62 min
80
2 0.4 I - 4,5
1 0.2 I - 3,4
3 0.7 II 1 6,7,8
4 0.1 II 1,2 8
5 0.3 II / III 2 10
8 0.6 III 3,4 9,1
7 0.32 III 3 9
6 0.11 III 3 9
10 0.38 III / IV 5,8 11
9 0.27 IV 6,7,8 11
11 0.5 V 9,1 12
12 0.12 VI 11 -

Final Solution:
81
2 0.4 I - 4,5
1 0.2 I - 3,4
3 0.7 II 1 6,7,8
4 0.1 II 1,2 8
5 0.3 II / III 2 10
8 0.6 III 3,4 9,1
7 0.32 III 3 9
6 0.11 III 3 9
10 0.38 III / IV 5,8 11
9 0.27 IV 6,7,8 11
11 0.5 V 9,1 12
12 0.12 VI 11 -

Example: Solution
82
Workstatio
n
No Tek (min) ΣTek
1 1 0.2 1.0
2 0.4
4 0.1
5 0.3
2 3 0.7 0.81
6 0.11
3 7 0.32 0.92
8 0.6
4 9 0.27 0.65
10 0.38
5 11 0.5 0.62
12 0.12

Example: Solution
83

Solution for Kilbridge and Wester Method
84
elements to stations using the Kilbridge and Wester Method

A ranked position weight (RPW) is calculated for
each work element
RPW for element k is calculated by summing the
Tek values for all of the elements that follow
element k in the diagram plus Tek itself
Work elements are then organized into a list
according to their RPW values, starting with the
element that has the highest RPW value
Proceed with same steps 1, 2, and 3 as in the
largest candidate rule
Ranked Positional Weights Method
85

Example: Ranked Positional Weights Method
86

87
Ranked Positional Weights Method. (Ts = 1 min)
No Tek (min) RPW Predecessors Successors Elements of RPW
1 0.2 3.30 - 3,4 1, 3, 4, 6, 7, 8, 9, 10, 11, 12
3 0.7 3.00 1 6,7,8 3, 6, 7, 8, 9, 10, 11, 12
2 0.4 2.67 - 4,5 2, 4, 5, 8, 9, 10, 11, 12
4 0.1 1.97 1,2 8 4, 8, 9, 10, 11, 12
8 0.6 1.87 3,4 9,1 8, 9, 10, 11, 12
5 0.3 1.30 2 10 5, 10, 11, 12
7 0.32 1.21 3 9 7, 9, 11, 12
6 0.11 1.00 3 9 6, 9, 11, 12
10 0.38 1.00 5,8 11 10, 11, 12
9 0.27 0.89 6,7,8 11 9, 11, 12
11 0.5 0.62 9,1 12 11, 12
12 0.12 0.12 11 - 12

Iteration 1:
ΣTek = 0.2 + 0.7 = 0.9 min
2, 7 and 6 can not be assigned,
since ΣTek > Ts
88
No Tek (min) RPW Predecessors Successors
1 0.2 3.30 - 3,4
3 0.7 3.00 1 6,7,8
2 0.4 2.67 - 4,5
4 0.1 1.97 1,2 8
8 0.6 1.87 3,4 9,1
5 0.3 1.30 2 10
7 0.32 1.21 3 9
6 0.11 1.00 3 9
10 0.38 1.00 5,8 11
9 0.27 0.89 6,7,8 11
11 0.5 0.62 9,1 12
12 0.12 0.12 11 -

Iteration 2:
Alternatives are 4, 5, 7 and 6.
ΣTek = 0.4 + 0.1 = 0.5 min
Alternatives are 8, 5, 7 and 6
> Ts
ΣTek = 0.5 + 0.3 = 0.8 min
ΣTek > Ts
ΣTek = 0.8 + 0.11 = 0.91 min 89
1 0.2 3.30 - 3,4
3 0.7 3.00 1 6,7,8
2 0.4 2.67 - 4,5
4 0.1 1.97 1,2 8
8 0.6 1.87 3,4 9,1
5 0.3 1.30 2 10
7 0.32 1.21 3 9
6 0.11 1.00 3 9
10 0.38 1.00 5,8 11
9 0.27 0.89 6,7,8 11
11 0.5 0.62 9,1 12
12 0.12 0.12 11 -

Iteration 3:
ΣTek = 0.6 + 0.32 = 0.92 min
ΣTek > Ts
90
1 0.2 3.30 - 3,4
3 0.7 3.00 1 6,7,8
2 0.4 2.67 - 4,5
4 0.1 1.97 1,2 8
8 0.6 1.87 3,4 9,1
5 0.3 1.30 2 10
7 0.32 1.21 3 9
6 0.11 1.00 3 9
10 0.38 1.00 5,8 11
9 0.27 0.89 6,7,8 11
11 0.5 0.62 9,1 12
12 0.12 0.12 11 -

Iteration 4:
Alternative is 9.
ΣTek = 0.38 + 0.27 = 0.65 min
Alternative is 11
> Ts
91
1 0.2 3.30 - 3,4
3 0.7 3.00 1 6,7,8
2 0.4 2.67 - 4,5
4 0.1 1.97 1,2 8
8 0.6 1.87 3,4 9,1
5 0.3 1.30 2 10
7 0.32 1.21 3 9
6 0.11 1.00 3 9
10 0.38 1.00 5,8 11
9 0.27 0.89 6,7,8 11
11 0.5 0.62 9,1 12
12 0.12 0.12 11 -

Iteration 5:
Alternative is 12.
ΣTek = 0.5 + 0.12 = 0.62 min
92
1 0.2 3.30 - 3,4
3 0.7 3.00 1 6,7,8
2 0.4 2.67 - 4,5
4 0.1 1.97 1,2 8
8 0.6 1.87 3,4 9,1
5 0.3 1.30 2 10
7 0.32 1.21 3 9
6 0.11 1.00 3 9
10 0.38 1.00 5,8 11
9 0.27 0.89 6,7,8 11
11 0.5 0.62 9,1 12
12 0.12 0.12 11 -

Final Solution:
93
1 0.2 3.30 - 3,4
3 0.7 3.00 1 6,7,8
2 0.4 2.67 - 4,5
4 0.1 1.97 1,2 8
8 0.6 1.87 3,4 9,1
5 0.3 1.30 2 10
7 0.32 1.21 3 9
6 0.11 1.00 3 9
10 0.38 1.00 5,8 11
9 0.27 0.89 6,7,8 11
11 0.5 0.62 9,1 12
12 0.12 0.12 11 -

Example: Solution
94
Workstatio
n
No Tek (min) ΣTek
1 1 0.2 0.9
3 0.7
2 2 0.4 0.91
4 0.1
5 0.3
6 0.11
3 7 0.32 0.92
8 0.6
4 9 0.27 0.65
10 0.38
5 11 0.5 0.62
12 0.12

Example: Solution
95

Solution for Ranked Positional Weights Method
96
elements to stations using the Ranked Positional Weights
Method

Balance efficiency
97
( )
8.0
15
4
==bE
( )
8.0
15
4
==bE

Ranked Positional Weights Method
Balance efficiency
98
( )
( ) 6.5796.060
60
1
6060
108.092.0
87.0
92.05
4
===
===
=+=+=
==
ERR
T
R
TTT
E
cp
c
c
rsc
b

A manual production line capable of producing
a variety of different product models
simultaneously and continuously (not in
batches)
Problems in designing and operating a
MMAL:
¤ Determining number of workers on the line
¤ Line balancing - same basic problem as in SMAL
except differences in work elements among
models must be considered
¤ Model launching - determining the sequence in
which different models will be launched onto
the line
Mixed Model Assembly Lines
99

Line efficiency
¤ Management is responsible to maintain line
operation at efficiencies (proportion uptime)
close to 100%
• Implement preventive maintenance
• Well-trained emergency repair crews to quickly fix
breakdowns when they occur
• Avoid shortages of incoming parts to avoid forced
downtime
• Insist on highest quality components from
suppliers to avoid downtime due to poor quality
parts
Other Considerations in Line Design
100

Methods analysis
¤ To analyze methods at bottleneck or other
troublesome workstations
Subdividing work elements
¤ It may be technically possible to subdivide some work
elements to achieve a better line balance
Sharing work elements between two adjacent
stations
¤ Alternative cycles between two workers
Other Considerations - continued
101

Utility workers
¤ To relieve congestion at stations that are
temporarily overloaded
Changing workhead speeds at mechanized
stations
¤ Increase power feed or speed to achieve a better
line balance
Preassembly of components
¤ Prepare certain subassemblies off-line to reduce
work content time on the final assembly line
102

Storage buffers between stations
¤ To permit continued operation of certain sections
of the line when other sections break down
¤ To smooth production between stations with large
task time variations
Parallel stations
¤ To reduce time at bottleneck stations that have
unusually long task times
Worker (Labor) Shifting with crosstraining
¤ Temporary (or periodic) relocation to expedite or
to reduce subassembly stocks
103

Zoning constraints - limitations on the
grouping of work elements and/or their
allocation to workstations
¤ Positive zoning constraints
• Work elements should be grouped at same station
• Example: spray painting elements
¤ Negative zoning constraints
• Elements that might interfere with each other
• Separate delicate adjustments from loud noises
104

Position constraints
¤ Encountered in assembly of large products
such as trucks and cars, making it difficult for
one worker to perform tasks on both sides of
the product
¤ To address, assembly workers are positioned
on both sides of the line
105

Arif Rahman – The Production Systems 106
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07 Line Balancing

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