Find introduction to cellular manufacturing group technology systems by Nilesh Arora - founder of AddValue consulting.

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2. ORIGINS
• FLANDERS’ PRODUCT ORIENTED
DEPARTMENTS FOR STANDARIZED PRODUCTS
WITH MINIMAL TRANSPORTATION (1925)
• SOKOLOVSKI/MITROFANOV: PARTS WITH
SIMILAR FEATURES MANUFACTURED
TOGETHER

3. BASIC PRINCIPLE
• SIMILAR “THINGS” SHOULD BE DONE
SIMILARLY
• “THINGS “
– PRODUCT DESIGN
– PROCESS PLANNING
– FABRICATION &ASSEMBLY
– PRODUCTION CONTROL
– ADMINISTRATIVE FUNCTIONS

4. TENETS OF GROUP TECHNOLOGY
• DIVIDE THE MANUFACTURING FACILITY
INTO SMALL GROUPS OR CELLS OF
MACHINES (1-5)
• THIS IS CALLED CELLULAR
MANUFACTURING

5. SYMPTOMS FOR RE-LAYOUT
• Symptoms that allow us to detect the need for a re-layout:
– Congestion and bad utilization of space.
– Excessive stock in process at the facility.
– Long distances in the work flow process.
– Simultaneous bottle necks and workstations with idle time.
– Qualified workers carrying out too many simple operations.
– Labor anxiety and discomfort. Accidents at the facility.
– Difficulty in controlling operations and personnel.

6. What is Group Technology (GT)?
• GT is a theory of management based on the
principle that similar things should be done similarly
• GT is the realization that many problems are similar,
and that by grouping similar problems, a single
solution can be found to a set of problems thus
saving time and effort
• GT is a manufacturing philosophy in which similar
parts are identified and grouped together to take
advantage of their similarities in design and
production

7. Implementing GT
Where to implement GT?
• „Plants using traditional batch production
and „process type layout
• „If the parts can be grouped into part families
„How to implement GT?
• „Identify part families
• „Rearrange production machines into machine
cells

8. Types of Layout
In most of today’s factories it is possible to
divide all the made components into families
and all the machines into groups, in such a way
that all the parts in each family can be
completely processed in one group only.
The three main types of layout are:
• Line (product) Layout
• Functional Layout
• Group Layout

9. Line (product) Layout
•It involves the arrangements of machines in one line, depending on
the sequence of operations. In product layout, if there is a more than
one line of production, there are as many lines of machines.
•Line Layout is used at present in simple process industries, in
continuous assembly, and for mass production of components
required in very large quantities.

10. Functional Layout
•In Functional Layout, all machines of the
same type are laid out together in the same
section under the same foreman. Each foreman
and his team of workers specialize in one
process and work independently. This type of
layout is based on process specialization.

11. Group Layout
•In Group Layout, each foreman and his team
specialize in the production of one list of parts
and co-operate in the completion of common
task. This type of layouts based on component
specialization.

12. The Difference between group and functional layout:

13. Evaluations of cell system design are incomplete
unless they relate to the Cell Design.
A few typical performance variables related to
system operation are:
• Equipment utilization (high)
• Work-in-process inventory (low)
• Queue lengths at each workstation (short)
• Job throughput time (short)
• Job lateness (low)
Evaluation criteria of Cell Design

14. Cell Formation Approach
Machine - Component Group Analysis:
Machine - Component Group Analysis is based
on production flow analysis

15. Production flow analysis involves four stages:
Stage 1: Machine classification.
Machines are classified on the basis of
operations that can be performed on them. A
machine type number is assigned to machines
capable of performing similar operations.
Machine - Component
Group Analysis

16. Stage 2: Checking parts list and
production route information.
For each part, information on the operations
to be undertaken and the machines required
to perform each of these operations is
checked thoroughly.
Production flow analysis involves four stages:
Machine - Component
Group Analysis

17. Stage 3: Factory flow analysis.
This involves a micro-level examination of flow
of components through machines. This, in
turn, allows the problem to be decomposed
into a number of machine-component groups.
Production flow analysis involves four stages:
Machine - Component
Group Analysis

18. Stage 4: Machine-component group
analysis.
An intuitive manual method is suggested to
manipulate the matrix to form cells. However,
as the problem size becomes large, the manual
approach does not work. Therefore, there is a
need to develop analytical approaches to
handle large problems systematically.
Production flow analysis involves four stages:
Machine - Component
Group Analysis

19. Example: Consider a problem of 4 machines and 6
parts. Try to group them.
Machines 1 2 3 4 5 6
M1 1 1 1
M2 1 1 1
M3 1 1 1
M4 1 1 1
Components
Machine - Component
Group Analysis

20. Machines 2 4 6 1 3 5
M1 1 1 1
M2 1 1 1
M3 1 1 1
M4 1 1 1
Components
Machine - Component
Group Analysis
Solution

21. Cellular Layout
Process (Functional) Layout Group (Cellular) Layout
Similar resources placed
together
Resources to produce similar
products placed together
T T T
M
M M T
M
SG CG CG
SG
D D D
D
T T T CG CG
T T T SG SG
M M D D D
M M D D D
A cluster
or cell

22. Group Technology (CELL)
Layouts
• One of the most popular hybrid layouts uses Group Technology (GT)
and a cellular layout
• GT has the advantage of bringing the efficiencies of a product layout
to a process layout environment

23. Process Flows before the Use of GT Cells

24. Process Flows after the Use of GT Cells

25. Designing Product Layouts
• Designing product layouts requires consideration
of:
– Sequence of tasks to be performed by each
workstation
– Logical order
–Speed considerations – line balancing

26. Designing Product Layouts –
con’t
Step 1: Identify tasks & immediate predecessors
Step 2: Determine TAKT TIME
Step 3: Determine cycle time
Step 4: Compute the Theoretical Minimum number of
Stations
Step 5: Assign tasks to workstations (balance the
line)
Step 6: Compute efficiency, idle time & balance delay

27. Step 1: Identify Tasks &
Immediate Predecessors
Example 10.4 Vicki's Pizzeria and the Precedence Diagram
Immediate Task Time
Work Element Task Description Predecessor (seconds
A Roll dough None 50
B Place on cardboard backing A 5
C Sprinkle cheese B 25
D Spread Sauce C 15
E Add pepperoni D 12
F Add sausage D 10
G Add mushrooms D 15
H Shrinkwrap pizza E,F,G 18
I Pack in box H 15
Total task time 165

28. Layout Calculations
• Step 2: Determine TAKT TIME
– Vicki needs to produce 60 pizzas per hour
– TAKT TIME= 60 sec/unit
• Step 3: Determine cycle time
– The amount of time each workstation is allowed to
complete its tasks
– Limited by the bottleneck task (the longest task in a
process):
 
 
sec./unit60
units/hr60
sec/min60min/hr x60
units/hroutputdesired
sec./daytimeavailable
)(sec./unittimeCycle 
hourperpizzasorunits/hr,72
sec./unit50
sec./hr.3600
timetaskbottleneck
timeavailable
outputMaximum 

29. Layout Calculations
• Step 4: Compute the theoretical minimum
number of stations
– TM = number of stations needed to achieve 100%
efficiency (every second is used)
– Always round up (no partial workstations)
– Serves as a lower bound for our analysis
 
stations3or2.75,
nsec/statio60
seconds165
timecycle
timestask
TM 


30. Layout Calculations
• Step 5: Assign tasks to workstations
– Start at the first station & choose the longest eligible task following precedence
relationships
– Continue adding the longest eligible task that fits without going over the desired
cycle time
– When no additional tasks can be added within the desired cycle time, begin
assigning tasks to the next workstation until finished
Workstation Eligible task Task Selected Task time Idle time
A A 50 10
B B 5 5
C C 25 35
D D 15 20
E, F, G G 15 5
E, F E 12 48
F F 10 38
H H 18 20
I I 15 5
1
2
3

31. Last Layout Calculation
• Step 6: Compute efficiency and balance delay
– Efficiency (%) is the ratio of total productive time divided
by total time
– Balance delay (%) is the amount by which the line falls
short of 100%
  91.7%100
sec.60xstations3
sec.165
NC
t
(%)Efficiency 

8.3%91.7%100%delayBalance 

32. Other Product Layout Considerations
• Shape of the line (S, U, O, L):
– Share resources, enhance communication & visibility,
impact location of loading & unloading
• Paced versus Un-paced lines
– Paced lines use an automatically enforced cycle time
• Number of Product Models produced
– Single
– Mixed-model lines

33. LINE BALANCING

34. The Line Balancing Problem
• The problem is to arrange the individual
processing and assembly tasks at the
workstations so that the total time required at
each workstation is approximately the same.
• Nearly impossible to reach perfect balance

35. Things to consider
• Sequence of tasks is restricted, there is a
required order
• Called precedence constraints
• There is a production rate needed, i.e. how
many products needed per time period
• Design the line to meet demand and within
constraints

36. Terminology and Definitions
• Minimum Work Element
• Total Work Content
• Workstation Process time
• Cycle Time
• Precedence Constraints
• Balance Delay

37. Minimum Work Element
• Dividing the job into tasks of a rational
and smallest size
• Example: Drill a hole, can’t be divided
• Symbol – Time for element j:
• is a constant ejT
ejT

38. Total Work Content
• Aggregate of work
elements


n
j
ejwc TT
1

39. Workstation Process time
• The amount of time for an individual
workstation, after individual tasks have been
combined into stations
• Sum of task times = sum of workstation times

40. Cycle time
• Time between parts coming off the line
• Ideally, the production rate, but may need to
be adjusted for efficiency and down time
• Established by the bottleneck station, that is
station with largest time

41. Precedence Constraints
• Generally given, determined by the required
order of operations
• Draw in a network style for understanding
• Cannot violate these, an element must be
complete before the next one is started

42. Balance Delay
• Measure of line
inefficiency due to
imbalances in
station times
c
wcc
nT
TnT
d



43. Line Balancing Example
EXAMPLE
Green Grass’s plant manager just received marketing’s latest forecasts of
fertilizer spreader sales for the next year. She wants its production line to
be designed to make 2,400 spreaders per week. The plant will operate 40
hours per week.
a. What should be the line’s cycle time or throughput rate per hour be?
Throughput rate/hr = 2400 / 40 = 60 spreaders/hr
Cycle Time = 1/Throughput rate= 1/60 = 1 minute = 60 seconds

44. Line balancing Example
Assume that in order to produce the new fertilizer spreader on the assembly
line requires doing the following steps in the order specified:
b. What is the total number of stations or machines required?
TM (total machines) = total production time / cycle time = 244/60 = 4.067 or 5
Work
Element
Description
Time
(sec)
Immediate
Predecessor(s)
A Bolt leg frame to hopper 40 None
B Insert impeller shaft 30 A
C Attach axle 50 A
D Attach agitator 40 B
E Attach drive wheel 6 B
F Attach free wheel 25 C
G Mount lower post 15 C
H Attach controls 20 D, E
I Mount nameplate 18 F, G
Total 244

45. Draw a Precedence Diagram
SOLUTION
The figure shows the complete diagram. We begin with work element A,
which has no immediate predecessors. Next, we add elements B and C, for
which element A is the only immediate predecessor. After entering time
standards and arrows showing precedence, we add elements D and E,
and so on. The diagram simplifies
interpretation. Work element F,
for example, can be done
anywhere on the line after
element C is completed.
However, element I must
await completion of
elements F and G.
D
40
I
18
H
20
F
25
G
15
C
50
E
6
B
30
A
40
Precedence Diagram for Assembling
the Big Broadcaster

46. Allocating work or activities to
stations or machines
• The goal is to cluster the work elements into workstations
so that
1. The number of workstations required is minimized
2. The precedence and cycle-time requirements are not violated
 The work content for each station is equal (or nearly so, but less than) the
cycle time for the line

47. Finding a Solution
• The minimum number of workstations is 5 and the cycle
time is 60 seconds, so Figure 5 represents an optimal
solution to the problem
Firtilizer Precedence Diagram Solution
D
40
I
18
H
20
F
25
C
50
E
6
B
30
A
40
G
15

48. Calculating Line Efficiency
c. Now calculate the efficiency measures of a five-station solution:
Efficiency = (100) =
t
nc
244
5(60)
= 81.3%
Idle time = nc – t = 5(60) – 244 = 56 seconds
Balance delay (%) = 100 – Efficiency = 100% - 81.3% = 18.7%

49. A Line Process
• The desired output rate is matched to the
staffing or production plan
• Line Cycle Time is the maximum time
allowed for work at each station is
c =
1
r
where
c = cycle time in hours
r = desired output rate

50. A Line Process
• The theoretical minimum number of
stations is
TM =
t
c
where
t = total time required to assemble each unit

51. A Line Process
• Idle time, efficiency, and balance delay
Idle time = nc – t
where
n = number of stations
Efficiency (%) = (100)
t
nc
Balance delay (%) = 100 – Efficiency

52. Solved Problem 2
A company is setting up an assembly line to produce 192 units per 8-hour
shift. The following table identifies the work elements, times, and
immediate predecessors:
Work Element Time (sec) Immediate Predecessor(s)
A 40 None
B 80 A
C 30 D, E, F
D 25 B
E 20 B
F 15 B
G 120 A
H 145 G
I 130 H
J 115 C, I
Total 720

53. Solved Problem 2
a. What is the desired cycle time (in seconds)?
b. What is the theoretical minimum number of stations?
c. Use trial and error to work out a solution, and show your solution on a
precedence diagram.
d. What are the efficiency and balance delay of the solution found?
SOLUTION
a. Substituting in the cycle-time formula, we get
c = =
1
r
8 hours
192 units
(3,600 sec/hr) = 150 sec/unit

54. Solved Problem 2
b. The sum of the work-element times is 720 seconds, so
TM =
t
c
= = 4.8 or 5 stations
720 sec/unit
150 sec/unit-station
which may not be achievable.

55. Solved Problem 2
c. The precedence diagram is shown in Figure 7.6. Each row in the
following table shows work elements assigned to each of the five
workstations in the proposed solution.
J
115
C
30
D
25
E
20
F
15
I
130
H
145
B
80
G
120
A
40
Figure 7.6 – Precedence Diagram
Work
Element
Immediate
Predecessor(s)
A None
B A
C D, E, F
D B
E B
F B
G A
H G
I H
J C, I

56. Solved Problem 2
J
115
C
30
D
25
E
20
F
15 I
130
H
145
B
80
G
120
A
40
A A 40 40 110
B B 80 120 30
D, E, F D 25 145 5
E, F, G G 120 120 30
E, F E 20 140 10
F, H H 145 145 5
F, I I 130 130 20
F F 15 145 5
C C 30 30 120
J J 115 145 5
Station Candidate(s) Choice
Work-Element
Time (sec)
Cumulative
Time (sec)
Idle Time
(c= 150 sec)
S1
S2
S3
S4
S5

57. Solved Problem 2
d. Calculating the efficiency, we get
Thus, the balance delay is only 4 percent (100–96).
Efficiency (%) = (100)
t
nc
=
720 sec/unit
5(150 sec/unit)
= 96%

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