1. 1
Facility Layout

2. 2
OBJECTIVES
•
Facility Layout and Basic Formats
•
Process Layout
•
Layout Planning
•
Assembly Line balancing
•
Service Layout

3. 3
Facility Layout
Defined
Facility layout can be defined as the process by which the
placement of departments, workgroups within
departments, workstations, machines, and stockholding points within a facility are determined
This process requires the following inputs:
•
Specification of objectives of the system in terms of output
and flexibility
•
•
Estimation of product or service demand on the system
•
•
Space requirements for the elements in the layout
Processing requirements in terms of number of operations
and amount of flow between departments and work
centers
Space availability within the facility itself

4. 4
Criteria for a good Layout
– designing of a layout is a creative
exercise
• Maximum Flexibility
• Maximum Co-ordination
• Maximum visibility
• Maximum Accessibility
• Minimum distance
• Minimum handling
• Minimum discomfort
• Inbuilt safety
• Efficient process flow
• Identification

5. Basic Production Layout
Formats
5
• Process Layout (also called job-shop or
functional layout)
• Product Layout (also called flow-shop layout)
• Group Technology (Cellular) Layout
• Fixed-Position Layout (Static Layout)
• Service Layouts

6. Process Layout: Interdepartmental6
Flow
•
Given
• The flow (number of moves) to and from all
•
•
•
departments
The cost of moving from one department to
another
The existing or planned physical layout of
the plant
Determine
• The “best” locations for each department,
where best means maximizing flow, which
minimizing costs

7. 7
Process Layout
Advantages :
• Reduced investment on m/cs – as they are general purpose
• Production flexibility
• Maximum utilization of men and machines
• Easier maintenance
Disadvantages
• Difficulty in materials movement
• Layout requires more space
• Production time is more as WIP has to travel from point to point in
search of the machines
• Accumulation of WIP at different locations
Developing a process layout
• Graphic and schematic analysis – templates and two – dimension
cutouts of equipments drawn to scale are the layout planning tools.
• Computer models – CRAFT (Computerized Relative Allocation of
Facilities Technique)
• Load Distance Model- used to minimize the material flow

8. Process Layout: CRAFT
Approach
•
It is a heuristic program; it uses a simple
rule of thumb in making evaluations:
• "Compare two departments at a time and
exchange them if it reduces the total cost of
the layout."
•
It does not guarantee an optimal solution
•
CRAFT assumes the existence of variable
path material handling equipment such as
forklift trucks
(Computerized Relative Allocation of
Facilities Technique)
•
8

9. 9
Product Layout
Advantages
• Avoids production bottlenecks
• Economy in manufacturing time
• Requires less floor area per unit of production
• WIP reduced – consequently less inventories.
• Greater incentive to group of workers to raise their level of
production.
Disadvantages
• Layout is known for inflexibility
• Expensive layout
• Lesser scope for expansion
• M/c breakdown along the production line can disrupt the whole
system.
Developing a product layout
• Line Balancing
• Mixed – model Line balancing

10. Assembly Lines – different configurations
• Materials Handling Devices – Belt or Roller Conveyors /
•
•
•
•
•
Overhead Cranes
Line Configuration – Straight Line / U-shaped /
Branching
Pacing – Mechanical / Human
Product Mix – One Product / Multiple Products
Workstation Characteristics – Workers may sit / stand /
walk with the line
Length of the Line – Few Workers / Many Workers

11. Assembly Lines Balancing
Concepts
Question: Suppose you load work into the three work
Question: Suppose you load work into the three work
stations below such that each will take the corresponding
stations below such that each will take the corresponding
number of minutes as shown. What is the cycle time of
number of minutes as shown. What is the cycle time of
this line?
this line?
Station 1
Station 2
Station 3
Minutes
6
7
3
per Unit
Answer: The cycle time of the line is always
Answer: The cycle time of the line is always
determined by the work station taking the longest
determined by the work station taking the longest
time. In this problem, the cycle time of the line is 7
time. In this problem, the cycle time of the line is 7
minutes. There is also going to be idle time at the
minutes. There is also going to be idle time at the
other two work stations.
other two work stations.
11

12. 12
Example of Line Balancing

You’ve just been assigned the job of setting
up an electric fan assembly line with the
following tasks:
Task
A
B
C
D
E
F
G
H
Time (Mins)
2
1
3.25
1.2
0.5
1
1
1.4
Description
Assemble frame
Mount switch
Assemble motor housing
Mount motor housing in frame
Attach blade
Assemble and attach safety grill
Attach cord
Test
Predecessors
None
A
None
A, C
D
E
B
F, G

13. 13
Example of Line Balancing:
Structuring the Precedence Diagram
Task Predecessors
A
None
B
A
C
None
D
A, C
A
Task Predecessors
E
D
F
E
G
B
H
F, G
B
G
H
C
D
E
F

14. 14
Example of Line Balancing:
Precedence Diagram
Question: Which process step defines the maximum
Question: Which process step defines the maximum
rate of production?
rate of production?
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
1.4
H
Answer: Task C is the cycle time of the line and
Answer: Task C is the cycle time of the line and
therefore, the maximum rate of production.
therefore, the maximum rate of production.

15. 16
Example of Line Balancing :
Determine Cycle Time
Question: Suppose we want to assemble
Question: Suppose we want to assemble
100 fans per day. What would our cycle
100 fans per day. What would our cycle
time have to be?
time have to be?
Answer:
Answer:
Production time per period
Required Cycle Time, C =
Required output per period
420 mins / day
C=
= 4.2 mins / unit
100 units / day

16. 17
Example of Line Balancing:
Determine Theoretical Minimum
Number of Workstations
Question: What is the theoretical minimum number of
Question: What is the theoretical minimum number of
workstations for this problem?
workstations for this problem?
Answer:
Answer:
Theoretical Min. Number of Workstations, N t
Sum of task times (T)
Nt =
Cycle time (C)
11.35 mins / unit
Nt =
= 2.702, or 3
4.2 mins / unit

17. 18
Example of Line Balancing: Rules
To Follow for Loading
Workstations

Assign tasks to station 1, then 2, etc. in sequence. Keep
assigning to a workstation ensuring that precedence is
maintained and total work is less than or equal to the
cycle time. Use the following rules to select tasks for
assignment.

Primary: Assign tasks in order of the largest number of
following tasks

Secondary (tie-breaking): Assign tasks in order of the
longest operating time

18. 19
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
Followers
6
4
3
2
2
1
1
0
Time (Mins)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3

19. 20
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
A (4.2-2=2.2)
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
Followers
6
4
3
2
2
1
1
0
Time (Mins)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3

20. 21
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
A (4.2-2=2.2)
B (2.2-1=1.2)
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
Followers
6
4
3
2
2
1
1
0
Time (Mins)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3

21. 22
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
Idle= .2
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
Followers
6
4
3
2
2
1
1
0
Time (Mins)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3

22. 23
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
Idle= .2
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
C (4.2-3.25)=.95
Followers
6
4
3
2
2
1
1
0
Time (Mins)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3

23. 24
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
C (4.2-3.25)=.95
Idle= .2
Idle = .95
Followers
6
4
3
2
2
1
1
0
Time (Mins)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3

24. 25
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
C (4.2-3.25)=.95
Idle= .2
Idle = .95
Followers
6
4
3
2
2
1
1
0
Time (Mins)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
D (4.2-1.2)=3

25. 26
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
C (4.2-3.25)=.95
Idle= .2
Idle = .95
Followers
6
4
3
2
2
1
1
0
Time (Mins)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
D (4.2-1.2)=3
E (3-.5)=2.5

26. 27
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
C (4.2-3.25)=.95
Idle= .2
Idle = .95
Followers
6
4
3
2
2
1
1
0
Time (Mins)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
D (4.2-1.2)=3
E (3-.5)=2.5
F (2.5-1)=1.5

27. 28
2
A
1
B
1
G
C
D
E
F
3.25
1.2
.5
1
Station 1
1.4
H
Task
A
C
D
B
E
F
G
H
Station 2
A (4.2-2=2.2)
B (2.2-1=1.2)
G (1.2-1= .2)
C (4.2-3.25)=.95
Idle= .2
Idle = .95
Followers
6
4
3
2
2
1
1
0
Time (Mins)
2
3.25
1.2
1
0.5
1
1
1.4
Station 3
D (4.2-1.2)=3
E (3-.5)=2.5
F (2.5-1)=1.5
H (1.5-1.4)=.1
Idle = .1
Which station is the bottleneck? What is the effective cycle time?

28. 29
Example of Line Balancing: Determine
the Efficiency of the Assembly Line
Sum of task times (T)
Efficiency =
Actual number of workstations (Na) x Cycle time (C)
11.35 mins / unit
Efficiency =
=.901
(3)(4.2mins / unit)

29. 30
Cellular Manufacturing
layout
• Machines are grouped into cells and the cells
•
•
•
•
function somewhat like a product layout within a
larger process layout.
Each cell in this layout is formed to produce a
single parts family
Lower WIP inventories, reduced materials
handling costs, simplified production planning
Improved on time delivery
Improved Quality.

30. Cellular Manufacturing layout
Cell # 1
1
5
2
Cell # 2
1
2
Part
D
3
Part
X
Part
Y
4
1
Part
A
Part
B
3
1
2
2
3
Cell # 4
4
Production
operation
Cell # 3
Product or materials

31. 32
Group Technology:
Benefits
1. Better human relations
2. Improved operator expertise
3. Less in-process inventory and
material handling
4. Faster production setup

32. 33
Group Technology:
Transition from Process Layout
1. Grouping parts into families that
follow a common sequence of steps
2. Identifying dominant flow patterns
of parts families as a basis for
location or relocation of processes
3. Physically grouping machines and
processes into cells

33. 34
Fixed Position Layout
Question: What are our primary considerations
Question: What are our primary considerations
for a fixed position layout?
for a fixed position layout?
Answer: Arranging materials and equipment
Answer: Arranging materials and equipment
concentrically around the production point in
concentrically around the production point in
their order of use.
their order of use.

34. 35
Fixed position layout or
Static layout

Involves the movement of men and
machines to the product which remains
stationery.
Raw materials
Machines
Equipments
Labour
Finished
Product
(Aircraft)

35. 36
Service facility layout



Two different types of service facility layouts
exist – based on the degree of customer
contact.
A) Layout which is totally designed around the
customer receiving service functions. (Eg Banks,
Fast food joints etc.)
B) Layout which is designed around technology,
processing of materials and production (Eg.
Hospitals, Restaurants etc.)

36. Hospital Wards
Hospital wards
Hospital Wards
Suregery, Radiology,
Intensive Care and
Technical Services
Doctors
Lounge/
Offices
Hospital Wards
Parking Area
Parking Area
Casualty Department
and Inpatient Department
Nureses
Lounge/
Offices
Cafetaria
Parking Area
Entrance
Exit
Service facility layout (Hospital Layout)
Parking
Casualty Department
and Inpatient Department
Aisles/Gangways
Parking Area
Exit
Parking
Entrance

37. 38
Retail Service Layout
Goal--maximize net profit per
square foot of floor space
 Servicescapes

– Ambient Conditions
– Spatial Layout and Functionality
– Signs, Symbols, and Artifacts

38. 39
Question Bowl
Which of the following are distinguishing
features of CRAFT?
a. It is an optimization methodology
b. Does not require any assumptions
about the layout or inter-relationships
of departments
c. Can handle over 50 departments
d. All of the above
e. None of the above
Answer: e. None of the above (It is a heuristic program, does not
guarantee an optimal solution, requires layout assumptions and
can handle up to 40 departments.)

39. 40
Question Bowl
Which of the following is a process that involves
developing a relationship chart showing the
degree of importance of having each
department located adjacent to every other
department?
a.
Systematic layout planning
b.
Assembly-line balancing
c.
Splitting tasks
d.
U-shaped line layouts
e.
None of the above
Answer: a. Systematic layout planning

40. 41
Question Bowl
If the production time per day is 1200 minutes and
the required output per day is 500 units, which of
the following will be the required workstation
cycle time for this assembly line?
a.
2.4 minutes
b.
0.42 minutes
c.
1200 units
d.
500 units
e.
None of the above
Answer: a. 2.4
minutes
(1200/500=2.4
minutes)

41. 42
Question Bowl
You have just finished determining the cycle time
for an assembly line to be 5 minutes. The
sum of all the tasks required on this
assembly is is 60 minutes. Which of the
following is the theoretical minimum number
of workstations required to satisfy the
workstation cycle time?
a.
1 workstation
b.
5 workstations
Answer: c. 12
c.
12 workstations
workstations
d.
60 workstations
(60/5=12)
e.
None of the above

42. 43
Question Bowl
If the sum of the task times for an assembly line
is 30 minutes, the actual number of
workstations is 5, and the workstation cycle
time is 10 minutes, what is the resulting
efficiency of this assembly line?
a.
0.00
b.
0.60
c.
1.00
d.
1.20
e.
Can not be computed from the data above
Answer: b. 0.60 (30/(5x10)=0.60)

43. 44
Question Bowl
Which of the following are ways that we can
accommodate a 20 second task in a 18
second cycle time?
a.
Share the task
b.
Use parallel workstations
c.
Use a more skilled worker
d.
All of the above
e.
None of the above
Answer: d. All of the above

44. 45
Question Bowl
Which of the following are “ambient conditions” that
should be considered in layout design?
a.
Noise level
b.
Lighting
c.
Temperature
d.
Scent
e.
All of the above
Answer: e. All of the above

45. 46
Solved Problems – OPERATIONS MANAGEMENT
(Class of 2010)
Q2. A toaster assembly line has to turn out 50 toasters per hour. The tasks,
their times and their immediate predecessors are shown in the table below.
(10 Marks)
Task
A
B
C
D
E
F
G
H
I
J
K
L
M
Time, secs Imm Pred
28
35
29
32
28
45
37
31
36
42
49
52
37
---A
A
C
C
C
E,F
D,G
H
H
H
J,K
L

46. 47
Design a line to meet the above production rate, and show your design as a diagram
in the template given below, writing task letter above the work-station numbers.
Remember to fill in the cycle time and efficiency boxes.
Cycle time, secs:
Work Station: 1
Answer.
a)
2
Line efficiency, %:
3
4
5
6
Assembly Line Balancing
Expected Cycle Time =
1
Throughput Rate
=
60 minutes
50
= 1.2 minutes
= 72 secs. / Toaster
7
8
9
10
11

47. 48
b)
Now, workout the Total Task Time
Task
A
B
C
D
E
F
G
H
I
J
K
L
M
Total
c)
Time, secs Imm Pred
28
35
29
32
28
45
37
31
36
42
49
52
37
481
---A
A
C
C
C
E,F
D,G
H
H
H
J,K
L
Theoretical no of Workstations Required (for maximum efficiency)
= Total Task Time
Cycle Time

48. 49
= 481
72
= 6.68
= 7 workstations
d)
Now, draw the Precedence Diagram (Note:- Let us denote “O” as an Activity)
B
35
A
28
C
29
D
32
F
45
H
31
E
28
G
37
I
36
J
42
K
49
Terminal
Point
L
52
M
37

49. 50
Note:- As there are no successor of activities B and I, we assume that all three activities
B,I and M will meet at a Termination Point. So, as soon as activity M gets
completed, all of them will meet at a Termination Point
e)
Now, sequencing as per no of followers
Task (Activity)
A
B
C
D
E
F
G
H
I
J
K
L
M
No. of followers
12
0
10
6
7
7
6
5
0
2
2
1
0

50. 51
f)
Now, grouping them in descending order of their follower numbers
Tasks
A
C
E,F
D,G
H
J,K
L
B,I,M
Followers
12
10
7
6
5
2
1
0

51. 52
g)
Now, realign them on the basis of maximum time taken
Task
A
C
F
E
G
D
H
K
J
L
M
I
B
Time
28
29
45
28
37
32
31
49
42
52
37
36
35

52. 53
h)
Now, we have to regroup them such that their total time does not exceed the cycle
time
Workstation
W1
W2
W3
W4
W5
W6
W7
W8
W9
Task
A,C
F
E,G
D,H
K
J
L
M
I,B
Total Time (secs)
57
45
65
63
49
42
52
37
71

53. 54
i)
Diagram in the template Given
100
90
80
70
60
Time 50
(secs)
40
30
57
45
65
63
49
42
52
37
71
20
10
WS1 WS2
WS3 WS4 WS5 WS6 WS7 WS8 WS9
Workstation (WS)

54. 55
j)
Therefore the no. of Workstations required (N) = 9
However, we found out initially, for maximum efficiency, we need 7 (approx.)
Workstations
Therefore Line Efficiency =
(where ∑ T = 7
C
and
∑T
C×N
N = 9)
Therefore Line Efficiency = 7 ×100 = 77.78 %
9
k)
Cycle time, secs = 72
Line Efficiency, % = 77.78

55. 56
Alternate Answer to Q.2
Task
A
C
E
F
D
G
H
J
K
L
B
I
M
No.of
Followers
(descending
order)
12
10
7
7
6
6
5
2
2
1
0
0
0
Rank
Order of
Task
1
2
3
3
4
4
5
6
6
7
8
8
8
A
C
F
E
G
D
H
K
J
L
M
I
B
Time
Required
(secs)
28
29
45
28
37
32
31
49
42
52
37
36
35
Workstation
Grouping
(Time should
not exceed
72 secs)
W1
W2
W3
W4
W5
W6
W7
W8
W9

56. 57
Note : When there are two tasks with the same ranking, then place the task with the
higher cycle time first
Diagram in the Given Template
71
65
57
72
secs
45
63
52
49
G
42
H
37
C
F
A
E
D
K
J
WS1 WS2 WS3 WS4 WS5 WS6
Workstation
(WS)
I
L
M B
WS7 WS8 WS9