Download - Tn6 facility layout
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Facility Layout
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• Facility Layout and Basic Formats
• Process Layout
• Layout Planning
• Assembly Line balancing
• Service Layout
OBJECTIVES
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Facility Layout Defined
Facility layout can be defined as the process by which the placement of departments, workgroups within departments, workstations, machines, and stock-holding 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
• Processing requirements in terms of number of operations and amount of flow between departments and work centers
• Space requirements for the elements in the layout
• Space availability within the facility itself
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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
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Basic Production Layout Formats
• 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
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Process Layout: Interdepartmental 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
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Process Layout
Advantages : • Reduced investment on m/cs – as they are general purpose• Production flexibility • Maximum utilization of men and machines• Easier maintenanceDisadvantages• 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 locationsDeveloping 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
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8Process 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)
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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
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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
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Station 1
Minutes per Unit 6
Station 2
7
Station 3
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Assembly Lines Balancing Concepts
Question: Suppose you load work into the three work stations below such that each will take the corresponding number of minutes as shown. What is the cycle time of this line?
Question: Suppose you load work into the three work stations below such that each will take the corresponding number of minutes as shown. What is the cycle time of this line?
Answer: The cycle time of the line is always determined by the work station taking the longest time. In this problem, the cycle time of the line is 7 minutes. There is also going to be idle time at the other two work stations.
Answer: The cycle time of the line is always determined by the work station taking the longest time. In this problem, the cycle time of the line is 7 minutes. There is also going to be idle time at the other two work stations.
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Example of Line Balancing
You’ve just been assigned the job of setting up an electric fan assembly line with the following tasks:
Task Time (Mins) Description PredecessorsA 2 Assemble frame NoneB 1 Mount switch AC 3.25 Assemble motor housing NoneD 1.2 Mount motor housing in frame A, CE 0.5 Attach blade DF 1 Assemble and attach safety grill EG 1 Attach cord BH 1.4 Test F, G
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Example of Line Balancing: Structuring the Precedence Diagram
Task PredecessorsA None
A
B A
B
C None
C
D A, C
D
Task PredecessorsE D
E
F E
F
G B
G
H F, G
H
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Example of Line Balancing: Precedence Diagram
A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
Question: Which process step defines the maximum rate of production?
Question: Which process step defines the maximum rate of production?
Answer: Task C is the cycle time of the line and therefore, the maximum rate of production.
Answer: Task C is the cycle time of the line and therefore, the maximum rate of production.
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Example of Line Balancing: Determine Cycle Time
Required Cycle Time, C = Production time per period
Required output per periodRequired Cycle Time, C =
Production time per period
Required output per period
C = 420 mins / day
100 units / day= 4.2 mins / unitC =
420 mins / day
100 units / day= 4.2 mins / unit
Question: Suppose we want to assemble 100 fans per day. What would our cycle time have to be?
Question: Suppose we want to assemble 100 fans per day. What would our cycle time have to be?
Answer: Answer:
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Example of Line Balancing: Determine Theoretical Minimum Number of
Workstations
Question: What is the theoretical minimum number of workstations for this problem?
Question: What is the theoretical minimum number of workstations for this problem?
Answer: Answer: Theoretical Min. Number of Workstations, N
N = Sum of task times (T)
Cycle time (C)
t
t
Theoretical Min. Number of Workstations, N
N = Sum of task times (T)
Cycle time (C)
t
t
N = 11.35 mins / unit
4.2 mins / unit= 2.702, or 3t
N = 11.35 mins / unit
4.2 mins / unit= 2.702, or 3t
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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
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A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
Station 1 Station 2 Station 3
Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4
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A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
Station 1 Station 2 Station 3
A (4.2-2=2.2)
Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4
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A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
A (4.2-2=2.2)B (2.2-1=1.2)
Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4
Station 1 Station 2 Station 3
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A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4
Station 1 Station 2 Station 3
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A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
C (4.2-3.25)=.95
Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
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C (4.2-3.25)=.95
Idle = .95
A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
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C (4.2-3.25)=.95
Idle = .95
A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
D (4.2-1.2)=3
Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
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A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
C (4.2-3.25)=.95
Idle = .95
D (4.2-1.2)=3E (3-.5)=2.5
Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
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A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
C (4.2-3.25)=.95
Idle = .95
D (4.2-1.2)=3E (3-.5)=2.5F (2.5-1)=1.5
Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
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A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
C (4.2-3.25)=.95
Idle = .95
D (4.2-1.2)=3E (3-.5)=2.5F (2.5-1)=1.5H (1.5-1.4)=.1Idle = .1
Task Followers Time (Mins)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4
A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)
Idle= .2
Station 1 Station 2 Station 3
Which station is the bottleneck? What is the effective cycle time?
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Example of Line Balancing: Determine the Efficiency of the Assembly Line
Efficiency =Sum of task times (T)
Actual number of workstations (Na) x Cycle time (C)Efficiency =
Sum of task times (T)
Actual number of workstations (Na) x Cycle time (C)
Efficiency =11.35 mins / unit
(3)(4.2mins / unit)=.901Efficiency =
11.35 mins / unit
(3)(4.2mins / unit)=.901
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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.
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Cellular Manufacturing layout
1 23
5 4
1 23
4
Part APart B
1 2 3 Part D
1 2
Part XPart Y
Cell # 3Cell # 4
Cell # 1 Cell # 2
Production operation
Product or materials flow
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Group Technology:Benefits
1. Better human relations
2. Improved operator expertise
3. Less in-process inventory and material handling
4. Faster production setup
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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
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Fixed Position Layout
Question: What are our primary considerations for a fixed position layout?
Question: What are our primary considerations for a fixed position layout?
Answer: Arranging materials and equipment concentrically around the production point in their order of use.
Answer: Arranging materials and equipment concentrically around the production point in their order of use.
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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)
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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.)
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Parking AreaCasualty Department
and Inpatient Department Parking Area
Hospital Wards
Hospital wards
Hospital Wards
Suregery, Radiology, Intensive Care and Technical Services
CafetariaNuresesLounge/Offices
Doctors Lounge/Offices
Hospital Wards
Parking AreaCasualty Department
and Inpatient Department Parking Area
Entrance Exit
Ais
les/
Gan
gway
s
Par
king
Par
king
Entrance Exit
Service facility layout (Hospital Layout)
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Retail Service Layout
Goal--maximize net profit per square foot of floor space
Servicescapes– Ambient Conditions– Spatial Layout and Functionality– Signs, Symbols, and Artifacts
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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 departmentsd. All of the abovee. 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.)
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Question BowlWhich 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
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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)
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Question BowlYou 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 workstationb. 5 workstationsc. 12 workstationsd. 60 workstationse. None of the above
Answer: c. 12 workstations (60/5=12)
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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)
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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
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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
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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 Time, secs Imm Pred
A 28 ----B 35 AC 29 AD 32 CE 28 CF 45 CG 37 E,FH 31 D,GI 36 HJ 42 HK 49 HL 52 J,KM 37 L
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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: Line efficiency, %:
Work Station: 1 2 3 4 5 6 7 8 9 10 11
Answer. Assembly Line Balancing
a) Expected Cycle Time = 1
Throughput Rate
= 60 minutes
50
= 1.2 minutes
= 72 secs. / Toaster
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b) Now, workout the Total Task Time
c) Theoretical no of Workstations Required (for maximum efficiency)
= Total Task Time Cycle Time
Task Time, secs Imm Pred
A 28 ----B 35 AC 29 AD 32 CE 28 CF 45 CG 37 E,FH 31 D,GI 36 HJ 42 HK 49 HL 52 J,KM 37 L
Total 481
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= 481
72
= 6.68
= 7 workstations
d) Now, draw the Precedence Diagram (Note:- Let us denote “O” as an Activity)
A28
B35
C29
F45
D32
E28
J42
H31
G37
I36
Terminal Point
M37
L52
K49
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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) No. of followers A 12 B 0 C 10 D 6 E 7 F 7 G 6 H 5 I 0 J 2 K 2 L 1 M 0
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f) Now, grouping them in descending order of their follower numbers
Tasks Followers
A 12
C 10
E,F 7
D,G 6
H 5
J,K 2
L 1
B,I,M 0
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g) Now, realign them on the basis of maximum time taken
Task Time
A 28
C 29
F 45
E 28
G 37
D 32
H 31
K 49
J 42
L 52
M 37
I 36
B 35
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h) Now, we have to regroup them such that their total time does not exceed the cycle time
Workstation Task Total Time (secs) W1 A,C 57
W2 F 45
W3 E,G 65
W4 D,H 63
W5 K 49
W6 J 42
W7 L 52
W8 M 37
W9 I,B 71
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i) Diagram in the template Given
10
100
90
80
70
60
50
40
30
20
WS1 WS2
WS5
WS3
WS7
WS4
WS6
WS9
WS8
57 45 65 63 5242 3749 71
Time (secs)
Workstation (WS)
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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 = ∑ T C × N
(where ∑ T = 7 and N = 9)
C
Therefore Line Efficiency = 7 ×100 = 77.78 %
9
k) Cycle time, secs = 72
Line Efficiency, % = 77.78
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Alternate Answer to Q.2
Task No.of Rank Order of Time Workstation Followers Task Required Grouping (descending (secs) (Time should order) not exceed
72 secs)A 12 1 A 28C 10 2 C 29E 7 3 F 45 W2F 7 3 E 28D 6 4 G 37G 6 4 D 32H 5 5 H 31J 2 6 K 49 W5K 2 6 J 42 W6L 1 7 L 52 W7B 0 8 M 37 W8I 0 8 I 36M 0 8 B 35
W1
W3
W4
W9
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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
5745
6563
4942
52
37
71
C
AF
G
E
H
DK J L M
I
B
Workstation (WS)
WS1
WS2
WS3
WS4
WS5
WS6
WS7
WS8
WS9
72 secs