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Slide 1 of 96
Manufacturing Facility Layout
Slide 2 of 96
Basic Layout Forms
• Process
• Product
• Cellular
• Fixed position
• Hybrid
Slide 3 of 96
Sistem produksi IntermittenLay out by process
Saw
Mill
Grind
Lathe
Paint
Drill
Ass
emb
ly
War
ehou
se
Sto
res
Slide 4 of 96
Tata Letak Proses (Manufacturing Process Layout)
L
L
L
L
L
L
L
L
L
L M
M
M
M
D
D
D
D
D
D
D
D
G
G
G
G
G
G
P
P
A A AReceiving andShipping
Assembly
Painting Department
Lathe Department
Milling Department
Drilling Department
Grinding Department
Slide 5 of 96
Sistem produksi ContinousLay out by product
Saw
Saw
Grind
Weld
Lathe
Mill
Mill
Grind
Mill
Drill
Drill
Lathe
Drill
Paint
Paint
Drill
Sto
res
Ass
e mb
ly
War
ehou
s e
Slide 6 of 96
Process (Job Shop) Layouts
• Equipment that perform similar processes are grouped together
• Used when the operations system must handle a wide variety of products in relatively small volumes (i.e., flexibility is necessary)
Slide 7 of 96
Characteristics of Process Layouts
• General-purpose equipment is used
• Changeover is rapid
• Material flow is intermittent
• Material handling equipment is flexible
• Operators are highly skilled
• . . . more
Slide 8 of 96
Characteristics of Process Layouts
• Technical supervision is required
• Planning, scheduling and controlling functions are challenging
• Production time is relatively long
• In-process inventory is relatively high
Slide 9 of 96
Product (Assembly Line) Layouts
• Operations are arranged in the sequence required to make the product
• Used when the operations system must handle a narrow variety of products in relatively high volumes
• Operations and personnel are dedicated to producing one or a small number of products
Slide 10 of 96
Characteristics of Product Layouts
• Special-purpose equipment are used
• Changeover is expensive and lengthy
• Material flow approaches continuous
• Material handling equipment is fixed
• Operators need not be as skilled
• . . . more
Slide 11 of 96
Characteristics of Product Layouts
• Little direct supervision is required
• Planning, scheduling and controlling functions are relatively straight-forward
• Production time for a unit is relatively short
• In-process inventory is relatively low
Slide 12 of 96
Product Layout-Advantages/Disadvanta
gesAdvantages:Low cost variable cost
per unitLower material
handling costs reduction in work in-
process inventorieseasier training and
supervision
Disadvantages:High volume required
because of large initial investment
Lack of flexibility in handling variety of products or production rates
Slide 13 of 96
Tata Letak Proses (Manufacturing Process Layout)
L
L
L
L
L
L
L
L
L
L M
M
M
M
D
D
D
D
D
D
D
D
G
G
G
G
G
G
P
P
A A AReceiving andShipping
Assembly
Painting Department
Lathe Department
Milling Department
Drilling Department
Grinding Department
Slide 14 of 96
Part Routing MatrixReordered To Highlight Cells
MACHINESPARTS 1 2 4 8 10 3 6 9 5 7 11 12 A x x x x x D x x x x x F x x x C x x x G x x x x B x x x H x x x E x x x
Slide 15 of 96
Cellular Layout Solution
12
12 3
4
5
6
7
8 910
11
A BCRaw materials
Cell1 Cell 2 Cell 3
Assembly
Slide 16 of 96
Designing and Analyzing a Product Layout
• Line Balancing
Slide 17 of 96
Designing and Analyzing a Product Layout
• Characteristics
• Inputs
• Design Procedure
• How Good Is The Layout?
Slide 18 of 96
Line Balancing Problem• Work stations are arranged so that the output of one
is an input to the next, i.e., a series connection
• Layout design involves assigning one or more of the tasks required to make a product to work stations
• . . . more
Slide 19 of 96
Line Balancing Problem
• The objective is to assign tasks to minimize the workers’ idle time, therefore idle time costs, and meet the required production rate for the line
• In a perfectly balanced line, all workers would complete their assigned tasks at the same time (assuming they start their work simultaneously)
• This would result in no idle time• . . . more
Slide 20 of 96
Inputs
• The production rate required from the product layout or the cycle time.– The cycle time is the reciprocal of the
production rate and visa versa
• All of the tasks required to make the product– It is assumed that these tasks can not be
divided further
• . . . more
Slide 21 of 96
Inputs
• The estimated time to do each task
• The precedence relationships between the tasks– These relationships are determined by the
technical constraints imposed by the product– These relationships are displayed as a
network known as a precedence diagram
Slide 22 of 96
Design Procedure
1. If not provided, find the cycle time for the line. Remember the cycle time is the reciprocal of the production rate. Make sure the cycle time is expressed in the same time units as the estimated task times.
2. Select the line-balancing heuristic that may be used to help with the assignments. (Two heuristics are described at the end of this procedure.)
. . . more
Slide 23 of 96
Design Procedure
3. Open a new work station with the full cycle time remaining.
4. Determine which tasks are feasible, i.e., can be assigned to this work station at this time. For a task to be feasible, two conditions must be met:– All tasks that precede that task must have already
been assigned– The estimated task time must be less than or equal to
the remaining cycle time for that work station.
Slide 24 of 96
Line-Balancing HeuristicsLine-Balancing Heuristics
• Heuristic methods, based on simple rules, have been used to develop very good, not optimal, solutions to line balancing problems.
Slide 25 of 96
How Good Is the Design?
• Utilization is one way of objectively determining how near perfectly balanced an assignment scheme is.
• Utilization is the percentage of time that a production line is working.
• Utilization is calculated as:
or
100 stations) work ofnumber (Actual Time) (Cycle
s task timeall of Sumx
x
100ons workstatiofnumber Actual
ons workstatiofnumber Minimumx
Slide 26 of 96
Station 1
Min/Unit 6
Station 2
7
Station 3
3
Why is Balancing the Line Important?
What’s Going to Happen?
Slide 27 of 96
Example 1: The ALB Problem• You’ve just been assigned the job a 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
Slide 28 of 96
Example 1: The ALB Problem The Precedence Diagram
A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
• Which process step defines the maximum rate of production?
Slide 30 of 96
Example 1: The ALB Problem We want to assemble 100 fans per day
Required Cycle Time, C = Production time per period
Required output per period
C = 420 mins / day
100 units / day= 4.2 mins / unit
What do these numbers this represent?
Slide 31 of 96
Example 1: The ALB Problem We want to assemble 100 fans per day
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
Why should we always round up?Why should we always round up?
Slide 32 of 96
Example 1: The ALB ProblemSelected Task Selection Rules
• Primary: Assign tasks in order the the largest number of following tasks.
• Secondary (tie-breaking): Assign tasks in order of the longest operating time
Slide 33 of 96
Example 1: The ALB ProblemSelected Task Selection Rules
Precedence Diagram
A
C
B
D E F
GH
2
3.25
1
1.2 .5
11.4
1
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 (Min)A 6 2C 4 3.25D 3 1.2B 2 1E 2 0.5F 1 1G 1 1H 0 1.4
Slide 34 of 96
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 (Min)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)
Slide 35 of 96
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 (Min)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)
Slide 36 of 96
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 (Min)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
Slide 37 of 96
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 (Min)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
C (4.2-3.25)=.95
Idle = .95
Slide 38 of 96
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 (Min)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
C (4.2-3.25)=.95
Idle = .95
D (4.2-1.2)=3
Slide 39 of 96
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 (Min)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
C (4.2-3.25)=.95
Idle = .95
D (4.2-1.2)=3E (3-.5)=2.5
Slide 40 of 96
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 (Min)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
C (4.2-3.25)=.95
Idle = .95
D (4.2-1.2)=3E (3-.5)=2.5F (2.5-1)=1.5
Slide 41 of 96
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 (Min)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)
C (4.2-3.25)=.95 D (4.2-1.2)=3E (3-.5)=2.5F (2.5-1)=1.5H (1.5-1.4)=.1
Idle=.2 Idle=.95 Idle=.1
Slide 42 of 96
Slide 43 of 96
Example 1: The ALB Problem
• Which station is the bottleneck?
• What is the effective cycle time?
Efficiency =Sum of task times (T)
Actual number of workstations (Na) x Cycle time (C)
Efficiency =11.35 mins / unit
(3)(4.2mins / unit)=.901