“Cellular Manufacturing� by
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ORIGINS • FLANDERS’ PRODUCT ORIENTED DEPARTMENTS FOR STANDARIZED PRODUCTS WITH MINIMAL TRANSPORTATION (1925) • SOKOLOVSKI/MITROFANOV: PARTS WITH SIMILAR FEATURES MANUFACTURED TOGETHER
BASIC PRINCIPLE • SIMILAR “THINGS” SHOULD BE DONE SIMILARLY • “THINGS “ – PRODUCT DESIGN – PROCESS PLANNING – FABRICATION &ASSEMBLY – PRODUCTION CONTROL – ADMINISTRATIVE FUNCTIONS
TENETS OF GROUP TECHNOLOGY
• DIVIDE THE MANUFACTURING FACILITY INTO SMALL GROUPS OR CELLS OF MACHINES (1-5) • THIS IS CALLED CELLULAR
MANUFACTURING
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.
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
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
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
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.
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.
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.
The Difference between group and functional layout:
Evaluation criteria of Cell Design 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)
Cell Formation Approach Machine - Component Group Analysis: Machine - Component Group Analysis is based on production flow analysis
Machine - Component Group Analysis 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 Production flow analysis involves four stages: 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.
Machine - Component Group Analysis Production flow analysis involves four stages: 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.
Machine - Component Group Analysis
Production flow analysis involves four stages: 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.
Machine - Component Group Analysis Example: Consider a problem of 4 machines and 6 parts. Try to group them. Components
Machines
1
2
3
4
5
6
M1
1
1
1
M2
1
1
1
M3
1
1
1
M4
1
1
1
Machine - Component Group Analysis Solution
Components
Machine s M1
2
4
6
1
3
5
1
1
1
M2
1
1
1
M3
1
1
1
M4
1
1
1
Cellular Layout Process (Functional) Layout
Group (Cellular) Layout A cluster or cell
T T M M
T T M M
T T D D
CG SG D D
Similar resources placed together
CG
T
T
T
SG
M
M
T
D
D
M
D
SG
CG
CG
D
D D
M
SG
Resources to produce similar products placed together
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
Process Flows before the Use of GT Cells
Process Flows after the Use of GT Cells
Designing Product Layouts • Designing product layouts requires consideration of: – Sequence of tasks to be performed by each workstation – Logical order
– Speed considerations – line balancing
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
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 B C D E F G H I
Roll dough Place on cardboard backing Sprinkle cheese Spread Sauce Add pepperoni Add sausage Add mushrooms Shrinkwrap pizza Pack in box
None A B C D D D E,F,G H Total task time
50 5 25 15 12 10 15 18 15 165
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 Cycle time (sec./unit ) =
–
available time ( sec./day ) 60 min/hr x 60 sec/min = = 60 sec./unit desired output ( units/hr ) 60 units/hr
Limited by the bottleneck task (the longest task in a process):
Maximum output =
available time 3600 sec./hr. = = 72 units/hr, or pizzas per hour bottleneck task time 50 sec./unit
Layout Calculations • Step 4: Compute the theoretical minimum
number of stations
– TM = number of stations needed to achieve 100% efficiency (every second is used)
TM =
∑ ( task times ) = cycle time
165 seconds = 2.75, or 3 stations 60 sec/station
– Always round up (no partial workstations) – Serves as a lower bound for our analysis
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 1
2
3
Eligible task A B C D E, F, G E, F F H I
Task Selected A B C D G E F H I
Task time 50 5 25 15 15 12 10 18 15
Idle time 10 5 35 20 5 48 38 20 5
Last Layout Calculation • Step 6: Compute efficiency and balance delay – Efficiency (%) is the ratio of total productive time divided by total time Efficiency (%) =
∑t = NC
165 sec. ( 100) = 91.7% 3 stations x 60 sec.
– Balance delay (%) is the amount by which the line falls short of 100% Balance delay = 100% − 91.7% = 8.3%
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
LINE BALANCING
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
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
Terminology and Definitions • Minimum Work Element • Total Work Content • Workstation Process time • Cycle Time • Precedence Constraints • Balance Delay
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: Tej • is a constant
Tej
Total Work Content • Aggregate of work elements
n
Twc = ∑ Tej j =1
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
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
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
Balance Delay • Measure of line inefficiency due to imbalances in station times
nTc − Twc d= nTc
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
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:
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
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
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 D interpretation. Work element F, H 40 B for example, can be done 20 E 30 anywhere on the line after 6 element C is completed. A F However, element I must 40 C 25 await completion of 50 I elements F and G. Precedence Diagram for Assembling the Big Broadcaster
G 15
18
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
ď Ź
The work content for each station is equal (or nearly so, but less than) the cycle time for the line
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 D B
E
30 F C
25
50
I G
Firtilizer Precedence Diagram Solution
20
6
A 40
H
40
15
18
Calculating Line Efficiency c.
Now calculate the efficiency measures of a five-station solution:
Efficiency Σt = nc
(100) =244 = 81.3% 5(60)
Balance delay (%) = 100 – Efficiency = 100% - 81.3% = 18.7%
Idle time = nc – Σt = 5(60) – 244 = 56 seconds
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=
where
1 r
c = cycle time in hours r = desired output rate
A Line Process • The theoretical minimum number of stations is TM =
Σt c
where each unit
Σt = total time required to assemble
A Line Process • Idle time, efficiency, and balance delay Idle time = nc – Σt where n = number of stations
Efficiency (%) =
Σt nc(100)
Balance delay (%) = 100 – Efficiency
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 A B C D E F G H
Time (sec) 40 80 30 25 20 15 120 145
Immediate Predecessor(s) None A D, E, F B B B A G
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 1 8 hours (3,600 sec/hr) = 150 sec/unit c= = r 192 units
Solved Problem 2 b.
The sum of the work-element times is 720 seconds, so
ÎŁt TM = c
720 sec/unit = 150 sec/unit-station
which may not be achievable.
= 4.8
or 5 stations
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. D 25 B
E
C
80
20
30
A
40 G gure 7.6 – Precedence Diagram 120
F
J
15
115
H
I
145
130
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
Solved Problem 2 Cumulative Time (sec)
Idle Time (c= 150 sec)
Candidate(s)
S1
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
S2
S3 S4
Choice
Work-Element Time (sec)
Station
D 25
S5
B
E
C
80
20
30
40
115
F
A G 120
15 H 145
J
I 130
Solved Problem 2 d.
Calculating the efficiency, we get
ÎŁt Efficiency (%) = nc
(100) =
720 sec/unit 5(150 sec/unit)
= 96% Thus, the balance delay is only 4 percent (100–96).
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