Introduction to Cellular Manufacturing - ADDVALUE - Nilesh Arora

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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 LayoutIn 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:

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

Cell Formation Approach

Machine - Component Group Analysis:

Machine - Component Group Analysis is based on production flow 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

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:


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.



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.



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


Machines

2 4 6 1 3 5

M1 1 1 1

M2 1 1 1

M3 1 1 1

M4 1 1 1

Components


Solution

Cellular Layout

Process (Functional) Layout Group (Cellular) Layout

Similar resources placed together

Resources to produce similar products placed together

T T TM

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

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 predecessorsStep 2: Determine TAKT TIMEStep 3: Determine cycle timeStep 4: Compute the Theoretical Minimum number of

StationsStep 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 DiagramImmediate Task Time

Work Element Task Description Predecessor (secondsA Roll dough None 50B Place on cardboard backing A 5C Sprinkle cheese B 25D Spread Sauce C 15E Add pepperoni D 12F Add sausage D 10G Add mushrooms D 15H Shrinkwrap pizza E,F,G 18I Pack in box H 15

Total task time 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

– Limited by the bottleneck task (the longest task in a process):

sec./unit 60

units/hr 60sec/min 60min/hr x 60

units/hroutput desired sec./day timeavailable)(sec./unit timeCycle

hourper pizzasor units/hr, 72sec./unit 50

sec./hr. 3600time task bottleneck

time availableoutput Maximum

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

stations 3or 2.75,

nsec/statio 60seconds 165

time cycletimes task

TM

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 timeA A 50 10B B 5 5C C 25 35D D 15 20

E, F, G G 15 5E, F E 12 48

F F 10 38H H 18 20I I 15 5

1

2

3

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%100sec. 60x stations 3

sec. 165NC

t (%) Efficiency

8.3%91.7%100%delay Balance

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:• is a constant ejT

ejT

Total Work Content

• Aggregate of work elements

n

jejwc TT

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

c

wcc

nTTnTd

Line Balancing Example

EXAMPLEGreen 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/hrCycle Time = 1/Throughput rate= 1/60 = 1 minute = 60 seconds

Line balancing ExampleAssume 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 NoneB Insert impeller shaft 30 AC Attach axle 50 AD Attach agitator 40 BE Attach drive wheel 6 BF Attach free wheel 25 CG Mount lower post 15 CH Attach controls 20 D, EI Mount nameplate 18 F, G

Total 244

Draw a Precedence DiagramSOLUTIONThe 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

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 minimized2. 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

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

25C

50

E

6

B

30

A

40

G

15

Calculating Line Efficiency

c. Now calculate the efficiency measures of a five-station solution:

Efficiency = (100) =tnc

2445(60)

= 81.3%

Idle time = nc – t = 5(60) – 244 = 56 seconds

Balance delay (%) = 100 – Efficiency = 100% - 81.3% = 18.7%

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 = 1r

wherec = cycle time in hoursr = desired output rate

A Line Process

• The theoretical minimum number of stations is

TM =tc

wheret = total time required to assemble each unit

A Line Process

• Idle time, efficiency, and balance delay

Idle time = nc – t

wheren = number of stations

Efficiency (%) = (100)tnc

Balance delay (%) = 100 – Efficiency

Solved Problem 2A 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

Solved Problem 2a. 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?

SOLUTIONa. Substituting in the cycle-time formula, we get

c = =1r

8 hours192 units

(3,600 sec/hr) = 150 sec/unit

Solved Problem 2b. The sum of the work-element times is 720 seconds, so

TM =tc = = 4.8 or 5 stations

720 sec/unit150 sec/unit-station

which may not be achievable.

Solved Problem 2c. 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

130H

145

B

80

G

120

A

40Figure 7.6 – Precedence Diagram

Work Element

Immediate Predecessor(s)

A NoneB AC D, E, FD BE BF BG AH GI HJ C, I

Solved Problem 2

J115

C

30

D25

E

20

F15 I

130H145

B

80

G

120

A40

A A 40 40 110B B 80 120 30D,

E, F D 25 145 5E,

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

S3S4

S5

Solved Problem 2d. Calculating the efficiency, we get

Thus, the balance delay is only 4 percent (100–96).

Efficiency (%) = (100)tnc = 720 sec/unit

5(150 sec/unit)

= 96%

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