Chapter 6Chapter 6

Process Selection and Process Selection and Facility LayoutFacility Layout

6-1

P S l tiProcess Selection

• How an organization chooses to produce its goods or provide its produce its goods or provide its services

• Key aspects– make or buy decisionsy– capital intensity

process flexibility– process flexibility

6-2

• M k B ?• Make or Buy?

– Available capacityAvailable capacity

– The nature of demand

– Expertise

– Quality consideration

– Cost

6-3

Types of Processes• Continuous Processing

– A highly standardized (uniform) product ( i t)(e.g., newsprint)

• Repetitive/AssemblyAll i ( TV)– Allow some variety (e.g., auto, TV)

• Intermittent/Batch Processingh dl i f d– handle a variety of products

• Job Shops– Small runs, custom products

6-4

Product-Process Matrix --Product Life Cycle

Product variety

High Moderate Low Very low

Equipment High Moderate Low Very lowEquipment Flexibility

High Moderate Low Very low

Low volume Job shop Flexibility-

Moderate volume

Batch

yQuality

volume

High volume Repetitive

Very high volume

Continuous Flow Dependability-

Cost

6-5

Flexibility-Quality Dependability-Cost

A iAutomation• Processes that have sensing and control • Processes that have sensing and control

devices that enable it to operate automaticallyautomatically– Numerically controlled (NC) machines (CNC,

DNC)DNC)– Computer-aided design and manufacturing

systems (CAD/CAM)systems (CAD/CAM)– Computer-integrated manufacturing (CIM)

Manufacturing cell – Manufacturing cell – Flexible manufacturing systems (FMS)

6-6

Manufacturing CellManufacturing Cell

• It consists of one or a small number of NC It consists of one or a small number of NC machines that produce a family of similar parts.parts.

Manufacturing Manufacturing Cell

6-7

Flexible Manufacturing SystemFlexible Manufacturing System

• It is a group of machines that include It is a group of machines that include supervisory computer control, automatic, and robots or other automated processing and robots or other automated processing equipment. (It is automated manufacturing cell)cell)

6-8

Automation requires VolumeAutomation requires VolumeVariable cost

Fixed cost

Before After

• Automation has been justified by volume through economy of scaleg y

• However, FMS allows a variety of products to be produced and still achieve volume

6-9

to be produced and still achieve volume (economy of scope)

Facility LayoutFacility Layout

6-10

L ?• Layout - the arrangement or configuration of

Layout?• Layout - the arrangement or configuration of

departments, work centers, and equipment within the confines of a building or areat t e co es o a bu d g o a ea

• Focuses on the movement of work (customers • Focuses on the movement of work (customers or material) through the system

• May involve designing a completely new layout or redesigning an existing layoutlayout or redesigning an existing layout

6-11

Th N d f L D i iThe Need for Layout Decisions

I ffi i t ti hi h t b ttl k• Inefficient operations: e.g., high cost bottleneck• Changes in the design of products and services• Introduction of new product and services• Accidents and safety hazards• Changes in environment or other legal

requirement• Changes in volume of output or mix of products• Changes in methods and equipmentg q p• Morale problems

6-12

B i L t TBasic Layout Types

• ProductP • Process

• Cellular• Fixed Position

6-13

Product Layout (1 of 3)Product Layout (1 of 3)

L t th t t d di d i • Layout that uses standardized processing operations to achieve smooth, rapid, high-volume flow

• Work stations or areas are arranged di t fi d f t k according to a fixed sequence of tasks

required to produce a product

6-14

Product Layout (2 of 3)Product Layout (2 of 3)

Raw materials Finished Station Station StationStationa ate a sor customer

s ed item

Stat o2

Stat o3

Stat o4

M t i l

Stat o1

M t i l M t i l M t i lMaterialand/or labor

Materialand/or labor

Materialand/or labor

Materialand/or labor

Layout design involves assigning one or more y g g gof the tasks (time) required to make a product at work stations

6-15

Product Layout (3 of 3)Advantages

Hi h t f t tDisadvantages

Product Layout (3 of 3)

• High rate of output• Low unit cost• Labor specialization

• Dull, repetitive jobs• Unskilled workers

• Labor specialization• Low material-handling cost• High utilization of labor and

• Fairly inflexible• Susceptible to shutdowns• High utilization of labor and

equipment• Routing and scheduling

p• PM, quick repairs, and

spare equipment partsg gdesigned initially

• Accounting, purchasing, d i t t l f i l

• Incentive plan impractical

and inventory control fairly routine

6-16

InOutA U-Shaped Production Line

110

Out• More compact than the straight

production line

29

p• Permits increased communication

among workers

38Workers

among workers• Increased flexibility in work

assignments

47

assignments• Materials enter at about the same

l h fi i h d d i 47place that finished products exit

6-17

Process Layout (1 of 3)Process Layout (1 of 3)

• Layout that can handle varied processing requirementsp g q– each product/service may require a

different sequence of operations (routing)different sequence of operations (routing)– wide variety of products/services in small

olumesvolumes• Work areas are arranged according to

the processes being performed, i.e., like processes are located together

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processes are located together

Process Layout (2 of 3)Process Layout (2 of 3)

Milling

Assembly& Test Grinding

Drilling Plating

Process Layout - work travels to dedicated process centers

6-19

Process Layout (3 of 3)Advantages Disadvantages

Process Layout (3 of 3)g

• Handles a variety of processing requirements

• In-process inventory can be high

• routing and scheduling g

• Not vulnerable to equipment failure

• routing and scheduling challenging

• Equipment utilization low• General-purpose

equipment

Equipment utilization low• Material-handling cost high• High supervision costs

• Individual incentive systems possible

g p• High unit cost • Accounting, purchasing, and

inventory control challenging

6-20

C ll l L tCellular Layouts

• Cellular ManufacturingL t i hi h hi d i t – Layout in which machines are grouped into a cell that can process items that have i il i i tsimilar processing requirements

• Group TechnologyG oup ec o ogy– The grouping into part families of items

with similar design or manufacturing with similar design or manufacturing characteristics

6-21

Figure 6-5Cellular Manufacturing Layout

Figure 6 5

-1111 -1111Lathe Mill Drill Heat treat

Gearcut

222222222 - 2222 lyMill Drill Heat t t Grind222222222 2222

ssem

bl

3333Heat

treat

As

3333333333 - 3333Lathe Mill Heat treat Grind

44444444444444 - 4444Mill Drill Gearcut

6-22

W k C ll I C ll l l tWork Cells In Cellular layouts

• A h b id b t i t d d • A hybrid between a process-oriented and product-oriented layouts

• G t h l id tifi f il f • Group technology identifies a family of items that require similar processing

• Th h d b th b f • The processes shared by the members of a family form a small product layout

• V lu f th ti family ju tifi th • Volume of the entire family justifies the dedication of resources

• Requires a high le el of training and • Requires a high level of training and flexibility on the part of the employees

6-23

Cellular Layouts Some AdvantagesCellular Layouts — Some Advantages

Reduced work-in-process inventoryp yfloor spaceraw material and finished goods inventoriesraw material and finished goods inventoriesdirect labor costs

Heightened sense of employee Heightened sense of employee participationIncreased utilization of equipment machinery

6-24

y

Fixed Position Layout

L t i hi h th d t j t

Fixed Position Layout

• Layout in which the product or project remains stationary, and workers, materials, and equipment are moved as needed

6-25e.g., Shipbuilding

Back to Product LayoutBack to Product Layout..

Raw materialsor customer

Finished item

Station2

Station3

Station4

Station

1Materialand/or

1Materialand/or

Materialand/or

Materialand/or

labor labor labor labor

Layout design involves assigning one or more of the tasks (time) required to make a product of the tasks (time) required to make a product to work stations

6-26

Cycle Time

Cycle time is the maximum time ll d t h k t ti tallowed at each workstation to

complete its set of tasks on a unit.

6-27

Li B l i R l ti hiOTdaypertimeoperating

Line Balancing Relationships

OTdaypertimeoperating

CTOT

timecycledayper timeoperatingcapacityOutput ==

DOTCT;

day)per (units rate demand desireddayper timeoperating= timeCycle =

tN;ifor task time

=onsworkstatiofnumberMinimum min1 Σ=

∑=

m

i

CTN;

timecycleonsworkstatiofnumber Minimum min

timesidleofSum x100 time)(cyclestations)x ofnumber (actual

timesidleof Sum= timeidle of Percentage

6-28

Inputs Needed to Design a Product Layout (1 of 2)

• The production rate required from the product layout or the cycle time.

• List of all of the tasks required to make the product (e.g., a, b, c, d, e)p ( g , , , , , )

• The estimated time to do each taskTask Task timea 0.1 minb 1 0 minb 1.0 minc 0.7 mind 0.5 min

6-29e 0.2 min

Inputs Needed to Design a Product Layout (2 of 2)

• The precedence relationships between the tasks

0.1 min. 1.0 min.

a b

c d ec d e0.7 min. 0.5 min. 0.2 min.

6-30

D i i P d L (1 f 2)Designing Product Layouts (1 of 2)

Station 1 Station 2 Station 3 Station 4

• Required tasks are assigned to workstations• The objective is to minimize the workers’ idle time

therefore idle time costs, and meet the required , qproduction rate for the line Task Task time

a 0.1 minb 1.0 minc 0.7 min

6-31

d 0.5 mine 0.2 min

D i i P d L (2 f 2)Designing Product Layouts (2 of 2)

• Ideally, each worker would be assigned the same amount of work (time), i.e., the required work ( ), , qwould be balanced among the work stations

• Unfortunately there are conditions that can prevent the achievement of a perfectly balanced linethe achievement of a perfectly balanced line– The estimated times for tasks

Th d l ti hi f th t k– The precedence relationships for the tasks– The combinatoric nature of the problem

6-32

Li b l i P d (1/3)

If id d fi d h l i

Line balancing Procedure (1/3)

• If not provided, find the cycle time.– Remember the cycle time is the reciprocal of the

d i Th l i i d i h production rate. The cycle time is expressed in the same time units as the estimated task times

• Determine the selection heuristic rule (in case of tie)– This is a rule of thumb that may be used to help with the

assignments

6-33

Some Heuristic (Intuitive) RulesA k d f l k• Assign tasks in order of longest task time

• Assign tasks in order of most number of following g gtasks

• Assign tasks in order of greatest positional weightg g p g– Positional weight is the sum of each task’s time

and the times of all following tasksand the times of all following tasks

6-34

Li b l i P d (2/3)Line balancing Procedure (2/3)

Step 1: Open a new station with the full cycle time remaining

Step 2: Determine which unassigned tasks are Step 2: Determine which unassigned tasks are feasible, i.e., can be assigned to this station at this timetime– For a task to be feasible, two conditions must be met

• All tasks that precede that task must have already been assignedAll tasks that precede that task must have already been assigned• The estimated task time must be equal to or less than the

remaining cycle time for that work station

6-35

Li b l i P d (3/3)Step 3: If there is only one feasible task, assign it to the

Line balancing Procedure (3/3)p y g

station. If there is more than one feasible task, use the heuristic (step 2) to determine which task to ( p )assign. Reduce the station’s remaining cycle time by the estimated time for the assigned task. Return to gstep 2.

Step 4: If there are no feasible tasks, assignments to that station are complete Go back to step 1 that station are complete. Go back to step 1.

6-36

Step 5: Stop when all tasks have been assigned to stations

E ample 1 Assembl Line BalancingExample 1: Assembly Line Balancing

Precedence Precedence Diagram

a bb0.1 min. 1.0 min.

c d e0 7 min 0 5 min 0 2 min0.7 min. 0.5 min. 0.2 min.

6-37

E ample 1 Assembl Line BalancingExample 1: Assembly Line Balancing

• Arrange tasks shown in the previous slide into workstations.– OT = 480 minutes per day– D = 480 units per day– Use a cycle time of 1.0 minute– What is the minimum number of workstation?

315.2

timecycle

ifor task time1 ≅=∑=

m

i

– Heuristic rule to be used: Assign tasks in order of the most number of followers

y

6-38

a bSolution to Example 1

a b0.1 min. 1.0 min.

c d e0.7 min. 0.5 min. 0.2 min.0.5 min. 0.2 min.

Work-Station

TimeRemaining Eligible

AssignTask

StationIdle TimeStation Remaining Eligible Task Idle Time

1 1.0.9

a, cc

ac

.2 none - .22 1.0

0b

noneb- 00 none - 0

3 1.0.53

de

de

36-39

.3 - - .3.5

a bSolution to Example 1

a b0.1 min. 1.0 min.

c d e0.7 min. 0.5 min. 0.2 min.0.5 min. 0.2 min.

Work-Station

TimeRemaining Eligible

AssignTask

StationIdle TimeStation Remaining Eligible Task Idle Time

1 1.0.9

a, cc

ac

.2 none - .22 1.0

0b

noneb- 00 none - 0

3 1.0.53

de

de

36-40

.3 - - .3.5

a bSolution to Example 1

a b0.1 min. 1.0 min.

c d e0.7 min. 0.5 min. 0.2 min.0.5 min. 0.2 min.

Work-Station

TimeRemaining Eligible

AssignTask

StationIdle TimeStation Remaining Eligible Task Idle Time

1 1.0.9

a, cc

ac

.2 none - .22 1.0

0b

noneb- 00 none - 0

3 1.0.53

de

de

36-41

.3 - - .3.5

a bSolution to Example 1

a b0.1 min. 1.0 min.

c d e0.7 min. 0.5 min. 0.2 min.0.5 min. 0.2 min.

Work-Station

TimeRemaining Eligible

AssignTask

StationIdle TimeStation Remaining Eligible Task Idle Time

1 1.0.9

a, cc

ac

.2 none - .22 1.0

0b

noneb- 00 none - 0

3 1.0.53

de

de

36-42

.3 - - .3.5

a bSolution to Example 1

a b0.1 min. 1.0 min.

c d e0.7 min. 0 5 min 0 2 min0.5 min. 0.2 min.

Work-Station

TimeRemaining Eligible

AssignTask

StationIdle TimeStation Remaining Eligible Task Idle Time

1 1.0.9

a, cc

ac

.2 none - .22 1.0

0b

noneb- 00 none - 0

3 1.0.53

de

de

36-43

.3 - - .3.5

a bSolution to Example 1

a b0.1 min. 1.0 min.

c d e0.7 min. 0 5 min 0 2 min0.5 min. 0.2 min.

Work-Station

TimeRemaining Eligible

AssignTask

StationIdle TimeStation Remaining Eligible Task Idle Time

1 1.0.9

a, cc

ac

.2 none - .22 1.0

0b

noneb- 00 none - 0

3 1.0.53

de

de

36-44

.3 - - .3.5

Percentage of idle time = {0.5/(1*3)} x 100 = 16.7%

Problem 2

1.4a

0.5b 0 5

c f0.6 0.5

a b 0.5

hd0.7

1.00.8

e g1.0

A shop works 60-minute, and expect hourly output of 33.333 units.

Cycle time? 60/33.33 = 1.8

Mi i b f t ti ? 6/1 8 3 33 4Minimum number of stations? = 6/1.8 = 3.33 ≅ 4

Assign tasks to work stations (rule: most number of

6-45

followers, tie breaker: great positional weight)

Problem 2(with most number of followers)(with most number of followers)

Work t ti

Time i i

Eligible t k

Assigned t k

Station idl tistation remaining tasks tasks idle time

1 1.8 a a 0.4 none none 0.4

6-46

Problem 2(with most number of followers)(with most number of followers)

Work t ti

Time i i

Eligible t k

Assigned t k

Station idl tistation remaining tasks tasks idle time

1 1.8 a a 0.4 none none 0.4

2 1.8 b b 1.3 c, d, e e

0.5 none none 0.5

6-47

Problem 2(with most number of followers)

Work t ti

Time i i

Eligible t k

Assigned t k

Station idl ti

(with most number of followers)station remaining tasks tasks idle time

1 1.8 a a 0.4 none none 0.4

2 1.8 b b 1.3 c, d, e e

0.5 none none 0.53 1.8 c, d d 1.1 c, g c

0 5 f f 00.5 f f 0

6-48

Problem 2(with most number of followers)(with most number of followers)

Work t ti

Time i i

Eligible t k

Assigned t k

Station idl tistation remaining tasks tasks idle time

1 1.8 a a 0.4 none none 0.4

2 1.8 b b 1.3 c, d, e e 0.5 none none 0.5

3 1.8 c, d d 1.1 c, g c

0.5 f f 00.5 f f 04 1.8 g g 0.8 h h 0.3

Total idle time per unit = 0.4 + 0.5 + 0 + 0.3 = 1.2

6-49Percentage of idle time (efficiency) = 1.2/ (1.8*4) = 16.7%

Minimum Cycle time is yequal to the longest task time

1 min2 min1 min1 min 1 min.2 min.1 min.1 min.

Task 1 Task 2 Task 3 Task 4

Minimum Cycle Time = 2 minutesOutput = 30 per Hour

6-50

Maximum Cycle time is l h f equal to the sum of

the task times

1 min.2 min.1 min.1 min.

1 + 1 + 2 + 1 = 5

Maximum Cycle Time = 5 minutesOutput = 12 per Hour

6-51

Output 12 per Hour

P dProductLayout!!!Process

Layout!!!

Layout!!!

Layout!!!

6-52

Chapter 6 Chapter 6 Facilities Layouty

• Solved problem– 1

• Discussion questions• Discussion questions– 1,3,4,5,8,11-13,15,23,24, , , , , , , ,

• Assigned problems- 1, 3, 4-a

6-53