Facility Location. BMW In the late 1980’s fluctuating exchange rates and rising costs convinced...

Facility Location

BMW• In the late 1980’s fluctuating exchange rates and rising costs

convinced BMW that it was time to consider operating a new production facility outside the European borders.

• A “blank page” approach was used to compile a list of 250 potential worldwide sites. Analysis pared the list down to 10 options; a location in the United States was preferred due to its proximity to a large market segment for BMW’s automobiles.

• BMW spent 3 1/2 years considered the labor climate, port and road access, geographical requirements and constraints, airport access, and its relations with the governments.

• The plant was located in Spartanburg, SC, and now employs approximately 4,700 workers who produce more than 500 vehicles a day.

LI & FUNG

European retailer order

10,000 garments

Buy the best Japanese zipper & button from

Chaina

Weave & dye in Taiwan

Manufacture garments in

Thailand

Buy yarn from a Korean mfgr.

5 weeks later 10,000 garments

reach Europe

Ellora Times• In 2001, Ellora Time Pvt. Ltd. (Ellora), a company based in Gujarat,

India, was the world's largest manufacturer of clocks. It also manufactured calculators, telephones, timepieces and educational toys. Ajanta and Orpat were closely held Ellora

companies with a combined investment of Rs 2 billion. • Almost all their products, marketed through a countrywide

network of 25,000 dealers and 180 service stations, were leaders in their respective categories. For the year 1999-00, the group recorded a combined turnover of over Rs 2.50 billion. Both Ajanta and Orpat received awards by the Government of India for superior exports performance throughout the 1990s. Ajanta, an ISO 9002 certified company, had even received the ‘Best

Electronics Industry'award many times.

Ellora Times

• In early 2001, Ellora shocked the corporate world by announcing its decision to shift its manufacturing activities to China.

• Mitshubishi at Haldia• Hero Hona at Uttranchal• Wipro in Himachal• Tata in Singur• Bio-con in Andhrapradesh

Factors Affecting the Location Decision


• Economic – Site acquisition, preparation and construction

costs– Labor costs, skills and availability– Utilities costs and availability– Transportation costs– Taxes



• Non-economic– Labor attitudes and traditions– Training and employment services– Community’s attitude– Schools and hospitals– Recreation and cultural attractions– Amount and type of housing available

Facility Types and TheirDominant Locational Factors

• Heavy Manufacturing– Near their raw material sources– Abundant supply of utilities– Land and construction costs are inexpensive

• Light Manufacturing– Availability and cost of labor

• Warehousing– Proximity to transportation facilities– Incoming and outgoing transportation costs

Facility Types and TheirDominant Locational Factors

• R&D and High-Tech Manufacturing– Ability to recruit/retain scientists, engineers, etc.– Near companies with similar technology interests

• Retailing and For-Profit Services– Near concentrations of target customers

• Government and Health/Emergency Services– Near concentrations of constituents

Some Reasons the Facility Location Decision Arises

• Changes in the market– Expansion– Contraction– Geographic shift

• Changes in inputs– Labor skills and/or costs– Materials costs and/or availability– Utility costs

Some Reasons the Facility Location Decision Arises

• Changes in the environment– Regulations and laws– Attitude of the community

• Changes in technology

Analyzing Industrial Facility Locations

• Locating a Single Facility– A simple way to analyze alternative locations is

conventional cost analysis

• Locating Multiple Facilities– More sophisticated techniques are often used:

• Linear programming, computer simulation, network analysis, and others

Facility Location Models

Location Analysis Methods

• Factor rating method• Load-distance model• Center of gravity approach• Break-even analysis• Transportation method• Dimensional Analysis

Location Rating Factor

Identify important factorsIdentify important factors Subjectively score each factor (0 - 100)Subjectively score each factor (0 - 100) Weight factors (0.00 - 1.00)Weight factors (0.00 - 1.00) Sum weighted scoresSum weighted scores

Location Factor Rating: Example

Labor pool and climateLabor pool and climateProximity to suppliersProximity to suppliersWage ratesWage ratesCommunity environmentCommunity environmentProximity to customersProximity to customersShipping modesShipping modesAir serviceAir service

LOCATION FACTORLOCATION FACTOR RatingRating

9090606045454545303015151515

NormalizedNormalized

SCORES (0 TO 100)SCORES (0 TO 100)



LOCATION FACTORLOCATION FACTOR

.30.30

.20.20

.15.15

.15.15

.10.10

.05.05

.05.05

RatingRating

9090606045454545303015151515

NormalizedNormalized





8080100100

60607575656585855050

Site 1Site 1





8080100100

60607575656585855050

Site 1Site 1

6565919195958080909092926565

Site 2Site 2





8080100100

60607575656585855050

Site 1Site 1

6565919195958080909092926565

Site 2Site 2

9090757572728080959565659090

Site 3Site 3





.30.30

.20.20

.15.15

.15.15

.10.10

.05.05

.05.05

WEIGHTWEIGHT

8080100100

60607575656585855050

Site 1Site 1

6565919195958080909092926565

Site 2Site 2

9090757572728080959565659090

Site 3Site 3


Location Factor Rating

24.0020.00

9.0011.25

6.504.252.50

77.50

Site 1

19.5018.2014.2512.00

9.004.603.25

80.80

Site 2

27.0015.0010.8012.00

9.503.254.50

82.05

Site 3

WEIGHTED SCORES

Site 3 has the highest factor rating

Example

Example

Nearness to the portNearness to the portProximity to suppliersProximity to suppliersAvailability of skilled laborAvailability of skilled laborGovt. PoliciesGovt. PoliciesProjected cost of operationProjected cost of operationShipping modesShipping modesEducational InfrastructureEducational Infrastructure


8080707090905050606070704040

ScoreScore

7575606050507070707080806060

Site 1Site 1

7070808070704545606090908080

Site 2Site 2

9090858585856060808070708080

Site 3Site 3


26

Center of Gravity Method

• The Center of Gravity Method is a tool that seeks to compute geographic coordinates for a potential single new facility that will minimize costs.

• The Center of Gravity Method takes many factors into account including:MarketsVolume of goods shippedShipping costs

Locate facility at center of geographic areaLocate facility at center of geographic area

Based on weight and distance traveled Based on weight and distance traveled

establish grid-map of areaestablish grid-map of area

Identify coordinates and weights shipped Identify coordinates and weights shipped for each locationfor each location

Center-of-Gravity Technique

Grid-Map Coordinates

xxii, y, yii ==coordinates of existing coordinates of existing

facility facility iiWWii = =annual weight shipped annual weight shipped

from facility from facility ii

xx11 xx22 xx33 xx

yy22

yy

yy11

yy33

1 (1 (xx11, , yy11), ), WW11

2 (2 (xx22, , yy22), ), WW22

3 (3 (xx33, , yy33), ), WW33

Grid-Map Coordinates

where,where,xx, , y y ==coordinates of new coordinates of new facility at center of gravityfacility at center of gravityxxii, y, yii ==coordinates of existing coordinates of existing

facility facility iiWWii = =annual weight shipped annual weight shipped

from facility from facility ii

nn

WWii

i = i = 11

xxiiWWii

i = i = 11

nn

x =x =

nn

WWii

i = i = 11

yyiiWWii

i = i = 11

nn

y =y =

xx11 xx22 xx33 xx

yy22

yy

yy11

yy33

1 (1 (xx11, , yy11), ), WW11

2 (2 (xx22, , yy22), ), WW22

3 (3 (xx33, , yy33), ), WW33

Example

AA BB CC DD

xx 200200 100100 250250 500500yy 200200 500500 600600 300300WtWt 7575 105105 135135 6060

Center-of-Gravity Technique: Example

AA BB CC DD

xx 200200 100100 250250 500500yy 200200 500500 600600 300300WtWt 7575 105105 135135 6060

yy

700700

500500

600600

400400

300300

200200

100100

00 xx700700500500 600600400400300300200200100100

AA

BB

CC

DD

(135)(135)

(105)(105)

(75)(75)

(60)(60)

MilesMiles

Mil

esM

iles

Center-of-Gravity Technique: Example (cont.)

x = = = 238n Wi

i = 1

xiWii = 1

n

n

Wii = 1

yiWii = 1

n

y = = = 444(200)(75) + (500)(105) + (600)(135) + (300)(60)75 + 105 + 135 + 60

(200)(75) + (100)(105) + (250)(135) + (500)(60)75 + 105 + 135 + 60

Center-of-Gravity Technique: Example (cont.)

AA BB CC DD

xx 200200 100100 250250 500500yy 200200 500500 600600 300300WtWt 7575 105105 135135 6060

yy

700700

500500

600600

400400

300300

200200

100100

00 xx700700500500 600600400400300300200200100100

AA

BB

CC

DD

(135)(135)

(105)(105)

(75)(75)

(60)(60)

MilesMiles

Mil

esM

iles

Center of gravity Center of gravity (238, 444)(238, 444)

Copyright 2006 John Wiley & Sons, Inc. Supplement 7-34

Center of

Gravity with

Excel

Example

0.8

301520

43012151120

ii

iii

l

xlx

5.5

301520

5.1305.9155.820

ii

iii

l

yly

© 2007 Pearson Education

Finding the Center of Gravity for Health Watch

Census Tract Population Latitude Longitude 15 2,711 42.134 -80.041 16 4,161 42.129 -80.023 17 2,988 42.122 -80.055 25 2,512 42.112 -80.066 26 4,342 42.117 -80.052 27 6,687 42.116 -80.023 28 6,789 42.107 -80.051

Total 30,190

ExampleExisting Facility Annul load Cost of moving

one unitCoordinate locations

W 279 10 (20,30)

X 473 10 (70,10)

Y 350 10 (50,40)

z 266 10 (10,80)

Load Distance Method

Euclidean or rectilinear distance measure may be used.

Distance Measures

Rectilinear DistanceRectilinear Distance

ddABAB = |20= |20 –– 80| + |10 – 60| = 11080| + |10 – 60| = 110

Euclidian DistanceEuclidian Distance

ddABAB = (20= (20 –– 80)80)22 + (10 – 60) + (10 – 60)22

= 78.1= 78.1

What is the distance between (20,10) and (80,60)?

Load-Distance Technique

• Compute (Load x Distance) for each site• Choose site with lowest (Load x Distance)• Distance can be actual or straight-line

Load-Distance Calculations

li di

i = 1

nLD =

LD = load-distance value

li = load expressed as a weight, number of trips or unitsbeing shipped from proposed site and location i

di = distance between proposed site and location i

di = (xi - x)2 + (yi - y)2

(x,y) = coordinates of proposed site(xi , yi) = coordinates of existing facility

where,

where,

or Ixi - xI + Iyi - yI

Load-Distance: Example

Potential SitesPotential SitesSiteSite XX YY11 360360 18018022 420420 45045033 250250 400400

SuppliersSuppliersAA BB CC DD

XX 200200 100100 250250 500500YY 200200 500500 600600 300300WtWt 7575 105105 135135 6060

Load-Distance: Example

Potential SitesPotential SitesSiteSite XX YY11 360360 18018022 420420 45045033 250250 400400

SuppliersSuppliersAA BB CC DD

XX 200200 100100 250250 500500YY 200200 500500 600600 300300WtWt 7575 105105 135135 6060

Compute distance from each site to each supplierCompute distance from each site to each supplier

= (200-360)= (200-360)22 + (200-180) + (200-180)22ddAA = (x = (xAA - x - x11))22 + (y + (yAA - y - y11))22Site 1Site 1 = 161.2= 161.2

= (100-360)= (100-360)22 + (500-180) + (500-180)22ddBB = (x = (xBB - x - x11))22 + (y + (yBB - y - y11))22 = 412.3= 412.3

ddCC = 434.2 = 434.2 ddDD = 184.4 = 184.4

Load-Distance: Example (cont.)Site 2Site 2 ddAA = 333 = 333 ddCC = 226.7 = 226.7ddBB = 323.9 = 323.9 ddDD = 170 = 170

Site 3Site 3 ddAA = 206.2 = 206.2 ddCC = 200 = 200ddBB = 180.4 = 180.4 ddDD = 269.3 = 269.3

Compute load-distanceCompute load-distance

i = 1i = 1

nn

lli i ddiiLD =LD =

Site 1 = (75)(161.2) + (105)(412.3) + (135)(434.2) + (60)(434.4) = 125,063Site 1 = (75)(161.2) + (105)(412.3) + (135)(434.2) + (60)(434.4) = 125,063

Site 2 = (75)(333) + (105)(323.9) + (135)(226.7) + (60)(170) = 99,791Site 2 = (75)(333) + (105)(323.9) + (135)(226.7) + (60)(170) = 99,791

Site 3 = (75)(206.2) + (105)(180.3) + (135)(200) + (60)(269.3) = 77,555*Site 3 = (75)(206.2) + (105)(180.3) + (135)(200) + (60)(269.3) = 77,555*

* * Choose site 3Choose site 3

Example

Example: Matrix Manufacturing is considering where to locate its warehouse in order to service its four Ohio stores located in Cleveland, Cincinnati, Columbus, Dayton. Two sites are being considered; Mansfield and Springfield, Ohio. Use the load-distance model to make the decision.

Break-Even Analysis• Break-even analysis can help a manager compare

location alternatives on the basis of quantitative factors that can be expressed in terms of total cost.1. Determine the variable costs and fixed costs for each

site.2. Plot the total cost lines—the sum of variable and fixed

costs—for all the sites on a single graph 3. Identify the approximate ranges for which each location

has the lowest cost.4. Solve algebraically for the break-even points over the

relevant ranges.

Break-Even Analysis • An operations manager has narrowed the search for a new

facility location to four communities.• The annual fixed costs (land, property taxes, insurance,

equipment, and buildings) and the variable costs (labor, materials, transportation, and variable overhead) are shown below.

• Total costs are for 20,000 units.Fixed CostsFixed Costs Variable CostsVariable Costs Total CostsTotal Costs

CommunityCommunity per Yearper Year per Unitper Unit (Fixed + Variable)(Fixed + Variable)

AA $150,000$150,000 $62$62 $1,390,000$1,390,000BB $300,000$300,000 $38$38 $1,060,000$1,060,000CC $500,000$500,000 $24$24 $ 980,000$ 980,000DD $600,000$600,000 $30$30 $1,200,000$1,200,000

Example

The operations manager for Mile-High Beer has narrowed the search for a new facility location to seven communities. Annual fixed costs (land, property taxes, insurance, equipment, and buildings) and variable costs (labor, materials, transportation, and variable overhead) are shown below.

Mile-High Beer

Example • Santro Electronics is considering 2 locations

for the audio equipment factory Ahmedabad & Chennai. At Ahmedabad fixed cost is estimated at Rs.1 million and the variable cost at Rs.1,200 per audio equipment. At Chennai fixed cost is Rs. 1.2 million and variable cost is Rs. 1100 per audio equipment. The selling price of the equipment will be Rs. 3000 per unit irrespective of the location. Decide which location is the best.

The Transportation Method

• The transportation method is a quantitative approach that can help solve multiple-facility location problems.

• The transportation method does not solve all facets of the multiple-facility location problem.

• It utilizes linear programming to minimize the cost of shipping products from two or more plants, or sources of supply, to two or more warehouses, or destinations.

The Transportation Method

The Sunbelt Pool Company has a plant in Phoenix and three warehouses. It is considering building a new 500-unit plant because business is booming. One possible location is Atlanta.

The cost to ship one unit from Atlanta to San Antonio.

The cost to ship one unit from Atlanta to San Antonio.

Initial Tableau

Example

Dimensional Analysis

Considers both tangible and intangible costs Intangibles could include (lack of ) facilities e.g. for

education, shopping, recreation, social life. Intangibles could be quantified on a scale. Weightages could be assigned to each ‘cost.’ A pair of sites is compared by a ratio.

Formula for Dimensional Analysis

If ‘C’ are costs, M & N are the two sites, and ‘w’ are weightages, the relative demerit of site M to N is:

(C1M / C1

N) X (C2M / C2

N) X ….. X (CzM / Cz

N)

where, CzM is the cost ‘z’ for site M.

If the above is >1, site N is superior.

W1 W2 Wz

Example

. Costs

Site

Labor Power Educational Facilities for children's(Score)

Recreational Facilities(Score)

M Rs.1.50.000 Rs.40,00,000 2 2

N Rs.1.00.000 Rs.25,00000 6 4

Weightage 1 1 2 2

FACILITY LAYOUT PROBLEM


Once a firm has decided where a facility will be located, the next important decision is the Arrangement of people and Equipment within the facility.


Facility Layout problem involves the location of departments (or sections) within the facility AND the arrangement of people and equipment within each department.

.


The layout decision will certainly affect the • Flow of materials• In-plant Transportation cost• Equipment utilization• General productivity and effectiveness of the

business.


Usually the layout is planned to minimize a particular criterion:•Minimizing total traveling time, total cost, total delays, etc.There are also situations in which the layout may be designed to maximize a criterion:•Maximize quality, flexibility, or space utilization.

Costs associated with a plant layout

Costs of customer dissatisfaction due to poor service (delivery, responsiveness, quality, flexibility)

Costs of movement of materials Costs of space Costs of spoilage of materials Costs of employee dissatisfaction Costs of changes required with operational

changes

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

Process Layout

Similar pieces of equipment that perform similar functions are grouped together. For example; all drill machines are grouped and placed together.

Process Layout An example

L L L L

M M

M M

D D

D D

D D

G G G

L L L L

Product A

Product C

Product B

Product Layout

The pieces of equipment required to make a Particular product are grouped together, as in an Automobile assembly line.

Product Layout

L

L D L

D M GProduct A

Product B

Product C

Step 1

Step 1

Step 1

Step 2

Step 2

Step 2

Step 3

Step 3

Step 3

Step 4

Step 4

Step 4

7-14

G

L D M L G

Step 5

Fixed Layout

The equipment is brought to the object being processed, and the object does not move. Example; house construction.

Cellular Manufacturing (CM) Layouts

• Cellular manufacturing is a type of layout in which machines are grouped into what is referred to as a cell.

• Groupings are determined by the operations needed to perform work for a set of similar items, or part families that require similar processing.

Process Layout Example

• Frontec Company wants to arrange Four of its departments in a Row so that the Total Distance Traveled between Departments is minimized.

• This part of the building will contain four departments arranged in a row.

• Frontec wishes to minimize the total daily inter-departmental distance traveled.

• The number of daily communications between each pair of department is shown below:

Example

Assume that adjacent departments are 20 feet apart.

Example

• We will use a trial-and-error approach to this problem.

• Assume that we selected the following configuration for the departments: A-B-C-D.

• For this configuration, Total communication cost (based on distance) is as follows:

Example

Example

• In terms of total daily communication distance, (B-A-C-D) is the preferred alternative.

• But the firm has to consider all of the 24 (4! = 4x3x2x1) possible configurations before it knows if this is the optimal configuration.

Example

• This trial-and-error approach becomes time-consuming as the number of departments increases AND It also becomes complex when the cost of communications vary between departments.

Product Layout

LINE BALANCING

• Essentially ,the layout design seeks to identify minimum number of resources required to meet a targeted production rate and the order in which this sequences are to be used. In the process it seeks to establish a balance among the resources

• Line Balancing is a method by which tasks are optimally combined without violating the precedence constraint and a certain number of workstation is designed to complete the task.

Station 1Minutes per Unit 6

Station 2

7

Station 3

3

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? Determine the interval between producton of two successive components?


Station 1Minutes per Unit 6

Station 2

7

Station 3

3

Assembly Lines Balancing Concepts



Answer: One component will come out of the system only every 7 minutes. This measure is known as cycle time.

Answer: One component will come out of the system only every 7 minutes. This measure is known as cycle time.

Assembly Line Balancing

1. Determine cycle time: Cycle time could be actual or desires

1. Determine required workstations (theoretical minimum)

cycle_timetask_times

N

t

units/hroutput tualdesired/ac

sec./day timeavailable)(sec./unit timeCycle

Assembly Line Balancing

5. Evaluate line efficiency:

kstationsactual_worN;CN

TE a

a

task timebottleneck

timeavailableoutput Maximum

Balance delay (%) is the amount by which the line falls short of 100%

Example

• A factory working two shifts each of eight hour produces 24000 electric bulbs using a set of workstations. Compute the actual cycle time of the plant operation. There are 8 tasks required to manufacture the bulbs. The sum of all task time is equal to 12 seconds. How many workstations are required to maintain the level of production.

Example of Line Balancing

• 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

Example of Line Balancing: Structuring the Precedence Diagram

A B G

H

C D E F

Task PredecessorsA None

B AC None

D A, C

Task PredecessorsE D

F E

G B

H F, G

Example of Line Balancing: Precedence Diagram

Question: Which process step defines the maximum rate of production?

Question: Which process step defines the maximum rate of production?

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

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.

Example of Line Balancing: Determine Cycle Time

Question: Suppose we want to assemble 100 fans per day. What would our cycle time have to be?


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

Answer: Answer:

Example of Line Balancing: Determine Theoretical Minimum Number of Workstations

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

Example of Line Balancing: Determine Theoretical Minimum Number of Workstations



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

A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1


A (4.2-2=2.2)


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



A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

C (4.2-3.25)=.95


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

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


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


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


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


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


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


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2


A

C

B

D E F

GH

2

3.25

1

1.2 .5

11.4

1

C (4.2-3.25)=.95

Idle = .95Efficiency=77%

D (4.2-1.2)=3E (3-.5)=2.5F (2.5-1)=1.5H (1.5-1.4)=.1Idle = .1Efficiency=98%


A (4.2-2=2.2)B (2.2-1=1.2)G (1.2-1= .2)

Idle= .2Efficiency=95%


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

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

Step 1: Identify Tasks & Immediate Predecessors

Layout Calculations

• Step 2: Determine output rate– Vicki needs to produce 60 pizzas per hour

• 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 60

sec/min 60x min/hr 60

units/hroutput desired

sec./day time available)(sec./unit time Cycle

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

sec./hr. 3600

time task bottleneck

time availableoutput Maximum

Layout Calculations (continued)

• 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 60

seconds 165

time cycle

times taskTM

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. 165

NC

t (%) Efficiency

8.3%91.7%100%delay Balance

Problem

– Draw precedence diagram– Determine cycle time—demand = 50 units/hr– Theoretical minimum no. of work stations– Assign tasks to workstations using cycle time– Efficiency and balance delay of line?– Bottleneck?– Maximum output?

Task Imm. predecessor Task time (sec)

A None 55

B A 30

C A 22

D B 35

E B, C 50

F C 15

G F 5

H G 10

TOTAL 222

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

Product Layout An example

LD

L M D

G

L

L

M

D

LD

L MD

G

L

L D

GM

Machine – Component Incident Matrix(MCIM)

Before Grouping

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20A 1 1 1B 1 1 1 1C 1 1 1 1D 1 1 1E 1 1 1 1 1 1F 1 1 1G 1 1 1 1 1 1H 1 1 1 1 1 1I 1 1 1 1 1 1J 1 1 1 1 1 1

Mac

hine

s

Components

Machine – Component Incident MatrixAfter Grouping

2 3 5 8 10 1 4 7 20 18 17 15 14 13 6 9 11 12 16 19B 1 1 1 1 1C 1 1 1 1 1D 1 1 1A 1 1 1F 1 1 1E 1 1 1 1 1 1I 1 1 1 1 1 1G 1 1 1 1 1 1H 1 1 1 1 1 1J 1 1 1 1 1 1

Mac

hine

s

Components

Rank Order Clustering(ROC)

1 2 3 4 5 6A 0 0 1 0 1 0B 0 1 1 0 0 1C 1 0 0 1 0 0D 0 1 1 0 1 0E 1 0 0 1 0 1

Mac

hine

s

Components


1. Read each row of the MCIM as binary number. Rank the rows in descending order.

2. If there is no change stop. Otherwise go to next step.

3. Rearrange the rows based on ranking.

4. Read each column of the MCIM as binary number. Rank the rows in descending order.

5. If there is no change stop. Otherwise go to next step.

6. Rearrange the rows based on ranking. Go to step 1.


Value of the binary number Rank

ROW1 10 5

ROW2 25 4

ROW3 36 2

ROW4 26 3

ROW5 37 1

Rows as binary numbers


1 2 3 4 5 6E 1 0 0 1 0 1C 1 0 0 1 0 0D 0 1 1 0 1 0B 0 1 1 0 0 1A 0 0 1 0 1 0

Mac

hine

s

Components


Value of the binary number Rank

Column1 24 1

Column2 12 5

Column3 13 4

Column4 24 1

Column5 5 6

Column6 18 3

Columns as binary numbers


1 4 6 3 2

E 1 1 1 0 0C 1 1 0 0 0

D 0 0 0 1 1B 0 0 1 1 1A 0 0 0 1 0

Mac

hine

s

Components

Example)

Mac

hine

s

Components

1 2 3 4 5 6 7 8 9 10

A 1 1 1 1 1

B 1 1 1

C 1 1 1 1 1

D 1 1 1

E 1 1 1

F 1 1

G 1 1 1

1.Use ROC to rank families and machine groups2.What will happen if we did column sorting first and then row?

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