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FACILITY LOCATION Prof. Kaushik Paul Associate Professor Operations Area E-Mail: [email protected] Phone: 43559308

Facility Location

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Page 1: Facility Location

FACILITY LOCATION

Prof. Kaushik PaulAssociate ProfessorOperations AreaE-Mail: [email protected]: 43559308

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Issues in Facility Location

Various Plant Location Methods

OBJECTIVES

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COMPETITIVE IMPERATIVES IMPACTING LOCATION

The need to produce close to the customer due to time-based competition, trade agreements, and shipping costs

The need to locate near the appropriate labor pool to take advantage of low wage costs and/or high technical skills

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ISSUES IN FACILITY LOCATION

Proximity to Customers

Business Climate

Total Costs

Infrastructure

Quality of Labor

Suppliers

Other Facilities

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ISSUES IN FACILITY LOCATION

Free Trade Zones

Political Risk

Government Barriers

Trading Blocs

Environmental Regulation

Host Community

Competitive Advantage

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PLANT LOCATION METHODOLOGY: FACTOR RATING METHOD EXAMPLE

Fuels in region 0 to 330Power availability and reliability 0 to 200Labor climate 0 to 100Living conditions 0 to 100Transportation 0 to 50Water supply 0 to 10Climate 0 to 50Supplies 0 to 60Tax policies and laws 0 to 20

Two refineries sites (A and B) are assigned the following range of point values and respective points, where the more points the better for the site location.

1231505424454855

Major factors for site location

Pt. Range 15

6100639650545020

SitesA B

Total pts. 418 544

Best Site is B

Best Site is B

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PLANT LOCATION METHODOLOGY: TRANSPORTATION

METHOD OF LINEAR PROGRAMMING

Transportation method of linear programming seeks to minimize costs of shipping from n units to m destinations or its seeks to maximize profit of shipping n units to m destinations

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PLANT LOCATION METHODOLOGY: CENTROID METHOD

The centroid method is used for locating single facilities that considers existing facilities, the distances between them, and the volumes of goods to be shipped between them

This methodology involves formulas used to compute the coordinates of the two-dimensional point that meets the distance and volume criteria stated above

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PLANT LOCATION METHODOLOGY: CENTROID

METHOD FORMULAS

C = d V

V x

ix i

i

C = d V

V x

ix i

i

Where:Cx = X coordinate of centroidCy = X coordinate of centroiddix = X coordinate of the ith locationdiy = Y coordinate of the ith locationVi = volume of goods moved to or from ith location

C = d V

Vy

iy i

i

C = d V

Vy

iy i

i

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PLANT LOCATION METHODOLOGY: EXAMPLE OF CENTROID METHOD

Question: What is the best location for a new Z-Mobile warehouse/temporary storage facility considering only distances and quantities sold per month?

Question: What is the best location for a new Z-Mobile warehouse/temporary storage facility considering only distances and quantities sold per month?

Centroid method example Several automobile showrooms are located

according to the following grid which represents coordinate locations for each showroom

S howroom No o f Z-Mo b ile s s o ld p e r mo nth

A 1250

D 1900

Q 2300X

Y

A(100,200)

D(250,580)

Q(790,900)

(0,0)

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EXAMPLE OF CENTROID METHOD (CONTINUED): DETERMINING THE COORDINATES OF THE EXISTING FACILITY

To begin, you must identify the existing facilities on a two-dimensional plane or grid and determine their coordinates.

To begin, you must identify the existing facilities on a two-dimensional plane or grid and determine their coordinates.

X

Y

A(100,200)

D(250,580)

Q(790,900)

(0,0)

You must also have the volume information on the business activity at the existing facilities.

You must also have the volume information on the business activity at the existing facilities.

S ho wro o m No o f Z-Mo b ile s s o ld p e r mo nth

A 1250

D 1900

Q 2300

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EXAMPLE OF CENTROID METHOD (CONTINUED): DETERMINING THE

COORDINATES OF THE NEW FACILITY

C = 100(1250) + 250(1900) + 790(2300)

1250 + 1900 + 2300 =

2,417,000

5,450 = x 443.49C =

100(1250) + 250(1900) + 790(2300)

1250 + 1900 + 2300 =

2,417,000

5,450 = x 443.49

C = 200(1250) + 580(1900) + 900(2300)

1250 + 1900 + 2300 =

3,422,000

5,450 = y 627.89C =

200(1250) + 580(1900) + 900(2300)

1250 + 1900 + 2300 =

3,422,000

5,450 = y 627.89

S ho wro o m No o f Z-Mo b ile s s o ld p e r mo nth

A 1250

D 1900

Q 2300X

Y

A(100,200)

D(250,580)

Q(790,900)

(0,0)

You then compute the new coordinates using the formulas:You then compute the new coordinates using the formulas:

ZZ

New location of facility Z about (443,627)

New location of facility Z about (443,627)

You then take the coordinates and place them on the map:You then take the coordinates and place them on the map:

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Reference: Operations Management for Competitive AdvantageBy Chase, Jacobs & Aquilano, 10e

HOPE YOU ENJOYED THE CLASS. QUESTIONS PLEASE

THANK YOU