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FACILITY LOCATION
1. NEED & DEFINITION2. DECISION PROCESS FOR
CHOICE OF LOCATION3. STRATEGIES FOR FACILITY
LOCATION4. CRITERIA FOR SELECTION and5. LOCATION MODELS
1. NEED OF PLANT LOCATION?
NEED OF PLANT LOCATION ARISES IN THE FOLLOWING SITUATIONS:-
• STARTING NEW BUSINESS FOR THE FIRST TIME.
• EXPANSION OF BUSINESS.• DIVERSIFICATION OF BUSINESS.• RE-LOCATION DUE TO SCARCITY / NON –
AVAILABILITY OF CRITICAL INPUTS OR DUE TO A CRITICAL INPUT BECOMING UN – ECONOMICAL AT THE PRESENT LOCATION.
1A. Facility Location-Defined• PLANT LOCATION MEANS DECIDING A
SUITABLE PLACE WHERE THE PLANT WILL BE HOUSED.
• BEST LOCATION IS ONE THAT RESULTS IN LOWEST COST OF PRODUCTION PER UNIT & HENCE MAXIMUM RATE OF RETURN.
• IDEAL LOCATION?
• IMPORTANCE OF LOCATION DECISION.
2. DECISION PROCESS FOR CHOICE OF LOCATION
• Identify set of alternatives for various regional locations
• Reduce set using evaluation procedures
• Decide the exact location of site based upon cost considerations
Regional Decision• Organizational Strategies• Management Control Considerations• Region Specific Business and Economic Considerations
– Transportation network– Geographical Environment– Compatible industry– Attractiveness of community– Taxes– Economic incentives– Legal restrictions– Cultural factors and
• Cost considerations
2a: 2a: Factors That Affect The Site Factors That Affect The Site SelectionSelection
Land requirements VS availabilityLand requirements VS availabilityTopographyTopographyRaw materialRaw materialMarketMarketTransportation costTransportation costLaborLaborPowerPowerWaterWaterClimateClimateProximity of other industryProximity of other industryGovt. restriction and benefits Govt. restriction and benefits andandCultural and cost factors.Cultural and cost factors.
3. STRATEGIES FOR FACILITY LOCATION
3.1 SEPARATE FACILITY FOR DIFFERENTPRODUCTS/SERVICES
3.2 SEPARATE FACILITY TO SERVE DIFFERENT GEOGRAPHICAL AREAS
3.3 SEPARATE FACILITY FOR MAJOR / CRITICAL INPUTS
3.1 SEPARATE FACILITY FOR DIFFERENTPRODUCTS/SERVICES
• Effective Management• Concept of specialization, hence higher
productivity• Profit Center ConceptEXAMPLES:-i. companies like LG, BPL, Videocon have separate
manufacturing units for TVs, Washing Machines, Microwave Ovens & Refrigerators.
ii. ICICI have separate offices for Banking, Insurance & Mutual Funds.
iii. Mahindra & Mahindra have separate units for mfr. Of Tractors, Telecom & Rural Finance
3.2 SEPARATE FACILITY TO SERVE DIFFERENT GEOGRAPHICAL AREAS
• Transportation Cost is Minimized• Lead Time for supply of goods & services
is reduced.
EXAMPLES:-i. Pepsi & Coca – Colaii. Banks – PNB, SBI, ICICI, HDFC, HSBCiii. Educational Institutions – IIT, BITS, DAViv. Vehicle Service Stations etc.
3.3 SEPARATE FACILITY FOR MAJOR / CRITICAL INPUTS
Generally adopted by mfrs. who plan for backward integration.
EXAMPLES:-i. Cement / sugar manufacturing plant
planning to have a mfg. unit for woven sacks.
ii. Tyre mfg. Unit planning to establish a unit for mfr of Carbon Black.
4. CRITERIA FOR SELECTION
• Maximization of Profit / Maximization of Revenue / Minimization of Cost
• Dominant Factor and• Proximity to Potential Customers
For Service Facilities:-Cost of access to service facilityCost of delivery of serviceResponse Time
Water as a Dominant factor
• One Tonne of Aluminium 40,000 ltrs• One Tonne of Steel Plate 20,000 ltrs• One Tonne of Paper 15,000 ltrs• One Tonne of Sugar 12,000 ltrs• Brew One Barrel of Beer 300 ltrs• One Kwh of Energy 20 ltrs• Tarapur Atomic Power Hs/day 2200 million
ltrs
5. LOCATION MODELS1. MEDIAN MODEL2. CENTROID MODEL3. LOCATION FACTOR RATING MODEL4. BREAK EVEN POINT MODEL5. FACTORS’ COST ANALYSIS MODEL6. TRANSPORTATION MODEL7. BRIDGEMAN’S DIMENSIONAL ANALYSIS
MODEL8. BROWN & GIBSON’S MODEL9. MODIFIED BROWN & GIBSON’S MODEL and10. ARDALAN HEURISTIC MODEL
5.1 MEDIAN MODEL
Used to locate a new facility such that Used to locate a new facility such that the total transportation cost between the total transportation cost between the new facility and the existing the new facility and the existing facilities of the organization is facilities of the organization is minimum. minimum.
Based upon the assumption of Based upon the assumption of movement of goods along rectilinear movement of goods along rectilinear direction.direction.
FACILITYFACILITY
(F)(F)
LOCATION LOCATION COORDINATESCOORDINATES
(ai & bi)
COSTCOST
(Cui)
ANNUAL LOADANNUAL LOAD
(Li)
BareillyBareilly (10,80)(10,80) 1010 452452
Shah - Shah - jahanpur jahanpur
(30,60)(30,60) 1010 678678
Gonda Gonda (80,50)(80,50) 1010 483483
Kanpur Kanpur (50,10)(50,10) 1010 711711
Sultanpur Sultanpur (80,10)(80,10) 1010 539539
2,8632,863
An Example:
FACILITYFACILITY
(F)(F)
LOCATION LOCATION COORDINATESCOORDINATES
(ai & bi)
COSTCOST
(Cui)
ANNUAL LOADANNUAL LOAD
(Li)
BareillyBareilly (10,80)(10,80) 1010 452452
Shah - Shah - jahanpur jahanpur
(30,60)(30,60) 1010 678678
Gonda Gonda (80,50)(80,50) 1010 483483
Kanpur Kanpur (50,10)(50,10) 1010 711711
Sultanpur Sultanpur (80,10)(80,10) 1010 539539
2,8632,863
STEP 1: Find Median Load = (2,863 + 1) 2,863 + 1) ÷ 2 = 1432
STEP 1: Find Median Load
STEP 2: Arrange F according to increasing X coordinates and identify for median load being handled
FF ai Li Cum. Load
BareillyBareilly 10 452452 452452
Shah - Shah - jahanpurjahanpur
30 678678 1130
Kanpur Kanpur 50 711711 1841
Gonda Gonda 80 483483 2324
SultanpurSultanpur 80 539539 2863
Median Load = 1432
STEP 1: Find Median LoadSTEP 2: Arrange F according to increasing X coordinates and identify
for median load being handled
FF bi Li Cum. Load
Kanpur Kanpur 10 711711 711711
Sultanpur Sultanpur 10 539539 1250
Gonda Gonda 50 483483 1733
Shah Jahanpur Shah Jahanpur 60 678678 2411
BareillyBareilly 80 452452 2863
Median Load = 1432
STEP 3: Arrange F according to increasing Y coordinates and identify for median load being handled
5.1 MEDIAN MODEL- steps1. FIND THE MEDIAN OF THE LOAD.2. ARRANGE THE FACILITIES IN THE INCREASING
ORDER OF ‘X’ COORDINATES. FIND THE VALUE OF THE COORDINATE OF
EXISTING FACILITY WHICH SENDS / RECEIVES MEDIAN LOAD IN ‘X’ DIRECTION. THIS IS THE ‘X’ COORDIATE OF THE DESIRED LOCATION.
3. SIMILARLY, FIND THE VALUE OF THE COORDINATE OF EXISTING FACILITY WHICH SENDS / RECEIVES MEDIAN LOAD IN ‘Y’ DIRECTION. THIS IS THE ‘Y’ COORDIATE OF THE DESIRED LOCATION.
4. FIND THE TOTAL COST OF TRANSPORTATION ASSUMING ‘RECTILINEAR MOVEMENT OF MATERIALS / PRODUCTS’.
(F)(F) (ai , bi) DistanceDistance
(D)(D)(Cui) (Li) COST=COST=
Dx Dx Cuix Li
BareillyBareilly (10,80)(10,80) [50-10]+[50-10]+
[50- 80] = [50- 80] = 7070
1010 452452 3,16,4003,16,400
ShahShah
Jahanpur Jahanpur
(30,60)(30,60)1010 678678
GondaGonda (80,50)(80,50) 1010 483483
Kanpur Kanpur (50,10)(50,10) 1010 711711
Sultanpur Sultanpur (80,10)(80,10) 1010 539539
Step 4: Find the total cost of transportation for the proposed location (50, 50)
(F)(F) (ai , bi) DistanDistancece
(D)(D)
Cost Cost (Cui)
Annual Annual loadload
(Li)
COST=COST=
Dx Dx Cuix Li
BareillyBareilly (10,80)(10,80) 7070 1010 452452 3,16,4003,16,400
ShahShah
Jahanpur Jahanpur
(30,60)(30,60)3030 1010 678678 2,03,4002,03,400
GondaGonda (80,50)(80,50) 3030 1010 483483 1,44,9001,44,900
Kanpur Kanpur (50,10)(50,10) 4040 1010 711711 2,84,4002,84,400
Sultanpur Sultanpur (80,10)(80,10) 7070 1010 539539 3,77,3003,77,300
13,26,40013,26,400
Step 4: Find the total cost of transportation for the proposed location (50, 50)
5.2 CENTROID MODEL• Let ai & bi be the coordinates of the existing
facilities and Li be the load being handled by the ith facility.
• Let (X,Y) be the coordinates of the plant being proposed to be located.
• Find the value of the coordinates using the following formulae:-
X = [∑ (Li x ai)]÷ ∑ Li for values of i =1 to m Y = [∑ (Li x bi )]÷ ∑ Li for values of i =1 to m
• Find the total cost of transportation using the following formulae:-
Total Cost = ∑ [Li x √(X-ai)2+(Y- bi)2 . Cui]
The centre of gravity modelThe centre of gravity model
X = [∑ (Li x ai)]÷ ∑ Li
=[(10*452)+(30*678)+(80*483)+(50*711)+=[(10*452)+(30*678)+(80*483)+(50*711)+
(80*539)] (80*539)] ÷ 2863 ÷ 2863
= = [4,520+20,340+38,640+35,550+43,120][4,520+20,340+38,640+35,550+43,120] ÷ 28632863
= 1,42,170 = 1,42,170 ÷ 28632863
= = 49.66 49.66
Y = [∑ (Li x bi )]Y = [∑ (Li x bi )]÷ ÷ ∑ Li∑ Li
Y = [(80*452)+(60*678)+(50*483)+Y = [(80*452)+(60*678)+(50*483)+ (10*711)+(10*539)](10*711)+(10*539)] = [36,160+40,680+24,150+= [36,160+40,680+24,150+ 7,110+5,390] 7,110+5,390] ÷ 2,8632,863
= 39.64 = 39.64
5.3 LOCATION FACTOR RATING MODEL
• LIST THE FACTORS TO BE CONSIDERED• RATE THE FACTORS ON A SCALE OF 1 TO 5
(RF)• PREPARE LOCATION RATING (RL) ON
A SCALE OF 1 TO 10 FOR EACH OF THESE FACTORS ACCORDING TO THE BENEFITS EACH LOCATION OFFERS.
• FIND ∑ (RF x RL) FOR EACH LOCATION.• SELECT THE LOCATION WITH THE HIGHEST
SCORE.
Site 1 Site 2 Site 3
Climate condition 3 8 6 9proximity to suppliers 2 10 9 8wage rates 4 6 8 7labour conditions 3 7 8 6Proximity to customers 4 6 9 10Shipping modes 3 9 9 7Air service 2 5 7 9
Location Rating RL
Location fcator Rating RF
Location Factor Rating
FIND ∑ (RF x RL) FOR EACH LOCATION.
Suggest a suitable site from the three given options
Site 1 Site 2 Site 3
Climate condition 3 24 18 27proximity to suppliers 2 20 18 16wage rates 4labour conditions 3Proximity to customers 4Shipping modes 3Air service 2
Measure RF x RL
Location fcator Rating RF
Location Factor Rating
Site 1 Site 2 Site 3
Climate condition 3 24 18 27proximity to suppliers 2 20 18 16wage rates 4 24 32 28labour conditions 3 21 24 18Proximity to customers 4 24 36 40Shipping modes 3 27 27 21Air service 2 10 14 18
Measure RF x RL
Location fcator Rating RF
Location Factor Rating
∑ (RF x RL) = 150, 169, 168 for Site 1, 2 and 3 respectively
Site 1 Site 2 Site 3
Climate condition 0.3 80 65 90proximity to suppliers 0.2 100 91 75wage rates 0.15 60 95 72labour conditions 0.15 75 80 80Proximity to customers 0.1 65 90 95Shipping modes 0.05 85 92 65Air service 0.05 50 65 90
Score 0 to 100Location fcator Weight
Location Factor Weighted Score Model
Site 1 Site 2 Site 3
Climate condition 0.3 24 19.5 27proximity to suppliers 0.2 20 18.2 15wage rates 0.15labour conditions 0.15Proximity to customers 0.1Shipping modes 0.05Air service 0.05
Total weighted score
Location fcator WeightWeighted Score
Location Factor Weighted Score Model
Site 1 Site 2 Site 3
Climate condition 0.3 24 19.5 27proximity to suppliers 0.2 20 18.2 15wage rates 0.15 9 14.25 10.8labour conditions 0.15 11.25 12 12Proximity to customers 0.1 6.5 9 9.5Shipping modes 0.05 4.25 4.6 3.25Air service 0.05 2.5 3.25 4.5
Total weighted score
Location fcator WeightWeighted Score
Location Factor Weighted Score Model
Site 1 Site 2 Site 3
Climate condition 0.3 24 19.5 27proximity to suppliers 0.2 20 18.2 15wage rates 0.15 9 14.25 10.8labour conditions 0.15 11.25 12 12Proximity to customers 0.1 6.5 9 9.5Shipping modes 0.05 4.25 4.6 3.25Air service 0.05 2.5 3.25 4.5
77.5 80.8 82.05Total weighted score
Location fcator WeightWeighted Score
Location Factor Weighted Score Model
Site 1 Site 2 Site 3
Climate condition 0.3 24 19.5 27proximity to suppliers 0.2 20 18.2 15wage rates 0.15 9 14.25 10.8labour conditions 0.15 11.25 12 12Proximity to customers 0.1 6.5 9 9.5Shipping modes 0.05 4.25 4.6 3.25Air service 0.05 2.5 3.25 4.5
77.5 80.8 82.05Total weighted score
Location fcator WeightWeighted Score
Location Factor Weighted Score Model
Load – Distance Technique
This is a modified form of the CG method.
In this method a single set of site coordinates is not identified. Instead, various sites are evaluated using a load distance value that is a measure of weight and distance (the inputs could come from another method like location factor rating method).
For a single potential site, a load distance value is computed as follows.
1
n
i ii
LD l d
LD = the load distance value
li = load expressed as weight, no. of trips or no. of units being shipped to from the site to the location i
di = the distance between the proposed site and location i
Load – Distance Technique
1
n
i ii
LD l d
2 2( ) ( )i i id x x y y
x, y = coordinates of proposed site
xi, yi = coordinates of existing locations
Load – Distance TechniqueT
her
e ar
e th
ree
po
ten
tial
sit
es t
o b
e ev
alu
ated
.
ai biSite 1 360 180Site 2 420 450Site 3 250 400
A B C Dai 200 100 250 500
bi 200 500 600 300
Li 75 105 135 60
Consider the following example
2 2(200 360) (200 180) 161.2Ad
2 2(100 360) (500 180) 412.3Bd
2 2(250 360) (600 180) 434.2Cd
2 2(500 360) (300 180) 184.4Dd
1
125063n
i ii
LD l d
Load – Distance Technique
(161.2 x 75 + 412.3 x 105 + 434.2 x 135 + 184.4 x 60)
1
99789n
i ii
LD l d
1
77555n
i ii
LD l d
Site 2:
Site 3:
Since site 3 has the lowest load distance value, this location would also minimize the transportation costs.
Quiz
A B C
X 150 300 400Y 250 100 500W 140 110 170
Find the best possible location of a warehouse to cater to these locations (distances in Km).
It was observed that a warehouse cannot be built within 10 Km radius of the location found using CG method. Compare at least four sites.