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FACILITY LOCATION Group- 4 Akhil Murali Akhil Nandakumar Aneesh K Saju Anila P Jestine Arun Mathews Jacob George K. J. Ghosh Arpan Yoshi Janssens

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Page 1: Layout ppc group 4

FACILITY LOCATION

Group- 4Akhil Murali

Akhil NandakumarAneesh K SajuAnila P Jestine

Arun Mathews JacobGeorge K. J.

Ghosh ArpanYoshi Janssens

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ALDEP Automated Layout Design Program

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Initial LayoutRelationship chart

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Relationship Chart

Diagram

Rooms Area in Sq. mtr

Abbreviation

Bathroom 100 BABedroom 200 BDDining Room

200 DR

Living Room

150 LR

Kitchen 250 KT

Relationship and Score

Relationship Score ScoreA 4^3 64E 4^2 16I 4^1 4O 4^0 1U 0 0X -4^5 -1024

1 Block= 25 sq. ft.Sweep = 2 blocks= 10 ft

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Layout 1

Relationship Chart

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Layout 2

Relationship Chart

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Layout 3

Relationship Chart

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Layout 4

Relationship Chart

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Layout 5

Relationship Chart

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Layout 6

Relationship Chart

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Layout 7

Relationship Chart

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Layout 8

Relationship Chart

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Final Optimal Layout Using ALDEP

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CRAFT Computerized Relative Allocation of Facilities Technique

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Initial Layout

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Material Handing Trips Matrix

Bed. Bth. Dng. Lvg. Ktn.Bed. 21 5 9 3

Bth. 20 20 5 1

Dng. 6 21 9 21

Lvg. 10 6 10 6

Ktn. 2 1 20 3

Abbreviations:-•Bed. – Bedroom•Bth. – Bathroom•Dng. – Dining room•Lvg. – Living room•Ktn. – Kitchen

From To

Material Movement Scores

A 20 21

E 14 15

I 9 10

O 5 6

U 2 3

X 1 1

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Total area:900 sq. feet

Distance calculation

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Calculation of Centroid

• Equation for Centroid :

• Centroid of Bathroom : (5,30)• Centroid of Dining room : (20,30)• Centroid of Bedroom : (5,15)• Centroid of Living room : (15,17.5)• Centroid of Kitchen : (25,12.5)

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Calculation of distances between the rooms

• Equation for distance:

• Distance between Bathroom & Dining room: 15 ft.• Distance between Bathroom & Bedroom: 15 ft.• Distance between Bathroom & Living room: 22.5 ft.• Distance between Bathroom & Kitchen: 37.5 ft.• Distance between Dining room & Bedroom: 30 ft.• Distance between Dining room & Living room: 17.5 ft.• Distance between Dining room & Kitchen: 22.5 ft.• Distance between Bedroom & Living room: 12.5 ft.• Distance between Bedroom & Kitchen: 22.5 ft.• Distance between Living room & Kitchen: 15 ft.

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Distance Matrix

Bed. Bth. Dng. Lvg. Ktn.Bed. 15 30 12.5 22.5

Bth. 15 15 22.5 37.5

Dng. 30 15 17.5 22.5

Lvg. 12.5 22.5 17.5 15

Ktn. 22.5 37.5 22.5 15

From To

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Total Distance Matrix

Bed. Bth. Dng. Lvg. Ktn.Bed. 315 150 112.5 67.5

Bth. 300 300 112.5 37.5

Dng. 180 315 157.5 472.5

Lvg. 125 135 175 90

Ktn. 45 37.5 450 75

Total distance travelled = 3652.5 ft.

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Interchange Possibility• Possible:

– Bathroom & Dining room (share a common boundary)– Bathroom & Bedroom (share a common boundary)– Dining room & Living room (share a common boundary)– Dining room & Kitchen (share a common boundary)– Dining room & Bedroom (same area)– Bedroom & Living room (share a common boundary)– Living room & Kitchen (share a common boundary)

• Not Possible:– Bathroom & Kitchen– Bathroom & Living room– Bedroom & Kitchen

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Checking the possible cases– Bathroom & Dining room (Defaults relationship chart)– Bathroom & Bedroom (Defaults relationship chart)– Dining room & Living room (Defaults relationship chart)– Dining room & Kitchen (Defaults relationship chart)– Dining room & Bedroom (Feasible, but may get an irregular

shaped layout)– Bedroom & Living room (Defaults relationship chart)– Living room & Kitchen (Defaults relationship chart)

– Therefore we have decided to interchange Dining room and Bedroom

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New Centroids

• Centroid of Bathroom : (5,30)• Centroid of Dining room : (5,15) [Previous Centroid

of Bedroom]• Centroid of Bedroom : (20,30) [Previous Centroid of

Dining room]• Centroid of Living room : (15,17.5)• Centroid of Kitchen : (25,12.5)

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New Distance Matrix

Bed. Bth. Dng. Lvg. Ktn.Bed. 15 30 12.5 22.5

Bth. 15 15 22.5 37.5

Dng. 30 15 17.5 22.5

Lvg. 12.5 22.5 17.5 15

Ktn. 22.5 37.5 22.5 15

Solution:Here we can see that the new distance matrix remains the same. So the new total distance matrix will also be the same. So there won’t be any savings in distance by interchanging Dining room and Bedroom. This implies that the original layout itself is the optimal layout. And there is no improvement that can be done to this layout through CRAFT.

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Proof: Interchanging Dining room & Living room

• The new distance matrix is as given below:

Bed. Bth. Dng. Lvg. Ktn.Bed. 15 12.5 30 22.5

Bth. 15 22.5 15 37.5

Dng. 12.5 22.5 17.5 22.5

Lvg. 30 15 17.5 15

Ktn. 22.5 37.5 22.5 15

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New Total Distance MatrixBed. Bth. Dng. Lvg. Ktn.

Bed. 315 62.5 270 67.5

Bth. 300 450 75 37.5

Dng. 75 472.5 157.5 472.5

Lvg. 300 90 175 90

Ktn. 45 37.5 450 75

Total distance= 4017.5 ft.Savings in total distance= 3652.5 - 4017.5 = -365 ft.

Therefore the savings is negative which means the layout obtained from this interchange is not a feasible solution . This implies that the original layout developed itself is the optimal layout.

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Final Optimal Layout Using CRAFT

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THANK YOU