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FACILITY LOCATION
Group- 4Akhil Murali
Akhil NandakumarAneesh K SajuAnila P Jestine
Arun Mathews JacobGeorge K. J.
Ghosh ArpanYoshi Janssens
ALDEP Automated Layout Design Program
Initial LayoutRelationship chart
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
Layout 1
Relationship Chart
Layout 2
Relationship Chart
Layout 3
Relationship Chart
Layout 4
Relationship Chart
Layout 5
Relationship Chart
Layout 6
Relationship Chart
Layout 7
Relationship Chart
Layout 8
Relationship Chart
Final Optimal Layout Using ALDEP
CRAFT Computerized Relative Allocation of Facilities Technique
Initial Layout
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
Total area:900 sq. feet
Distance calculation
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)
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.
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
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.
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
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
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)
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.
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
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.
Final Optimal Layout Using CRAFT
THANK YOU