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Multiple Depot Vehicle Routing Problem & Transportation Problem
Vehicle Routing Problem (VRP) Chinese Postman Problem   Arc-Covering Problems   Routing Problems Traveling Salesman Problem Multiple Depot VRP Node-Covering Problems
How Many Problems Lead To MDVRP
 
STEPS INVOLVED IN SOLVING MDVRP
D
D D D
D
 
 
 
 
 
 
North West Corner Method From To D1 D2 D3 Availability S1 0 2 1 6 S2 2 1 5 9 S3 2 4 3 5 DEMAND 5 5 10
5 X X X 5 X 5 4 1 Z = 5X0+1X1+5X1+4X5+5X3 = 41 Thus total cost spent shall be Rs.41 From To D1 D2 D3 Availability S1 2 1 6 S2 2 1 5 9 S3 2 4 3 5 DEMAND 5 5 10
[object Object],[object Object],[object Object],To From D1 D2 D3 Availability S1 2 1 6 S2 2 1 5 9 S3 2 4 3 5 DEMAND 5 5 10
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],(S1,D1) 5 (S2,D1) -2 (S3,D1) 0 (S3,D2) 5
[object Object],[object Object],[object Object],[object Object],[object Object],5 5 4 1 Final Solution D1 D2 D3 Availability  Row No. - + 6 u1=0 + - - 9 u2=4 X 5 u3=4 Demand 5 5 10 Column No. v1=0 v2=-3 v3=1
NUMERICAL EXAMPLE
Distance Table 1 2 3 4 0 3 3 4 3 1 3 4 6 2 5 8 3 2 GIVEN SITUATION 1 2 0 3 4 3 3 4
1 2 3 4 0 … .. … .. … .. … .. 1 3 2 3 Net Savings of going 0 to 1 to 2 to 0 (rather than 0 to 1 and 0 to 2 and back to 0) is: =  D 0i  + D 0j  – D ij =  3 + 3 – 3  = 3 Distance between 0 and 1 = D0i = D01 = 3 Distance between 0 and 2 = D0j = D02 = 3 Distance between 1 and 2 = Dij = D12 = 3
Net Savings per Route 1 to 3 3 + 4 – 4 = 3 1 to 4 3 + 3 – 6 = 0 2 to 3 3 + 4 – 5 = 2 2 to 4 3 + 3 – 8 = -2 or 0 3 to 4 4 + 3 -2 = 5
[object Object],1 2 0 3 4 3 3 4 Customers Depot(s)
[object Object],1 2 3 4 0 2 2 2 2 1 3 3 0 2 2 0 3 5 ,[object Object],[object Object]
[object Object],[object Object],[object Object],[object Object]
Reroute the original trip from:  0 – 3 – 0 – 4 - 0  to: 0 – 3 – 4 – 0  1 2 0 3 4
1 2 3 4 0 2 2 2 2 1 3 3 0 2 2 0 3 1
[object Object],1 2 3 4 0 2 2 2 2 1 3 3 0 2 2 0 3 1
[object Object],[object Object],[object Object],[object Object]
[object Object],1 2 0 3 4
1 2 3 4 0 2 2 2 2 1 1 3 0 2 2 0 3 1
[object Object],1 2 3 4 0 2 2 2 2 1 1 3 0 2 2 0 3 1
[object Object],[object Object],[object Object],[object Object],[object Object]
[object Object],1 2 0 3 4
1 2 3 4 0 2 2 2 2 1 1 1 0 2 2 0 3 1
[object Object],[object Object]
1 2 0 3 4 1 2 0 3 4
[object Object]
 

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Maths Saves Money

  • 1. Multiple Depot Vehicle Routing Problem & Transportation Problem
  • 2. Vehicle Routing Problem (VRP) Chinese Postman Problem Arc-Covering Problems Routing Problems Traveling Salesman Problem Multiple Depot VRP Node-Covering Problems
  • 3. How Many Problems Lead To MDVRP
  • 4.  
  • 5. STEPS INVOLVED IN SOLVING MDVRP
  • 6. D
  • 8. D
  • 9.  
  • 10.  
  • 11.  
  • 12.  
  • 13.  
  • 14.  
  • 15. North West Corner Method From To D1 D2 D3 Availability S1 0 2 1 6 S2 2 1 5 9 S3 2 4 3 5 DEMAND 5 5 10
  • 16. 5 X X X 5 X 5 4 1 Z = 5X0+1X1+5X1+4X5+5X3 = 41 Thus total cost spent shall be Rs.41 From To D1 D2 D3 Availability S1 2 1 6 S2 2 1 5 9 S3 2 4 3 5 DEMAND 5 5 10
  • 17.
  • 18.
  • 19.
  • 21. Distance Table 1 2 3 4 0 3 3 4 3 1 3 4 6 2 5 8 3 2 GIVEN SITUATION 1 2 0 3 4 3 3 4
  • 22. 1 2 3 4 0 … .. … .. … .. … .. 1 3 2 3 Net Savings of going 0 to 1 to 2 to 0 (rather than 0 to 1 and 0 to 2 and back to 0) is: = D 0i + D 0j – D ij = 3 + 3 – 3 = 3 Distance between 0 and 1 = D0i = D01 = 3 Distance between 0 and 2 = D0j = D02 = 3 Distance between 1 and 2 = Dij = D12 = 3
  • 23. Net Savings per Route 1 to 3 3 + 4 – 4 = 3 1 to 4 3 + 3 – 6 = 0 2 to 3 3 + 4 – 5 = 2 2 to 4 3 + 3 – 8 = -2 or 0 3 to 4 4 + 3 -2 = 5
  • 24.
  • 25.
  • 26.
  • 27. Reroute the original trip from: 0 – 3 – 0 – 4 - 0 to: 0 – 3 – 4 – 0 1 2 0 3 4
  • 28. 1 2 3 4 0 2 2 2 2 1 3 3 0 2 2 0 3 1
  • 29.
  • 30.
  • 31.
  • 32. 1 2 3 4 0 2 2 2 2 1 1 3 0 2 2 0 3 1
  • 33.
  • 34.
  • 35.
  • 36. 1 2 3 4 0 2 2 2 2 1 1 1 0 2 2 0 3 1
  • 37.
  • 38. 1 2 0 3 4 1 2 0 3 4
  • 39.
  • 40.