maths saves money

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

Traveling Salesman Problem(TSP)

Multiple Salesmen The Capacity -constraint

Vehicle Routing Problem(VRP)

Multiple Depot

Multiple DepotVehicle Routing Problem

(MDVRP)

How Many Problems Lead To MDVRP

STEPS STEPS INVOLVED IN INVOLVED IN

SOLVING SOLVING MDVRPMDVRP

D

D

D

D

D

To D1 D2 D3 Availability

S1 0 2 1 6

S2 2 1 5 9

S3 2 4 3 5

DEMAND 5 5 10

North West North West Corner MethodCorner Method

From

To D1 D2 D3 Availability

S1 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

Aplly formulae (m+n)-1 whre m = no. of rows and n = no. of clolumns

Here m= 5 & n = 3 thus we have 5 allocations Thus we can find optimal solution by MODI method

ToFrom

D1 D2 D3 Availability

S1 2 1 6

S2 2 1 5 9

S3 2 4 3 5

DEMAND 5 5 10

Let u1,u2,u3 denotes the three rows and v1,v2,v3 denote the three columns

Therefore u1+v1=0, u1+v3 =1,u2+v2=1 u2+v3=5,u3+v3=3

Let u1=0 therefore we get other variables i.e u1=0,u2=4,u3=2,v1=0,v2=-3,v3=1

Now we shall calculate the opportunity cost by the relation i.e

for each occupied cell Here the most negative

opportunity cost is the cell (S2,D1)

( )ij cij ui vj= − +V

(S1,D1) 5

(S2,D1) -2

(S3,D1) 0

(S3,D2) 5

Final Solution Now we shall make the plus and

minus adjustment in the assigned cell on the closed path.

The closed path therefore starts from the cell (S2,D1) passing through the assigned cells (S1,D1),(S1,D3),S2,D1) as shown

In the closed path at the corner points minimum assignments is 1. now add this sign value to the cells having positive sign in the closed path and subtract having negative sign

Thus transportation cost in this case is

Z= 4x0+2x1+1x2+5x1+3x5+5x3 = 39

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

5

5

4

1

NUMERICAL NUMERICAL EXAMPLEEXAMPLE

1 2 3 4

0 3 3 4 3

1 3 4 6

2 5 8

3 2

Distance Table

GIVEN SITUATION

12

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:= D0i + D0j – Dij

= 3 + 3 – 3 = 3

Distance between 0 and 1 = D0i = D01 = 3Distance between 0 and 2 = D0j = D02 = 3Distance between 1 and 2 = Dij = D12 = 3

Net Savings per Route1 to 31 to 3 3 + 4 – 4 = 33 + 4 – 4 = 3

1 to 41 to 4 3 + 3 – 6 = 03 + 3 – 6 = 0

2 to 32 to 3 3 + 4 – 5 = 23 + 4 – 5 = 2

2 to 42 to 4 3 + 3 – 8 = -2 or 03 + 3 – 8 = -2 or 0

3 to 43 to 4 4 + 3 -2 = 54 + 3 -2 = 5

Start with the most inefficient route 3 + 3 + 3 + 3 + 4 + 4 + 3 + 3 = 26kmStart with the most inefficient route 3 + 3 + 3 + 3 + 4 + 4 + 3 + 3 = 26km

12

0

3

4

3

3

4

Customers

Depot(s)

Original Net Savings Matrix with the bold values being the original “T” values.Original Net Savings Matrix with the bold values being the original “T” values.

1 2 3 4

0 2 2 2 2

1 3 3 0

2 2 0

3 5

Find the highest number on Find the highest number on the Net Savings Matrix and see if the Net Savings Matrix and see if the route satisfies our the route satisfies our assumptions.assumptions.

Route 3 to 4 has a savings of 5 Route 3 to 4 has a savings of 5 miles if we don’t go back to the miles if we don’t go back to the warehouse.warehouse.

1st Assumption is that the D0i and D0j do not have a “T” value of 0

Answer: This is true. In this model, D03 currently has T = 2 and D04 has T = 2

2nd Assumption is that D0i and D0j are not on the same path

Answer: This is true.

Reroute the original trip from: 0 – 3 – 0 – 4 - 0 to: 0 – 3 – 4 – 0

12

0

3

4

1 2 3 4

0 2 2 2 2

1 3 3 0

2 2 0

3 1

Go to the next highest value on the Net Savings Matrix. It is 3 (doesn’t matter which 3 you choose)

1 2 3 4

0 2 2 2 2

1 3 3 0

2 2 0

3 1

1st Assumption is that the D0i and D0j do not have a “T” value of 0

Answer: This is true. In this model, D01 currently has T = 2 and D02 has T = 2

2nd Assumption is that D01 and D02 are not on the same path

Answer: This is true.

Reroute the original trip from: 0 – 1 – 0 – 2 - 0 to: 0 – 1 – 2 – 0

12

0

3

4

1 2 3 4

0 2 2 2 2

1 1 3 0

2 2 0

3 1

Go to the next highest value on the Net Savings Matrix. It is 3 (trips between 1 and 3).

1 2 3 4

0 2 2 2 2

1 1 3 0

2 2 0

3 1

1st Assumption is that the D0i and D0j do not have a “T” value of 0

Answer: This is true. In this model, D01 currently has T = 1 and

D03 has T = 1 2nd Assumption is that D01 and D02 are not on the

same path Answer: This is true.

Reroute the original trip from: 0 – 1 – 0 – 3 - 0 to: 0 – 1 – 3 – 0

12

0

3

4

1 2 3 4

0 2 2 2 2

1 1 1 0

2 2 0

3 1

Take a look at the model. We can eliminate the trip between 0 and 3 because it serves no purpose. The most efficient route becomes 0 – 2 – 1 – 3 – 4 – 0 or

3 + 3 + 4 + 2 + 3 = 15km

12

0

3

41

2

0

3

4

We can apply this method for other problems, e.g., the VRP, and VRPTW, have not be solved by this algorithm.