# maths saves money

Post on 09-Jul-2015

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I used some help from net from my teachers to present this to u hope u adore it

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• Multiple Depot Vehicle Routing Problem & Transportation Problem

• Vehicle Routing Problem (VRP)Chinese Postman Problem Arc-Covering Problems Routing ProblemsTraveling Salesman ProblemMultiple Depot VRPNode-Covering Problems

• Traveling Salesman Problem(TSP)

Multiple Salesmen

The Capacity-constraint

Vehicle Routing Problem(VRP)

Multiple Depot

Multiple DepotVehicle Routing Problem(MDVRP)

• STEPS INVOLVED IN SOLVING MDVRP

• D

• DDD

• D

• Chart1

0.13

0.87

13%

Sheet1

Logistic Cost0.13

Other Items0.87

Sheet1

13%

Sheet2

Sheet3

• North West Corner MethodFrom

ToD1D2D3AvailabilityS10216S22159S32435DEMAND5510

• 5XXX5X541Z = 5X0+1X1+5X1+4X5+5X3 = 41Thus total cost spent shall be Rs.41From

ToD1D2D3AvailabilityS1216S22159S32435DEMAND5510

• Aplly formulae (m+n)-1 whre m = no. of rows and n = no. of clolumnsHere m= 5 & n = 3 thus we have 5 allocationsThus we can find optimal solution by MODI method

ToFromD1D2D3AvailabilityS1216S22159S32435DEMAND5510

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

(S1,D1)5(S2,D1)-2(S3,D1)0(S3,D2)5

• 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 = 395541

D1D2D3Availability Row No.-+6u1=0+--9u2=4X5u3=4Demand5510ColumnNo.v1=0v2=-3v3=1

• Distance Table

123403343134625832

• 12340........1323

• Net Savings per Route1 to 33 + 4 4 = 31 to 43 + 3 6 = 02 to 33 + 4 5 = 22 to 43 + 3 8 = -2 or 03 to 44 + 3 -2 = 5

• 123402222133022035

• 1st Assumption is that the D0i and D0j do not have a T value of 0Answer: 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 pathAnswer: This is true.

• 123402222133022031

• Go to the next highest value on the Net Savings Matrix. It is 3 (doesnt matter which 3 you choose)

123402222133022031

• 1st Assumption is that the D0i and D0j do not have a T value of 0Answer: 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 pathAnswer: This is true.

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

• 123402222113022031

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

123402222113022031

• 1st Assumption is that the D0i and D0j do not have a T value of 0Answer: 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 pathAnswer: This is true.

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

• 123402222111022031

• 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

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

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