prof.dr. füsun Ülengin vehicle routing: coincident origin and destination points travelling...

Prof.Dr. Füsun Ülengin

Vehicle Routing:

Coincident Origin and Destination Points

Travelling Salesman problem Heuristics Little et al. Algorithm

Vehicle Scheduling Clark and Wright Saving Approach


Heuristics Some of the well-known heuristics

Nearest- neighbor Heuristics (NNH) Cheapest insertion Heuristics (CIH) Neural networks (meta heuristics) Lets analyse the first two through an

example (see the next slide)


Distance Between the cities(km)

City 1 City 2 City 3 City 4 City 5

City 1 0 132 217 164 58

City 2 132 0 290 201 79

City 3 217 290 0 113 303

City 4 164 201 113 0 196

City 5 58 79 303 196 0


Nearest- Neighbor Heuristics (NNH)

We begin at any city and then “visit” the nearest city.

Then we go to the unvisited city closest to the city we have most recently visited

We continue in this fashion until a tour is obtained

A popular heuristic is to apply the NNH beginning at each city and then take the best tour obtained.


Cheapest Insertion Heuristics (CIH) We begin at any city and find its closest neigbor The we create a subtour joining those two cities Next, we replace an arc in the subtour (say, arc (i,j)) by

the combination of two arcs- (i,k) and (k,j), where k is not in the current subtour-that will increase the length of the subtour by the smallest (or cheapest amount).

Let cij be the lenght of arc (i,j). Note that if arc (i,j) is replaced by arc (i,k) and (k,j), a

length cik + ckj –cij is added to the subtour. Then we continue with this procedure until a tour is

obtained.


1.DETERMINING WHICH ARC OF (1,5)-(5,1) IS REPLACED

Arc Replaced Arc Added to subtour Added length(1,5)* (1,2)-(2,5) c12+c25-c15=153(1,5) (1,3)-(3,5) c13+c35-c15=462(1,5) (1,4)-(4,5) c14+c45-c15=302(5,1)* (5,2)-(2,1) c52+c21-c51=153(5,1) (5,3)-(3,1) c53+c31-c51=462(5,1) (5,4)-(4,1) c54+c41-c51=302

2.DETERMINING WHICH ARC OF (1,2)-(2,5)-(5,1) IS REPLACED

Arc Replaced Arc Added to subtour Added length(1,2) (1,3)-(3,2) c13+c32-c12=375(1,2)* (1,4)-(4,2) c14+c42-c12=233(2,5) (2,3)-(3,5) c23+c35-c25=514(2,5) (2,4)-(4,5) c24+c45-c25=318(5,1) (5,3)-(3,1) c53+c31-c51=462(5,1) (5,4)-(4,1) c54+c41-c51=302

3. DETERMINING WHICH ARC OF (1,4)-(4,2)-(2,5)-(5,1) IS REPLACED

Arc Replaced Arc Added to subtour Added length(1,4)* (1,3)-(3,4) C13+C34-C14=166(4,2) (4,3)-(3,2) C43+C32-C42=202(2,5) (2,3)-(3,5) C23+C35-C25=514(5,1) (5,3)-(3,1) C53+C31-C51=462


TSP Model Formulation

Minimize Subject to:

c xij ijj Ji I

x i I

x j J

x U U N

x i I j J

ijj J

iji I

iji j E U

ij

1

1

1

,

,

,

{0,1}, ( , ) ( )

,


Little et al. Algorithm 1. Begin by transforming the T matrix into a reduced form in

such a way that the matrix will have at least one zero in each row and in each column

The sum of the subtracted elements will constitute an upper bound for the objective function

2. For each pair of nodes (i, j) calculate the regret of not going from i to j

For example the regret of (B,A) = rBA = minj≠ A tBj

0 + min j ≠ B tiA 0

3. Among the elements having zero balue, select the one having the highest regret; i.e the one that will make the highest increase in the objective function if it is not included in the Hamiltonian cycle

4. At each iteration, make the necessary arrangements to avoid subtours and be sure take the resulting matrix is in the reduced form. matrix


Example A B C D E A - 2 1 3 4 B 1 - 6 5 7 C 4 3 - 8 2 D 5 7 4 - 1 E 2 3 5 2 - 1.Each value in the matrix is given in “half- day”. 2.It takes a half-day to visit a customer 3.Assuming that the traveling salesman does not work on Sunday and that his first day of travel is a Wednesday , which is the first day of the month, what is his earliest possible return date to home (A)4.HOMEWORK :Suppose that the customer D can accept the salesman only on Friday morning. What delay this will cause in the current solution?


THE REDUCED MATRIX ( T0 )

A B C D E

A - 0 0 2 3

B 0 - 5 4 6

C 2 3 - 6 0

D 4 5 3 - 0

E 0 0 3 0 -


A B C D EA - 11 7 13 11B 5 - 13 15 15C 13 15 - 23 11D 9 13 5 - 3E 3 7 7 7 -

EXAMPLE OF LITTLE ET AL. ALGORITHM: MATRIX T


A B C D EA - 0 0 2 4B 0 - 8 6 10C 2 0 - 8 0D 6 6 2 - 0E 0 0 4 0 -

REDUCED MATRIX T0


Vehicle Scheduling :Clarke-Wright savings approach

1.Initially, assume that enough vehicles are available and allocate one to a customer. For our example we will assume that we have 3 tucks of 5,000- gallon capacity, 4 trucks of 6,000-gallon capacity and an unlimited supply of 4,000- gallon capacity. One truck of the smallest capacity is initially allocated to each customer and provides an initial feasible solution of the problem

2. For hand computation, set up a matrix (see the distributed sheet)


Clarke-Wright Savings Approach(cnt.)

The load to be delivered to each customer Pi is listed in column q. The right-hand value in each cell is the distance dy,z between Py and Pz, where y and z are specific customers. The left-hand value represents the savings Sy,z in distance associated with Py and Pz when Py enters the tour.

The value in the middle of the cell ty,z indicates whether the customer combinations Py and Pz are in the tour. The dsignatorhas the following values:

t y,z = 1 is two customers are linked on a truck route t y,z = 0 if the customers are not linked on a truck route t y,z = 2 if the customer is served exclusively by a single

truck For ease of computation, the matrix is ordered from left to

right on the basis of increasing savings S y,z


Clarke-Wright Savings approach(cnt.)

3. Search the matrix for the largest savings subject to the following conditions for any cell (y,z,)A) t y,0 and t z,o are >0

B) Py and Pz are not already on the same truck run

C) By this allocation you do not exceed the capacity of the trucks available

4. Make the necessary changes in the t values of the combined tours

See the distributed sheet

prof.dr. füsun Ülengin vehicle routing: coincident origin and destination points travelling...

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