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Vehicle Routing: Coincident Origin and Destination Points Travelling Salesman problem Heuristics Little et al. Algorithm Vehicle Scheduling Clark and Wright Saving Approach Prof.Dr. Füsun Ülengin 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) Prof.Dr. Füsun Ülengin 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 Prof.Dr. Füsun Ülengin 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. Prof.Dr. Füsun Ülengin 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. Prof.Dr. Füsun Ülengin 1.DETERM INING WHICH ARC OF (1,5)-(5,1) IS REPLACED Arc Replaced (1,5)* (1,5) (1,5) (5,1)* (5,1) (5,1) Arc Added to subtour (1,2)-(2,5) (1,3)-(3,5) (1,4)-(4,5) (5,2)-(2,1) (5,3)-(3,1) (5,4)-(4,1) Added length c12+c25-c15=153 c13+c35-c15=462 c14+c45-c15=302 c52+c21-c51=153 c53+c31-c51=462 c54+c41-c51=302 2.DETERM INING WHICH ARC OF (1,2)-(2,5)-(5,1) IS REPLACED Arc Replaced (1,2) (1,2)* (2,5) (2,5) (5,1) (5,1) Arc Added to subtour (1,3)-(3,2) (1,4)-(4,2) (2,3)-(3,5) (2,4)-(4,5) (5,3)-(3,1) (5,4)-(4,1) Added length c13+c32-c12=375 c14+c42-c12=233 c23+c35-c25=514 c24+c45-c25=318 c53+c31-c51=462 c54+c41-c51=302 3. DETERM INING WHICH ARC OF (1,4)-(4,2)-(2,5)-(5,1) IS REPLACED Arc Replaced (1,4)* (4,2) (2,5) (5,1) Arc Added to subtour (1,3)-(3,4) (4,3)-(3,2) (2,3)-(3,5) (5,3)-(3,1) Added length C13+C34-C14=166 C43+C32-C42=202 C23+C35-C25=514 C53+C31-C51=462 Prof.Dr. Füsun Ülengin TSP Model Formulation cij xij Minimize i I j J Subject to: x 1, i I x 1, j J ij j J ij i I x ij ( i , j ) E (U ) U 1, U N xij {0,1}, i I , j J Prof.Dr. Füsun Ülengin 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 Prof.Dr. Füsun Ülengin Example A B C D E A 1 4 5 2 B 2 3 7 3 C 1 6 4 5 D 3 5 8 2 E 4 7 2 1 - 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? Prof.Dr. Füsun Ülengin 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 - Prof.Dr. Füsun Ülengin EXAMPLE OF LITTLE ET AL. ALGORITHM: MATRIX T A B C D E A 11 7 13 B 5 13 15 C 13 15 23 D 9 13 5 E 3 7 7 7 Prof.Dr. Füsun Ülengin 11 15 11 3 - REDUCED MATRIX T0 A B A 0 B 0 C 2 0 D 6 6 E 0 0 C D 0 8 2 4 Prof.Dr. Füsun Ülengin E 2 6 8 0 4 10 0 0 - 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) Prof.Dr. Füsun Ülengin 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 Prof.Dr. Füsun Ülengin 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