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The 2 Period Travelling Salesman Problem Applied to Milk Collection in Ireland By

The 2 Period Travelling Salesman Problem Applied to Milk Collection in Ireland By Professor H P Williams,London School of Economics Dr Martin Butler, University College Dublin. Appears in: M. Butler, H Paul Williams & l-A Yarrow

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The 2 Period Travelling Salesman Problem Applied to Milk Collection in Ireland By

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  1. The 2 Period Travelling Salesman Problem Applied to Milk Collection in Ireland By Professor H P Williams,London School of Economics Dr Martin Butler, University College Dublin Appears in: M. Butler, H Paul Williams & l-A Yarrow Computational Optimization and Applications, 7(1997) 291-306

  2. Ireland Farmer Catchment Area

  3. Location of Suppliers

  4. A MILK DISTRIBUTION PROBLEM • Milk is to be collected from 41 farms using a vehicle based at a central depot. • For 12 of the farms there is to be a daily collection. • For the other 29 farms collection is to be every other day. • 1. Decide which farms are to be visited on which days • 2. Route the vehicle on these days. • The objective is to minimise total distance travelled

  5. 42 Node Problem + Every Day * Second Day Nodes 1-13 (+) Visited Both days Nodes 14-42 Visited Every Other Day

  6. Extension of the Symmetric Travelling Salesman Problem Number of Nodes visited every other day Number of Nodes visited every day 1 – Period Problem Number of Tours 2 – Period Problem Number of Pairs of Tours

  7. A HEURISTIC SOLUTION PROCEDURE(gives upper bound for total distance) • Create a tour around “every day” farms (e.g. nearest neighbour heuristic) • Apply an improvement heuristic (e.g. 2-interchange method) • Duplicate tours • Insert “every other day” farms into tours by an insertion heuristic (e.g. cheapest insertion method)

  8. 42 Node Problem Nearest Neighbour Solution for Every Day Farms + Every Day * Second Day Nodes 1-13 (+) Visited Both days Nodes 14-42 Visited Every Other Day

  9. 42 Node Problem Nearest Neighbour +2-Interchange for Every Day Farms (Length = 687 Optimal) + Every Day * Second Day Nodes 1-13 (+) Visited Both days Nodes 14-42 Visited Every Other Day

  10. 42 Node Problem Heuristic Solution Length 1750 + Every Day * Second Day Nodes 1-13 (+) Visited Both days Nodes 14-42 Visited Every Other Day

  11. OTHER HEURISTIC SOLUTION PROCEDURES (give upper bounds for total distance) • (i) Partition Farms into 2 sets • (e.g. Clustering, Distance from Depot etc) • (ii) Apply 1-Period TSP heuristic to each cluster +depot • 2. (i) Grow 2 Spanning Trees from Depot. Each “every day” farm included in both trees • Each “every other day” farm included in one tree • (ii) Match odd degree nodes in each tree • (iii) Create Eulerian tours • (iv) “Short circuit” farms visited twice

  12. INTEGER PROGRAMMING FORMULATION every other day visits both day visits every other day visits This is the 2-Matching Relaxation

  13. SOLUTION APPROACH • Solve Linear Programming Relaxation of 2-Matching Relaxation. (2 secs, 993 iterations) • (Could then apply the Branch and Bound alogrithm to try to obtain optimal integer solution. This • (i) takes a prohibitive amount of time – weeks – to solve • (ii) only produces subtours) • 2. Identify violated “VUB” Constraints (61) • Identify violated single day subtours (14) • Identify violated single day combs (1) • Identify violated aggregated combs (1) • Append these constraints (and those for other day) • Resolve LP Relaxation (2 secs, 536 iterations from starting basis) • Append violated VUB constraints (63) • Resolve LP Relaxation (13 secs, 2365 iterations) • 7. Repeat 3,4,5

  14. 42 Node Problem ½ ½ ½ ½ ½ LP Relaxation of 2 – Matching Relaxation ½ ½ ½ ½ ½ ½ ½ ½ Length =1570 ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ + Every Day * Second Day ½ Nodes 1-13 (+) Visited Both days Nodes 14-42 Visited Every Other Day

  15. VARIABLE UPPER BOUND CONSTRAINTS(CUTS) Second pair of constraints may be violated by fractional solutions which satisfy first constraint. Would need to append such constraints. Will only append if violated.

  16. 42 Node Problem ½ ½ ½ ½ ½ LP Relaxation after violated VUB constraints appended Length = 1696.5 ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ ½ + Every Day * Second Day Nodes 1-13 (+) Visited Both days Nodes 14-42 Visited Every Other Day

  17. SUBTOURS * + * + + * COMBS + + * * * + * * + + * + *

  18. SUBTOUR CONSTRAINTS all Where is set of farms visited every other day is set of farms visited every day Is set of edges with both ends in

  19. SUBTOURS, COMBS AND AGGREGATED COMBS A violated subtour constraint 34 * 10 + * 39 A violated aggregated comb constraint 2.5 * 5 8 + + * 2.8 + 6 * 26 * 27 (handle) (tooth 1) (tooth 2) (tooth 3)

  20. 42 Node Problem Length = 1711 + Every Day * Second Day Nodes 1-13 (+) Visited Both days Nodes 14-42 Visited Every Other Day

  21. 42 Node Problem LP Relaxation after VUBs 2nd set of subtours & Comb Constarints Length 1720 + Every Day * Second Day Nodes 1-13 (+) Visited Both days Nodes 14-42 Visited Every Other Day

  22. 42 Node Problem Optimal Solution Length 1725 + Every Day * Second Day Nodes 1-13 (+) Visited Both days Nodes 14-42 Visited Every Other Day

  23. SOLUTION RESULTS Objective (1/10 miles) LP Relaxation 1570 (B & B takes weeks and produces subtours) LP Relaxation after VUB constraints 1696.5 (B&B TAKES 40 MINUTES AND PRODUCES SUBTOURS) LP Relaxation after VUB, Subtour and Comb 1711.5 Constraints (B & B takes 7 minutes to produce subtours) LP Relaxation after VUB and further Subtour 1720.83 Constraints B & B takes 13 seconds to produce optimal solution 1725 Heuristic Solution 1750 A total of 63 VUB, 18 Subtour, 1Comb Constraint and 2 Aggregated Comb Constraints were used. Resultant model has 149 Constraints (excluding (VUB’s), 1778 Variables and solves in a total of 26 seconds (On a 486 PC)

  24. Solution Approach SOLVE LP RELAXATION AUTOMATICALLY APPEND VIOLATED VUB CONSTRAINTS APPEND VIOLATED CONSTRAINTS SOLVE LP RELAXATION APPEND VIOLATED CONSTRAINTS DRAW GRAPH OF SOLUTION Yes IS SOLUTION INTEGER? Yes ARE THERE VIOLATED SUBTOURS? No No IDENTITY INEQUALITIES VIOLATED BY SOLUTION STOP AN OPTIMAL 2-TOUR FOUND Yes VIOLATED INEQUALITIES IDENTIFIED APPLY BRANCH AND BOUND ALGORITHM No

  25. FURTHER CONSIDERATIONS • Could identify further cuts (facets?) and avoid use of Branch and Bound. • . Can include capacity constraints limiting each day’s collection • (Constraints on for each 3. Could generalise to more than 2 time periods.

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