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Worst-case Analysis for the Split Delivery VRP with Minimum Delivery Amounts

Worst-case Analysis for the Split Delivery VRP with Minimum Delivery Amounts. By Yupei Xiong, Damon Gulczynski, Bruce Golden, and Edward Wasil. Presented August 2010. SDVRP. There has been a lot of work published on the SDVRP in the last 5 years

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Worst-case Analysis for the Split Delivery VRP with Minimum Delivery Amounts

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  1. Worst-case Analysis for the Split Delivery VRP with Minimum Delivery Amounts By Yupei Xiong, Damon Gulczynski, Bruce Golden, and Edward Wasil Presented August 2010

  2. SDVRP • There has been a lot of work published on the SDVRP in the last 5 years • Archetti, Savelsbergh, and Speranza published a nice paper on worst-case analysis in 2006 • They asked the question: In the worst case, how badly can the VRP perform relative to the SDVRP? • In the best case, how much better can you do with split deliveries vs. no split deliveries?

  3. Q/2 + 1 Є Є Є Q/2 + 1 1 SDVRP z(VRP) z(SDVRP) • Archetti et al. show that and the bound is tight • Key point: You can do 50% better, if you allow split deliveries • The bound is tight ≤2

  4. SDVRP-MDA • Next, Damon, Ed, and I looked at a generalization of the SDVRP motivated by practical concerns: SDVRP-MDA • Deliveries take time and are costly to customers and distributors • How can we model this? • We use a percentage (e.g., 10%) • We published a paper on this in Trans. Res. E • Given an instance, solve it to near-optimality

  5. SDVRP-MDA • Let p be the minimum percentage delivered • When p=0, we have the SDVRP • When p>0.5, we have the VRP • We asked the question: In the best case, how much better can you do with SDVRP-MDA vs. VRP, as a function of p? • What did we expect?

  6. SDVRP-MDA Example SDVRP p = 0 Total Distance = 22 SDVRP-MDA p = .3 Total Distance = 24 VRP Total Distance = 30 (100) (100) (100) 2 2 2 1 5 3 Depot Depot Depot (80) (80) (60) (20) (60) (20) 5 5 1 1 1 3 3 3 1 (60) (40) 6 Vehicle capacity is 120 units.

  7. SDVRP-MDA • Let’s look at an instance • In general, as p increases from 0 (within 0 < p ≤0.5), the cost (distance) increases • In other words, as p increases, split deliveries buy you less • We expected something like z(VRP) ≤ 2 – p for 0 < p ≤ 0.5 z(SDVRP-MDA)

  8. Є Є 2 2 2 1 1 1 Cap = 3 SDVRP-MDA • Example for p=.5 • On the other hand, we were able to prove a very surprising result • Assume all customer demands are equal and not larger than vehicle capacity z(VRP) 6 z(SDVRP-MDA) 4 ≈

  9. SDVRP-MDA • z(VRP) ≤ 2 for 0 < p < 0.5 z(SDVRP-MDA) and the bound is tight • Corollary. For arbitrary demands (no larger than vehicle capacity), the above result still holds • Now, what happens when p=0.5?

  10. SDVRP-MDA • First, assume demands are equal and not larger than vehicle capacity • We were able to prove that • The only case not addressed is when p=0.5 and demands are arbitrary (no larger than vehicle capacity) • Our conjecture is that the bound of 1.5 still applies z(VRP) z(SDVRP-MDA) ≤1.5

  11. The Last Case • We are working to settle this last case • If anyone in the audience can settle it before us, please let me know • We’ll gladly add you as a co-author

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