By Yupei Xiong, Damon Gulczynski, Bruce Golden, and Edward Wasil 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 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?

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

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

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?

Vehicle capacity is 120 units. 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) Vehicle capacity is 120 units. 6

SDVRP-MDA Let’s look at an instance We expected something like 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)

SDVRP-MDA Example for p=.5 z(VRP) 6 2 1 Є Cap = 3 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 ≈

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?

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

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