1. 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

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. 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

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. 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

7. 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)

8. SDVRP-MDA Example for p=.5 z(VRP) 62 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)
z(SDVRP-MDA) ≈

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