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

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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

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?

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

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?
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.

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)

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

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