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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

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
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?
sdvrp3

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
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 mda5
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-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 mda7
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)

sdvrp mda8

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 mda9
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 mda10
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
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