[PPT] - The Split Delivery Vehicle Routing Problem Hande Yaman Joint work PowerPoint Presentation

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The Split Delivery Vehicle Routing Problem

Hande Yaman

Joint work with Gizem ¨ Ozbaygın, Oya E. Kara¸ san and Barbaros C ¸. Tansel Bilkent University, Department of Industrial Engineering

Hande Yaman (Bilkent University) SDVRP 1 / 26

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Split Delivery VRP

Capacitated Vehicle Routing Problem (CVRP)
given a set of customers and a depot
capacitated vehicles
tours for vehicles that start and end at the depot
visit each customer exactly once
respect the vehicle capacities
minimize the total transportation cost
many variants: time windows, heterogeneous fleet, periodic VRP,

pickup and delivery ...

Split Delivery Vehicle Routing Problem (SDVRP)
relax “visit each customer exactly once”

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History - exact approaches

Dror and Trudeau (1989): split deliveries can lead to considerable cost

savings

Dror and Trudeau (1990): SDVRP is NP-hard, k-split cycles
Dror et al. (1994): valid inequalities for a vehicle index formulation
Belenguer et al. (2000): polyhedral study, cutting plane algorithm
Lee et al. (2006): dynamic programming
Jin et al. (2007): iterative two stage method, cluster and route with lower

bounds on route lengths in clustering

Jin et al. (2008): column generation
Moreno et al. (2010): extended formulation with load and quantity

delivered, generate columns and cuts

Archetti et al. (2011): branch-and-price-and-cut
Archetti et al. (2013): branch-and-cut

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Notation

N = {0, 1, . . . , n}, 0 depot, 1, . . . , n customers
G = (N, A) directed graph, complete
ca: cost of travelling on arc a
m identical vehicles, each with capacity Q
di: demand of customer i
δ−(S), δ+(S): set of incoming and outgoing arcs of S ⊂ N, resp.
Assumption: costs are symmetric and satisfy triangle inequality

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Model

SDVRP can be modeled using variables with vehicle index

min

a∈A

m

k=1

caxk

a

s.t. xk(δ+(0)) = 1 k = 1, . . . , m, xk(δ−(i)) − xk(δ+(i)) = 0 i ∈ N \ {0} , k = 1, . . . , m,

m

k=1

wik = 1 i ∈ N \ {0} ,

n

i=1

diwik ≤ Q k = 1, . . . , m, wik ≤ xk(δ−(i)) i ∈ N \ {0} , k = 1, . . . , m, subtour elimination constraints, xk

a ∈ {0, 1}

a ∈ A, k = 1, . . . , m, wik ≥ 0 i ∈ N \ {0} , k = 1, . . . , m.

If wik ∈ {0, 1} then CVRP.

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Without vehicle index

aim : to come with a model without vehicle index
R-SDVRP

min

a∈A

caxa s.t. f (δ−(i)) − f (δ+(i)) = di i ∈ N \ {0} , x(δ+(0)) = m, x(δ−(i)) − x(δ+(i)) = 0 i ∈ N \ {0} , 0 ≤ fa ≤ Qxa a ∈ A, xa ∈ Z+ a ∈ A.

remove second and third constraints: network loading

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An example: eil30

30 nodes and 3 vehicles
at node 18, incoming 2700 + 625, outgoing 3175 + 0
virtual depot: customer node where unloading and loading take place

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Belenguer, Martinez and Mota (2000)

undirected graph G = (N, E)

min

e∈E

cexe s.t. x(δ(0)) ≥ 2m and even, x(δ(i)) ≥ 2 and even i ∈ N \ {0} , x(δ(S)) ≥ 2 d(S) Q

S ⊂ N \ {0} : 2 ≤ |S| ≤ n − 1,

xe ∈ Z+ e ∈ E.

relaxation
eliminate solutions that are not feasible with cuts
they obtain the same solution as ours for eil30

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Comparing the two relaxations

Projecting out the flow variables in the LP relaxation gives the

fractional capacity inequalities x(δ−(S)) ≥ d(S) Q S ⊆ N \ {0}

In the integer problem, the rounded capacity inequalities

x(δ−(S)) ≥ d(S) Q

S ⊆ N \ {0}

are satisfied.

NP-hard to check whether there exists a feasible routing when x is

fixed (x0i = 2 for all i ∈ N \ {0}, binpacking).

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Checking feasibility

A solution (f , x) is called regular if at each customer node, one can

match each incoming arc with an outgoing arc so that the flow of the incoming arc is greater than or equal to the one on its outgoing arc.

A solution (f , x) of R-SDVRP is regular iff it is feasible for the

SDVRP.

Regularity can be checked in polynomial time.

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Archetti, Bianchessi, Speranza (2013)

Similar formulations to the one of Belenguer et al. (2000) and to

R-SDVRP.

Without flows: build the routes for given x and choose the best by

solving a model.

With flows: same idea as regularity check.
Branch and cut: eliminate vectors x for which there are no feasible

routes by branching

The formulation without flows is better.

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Aim

To solve the R-SDVRP more quickly
Eliminate virtual depots

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k-splits by Dror and Trudeau

There exists an optimal solution to SDVRP with no k-splits.

⇒ There exists an optimal solution to SDVRP where xa ∈ {0, 1} for all a ∈ A \ (δ−(0) ∪ δ+(0)).

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Improvements for R-SDVRP

There exists an optimal solution to SDVRP where xa ∈ {0, 1} for all
utgoing arcs of the depot.
For i ∈ V \ {0}, x(δ−(i)) ≥ 1 is often used to strengthen.
We use the cutset inequalities for singletons at the root node.

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Cutset inequalities

Let S ⊆ N \ {0}.

f (δ−(S)) − f (δ+(S)) = d(S), 0 ≤ fa ≤ Qxa, xa ∈ Z+ a ∈ δ−(S) ∪ δ+(S).

Atamturk (2002): convex hull = trivial + cutset.

A− ⊆ δ−(S), A+ ⊆ δ+(S) η = d(S) Q

, r = d(S) −

d(S) Q

Q

The cutset inequality f (δ−(S) \ A−) + rx(A−) + (Q − r)x(A+) − f (A+) ≥ rη is valid.

A− = δ−(S) and A+ = ∅, rounded capacity inequality.

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Eliminating virtual depots: Patching

solve R-SDVRP
add vehicle indexed variables at a node where regularity is not

satisfied, iterate f k(δ−(i)) − f k(δ+(i)) ≥ 0 k = 1, . . . , m, xk(δ−(i)) − xk(δ+(i)) = 0 k = 1, . . . , m, xk(δ−(i)) ≤ 1 k = 1, . . . , m, 0 ≤ f k

a ≤ Qxk a

a ∈ δ−(i) ∪ δ+(i), k = 1, . . . , m, xk

a ∈ {0, 1}

a ∈ δ−(i) ∪ δ+(i), k = 1, . . . , m, xa =

m

k=1

xk

a , fa = m

k=1

f k

a

a ∈ δ−(i) ∪ δ+(i).

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Eliminating virtual depots: Node splitting

Idea: split nodes as necessary to force regularity at every customer

node

solve R-SDVRP
create a duplicate of the node violating regularity, enlarge the node &

arc sets accordingly

Ni: the set of nodes containing the original customer i and its

duplicates.

Order Ni so that a node j ∈ Ni is represented with (i, l) where l is the
rder of j in Ni.
Let N′ = ∪i∈N\{0}Ni. Define A′ to be the set of all arcs except arcs

between duplicates of the same node.

vi,l is 1 if node (i, l) is visited; 0 otherwise.

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Eliminating virtual depots: Node splitting

min

a∈A′

caxa s.t.

j∈Ni
f (δ−(j)) − f (δ+(j))
= di

i ∈ N \ {0} , f (δ−(j)) − f (δ+(j)) ≥ 0 j ∈ N′, x(δ+(0)) = m, x(δ−(j)) − x(δ+(j)) = 0 j ∈ N′, x(δ−(i, l)) = vi,l (i, l) ∈ N′ : |Ni| ≥ 2, l = |Ni|, x(δ−(i, |Ni|)) ≤ (m − |Ni| + 1)vi|Ni | i ∈ N \ {0} : |Ni| ≥ 2, vi,l ≥ vi,l+1 (i, l) ∈ N′ : |Ni| ≥ 2, l = |Ni|, 0 ≤ fa ≤ Qxa a ∈ A′, vi,l ∈ {0, 1} (i, l) ∈ N′ : |Ni| ≥ 2, xa ∈ {0, 1} a ∈ A′ \ δ−(0), xa ∈ Z+ a ∈ δ−(0).

If |Ni| = m, same as patching but we break symmetry with vi,l ≥ vi,l+1.

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Eliminating virtual depots: Cutting

H ⊆ V \ {0} and S1, . . . , St disjoint nonempty subsets of H with

d(Su) ≤ Q for u = 1, . . . , t.

b(S1, . . . , St): optimal value of binpacking problem with items

1, . . . , t of size d(S1), . . . , d(St).

The framed capacity inequality (Pochet (1998), Augerat (1995),

Naddef and Rinaldi (2002)) x(δ−(H)) +

t

u=1

x(δ−(Su)) ≥ t + b(S1, . . . , St) (1) is valid.

The inequality used by Belenguer et al. is a special case where

H = V \ {0} (known as the generalized capacity inequality).

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Results: Patching

instance with x(δ−(i)) ≥ 1 with cutset inequalities name nodes vehicles gap iters cpu gap iters cpu eil22 22 4 1 4.09 1 2.60 eil23 23 3 1 0.56 1 0.57 eil30 30 3 3 1804.22 4 1818.54 eil33 33 4 0.22 3 13540.53 1 1037.77 eil51 51 5 1 2951.48 1 1383.03 S51D1 51 3 1 18.71 1 23.51 S51D2 51 9 2.86 N/A 5210.24 3.08 N/A 5226.98 p01-110 51 3 1 19.43 1 15.15 p01-1030 51 11 2.41 N/A 6100.37 2.41 N/A 5946.72 p01-1050 51 16 1.37 N/A 4817.89 1.38 N/A 4621.67

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Results: Node splitting

instance with x(δ−(i)) ≥ 1 with cutset inequalities name nodes vehicles gap iters cpu gap iters cpu eil22 22 4 1 4.06 1 2.58 eil23 23 3 1 0.64 1 0.58 eil30 30 3 4 248.4 5 2403.13 eil33 33 4 3 7129.49 1 1033.82 eil51 51 5 1 2953.74 1 1382.24 S51D1 51 3 1 18.77 1 23.58 S51D2 51 9 2.86 N/A 5296.85 2.82 N/A 5098.22 p01-110 51 3 1 19.70 1 15.04 p01-1030 51 11 2.41 N/A 6111.64 2.41 N/A 6050.16 p01-1050 51 16 1.37 N/A 4983.22 1.38 N/A 4767.24

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Comparison

patching with node splitting with instance nodes vehicles x(δ−(i)) ≥ 1 cutsets x(δ−(i)) ≥ 1 cutsets eil30 30 3 1804.22 (3) 1818.54 (4) 248.4 (4) 2403.13 (5) eil33 33 4 13540.53 (3) 1037.77 (1) 7129.49 (3) 1033.82 (1)

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eil30 with cutting planes

Figure : Initial solution with objective 508

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eil30 with cutting planes

Figure : Seventh solution with objective 510

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eil30 with cutting planes

Figure : Ninth solution with objective 510: optimal!

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To be done

Strengthen R-SDVRP
Properties of optimal solutions
Separation of framed capacity constraints
Complete model ?
Combine different approaches
Variants: at most 2 splits, splitting and/or stocking costs ...

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