Fiber Cable Network Design with

Operations Administration & Maintenance Constraints

Vincent Angilella

1,2

, Matthieu Chardy

and Walid Ben-Ameur

Orange Labs, 40-44 Avenue de la Republique, 92320, Chatillon, France

SAMOVAR, Telecom SudParis, CNRS, Paris-Saclay University, 9 Rue Charles Fourier, 91011 Evry Cedex, France

Keywords:

Optical Networks, Network Design, Mixed Integer Programming, Dynamic Programming.

Abstract:

We introduce two speciﬁc design problems of optical ﬁber cable networks that differ by a practical mainte-

nance constraint. An integer programming based method including valid inequalities is introduced for the

unconstrained problem. We propose two exact solution methods to tackle the constrained problem: the ﬁrst

one is based on mixed integer programming including valid inequalities while the second one is built on dy-

namic programming. The theoretical complexities of both problems in several cases are proven and compared.

Numerical results assess the efﬁciency of both methods in different contexts including real-life instances, and

evaluate the effect of the maintenance constraint on the solution quality.

1 INTRODUCTION

Fiber To The Home (FTTH) networks are currently

deployed by telecommunications operators, and re-

quire a huge capital expenditure (see (Europe, 2016),

it can cost several billion euros to connect one mil-

lion households). The technological architecture cho-

sen by a majority of operators is to deploy passive

optical networks, which are based on passive optical

splitters. A passive optical splitter connects several

ﬁbers on one of its sides to one at the other side (di-

vides or gathers the signal depending on its origin),

which leads to a tree topology of the FTTH networks

(illustrated in Fig. 1). The design of such networks

includes to decide the splitter locations, the civil en-

gineering infrastructure used (see (Bley et al., 2013),

(Chardy et al., 2013), (Gollowitzer et al., 2013), (Con-

treras and Fernandez, 2012)). Finally, the ﬁber cable

network has to be designed to connect these equip-

ment (see Fig. 1). These decisions are usually taken

in different steps.

This paper focuses on the problem of ﬁber cable

network design. This problem is highlighted in the

survey (Gr

otschel et al., 2013) as an incomplete ﬁeld

of study, especially when cable separation techniques

are considered. The work from (Angilella et al., 2016)

tackles the issue including the selection of civil engi-

neering infrastructure, but faces computational limits

on real-life instances. The paper (Mateus et al., 2000)

excludes weld costs, which are a signiﬁcant expense

source. The work from (Angilella et al., 2017) deals

with the issue of cable backfeed, speciﬁc to the prob-

lem, but restricts the possible ways to serve the de-

mand. In the following we include several ways to

serve the demand (with ﬁber cables or ﬁber modules),

and introduce a maintenance constraint which, to our

knowledge, is novel.

The next section introduces two problems which

differ by the introduction of an Operation Adminis-

tration & Maintenance constraint. We introduce an al-

gorithm based on integer programming for the uncon-

strained problem in Section 3.1. Two solution meth-

ods are then proposed for the constrained problem, an

integer programming based solution in Section 3.2,

and a dynamic programming based solution in Sec-

tion 4. The theoretical complexities of the problems

are argued in Section 5. All solution methods are as-

sessed numerically in Section 6.

2 PROBLEM DESCRIPTION

The general problem tackled in this paper consists

in connecting one splitter location to several client

groups, using ﬁber cables, with minimal cost. It arises

several times in a given FTTH network, notably once

for each splitter location.

Angilella, V., Chardy, M. and Ben-Ameur, W.

Fiber Cable Network Design with Operations Administration & Maintenance Constraints.

DOI: 10.5220/0006621700940105

In Proceedings of the 7th International Conference on Operations Research and Enterprise Systems (ICORES 2018), pages 94-105

ISBN: 978-989-758-285-1

Figure 1: Underlying optical architecture example. It has

a tree topology; the splitter location is connected to every

client group.

Figure 2: Underlying civil engineering tree example. The

ducts, cabinets, demands and number of ﬁber modules are

known.

2.1 Unconstrained Problem

Cables are to be laid out in a civil engineering in-

frastructure (usually the one used for the legacy cop-

per network) with a tree topology, assumed chosen

within previous decision steps. The cables have an

arborescent structure from the splitter location to the

client groups. Along the ducts of this infrastructure

are located street cabinets, in which the demand lies.

The civil engineering structure used is supposed to be

known due to previous decision making, as well as the

demand in each cabinet.

Fiber cables contain several ﬁber modules, and

each ﬁber module contains several ﬁbers. Due to op-

erational constraints, modules are not dividable, and

all modules on a given network are supposed to be

identical. This allows us to consider only ﬁber mod-

ules, and ignore the ﬁber level. Some of the mod-

ules are connected to the ﬁber source on one of their

ends, and on the ﬁber demand on the other end. These

are actually used, and are called ”active modules”, the

other ones are called ”dead modules”. The latter can

arise due to cables not matching exactly the demand

or in the operations described below (example: a 4

module cable serving a cabinet which requires 3 mod-

ules). Since all the demand is known and there is only

one path from the source to a given demand point,

the number of active modules that must be deployed

through a given duct is known (see Fig 2).

At a cabinet, cables can endure a splicing opera-

tion, which leads to two basic conﬁgurations (see Fig.

3):

• All cables are continued. One only has to pay for

the cost of laying out cables.

• One cable is spliced. It is cut at the cabinet, and

its active modules are welded to active modules of

new cables, referred to as ”born cables”. A protec-

tive box, the size of which depends on the spliced

cable size, is installed. One has to pay for the ca-

bles, the box and the welds.

There are two different ways to serve the demand

that cannot be combined (see Fig. 4):

• Cable-served. In this case, a single cable brings

all the required active modules to the demand cab-

inet.

• Module-served. In this case, a splicing operation

is done in the cabinet, and some modules from the

spliced cable are used to serve the demand. No

welds are done on these modules.

Additional engineering rules have to be taken into

account:

• At most one cable can be spliced at a street cab-

inet. This is due to space restrictions and regula-

tory constraints (protective boxes are large).

• The demand of a given cabinet must be served by

at most one cable.

The cost elements are as follows:

• The cost of a cable is linear with respect to its

length, and concave with respect to its size (i.e.

its number of modules). This derives from the

catalogues of cable manufacturers, who propose

a ﬁxed price per length unit for each cable size.

• The cost of a protective box depends on the size

of the cable being spliced. It is a piecewise con-

stant function. This derives from the number of

different boxes sold by manufacturers.

• The cost of welds depends on the number of welds

to be done in a given cabinet. It is piecewise linear

concave, and derives from manpower cost consid-

erations.

Fiber Cable Network Design with Operations Administration & Maintenance Constraints

Figure 3: Left: Continued cables; Right: Splicing operation.

Figure 4: Left: Module-served demand node; Right: Cable-

served demand node.

This decision problem, referred to as FCNDA

(Fiber Cable Network Design in an Arborescence) in

the following, can be formulated as follows: given a

civil engineering arborescence, demand nodes, a set

of available cables and the associated costs, design a

minimum cost optical ﬁber cable network satisfying

the engineering rules listed above.

Section 2.2 introduces a restriction of the FCNDA

problem.

2.2 Constrained Problem

We restrict the problem by imposing that all cables

going through a given duct are born in the same cab-

inet (eventually the ﬁber source). This restriction is

illustrated in Fig. 5. It is motivated by operations and

maintenance considerations. Indeed, assuming all the

cables of a given duct are damaged, then an interven-

tion has to be done at the cabinets where each of these

cables is born. If the rule is respected, an intervention

is necessary in only one cabinet.

The constrained decision problem, referred to as

EFCNDA (Easy-maintenance Fiber Cable Network

Design in an Arborescence) in the following consists

in designing a FCNDA solution where cables on a

same duct are born in the same cabinet with minimal

cost.

3 INTEGER PROGRAMMING

3.1 FCNDA

3.1.1 Notation and Formulation

The following notation will also be used in section 4.

An arborescence G = (V, A) describes the civil

engineering infrastructure, V the cabinets and A the

ducts, and its root r ∈ V denotes the ﬁber source.

For any i ∈ V, D

∈ N denotes the demand (number

of active modules required) in node i. We deﬁne

∗

= V \ {r}. The set of demand nodes is noted

= {v ∈ V, D

> 0}, the set of nodes without de-

mand V

= V

∗

. Each arc (i, j) ∈ A has a length

i, j

and must contain m

act

i, j

active modules. For i ∈ V ,

let Γ

(i) denote the set of successors of i and let γ(i)

be its predecessor.

The set of cable types is denoted by L = {1, ..,L}

where L is the number of different cable types avail-

able. Cables of type l ∈ L have a size of M

∈ N mod-

ules, and for l ∈ L, we denote M

= {1, .., M

} the

range of possible number of active modules in a cable

ICORES 2018 - 7th International Conference on Operations Research and Enterprise Systems

Figure 5: Left: Allowed splicing conﬁguration for EFCNDA. On all edges, cables are born in the same cabinet; Right:

Forbidden splicing conﬁguration for EFCNDA. On the bottom-right duct, two different cables are born in different cabinets.

of type l. The cable sizes (M

)

are supposed to be

ordered with respect to l. The largest cable has a size

of M

, which corresponds to the maximal number of

welds done in a node.

For l ∈ L, let us deﬁne C

the cost per length unit

of a cable of type l, and PB

the cost of a box of type l.

For m ∈ {0, .., M

}, let us deﬁne the cost of the small-

est cable able to contain m active modules C

min

= C

where l

= min{l ∈ L, m ≤ M

} (which is also the

cheapest as cable costs are increasing with respect to

the cable size), and PW

the cost for welding m mod-

ules.

We introduce P the set of oriented paths of G, and

for p ∈ P , we denote by s(p) its source node, t(p) its

target node, and d

its length (which extends d from

A to P ).

We deﬁne the following variables:

• ∀l ∈ L, ∀p ∈ P , k

spl

p,l

∈ {0, 1} the binary variable

equal to 1 iff there is a cable of size l on path p

spliced in t(p).

• ∀p ∈ P ,k

dem

∈ {0, 1} the binary variable equal to

1 iff there is a cable on path p serving the demand

in t(p) in a cable-served way; its type is known, it

is min{l ∈ L |M

≥ D

t(p)

• ∀p ∈ P , m

spl

∈ {0, .., M

} the number of active

modules of the cable on path p spliced in t(p).

• ∀i ∈ V

∗

, ∀m ∈ M

, w

i,m

the binary variable equal

to 1 iff m welds are done in node i.

The problem can be formulated as follows:

min

∑

p∈P

∑

l∈L

· k

spl

p,l

) +C

min

t(p)

· k

dem

∑

i∈V

∑

m∈M

· w

i,m

∑

p∈P

∑

l∈L

· k

spl

p,l

such that

∀i ∈ V

∗

∑

p∈P |t(p)=i

∑

l∈L

spl

p,l

≤ 1 (1)

∀i ∈ V

∑

p∈P |t(p)=i

dem

≤ 1 (2)

∀p ∈ P ,

∑

l∈L

· k

spl

p,l

≥ m

spl

(3)

∀i ∈ V

∗

∑

p∈P |t(p)=i

spl

· (1 −

∑

p∈P |t(p)=i

dem

)

∑

p∈P |s(p)=i

spl

+ D

t(p)

dem

) (4)

∀i ∈ V

∗

∑

m∈M

m · w

i,m

∑

p∈P |i=s(p)

spl

+ D

t(p)

· k

dem

) (5)

∀i ∈ V

∑

m∈M

i,m

≤ 1 (6)

dem

∈ {0, 1}

|P |

, k

spl

∈ {0,1}

|P |×L

w ∈ {0, 1}

∗

|×M

, m

spl

∈ {0,.., M

}

|P |

In the cost function, the ﬁrst term stands for the

cost of cables, the second term for the cost of welds,

and the last term for the cost of boxes. Equations

(1) ensure that at most one cable is spliced in a node.

Constraints (2) state that at most one cable serves the

demand in a cable-served way. Equations (3) make

sure that spliced cables are large enough to contain

their number of active modules. Constraints (4) are

active module conservation equations. The left-hand

side term stands for the number of modules of the

spliced cable. The ﬁrst right-hand side term is the

number of modules necessary to serve the demand, in

Fiber Cable Network Design with Operations Administration & Maintenance Constraints

case it is not cable-served. The last term is the number

of active modules of born cables. Finally, constraints

(5) and (6) ensure that w counts the number of welds

to be done in each node.

Remark 3.1. It is possible to ﬁx the value of some

variables. First, notice that leaf nodes are demand

nodes. These nodes will be served in a cable-served

way, and no operation will be done inside them. This

gives, for all nodes i ∈ V

such that |Γ

(i)| = 0:

∀m ∈ M

, w

i,m

= 0

∀p ∈ P |t(p) = i, ∀l ∈ L , k

spl

p,l

= 0

Furthermore, the number of welds done in a node

cannot exceed the number of active modules going out

of this node. This gives:

∀i ∈ V

∗

, ∀m ∈ M

, m >

∑

j∈Γ

(i)

act

i, j

=⇒ w

i,m

= 0

3.1.2 Valid Inequalities

We propose here several valid inequalities to tighten

the linear programming continuous relaxation the for-

mulation.

Let us deﬁne, for all m ∈ N, the minimum cost

per length unit of a set of cables able to contain m

active modules denoted by LB(m). For a given m,

LB(m) = {min

∑

l∈L

·n

∑

l∈L

·n

≥ m, n ∈ N

The following inequalities are valid for the FCNDA

problem:

∀(i, j) ∈ A,

∑

p∈P |(i, j)∈p



∑

l∈L

· k

spl

p,l

) +

min

t(p)

· k

dem



≥ LB(m

act

i, j

) (7)

The left hand side is the cost per length unit of the

cables going through arc (i, j).

Let us consider a path p ∈ P such that t(p) ∈ V

and s(p) 6= r. If there is a cable deployed on p, born

in s(p) and serving the demand in t(p), then we know

that there is a splicing operation done in s(p). Fur-

thermore, there are at least D

t(p)

welds in this opera-

tion, since the cable serving t(p) contains D

t(p)

active

modules. This gives the following valid inequalities

for the FCNDA problem:

∀p ∈ P |t(p) ∈ V

and s(p) 6= r,

dem

≤

∑

m≥D

t(p)

s(p),m

(8)

3.2 EFCNDA

EFCNDA can be solved by using the same variables

as in Section 3.1. The cost function is the same, the

set of feasible solutions is described by constraints (1)

to (6) to which we add the maintenance constraints

described below:

∀(p, p

) ∈ P

such that s(p) 6= s(p

)

and ∃a ∈ A, a ∈ p, a ∈ p

dem

+ k

dem

≤ 1 (9)

∑

l∈L

spl

p,l

∑

l∈L

spl

≤ 1 (10)

∑

l∈L

spl

p,l

+ k

dem

≤ 1 (11)

These constraints ensure that on two paths which

have different origins but an edge in common, there

can be only one cable.

The next section introduces an alternative mixed

integer programming approach for EFCNDA, based

on arcs rather than paths. It uses the properties of the

problem, and has less variables and less constraints.

3.2.1 Notation and Formulation

We keep the same notation for the problem instance.

In addition, let us deﬁne for (i, j) ∈ A,U

i, j

an upper

bound of the cost per length unit of the cables going

through (i, j).

We deﬁne the following variables:

• ∀(i, j) ∈ A, x

i, j

∈ {0, 1} the binary variable equal

to 1 iff the cables on edge (i, j) are born in i.

• ∀(i, j) ∈ A, c

i, j

∈ R the continuous variable equal

to the cost per length unit of the cables on edge

(i, j) (when all constraints are satisﬁed, including

integrality constraints).

• ∀(i, j) ∈ A, z

i, j

∈ R the continuous variable equal

to x

i, j

· c

i, j

• ∀i ∈ V

, u

∈ {0, 1} the binary variable equal to 1

iff the node i is module-served.

• ∀i ∈ V

∗

, ∀m ∈ M

, w

i,m

the binary variable equal

to 1 iff m welds are done in node i (since its mean-

ing is identical to Section 3.1, we keep the same

name).

• ∀i ∈ V

∗

, ∀l ∈ L, y

i,l

the binary variable equal to 1

iff a cable of size l is spliced in i.

The problem can be formulated as follows:

min

∑

i∈V

∗

∑

m∈M

· w

i,m

∑

(i, j)∈A

i, j

· c

i, j

∑

i∈V

∗

∑

l∈L

· y

i,l

such that ∀i ∈ V

ICORES 2018 - 7th International Conference on Operations Research and Enterprise Systems

γ(i),i

∑

l∈L

i,l

∑

j∈Γ

(i)

i, j

−

∑

j∈Γ

(i)

i, j

+ (1 − u

) ·C

min

(12)

∀i ∈ V

, c

γ(i),i

∑

l∈L

i,l

∑

j∈Γ

(i)

i, j

−

∑

j∈Γ

(i)

i, j

(13)

∀i ∈ V

∑

l∈L

· y

i,l

≥

· u

∑

j∈Γ

(i)

act

i, j

· x

i, j

(14)

∀i ∈ V

∗

∑

l∈L

· y

i,l

≥

∑

j∈Γ

(i)

act

i, j

· x

i, j

(15)

∀i ∈ V

∗

∑

l∈L

i,l

≤ 1 (16)

∀i ∈ V

∗

∑

m∈M

m · w

i,m

∑

j∈Γ

(i)

act

i, j

· x

i, j

(17)

∀i ∈ V

∗

∑

m∈M

i,m

≤ 1 (18)

∀(i, j) ∈ A, z

i, j

≥ c

i, j

−U

i, j

· (1 − x

i, j

) (19)

∀(i, j) ∈ A, z

i, j

≤ U

i, j

· x

i, j

(20)

∀(i, j) ∈ A, z

i, j

≤ c

i, j

(21)

u ∈ {0, 1}

, w ∈ {0, 1}

∗

|×M

x ∈ {0, 1}

|A|

, y ∈ {0, 1}

∗

|×L

c ∈ R

|A|

, z ∈ R

|A|

The ﬁrst term of the cost function denotes the cost

of welds, the second term stands for the cost of ca-

bles, and the last term stands for the cost of boxes.

Equations (12) ensure that the cost per length unit of

any arc is properly counted. The term

∑

l∈L

i,l

stands for the cost of the cable spliced in i, if any.

If for some arc (i, j) ∈ A such that j ∈ Γ

(i) we have

i, j

= 0, then the cables on arc (i, j) come from arc

(γ(i), i) unchanged. Otherwise, they come from the

splicing operation done in i. The last term stands for

the cost of the cable serving the demand in i. Equa-

tions (13) are the equivalent concerning nodes without

demand. Equations (14), (15) and (16) ensure that the

cable spliced in i is large enough to contain its active

modules. The ﬁrst term of the right-hand side of (14)

stands for modules serving the demand, the second

term for modules of born cables. Constraints (17) and

(18) ensure that the variable w

i,m

is equal to 1 iff there

are m welds done in node i. Finally, constraints (19),

(20) and (21) ensure that ∀(i, j) ∈ A, z

i, j

= x

i, j

· c

i, j

(these are Mc Cormick linearisation equations).

Remark 3.2. It is possible to ﬁx the value of some

variables. Assuming there exists i ∈ V

∗

and m

∈

such that w

i,m

= 1, then by (17), we know that

there exists S ⊆ Γ

(i) such that m

∑

j∈S

act

i, j

This gives by contraposition ∀i ∈ V

∗

, ∀m ∈ M

m 6∈ {

∑

j∈S

act

i, j

|S ⊆ Γ

(i)} then w

i,m

= 0. It can be

computed either in O(2

|Γ

(i)|

) or in O(|Γ

(i)| × M

)

(which is not a polynomial with respect to the instance

size, provided M

is not coded in an unary system).

3.2.2 Valid Inequalities

The continuous relaxation of the formulation above

does not seem to be very tight in practice (see Table

4 from Section 6). We propose here several valid in-

equalities to tighten the continuous relaxation of the

formulation.

In nodes without demand, if a cable of size l is

spliced, then it has a number of active modules be-

tween M

and M

l−1

+ 1; otherwise one could install a

smaller cable and obtain a cheaper solution. With the

convention M

= 0 and M

0, every optimal solu-

tion of the EFCNDA problem veriﬁes the following:

∀i ∈ V

, ∀l ∈ L , y

i,l

∑

m∈M

l−1

i,m

(22)

With a reasoning similar to the one from 3.1.2 (see

deﬁnition of LB), we can get a lower bound of the cost

per length unit of the cables on each arc. The follow-

ing inequalities are valid for the EFCNDA problem:

∀(i, j) ∈ A, c

i, j

≥ LB(m

act

i, j

) (23)

If the cables on some arc (i, j) ∈ A are born in

i, then at least m

act

i, j

welds are done in node i. This

implies that the following inequalities are valid for the

EFCNDA problem.

∀(i, j) ∈ A, x

i, j

≤

∑

m∈M

|m≥m

act

i, j

i,m

(24)

4 DYNAMIC PROGRAMMING

FOR EFCNDA

For any node i ∈ V

∗

, we introduce the additional no-

tation V

(i), which refers to the set of nodes on the

path from the root to i, excluding i and including r.

The EFCNDA problem can be solved by Algo-

rithm 1. To each node i ∈ V

∗

, and for each node

j ∈ V

(i), we associate to i a label < j,C(i, j) >∈

(i) × R where C(i, j) is the minimum cost of the

Fiber Cable Network Design with Operations Administration & Maintenance Constraints

Algorithm 1: Exact Resolution Algorithm for EFCNDA.

1: procedure INITIALISATION()

2: for i ∈ V

|Γ

(i) =

0 do

3: for j ∈ V

(i) do

4: Add to i the label < j,C

min

· d

> where p ∈ P is the only path s.t. s(p) = j and t(p) = i.

5: end for

6: Declare i labeled.

7: end for

8: end procedure

9: procedure RECURSION()

10: while ∃r

∈ Γ

(r) such that r

has not been labeled do

11: for every node i ∈ V

∗

such that all nodes in Γ

(i) have been labeled do

12: for j ∈ V

(i) do

13:  We select the operation in i minimizing the network cost.

14: Add the label < j,C(i, j) > to node i where

C(i, j) = min

S⊆Γ

(i),b∈{0,1}

∑

k∈S

C(k, i) +

∑

k∈Γ

(i)\S

C(k, j)

+PW

+ d

·C

+ d

·C

min

· (1 − b) (25)

with



m =

∑

k∈S

act

i,k

= min{l ∈ L |M

≥ b · D

∑

k∈S

act

i,k

}

p ∈ P is the only path such that s(p) = j,t(p) = i

15: end for

16: Declare i labeled.

17: end for

18: end while

19: end procedure

20: procedure TERMINATION()

21: return

∑

∈Γ

(r)

C(r

, r)

22: end procedure

network rooted in i plus the cost of the cables on the

path from j to i, assuming these are born in node j.

The algorithm is initialized at leaf nodes (line 4),

which are cable-served demand nodes, and where the

size of the cable serving the demand is known.

For a node i such that all nodes in Γ

(i) have been

labeled, and for j ∈ V

(i), (25) computes the mini-

mum cost operation if the next operation is done in j.

For i ∈ V

∗

and k ∈ Γ

(i), k ∈ S iff the cables going

through arc (i, k) are born in node i. Similarly, the

boolean b is equal to 1 iff the node i is module-served

(its meaning is similar than the variable u

in Section

3.2).

We propose to compute it with a brute-search al-

gorithm on the set S and on b. For given nodes i ∈

∗

, j ∈ V

(i), it can be done in O(P

∗

(ln(M

), L) ×

|Γ

(i)|+1

) where P

∗

is a time sufﬁcient to compute C

∗

(sums and minimums can be computed in polynomial

time).

Lemma 4.1. Algorithm 1 runs in time

O(P

∗

(ln(M

), L) × 2

maxΓ

× |V |

) where maxΓ

denotes the maximal degree (number of successors)

of a node in the graph and P

∗

is a polynomial.

Indeed, in each loop a number of iterations smaller

than V

∗

is done.

Proposition 4.2. Let us consider i ∈ V

∗

. When i is

declared labeled in algorithm 1, there exists a node

j ∈ V

(i) such that in the label < j,C(i, j) >, C(i, j)

describes the cost of the minimum cost network in the

arborescence rooted in node i plus the cost of the ca-

bles on the path from j to i.

This proposition shows the validity of the algo-

rithm. We will start to prove it for leaf nodes, then

recursively on higher nodes.

Proof.  Let us consider a leaf node i. In the mini-

mum cost network, it is served in a cable-served way

with a cable of type l

= min{l ∈ L |M

≥ D

}. This

cable is born in some node j ∈ V

(i), eventually the

root. The minimum cost network has a cost 0 in

the arborescence rooted in j. Let us call p ∈ P the

only path such that s(p) = j and t(p) = i. The label

< j,C(i, j) > of i has a cost of C

min

· d

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100

 Let us consider a non-leaf node i ∈ V

∗

such that

all nodes in Γ

(i) have been labeled. In the minimal

cost network, the cables going through arc (γ(i), i) are

all born in a node j ∈ V

(i). Thanks to the main-

tenance constraint, we know that they are all born

in the same node. Since all nodes k ∈ Γ

(i) have

been labeled, for each of these nodes, there is a node

∈ V

(k) such that in the label < j

,C(k, j

) >,

C(k, j

) describes the cost of the minimum cost net-

work in the arborescence rooted in k plus the cost of

the cables on the path from j

to k. Furthermore, since

the cables going through arc (γ(i), i) are all born in j,

we have either j

= j or j

= i. Let us consider the

label < j,C(i, j) > of node i. If in the minimal net-

work i is module-served, then we will have b = 0 in

the computation of (25). Furthermore, let us consider

k ∈ Γ

(i). If j

= i, we will have k ∈ S in the compu-

tation of (25), and k ∈ Γ

(i) \ S otherwise. Hence the

result.

The termination of the algorithm derives from

Proposition 4.2. For each node r

∈ Γ

(r), we have

) = {r}. This implies, using this proposition,

that in the label < r,C(r

, r) >, C(r

, r) is the cost of

the minimum network cost in the arborescence rooted

in r

plus the cost of the cables on (r, r

). Summing

these values gives the minimum network cost.

The next section assesses the complexity of FC-

NDA and EFCNDA.

5 COMPLEXITY

We show in Section 5.1 that FCNDA is NP-hard even

with 1 cable size and an upper bound on the node de-

gree of 2, and in Section 5.2 that EFCNDA is NP-

hard.

5.1 FCNDA

Let us consider the Number Partitioning Problem

(NPP), which is shown to be NP-complete in (Karp,

1972).

Instance: A set of N strictly positive integers {n

∈

N|i ∈ {1, .., N}}.

Question: Is there a subset S ⊆ {1, ..,N} such that

∑

i∈S

∑

i6∈S

We consider an instance of the NPP and reduce

it to the following FCNDA instance. Let G = (V, A)

be an arborescence describing the civil engineer-

ing structure, (V = {r, 0, 1} ∪ {v

|i ∈ {1, .., N}}, A =

{(r, 0); (0, 1);(1, v

);(v

i−1

, v

)|i ∈ {2, .., N}}) (G is a

chain graph), r is the root. The demand nodes are

, i ∈ {1, .., N}} and have respective demands n

, i ∈

{1, .., N} modules. Only one type of cable is avail-

able, with size M

∑

i∈{1,..,N}

. Its cost per length

unit is C

= 1. The lengths of all arcs of the arbores-

cence are null, except (r, 0) which is of length 1. This

means the cost of a cable born in r is 1, and the cost

of the other ones is 0. The cost of welds and boxes is

null.

The question associated to this FCNDA instance

is ”Is there a cabling solution cheaper than 2 ?”.

 If NPP is feasible: ∃S ⊆ {1, .., N} such that

∑

i∈S

∑

i6∈S

. We then build the following cabling

solution:

• Two cables holding only active modules are in-

stalled on link (r, 0).

• In node 0, one incoming cable is spliced into N −

|S| born cables. The born cables have a number of

active modules n

, i 6∈ S and serve respectively the

demand nodes (v

)

i6∈S

• In node 1, the cable coming from the root with

only active modules is spliced into |S| born cables.

The born cables have n

active modules and serve

the demand nodes (v

)

i∈S

Since the number of active modules is conserved in

each splicing, the cabling solution described above is

feasible (it is illustrated in Fig. 6, as well as the in-

stance). Its cost is equal to 2.

Figure 6: Instance and solution used in the complexity

proof.

 If NPP is not feasible. Then, the solution de-

scribed above is not possible anymore. One cable is

not large enough to cover link (r, 0). Two cables can-

not cover (r, 0) either, since they would both have only

active modules, which would mean that the NPP prob-

lem was feasible. Consequently, at least 3 cables need

to be installed on arc (r, 0), and such a solution has a

cost of a least 3.

Remark 5.1. The solution illustrated in Fig. 6 is not

valid for EFCNDA, the maintenance rule is not re-

spected in nodes 0 and 1.

5.2 EFCNDA

EFCNDA can be shown to be NP-complete by re-

duction from the NPP. With the same notation, let

us consider an instance of the NPP and reduce it

Fiber Cable Network Design with Operations Administration & Maintenance Constraints

101

to the following EFCNDA instance. The civil engi-

neering structure is described by the set of nodes is

V = {r, 0} ∪ {v

|i ∈ {1, .., N}}; the set of arcs A =

{(0, v

)|i ∈ {1, .., N}} ∪ {(r, 0)}; r is the root, the

nodes {v

|i ∈ {1, .., N}} have a demand of n

mod-

ules. The length of all arcs except (r, 0) is null. We

have N + 1 available cables:

• N cables of sizes n

modules and cost per length

unit n

• A cable of size

∑

i=1

and cost per length unit

∑

i=1

− 1

The cost of welds and boxes is null.

The question we ask is ”is there a solution of cost

at most

∑

i=1

− 1”?

 If NPP is feasible. Then, we have S ⊆ {1, ..,N}

such that

∑

i∈S

∑

i6∈S

. We consider the solution

of EFCNDA where

• For i ∈ {1, .., N}, on each arc (0, v

), we lay down

a cable of size n

• In the node 0, a cable of size

∑

i∈{1,..,N}

spliced. Cables of size n

, i ∈ S are born, and serve

the demand of nodes v

, i ∈ S.

• On the arc (r, 0), a cable of size

∑

i∈{1,..,N}

holding only active modules is deployed (the one

spliced in 0); as well as N − |S| cables of sizes

, i 6∈ S which serve the demand in nodes v

, i 6∈ S.

The cost of this solution is the cost of cables on

arc (r, 0) which is

∑

i∈{1,..,N}

− 1. It is illustrated in

Fig. 7.

Figure 7: Instance and solution used in the complexity proof

for EFCNDA.

 If NPP is not feasible. In a minimal cost solu-

tion, the size of cables serving the demand is known.

For a given i ∈ {1, .., N}, v

is served by a cable of size

. Which leaves three types of solutions to consider.

The solution without splicing has a cost

∑

i∈{1,..,N}

. Each demand node is served by a

cable coming directly from the root r.

Any solution where a cable of size

∑

i∈{1,..,N}

is spliced in 0 has a cost at least equal to

∑

i∈{1,..,N}

Indeed, let us note E ⊆ {1, .., N} the set such that

cables of sizes n

, i ∈ E are born in 0. Since the

NPP instance is not feasible, we have

∑

i∈E

∑

i∈{1,..,N}

, so the cost of cables which are con-

tinued in 0 is

∑

i6∈E

∑

i∈{1,..,N}

, and the total

cost of the network is

∑

i6∈E

∑

i∈{1,..,N}

− 1 ≥

∑

i∈{1,..,N}

Any solution where a smaller cable is spliced in 0

has a cost at least equal to

∑

i∈{1,..,N}

. Indeed, in any

splicing of a cable of size n

for a given i ∈ {1, .., N},

the spliced cable is at least as expensive as the born

cables.

5.3 Synthesis

To the results proven in Sections 5.1 and 5.2, we

can add those deducible from Section 4. The restric-

tion of EFCNDA where there is an upper bound on

the node degree can be solved in polynomial time,

since in that case the computation of (25) can be done

in polynomial time (straightforward consequence of

Lemma 4.1). This implies that it is also polynomial

when more parameters are ﬁxed. As for FCNDA,

its NP-hardness in a restricted setting implies its NP-

hardness in the more general cases. These results are

summed up in Table 1 (recall that L is the number of

cable sizes available, and maxΓ stands for the maxi-

mum degree of a node in the graph).

Table 1: Complexity of the two problems in different con-

texts.

Fixed

elements

none

maxΓ

and L

EFCNDA NP-hard Polynomial Polynomial

FCNDA NP-hard NP-hard NP-hard

Table 1 shows a theoretical difference in the com-

plexities of the two problems EFCNDA and FCNDA.

Fixing the maximum degree of the node in the in-

stances makes EFCNDA polynomial. It seems like a

very important factor of the complexity of this prob-

lem. Meanwhile, FCNDA stays NP-complete even

with a ﬁxed maximum degree and number of cable

sizes.

We assess the numerical aspect of this complexity

difference in the next section.

6 RESULTS

We assessed the solution methods on real-life in-

stances taken from the city of Arles (France). Some

of their features are displayed in Table 2.

The cables available have a size of 1, 2, 4, 6, 8, 12,

18 or 24 modules. The resolution algorithm for the

MIPs was the Cplex 12.6 default branch-and-bound

algorithm. The dynamic algorithm was implemented

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102

in Java. Both were run on a machine composed of 16

processors Intel Xeon of CPU 5110 and clocked at 1.6

GHz each.

Table 2: Key features of the real-life instances.

instance features

max

degree

arcs

demand

nodes

total

demand

Ar 1 4 113 45 61

Ar 2 6 103 38 55

Ar 3 5 103 35 66

Ar 4 6 123 43 80

Ar 5 7 129 44 68

Ar 6 6 137 43 67

Ar 7 4 139 35 68

Ar 8 5 163 41 63

Ar 9 4 219 68 78

6.1 Models Comparison

The results of the numerical experiments regarding

the FCNDA and EFCNDA problems are displayed

respectively in Tables 3 and 4, ”base model” always

refers to the MIP without valid inequalities, and ”en-

hanced model” to the MIP with valid inequalities.

The columns of both tables are labeled as follows:

”time” stands for the computation time; ”CR” stands

for the ratio between the continuous relaxation value

and the optimal solution; ”DP” stands for dynamic

programming (Table 4 only).

Table 3: Results for FCNDA.

instance base model

enhanced

model

time

(s)

(%)

time

(s)

(%)

Ar 1 8 90.3 16 91.0

Ar 2 9 83.7 24 92.4

Ar 3 17 92.2 22 93.3

Ar 4 19 89.2 46 90.0

Ar 5 1 94.9 2 95.2

Ar 6 2 92.5 3 94.7

Ar 7 13 92.4 29 93.7

Ar 8 8 89.6 12 91.7

Ar 9 4837 89.4 408 91.6

Regarding FCNDA (Table 3), the valid inequali-

ties have had a positive effect on the average compu-

tation time, which went down from 546 to 62 seconds.

However, on most instances (8 out of 9), the MIP is

solved faster without the valid inequalities. This sug-

gest that they are more useful for instances that are

hard to solve. Regarding the algorithm, ratio CR goes

from an average of 90.5 % to 92.6 %. The relatively

high CR of the base model can explain the mitigated

impact of the inequalities on the performances.

Regarding EFCNDA (Table 4), all instances were

easier to solve (computation times are displayed in

milliseconds). The valid inequalities have had a bene-

ﬁcial effect on the computation time, all instances are

solved faster with the enhanced formulation. The av-

erage computation time goes from 1730 to 329 ms.

On an algorithmic level, the ratio CR goes from an

average of 13.2 % to 87.3 % of the optimal solution

cost. The dynamic programming approach was more

efﬁcient than the enhanced integer programming for-

mulation, it solved 7 out of 9 instances faster.

Table 4: Results for EFCNDA.

instance base model

enhanced

model

time

(ms)

(%)

time

(ms)

(%)

time

(ms)

Ar 1 1457 14.0 305 89.2 324

Ar 2 1174 17.8 239 86.6 239

Ar 3 1317 13.6 318 81.7 66

Ar 4 742 15.7 268 86.8 87

Ar 5 746 18.2 477 89.2 88

Ar 6 1477 15.5 238 91.8 110

Ar 7 1667 9.7 190 80.1 121

Ar 8 1786 9.4 344 89.8 103

Ar 9 5204 5.3 507 90.8 306

6.2 Sensitivity Analysis

Section 5 highlights the maximal node degree as a key

element of the problems complexity. Since the high-

est node degree of all real-life instances is between 4

and 7, we used ﬁctive (simulated) instances to assess

the performances of each resolution technique when

some of the nodes have a high degree. Their features

are displayed in Table 5.

Table 5: Key features of the ﬁctive instances.

instance features

max

degree

arcs

demand

nodes

total

demand

Fi 10 11 20 15 71

Fi 11 12 22 16 84

Fi 12 13 24 18 97

Fi 13 14 26 19 112

Fi 14 15 28 21 112

Fi 15 16 30 22 127

Fi 16 17 32 24 144

As expected, the dynamic programming algorithm

was very sensitive to the node degree, the computa-

tion time growing exponentially (see Table 6). The

enhanced MIP formulation for EFCNDA was able to

solve all instances in less than one second, with an av-

erage of 200 ms. This is the opposite of the results ob-

tained on real-life instances, where the dynamic pro-

Fiber Cable Network Design with Operations Administration & Maintenance Constraints

103

Table 6: Computation time on ﬁctive instances (ms).

instance

enhanced

model

FCNDA

enhanced

model

EFCNDA

dynamic

programming

Fi 10 205 166 322

Fi 11 327 77 652

Fi 12 993 332 1409

Fi 13 1130 120 3800

Fi 14 1369 347 12 403

Fi 15 1450 98 39 654

Fi 16 2691 280 164 243

gramming was more efﬁcient. As for FCNDA, the

MIP formulation proved to be efﬁcient, with an av-

erage computation time of 900 ms. Although the in-

stances with a higher degree are harder to solve, these

instances stay tractable in practice. One should fa-

vor a MIP based approach, regardless of the problem,

when dealing with high degree nodes. As the ﬁctive

instances have less arcs, the MIP approaches seem

more sensible to the overall number of arcs than to

the maximum degree of the instances.

6.3 Operational Considerations

We compared the optimal solutions of both problems.

Results are displayed in Table 7, the column labeled

”arcs with rule broken” denotes the number of arcs

where the maintenance rule (illustrated in ﬁgure 5) is

violated when FCNDA is solved.

Table 7: Optimal solution costs and characteristics.

instance

Solution

EFCNDA

Solution

FCNDA

arcs with

rule broken

Ar 1 6156.6 6087.3 6

Ar 2 10 357.3 9870.0 8

Ar 3 6546.2 6125.8 14

Ar 4 6720.8 6461.9 14

Ar 5 5081.8 5081.8 0

Ar 6 6546.5 6544.2 1

Ar 7 9348.0 8638.6 18

Ar 8 12 328.3 12 248.4 4

Ar 9 25 619.1 24 422.8 15

An optimal EFCNDA solution is on average 3.7 %

more expensive than a FCNDA optimal solution (see

Table 7). This can be seen as an acceptable capital

expenditure over-cost if it is compensated by future

easier maintenance activities.

The maintenance rule is violated in almost every

real-life instance we tried (8 out of 9). On average,

it is not respected in 6.2 % of the arcs, which is sig-

niﬁcant. This suggests that optimal FCNDA solutions

will be much harder to repair in case of failure on one

duct.

7 CONCLUSION

We introduced two combinatorial problems related to

FTTH network design, one unconstrained by main-

tenance consideration and the other one constrained.

Regarding the unconstrained problem, one integer

programming based solving algorithm was proposed.

Adding valid inequalities leads to a more tractable

problem. We proposed two solution methods for the

constrained problem. These methods are complemen-

tary, as they prove efﬁcient in different contexts: the

dynamic programming approach is generally faster

in graphs where nodes have a small degree, whereas

the mixed integer programming, embedding efﬁcient

valid inequalities, is generally faster otherwise.

On a complexity level, the unconstrained problem

seems harder to solve than the constrained problem.

Our numerical experiments conﬁrmed this tendency

on real-life instances. From the operational point of

view, the maintenance rule can be considered as a

reasonable compromise between capital expenditure

over-costs for the network deployment and mainte-

nance savings.

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