Return to Article Details A numerical comparison between two exact simplicial methods for solving a capacitated 4-index transportation problem

A numerical comparison between two exact simplicial methods for solving a capacitated 4-index transportation problem

Rachid ZITOUNI Mohamed ACHACHE§

March 26, 2017.

Laboratoire de Mathématiques Fondamentales et Numériques, Université Ferhat Abbas- Sétif 1, Sétif, Algérie, e-mail: rzitou@yahoo.fr.

§Laboratoire de Mathématiques Fondamentales et Numériques, Université Ferhat Abbas- Sétif 1, Sétif, Algérie, e-mail: achache_m@yahoo.fr.

In this paper, we deal with a numerical comparison between two exact simplicial methods for solving a capacitated four-index transportation problem. The first method was developed by R. Zitouni and A. Keraghel for solving this problem [Resolution of a capacitated transportation problem with four subscripts, Kybernetes, Emerald journals, 32, 9/10: 1450-1463 (2003)]. The second approach is the well-known simplex method. We show across some obtained numerical results that the first algorithm competes well with the simplex method.

MSC. 90C05, 90C08

Keywords. Capacitated transportation problem, four-index transportation problem, linear programming, simplicial algorithms.

1 Introduction

The transportation problem is an important subject in real-world life that covers many important problems in economy, telecommunication and localization, among others. As a special case, the transportation problem with two-dimensional index has been extensively studied and solved by L. V. Kantorovich and M. K. Gavurin (1949, [ 4 ] ) and G. B. Dantzig (1951, [ 2 ] ). Next, the study and the solution have been extended to transportation problems where the dimension of the index is higher than two. Since the sixties, several papers have been published for uncapacitated problems with a three-dimensional index and a general multi-dimensional index, see for instance [ 3 ] , [ 5 ] , [ 6 ] and [ 7 ] .

Recently, Zitouni [ 10 ] , first introduced an algorithm for solving a capacitated transportation problem with 4-index (four subscripts). The choice for the index transportation problem allows us to getting an idea for the more general multi-dimensional transportation problem case.

Our aim in this paper is to examine two exact simplicial methods for the solution of the capacited 4-index transportation problem, the method developed in [ 10 ] and the simplex method. Because the algorithm developed in [ 10 ] , tackles directly the problem, and shows its elegant computation and its rapidity convergence to provide a solution of this problem. For the comparison purpose, the obtained numerical results are compared with those obtained by the classical simplex approach applied to the reformulation of this problem as a linear program.

The outline of the paper is as follows. In Section 2, the 4-index capacited transportation problem formulation is stated. In section 3, the transportation problem solution is given. In section 4, the detailed description of two exact simplicial algorithms and the convergence of the second algorithm are presented. In section 5, some computational results are reported, followed by an important numerical comparison between their performances. Finally, a conclusion and future researchers are drawn in the last section.

Some notation used throughout the paper is as follows. Rr denotes the space of r-dimensional real vectors whereas the set of all matrices with type (r1,r2) is denoted by Rr1×r2. If x,z Rr, then xTz denote their usual inner product. If SRr1×r2, then its rank is defined as rank(S)=rmin(r1,r2).

2 The capacitated 4-index transportation problem formulation

The capacitated transportation problem with a four-dimensional index, denoted by (T4C), is formulated as the following constrained optimization problem:

MinimizeZ=i=1mj=1nk=1pl=1qcijklxijkl
1

          subject to the constraints:

j=1nk=1pl=1qxijkl=αifor all\qquad i=1,...,mi=1mk=1pl=1qxijkl=βjfor all\qquad j=1,...,ni=1mj=1nl=1qxijkl=γkfor all\qquad k=1,...,pi=1mj=1nk=1pxijkl=δl for all\qquad l=1,...,q

0xijkl dijklfor all\qquad (i,j,k,l),
6

where αi, βj, γk, δl, dijkl and cijkl are given and such that for all i, j, k, and l, αi>0, βj>0, γk>0, δl>0, dijkl>0 and cijkl0.

The T4C problem can be also reformulated as the following linear program:

minxZ=cTxs.t. Ax=b,0xd,
7

with

  • x=(x1111,...,xmnpq)TRN,

  • c=(c1111,...,cmnpq)TRN,

  • d=(d1111,...,dmnpq)TRN,

  • b=(α1,...,αm,β1,...,βn,γ1,...,γp,δ1,...,δq)TRM,

  • ARM×N, whereM=m+n+p+q and N=mnpq.

In this representation x=(x1111,x1211,...,xmnpq) has been associated to a vector xRN. To do that, we associate to each (i,j,k,l){1,..,m}×{1,..,n}×{1,..,p}×{1,..,q} a vector PijklRM. Only four entries of the vector Pijkl are nonzero, they are located on lines i, m+j, m+n+k and m+n+p+l, and their common value is 1. Note that Pijkl are the columns of A, they are called coefficients vectors.

In the sequel, we quote the following useful definitions (see [ 13 ] ).

Definition 1

A feasible solution x of T4C is called basic solution if the columns of the sub-matrix Ax obtained from A by keeping only the columns corresponding to the variables xijkl such that

0<xijkl<dijkl

are linearly independent.

Definition 2

A basic feasible solution is said to be non degenerate if

rank(Ax)=rank(A).
Definition 3

Given a basic feasible solution x=(xijkl), the 4-tuple (i,j,k,l) is called interesting if

0<xijkl<dijkl.

Throughout the paper, we assume that the following feasibility assumption for T4C,

i=1mαi=j=1nβj=k=1pγk=l=1qδl=H
8

holds.

As a consequence, it results that

rank(A)=M3.

Unlike the transportation problem with two-dimensional index, the matrix A is not totally unimodular since some of its minors do not belong to {1,0,1}.

It is useful to present the data of the problem thanks to the following transportation table. It consists of an array of M rows and N columns, three additional rows and an additional column. The entries of these N columns of the first, second, and third additional rows are reserved for the data of the quantities dijkl, cijkl, and xijkl, respectively. The additional column is for the data of quantities αi, βj, γk, and δl, respectively. Finally, the entry of the array on the line corresponding to αi and the column Pijkl is 1 if i=i and 0 otherwise. Same things for βj, γk, and δl. We illustrate that by the following example.

Unknown environment 'tabular'

d1111

.

Misplaced &

d1211

.

Misplaced &

dmnpq

.

Misplaced &

c1111

.

c1211

.

Misplaced &

cmnpq

.

Misplaced &

x1111

.

x1211

.

Misplaced &

xmnpq

.

Misplaced &

α1

.


1|c|: : ... : 1|c|:
1|c|0 0 ... 1 1|c|

αm

.


1|c|1 0 ... 0 1|c|

β1

.


1|c|0 1 ... 0 1|c|

β2

.


1|c|: : ... : 1|c|:
1|c|0 0 ... 1 1|c|

βn

.


1|c|1 1 ... 0 1|c|

γ1

.


1|c|: : ... : 1|c|:
1|c|0 0 ... 1 1|c|

γp

.


1|c|1 1 ... 0 1|c|

δ1

.


1|c|: : ... : 1|c|:
1|c|0 0 ... 1 1|c|

δq

.


None
T4C - Transportation table.

3 Transportation problem solution

3.1 Feasibility conditions

We begin first to give a useful theorem that ensures the feasibility of T4C problem (see [ 13 ] ).

Theorem 4

(Feasibility)
1. A necessary condition for the feasibility of the problem T4C i.e., it has a feasible solution if the condition 8 holds and the following conditions

{αij=1nk=1pl=1qdijklfori=1,...,m,βji=1mk=1pl=1qdijklforj=1,...,n,γki=1mj=1nl=1qdijklfork=1,...,p,δli=1mj=1nk=1pdijklforl=1,...,q,
9

are satisfied.

2. A sufficient condition for the feasibility of the problem T4C, i.e., it has a feasible solution if the condition 8 holds and the following conditions

αiβjγkδlH3dijkl, for all(i,j,k,l)
10

are satisfied.

Proof â–¼
1. It is clear that if x=(xijkl) is a feasible solution for the problem T4C, then conditions 8 and 9 hold.
2. Assume that x=(xijkl) is a vector of RN such that
xijkl=αiβjγkδldijkl, for all \, \, (i,j,k,l).

We can easily verify that x is a feasible solution for the problem T4C. Note that if the set of feasible solutions of the problem T4C is non empty, then it is a polytopes. Since the objective function is continuous, then the set of the solution of the problem T4C is non empty, i.e., there exists at least an optimal solution of it.

Proof â–¼

3.2 Optimality conditions

In this subsection, we give a theorem that ensures when a feasible solution, is an optimal solution of T4C.

Theorem 5

Assume that the problem T4C is feasible. Then a feasible solution x of T4C is optimal if and only if there exists a vector

y=(u1,,um,v1,,vn,w1,,wp,t1,,tq)TRM

such that:

{ui+vj+wk+tlcijklifxijkl=0,ui+vj+wk+tl=cijklif0<xijkl<dijkl,ui+vj+wk+tlcijklifxijkl=dijkl.

Proof â–¼

For that, we consider the following formulation of the problem T4C:

minx[cTx:Ax=b,xd,x0],
11

and its dual problem is given by

max(y,z)[bTydTz:ATyzc,z0].
12

Let (y,z) an optimal solution of the problem 12. Then a feasible solution x of the problem 11 is optimal if and only if the two following complementarity conditions are satisfied

(ATyzc)ijklxijkl=0,
13

and

(dx)ijklzijkl=0.
14
  • If  xijkl=0,   then   xijkl<dijkl  and 14 imply   zijkl=0. So   (ATy)ijklcijkl.

  • If   0<xijkl<dijkl   then  14 implies   zijkl=0   and 13 leads to   (ATyzc)ijkl=0. So (ATy)ijkl=cijkl.

  • If xijkl=dijkl then xijkl>0 and 13 imply (ATyzc)ijkl=0, consequently (ATyc)ijkl=zijkl. Finally, we get (ATy)ijklcijkl.

Proof â–¼

4 Description of two simplicial algorithms for T4C

4.1 The simplex algorithm

In this subsection, in order to apply the technical of the simplex algorithm for solving T4C problem, we need first to put T4C in the framework of a special linear program. The simplex algorithm, is referred to us as Algorithm 1. It is well-known that the problem T4C, according to 7, can be easily reformulated as the following linear program:

minx^Z=c^Tx^s.t.A^x^=b^,x^0,
15

in adding to the constraints those of the form: xijkl+yr=dijkl, (see 6 above), for i=1,,m, j=1,,n, k=1,,p, l=1,,q, r=1,...,N, where

  • yr is the gap variable,

  • x^=(xT,yT)TR2N with y=(y1,,yN)T,

  • c^=(cT,0T)TR2N where 0 is a N-null vector,

  • b^=(bT,dT)TRM+N, with d is the vector of N components dijkl.

  • Unknown environment 'tabular'AMisplaced &0Misplaced \hlineI_NMisplaced &I_N

R^M+N×2N  with rank(A^)=M+N3. INRN×N is the identity matrix.

4.2 Algorithm 2

According to the particularities of T4C problem, Zitouni [ 10 ] , proposed a modification of the simplex algorithm. This algorithm is referred here as Algorithm 2, which shares with the simplex algorithm, a structure consists also of two phases such as the finite convergence and the use of the pivot principle. The advantage of this new algorithm is that it tackles the T4C directly without passing by other reformulations. For more comprehension of this new algorithm, we detail its fundamental ingredients (steps) as follows.

Phase 1. (It finds a basic feasible solution or declare that T4C is not solvable)

Step 1:

Initialization: for all (i,j,k,l), α^i=αi, β^j=βj, γ^k=γk, δ^l=δl and bijkl=0, (bijkl is a boolean variable equal to 1 if xijkl has already been determined and 0 if not yet),

E={(i,j,k,l), such that bijkl=0}.

Iteration:

While E is non empty do

  • Choose an 4-tuple (i¯,j¯,k¯,l¯)E, such that ci¯j¯k¯l¯=min(i,j,k,l)Ecijkl, (see Remark 6)

  • Take xi¯j¯k¯l¯=min(α^i¯,β^j¯,γ^k¯,δ^l¯,di¯j¯k¯l¯), and bi¯j¯k¯l¯=1, (i.e., xi¯j¯k¯l¯ is determined),

  • Update α^i¯,β^j¯,γ^k¯, and δ^l¯ by the procedure (P1) below.

Step 2:

a) Determine ξ as shown in the procedure (P2) below.
If ξ=0, then x=(xijkl) is an initial basic feasible solution for the problem T4C, we denote it by x(0). Go to Phase 2.

Else, construct the problem T4C(M~) (as shown in the procedure (P3) below) and determine for it an initial basic feasible solution x(0) as in step 1 by taking x¯m+1,n+1,p+1,q+1(0)=0. Note that x(0)=(xijkl(0)), with i=1,,m+1, j=1,,n+1, k=1,,p+1 and l=1,,q+1. If x(0) is optimal then the problem T4C is not solvable. Stop.

b) Improvement of a basic feasible solution for T4C(M~).

Initialization: r=1, ξ>0 is known,

1) Determine x(r) as in Phase 2.

2) If  x¯m+1,n+1,p+1,q+1(r)=ξ, then x(r)=(xijkl(r)) with i=1,,m, j=1,,n, k=1,,p, and l=1,,q, is an initial basic feasible solution for the problem (T4C). Go to Phase 2.

3) If x(r) is optimal (Phase 2), then the problem T4C is not solvable. Stop. 4) Do r=r+1 and repeat 1), to 3).

Next, we describe the second phase.

Phase 2. (Research of an optimal solution for T4C)

When Phase 2 starts, we know an initial basic feasible solution x(0). Take r=0.

a) Determine the set I(r) of the interesting 4-tuple (i,j,k,l), (see Remark 7).

b) For all (i,j,k,l)I(r), solve the linear system

ui(r)+vj(r)+wk(r)+tl(r)=cijkl.
c) For all (i,j,k,l)I(r) determine

ijkl(r)=cijkl(ui(r)+vj(r)+wk(r)+tl(r)).

d) Apply the procedure described in (P4) below.

If the optimality condition holds then the feasible solution x(r) is optimal. stop.

Else determine a vector Pi0j0k0l0 entering at the base, it corresponds to i0j0k0l0(r).

e) Construct a cycle μ(r) via the procedure described in (P5) and determine a new feasible solution as the procedure (P6) shows.

f) Do r=r+1 and repeat a), to e) until the optimality condition holds.

Remark 6

If there are several elements corresponding to the minimum of cijkl, we choose one, for instance the first found in the transportation table by going from the left to the right. â–¡

Remark 7

If the feasible solution is degenerate, i.e., the number of columns of Ax is strictly less than rank(A), we complete Ax with additional columns so that we obtain a matrix having rank(A) linearly independent columns. Next I(r) can be determined. Thus will be done in the procedure (P7). â–¡

Also, the algorithm makes appeal to the following procedures.

(P1) - Updating of α^i¯,β^j¯,γ^k¯, and δ^l¯.

1) α^i¯=α^i¯xi¯j¯k¯l¯,

if α^i¯=0 then take xi¯jkl=0 for all (j,k,l)(j¯,k¯,l¯) and bi¯jkl=1 for all (j,k,l),

2) β^j¯=β^j¯xi¯j¯k¯l¯,

if β^j¯=0 then take xij¯kl=0 for all (i,k,l)(i¯,k¯,l¯) and bij¯kl=1 for all (i,k,l),

3) γ^k¯=γ^k¯xi¯j¯k¯l¯,

if γ^k¯=0 then take xijk¯l=0 for all (i,j,l)(i¯,j¯,l¯) and bijk¯l=1 for all (i,j,l),

4) δ^l¯=δ^l¯xi¯j¯k¯l¯,

if δ^l¯=0 then take xijkl¯=0 for all (i,j,k)(i¯,j¯,k¯) and bijkl¯=1 for all (i,j,k).

(P2) - Determination of ξ.

ξ=i=1mai=j=1nbj=k=1pek=l=1qfl, such that:

ai=αij=1nk=1pl=1qxijklwithi=1,...,m,bj=βji=1mk=1pl=1qxijklwithj=1,...,n,ek=γki=1mj=1nl=1qxijklwithk=1,...,p,fl=δli=1mj=1nk=1pxijkl withl=1,...,q.

Note that the numbers ai, bj, ek and fl are nonnegative.

(P3) - Construction of the problem T4C(M~).

The problem T4C(M~) is obtained from problem T4C by adding four fictitious points with indices m+1, n+1, p+1, and q+1 such that:  cm+1,n+1,p+1,q+1=0, and  cm+1,jkl=ci,n+1,kl=cij,p+1,l=cijk,q+1=M~, where M~ is a very large number and there are no limitation on the capacities for the paths involving a fictitious point.

(P4) - Determination of a vector Pi0j0k0l0 entering at the base or confirming that the feasible solution x(r) is optimal.

Take Γ0(r) and Γd(r) as two tables such that

Γ0(r)={ijkl(r)\quad such that\quad xijkl(r)=0},

Γd(r)={ijkl(r)\quad such that\quad xijkl(r)=dijkl},

and elements ijkl(r) are represented as variables xijkl in the transportation table.
By going from the left to the right in Γ0(r), choose
i0j0k0l0(r) as the first element ijkl(r)<0 found,
if all elements of Γ0(r) are nonnegative then choose in Γd(r) similarly,

i0j0k0l0(r) as the first element ijkl(r)>0 found.

If all elements of Γd(r) are non positive, then the feasible solution x(r) is optimal. Stop.

(P5) - Determination of a cycle.

A cycle μ(r) containing some interesting 4-tuple (i,j,k,l) and the non interesting 4-tuple (i0,j0,k0,l0) corresponding to i0j0k0l0(r) is determined by solving the linear system (i,j,k,l)I(r)αijkl(r)Pijkl=Pi0j0k0l0
The non null solutions αijkl(r) are called coefficients of the cycle μ(r).

(P6) - Determination of a new feasible solution.

Take

σ(r)={(i,j,k,l) such that (i,j,k,l) is a case of the cycle μ(r)}

σ(r)={(i,j,k,l) such that (i,j,k,l)σ(r), with αijkl(r)<0}

σ(r)+={(i,j,k,l) such that (i,j,k,l)σ(r), with αijkl(r)>0}

If i0j0k0l0(r)Γ0(r), determine

θ1(r)=min(i,j,k,l)σ(r)(xijkl(r)/αijkl(r)),θ2(r)=min(i,j,k,l)σ(r)+((dijklxijkl(r))/αijkl(r)),θ(r)=min(θ1(r),θ2(r)).

Next, take

x(r+1)={xijkl(r),\hspace{0.2cm}(i,j,k,l)σ(r)}{xijkl(r)+αijkl(r)θ(r),\hspace{0.2cm}(i,j,k,l)σ(r)}.

Else (i0j0k0l0(r)Γd(r)), determine

θ1(r)=min(i,j,k,l)σ(r)+(xijkl(r)/αijkl(r)),θ2(r)=min(i,j,k,l)σ(r)((dijklxijkl(r))/αijkl(r)),θ(r)=min(θ1(r),θ2(r)).

Next, take

x(r+1)={xijkl(r),\hspace{0.2cm}(i,j,k,l)σ(r)}{xijkl(r)αijkl(r)θ(r),\hspace{0.2cm}(i,j,k,l)σ(r)}.

(P7) - Determination of a set I(r) in a degenerate case.

1. Take

  • Eb the set of vectors corresponding to variables xijkl(r) verifying

    0<xijkl(r)<dijkl

    and Nb is its element number.

  • Eh the set of vectors corresponding to variables xijkl(r) verifying

    xijkl(r)=0 or xijkl(r)=dijkl

    and Nh is its element number. We are Nh=NNb,  with N=mnpq.

  • Es any subset with s=rank(A)Nb elements of Eh, for example the first s elements in Eh by going from the left to the right in the transportation table.

2. At the beginning of this procedure, we know a subset Es of Eh.

i) If the set(EsEb) is a free base, then a set I(r) is determined. Stop.

ii) Replace the first (the second, the third,..., or the sth) element in Es by the (s+1)th element in Eh, or take any other subset Es of Eh, and repeat i) until a set I(r) will be determined. Stop.

4.3 The convergence of Algorithm 2

Assume that the problem T4C is non degenerate. Observe that if x(r) and x(r+1) are two successive feasible solutions of the problem T4C with Z(r) and Z(r+1) are respectively the corresponding objective values then

Z(r+1)=Z(r)(1)ηθ(r)i0j0k0l0(r+1)

with

Unknown environment 'tabular'

So Z(r+1)<Z(r).
Hence the Algorithm 2 guaranties that the same base never appear in two distinct iterations and since the number of the visited vertices is necessarily finite (at most CNM3), then it converges finitely.

5 Computational results

The implementation of the two algorithms is running on a Pentium IV with a Microsoft Windows Environment, they are totally written with Borland Delphi. Table 1 summarizes the results obtained for a few numerical experiments.

Ex.

Dimension of

Number

Optimal

Time in

no.

the problem

of iterations

value (Z)

seconds

   

Alg. 1

Alg. 2

Alg. 1

Alg. 2

Alg. 1

Alg. 2

1

22×32

22

6

41.6025

41.6025

0.078

0.016

2

46×72

26

8

1317.79

1317.79

0.391

0.031

3

121×216

38

10

369.536

369.536

0.172

0.078

4

149×270

46

11

22.25

22.25

0.282

0.125

5

167×300

55

16

40

40

0.375

0.172

6

198×360

67

17

10.25

10.25

0.609

0.266

7

257×480

81

26

10.4375

10.4375

2.281

0.329

8

318×600

54

15

38.75

38.75

1.282

0.485

9

378×720

102

35

95.0625

95.0625

3.609

0.687

10

469×900

178

55

11.3125

11.3125

6.437

1.000

11

560×1080

257

83

46.25

46.25

10.438

1.406

12

741×1440

383

117

48.75

48.75

22.890

8.672

Table 1 Comparison between Algorithm 1 and Algorithm 2.

6 Conclusion and future researches

In this paper, we presented two exact simplicial methods for solving the capacited 4-index transportation problem. The numerical experiments showed that Algorithm 1 is more efficient than Algorithm 8 with respect to the number of iterations and also to the computational time. Moreover, as we have successfully done in some experiments, the arrangements made on Algorithm 1 have appropriately treated the degeneracy problems and led to a considerable reduction of the computational results.

Finally, we point out that our obtained results are independent of the number of indices of the problem, so we can extend Algorithm 1 for solving capacitated transportation problems with the number of indices that are greater than four.

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13

R. Zitouni, Etude qualitative de modèles de transport et localisation, Thèse de Doctorat d’état, Université ferhat Abbas-Sétif 1, Sétif, Algérie. 2007.