PP Pk?1 - CiteSeerX
Approximation algorithms for multi-index transportation problems with decomposable costs M. Queyranne and F.C.R. Spieksmay December 22, 1998
Abstract
The axial multi-index transportation problem is de ned as follows. Given are k sets Ar , each set having nr elements, r = 1; : : : ; k. The cartesian product of the sets Ar is denoted by A. To each element a A a certain cost, ca , is associated. Further, a nonnegative demand eri is associated to each set Ari = a A : a(r) = i . The problem nonnegative real numbers xa such that each demand is satis ed P is tox nd = e for r = 1; : : : ; k; i = 1; : : : ; nr ) and such that total cost (that (that is a ri P a2Ari is a2A ca xa ) is minimized. In this paper we deal with a special case of this problem where the costs ca are decomposable, that is, given a real-valued function f and a distance dijAr As between element i of Ar and element j of As , we assume that ca = f (daA(1);aA(2); : : : ; daA(kk??1);aAk(k) ) for all a A. We present two algorithms for this problem, and we analyze their worstcase behavior without requiring explicit knowledge of the cost-function f . Next, we use these results to derive explicit bounds in the case where f is the diameter costfunction (that is ca = maxr;s dAa(rr);aA(ss)), and in the case where f is the Hamiltonian P ?1 dA i A i : is a cyclic permutation path cost-function (that is ca = min ik=1 a((i));a((i+1)) of 1; : : : ; k ). 2
2 IR f
2
g
1
2
1
2
( )
f
f
( +1)
g
University of British Columbia, Faculty of Commerce and Business Administration, Vancouver, Canada V6T 1Z2. y Maastricht University, Department of Mathematics, P.O. Box 616, 6200 MD Maastricht, the Netherlands.
1
1 Introduction The axial multi-index transportation problem can be formulated as follows. Given are k sets Ar with Ar =: f1; : : : ; nr g, for r 2 K with K =: f:1; : : : ; kg. The cartesian product of these sets Ar , 1 r k, is denoted as A, that is A = fa : a 2 A1 A2 Ak g. We will refer to an element a 2 A as cluster a, with a(r) denoting the r-th entry of vector a. For each cluster a :2 A a certain cost ca 2 is speci ed. Further, for r 2 K , i 2 Ar , we de ne a section Ari = fa 2 A : a(r) = ig. To each section Ari , a nonnegative demand eri 2 is associated. The problem is now to nd nonnegative real numbers xa , a 2 A, such that the sum of those numbers xa for which a(r) = i, is equal to eri for each r 2 K , i 2 Ar , and such that total cost, summed over all clusters, is minimized. Mathematically, the problem can be described as follows: IR
IR
(kTP ) minimize such that
X a2A
ca xa
X
xa = eri
a2A xa 0 ri
for a 2 A:
P
P
for r 2 K ; i 2 Ar ;
We assume that i2A eri = j 2A esj for all r; s 2 K . It is not dicult to show that this assumption is a necessary and sucient condition for the existence of a feasible solution to (kTP ). Notice that for k = 2 the familiar (2-index) transportation problem arises. Axial k-index transportation problems have not been widely studied for k 3. An early reference to the axial 3-index transportation problem (3TP ) is Schell (1955). Other early references to (3TP ) are Haley (1963) and Corban (1964, 1966). More recently, in Queyranne, Spieksma and Tardella (1993), the special case of (kTP ) is investigated, where all elements of the sets Ar , r 2 K , lie on a single line, and the cost of each cluster a depends on the distances between elements of that cluster. A problem closely related to (kTP ) is the axial k-index assignment problem. This problem arises when the sets Ar have equal size, all demands eri are equal to 1, and xa is restricted to be either 0 or 1 for all a 2 A (see Pierskalla, 1968). Multi-index assignment problems occur in various real-world situations (see Balas and Saltzman (1991) for a recent overview). For instance, the scheduling of classes, teachers and rooms (see Frieze and Yadegar, 1981) as well as the manufacturing of printed circuit boards (see Crama and Spieksma, 1992), may give rise to (specially structured) instances of the 3-index assignment problem. Another interesting application of a problem related to the k-index assignment problem can be found in computational molecular biology (see Gus eld, 1993; and Pevzner, 1992). This paper deals with the special case of (kTP ) where the costs ca , a 2 A, are not arbitrarily given numbers, but are in some sense decomposable. More precisely, we will assume that there exist for each pair of sets Ar , As , r 6= s, nonnegative numbers dAij A , representing a distance between each element i of Ar and each element j of As , which determine the cost of a cluster a 2 A. Formally, the costs ca , a 2 A, are de ned to be r
s
r
2
s
decomposable if there exists a function f : IR( ) ! IR (called the cost-function) such that k
2
ca = f dAa r;aA s ; r; s 2 K; r 6= s; for all a 2 A. r
s
( )
( )
We will further assume that the distance (or length) function d is symmetric, that is, dAij A = djiA A for all r; s 2 K; i 2 Ar ; j 2 As : The motivation for investigating decomposable costs is that often, in practical applications, some structure in the cost-coecients ca can be found. A potential way to capture this structure is by using the concept of decomposable costs. Indeed, in a number of applications these decomposable costs arise naturally (see Frieze and Yadegar, 1981; Gus eld, 1993; and Crama and Spieksma, 1992). In the latter paper it is shown that for k 3, and for some simple, decomposable cost-functions the k-index assignment problem is NP -hard. Bandelt, Crama and Spieksma (1994) present heuristics for the multi-index assignment problem with decomposable costs, along with worst-case analyses for di erent speci cations of the cost-function (see also Gus eld, 1993). For instance, in the case where the cost of a cluster is equal to the sum of all distances in the cluster, Bandelt et al. (1991) propose an algorithm which is guaranteed to nd a solution with a cost bounded by twice the cost of an optimal solution for arbitrary k 2. These worst-case analyses depend on the assumption that d satis es the triangle inequality (see Section 4). The purpose of this paper is to present a general, unifying framework for such worstcase analyses. The new contributions are (i) an extension to multi-index transportation problems; (ii) an elucidation of the role of the triangle inequality in deriving such results; and (iii) the treatment of new cost-functions, the diameter and shortest Hamiltonian path. In the next section, the heuristics are introduced, and in Section 3 a general worst-case analysis is presented. Finally, in Section 4, two speci c cost-functions are investigated. r
s
s
r
2 Single-Hub and Multiple-Hub Heuristics First, consider the following two-step heuristic de ned for a xed index h. The rst step amounts to solving k ? 1 ordinary (that is, 2-index) transportation problems with respect to Ah and Ar for all r 2 K , r 6= h. In the second step, a solution to the k-index transportation problem is constructed based on the solutions found in the rst step. We will refer to the heuristic as the Single-Hub heuristic (cf. Bandelt et al., 1994). A formal description is as follows, where we denote an optimal solution to an (ordinary) 2-index transportation problem de ned by demands eri and esj and distance function dA A , by yA A for some r; s 2 K with r 6= s. The solution to the k-index transportation problem is given by fzah g. r
3
s
r
s
Step 1 Fix h, 1 h k. For all r 2 K n fhg, compute yA A . Set y~A A = yA A for r 2 K n fhg. Step 2 For j := 1 to nh do r
begin
r
h
h
r
h
q := 0 a(h) := j a (i) := 1 for all i 2 K n fhg 8 while q < eh;a h do > >
> begin > > > let ` be such that y~aA `;aA h = minr6 h y~aA r;aA h > > zah := y~aA `;aA h < inner loop > y~aA r;aA h := y~aA r;aA h ? zah for all r 2 K n fhg > > > a(`) := a(`) + 1 > > > q := q + zah > : end; ( )
`
h
( )
`
( )
r
( )
r
=
( )
( )
h
( )
h
( )
r
h
( )
( )
h
( )
end.
We now illustrate the algorithm on an instance of (3TP ) as presented in the following example. The feasibility of the algorithm is given later in Theorem 2.1.
Example 2.1
Let k = 3, n1 = n2 = 3, n3 = 4, 0 0 0 1 1 1 2 3 5 2 7 6 3 2 5 6 4 dA A = B @ 4 5 6 CA ; dA A = B@ 4 5 2 6 CA ; dA A = B@ 3 4 4 3 C A; 8 4 4 6 3 5 7 6 5 4 3 1
2
1
3
2
3
e = 12 5 4 ; e = 7 5 9 ; and e = 3 5 7 6 : 1
2
3
Fix h = 1. In Step 1 of the algorithm we nd that 0 0 1 1 7 5 0 3 1 2 6 yA A = B @ 0 0 5 CA and yA A = B@ 0 0 5 0 CA : 0 0 4 0 4 0 0 In Step 2, the inner loop is entered with a(1) = a(2) = a(3) = 1, and q = 0. Then A A ; y~A A ) = min(7; 3) = 3, y~A A := 7 ? 3 = 4, y~A A := 0, and, since 1 := min(~y11 z111 11 11 11 l = 3, a(3) := 1 + 1 = 2, and q := 3. Since q = 3 < 12 = e11 , we enter the inner loop again. A A ; y~A A ) = min(4; 1) = 1, y~A A := 3, y~A A := 0, and, since 1 := min(~y11 Then z112 12 11 12 l = 3, a(3) := 3, and q := 4. Again, since q = 4 < 12 = e11 , we enter the inner loop. A A ; y~A A ) = min(3; 2) = 2, y~A A := 1, y~A A := 0, and, since 1 := min(~y11 Then z113 13 11 13 l = 3, a(3) := 4, and q := 6. Since q = 6 < 12 = e11 , we enter the inner loop. 1
2
1
1
2
1
3
3
1
2
1
3
1
2
1
3
1
2
1
3
1
2
1
3
1
2
1
3
4
A A ; y~A A ) = min(1; 6) = 1, y~A A := 0, y~A A := 5, and, since 1 := min(~y11 Then z114 14 11 11 l = 2, a(2) := 2, and q := 7. Since q = 7 < 12 = e11 , we enter the inner loop. A A ; y~A A ) = min(5; 5) = 5, y~A A := y~A A := 0, q := 12. Since 1 := min(~y12 Then z124 14 12 14 q = 12 12 = e11 , we leave the inner loop. Step 2 proceeds by setting a(1) = 2, and a(2) = a(3) = 1, and by entering the inner loop 1 with q = 0. After four iterations of the inner loop we nd that z233 = 5. Then, we leave the inner loop, set a(1) = 3, and a(2) = a(3) = 1, and q = 0, and enter the inner loop again. 1 = 4, and the algorithm stops. After three iterations of the inner loop we nd that z332 1 1 1 1 1 1 = 5, = 1, z124 = 2, z114 Summarizing, the solution found is z with z111 = 3, z112 = 1, z113 1 1 1 z233 = 5, and z332 = 4, and all other za = 0. This is a feasible solution, as is easily veri ed. 2 1
2
1
3
1
2
1
3
1
2
1
1
2
1
3
3
Notice that, for a given h, the solution constructed by the Single-Hub heuristic depends only on the distance function d and not on the actual costs ca . We now establish the correctness of the algorithm.
Theorem 2.1 The solution zh found by the Single-Hub heuristic is a feasible solution to
(kTP ).
Proof: Consider Step 2 of the Single-Hub heuristic. Suppose a(h) = j , and a(r) := i, for some r = 6 h, 1 i nr . Since for a speci c value of a(h), a(r) is increasing in value, no clusters a 2 A with a(h) = j and a(r) = i, have been considered before. Hence, let us
now focus on the next L consecutive iterations for which a(h) = j and a(r) = i is the case. Thus, in each iteration, some at is considered with at (h) = j , at (r) = i for t = 1; : : : ; L. The value zah assigned to each cluster at 2 A is subtracted from y~ijA A , and nally, in the L-th iteration, when the minimum is attained for ` = r, y~ijA A is reduced to 0. In other words, the value of yijA A is distributed over all zah , t = 1; : : : ; L. Moreover, aL (r) := i + 1, and no clusters a 2 A with a(h) = j , a(r) = i will be considered in all next iterations. Hence for each i 2 Ar , j 2 Ah , we have: r
h
t
r
r
h
h
t
L X t=1
zah =
X
t
a:a(h)=j
zah = yijA A r
for all r 2 K n fhg:
h
a(r )=i
Summing over j yields: n X X h
j =1
a:a(h)=j
a(r )=i
zah =
n X h
j =1
yijA A = eri; r
for all r 2 K n fhg; i 2 Ar ;
h
where the last equality follows since yA A is a feasible solution to the two-index transportation problem between P APr and Azhh. = Pn yA A = e for j 2 A . Therefore, zh is Similarly, we have ni=1 hj h a a i=1 ij a feasible solution to (kTP ). 2 r
r
h
r
a:a(h)=j
a(r )=i
5
r
h
Regarding the complexity of the algorithm, notice that one iteration of the inner loop takes The number of iterations of the inner loop is bounded by the sum P O(kn) time. of the number of Pelements in the sets Ar , (r 6= h). Therefore, the comr2K nfhg r plexity of Step 2 equals nh r2K nfhg nr k. This is O(k2 n2 ) if nr = O(n) for all r 2 K . Since in Step P 1, O(k) transportation problems have to be solved, the overall complexity is O( r2K nfhg(T (nh ; nr ) + nh nr k)), where T (p; q) is the time needed to solve a p q transportation problem. With nr = O(n) for all r 2 K and T (p; q) = O(pq log(p + q)(pq +(p + q)log(p + q))) (see Orlin, 1988), we obtain an overall complexity of O(kn4 log n + k2n2) for the Single-Hub heuristic. Notice that this complexity is polynomial in k and n, in spite of the fact that the number of variables in the formulation of (kTP ) in Section 1 is O(nk ). The Multiple-Hub heuristic is derived from the Single-Hub heuristic in the following straightforward way: apply the Single-Hub heuristic for h = 1; : : : ; k and pick the best solution. Its complexity is equal to k times the complexity of the Single-Hub heuristic.
3 A general worst-case analysis In this section we will establish upper bounds on the ratio between the cost of solutions found by the heuristics and the cost of an optimal solution. Notice that these bounds remain valid when xa is restricted to be integer for all a 2 A. This is due to the fact that the solution found by the Single-Hub heuristic is composed from k ?1 solutions to ordinary (that is, two-index) transportation problems, using only additions and subtractions. Since the transportation problem has an optimal solution which is integral when all eri are integral, (see for instance Nemhauser and Wolsey, 1988), adding the integrality constraint, will only increase the cost of an optimal solution, which implies that the bounds remain valid. Obviously, the problem with integer decision variables is a direct generalization of the multiindex assignment problem with decomposable costs, dealt with in Bandelt et al. (1994). We will show that their bounds remain valid in this more general setting. In the sequel of this paper, the superscripts of the length function d are omitted when no confusion is likely to arise. De ne, for some h with 1 h k, and for each a 2 A:
Ha =
X r2K nfhg
da r ;a h ; ( )
( )
referred to as a hub. For a given cost-function f , it may be possible to bound ca from above in terms of the hub Ha . More precisely, instances arising in practical applications often admit a certain structure which can be captured by introducing a parameter 1 (k), k 2, such that the following inequality holds for all a 2 A:
ca (k) Ha 1
To illustrate this, consider the following example.
Example 3.1 Suppose f is the sum-cost function, that is 6
ca =
k X X s=2 r<s
da r ;a s for a 2 A: ( )
( )
Bandelt et al. (1994) show that, when d satis es the triangle inequality (see 16), then ca (k ? 1) Ha for all a 2 A: Thus, 1 (k) = k ? 1, for k 2. 2 Of course, depending on a speci c situation, similar analyses can be done for other cost-functions, and other restrictions on the length function. In a similar way, a parameter 2 (k) is introduced such that the following inequality holds for all a 2 A:
Ha (k) ca : 2
Obviously, for Example 3.1, 2 (k) = 1 for all k 2. Now, let c(SHh ) denote the cost of the solution found by the Single-Hub heuristic, and let OPT denote the cost of an optimal solution to the k-index transportation problem with decomposable costs. Theorem 3.1 c(SHh) 1(k) 2 (k) OPT for all h. Proof: First, ? we show how to `decompose' a feasible solution to (kTP ) into feasible solutions to k2 two-index transportation problems de ned by the sets Ar and As with demands eri , esj and length function dAij A , r; s 2 K , r 6= s, i 2 Ar , j 2 As . Let xa be any feasible solution to (kTP ). We claim that, for any r; s 2 K , r 6= s, the numbers r
X
a:a(r )=i;
s
xa ; i 2 A r ; j 2 A s ;
a(s)=j
constitute a feasible solution to the 2-index transportation problem between Ar and As . Indeed, since for any r; s 2 K , r 6= s,
X X
j 2A
s a:a(r )=i; a(s)=j
xa =
and, similarly
X X
i2A
r a:a(s)=j; a(r )=i
xa =
X
a:a(r)=i
X a2A
xa =
X
a2A
xa = eri for i 2 Ar ;
ri
xa = esj for j 2 As;
sj
the claim is true. In the remainder of the proof xa , a 2 A, denotes an optimal solution to (kTP ). 7
Now, we can prove the desired result: X h ca z a c(SHh ) = a2A
=
0 X@ X
1 (k) dAa r;aA h A zah a2A r2K nfhg X X X A A X h (k) dij za r2K nfhg i2A j 2A X X X A A A A (k) dij yij r2K nfhg i2A j 2A X X X A A X (k) dij xa r2K nfhg i2A j 2A 0 1 X @ X A A A (k) da r ;a h xa a2A r2K nfhg X (k) (k) ca xa = (k) (k) OPT: 1
r
r
r
h
h
r
h
r
h
(4) (5)
a:a(r )=i;
h
a(h)=j
r
2
r
h
h
( )
1
(3)
a(h)=j
1
1
h
(2)
a:a(r )=i;
1
r
=
( )
1
r
=
h
( )
(1)
( )
1
a2A
2
(6) (7)
(2) and (7) hold by de nition of 1 (k) and 2 (k) respectively, (3) and (6) are a rearrangement of terms, (4) follows from Theorem 2.1, and (5) follows from the rst part of this proof and the fact that yA A is an optimal solution to the transportation problem between Ar and Ah . 2 r
h
Evidently, the bound 1 (k) 2 (k), k 2, is also a valid upper bound for the ratio between the cost of the solution found by the Multiple-Hub heuristic and the cost of an optimal solution. In order to be able to derive a possibly better bound for the Multiple-Hub heuristic, we introduce a parameter 3 (k), k 2 such that the following inequality holds for all a 2 A: k X X s=2 r<s
da r ;a s (k) ca : ( )
3
( )
For example, 3 (k) is easily seen to equal 1 for Example 3.1. With c(MH ) denoting the cost of the solution found by the Multiple-Hub heuristic, we have the following theorem: Theorem 3.2 c(MH ) 2 (kk) (k) OPT . Proof: c(MH ) = h=1 min c(SHh) (8) ;:::;k 1
k X 1 k c(SHh) h=1 k 1X
3
0 X@ X
k (k) a2A h =1
1
1 dAa r;aA h A zah r
r2K nfhg
8
( )
h
( )
(9) (10)
k X X X X = 1k(k) h=1 r2K nfhg i2A j 2A k X X X X 1 (k ) r
k
dAij A yijA A r
h
r
dAij A
h
X
h
h=1 r2K nfhg i2A j 2A
r
(11)
h
a:a(r )=i;
xa
0k 1 X @X X A A A ( k ) = k da r ;a h xa a2A h r2K nfhg 0 k 1 X@ X X A A A ( k ) 2 da r ;a h xa = k a2A h r 2),
c(MH ) (k ? 1 ? k2 ) OPT if k odd,
for all instances satisfying the triangle inequality. Moreover, there exist instances with k = 3, satisfying the triangle inequality, for which these bounds are tight.
Proof: In order to prove the theorem, we will derive consecutively the values of (k); (k) and (k). Next, we can apply theorems 3.1 and 3.2. Consider a cluster a 2 A with path 1
3
11
2
i ?L@L ? ? , L j@e@ ? ,, e @ ? ? ,, ?k@ ee@@ ? , ? @ e @ 1
3
5
? ? ? ,,, ? ??,, ?? ?, ? ? , ?? k , ? j ? !!!aaa?i 4
1
5
@
e @ @ e @ @ @ ee @ @ @ e @ @ @ ee @ @ k j@aeaa e!!i !@ 6
1
3
i ! aa j i aa !! j k ! !a @ ee @ ? k a,! ? @ e @ ? ? , @@ ee @@ ? ? ? ,, ? @ @ ee @ @ ? , ?? , ? @ e @ ? , ? @@ ee @@ k ?? ,, ?? @@ee @ ?,,, ?? @ eej , , ? @@LL ?? @Li? 4
2
6
5
2
3
6
4
2
Figure 1: Worst-case instance for the Single- and Multiple-Hub heuristic for k = 3 (diameter costs).
12
costs:
ca = min
(kX ?
1
i=1
da i
;a((i+1)) : is a cyclic permutation of f1; : : : ; kgg:
( ( ))
Using the triangle inequality (16), it is not dicult to show that X ca 2 da(h);a(j) ? da(h);a(r) ? da(h);a(s) for any r; s 2 K n fhg; r 6= s:
(18)
j 2K nfhg
Now, consider the k ? 1 distances da(h);a(j ) , j 2 K n fhg, determining the hub Ha . We order the elements of the cluster a by j1 ; j2 ; : : : ; jk?1 such that da(h);a(j ) da(h);a(j ) for i = 1; : : : ; k ? 2. For the two largest distances in the hub Ha we have (19) da(h);a(j ) + da(h);a(j ) k ?2 1 Ha : The following argument shows, by contradiction, that (19) is true. Suppose that (19) is not true, then da(h);a(j ) da(h);a(j ) < k ?1 1 Ha for i 3: But then kX ?1 X 3 H =H ; da(h);a(j) = da(h);a(j ) + da(h);a(j ) + da(h);a(j ) < k ?2 1 Ha + kk ? Ha = a ? 1 a i=3 j 2K nfhg i
1
2
i
2
1
2
i+1
i
which is a contradiction. Since (18) holds for any r; s 2 K nfhg, we may take in (18) r = j1 and s = j2 . Together with (19), we arrive at ca 2 (1 ? k ?1 1 ) Ha ; (20) thus showing that 1 (k) = 2(1 ? k?1 1 ). Regarding 2 (k), since da(h);a(j ) ca for all j , it follows easily that 2 (k) k ? 1 for k 2. Let us now derive an estimate for 3(k). To do this, we need to refer to the cost of the shortest Hamiltonian tour through a cluster, and to its corresponding 3 (k) function. tour Thus let ctour a refer to these shortest tour costs and let 3 (k) refer to the corresponding function (where ca and 3 (k) are still de ned with respect to the shortest Hamiltonian path costs). Since the cost of a shortest Hamiltonian tour is smaller or equal to twice the length of a path we derive: X da(r)a(s) tour (k) ctour 3 a r<s
2 (k) ca for all a 2 A: tour 3
Table 1 gives:
(k) 2 (k) = 41 k if k even, and 1 (k + 1)(k ? 1) if k odd. 4 3
tour 3
2
13
Now, applying Theorem 3.1 and Theorem 3.2 to the estimates derived here for 1 (k),
(k) and (k) results in the bounds for the Single- and Multiple-Hub heuristic. 2
3
In order to show that these bounds are tight for both the Single-Hub and Multiple-Hub heuristic if k = 3, consider the following two instances of the 3-index assignment problem with path costs, depicted in Figures 2 and 3 hereunder (cf. with Bandelt et al., 1994).
i =i =i ?@ 1
?
? j =k 2
?? ?
2
? @
j =k 3
3
3
@@ @
@@
j =k 1
1
2
Figure 2: Worst-case instance for the Single-Hub heuristic for k = 3 (path costs).
i =j =k ?@ 1
1
1
? @ ? @@ ? @@ ?? @ ?
j =k 3
2
i =j 2
2
i =k 3
3
Figure 3: Worst-case instance for the Multiple-Hub heuristic for k = 3 (path costs). A drawn edge in Figure 2 or 3 indicates a distance of 1; any other distance is equal to 2. Further, the demand associated to each node equals 1. Let A1 = fi1 ; i2 ; i3 g, A2 = fj1 ; j2 ; j3 g, A3 = fk1 ; k2 ; k3 g in both gures. Now, consider the instance for the Single-Hub heuristic, depicted in Figure 2. The optimal solution is found by setting xa = 1 for a 2 f(i1 ; j2 ; k3 ), (i2 ; j3 ; k1 ), (i3 ; j1 ; k2 )g, and xa = 0 otherwise. This solution has a total cost of 3. The Single-Hub heuristic with h = 1 may nd as optimal solutions to the 2 two-index assignment problems de ned by A1 and A2 , and A1 and A3 , yiAj A = 1 and yiAkA = 1 for ` = 1; 2; 3, and 0 otherwise. Next, in Step 2 of the Single-Hub heuristic, we nd the solution za = 1 for a = (il ; jl ; kl ), l = 1; 2; 3, and za = 0 otherwise, which has total cost equal to 6, thus achieving the desired ratio for k = 3. In fact, it is not dicult to generalize this instance in such a way that the Single-Hub heuristic produces solutions bounded by (2k ? 4) OPT , showing that the bound proven here is tight for any k 3, (see Bandelt et al., 1994). Let us now consider the Multiple-Hub heuristic and the corresponding instance depicted in Figure 3. The symmetry of this instance implies that we may restrict ourselves to investigating the performance of the Single-Hub heuristic for this instance. The optimal solution is found by setting xa = 1 for a 2 f(i1 ; j3 ; k2 ), (i2 ; j2 ; k1 ), (i3 ; j1 ; k3 ), g and xa = 0 otherwise. 1
` `
14
2
1
`
3
`
This solution has a total cost of 3. Now, consider the Single-Hub heuristic with h = 1. Optimal solutions to the 2 two-index assignment problems de ned by A1 and A2 , and A1 and A3 , are given by respectively yiAj A = 1 and yiAkA = 1 for ` = 1; 2; 3, and 0 otherwise. Next, in Step 2 of the Single-Hub heuristic, we nd the solution za = 1 for a = (il ; jl ; kl ), l = 1; 2; 3, and za = 0 otherwise, which has total cost equal to 4, thus achieving the desired ratio for k = 3. 2 1
1
2
` `
`
3
`
Acknowledgements
The second author wishes to acknowledge the support of the Landelijk Netwerk Mathematische Besliskunde.
References Balas, E. and M.J. Saltzman (1991), \An algorithm for the three-index assignment problem", Operations Research , 39, 150{161. Bandelt, H.-J., Y. Crama and F.C.R. Spieksma (1994), \Approximation algorithms for multidimensional assignment problems with decomposable costs", Discrete Applied Mathematics , 49, 25{50. Corban, A. (1964), \A multidimensional transportation problem", Revue Roumaine de Mathematiques pures et appliquees , 9, 721{735. Corban, A. (1966), \On a three-dimensional transportation problem", Revue Roumaine de Mathematiques pures et appliquees , 11, 57{75. Crama, Y. and F.C.R. Spieksma (1992), \Approximation algorithms for three-dimensional assignment problems with triangle inequalities", European Journal of Operational Research , 60, 273{279. Frieze, A.M. and J. Yadegar (1981), \An algorithm for solving 3-dimensional assignment problems with application to scheduling a teaching practice", Journal of the Operational Research Society , 32, 989{995. Gus eld, D. (1993), \Ecient methods for multiple sequence alignment with guaranteed error bounds", Bulletin of Mathematical Biology , 55, 141{154. Haley, K.B. (1963), \The multi-index problem", Operations Research , 11, 368{379. Nemhauser, G.L. and L.A. Wolsey (1988), \Integer and Combinatorial Optimalization ", Wiley, New York. Orlin, J.B. (1988), \A faster strongly polynomial minimum cost ow algorithm", in: Proceedings of the 20th ACM Symposium on the Theory of Computing, 377{387. Pevzner, P.A. (1992), \Multiple alignment, communication cost and graph matching", SIAM Journal of Applied Mathematics , 52, 1763-1779. 15
Pierskalla, W.P. (1968), \The multi-dimensional assignment problem", Operations Research , 16, 422{431. Queyranne, M., F.C.R. Spieksma and F. Tardella (1993), \A general class of greedily solvable linear programs", in: Proceedings of the third IPCO Conference , 385{399. Schell, E.D. (1955), \Distribution of a product by several properties", in: Proceedings of the Second Symposium in Linear Programming, 615{642.
16
Abstract
The axial multi-index transportation problem is de ned as follows. Given are k sets Ar , each set having nr elements, r = 1; : : : ; k. The cartesian product of the sets Ar is denoted by A. To each element a A a certain cost, ca , is associated. Further, a nonnegative demand eri is associated to each set Ari = a A : a(r) = i . The problem nonnegative real numbers xa such that each demand is satis ed P is tox nd = e for r = 1; : : : ; k; i = 1; : : : ; nr ) and such that total cost (that (that is a ri P a2Ari is a2A ca xa ) is minimized. In this paper we deal with a special case of this problem where the costs ca are decomposable, that is, given a real-valued function f and a distance dijAr As between element i of Ar and element j of As , we assume that ca = f (daA(1);aA(2); : : : ; daA(kk??1);aAk(k) ) for all a A. We present two algorithms for this problem, and we analyze their worstcase behavior without requiring explicit knowledge of the cost-function f . Next, we use these results to derive explicit bounds in the case where f is the diameter costfunction (that is ca = maxr;s dAa(rr);aA(ss)), and in the case where f is the Hamiltonian P ?1 dA i A i : is a cyclic permutation path cost-function (that is ca = min ik=1 a((i));a((i+1)) of 1; : : : ; k ). 2
2 IR f
2
g
1
2
1
2
( )
f
f
( +1)
g
University of British Columbia, Faculty of Commerce and Business Administration, Vancouver, Canada V6T 1Z2. y Maastricht University, Department of Mathematics, P.O. Box 616, 6200 MD Maastricht, the Netherlands.
1
1 Introduction The axial multi-index transportation problem can be formulated as follows. Given are k sets Ar with Ar =: f1; : : : ; nr g, for r 2 K with K =: f:1; : : : ; kg. The cartesian product of these sets Ar , 1 r k, is denoted as A, that is A = fa : a 2 A1 A2 Ak g. We will refer to an element a 2 A as cluster a, with a(r) denoting the r-th entry of vector a. For each cluster a :2 A a certain cost ca 2 is speci ed. Further, for r 2 K , i 2 Ar , we de ne a section Ari = fa 2 A : a(r) = ig. To each section Ari , a nonnegative demand eri 2 is associated. The problem is now to nd nonnegative real numbers xa , a 2 A, such that the sum of those numbers xa for which a(r) = i, is equal to eri for each r 2 K , i 2 Ar , and such that total cost, summed over all clusters, is minimized. Mathematically, the problem can be described as follows: IR
IR
(kTP ) minimize such that
X a2A
ca xa
X
xa = eri
a2A xa 0 ri
for a 2 A:
P
P
for r 2 K ; i 2 Ar ;
We assume that i2A eri = j 2A esj for all r; s 2 K . It is not dicult to show that this assumption is a necessary and sucient condition for the existence of a feasible solution to (kTP ). Notice that for k = 2 the familiar (2-index) transportation problem arises. Axial k-index transportation problems have not been widely studied for k 3. An early reference to the axial 3-index transportation problem (3TP ) is Schell (1955). Other early references to (3TP ) are Haley (1963) and Corban (1964, 1966). More recently, in Queyranne, Spieksma and Tardella (1993), the special case of (kTP ) is investigated, where all elements of the sets Ar , r 2 K , lie on a single line, and the cost of each cluster a depends on the distances between elements of that cluster. A problem closely related to (kTP ) is the axial k-index assignment problem. This problem arises when the sets Ar have equal size, all demands eri are equal to 1, and xa is restricted to be either 0 or 1 for all a 2 A (see Pierskalla, 1968). Multi-index assignment problems occur in various real-world situations (see Balas and Saltzman (1991) for a recent overview). For instance, the scheduling of classes, teachers and rooms (see Frieze and Yadegar, 1981) as well as the manufacturing of printed circuit boards (see Crama and Spieksma, 1992), may give rise to (specially structured) instances of the 3-index assignment problem. Another interesting application of a problem related to the k-index assignment problem can be found in computational molecular biology (see Gus eld, 1993; and Pevzner, 1992). This paper deals with the special case of (kTP ) where the costs ca , a 2 A, are not arbitrarily given numbers, but are in some sense decomposable. More precisely, we will assume that there exist for each pair of sets Ar , As , r 6= s, nonnegative numbers dAij A , representing a distance between each element i of Ar and each element j of As , which determine the cost of a cluster a 2 A. Formally, the costs ca , a 2 A, are de ned to be r
s
r
2
s
decomposable if there exists a function f : IR( ) ! IR (called the cost-function) such that k
2
ca = f dAa r;aA s ; r; s 2 K; r 6= s; for all a 2 A. r
s
( )
( )
We will further assume that the distance (or length) function d is symmetric, that is, dAij A = djiA A for all r; s 2 K; i 2 Ar ; j 2 As : The motivation for investigating decomposable costs is that often, in practical applications, some structure in the cost-coecients ca can be found. A potential way to capture this structure is by using the concept of decomposable costs. Indeed, in a number of applications these decomposable costs arise naturally (see Frieze and Yadegar, 1981; Gus eld, 1993; and Crama and Spieksma, 1992). In the latter paper it is shown that for k 3, and for some simple, decomposable cost-functions the k-index assignment problem is NP -hard. Bandelt, Crama and Spieksma (1994) present heuristics for the multi-index assignment problem with decomposable costs, along with worst-case analyses for di erent speci cations of the cost-function (see also Gus eld, 1993). For instance, in the case where the cost of a cluster is equal to the sum of all distances in the cluster, Bandelt et al. (1991) propose an algorithm which is guaranteed to nd a solution with a cost bounded by twice the cost of an optimal solution for arbitrary k 2. These worst-case analyses depend on the assumption that d satis es the triangle inequality (see Section 4). The purpose of this paper is to present a general, unifying framework for such worstcase analyses. The new contributions are (i) an extension to multi-index transportation problems; (ii) an elucidation of the role of the triangle inequality in deriving such results; and (iii) the treatment of new cost-functions, the diameter and shortest Hamiltonian path. In the next section, the heuristics are introduced, and in Section 3 a general worst-case analysis is presented. Finally, in Section 4, two speci c cost-functions are investigated. r
s
s
r
2 Single-Hub and Multiple-Hub Heuristics First, consider the following two-step heuristic de ned for a xed index h. The rst step amounts to solving k ? 1 ordinary (that is, 2-index) transportation problems with respect to Ah and Ar for all r 2 K , r 6= h. In the second step, a solution to the k-index transportation problem is constructed based on the solutions found in the rst step. We will refer to the heuristic as the Single-Hub heuristic (cf. Bandelt et al., 1994). A formal description is as follows, where we denote an optimal solution to an (ordinary) 2-index transportation problem de ned by demands eri and esj and distance function dA A , by yA A for some r; s 2 K with r 6= s. The solution to the k-index transportation problem is given by fzah g. r
3
s
r
s
Step 1 Fix h, 1 h k. For all r 2 K n fhg, compute yA A . Set y~A A = yA A for r 2 K n fhg. Step 2 For j := 1 to nh do r
begin
r
h
h
r
h
q := 0 a(h) := j a (i) := 1 for all i 2 K n fhg 8 while q < eh;a h do > >
> begin > > > let ` be such that y~aA `;aA h = minr6 h y~aA r;aA h > > zah := y~aA `;aA h < inner loop > y~aA r;aA h := y~aA r;aA h ? zah for all r 2 K n fhg > > > a(`) := a(`) + 1 > > > q := q + zah > : end; ( )
`
h
( )
`
( )
r
( )
r
=
( )
( )
h
( )
h
( )
r
h
( )
( )
h
( )
end.
We now illustrate the algorithm on an instance of (3TP ) as presented in the following example. The feasibility of the algorithm is given later in Theorem 2.1.
Example 2.1
Let k = 3, n1 = n2 = 3, n3 = 4, 0 0 0 1 1 1 2 3 5 2 7 6 3 2 5 6 4 dA A = B @ 4 5 6 CA ; dA A = B@ 4 5 2 6 CA ; dA A = B@ 3 4 4 3 C A; 8 4 4 6 3 5 7 6 5 4 3 1
2
1
3
2
3
e = 12 5 4 ; e = 7 5 9 ; and e = 3 5 7 6 : 1
2
3
Fix h = 1. In Step 1 of the algorithm we nd that 0 0 1 1 7 5 0 3 1 2 6 yA A = B @ 0 0 5 CA and yA A = B@ 0 0 5 0 CA : 0 0 4 0 4 0 0 In Step 2, the inner loop is entered with a(1) = a(2) = a(3) = 1, and q = 0. Then A A ; y~A A ) = min(7; 3) = 3, y~A A := 7 ? 3 = 4, y~A A := 0, and, since 1 := min(~y11 z111 11 11 11 l = 3, a(3) := 1 + 1 = 2, and q := 3. Since q = 3 < 12 = e11 , we enter the inner loop again. A A ; y~A A ) = min(4; 1) = 1, y~A A := 3, y~A A := 0, and, since 1 := min(~y11 Then z112 12 11 12 l = 3, a(3) := 3, and q := 4. Again, since q = 4 < 12 = e11 , we enter the inner loop. A A ; y~A A ) = min(3; 2) = 2, y~A A := 1, y~A A := 0, and, since 1 := min(~y11 Then z113 13 11 13 l = 3, a(3) := 4, and q := 6. Since q = 6 < 12 = e11 , we enter the inner loop. 1
2
1
1
2
1
3
3
1
2
1
3
1
2
1
3
1
2
1
3
1
2
1
3
1
2
1
3
4
A A ; y~A A ) = min(1; 6) = 1, y~A A := 0, y~A A := 5, and, since 1 := min(~y11 Then z114 14 11 11 l = 2, a(2) := 2, and q := 7. Since q = 7 < 12 = e11 , we enter the inner loop. A A ; y~A A ) = min(5; 5) = 5, y~A A := y~A A := 0, q := 12. Since 1 := min(~y12 Then z124 14 12 14 q = 12 12 = e11 , we leave the inner loop. Step 2 proceeds by setting a(1) = 2, and a(2) = a(3) = 1, and by entering the inner loop 1 with q = 0. After four iterations of the inner loop we nd that z233 = 5. Then, we leave the inner loop, set a(1) = 3, and a(2) = a(3) = 1, and q = 0, and enter the inner loop again. 1 = 4, and the algorithm stops. After three iterations of the inner loop we nd that z332 1 1 1 1 1 1 = 5, = 1, z124 = 2, z114 Summarizing, the solution found is z with z111 = 3, z112 = 1, z113 1 1 1 z233 = 5, and z332 = 4, and all other za = 0. This is a feasible solution, as is easily veri ed. 2 1
2
1
3
1
2
1
3
1
2
1
1
2
1
3
3
Notice that, for a given h, the solution constructed by the Single-Hub heuristic depends only on the distance function d and not on the actual costs ca . We now establish the correctness of the algorithm.
Theorem 2.1 The solution zh found by the Single-Hub heuristic is a feasible solution to
(kTP ).
Proof: Consider Step 2 of the Single-Hub heuristic. Suppose a(h) = j , and a(r) := i, for some r = 6 h, 1 i nr . Since for a speci c value of a(h), a(r) is increasing in value, no clusters a 2 A with a(h) = j and a(r) = i, have been considered before. Hence, let us
now focus on the next L consecutive iterations for which a(h) = j and a(r) = i is the case. Thus, in each iteration, some at is considered with at (h) = j , at (r) = i for t = 1; : : : ; L. The value zah assigned to each cluster at 2 A is subtracted from y~ijA A , and nally, in the L-th iteration, when the minimum is attained for ` = r, y~ijA A is reduced to 0. In other words, the value of yijA A is distributed over all zah , t = 1; : : : ; L. Moreover, aL (r) := i + 1, and no clusters a 2 A with a(h) = j , a(r) = i will be considered in all next iterations. Hence for each i 2 Ar , j 2 Ah , we have: r
h
t
r
r
h
h
t
L X t=1
zah =
X
t
a:a(h)=j
zah = yijA A r
for all r 2 K n fhg:
h
a(r )=i
Summing over j yields: n X X h
j =1
a:a(h)=j
a(r )=i
zah =
n X h
j =1
yijA A = eri; r
for all r 2 K n fhg; i 2 Ar ;
h
where the last equality follows since yA A is a feasible solution to the two-index transportation problem between P APr and Azhh. = Pn yA A = e for j 2 A . Therefore, zh is Similarly, we have ni=1 hj h a a i=1 ij a feasible solution to (kTP ). 2 r
r
h
r
a:a(h)=j
a(r )=i
5
r
h
Regarding the complexity of the algorithm, notice that one iteration of the inner loop takes The number of iterations of the inner loop is bounded by the sum P O(kn) time. of the number of Pelements in the sets Ar , (r 6= h). Therefore, the comr2K nfhg r plexity of Step 2 equals nh r2K nfhg nr k. This is O(k2 n2 ) if nr = O(n) for all r 2 K . Since in Step P 1, O(k) transportation problems have to be solved, the overall complexity is O( r2K nfhg(T (nh ; nr ) + nh nr k)), where T (p; q) is the time needed to solve a p q transportation problem. With nr = O(n) for all r 2 K and T (p; q) = O(pq log(p + q)(pq +(p + q)log(p + q))) (see Orlin, 1988), we obtain an overall complexity of O(kn4 log n + k2n2) for the Single-Hub heuristic. Notice that this complexity is polynomial in k and n, in spite of the fact that the number of variables in the formulation of (kTP ) in Section 1 is O(nk ). The Multiple-Hub heuristic is derived from the Single-Hub heuristic in the following straightforward way: apply the Single-Hub heuristic for h = 1; : : : ; k and pick the best solution. Its complexity is equal to k times the complexity of the Single-Hub heuristic.
3 A general worst-case analysis In this section we will establish upper bounds on the ratio between the cost of solutions found by the heuristics and the cost of an optimal solution. Notice that these bounds remain valid when xa is restricted to be integer for all a 2 A. This is due to the fact that the solution found by the Single-Hub heuristic is composed from k ?1 solutions to ordinary (that is, two-index) transportation problems, using only additions and subtractions. Since the transportation problem has an optimal solution which is integral when all eri are integral, (see for instance Nemhauser and Wolsey, 1988), adding the integrality constraint, will only increase the cost of an optimal solution, which implies that the bounds remain valid. Obviously, the problem with integer decision variables is a direct generalization of the multiindex assignment problem with decomposable costs, dealt with in Bandelt et al. (1994). We will show that their bounds remain valid in this more general setting. In the sequel of this paper, the superscripts of the length function d are omitted when no confusion is likely to arise. De ne, for some h with 1 h k, and for each a 2 A:
Ha =
X r2K nfhg
da r ;a h ; ( )
( )
referred to as a hub. For a given cost-function f , it may be possible to bound ca from above in terms of the hub Ha . More precisely, instances arising in practical applications often admit a certain structure which can be captured by introducing a parameter 1 (k), k 2, such that the following inequality holds for all a 2 A:
ca (k) Ha 1
To illustrate this, consider the following example.
Example 3.1 Suppose f is the sum-cost function, that is 6
ca =
k X X s=2 r<s
da r ;a s for a 2 A: ( )
( )
Bandelt et al. (1994) show that, when d satis es the triangle inequality (see 16), then ca (k ? 1) Ha for all a 2 A: Thus, 1 (k) = k ? 1, for k 2. 2 Of course, depending on a speci c situation, similar analyses can be done for other cost-functions, and other restrictions on the length function. In a similar way, a parameter 2 (k) is introduced such that the following inequality holds for all a 2 A:
Ha (k) ca : 2
Obviously, for Example 3.1, 2 (k) = 1 for all k 2. Now, let c(SHh ) denote the cost of the solution found by the Single-Hub heuristic, and let OPT denote the cost of an optimal solution to the k-index transportation problem with decomposable costs. Theorem 3.1 c(SHh) 1(k) 2 (k) OPT for all h. Proof: First, ? we show how to `decompose' a feasible solution to (kTP ) into feasible solutions to k2 two-index transportation problems de ned by the sets Ar and As with demands eri , esj and length function dAij A , r; s 2 K , r 6= s, i 2 Ar , j 2 As . Let xa be any feasible solution to (kTP ). We claim that, for any r; s 2 K , r 6= s, the numbers r
X
a:a(r )=i;
s
xa ; i 2 A r ; j 2 A s ;
a(s)=j
constitute a feasible solution to the 2-index transportation problem between Ar and As . Indeed, since for any r; s 2 K , r 6= s,
X X
j 2A
s a:a(r )=i; a(s)=j
xa =
and, similarly
X X
i2A
r a:a(s)=j; a(r )=i
xa =
X
a:a(r)=i
X a2A
xa =
X
a2A
xa = eri for i 2 Ar ;
ri
xa = esj for j 2 As;
sj
the claim is true. In the remainder of the proof xa , a 2 A, denotes an optimal solution to (kTP ). 7
Now, we can prove the desired result: X h ca z a c(SHh ) = a2A
=
0 X@ X
1 (k) dAa r;aA h A zah a2A r2K nfhg X X X A A X h (k) dij za r2K nfhg i2A j 2A X X X A A A A (k) dij yij r2K nfhg i2A j 2A X X X A A X (k) dij xa r2K nfhg i2A j 2A 0 1 X @ X A A A (k) da r ;a h xa a2A r2K nfhg X (k) (k) ca xa = (k) (k) OPT: 1
r
r
r
h
h
r
h
r
h
(4) (5)
a:a(r )=i;
h
a(h)=j
r
2
r
h
h
( )
1
(3)
a(h)=j
1
1
h
(2)
a:a(r )=i;
1
r
=
( )
1
r
=
h
( )
(1)
( )
1
a2A
2
(6) (7)
(2) and (7) hold by de nition of 1 (k) and 2 (k) respectively, (3) and (6) are a rearrangement of terms, (4) follows from Theorem 2.1, and (5) follows from the rst part of this proof and the fact that yA A is an optimal solution to the transportation problem between Ar and Ah . 2 r
h
Evidently, the bound 1 (k) 2 (k), k 2, is also a valid upper bound for the ratio between the cost of the solution found by the Multiple-Hub heuristic and the cost of an optimal solution. In order to be able to derive a possibly better bound for the Multiple-Hub heuristic, we introduce a parameter 3 (k), k 2 such that the following inequality holds for all a 2 A: k X X s=2 r<s
da r ;a s (k) ca : ( )
3
( )
For example, 3 (k) is easily seen to equal 1 for Example 3.1. With c(MH ) denoting the cost of the solution found by the Multiple-Hub heuristic, we have the following theorem: Theorem 3.2 c(MH ) 2 (kk) (k) OPT . Proof: c(MH ) = h=1 min c(SHh) (8) ;:::;k 1
k X 1 k c(SHh) h=1 k 1X
3
0 X@ X
k (k) a2A h =1
1
1 dAa r;aA h A zah r
r2K nfhg
8
( )
h
( )
(9) (10)
k X X X X = 1k(k) h=1 r2K nfhg i2A j 2A k X X X X 1 (k ) r
k
dAij A yijA A r
h
r
dAij A
h
X
h
h=1 r2K nfhg i2A j 2A
r
(11)
h
a:a(r )=i;
xa
0k 1 X @X X A A A ( k ) = k da r ;a h xa a2A h r2K nfhg 0 k 1 X@ X X A A A ( k ) 2 da r ;a h xa = k a2A h r 2),
c(MH ) (k ? 1 ? k2 ) OPT if k odd,
for all instances satisfying the triangle inequality. Moreover, there exist instances with k = 3, satisfying the triangle inequality, for which these bounds are tight.
Proof: In order to prove the theorem, we will derive consecutively the values of (k); (k) and (k). Next, we can apply theorems 3.1 and 3.2. Consider a cluster a 2 A with path 1
3
11
2
i ?L@L ? ? , L j@e@ ? ,, e @ ? ? ,, ?k@ ee@@ ? , ? @ e @ 1
3
5
? ? ? ,,, ? ??,, ?? ?, ? ? , ?? k , ? j ? !!!aaa?i 4
1
5
@
e @ @ e @ @ @ ee @ @ @ e @ @ @ ee @ @ k j@aeaa e!!i !@ 6
1
3
i ! aa j i aa !! j k ! !a @ ee @ ? k a,! ? @ e @ ? ? , @@ ee @@ ? ? ? ,, ? @ @ ee @ @ ? , ?? , ? @ e @ ? , ? @@ ee @@ k ?? ,, ?? @@ee @ ?,,, ?? @ eej , , ? @@LL ?? @Li? 4
2
6
5
2
3
6
4
2
Figure 1: Worst-case instance for the Single- and Multiple-Hub heuristic for k = 3 (diameter costs).
12
costs:
ca = min
(kX ?
1
i=1
da i
;a((i+1)) : is a cyclic permutation of f1; : : : ; kgg:
( ( ))
Using the triangle inequality (16), it is not dicult to show that X ca 2 da(h);a(j) ? da(h);a(r) ? da(h);a(s) for any r; s 2 K n fhg; r 6= s:
(18)
j 2K nfhg
Now, consider the k ? 1 distances da(h);a(j ) , j 2 K n fhg, determining the hub Ha . We order the elements of the cluster a by j1 ; j2 ; : : : ; jk?1 such that da(h);a(j ) da(h);a(j ) for i = 1; : : : ; k ? 2. For the two largest distances in the hub Ha we have (19) da(h);a(j ) + da(h);a(j ) k ?2 1 Ha : The following argument shows, by contradiction, that (19) is true. Suppose that (19) is not true, then da(h);a(j ) da(h);a(j ) < k ?1 1 Ha for i 3: But then kX ?1 X 3 H =H ; da(h);a(j) = da(h);a(j ) + da(h);a(j ) + da(h);a(j ) < k ?2 1 Ha + kk ? Ha = a ? 1 a i=3 j 2K nfhg i
1
2
i
2
1
2
i+1
i
which is a contradiction. Since (18) holds for any r; s 2 K nfhg, we may take in (18) r = j1 and s = j2 . Together with (19), we arrive at ca 2 (1 ? k ?1 1 ) Ha ; (20) thus showing that 1 (k) = 2(1 ? k?1 1 ). Regarding 2 (k), since da(h);a(j ) ca for all j , it follows easily that 2 (k) k ? 1 for k 2. Let us now derive an estimate for 3(k). To do this, we need to refer to the cost of the shortest Hamiltonian tour through a cluster, and to its corresponding 3 (k) function. tour Thus let ctour a refer to these shortest tour costs and let 3 (k) refer to the corresponding function (where ca and 3 (k) are still de ned with respect to the shortest Hamiltonian path costs). Since the cost of a shortest Hamiltonian tour is smaller or equal to twice the length of a path we derive: X da(r)a(s) tour (k) ctour 3 a r<s
2 (k) ca for all a 2 A: tour 3
Table 1 gives:
(k) 2 (k) = 41 k if k even, and 1 (k + 1)(k ? 1) if k odd. 4 3
tour 3
2
13
Now, applying Theorem 3.1 and Theorem 3.2 to the estimates derived here for 1 (k),
(k) and (k) results in the bounds for the Single- and Multiple-Hub heuristic. 2
3
In order to show that these bounds are tight for both the Single-Hub and Multiple-Hub heuristic if k = 3, consider the following two instances of the 3-index assignment problem with path costs, depicted in Figures 2 and 3 hereunder (cf. with Bandelt et al., 1994).
i =i =i ?@ 1
?
? j =k 2
?? ?
2
? @
j =k 3
3
3
@@ @
@@
j =k 1
1
2
Figure 2: Worst-case instance for the Single-Hub heuristic for k = 3 (path costs).
i =j =k ?@ 1
1
1
? @ ? @@ ? @@ ?? @ ?
j =k 3
2
i =j 2
2
i =k 3
3
Figure 3: Worst-case instance for the Multiple-Hub heuristic for k = 3 (path costs). A drawn edge in Figure 2 or 3 indicates a distance of 1; any other distance is equal to 2. Further, the demand associated to each node equals 1. Let A1 = fi1 ; i2 ; i3 g, A2 = fj1 ; j2 ; j3 g, A3 = fk1 ; k2 ; k3 g in both gures. Now, consider the instance for the Single-Hub heuristic, depicted in Figure 2. The optimal solution is found by setting xa = 1 for a 2 f(i1 ; j2 ; k3 ), (i2 ; j3 ; k1 ), (i3 ; j1 ; k2 )g, and xa = 0 otherwise. This solution has a total cost of 3. The Single-Hub heuristic with h = 1 may nd as optimal solutions to the 2 two-index assignment problems de ned by A1 and A2 , and A1 and A3 , yiAj A = 1 and yiAkA = 1 for ` = 1; 2; 3, and 0 otherwise. Next, in Step 2 of the Single-Hub heuristic, we nd the solution za = 1 for a = (il ; jl ; kl ), l = 1; 2; 3, and za = 0 otherwise, which has total cost equal to 6, thus achieving the desired ratio for k = 3. In fact, it is not dicult to generalize this instance in such a way that the Single-Hub heuristic produces solutions bounded by (2k ? 4) OPT , showing that the bound proven here is tight for any k 3, (see Bandelt et al., 1994). Let us now consider the Multiple-Hub heuristic and the corresponding instance depicted in Figure 3. The symmetry of this instance implies that we may restrict ourselves to investigating the performance of the Single-Hub heuristic for this instance. The optimal solution is found by setting xa = 1 for a 2 f(i1 ; j3 ; k2 ), (i2 ; j2 ; k1 ), (i3 ; j1 ; k3 ), g and xa = 0 otherwise. 1
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This solution has a total cost of 3. Now, consider the Single-Hub heuristic with h = 1. Optimal solutions to the 2 two-index assignment problems de ned by A1 and A2 , and A1 and A3 , are given by respectively yiAj A = 1 and yiAkA = 1 for ` = 1; 2; 3, and 0 otherwise. Next, in Step 2 of the Single-Hub heuristic, we nd the solution za = 1 for a = (il ; jl ; kl ), l = 1; 2; 3, and za = 0 otherwise, which has total cost equal to 4, thus achieving the desired ratio for k = 3. 2 1
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Acknowledgements
The second author wishes to acknowledge the support of the Landelijk Netwerk Mathematische Besliskunde.
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