a solvable routing problem

8/3/2019 A Solvable Routing Problem

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A Solvable Routing ProblemE. N.GilbertA T&T Bell Laboratories, Murray Hill, New Jersey 07974

Stations 1,2, . . . ,n are interconnected; b,, channels join stations i and j . Channels may begrouped together in cables to reduce cost. A routing problem requires a channel layout (ornetwork of cables) that has minimum cost. The cost of a cable is taken to be independent ofits length but a functionf(k) of the cable size k. That kind of cost is unusual in practice, butmight be appropriate if the cable is actually a satellite link. Requiringf(k) to be concavegives a discount for large cables. That im poses a num ber of special properties on the minimizingnetwork. An extra assumption b,, = b , a constant for all i, j , produces a solvable problem.Depending on the shape off # ), the solution network is either a complete graph or a particularkind of tree.

1. INTRODUCTIONRouting problems arise in determining a customers bill for renting a large com-munications network. The bill is not based on the actual circuits supplied; these maybe roundabout and may change from time to time as other customers change theirdeman ds. Instead, the bill is obtained by applying some simple formula to an idealnetwork, that would provide all the required facilities at minimum cost. Finding thisminimizing network is a routing problem. It can be so difficult that only computersearches or heuristic methods apply [4]. For instance, a simplified kind of networkbilling leads to the Steiner minimal tree problem [2,3,5], problem known to be NP-complete [11 . One simplified billing problem, requiring a minimal spanning tree, canbe solved by an easy algorithm [6].This paper gives another routing problem that alsohas a direc t solution. The new routing problem is an oversimplification of currentbilling procedure s. However it may supply useful insight because it retains a commonfeature of most billing formu las, a discount for grouping many channels together intoa comm on cable.2. THE PROBLEM

There are n stations. A prescribed number b, of two-way chan nels join each pair( i , j ) of stations. The b, channels need not go directly from i to j. When routedNETWORKS, Vol. 19 (1989) 587-5940 1989 John Wiley & Sons, Inc. CCC 0028-3045/89/050587-08$04.00


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588 GILBERT

b. C.

Im 3

0.

e. f.>(D Ch.

FIG. 1 . Eight networks for 4 stations, one channel for each pair of stations.through other stations, these chan nels combine with other channels into large cables.Ordinarily the cost of a cable containing k channels is less than k times the cost of asingle channel. Then routing channels indirectly into large cables may be economical.Each way of com bining channels into cables determines a distinct network. Figure 1shows some networks with n = 4 and b ij = 1 for all (i, j ) . Each line represents acable. The numb er labeling a line is the size of the cable, i.e ., the number of channelsit contains. Some chan nels in a cable entering point P may not terminate at P but mayleave P via other cables. Figure 1 does not specify how channels of different cablesinterconnect at each point but it should be clear that cables of the sizes shown canprovide the required channels. Thus, in Figure lb the channel ( B , D ) an g o throughpoint C while the channel ( A , C ) goes through D. teiner points (S, S', in Figs. Ifand h) are extra points (not stations) where cables meet and exchange channels, n oneterminating. In some routing problems Steiner points can help to reduce cost [2,3,5].Ordinarily the cost of a cable would depend both on its size k and on its length.Here the cable cost will be a functionf(k) of size only. This cable cost might beappropriate if terminal equipment at the ends of the cable costs much more than thecable itself (e.g. . in satellite system s). If Steine r points are allowed , and have differen tterminal equipment from stations, it may not be appropriate to assume that cable costdepends only on cable size. U nder that assumption, Theorem 2 will show that Steinerpoints are uneconom ical. Ignoring cable length as a co st factor is the most unrealisticassumption that will be m ade. Other assump tions will be


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SOLVABLE ROUTING PROBLEM 589(iii) f ( k ) is an increasing (or at least non-decreasing) function of k,(iv) f ( k ) s concave.

Concavity off@)means that the m arginal costf(k + 1) - ( k ) of adding an extrachannel to a cable is a non-increasing function of the cable size k. Thus (iv) allows adiscount for using large cables. A typical concave function might bef(k) = a + (3kfor k > 0. There OL is a positive first cost of buying a right of way and digging atrench; p is a small positive constant incremental cost for each channel installed. Ifcables came only in one fixed size y, f ( k ) might increase stepwise at m ultiples of y;but that f ( k ) would not be concave.In a given network, let q ( k ) denote the number of cables of size k. The cost of thenetwork will be the sum of the costs of its cables,The problem is to find a network with m inimum C. he assumptions made above willensure that the minimizing network has some special properties (Theorems 1, 2, 3,and 4 below) and a further assumption about the form of bij will ensure a simplesolution (Theorems 5 and 6).Some of the networks in Figure 1 cannot possibly be solutions. Network f costsmore than network c, having two extra cables of size 3. Network h has a cable ofsize 3 in addition to those of b or g . The increasing property of@ ) eliminates networkc because, compared with d , c replaces a cable of size 3 by one of size 4. Comparisonsbetween the remaining networks a, b, d , e, and g must use concavity or specificnumerical details about f(k).3. BASIC PROPERTIES

The assumptions in Section 2 impose special properties on the solution.Theorem 1. There is a minimizing network in which no two cab les share the samepair of endpoints.

Proof. If two cables, of sizes r and s, have the same endpoints, combine theminto a single cable of size r + s. The net saving in cost isf( r) + f(s) - f(r + s) 2 0,Theorem 2.

Proof.

the inequality following from (iv). rnA minimizing network has no Steiner points.

If S is a Steiner point, let P be one of the points connected to S by a cable.Without breaking any channel, move S o coincide with P. his motion changes cablelengths but not their costs. But now the cable PS, f length zero, can be removed tosave its cost.For example, moving S to A and then S' to A improves Figure If to Figure Id.Removing a Steiner point may produce a network with cables in parallel (as in m oving

S to A in Fig. lh), in which case Theorem 1 will give a further saving (to Fig. Id).Next consider all the channels that pass through a given pair of stations (ij).Thesechannels include the b, channels that connect i and j . Other channels may visit i and


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IFIG. 2. Network with split routing.

j without terminating. Do all these channels follow the same route between i and j?If not, the network will be said to use split routing between i and j. Figure 2 illustratessplit routing between A and C. Unlike Figure 1, Figure 2 shows each channel as aline in order to show the channel terminations explicitly. The channels that visit bothA and C are AC (routed via B) and AB (routed via D and C).Theorem 3. There is a minimizing network that does not employ split routing.

The proof appears essentially in [3]. If some channe ls visiting both points i. j followdifferent routes U and V, two different modifications are possible. Either reroute thechannels of U to take route V or vice versa. The proof uses concavity of f ( k ) todemonstrate that the average of the changes in network cost from these two modifi-As a corollary to Theorem 3, the b, channels joining i and j can all follow the sameroute; otherwise split routing would occur between i and j . Also if there is a cablebetween i and j then the b, channels ( i , j ) may be sent directly over this cable toavoid split routing.In particular, if b, > 0 , here need be no cable i j containing only channels that donot terminate at i. If b, = 0 , a cable i j with no channel destined for i or for j mayactually be needed. Figure 3 illustrates this by an example with n = 8, bM = bAc

= bDE = bDF = bm = 1, and bij = 0 otherwise. Supposef(k) = CY + k with CYso large that the minimizing network must use the few est cables (a tree). Large cableswill be needed from A to B , C , D and from H to E , F , G . The remaining 9 channels canbe routed together via cable of size 9 connecting one of A,B,C ,D with one of E , F . G , H(Fig. 3a). The cheapest network (Fig. 3b) places the cable betweenA and H (bM = 0);but none of the channels in this cable terminate at A or at H.Although costs often decrease as channels are grouped into larger cables, there isa point of diminishing return.Theorem4. Let Bj denote the total number of channels that terminate at j, B j = Zibij.There is a minimizing network in which, for all j, no cable to station j has size more

Consider any cable ij , where Bj S B i . If this cable has size B > Bj, removeit. Move the endpoint of every other cable at j from j to i (at no change in cost). That

cations is not positive; then both changes cannot increase the cost.

= bm = 100, bEH = bFH = bGH = lo(), bBE = bBF = bBG = bCE = bcF = bCG

than Bj.Proof.


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SOLVABLE ROUTING PROBLEM 591

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a.

103 103FH,,9 " 03

103 103

4. SOLVABLE PROBLEMFurther results may be obtained when bijhas special form. In this section each point

i contains a population of p i > 0 individuals and each pair of individuals requires aseparate channel; then b , = pip, .

Regarding stations and cables as vertices and lines of a graph, and ignoring cablesizes, one may speak of networks that are trees (Figs. lc,d,f), complete graphs (Fig.la), etc. One may also distinguish between end vertices (stations having one cable)and infernal vertices (stations with two or more cables). The next theorem applies tot r e e s only.


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592 GILBERTTheorem 5. Let b,, = p p J for all ( i , j ) and let n 5 3. Among trees, one with leastcost has only one internal vertex, a point I with largest population. The cost of theminimizing tree is Z.f(B,), sum med for i f 1.

A tree with 2 or more internal vertices contains a pair i, jjo in ed by a cable.If this cable were broken, the tree w ould separate into two parts, say with populationsPI and PJ. Then the cable ij contains B = PQ, channels. Suppose P , is the smallerof PI andP, , i .e . , P, 6 P/2 , whereP = PI + P, = pI + p2 + * . + pn s thetotal population. Now

Proof.

B, = p,(P - p J )C P,(P - PI) = Bfollows becausep, < P, and the functionp(P - p) is strictly increasing in 0 6 p < P/2.Then Theorem 4 shows how to save cost by making j an end vertex. Moreover, thismodification again leaves a tree.Continuing in this way, one eventually obtains onlyone internal vertex I , which is connected by cables of sizes B, to the o ther vertices i.The cost of this tree is the sum Zf(B,) for i f I , which is minimized by taking B, aslargeaspossible. BecauseBI - B, = (p, - p,)(P - pI - p,), astationIwithlargest

The minimizing tree is star-shaped, w ith all lines radiating from I . Of course othershapes (Fig. 3) are possible if b, is not of the form plpJ.While Theorem 5 gives a least-cost tree, networks that are not trees may be evencheaper. The next theorem gives a complete solution when the populations all havethe same size, i.e., b,, = b, a constant for all (i,j). Because of Theorem 3 the bchannels ( i j ) can all take the same route. Each of the (d2) roups of b channels canbe regarded as a kind of superchannel and the problem reduces to finding a networkgiving one superchannel to each pair of stations. Then it suffices to solve the caseb = 1 .Theorem 6. Let b,, = b = 1 for all i j . If f(n - l ) / f ( l ) G n/2 , a tree of cost( n - l)f(n - 1 ) is a minimizing network. Otherwise a complete graph of cost (n/2)f(l) is a minimizing network.

Let 4 ( k ) denote the number of cables of size k as in (1 ). Define the lengthof the network to be

B, also has largest population pI. w

Proof.

L = Z 4 ( k ) k . (2)kIn ( 1 ) and (2) the sum mation extends over 1 G k G n - 1 because Theorem 4 makescables of size n or more unnecessary. In that range, k can be expressed as an averageof 1 and n - 1 with weights ( n - 1 - k ) / ( n - 2) and ( k - l ) / (n - 2)

k = { ( n - 1 - k)l + ( k - 1 ) ( n - l)}/(n - 2 )Then, becausef(k) is concave,

( n - 2)f(k) 3 ( n - 1 - kM 1 ) + ( k - l)f(n - 1).This inequality converts (1) into( n - 2) C 3 {(n - 1)Al) - A n - 1 )) & (k ) + Mn - 1 ) - A1))L. (3 )


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SOLVABLE ROUTING PROBLEM 593In (3) the sum, which extends over k = 1 , . . . , n - 1 , is the total number ofcables

( q ( 0 ) is the number of station pairs not joined by a cable). Moreover each pair ( j j )with no cable ij has a channel routed via two or more cables. Hence

Now (3) simplifies toc 3 (;) f l l ) - (nf(1) - 2fln - l)}q(O)/(n - 2). (4)

U f ( n - l)/f (1) 3 n/2 the coefficient of q(0) in (4) is negative or zero; the right-hand side of (4) is minimized by taking q(0 ) = 0. The bound (4) is achieved by acomplete graph, providing each pair (ij)with a direct channel in a cable of size 1 atminimum cost C = (n/2)f(l).Iff(n - l)/f(l) d n/2 he right-hand s ide of (4) is minimized by taking q(0) large,i.e., using a graph with as few lines as possible. The connected graphs with fewestlines are the trees, having n - 1 lines and henceq(0) = (;) - n + 1 = (n - l ) (n - 2)/2.

For trees (4) becomesc 3 ( n - 1)fln - 1) . ( 5 )

The bound ( 5 ) is achieved if the channels are routed into n - 1 cables of size n - 1to a single vertex, as in Theorem 5 .With more general p i the m inimizing network may be neither a complete graph nor

a tree. One such example has n = 6 stations with populations 1, 1, 1, 3, 3, and 3.Supposef(k) = k fo rk = 1,2, . . . , 11 andf(k) = 11 for k 3 11. The costs of thecomplete graph and cheapest tree (Theorem 5 ) are 57 and 55. However, the readerwill easily find other cheaper networks.REFERENCES[ l ] M. R. Garey, R. L. Graham, nd D. S. ohnson, The complexity of computing Steiner[2] E. N. Gilbert, Minimum cost com munications networks. BSTJ 46 (1967) 2209-2227.[3] E. N . Gilbert and H. . Pollak, Steiner minimal trees. SIAM J. Appl. Math 16 (1968) 1-[4] S . Lin,Heuristic programming as an aid to network design. Networks 5 (1975) 33-43 .

minimal trees. SIAM J. Appl . Math. 32 (1977) 835-859 .

29 .


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594 GILBERT[S] 2.A . Melzak, On the problem of Steiner. C a d . M a r k Bull. 4 (1961) 143-148.[a ] R . C. Prim, Shortest connecting networks and some generalizations. BSTJ 36 (1957) 1389-

1401.

Received August 1987Accepted August 1988

a solvable routing problem

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