Hudec, Oto

Zimmermann, Karel

Institute of Mathematics AS CR, v. v. i.

Acta Universitatis Carolinae. Mathematica et PhysicaOto Hudec; Karel ZimmermannA service points location problem with Min-Max distance optimality criterionActa Universitatis Carolinae. Mathematica et Physica, Vol. 34 (1993), No. 1, 105--112Persistent URL: http://dml.cz/dmlcz/142655Terms of use:© Univerzita Karlova v Praze, 1993Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document mustcontain these Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://project.dml.cz1993 ACTA UNrVERSITATIS CAROLINAE - MATHEMATICA ET PHYSICA VOL. 34. NO. 1 A Service Points Location Problem with Min-Max Distance Optimality Criterion OTO HUDEC*), KAREL ZIMMERMANN**) TU Kosice, MFF UK Praha Received 13 December 1991 1. In troduct ion The problem considered in this paper generalizes in a certain sense the problem of finding the absolute centre of a graph, which was studied e.g. in [1], [2], [3], [5], [6]. The problem studied in this paper can, similarly to the problem of the absolute centre of a graph, arise in e.g. the selection of sites for emergency service centres and involves the global minimization of certain piecewise-linear non-convex con-tinuous function. We shall bring a motivating example. Let us suppose that m points Yl9..., Ym are given, which have to be served (or supplied) from n points Sl9..., Sn. Each service point 57 is to be placed on a given segment AJBJ\ the points Aj9 Bj are given (for all j = 1,..., ri). The distances | YtAj\ = aij9 \ YtBj\ = a'ij9 \AJBJ\ = dj are known non-negative numbers (i = 1,..., m9 j = 1,..., ri). In planning emergency transport systems, it is natural to evaluate any given proposal by the worst service it provides. The proposal for which the worst service is as good as possible is then accepted. We shall use this idea in the process of formulating our optimization problem. Let Xj = \AJSJ\ for j = 1,..., n9 (i.e. Xj is the distance between Aj and Sj9 which has to be chosen). It is then obviously x} e [0, dj\. The function rv(Xj) ~ m i n (<*v + Xj> ah + dj - Xj) expresses thus the minimal distance we can choose if we serve the point Yt from a chosen point Sj e A}Bj with \AjS}\ = Xj (compare Figure 1). The value di(xl9...9 xn) = dt(x) = max rv(xj) 1 £ j * n *) Technical University, Department of Mathematics, Švermova 9, 042 00 Košice, Czechoslovakia **) Faculty of Mathematics and Physics, Charles University, 118 00 Praha 1, Malostranské nám. 25, Czechoslovakia 105 dj-xj FIG. 1. expresses thus the worst case, which can occur for the given point Yt (if S/s are chosen). Therefore we try to find xj9 j = 1,..., n in such a way that it solves the optimization problem subject to / ( * . , . . . , xn) = /(*) = max d,{x) -+ пàӣҳ l S . S m di(x) = max min (a l ; + Xj, bi} — Xj) i=l,...,m 0 = Xj=dj J=í,...,n,l (PO) where btj = a- + dj. The algorithm we are going to suggest looks for minima of certain piecewise-linear non-convex functions which requires 0(mn log m) time, m being the number of supplied points and n number of located service points. 2. Theore t i ca l background of the a lgor i thm We shall consider the following optimization problem: f(xl9...9xn) = max max r^xj) min subject to (PІ) ft < x < H Пj - Лj _ Пj j = 1, . . ., П9 where rtj(xj) = min (atj + xj9 btj — x}) for all i, j (aij9 bij9 hj9 H; are given real numbers). Note that in the objective the order of the operators is interchanged, what does not have effect upon the solution, as in both (PO) and (PI) the maximum distance between the service points and the supplied points is minimized. The problem (PI) is slightly more general than (PO), since aij9 bij9 hj9 Hj are arbitrary real numbers. Denote xf = (btj — ai})/2 for all i, j we have 106 rtJ(xj) = aif + Xj ^ bij - x} for *i ^ xf r^Xj) = bij - xj < axi + x} for Xy > xf Thus, rl7(jc;) is a piecewise-linear, concave function with slopes ± \ attaining its maximum in xfp. For Xj e [/*,, Hj], r/,(jc;) is a function *ti Xj either linear and attaining its maximum at one of the points hf or Hj9 o r piecewise-linear with a maximum at the point jr .̂ Along each segment [hj9 Hj], / = ! , . . . , « , the function s,{xj) = max r^Xj) is the piecewise-linear upper envelope of the functions r,7(*/), where rtj(Xj) are the one- or two-piece linear concave functions in Xj. A typ*cal plot of Sj(xf) along a segment is shown in the Figure 2. FIG. 2 We will solve the problem (PI) by taking the upper envelope sJ(Xj) of the functions r,,(*/), i = 1-..., w, and inspecting it to find a minimum point jt,. The optimum value then may be found by choosing the largest of the n values r//(Jty). The latter formulation suggests that the optimum value to the problem (PI) will be found by solving a sequence of the n problems (P2), where (P2) is defined for / — 1,..., n as follows: 107 sy(jty) = max r/y(jty) -• min l*.-*m J /p2) subject to hj = Jty = Hj. It is clear that any point fy where the function Sj(xf) attains its minimum Sj = s;(iy) is either hy or H}, or iy is a break point where two functions r/y(jty) and rkj(xj) intersect, 1 = i, k = m. Furthermore the slopes of such r/y(jty), rfcy(jty) have the opposite signs. There are at most m(m — l ) /2 break points and to determine Xj is possible to examine at most m(m — l ) /2 + 2 points on [h}, Hj] and select the best among them. 3 . An algorithm for the problem ( P 2 ) In this section, we present an algorithm for solving the problem (P2) for fixed /, j e {1,.. . , n}. The approach of Hakimi [2] or Cunninghame-Green [1] of finding the absolute centre of a network might be useful to solve the problem (P2). This section presents a simple new technique for determining the minimum value of the upper envelope of the piecewise-linear functions described in the Section 2. A numerical example and the complexity of an algorithm are given — this particular problem is solved in the same time, as in [1]. Let xf, 1 = i = m, be the point of the maximum of r/y(xy) and let us define cy as the minimum of the m values r^xf), i.e., Cj = min rij(xf). Let ij be the minimum value of the function sy(Jty). For any function r/y(jty), 1 = / = m, let us denote */ = CJ - a n tj^bij- Cj. Lemma 1. r/y(jtj) = r/y(JPj) = cy for i = 1,..., m. Proof. It is under our assumptions Cj = min r^xf) = min (atj + xf) = min (btj — xf) =* min (atJ + b/y)/2 . It follows from the definition of the functions r/y(jty): './(•*/) = m i n K + 4 bu " */) = m i n K + CJ - au> ba - Cj + aa) • Since cy = (atj + 6/y)/2, Z>/y - cy + n /y = 6 / y + fl/y - (fl/y + 6/y)/2 = (tf/y + 6/y)/2 = cy 108 and we have rtj(Xj) = Cj. Similarly, r/y(*j) = min (atj + x), btj - *j) = min (a/y + bif - cy, b/y - b/y + cy) = c y . Q.E.D. Remark 1. If r,y(*}°) = cy for some i e {1,..., m} then xj = Jtj. Let jt be an ordering of the set {1,..., m] such that for i = 1,..., m — 1 either x71^ < x^i+1) or Jtf'"> = xf+1) and Jcf° ^ Jcf'+1). According to the above ordering we can easily find an index which "takes an active part in determining the value 8y(*y)". We can suppose, without loss of generality, that no pair of indices (/, k), 1 ^ i, k S m exists that JcJ ^ x) and x) î  x). In that case the piecewise-linear function rkj(xj) corresponding to the index k does not have effect upon the solution and may be omitted, a) the case cy ^ fy: If we suppose cy ^ 8y then fy may be found as follows: Remark 2. (compare Fig. 3) The inequality cy ^ 8y holds if either Sj(hj) S c} or SJ(HJ) S cj OГ xf> Ú xfi+l) for some i e {1, ...,m - 1}. Along the interval [J^°, xfi+1)] the minimum of the function SJ(XJ) is attained at the point (xfi+1) + xf^/2 because the functions r^Xj), r4/+1)y(jcy) are there 109 linear with slopes — 1, + 1. The minimum solution value is then the minimum of the values sX(*f+1) + xf*)/2) for which xf* ^ xfi+1), s,(/i,) and s,<H,). The calculation of 4 */, for i: *- 1,..., m, requires O(m) time. Then the or-dering of the set {1, •••> m] needs 0(m\o%m) time. Finally, determining sf((xfi+1) + xf*)/2) for i = 1,..., m, Sj(hj), SJ(HJ) and minimum of them can be done in O(m) time. Hence, we can summarize: Theorem 1. / / c, ^ sj9 the minimum solution value to the problem (P2) can be found in 0(m log m) time. b) the case c, < s,: In the case c, < s,, i.e. s,(/i,) > c,, s,(H,) > c} and xf * > Jcf+1) for 1 .S i ^ m — 1, we will use a different approach to find the minimum of s,{jc,) along the interval [h}, / / , ] . A point of the minimum is either at the end points /i, or Hj, or at a point i, where two functions rJt(i)j(Xj), r-^+i^C*/) intersect, l S / ^ w - 1 . Notice, we have supposed that in the case Jtj ^ x) and JFj ̂  x) the index A: has been omitted. Then only the following case can occur: Lemma 2. xf* < xfi+1) and xf* < xfi+1) j b r l S / S « - l . Proof. According to the ordering n, either xf* < xfi+1) or xf * = jf*1""̂  and JP*(0 ^ Jp^/+1). If Jĉ ° < xfi+1) and xf* ^ xf/+1) then the index n(i + 1) might be omitted. Similarly, if xf* = xfi+1) and Jc,̂ 0 ^ xfi+1), then the index jr(i) might be omitted. This completes the proof. Q.E.D. After omitting of the needless indices we shall find the indices jr(s), n(t) such that si(hi) = r*s)i(h]) and Sj(H;) - r^f(Hj) . Thus, the indices JI(S), Ji(t) correspond to the functions active in sfxj) at the end points of the interval [hj9 //,-]. Then we inspect only the break points of the pairs of the functions [rn(s)j(X])> r7t(s+l)j(Xj)]i [rjt(s+l)j(Xj)> rn(s+2)j(Xj)]> •••- [r7t(t-l)j(Xj)> r*(t)j(Xj)] and the points hj9 Hj for finding the minimum s, of SJ(XJ) along the interval [/*,, H,]. It is obvious that the break point of any pair of the function [r^i)j(x]), r^i+1)J(Xj)] is (b^i)i ~ a<i+l)j)/2 for i = 1,..., m . Thus, in the case c, < s, the algorithm requires 0(m log m) time for ordering of the set {1,..., m], further 0(m) time for calculation of Jtj, x), i= 1,..., m. Also 0(m) time is required for omitting the needless indices, calculation of the above break points and selecting the minimum. Then the next result holds: Theorem 2. The minimum solution value to the problem (P2) can be found in 0(m log m) time. 110 We can use now the Theorems 1, 2 to construct an algorithm for solving (P2). The algorithm AP2. (optimal solution of (P2)). (1) determine order JI of the set {1,..., m}\ (2) omit the "needless" indices; (3) compute xf9 JtJ, Jj for i = 1,..., m\ (4) if Cj < Sj then go to 6; (5) determine the minimum value sf as in the Theorem 1, go to 7; (6) determine the minimum value Sj as in the Theorem 2; (7) end; It remains to give an algorithm for solving (PI): Theorem 3. The problem (PI) can be solved in 0(nm log rri) time. Proof. As we have mentioned in the Section 2, the optimum value s to the problem (PI) can be found as follows: Algorithm API. (optimal solution of (PI)). (1) compute xj9 Sj for j = 1,..., n by the algorithm AP2; (2) find the maximum s of s/s, / = 1,..., n\ (3) end; Trivially, we use the algorithm API n times what requires 0(nm log rri) time. Q.E.D. Now we consider the following example. Suppose the case of two intervals and six points which have to be served. The Table 1 gives the entries hj9 Hj9 aij9 bij9 for / = 1,...,6, / = 1,2. i TABLE 1 1 2 3 4 5 6 Й, = 4 a,i -3 3 6 -7 -8 -7 Яt = 20 bn 18 17 16 27 35 22 h2 3 "a 10 14 11 2 0 6 H2= 15 ba 10 11 20 24 28 23 The Table 2 gives the computed values j ^ 9 j?j9 $j9 ct for all i9 /'. Further according to the relation between c; and Sj is the minimum sx computed by the step (5) and s2 by the step (6) of the algorithm API. Note that si is realized at the inner point xx = 12.5 of the interval [4, 20] and ^ is realized at the inner point J^ = 0 as well as at the end point h^ = —3 of the interval [ — 3,15]. The optimum value to the problem (PI) is then s = max (7.5,10) = 10. ш i TABLE 1 1 2 3 4 5 6 c. = 7.5 xf 10.5 7 5 17 21.5 14.5 x, = 12.5 *í 10.5 4.5 1.5 14.5 15.5 7.5 /, = 5.5 *í 10.5 9.5 8.5 19.5 27.5 7.5 c2 = 10 4° 0 -1 .5 4.5 11 14 8.5 x2 = - 3 , 0 4 0 - 4 • - 1 8 10 4 s2 = 11 4 0 1 10 14 18 13 The optimum value: s = max (si, s2) = 11. Location: (xu x2) = (12.5, —3) or (12.5,0). R e f e r e n c e s [1] CUNNINGHAME - GREEN R. A: The absolute centre of a graph. Discгete Appl. Math. 7 (1984), 275-293. [2] HAKIMI S. L.: Optimum location of switching centers and the absolute centers and medians of a graph. Oper. Res. 12 (1964), 450-459. [3] HANDLER G. Y., MIRCHANDANI P. B.: Location on networks: Theory and algorithms. MГГ Press (1979). [4] HOOKER J. N., GARFINKEL R. S., CHEN C K.: Finite dominating sets foг network location problems. Oper. Res. 39, 1 (1991), 100-118. [5] KARIV O., HAKIMI S. L.: An algorithmic approach to network location problems. I. SIAM J. Appl. Math. 37, 3, (1979), 513-538. [6] MINIEKA E.: A polynomial time algorithm foг finding the absolute center of a netwoгk. Networks 11 (1981), 351-355. 112 

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A service points location problem with Min-Max distance optimality criterion

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A service points location problem with Min-Max distance optimality criterion

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