Robust Positioning of Service Units - CiteSeerX

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

STOCHASTIC MODELS Vol. 19, No. 1, pp. 125–147, 2003 1 2 3 4 5 6 7 8

Robust Positioning of Service Units

9 10 11

J. Puerto1,* and A. M. Rodrı´guez-Chı´a2,*

12 13 14 15 16

1

Dept. Estadı´stica e Investigacio´n Operativa, Universidad de Sevilla, Fac. de Matema´ticas, Sevilla, Spain 2 Dept. Estadı´stica e Investigacio´n Operativa, Universidad de Ca´diz, Fac. de Ciencias del Mar, Polı´gono Rı´o San Pedro, Ca´diz, Spain

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

ABSTRACT In this paper, we address the problem of locating mobile service units to cover random incidents. The model does not assume complete knowledge of the probability distribution of the location of the incident to be covered. Instead, only the mean value of that distribution is known. We propose the minimization of the maximum expected response time as an effectiveness measure for the model. Thus, the solution obtained is robust with respect to any probability distribution. The cases of one and two service units under the nearest allocation rule are studied in the paper. For both problems, the optimal solutions are shown to be degenerate distributions for the servers. Key Words:

Location theory; Stochastic problems.

32 33 34

1.

INTRODUCTION

35 36 37 38

In distribution systems and in continuous location models, a common problem is to find the optimal placement of one or more servers minimizing the distances to a given set of points. All the models considered so far in the literature assume that the positions of

39 40 41 42 43

*Correspondence: J. Puerto, Dept. Estadı´stica e Investigacio´n Operativa, Universidad de Sevilla, Fac. de Matema´ticas, C/Tarfia s/n, 41012 Sevilla, Spain; E-mail: [email protected]. A. M. Rodrı´guezChı´a, Dept. Estadı´stica e Investigacio´n Operativa, Universidad de Ca´diz, Fac. de Ciencias del Mar, Polı´gono Rı´o San Pedro, 11510 Puerto Real, Ca´diz, Spain; E-mail: [email protected].

44 45 46

125 DOI: 10.1081/STM-120018142 Copyright q 2003 by Marcel Dekker, Inc.

1532-6349 (Print); 1532-4214 (Online) www.dekker.com

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Puerto and Rodrı´guez-Chı´a

126 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77

these points are either deterministic or distributed according to a known probability distribution on the family of Borel sets in Rn (see for instance Anderson and Fontenot,[1] Carrizosa, Mun˜oz-Ma´rquez and Puerto,[2] Larson and Odoni,[5] Levine,[6] De Palma, Liu and Thisse,[7] among others). However, it is easy to find situations in the real-world where the hypothesis of complete knowledge of this probability distribution is unrealistic. In this paper, we propose a more general model where only the mean value of this distribution is known. This assumption is not really restrictive because we can obtain good estimates of the unknown mean value by sampling. Although more information can be obtained from the sample, our model only needs the estimation of the mean value, which is a very wellsolved problem in mathematical statistics. A real-world application of such models is, for instance, the problem of locating a read/write head of a computer hard-disk to easily access the stored data. Similarly, our framework includes the problem of positioning police-cars that must cover incidents where the law is being broken, and positioning idle elevators to minimize response time(see Vickson, Gerchak and Rotem[10] or Smith[9] for a different analyses assuming that the distribution of the data is known). Indeed, in these cases, usually the distribution of the places where the law will be broken, the data are stored, or the elevator is needed is not known. Nevertheless, it would be less restrictive to assume that either we know the mean value for these distributions or we may estimate it by means of an empirical study. When the probability distribution of the position of the incident is unknown, the classical minimization of the expected distances is not possible. Therefore, alternative approaches have to be considered. In this paper, we propose a robust alternative consisting of minimizing the maximum expected distance within the whole family of probability measures which model the incident (see Gallego[4] and Puerto and Ferna´ndez[8] for similar analyses applied to different problems in Operations Research). Let F(l) and G(m) be the families of random variables (r.v.) (given by their cumulative distribution functions (c.d.f.) defined on the n-dimensional hypercube [0,1]n with mean values l [ Rn for F(l) and m [ Rn for G(m), that is, Z n F ðlÞ ¼ {X : r:v: on ½0; 1 with c:d:f: F X ; x dF X ðxÞ ¼ l}; ½0;1n

78 79 80 81 82 83 84 85 86 87 88 89 90 91 92

n

GðmÞ ¼ {A : r:v: on ½0; 1 with c:d:f: GA ;

Z ½0;1n

a dGA ðaÞ ¼ m}:

Define F :¼ < n F ðlÞ: l[½0;1

The families F and G(m) are the sets of random variables which model the position of the server and the incident, respectively. It is worth noting that we have defined these random variables in the n-dimensional hypercube [0,1]n, but they can be extended to any hyperrectangle by a linear transformation. As previously mentioned, some authors have studied the problem of minimizing the expected distance to the random incident, i.e., Z Z min dðx; aÞ dGA ðaÞ dF X ðxÞ; X[F

½0;1n

½0;1n

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Robust Positioning of Service Units 93 94 95 96 97 98 99 100 101 102 103

where d is a measure of distance, X is a r.v. with c.d.f. FX, representing the position of the server, and A is a r.v. with c.d.f. GA, representing the position of the incident. Our model does not assume any a priori knowledge about the probability distribution of the incident apart from its mean value. That is to say, we have almost complete uncertainty about where the incident will take place, and we search for the policy that an emergency unit, X, has to follow to minimize the maximum expected distance to any random incident. Therefore, the problem is Z Z min max kx 2 ak1 dGA ðaÞ dF X ðxÞ; ð1Þ X[F A[GðmÞ

½0;1n

½0;1n

where k·k1 is the l1-norm in Rn , so that, for x ¼ (x1,. . .,xn) [ Rn we have that

104 105 106

127

kxk1 ¼

n X

jxi j:

i¼1

107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131

The readers should note that there are essentially two kinds of factors that influence the formulation of the problem: 1) the dimension n of the space where the incidents occur; and 2) the number of service units to be located. It is also worth noting that this problem formulation can be used to model the above mentioned real-world situations because: a) we do not need to know the distribution of the incident; and b) the read/write head only admits displacements following the directions of the coordinate axes; and both the highway and the trajectory of the elevator can be considered like line segments where displacements are linear. Thus, the l1-norm is an appropriate measure of distance. Finally, the formulation (1) gives us a new interpretation of the solutions obtained in terms of statistics. As we shall show in the paper, the optimal probability distributions for our problem are degenerate random variables. Since principal points of probability distributions are those points optimizing some effectiveness measure (see Flury[3]), we can see our solutions as a generalization of the principal points, but now we are optimizing over a family of distributions with a fixed mean rather than the values of a single probability distribution. The paper is organized as follows. In Section 2, we consider the problem of locating a single facility; we first study the problem considering the service unit as a degenerate random variable and then we extend these results to the general case with any random variable. In Section 3, we consider the two-facility problem under the nearest allocation rule and we follow a scheme similar to that followed in Section 2. In Section 4, we include some concluding remarks and possible extensions to the considered model. Finally, in the Appendix, we include, for the sake of readability, several technical results that have been used in the paper.

132 133 134

2.

THE SINGLE FACILITY PROBLEM

135 136 137 138

We begin this section by considering the one-dimensional case, then we proceed to the n-dimensional single facility problem. Let F1(l) and G1(m) be, respectively, the families of random variables F(l) and G(m) in the 1-dimensional case. For ease of

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Puerto and Rodrı´guez-Chı´a

128 139 140 141 142 143 144 145

understanding, we distinguish two cases. In the first case, the server is not allowed to patrol, i.e., we model the location of the server with a degenerate random variable. In the second case, the server is allowed to patrol, which means that it is any random variable in F1. For the first case, the mathematical formulation of the problem is: Z min max jx 2 aj dGA ðaÞ: ð2Þ x[½0;1 A[G1 ðmÞ

½0;1

146 147 148 149 150 151 152

Theorem 2.1 The optimal positioning policy in the hypothesis of Problem (2) is 8 0 if m ¼ 0:5 > > < x* ¼ y for any y [ ½0; 1 if m ¼ 0:5 > > :1 if m . 0:5:

ð3Þ

153 154

158

Remark 2.1 This result states that the optimal location for a fixed service unit when only the mean value m of the distribution of the incident is known, is on an extreme point of the interval of feasible locations for the incident. Further, when m ¼ 0:5 any point on the interval is an optimal location of the server.

159 160

Proof:

155 156 157

161 162 163

By Lemma A.2 in the Appendix we have that  Z x  Z min max jx 2 aj dGA ðaÞ ¼ min max 2 GA ðaÞ da 2 x þ m :

x[½0;1 A[G1 ðmÞ

½0;1

x[½0;1 A[G1 ðmÞ

ð4Þ

0

164 165 166 167 168 169 170

Hence, we prove that the maximum in the last expression is reached at the random variable A* with the following c.d.f. 8 0 if a , 0 > > < GA* ðaÞ ¼ 1 2 m if 0 # a , 1 > > :1 if a $ 1:

171 172 173 174 175 176 177 178 179 180 181 182 183 184

Indeed, since x and m are constants for the inner maximum in the right hand side of (4), we have to prove the following inequality Z x GA ðaÞ da 2 ð1 2 mÞx # 0; ;x [ ½0; 1; ;A [ G1 ðmÞ: ð5Þ 0

R1 But, since 0 GA ðaÞ da ¼ 1 2 m (Lemma A.2) and GA(·) is a distribution function, we are under the hypotheses of Lemma A.1 which proves the inequality (5). Therefore, the minimization Problem (2) reduces to the following problem: min xð1 2 2mÞ þ m:

x[½0;1

Hence, depending on the relative values of m, we obtain that the optimal positioning x* satisfies equation (3). A

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Robust Positioning of Service Units 185 186 187 188 189 190 191

129

In the second case, the service unit is also allowed to patrol. Initially, we permit the service unit to be distributed on the interval according to a random variable with the only condition that its mean value is fixed to l. Then, we solve the case when l is not fixed. For the first case, the problem is Z Z min max jx 2 aj dGA ðaÞ dF X ðxÞ: ð6Þ X[F 1 ðlÞ A[G1 ðmÞ

½0;1

½0;1

192 193 194

Theorem 2.2 Problem (6).

Any random variable X [ F 1 ðlÞ constitutes an optimal policy for

195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210

By Lemma A.4, we have that   Z 1 Z jx 2 aj dGA ðaÞ dF X ðxÞ ¼ 2 1 2 GA ðyÞ F X ðyÞ dy 2 l 2 m:

Proof: Z ½0;1

½0;1

Therefore, using that l and m are fixed,we can solve Problem (6) by solving the equivalent problem Z 1 max min GA ðyÞ F X ðyÞ dy: X[F 1 ðlÞ A[G1 ðmÞ

213 214 215

0

0

Considering t0 :¼ t0 ðAÞ [ ð0; 1Þ such that t0 ¼ inf{t [R : GA ðtÞ $ 1 2 m} we have the following inequalities: Z t0 Z 1 I A;X ¼ F X ðyÞðGA ðyÞ 2 ð1 2 mÞÞ dy þ F X ðyÞðGA ðyÞ 2 ð1 2 mÞ dy t0

0

216

Z

217 218

t0

F X ðt0 ÞðGA ðyÞ 2 ð1 2 mÞÞ dy þ

$

220

223 224 225 226

Z

0

219 221 222

0

In order to do this, we are going to prove that the inner minimum is achieved by the random variable A* such that P(A* ¼ 0) ¼ 1-m and P(A* ¼ 1) ¼ m. Z 1 Z 1 F X ðyÞGA ðyÞ dy 2 F X ðyÞð1 2 mÞ dy $ 0: I A;X :¼

211 212

0

¼ F X ðt0 Þ

1

F X ðt0 ÞðGA ðyÞ 2 ð1 2 mÞÞ dy t0

Z



1

ðGA ðyÞ 2 ð1 2 mÞÞ dy

¼ 0;

0

by Lemma A.2. Similarly, Lema A.2 implies that Z 1 Z 1 X aP½A ¼ aP½X ¼ a ¼ min GA ðyÞF X ðyÞ dy þ F X ðyÞð1 2 mÞ dy A[G1 ðmÞ

0

a[ð0;1Þ

0

227 228 229 230

¼ ð1 2 lÞð1 2 mÞ; regardless of the choice of X [ F 1 ðlÞ; and the result follows.

A

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Puerto and Rodrı´guez-Chı´a

130 231 232 233 234 235 236 237 238

Let us consider in the following that no assumptions are made on the mean value, l, of the random variable modelling the service unit. In this situation, the problem is Z Z min max jx 2 aj dGA ðaÞ dF X ðxÞ: ð7Þ X[F 1 A[G1 ðmÞ

½0;1

½0;1

Corollary 2.1 An optimal positioning policy of Problem (7) is the random variable X* such that P½X* ¼ x*  ¼ 1 where x* was defined in (3).

239 240 241 242 243 244 245

Note that Z min max

Proof:

X[F 1 A[G1 ðmÞ

Z

½0;1

jx 2 aj dGA ðaÞ dF X ðxÞ ½0;1

min

247 248 249 250 251 252 253 254 255 256 257

jx 2 aj dGA ðaÞ dF X ðxÞ:

max

l[½0;1 X[F 1 ðlÞ A[G1 ðmÞ

246

Z

Z

¼ min

R

Let HðlÞ ¼ min

½0;1

R

max X[F 1 ðlÞ A[G1 ðmÞ ½0;1 ½0;1

½0;1

jx 2 aj dGA ðaÞ dF X ðxÞ:

For each l [ ½0; 1; by the proof of Theorem 2.2 we have that HðlÞ ¼ 2ð1 2 ð1 2 lÞð1 2 mÞÞ 2 l 2 m ¼ ð1 2 2mÞl þ m: Thus, if we look for the minimum in l we obtain 8 {0} if m , 0:5 > > < arg min HðlÞ ¼ y for any y [ ½0; 1 if m ¼ 0:5 l[½0;1 > > : {1} if m . 0:5;

258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273

and the result follows. A This corollary shows that it is optimal to park the service unit when no hypotheses are made on the distribution of the service unit and only the mean value of the incident is known. Thus, although patrolling may be good for other reasons such as crime prevention, etc., it is not necessary in order to minimize the maximum expected distance to any random incident. Finally, we also solve the n-dimensional problem. Indeed, let us consider the problem: Z Z n X min max jxi 2 ai j dGA ðaÞ dF X ðxÞ; ð8Þ X[F

A[GðmÞ

½0;1n i¼1

where m ¼ ðm1 ; . . .; mn Þ; x ¼ ðx1 ; . . .; xn Þ and a ¼ ða1 ; . . .; an Þ: Problem (8) can be written equivalently as follows: Z n Z X min max jxi 2 ai j dGAi ðai Þ dF Xi ðxi Þ X[F

A[GðmÞ

274 275 276

½0;1n

# min X[F

i¼1

n X i¼1

½0;1

½0;1

Z

Z

max

Ai [G1 ðmi Þ ½0;1

½0;1

ð9Þ jxi 2 ai j dGAi ðai Þ dF Xi ðxi Þ;

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Robust Positioning of Service Units 277 278 279 280 281 282 283 284

where GAi and FXi are the marginal distributions of GA and FX respectively. Let A*1 ; . . .; A*n be the 1-dimensional random variables attaining the inner maxima and GA*1 ; . . .; GA*n their respective cumulative distribution functions. Consider dGA* ¼ dGA*1 £ . . . £ dGA*n ; the measure in the product space generated by the measures dGA*1 ; . . .; dGA*n ; and let A* be a n-dimensional random variable with cumulative distribution GA*. That means, A* is a random vector whose components are independent random variables. Since A* is feasible for the former maximum in (8), we have that (9) holds with equality. By a similar argument, we get

285 286 287

Z

Z

n X

min max

X[F

A[GðmÞ

½0;1n

½0;1n i¼1

288 289 290

131

¼

n X i¼1

291

Z min

jxi 2 ai j dGA ðaÞ dF X ðxÞ Z

max

X i [F 1 Ai [G1 ðmi Þ ½0;1

jxi 2 ai j dGAi ðai Þ dF Xi ðxi Þ:

½0;1

292 293 294 295 296 297 298

Thus, we have obtained that the n-dimensional problem can be solved by solving n different 1-dimensional problems. This reduction allows the resolution of Problem (8) by Corollary 2.1. In particular, the results in this section show that if the l1-norm is used, the optimal policy is to park (to fix) the service unit at some vertex of the region where the random incident takes place.

299 300

3.

301

THE TWO-FACILITY PROBLEM

302 303 304 305 306 307 308

In the previous section, we considered the problem of locating only one facility to cover a random incident. However, often more than one service unit is necessary, especially if the coverage region is large. In this section, we consider the case where two service facilities cover a random incident under the usual nearest allocation rule: the random incident is covered by the closest service unit. This allocation rule leads to the following formulation: Z

309 310 311

min

Z

max

X 1 ;X 2 [F 1 A[G1 ðmÞ

½0;12

minfjx1 2 aj; jx2 2 ajg dGA ðaÞ dF 1;2 ðx1 ; x2 Þ;

ð10Þ

½0;1

312 313 314 315 316 317 318 319 320 321 322

where F1, G1(m) were defined in Section 2, GA(·) is the c.d.f. of the random variable A and F1,2(·,·) is the joint c.d.f. of the random variables X1 and X2. It is worth noting that this is a non-trivial problem: 1) it is a minmax problem, and 2) the decision space is a functional space of random vectors. This formulation allows us to model different real-world situations where there are two-service units to cover a random incident. This is for example the case of highways with two patrolling vehicles so that each one covers the closest incident. In order to solve this problem, first we consider the case where the servers are not allowed to patrol, that is, X1 and X2 are degenerate random variables. After that, we deal with the general case: X1 and X2 are any random variables belonging to F1.

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Puerto and Rodrı´guez-Chı´a

132 323 324 325 326

The formulation of Problem (10) for the first case is given by the following expression Z min max minfjx1 2 aj; jx2 2 ajg dGA ðaÞ: ð11Þ

x1 ;x2 [½0;1 A[G1 ðmÞ

½0;1

327 328 329 330 331 332 333

Remark 3.1

Without loss of generality we can assume that x1 # x2 :

Before we proceed to obtain the solution of Problem (11), we define the following functions:

334 335 336

 1 ; x2 ; AÞ :¼ 2 dðx

Z

x1

GA ðaÞ da þ 0

Z

!

x2 x1 þx2 2

GA ðaÞ da

þ m 2 x2 ;

ð12Þ

337 338 339 340

  !! x1 þ x2 x1 þ x2 2 2 p1 x1 2 Cðx1 ; x2 ; p1 Þ :¼ 2 ð1 2 mÞ 2 x2 þ m; 2 2

ð13Þ

341 342 343 344

and the set Tðx1 ; x2 Þ :¼ {p ¼ ðp0 ; p1 ; p2 Þ $ 0 : p0 þ p1 þ p2 ¼ 1

345 346

and p1

347 348 349

x1 þ x2 þ p2 ¼ m}; 2

ð14Þ

where A is a random variable in G1(m) with distribution function GA and x1 ; x2 [ ½0; 1:

350 351 352

Theorem 3.1 The optimal positioning policy in the hypothesis of Problem (11) is x1 ¼ m 2 and x2 ¼ 2m 2 m 2 :

353 354 355 356 357 358 359 360

First, by Lemma A.5 and A.7, we have that Z min max min{jx1 2 aj; jx2 2 aj}dGA ðaÞ ¼ min

Proof:

0#x1 #x2 #1 A[G1 ðmÞ

½0;1

max Cðx1 ; x2 ; p1 Þ;

0#x1 #x2 #1 p[Tðx1 ;x2 Þ

where C and T(x1,x2) were defined in (13) and (14), respectively. By Lemma A.8 the optimal solution of this problem is x1 ¼ m 2 and x2 ¼ 2m 2 m 2 and the proof is concluded. A

361 362 363 364

Once we have studied the problem of locating two deterministic service units, we consider the general problem where the service units are random vectors. In this case we consider the original Problem (10).

365 366 367 368

Theorem 3.2 The optimal positioning policy

of Problem (10)

are the random variables X *1 and X *2 such that P X *1 ¼ m 2 ¼ 1 and P X *2 ¼ 2m 2 m 2 ¼ 1:

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Robust Positioning of Service Units 369 370 371 372 373 374

Proof: Using Lemma A.5, we can bound the expression in (10) as follows (recall that d¯ was defined in (12)): Z Z max min{jx1 2 aj; jx2 2 aj} dGA ðaÞ dF 1;2 ðx1 ; x2 Þ A[G1 ðmÞ

A[G1 ðmÞ

376

#

378 380

þ

381 382

 1 ; x2 ; AÞdF 1;2 ðx1 ; x2 Þ þ dðx

Z

½0;x2 £½0;1

  2 ; x1 ; AÞdF 1;2 ðx1 ; x2 Þ dðx

ðx2 ;1£½0;1

 1 ; x2 ; AÞ dF 1;2 ðx1 ; x2 Þ max dðx

½0;x2 £½0;1 A[G1 ðmÞ

379

385

½0;1

Z

Z

377

384

½0;12

¼ max

375

383

133

Z

 2 ; x1 ; AÞ dF 1;2 ðx1 ; x2 Þ: max dðx

ðx2 ;1£½0;1 A[G1 ðmÞ

ð15Þ Define S1 ¼ {ðx1 ; x2 Þ [ R2 : 0 # x1 # x2 # 1};

386 387 388 389 390 391 392

S2 ¼ {ðx1 ; x2 Þ [ R2 : 0 # x2 , x1 # 1}; and let XSj(·) denote the indicator function of the set Sj for j ¼ 1; 2: Now, Lemma A.7 allows to write the integrands in (15) as  1 ; x2 ; AÞ ¼ max Cðx1 ; x2 ; p1 Þ for ðx1 ; x2 Þ [ S1 ; max dðx

ð16Þ

 2 ; x1 ; AÞ ¼ max Cðx2 ; x1 ; p1 Þ for ðx1 ; x2 Þ [ S2 : max dðx

ð17Þ

A[G1 ðmÞ

p[Tðx1 ;x2 Þ

393 394 395

A[G1 ðmÞ

p[Tðx1 ;x2 Þ

396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411

Combining (16) and (17) we can rewrite the expression (15) as Z max ½Cðx1 ; x2 ; p1 ÞX S1 ðx1 ; x2 Þ þ Cðx2 ; x1 ; p1 ÞX S2 ðx1 ; x2 Þ dF 1;2 ðx1 ; x2 Þ: ½0;12 p[Tðx1 ;x2 Þ

Let p* ðx1 ; x2 Þ ¼ ðp*0 ðx1 ; x2 Þ; p*1 ðx1 ; x2 Þ; p*2 ðx1 ; x2 ÞÞ [ Tðx1 ; x2 Þ be the function where the expression above reaches its inner maximum. Notice that the expression of p*(x1,x2) can be obtained from the proof of Lemma A.8, and it is defined in a different way depending on the region that (x1,x2) belongs to. Now, for all ðx1 ; x2 Þ [ ½0; 12 ; let A*(x1,x2) be a random variable independent of (X1,X2), whose probability distribution is dGA*(x1,x2), defined by 8 * p0 ðx1 ; x2 Þ if a ¼ 0 > > < x1 þx2 * ð18Þ dGA* ðx1 ;x2 Þ ðaÞ ¼ p1 ðx1 ; x2 Þ if a ¼ 2 > > : p* ðx1 ; x2 Þ if a ¼ 1: 2

412 413 414

Notice that, for a fixed x1 and x2 belonging to the interval [0,1], A*(x1,x2) is a discrete * * 2 random variable taking the values 0, x1 þx 2 and 1 with probabilities p0 ðx1 ; x2 Þ; p1 ðx1 ; x2 Þ and

120018142_STM_019_001_R1_X0.ald

Puerto and Rodrı´guez-Chı´a

134 415 416 417 418 419 420 421 422 423 424 425

p*2 ðx1 ; x2 Þ respectively. Thus, by the definition of the functions d¯ and C (see the proof of Lemma A.7), A*(x1,x2) verifies that  1 ; x2 ; A* ðx1 ; x2 ÞÞ ¼ Cðx1 ; x2 ; p*1 ðx1 ; x2 ÞÞ with ðx1 ; x2 Þ [ S1 ; dðx

ð19Þ

 2 ; x1 ; A* ðx1 ; x2 ÞÞ ¼ Cðx2 ; x1 ; p* ðx1 ; x2 ÞÞ with ðx1 ; x2 Þ [ S2 : dðx 1

ð20Þ

Now, using (19), (20), and Lemma A.5, we have that Z max ½Cðx1 ; x2 ; p1 ÞX S1 ðx1 ; x2 Þ þ Cðx2 ; x1 ; p1 ÞX S2 ðx1 ; x2 Þ dF 1;2 ðx1 ; x2 Þ ½0;12 p[Tðx1 ;x2 Þ

426

Z

½Cðx1 ; x2 ; p*1 ðx1 ; x2 ÞX S1 ðx1 ; x2 Þ

427 428

¼

429

þ Cðx2 ; x1 ; p*1 ðx1 ; x2 ÞÞX S2 ðx1 ; x2 Þ dF 1;2 ðx1 ; x2 Þ Z  1 ; x2 ; A* ðx1 ; x2 ÞÞ dF 1;2 ðx1 ; x2 Þ dðx ¼

430 431 432

½0;12

½0;x2 £½0;1

433 434

þ

435 436 437

440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460

Z

 2 ; x1 ; A* ðx1 ; x2 ÞÞ dF 1;2 ðx1 ; x2 Þ dðx ½x2 ;1£½0;1

¼

Z

Z

438 439

8/2/2003—KALYAN—59071

½0;12

½0;1

min{jx1 2 aj; jx2 2 aj} dGA* ðx1 ;x2 Þ ðaÞ dF 1;2 ðx1 ; x2 Þ:

Therefore, using the inequality in (15), we have that Z Z max min{jx1 2 aj; jx2 2 aj} dGA ðaÞ dF 1;2 ðx1 ; x2 Þ A[G1 ðmÞ

½0;12

Z

½0;1

Z

# ½0;12

½0;1

min{jx1 2 aj; jx2 2 aj} dGA* ðx1 ;x2 Þ ðaÞ dF 1;2 ðx1 ; x2 Þ:

Moreover, since p* ðx1 ; x2 Þ [ Tðx1 ; x2 Þ; we have that the mean value of A*(x1,x2) is m for all ðx1 ; x2 Þ [ ½0; 12 : Therefore, A*(X1,X2) is also a random variable with mean value m, that is, A* ðX 1 ; X 2 Þ [ G1 ðmÞ: Thus, the inequality above has to be an equality and Problem (10) can be reformulated as follows: Z min ½Cðx1 ; x2 ; p*1 ðx1 ; x2 ÞX S1 ðx1 ; x2 Þ X 1 ;X 2 [F 1 ½0;12

þ Cðx2 ; x1 ; p*1 ðx1 ; x2 ÞÞX S2 ðx1 ; x2 Þ dF 1;2 ðx1 ; x2 Þ: Let us define the following function; Lðx1 ; x2 Þ :¼ Cðx1 ; x2 ; p*1 ðx1 ; x2 ÞX S1 ðx1 ; x2 Þ þ Cðx2 ; x1 ; p*1 ðx1 ; x2 ÞX S2 ðx1 ; x2 Þ:

ð21Þ

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Robust Positioning of Service Units 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476

135

Since, it always holds that, Z min Lðx1 ; x2 Þ dF 1;2 ðx1 ; x2 Þ ¼ min

x1 ;x2 [½0;1

X 1 ;X 2 [F 1 ½0;12

Lðx1 ; x2 Þ;

then the minimum in (21) is reached by two degenerate random variables. On the other hand, using that the function L(x 1,x 2) is defined in disjoint sets, Cðx1 ; x2 ; p*1 ðx1 ; x2 ÞÞX S1 ðx1 ; x2 Þ $ 0 and Cðx2 ; x1 ; p*1 ðx1 ; x2 ÞÞX S2 ðx1 ; x2 Þ $ 0 (see (19), (20) and Lemma A.5 to justify the non-negativity of these functions), we have that Lðx1 ; x2 Þ ¼ max {Cðx1 ; x2 ; p*1 ðx1 ; x2 ÞÞX S1 ðx1 ; x2 Þ; Cðx2 ; x1 ; p*1 ðx1 ; x2 ÞÞX S2 ðx1 ; x2 Þ}: Therefore, we can use the same arguments as in the deterministic case (Lemma A.8) in order to obtain that the optimal solutions are the random vectors (X1,X2) such that: P½ðX 1 ; X 2 Þ ¼ ðm 2 ; 2m 2 m 2 Þ ¼ 1

or

P½ðX 1 ; X 2 Þ ¼ ð2m 2 m 2 ; m 2 Þ ¼ 1:

A

477 478 479 480 481 482

In conclusion, Theorem 3.2 proves that it is optimal to park the service units when no hypotheses are made on their c.d.f.’s and only the mean value of the position of the random incident is known.

483

4.

484

CONCLUDING REMARKS

485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506

The results in this paper extend other previously known results about the optimal location of one or two service units to situations where no assumptions are made on the probability distribution of the random incident that these service units cover apart from the knowledge of its mean value (whereas all the previous papers require exact knowledge of this distribution). This is accomplished by minimizing the maximum expected response time (whereas the previous results minimize expected distances). In particular, we show that when the only available information is the mean value of the position of the incidents, then the optimal policy is to park the service units at concrete points. On the other hand, since our goal is to minimize the response time from the service unit to the incident, another interesting problem is to assume the same probability distribution for both the service unit and the incident. Notice that one interpretation of this policy is that we are fixing the location of the service unit at the location of the previous incident. The worst case for this policy is given by Z Z max kx 2 ak1 dFðaÞ dFðxÞ F[F 1 ðmÞ ½0;1

½0;1

(assuming that incidents occur independently). Using a similar argument to the one used in the proof of Theorem 2.2, we have that Z Z max kx 2 ak1 dFðaÞ dFðxÞ ¼ m þ mð1 2 2mÞ: F[F 1 ðmÞ ½0;1

½0;1

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Puerto and Rodrı´guez-Chı´a

136 507 508 509 510 511 512

Thus, the maximum objective value above is 1/2 and it is achieved by a c.d.f. F with mean value m ¼ 1=2: Moreover, by the proof of Corollary 2.1, we have that the objective value of the worst case for Problem (6) is m þ lð1 2 2mÞ: Thus, the best objective value when l varies, is: ( m if m # 0:5 and it is achieved at l* ¼ 0 if m . 0:5 and it is achieved at l* ¼ 1:

12m

513 514 515 516

Therefore, for a fixed m, the difference in worst case performance between the approach considered above and the approach of the paper is given by

517

(

518

jðl* 2 mÞð1 2 2mÞj ¼

519 520

mð1 2 2mÞ

if m # 0:5

ð1 2 mÞð21 þ 2mÞ

if m . 0:5:

521 522 523 524 525 526 527 528 529 530 531 532

Notice that there is no difference when m [ {0; 0:5; 1}; but that there is a difference that can be quantified for other choices of m. We can see that this difference is maximal when m ¼ 0.25 and m ¼ 0:75 which is reasonable because these points are at the maximum distances to the values of m where the difference is null. Finally, we can also study the natural extension of Problem (10) where we consider k service units instead of two. It should be noted that using similar arguments to those used for the case k ¼ 2; we can obtain that the worst case in the distribution of the xk21 þxk 2 x2 þx3 incident is given by a random variable taking the values 0; x1 þx ; 1 (see 2 ; 2 ; . . .; 2 Eq. (18) for the case when k ¼ 2). However, the complexity of the expressions obtained in the analysis does not allow us to present an explicit formula of the optimal solution of this problem.

533 534 535 536

APPENDIX

537

In this section, we include for the sake of completeness, several results and their proofs which have been used in the paper.

538 539 540 541 542 543 544

Lemma A.1 Let G(·) be a nondecreasing function such that G:R ! [0,1). If there exists M [ R such that Z b GðtÞ dt ¼ Mðb 2 aÞ with a; b [ R a

545 546

then

547 548

I G ðzÞ :¼

549

Z

z

GðtÞ dt 2 Mðz 2 aÞ # 0

;z [ ½a; b:

a

550 551 552

Proof:

Let t0 [ ½a; b be such that t0 ¼ inf {t : GðtÞ $ M};

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Robust Positioning of Service Units

137

i) If z , t0 we have that GðtÞ , M ii) If z $ t0 we obtain that,

553 554

;t # z thus, I G ðzÞ # 0:

555 556 557 558

I G ðzÞ ¼

Z

z

GðtÞ dt þ

Z

z

b

GðtÞ dt 2 Mðz 2 aÞ ¼ Mðb 2 zÞ 2

Z

z

561 562

b

GðtÞ dt z

# Mðb 2 zÞ 2 GA ðzÞðb 2 zÞ ¼ ðM 2 GðzÞðb 2 zÞ # 0;

563 565 566

b

GðtÞ dt 2 Mðz 2 aÞ

z

¼ Mðb 2 aÞ 2

560

Z

b

GðtÞ dt 2

a

559

564

Z

where we have used the fact that the function G(·) is nondecreasing. Thus, the lemma is proved. A

567 568 569 570 571 572

Lemma A.2 For any A [ G1 ðmÞ with c.d.f. GA(·), we have that: R1 i) R 0 GA ðaÞ da ¼ 1 2 m: R x ii) ½0;1 jx  aj dGA ðaÞ ¼ 2 0 GA ðaÞ da 2 x þ m ;x [ ½0; 1:

573 574 575 576 577 578 579

Proof: Denote DG the set of denumerable number of discontinuity points of GA(·) in the interval [0,1] union with the set {0,1}. Applying integration by parts to the interval (xi21,xi) where xi21 and xi are two consecutive points of DG, we have that Z

GA ðaÞ da ¼

580 581 582

ðxi21 ;xi Þ

x2 aGA ðaÞjxiþ i21

Z

a dGA ðaÞ

2 ðxi21 ;xi Þ

2 þ þ ¼ x2 i GA ðxi Þ 2 xi21 GA ðxi21 Þ 2

583 584 585

a dGA ðaÞ ðxi21 ;xi Þ

¼ xi ðGA ðxi Þ 2 P½A ¼ xi Þ 2 xi21 GA ðxi21 Þ 2

586

Z

a dGA ðaÞ ðxi21 ;xi Þ

587 588

¼ xi GA ðxi Þ 2 xi21 GA ðxi21 Þ 2

589 590 591

Z

Z

a dGA ðaÞ: ðxi21 ;xi Þ

If we sum the equality above for each element of DG we obtain

592 593 594 595

XZ xi [DG ðxi21 ;xi Þ

GA ðaÞ da ¼

X

¼12

a dGA ðaÞ ðxi21 ;xi 

xi [DG

596 597 598

xi GA ðxi Þ 2 xi21 GA ðxi21 Þ 2



Z

Z ð0;1

a dGA ðaÞ:

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Puerto and Rodrı´guez-Chı´a

138 599

Hence, since the integrant in the last expression is null at zero we have that

600 601 602

Z

603 604

1

GA ðaÞ da ¼ 1 2

Z

a dGA ðaÞ ¼ 1 2 m: ½0;1

0

605 606

Now we prove the second assertion. We have the following equalities

607 608 609 610

Z

jx 2 aj dGA ðaÞ ¼

Z

½0;1

ðx 2 aÞ dGA ðaÞ þ ½0;x

611 612

ða 2 xÞ dGA ðaÞ ðx;1

¼ xGA ðxÞ 2

Z

a dGA ðaÞ þ

Z

½0;x

613 614

¼ xð2GA ðxÞ 2 1Þ 2

615

a dGA ðaÞ 2 xð1 2 GA ðxÞÞ ðx;1

Z

adGA ðaÞ þ m 2 ½0;x

616 617

¼ 2ðxGA ðxÞ 2

618

Z

Z

adGA ðaÞ ½0;x

a dGA ðaÞÞ þ m 2 x: ½0;x

619 620 621

Z

Applying integration by parts using the arguments above we have that

622 623 624 625 626 627 628

Z

x

GA ðaÞ da ¼ xGA ðxÞ 2

Z

a dGA ðaÞ; ½0;x

0

and the result follows.

A

629 630 631 632 633

Lemma A.3 For any X [ F 1 ðlÞ and A [ G1 ðmÞ with c.d.f’s FX(·) and GA(·), respectively, we have that

634 635 636 637

Z ½0;1

yGA ðyÞ dF X ðyÞ þ

Z

yF X ðyÞ dGA ðyÞ ¼ 1 þ ½0;1

640

yP½X ¼ yP½A ¼ y

y[D

638 639

X

Z

1

GA ðyÞF X ðyÞ dy;

2 0

641 642 643 644

where D is the set of denumerable number of discontinuity points either of FX(·) or GA(·) (or both) union with the set {0,1}.

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Robust Positioning of Service Units

139

Proof: Applying integration by parts to the inteval (xi-1,xi) where xi-1 and xi are two consecutive points of D, we have the following equalities

645 646 647

Z

648

ðxi21 ;xi Þ yGA ðyÞ dF X ðyÞ

649 650 651

¼

652

x2 yGA ðyÞF X ðyÞjxiþ i21

þ

Z Z

¼

655

GA ðyÞF X ðyÞdy

2 ðxi21 ;xi Þ

653 654

yF X ðyÞdGA ðyÞ ðxi21 ;xi Þ

2 2 x2 i GA ðxi ÞF X ðxi Þ

2

þ þ xþ i21 GA ðxi21 ÞF X ðxi21 Þ

Z

656 657 658

xi

GA ðyÞF X ðyÞdy

2 xi21

¼ xi ðGA ðxi Þ 2 P½A ¼ xi ÞðF X ðxi Þ 2 P½X ¼ xi Þ 2 xi21 GA ðxi21 ÞF X ðxi21 Þ Z

659

xi

GA ðyÞF X ðyÞdy ¼ xi GA ðxi ÞF X ðxi Þ 2 xi21 GA ðxi21 ÞF X ðxi21 Þ

2

660

xi21

661 662

2 xi P½X ¼ xi GA ðxi Þ 2 xi P½A ¼ xi F X ðxi Þ þ xi P½A ¼ xi P½X ¼ xi 

663

Z

664

xi

GA ðyÞF X ðyÞdy:

2

665 666

xi21

667

The equality above can be rewritten as

668 669

Z

670 671

yGA ðyÞ dF X ðyÞ þ

ðxi21 ;xi 

672 673 674

Z

yF X ðyÞ dGA ðyÞ ðxi21 ;xi 

¼ xi GA ðxi ÞF X ðxi Þ 2 xi21 GA ðxi21 ÞF X ðxi21 Þ

675

þ xi P½A ¼ xi P½X ¼ xi  2

676

Z

xi

GA ðyÞF X ðyÞdy: xi21

677 678

If we sum the expression above for each element of D we have

679 680

X Z

681 682 683 684 685 686

Q1

yGA ðyÞ dF X ðyÞ þ ðxi21 ;xi 

xi [D

¼

X

yF X ðyÞ dGA ðyÞ ðxi21 ;xi 

" xi GA ðxi ÞF X ðxi Þ 2 xi21 GA ðxi21 ÞF X ðxi21 Þ 2 þxi P½A ¼ xi P½X ¼ xi 

xi [D

687

Z

688 689 690



Z

xi

2 xi21

 Z 1 X xi P½A ¼ xi P½X ¼ xi  2 GA ðyÞF X ðyÞdy ¼ 1 þ GA ðyÞF X ðyÞdy: xi [D

0

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Puerto and Rodrı´guez-Chı´a

140 691 692 693 694

Since, at zero the integrants of the first part of the equalities above are null, we have that Z Z X xi P½A ¼ xi P½X ¼ xi  yGA ðyÞ dF X ðyÞ þ yF X ðyÞ dGA ðyÞ ¼1 þ ½0;1

½0;1

xi [D

Z

695 696

699

GA ðyÞF X ðyÞ dy; 0

697 698

1

2 and the result follows.

A

700 701 702 703 704 705

Lemma A.4 For any X [ F 1 ðlÞ and A [ G1 ðmÞ with c.d.f’s FX(·) and GA(·), respectively, we have that   Z 1 jx 2 aj dGA ðaÞ dF X ðxÞ ¼ 2 1 2 GA ðyÞF X ðyÞ dy 2 l 2 m:

Z

Z ½0;1

½0;1

0

706 707 708 709 710 711 712 713

Proof: Z

We have that Z jx – aj dGA ðaÞ dF X ðxÞ

½0;1

½0;1

¼

Z

714 715

719 720 721

½0;x

Z

Z ½0;1

717

¼

Z

þ

723 724

¼2

727 728 729 730

þ

732

735 736

Z

ðm 2

½0;1

Z

Z

a dGA ðaÞÞ dF X ðxÞ ¼ 2

Z

xGA ðxÞ dF X ðxÞ ½0;1

adGA ðaÞ dF X ðxÞ 2 l þ m ¼ 2 Z

Z

½0;1

½a;1

Z

dF X ðxÞ dGA ðaÞ 2 2

a ða;1

xGA ðxÞ dF X ðxÞ ½0;1

dF X ðxÞ dGA ðaÞ 2 l þ m ¼ 2

a

Z

a dGA ðaÞ dF X ðxÞ

½0;x

½0;x

Z 22

½0;1

xð1 2 GA ðxÞÞ dF X ðxÞ

½0;x

Z

22

Z ½0;1

22

734

a dGA ðaÞ dF X ðxÞ

a dGA ðaÞ dF X ðxÞ 2

½0;1

Z

x dGA ðaÞ dF X ðxÞ

ðx;1

½0;x

½0;1

Z

733

Z

Z

xGA ðxÞ dF X ðxÞ 2 l 2

731

Z

ðx;1

½0;1

725 726

Z

Z

Z ½0;1

½0;1

½0;1

a dGA ðaÞ dF X ðxÞ

½0;x

ðx;1

½0;1

Z

Z

½0;1

a dGA ðaÞ dF X ðxÞ 2

xGA ðxÞ dF X ðxÞ 2

722

Z

x dGA ðaÞ dF X ðxÞ 2

½0;1

þ

716 718

Z

Z

xGA ðxÞ dF X ðxÞ ½0;1

Z

aP½X ¼ a dGA ðaÞ 2 l þ m: ½0;1

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Robust Positioning of Service Units 737 738

141

Denote by D the denumerable number of discontinuity points of FX or GA union with {0,1}. Then, we rewrite the expression above as

739

Z

740

2

741 742

xGA ðxÞdF X ðxÞ22m þ2

½0;1

¼2

Z

yGA ðyÞdF X ðyÞþ ½0;1

745

aF X ðaÞdGA ðaÞ22

½0;1

743 744

Z

X

yP½X ¼ yP½A ¼ y2 l þ m

y[D



Z

yF X ðyÞdGA ðyÞ 22 ½0;1

X

y[D

yP½X ¼ yP½A ¼ y2 l 2 m: ð22Þ

746 747

Now, using Lemma A.3 we have that

748

  Z 1 ð22Þ ¼ 2 12 GA ðyÞF X ðyÞdy 2 l 2 m;

749 750

0

751 752

and the result follows.

A

753 754 755 756

Lemma A.5 For any A [ G1 ðmÞ with c.d.f. GA(·), the function d¯ defined in (12) admits the following representation.

757 758

 1 ; x2 ; AÞ ¼ dðx

Z

min fjx1 2 aj; jx2 2 ajg dGA ðaÞ ;x1 # x2 [ ½0; 1:

½0;1

759 760 761 762

Proof:

763

Z

764 765 766 767

We have the following equalities:

min {jx1 2 aj; jx2 2 aj} dGA ðaÞ ¼

½0;1

Z x1 þx2 2 

¼

Z ½0;x1 

769

Z

þ

ð

771

ð

ðx1 2 aÞ dGA ðaÞ þ

768 770

jx1 2 aj dGA ðaÞ þ

½0;

775 776

779

x1 þx2 2 ;x2 

x1 þx2 2 

ða 2 x1 Þ dGA ðaÞ

Z

ðx2 2 aÞ dGA ðaÞ þ

ða 2 x2 Þ dGA ðaÞ

ðx2 ;1

  x1 þ x2 ðx1 þ x2 Þ þ 2x2 GA ðx2 Þ 2 Z adGA ðaÞ þ a dGA ðaÞ x þx

Z 2 x2 2 ½0;x1 

Z 2 ð

x1 þx2 2 ;x2 

ðx1 ;

a dGA ðaÞ þ

780 781 782

Z

jx2 2 aj dGA ðaÞ

¼ 2x1 GA ðx1 Þ 2 GA

777 778

x1 þx2 2 ;1

ðx1 ;

772 773 774

Z

¼ 2ð x1 GA ðx1 Þ 2

Z

1

2

2

a dGA ðaÞ ðx2 ;1

Z

a dGA ðaÞ þ x2 GA ðx2 Þ ½0;x1 

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Puerto and Rodrı´guez-Chı´a

142

!   Z x1 þ x2 x1 þ x2 2 x þx 2GA a dGA ðaÞ 2 2 ð 1 2 2 ;x2 

783 784 785 786 787 788

þ m 2 x2 ¼ 2

Z

x1

GA ðaÞ da þ 0

Z

!

x2 x1 þx2 2

GA ðaÞ da

789

 1 ; x2 ; AÞ; þ m 2 x2 ¼ dðx

790 791 792

and the result is proved.

A

793 794 795 796 797 798 799 800 801 802

Lemma A.6 For each random variable A [ G1 ðmÞ with distribution function GA(·) and each 0 # x1 # x2 # 1; there exists a discrete random variable A [ G1 ðmÞ defined by 8 p0 if a¼0 > > < 2 a ¼ x1 þx P½A ¼ a ¼ p1 if 2 > > : p2 if a ¼ 1; where( p0, p1, p2) satisfies that

803 804

p0 þ p1 þ p2 ¼ 1

805

 1 ; x2 ; AÞ # dðx  1 ; x2 ; AÞ  dðx

806 807 808

Z

809 810 811 812 813

x1 þx2 2

GA ðaÞ da ¼ p0

0

Z

x1 þ x2 2

  x1 þ x2 GA ðaÞ da ¼ ðp0 þ p1 Þ 1 2 : x1 þx2 2 2

ð23Þ

1

ð24Þ

814 815 816 817 818 819 820 821 822 823 824 825 826 827 828

Remark A.1 It should be noted that from this result and part i) of Lemma A.2, one obtains  1 ; x2 ; AÞ is attained in a discrete random that for each 0 # x1 # x2 # 1; the maxA[G1 ðmÞ dðx 2 variable with mean value m and defined only on the values 0, x1 þx and 1. Moreover, 2 p ¼ ðp0 ; p1 ; p2 Þ [ Tðx1 ; x2 Þ (where T was defined in (14)). Proof: First, we note that A [ G1 ðmÞ; see Remark A.1. Second, in order to complete the proof of the lemma it suffices to prove: Rx i) 0 1 GA ðaÞ da 2 p0 x1 # 0   Rx 2 # 0: ii) x12þx2 GA ðaÞ da 2 ðp0 þ p1 Þ x2 2 x1 þx 2 2

To this end, we apply Lemma A..1. Since (23) and (24) hold and GA(·) is a probability distribution function, we are under hypotheses of Lemma A.1 and thus, i) and ii) are proved. A

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Robust Positioning of Service Units 829 830 831

143

If x1, x2 [ [0,1], x1 # x2, then

Lemma A.7

 1 ; x2 ; AÞ ¼ max Cðx1 ; x2 ; p1 Þ; max dðx p[Tðx1 ;x2 Þ

A[G1 ðmÞ

832 833 834

where d¯, C, and T were defined in (12), (13), and (14), respectively.

835 836 837 838 839 840

We have, by the definition of d¯ in (12) and Lemma A.6, that; ! Z x1 Z x2  max dðx1 ; x2 ; AÞ ¼ max 2 GA ðaÞ da þ GA ðaÞ da þ m 2 x2

Proof:

A[G1 ðmÞ

A[G1 ðmÞ

841 842

845 846 847 848 849 850 851 852 853 854 855 856 857 858 859 860 861 862

x1 þx2 2

   x1 þ x2 ¼ max 2 p0 x1 þ ðp0 þ p1 Þ x2 2 þ m 2 x2 : p[Tðx1 ;x2 Þ 2

843 844

0

2 Using that p [ Tðx1 ; x2 Þ; (i.e., p0 þ p1 þ p2 ¼ 1 and x1 þx 2 p1 þ p2 ¼ m) we have that    x1 þ x2 max 2 p0 x1 þ ðp0 þ p1 Þ x2 2 þ m 2 x2 p[Tðx1 ;x2 Þ 2

 x1 þ x2 2 1ÞÞx1 m 2 x2 þ 2 ð1 2 m þ p1 ð p[Tðx1 ;x2 Þ 2  x1 þ x2 x1 þ x2 Þðx2 2 Þ þð1 2 m þ p1 2 2 ¼ max

  !! x1 þ x2 x1 þ x2 2 ¼ max m 2 x2 þ 2 ð1 2 mÞ 2 p1 x 1 2 p[Tðx1 ;x2 Þ 2 2 ¼ max

p[Tðx1 ;x2 Þ

Cðx1 ; x2 ; p1 Þ;

which proves the result.

A

863 864 865 866 867

Lemma A.8 min

The optimal solution for the problem max Cðx1 ; x2 ; p1 Þ

0#x1 #x2 #1 p[Tðx1 ;x2 Þ

ð25Þ

868 869

is x1 ¼ m 2 and x2 ¼ 2m 2 m 2 ; where C and T were defined in (13) and (14), respectively.

870 871 872 873 874

Proof: Since C(x1, x2, p1) is linear with respect to p1, we analyze the cases where the coefficient that multiplies p1 is positive or negative separately.

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Puerto and Rodrı´guez-Chı´a

144

x

1 þx2

2

875

Case 1: x1 $

876

In this case, the function Cðx1 ; x2 ; p1 Þ is decreasing in p1, thus the maximum is reached at p1 ¼ 0: That means that p0 ¼ 1-m and p2 ¼ m. Therefore, the expression that we have to consider is the following:   x1 þ x2 min C 1 ðx1 ; x2 Þ :¼ 2 ð1 2 mÞ ð26Þ 2 x2 þ m 0#x1 #x2 #1 2

877 878 879 880 881

2

882 883 884 885

  x1 þ x2 2 s:t: x1 $ : 2

886 887 888 889 890 891

It is clear that:

›C 1 ðx1 ; x2 Þ ¼ 1 2 m $ 0: ›x 1 Since the function C1(x1,x2) is increasing in x1 and x1 $

x

892 893 894 895 896 897 898 899 900 901

1 þx2

2

2

we have that

C 1 ðx1 ; x2 Þ $ C 1 ðt; x2 Þ;  2 pffi pffi 2 where t ¼ tþx : Thus, since x2 $ 0; this implies that x2 ¼ 2 t 2 t (notice that 2 t 2 2 t $ t for all t $ 0).Therefore, solving (26) is equivalent to solving the following problem; pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi min 2ð1 2 mÞ x1 2 2 x1 þ x1 þ m ¼ min x1 þ ð1 2 2 x1 Þm; x1 [½0;1

x1 [½0;1

and this problem reaches its minimum at the point x1 ¼ m 2 : Hence, x2 ¼ 2m 2 m 2 and the minimum objective value is m 2 m 2. A

902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920

 2 2 Case 2: x1 # x1 þx 2 Notice that in Case 1, we have already studied the points (0,0) and (1,1). Therefore, in what follows, we can assume without loss of generality that (x1,x2) is neither (0,0) nor (1,1). In this case, the function C(x1,x2,p1) is increasing in p1. Since p [ Tðx1 ; x2 Þ we have that x1 þx2 2 p0 ¼ 1 2 m þ p1 ðx1 þx 2 2 1Þ; p2 ¼ m 2 p1 2 ; 0 # p0 # 1 and 0 # p2 # 1 then, we have that, m 12m 2 a) 0 # ð1 2 mÞ 2 p1 ð1 2 x1 þx 2 Þ # 1; that is, 2 12x1 þx2 # p1 # 12x1 þx2 2 2 ð1; 1Þ: 12m m 2 b) 0 # m 2 p1 x1 þx 2 # 1; that is, 2 x1 þx2 # p1 # x1 þx2

Using that p1 $ 0; 2 a) p1 #

2

m 12

12m

x1 þx2 2

if ðx1 ; x2 Þ –

if ðx1 ; x2 Þ – ð0; 0Þ:

2

# 0 and 2 x1 þx2 # 0 the previous conditions reduce to; 2

12m x þx 12 1 2 2 m

b) p1 # x1 þx2 : 2  Hence, p1 # min

12m x þx 12 1 2 2

m

; x1 þx2 2

 and to study this minimum we distinguish two cases;

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Robust Positioning of Service Units 921 922 923 924 925 926

.

Case 2.1: If

.

Case 2.2: If

x1 þx2 2 x1 þx2 2

145

$ m then # m then

12m x þx 12 1 2 2 12m x þx 12 1 2 2

$ x1mþx2 ; thus p1 # x1mþx2 : 2

# x1mþx2 ; thus, p1 # 2

2

12m x þx 12 1 2 2

:

Since the function C(x1,x2,p1) is increasing in p1, in Case 2.1. its maximum in p1 is reached at p1 ¼ x1mþx2 and in Case 2.2 at p1 ¼ 12x1mþx2 : 12 2 Hence, to2find the maximum of the function C(x1,x2,p1) we have the following two cases:

927 928 929 930 931

Case 2.1: p1 ¼ x1mþx2 :

.

2

.

Case 2.2: p1 ¼

12m x þx 12 1 2 2

:

932 933 934 935 936 937 938

Case 2.1: p1 ¼ x1mþx2 2 In this case, Problem (25) reduces to the following optimization problem ! 2m min C 2 ðx1 ; x2 Þ :¼ x1 1 2 x1 þx2 þ m 0#x1 #x2 #1

2

939 940



941 942

s:t: : x1 #

x1 þ x2 2

2

m#

x1 þ x2 : 2

943 944 945 946 947

We obtain that

›C 2 ðx1 ; x2 Þ x1 ¼ m 2 $ 0: x þx 1 2 ›x 2 2

948 949 950 951 952 953 954 955 956 957 958 959 960 961

Therefore, C2(x1,x2) is a increasing function in x2. Since in this case, (x1,x2) satisfies that pffiffiffiffiffi x2 $ 2m 2 x1 and x2 $ 2 x1 2 x1 we have that C 2 ðx1 ; x2 Þ $ C2 ðx1 ; tÞ where t ¼ 2m 2 x1 if x1 # m 2 pffiffiffiffiffi . t ¼ 2 x1 2 x1 if x1 $ m 2 : Thus, .

a) If x1 # m 2 we have that min

x1 ;x2 [½0;1

C 2 ðx1 ; x2 Þ ¼ min m 2 x1 : x1 [½0;1

962 963 964 965 966

b) If x1 $ m 2 we have that   2m C 2 ðx1 ; x2 Þ ¼ min x1 1 2 pffiffiffiffiffi þ m: x1 ;x2 [½0;1 x1 [½0;1 x1 min

120018142_STM_019_001_R1_X0.ald

8/2/2003—KALYAN—59071

Puerto and Rodrı´guez-Chı´a

146 967 968

Both cases give us the same optimal solution x1 ¼ m 2 and x2 ¼ 2m 2 m 2 and its objective value is m-m 2.

969 970 971 972 973 974 975 976 977

Case 2.2:

p1 ¼

12m x þx 12 1 2 2

In this case, Problem (25) reduces to the following optimization problem ! 12m 12m min C3 ðx1 ; x2 Þ :¼ min x2 2 1 2 x1 þm 2 2 0#x1 #x2 #1 x1 ;x2 [½0;1 1 2 x1 þx 1 2 x1 þx 2 2

978



979 980

s:t:x1 #

x1 þ x2 2

2

981 982

m$

983 984 985

We obtain that,

›C 3 ðx1 ; x2 Þ ð1 2 mÞðx2 2 1Þ ¼ # 0: 2 2 ›x 1 ð1 2 x1 þx 2 Þ

986 987 988 989 990

That means 3(x1,x2) is a decreasing function in x1. Since, in this case, (x1,x2) satisfies  thatC 2 2 2 2 that x1 # x1 þx and m $ x1 þx 2 2 ; we have that x1 # m then

991 992 993 994 995 996 997

x1 þ x2 : 2

C 3 ðx1 ; x2 Þ $ C 3 ðm 2 ; x2 Þ:

 2 2 2 Thus, taking x1 ¼ m 2 we have that m $ m 2þx2 and m 2 # m 2þx2 ; that is, x2 # 2m 2 m 2 and x2 $ 2m 2 m 2 : (Notice that we do not have to consider the other solution x2 $ 22m 2 m 2 since 22m 2 m 2 # 0 and x2 $ 0). Therefore, x2 ¼ 2m 2 m 2 : Since, all the cases give us the same optimal solution, the optimal solution to Problem (25) is x1 ¼ m 2 and x2 ¼ 2m 2 m 2 :

998 999 1000

ACKNOWLEDGMENTS

1001 1002 1003 1004 1005 1006

The authors would like to thank the suggestions made by two anonymous referees and the editor who have improved the readability and the final presentation of the paper. This research has been partially financed by Spanish research grants BFM2001-2378 and BFM2001-4028.

1007 1008

REFERENCES

1009 1010 1011 1012

1.

Anderson, L.R.; Fontenot, R.A. Optimal positioning of service units along a coordinate line. Transportation Sci. 1992, 26 (4), 346 – 351.

120018142_STM_019_001_R1_X0.ald

Robust Positioning of Service Units 1013

2.

1014 1016

3. 4.

1017 1018

5.

1015

1019 1020 1021

6. 7.

1022 1023

8.

1024 1025 1026

9.

1027

10.

1028

1031 1032 1033 1034 1035 1036 1037 1038 1039 1040 1041 1042 1043 1044 1045 1046 1047 1048 1049 1050 1051 1052 1053 1054 1055 1056 1057 1058

147

Carrizosa, E.; Mun˜oz-Ma´rquez, M.; Puerto, J. A note on the optimal positioning of service units. Oper. Res. 1998, 46 (1), 155 –156. Flury, B.A. Principal points. Biometrika 1990, 77 (1), 33 – 41. Gallego, G. A minmax distribution free procedure for the (Q, R) inventory model. Oper. Res. Lett. 1992, 11, 55– 60. Larson, R.; Odoni, A. Urban Operations Research; Prentice-Hall Englewood Cliffs: New Jersey, 1981. Levine, A. A patrol problem. Math. Mag. 1986, 59, 159 –166. De Palma, A.; Liu, Q.; Thisse, J.F. Optimal location on a line with random utilities. Transportation Sci. 1994, 28 (1), 63 –69. Puerto, J.; Ferna´ndez, F.R. Pareto-optimality in classical inventory problems. Naval Res. Logistic 1998, 45, 83 –98. Smith, D.K. Police patrol policies on motorways with unequal patrol lengths. J. Oper. Res. Soc. 1997, 48, 996– 1000. Vickson, R.G.; Gerchak, Y.; Rotem, D. Optimal positioning of read/write head in mirrored disks. Location Sci. 1995, 3 (2), 125 –132.

1029 1030

8/2/2003—KALYAN—59071

Received April 20, 1999 Revised April 18, 2000 Accepted October 1, 2002