Appromate Crce Pacng n a Rectanguar Contaner: Integer Programmng Formuatons and Vad Inequates Igor Ltvnchev, Lus Infante, and Edth Lucero Ozuna Espnosa Department of Mechanca and Eectrca Engneerng Nuevo Leon State Unversty Monterrey, Nuevo Leon, Meco {gor,nfante,ucero}@yama.fme.uan.m Abstract. A probem of pacng a mted number of unequa crces n a fed sze rectanguar contaner s consdered. The am s to mamze the (weghted) number of crces paced nto the contaner or mnmze the waste. Frequenty, the probem s formuated as a nonconve contnuous optmzaton probem whch s soved by heurstc technques combned wth oca search procedures. A new formuaton s proposed usng a reguar grd appromatng the contaner and consderng the nodes of the grd as potenta postons for assgnng centers of the crces. The pacng probem s then stated as a arge scae near 0- optmzaton probem. The bnary varabes represent the assgnment of centers to the nodes of the grd. The resutng bnary probem s then soved by commerca software. Two fames of vad nequates are proposed to strengthen the formuaton. Numerca resuts are presented to demonstrate the effcency of the proposed approach. Keywords: Crce pacng, nteger programmng, arge scae optmzaton. Introducton Pacng probems consttute a famy of natura combnatora optmzaton probems, whch occur n many feds of study such as computer scence, ndustra engneerng, ogstcs and manufacturng and producton processes. For nstance, severa rea fe ndustra appcatons requre the aocaton of a set of peces to a arger standardzed rectanguar stoc unt. They generay consst of pacng a set of tems of nown dmensons nto one or more arge obects n order to mnmze a certan obectve (e.g. the unused part of the obects or waste). The crce pacng probem s a we studed probem [8, 3] whose am s the pacng of a certan number of crces, each one wth a fed nown radus (not necessary the same for each crce) nsde a contaner. The shape of the contaner may vary from a crce, a square, a rectanguar, etc. Ths probem has been apped n dfferent areas, such as the coverage of a geographca area wth ce transmtters, storage of a cyndrca drums nto contaners or R.G. Gonzáez-Ramírez et a. (Eds.) : ICCL 204, LNCS 8760, pp. 47 60, 204. Sprnger Internatona Pubshng Swtzerand 204
48 I. Ltvnchev, L. Infante, and E.L.O. Espnosa stocng them nto an open area, pacagng bottes or cans nto the smaest bo, pantng trees n a gven regon as to mamze the forest densty and the dstance between the trees, and so forth [2, 6, 9]. Other appcatons one can fnd n the motor cyce ndustry, crcuar cuttng, communcaton networs, facty ocaton and dashboard ayout [6,, 2]. In ths paper we address the probem of pacng a set of crcuar tems n a rectanguar contaner. There are two prncpa types of obectves that have been used n the terature: a) regardng the crces (not necessary equa) as beng of fed sze and the contaner as beng of varabe sze and b) regardng the crces and the contaner as beng of fed sze and mnmze waste. Eampes of the frst approach ncude [8]: For the square contaner mnmze the ength of the sde and hence mnmze the permeter and area of the square; Mnmze the permeter of the rectange; Mnmze the area of the rectange; Consderng one dmenson of the rectange as fed, mnmze the other dmenson. Probems of ths type are often referred to as strp pacng probems (or as crcuar open dmenson probems). For the second approach varous defntons of the waste can be used. The waste can be defned n reaton to crces not paced (e.g., the number of unpaced crces or the permeter/area of unpaced crces), or ntroducng a vaue assocated wth each crce that s paced (e.g., area of the crces paced), etc. Many varants of pacng crcuar obects n the pane have been formuated as nonconve (contnuous) optmzaton probems wth decson varabes beng coordnates of the centres. The nonconvety s many provded by no overappng condtons between crces. These condtons typcay state that the Eucdean dstance separatng the centres of the crces s greater than a sum of ther rad. The nonconve probems can be taced by avaabe nonnear programmng (NLP) sovers; however, most NLP sovers fa to dentfy goba optma. Thus, the nonconve formuaton of crcuar pacng probems requres agorthms whch m oca searches wth heurstc procedures n order to wdey epore the search space. We refer the reader to revew papers presentng the scope of technques and appcatons for the crce pacng probem (see, e.g., [, 4, 5, 8, 7-9] and the references theren). In ths paper we propose a new formuaton for appromate souton of crcuar pacng probems usng a reguar grd to appromate the contaner. The nodes of the grd are consdered as potenta postons for assgnng centers of the crces. The pacng probem s then stated as a arge scae near 0- optmzaton probem. Two casses of vad nequates are proposed to strengthenng the formuaton. Numerca resuts are presented to demonstrate effcency of the proposed approach. To the best of our nowedge, the dea to use a grd was frst mpemented by Beasey [3] n the contet of cuttng probems. Ths approach was recenty apped n [0, 4, 5] for pacng probems. Ths wor s a contnuaton of [5].
Appromate Crce Pacng n a Rectanguar Contaner 49 2 The Mode Suppose we have non-dentca crces C of nown radus R, K = {,2,... K}. Let at most M crces C be avaabe for pacng and at east m of them have to be paced. Denote by I = {,2..., n} the node ponts of a reguar grd coverng the rectanguar contaner. Let F I be the grd ponts yng on the boundary of the contaner. Denote by d the Eucdean dstance between ponts and of the grd. Defne bnary varabes = f the centre of a crce C s assgned to the pont ; = 0 otherwse. In order to the crce C assgned to the pont be non-overappng wth other crces beng paced, t s necessary that = 0 for I, K, such that d < R + R. For fed et, N = {,:, d < R + R }. Let n be the cardnaty of N : n = N. Then the probem of mamzng the area covered by the crces can be stated as foows: subect to 2 ma R () I K m M, K, (2) I, I \ F, (3) K R mn d, I, K, (4) F +, for I, K, (, ) N (5) {0,}, I, K (6) Constrants (2) ensure that the number of crces paced s between m and M ; constrants (3) guarantee that at most one centre s assgned to any grd pont; constrants (4) ensure that the pont cannot be a centre of the crce C f the dstance from to the boundary s ess than R ; par-wse constrants (5) guarantee that there s no overappng between the crces; constrants (6) represent the bnary nature of the varabes. By defnton, N = {,:, d < R + R } and hence f (, ) N, then (, ) N. Thus, haf of the constrants n (5) are redundant snce we have: +, for I, K, (, ) N,
50 I. Ltvnchev, L. Infante, and E.L.O. Espnosa +, for I, K, (, ) N We may emnate any (none) of these two constrants to get the reduced equvaent formuaton. Ths can be represented by mutpyng constrants (5) by fed λ {0,} : λ + λ λ, for I, K, (, ) N, (7) subect to λ + λ. Ths way ether one of the redundant constrants s emnated ( λ + λ = ) or none ( λ + λ = 2 ). Snce emnatng redundant constrants does not affect the feasbe set, the probem ()-(6) s equvaent to ()-(4), (6), (7) for any λ fufng the normazed condton λ Λ= { λ {0,}: λ + λ,(, ) N }. Smar to pant ocaton probems [20] we can state non-overappng condtons n a more compact form. Summng up constrants (7) over (, ) N we get λ + λ λ,, K. (8) for I (, ) N (, ) N (, ) N Proposton. For any λ Λ constrants (5), (6) are equvaent to constrants (6), (8). Proof. If constrants (5) are fufed, than obvousy constrants (8) hod by constructon. Now et constrants (8) be fufed. Defne 0 0 N = {(, ) N : λ = }, N = {(, ) N : λ = 0}, N N = N, N = n, N = n. 0 0 By (8) we have n + n (, ) N and hence, f =, then = 0 for (, ) N. (9) By defnton, f (, ) N, then (, ) N. Thus by (8) we have λ + λ λ, K. (0) for I (, ) N (, ) N (, ) N 0 0 In partcuar, (0) s fufed for (, ) N. Snce λ + λ, then for (, ) N a λ n (0) are postve ( λ = ). Then by (0) we have: 0 f = for at east one (, ) N, then = 0. ()
Appromate Crce Pacng n a Rectanguar Contaner 5 Note that constrants (5) can be nterpreted n two ways. Frst, f =, then = 0 for a (, ) N. Second, f = for at east one (, ) N, then = 0. Combnng (9) and () we may concude that f constrants (8) are fufed, then constrants (5) hod. Remar. In [0] the compact formuaton n + n for I (2) N was used to represent non-overappng condtons for the case of pacng dentca crces of radus R. Here n s the cardnaty of the set N = { :, d < 2 R}. Ths case corresponds to a sngeton set K and a mutpers λ equa to n (8). Remar 2. Proposton remans true for nonnegatve (not necessary bnary) mutpers λ subect to λ + λ 0. The proof s smar. Eampe. Consder the same eampe as n [0]. Suppose we want to pac the mamum possbe number of crces of radus nto a rectange of wdth 4 and heght 3. Tae a rectanguar unform grd of sze Δ= aong both sdes of the contaner. The boundary grd ponts can be emnated from the consderaton (see [0] for detas) eavng ony 6 nteror grd ponts to state the non-overappng condtons as shown n Fg.. 2 3 4 5 6 Fg.. Eampe. For =,2...,6 the correspondng sets N are as foows: 2 3 4 5 6 N 2,4,5,3,4,5,6 2,5,6,2,5,2,3,4,6 2,3,5 The correspondng par-wse non-overappng constrants (5) have the foowng form:
52 I. Ltvnchev, L. Infante, and E.L.O. Espnosa 2 + + 2 2 3 + 3 + 2 + 4, 2 + 4, 3 + 5, + 5 2 + 5 3 + 6 2 + 6 6 + 2 6 + 3 6 + 5 + +, + 4 4 2 4 5 + + +, + + 5 5 2 5 3 5 4 5 6 Summng up constrants n each group (a mutpers λ equa to n (8)) we get the same compact formuaton as n [0], whch s equvaent to (3) for bnary : 3 + + + 3 2 4 5 5 + + + + + 5 2 3 4 5 6 3 2 5 6 (3) 3 + + + 3 (4) 3 + + + 3 4 2 5 5 + + + + + 5 5 2 3 4 6 3 + + + 3. 6 2 3 5 We see that n (3) haf of the constrants are redundant. Constrant + 2 appears n the frst and n the second group of (3), constrant + 4 s presented n the frst and n the fourth, etc. Obvousy, emnatng redundant constrants does not change the set defned by (3). However, the way of emnatng redundant constrants affects the equvaent system obtaned by summng up the remanng constrants. Suppose that we w emnate redundant constrants begnnng from the frst group of (3) where redundancy appears. That s, + 2 w be emnated from the frst group, 2 + 3 from the second, etc. Dong t ths way we get a reduced set of constrants 5 + 4 + 6 + 2 5 + 2 { 2 + }, { 3 + 2 },,, 6 + 3 (5) 4 + 2 5 + 3 5 + 4 6 5 + The compact equvaent system obtaned by summng up constrants n each group n (5) s as foows: +, 2
Appromate Crce Pacng n a Rectanguar Contaner 53 +, 3 2 2 + + 2, (6) 4 2 4 + + + + 4, 5 2 3 4 3 + + + 3. 6 2 3 5 If we now w emnate redundant constrants begnnng from the ast group, the reduced set of constrants becomes 2 + 3 + 2 2 + 4 3 + 5 + 4,,, { 4 + 5 }, { 5 + 6 } (7) + 5 2 + 5 3 + 6 2 + 6 The equvaent compact system obtaned by summng up constrants n each group n (7) s as foows: 3 + + + 3, 2 4 5 4 + + + + 4, 2 3 4 5 6 2 + + 2, (8) 3 5 6 +, 4 5 +. 5 6 Emnatng redundant constrants wth the obectve to baance the number of constrants n the groups of (3) we get the reduced system + 4, + 5 2 +, 2 + 5 3 + 2, 3 + 5 4 + 2 4 + 5 +,, { } 5 6 +. (9) + 6 2 6 3 The correspondng compact system equvaent to (9) s as foows 2 + + 2, 4 5 2 + + 2, 2 5 2 + + 2, 3 2 5 2 + + 2, (20) 4 2 5
54 I. Ltvnchev, L. Infante, and E.L.O. Espnosa +, 5 6 2 + + 2. 6 2 3 As foows from Proposton, the non-overappng condtons can be stated n dfferent forms. We have a famy of formuatons equvaent to (5) and obtaned for dfferent mutpers λ n (8). To compare equvaent formuatons, et P = { 0: +, for I, K, (, ) N }, P2 = 0: λ + λ λ, I, K (, ) N (, ) N (, ) N, where mutpers λ n P2 fuf normazng condton stated n Proposton. Proposton 2. P P2. Proof. Snce constrants of P 2 are a near combnaton of those n P wth nonnegatve mutpers λ, then P P2. To show that P P2we need to fnd a pont n P2 that s not n P. Ths pont can be constructed as foows. Choose (, ) N (and hence (, ) N ) such that λ, λ 2. Set to zero a the varabes e- (, ) N (, ) N cept,. Obvousy a constrants n P 2 correspondng to zero varabes are fufed. Defne, to fuf the two remanng constrants as equates: Denote n λ + = λ, λ + = λ (, ) N (, ) N (, ) N (, ) N = λ, n = λ 2 (, ) N (, ) N n n. The correspondng souton of the two equatons above s n ( n ) n ( n ) = <, = < nn nn wth + nn n n + = + >. nn Ths pont voates correspondng constrant n P and hence P P2 as desred. As foows from Proposton 2, the parwse formuaton ()-(6) s stronger than the compact one ()-(4), (6), (8) n the sense of [20].
Appromate Crce Pacng n a Rectanguar Contaner 55 3 Vad Inequates We may epect that the near programmng reaaton of the probem ()-(6) provdes a poor upper bound for the optma obectve vaue. For eampe, for K = and sutabe M, m the pont = 0.5 for a I may be feasbe to the reaed probem wth the correspondng obectve growng neary wth respect to the number of grd ponts. To tghtenng the LP-reaaton we consder two fames of vad nequates. The frst ensures that no grd pont s covered by two crces, whe the second guarantees that there s at most one centre assgned to the area covered by a crce. To present the frst famy, defne matr α as foows. Let α = for d < R, α = 0, otherwse. By ths defnton, α = f the crce C centered at covers pont. The foowng constrants ensure that no ponts of the grd can be covered by two crces: α, I (2) K I Note that (2) s not equvaent to non-overappng constrants (5). Constrants (2) ensure that there s no overappng n grd ponts, whe (5) guarantee that there s no overappng at a. We w refer to (2) as pont-coverng vad nequates. The second famy of nequates s stated as foows: +, for I, K. (22) d : < R To demonstrate that (22) s vad for the probem ()-(6) assume that = n (22). That s, the centre of the crce C s assgned at. By (22) we have 0, for I, K and then t foows that = 0for : d < R. That s, d : < R there are no other centres assgned to ponts nsde the crce centred at. For = 0 we have. Ths means that among a grd ponts covered by the d : < R (magnary) crce centred at, at most one pont can be assgned as a centre. Ths s true snce the dstance between any par of these ponts s ess than 2R and assgnng the centres of C voates non overappng constrants. 4 Numerca Eperments In ths secton we numercay compare dfferent probem formuatons and study the mpact of ntroducng vad nequates for the case of pacng equa crces. A rectanguar unform grd of sze Δ aong both sdes of the contaner was used. The test
56 I. Ltvnchev, L. Infante, and E.L.O. Espnosa bed set of nne nstances from [0, Tabe 3] was used for pacng a mama number of crces nto a rectanguar contaner of wdth 3 and heght 6. A optmzaton probems were soved by the system CPLEX 2.6. The runs were eecuted on a destop computer wth CPU AMD FX 8350 8-core processor 4 Ghz and 32Gb RAM. The foowng four formuatons were compared: parwse formuaton ()-(6) (compete), reduced formuaton ()-(6) wthout redundant constrants (haf), compact formuaton (2) as n [0] (compact), and the compact formuaton obtaned by summng up constrants n the reduced formuaton ()-(6) (compact haf). The resuts of the numerca eperment are gven n Tabe. Here the frst fve coumns present nstance number, crce radus, sze of the grd Δ, number of bnary varabes and the number of crces paced. The ast four coumns gve CPU tme (n seconds) for dfferent formuatons. For a probem nstances mpgap = 0 was set for runnng CPLEX. In ths tabe and a tabes beow, an asters ndcates that the computaton was nterrupted after the computaton tme eceeded 2-hour CPU tme. For probem nstances 7 and 9, where optmaty was not acheved wthout vad nequates nether for compete nor for compact formuatons, the number of crces n coumn 5 corresponds to optma souton obtaned by usng vad nequates (see the tabes beow). As we can see from Tabe, CPU tme for compete formuatons s much ower than for the compact, especay for arge probems. Emnatng redundant constrants typcay (but not aways) reduces CPU tme. Athough emnatng redundancy does not change correspondng LP-reaaton, t may affect the path seected by branch and bound technque and thus resuts n ncrease/decrease of CPU tme. Instance Crce radus Tabe. Integer souton (gap 0%) wthout vad nequates Δ Probem dmenson Crce number compete haf compact compact haf 0.5 0.25 697 8 2 276 5 2 0.625 0.5625 403 0 7 4 040 50 3 0.5625 0.070325 2449 3 337 86 666 83 4 0.375 0.046875 425 32 6 4 2698 69 5 0.325 0.07825 239 45 96 4 * * 6 0.4375 0.0546875 3666 2 7473 7654 * * 7 0.25 0.0625 3649 74 * * * * 8 0.275 0.06875 2880 6 32 77 * * 9 0.875 0.046875 6897 40 * * * * In the second part of the eperment we study the effect of ntroducng vad nequates n the probem formuatons. Tabe 2 presents vaues of upper bounds obtaned by the LP reaaton for compete formuaton ()-(6) (coumn 4) and for the compete formuaton wth vad nequates (2) (coumn 5). The resuts for compact
Appromate Crce Pacng n a Rectanguar Contaner 57 formuatons and vad nequates (22) are qute smar and are omtted here. We see that a) LP-bounds wthout vad nequates are very poor and b) vad nequates mprove sgnfcanty the quaty of the bounds for a probem nstances. Instance Probem dmenson Tabe 2. LP-reaaton Crce number compete Compete + C 697 8 348.5 9 2 403 0 70.5 0 3 2449 3 224.5 4.07 4 425 32 72.5 36.33 5 239 45 069.5 53.4 6 3666 2 833.5 23.86 7 3649 74 824.5 90.98 8 2880 6 440 72 9 6897 40 3448.5 62 Tabe 3 presents the mpact of vad nequates (2) and (22) (C and C2, respectvey) on CPU tme requred to get nteger souton wth mpgap = 0 for parwse formuaton wth (compete) and wthout (haf) redundant constrants. Tabe 4 presents correspondng resuts for compact formuatons. We see that ntroducng vad nequates decreases CPU tme for a probem nstances and a probem formuatons. Athough ntroducng vad nequates sghty ncreases tme to sove the LP-reaaton, the effect of mprovng the quaty of the LP-bound becomes more mportant for the convergence of the overa branch and bound scheme. That s why CPU tmes decrease sgnfcanty for hard nstances 6, 7, 9, whe for easy nstances the decrease may be reatvey modest. Moreover, wth vad nequates CPU tmes to get provaby optma soutons ( mpgap = 0 ) s comparabe wth those reported n [0] for ther heurstc approach. Tabes 5, 6 are smar to Tabes 3, 4, but present CPU tme for mpgap = 5%. Here numbers n parenthess n the frst coumn ndcate the best nteger souton obtaned (crces paced). In the net coumn ths number s ndcated ony f t s dfferent from the prevous coumn. We see that the mpact of vad nequates s ess vsbe and even may be negatve (see, e.g., nstances 7-9 n Tabe 5). For arger vaues of the gap the quaty of bounds s ess mportant comparng wth ncreasng CPU tme to sove the LP-reaaton. Comparng vad nequates C and C2, we may concude that there s not much dfference; they both produce smar mpact on CPU tme.
58 I. Ltvnchev, L. Infante, and E.L.O. Espnosa Tabe 3. Integer souton (gap 0%) wth vad nequates, compete formuaton Instance compete Compete C Compete C2 Compete C + C2 haf Haf C Haf C2 Haf C+C2 2 2 2 2 7 5 2 8 4 9 0 3 337 82 8 86 86 75 72 73 4 6 9 6 8 4 4 3 4 5 96 63 57 50 4 89 80 9 6 7473 392 355 22 7654 379 364 776 7 * 353 3540 5542 * 378 386 2978 8 32 87 03 88 77 87 03 04 9 * 7437 793 24602 * * * 892 Instance Tabe 4. Integer souton (gap 0%) wth vad nequates, compact formuaton compact Compact C Compact C2 Compact C+C2 haf Haf C Haf C2 Haf C+C2 276 4 4 4 5 4 4 5 2 040 35 33 27 50 2 3 6 3 666 37 36 4 83 32 32 34 4 2698 29 28 3 69 92 94 87 5 * 89 88 09 * 027 029 849 6 * 39347 39260 454 * * * * 7 * * * * * * * * 8 * 2523 2525 246 * 2860 2850 288 9 * * * * * * * * Tabe 5. Integer souton (gap 5%) wth vad nequates, compete formuaton Instance Compete Compete Compete Compete Haf Haf Haf Haf C C2 C+C2 C C2 C+ C2 4 (8) 2 62 (0) 3 3 3 42 0 3 285 (3) 83 82 83 23 75 75 74 4 4 (32) 6 6 7 4 3 3 4 5 28 (45) 28 27 34 26 42 4 33 6 7447 (2) 586 579 538 7654 379 364 776 7 373 (73) 26 (74) 230 782 978 76 756 862 8 39 (60) 77 (59) 7 58 9 (6) 8 (60) 65 (59) 69 (59) 9 22 (35) 260 2590 236 3768 3458 355 394
Appromate Crce Pacng n a Rectanguar Contaner 59 Tabe 6. Integer souton (gap 5%) wth vad nequates, compact formuaton Instance compact Compact C Compact C2 Compact C+C2 Haf Haf C Haf C2 Haf C + C2 6 (8) 3 2 3 2 2 2 2 2 372 (0) 2 3 28 48 9 9 2 3 63 (3) 30 30 32 063 27 26 26 4 234 (32) 3 2 4 5 54 53 75 5 * (44) 47 (45) 424 043 * 634 630 568 6 * (20) 39347 (2) 39260 3332 * * * * 7 * (63) * (74) * * (73) * 734 734 7340 8 * (50) 89 (59) 809 325 (6) 3859 (60) 999 972 70 (59) 9 * (6) * (35) * * (36) * (35) * * * Summarzng the resuts of our numerca eperment, we may concude that parwse formuaton ()-(6), wth or wthout redundant constrants, s computatonay much more attractve than compact formuatons. Ths s n good correspondence wth Proposton 2. Vad nequates proposed n the paper are very effcent f a provaby optma souton wth zero gap s requred. 5 Concusons We studed nteger programmng formuatons obtaned by usng a grd to appromate a contaner n a pacng probem. The equvaence of these formuatons s demonstrated and a stronger formuaton n the sense of [20] s ndcated. Two casses of vad nequates are used to strengthen the formuatons. Numerca resuts are presented to demonstrate the usefuness of these nequates. An nterestng area for the future research s to use the LP-reaaton of nteger formuatons to get a feasbe souton. The LP-reaaton wthout vad nequates s trva a varabes are equa to 0.5. However, aggregatng vad nequates changes sgnfcanty the structure of the reaed souton and ths mpact becomes more vsbe for fne grds. Some resuts n ths drecton are n course. The other drecton for the future research s to study the use of Lagrangan reaaton/decomposton [6] to cope wth arge dmenson of the probem formuaton. Other vad nequates [7] can aso be used. References. Aeb, H., Hf, M.: Sovng the crcuar open dmenson probem usng separate beams and oo-ahead strateges. Computers & Operatons Research 40, 243 255 (203) 2. Batacogu, E., Moore, J.T., H, R.R.: The dstrbutor s three-dmensona paet-pacng probem: a human ntegence-based heurstc approach. Internatona Journa of Operatona Research, 249 266 (2006)
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