Cutting Stock by Iterated Matching. Andreas Fritsch, Oliver Vornberger. University of Osnabruck. D Osnabruck.

Transcription

1 Cutting Stock by Iterated Matching Andrea Fritch, Oliver Vornberger Univerity of Onabruck Dept of Math/Computer Science D-4909 Onabruck Abtract The combinatorial optimization problem conidered in thi paper i a pecial two dimenional cutting tock problem ariing in the wood, metal and gla indutry Given a demand of non oriented mall rectangle and a theoretically innite et of large tock rectangle of given length and width, the aim i to generate layout with minimal wate, pecifying how to cut the demand out of the tock rectangle Special retriction are choen with repect to the production condition of the at gla indutry An iterative algorithm for olving thi problem heuritically i developed By a maximum weight matching uitable rectangle are matched in every iteration tep and the matched pair are conidered a new, o called meta rectangle Thee meta rectangle can be treated in the ame way a ordinary rectangle Furthermore, by the ue of hape function the orientation of the demand rectangle are not xed until the layout ha been computed The algorithm, programmed in C, ha been teted with everal intance, containing between 52 and 11 rectangle, taken from real demand of a gla factory The reulting layout, calculated within a few minute (on a PC-48), have an average wate of le than percent 1 Introduction Becaue of the increaing cot of raw material and the need to avoid indutrial wate, olving cutting tock problem became of great interet in the area of operation reearch Cutting tock problem belong to the cla of combinatorial optimization problem, o a olution among all poible olution ha to be found, which optimize a criterion function ubject to a et of contraint Unfortunately the ize of the olution pace containing all feaible olution in general i enormou A a conequence an exhautive earch within the whole olution pace i impractical 1

2 and ophiticated heuritic which enure the calculation of reaonable olution within limited time are required The combinatorial optimization problem conidered in thi paper i a pecial two dimenional cutting tock problem ariing in the wood, metal and gla indutry Given a demand of mall rectangle and a et of large tock rectangle, layout have to be found which pecify how to cut the demand out of the tock rectangle with minimal wate Gilmore and Gomory ued linear programming to olve uch kind of a problem exactly [Gil 191] [Gil 195] Their uggetion were improved by Herz [Her 1972] Branch&Bound and tree earch algorithm were developed by Cani [Can 1979] and Whitlock & Chritode [Whi 1977] Thee algorithm fail if more than 20 rectangle have to be packed Mot of the heuritic algorithm for generating layout are baed on a greedy trategy After orting the demand rectangle they are placed in the tock rectangle and none of them can be repoitioned Level oriented packing algorithm were developed by Coman, Garey, Johnon and Trajan [Cof 1980] Thee algorithm are fat, but the performance bound are not good enough to tand the requirement of the indutry [Cof 1990] [Gar 1981] A weakne of the mot known algorithm i that at the beginning of the algorithm the alignment of all rectangle that have to be packed are xed In many domain, epecially in the at gla indutry, a xing of the alignment i unneceary and it only lead to the fact, that ome layout with probably minimal wate are not conidered Furthermore, the mot algorithm are unable to pack a high number of demand rectangle a dene a it i neceary for the at gla indutry Therefore, an algorithm i needed which i able to handle a et of many rectangle (up to 200) and which take every poible and uitable layout in conideration In the econd ection of thi paper a detailed decription of the conidered problem i given and the olution trategy i preented Since hape function are ued for olving the problem, an introduction to thi theme i given in ection three The fourth ection deal with licing tree In the fth ection the maximum weight matching problem i decribed The developed algorithm conit of four tage which are decribed in ection ix Computational reult are given in ection even 2 The Two Dimenional Cutting Stock Problem The problem conidered in thi paper can be formulated a follow: Given a nite number n of demand rectangle of length l i and width w i and a theoretically innite et of tock rectangle fr1; R2; : : :g of a given length and width Find layout for cutting all demand out of the tock rectangle o that the minimum number of tock rectangle i required Only guillotine cut are permitted A rotation of the demand rectangle of 90 i allowed The above contraint are choen with repect to the production condition in the at gla indutry A 90 rotation of the rectangle i poible becaue gla i an iotropic material that mean that it ha the ame tructure in every dimenion Only guillotine cut are permitted mean that the cut mut go from one edge of a rectangle to the oppoite edge in a traight line Thi 2

3 contraint i crucial, becaue mot factorie in the gla indutry ue a cutting machine which can only carry out thee cut It i aumed that the tock rectangle i large enough, o that every demand rectangle t in the tock rectangle in at leat one orientation Without lo of generality it i aumed that the length and width of all rectangle are integer There are two more retriction concerning the practicability of cutting in the gla indutry: the contruction of travere and the obervance of cut ditance For cutting the demand rectangle out of the tock rectangle the latter are placed on a cutting table There a carriage with a diamond placed at the bottom of it lice the gla heet in x- and y-dimenion Since the big gla heet are even wore to handle (the typical extenion are up to ix meter in length and three meter in width) the rt cut that have to be made are in y-direction and go directly from one point of the longer ide of the heet to the oppoite ide They divide the tock rectangle into ome ection We call thee cut travere cut and the reulting ection of the tock rectangle are called travere Notice that the travere cut are alo guillotine cut The x-dimenion of the travere hould not exceed a given value For example, a typical bound of the travere length i 2300 mm if a tock heet of 000 mm length i ued Since the maximal travere length depend on the dimenion of the ued tock and that of the ued cutting table, thi value hould be given together with the other parameter of an intance like the dimenion of tock and demand rectangle A nice property of the licing of gla i that no awdut arie a it doe by the cutting of wood But it i crucial to take care of the ditance between the cut The cutting of gla i done by licing the tu and then breaking it along the lice If two lice are too cloe, the gla can not be broken The minimal ditance between two cut i the double heet thickne, but for a gla heet of 4 mm thickne the typical value of 20 mm i choen Alo thi value hould be given in the intance Figure 1 how how a tock rectangle can be liced into the demand rectangle The number give information about a poible equence of the cut The rt two cut are travere cut which divide the gla heet into three travere Figure 1: Poible guillotine cut in a layout For olving the given two dimenional cutting tock problem of the at gla indutry an iterative algorithm i developed baed on the following idea: In every iteration tep ome rectangle are paired by a maximum weight matching and the matched pair are conidered a new, larger 3

4 rectangle, o called meta rectangle In the following iteration tep thee meta rectangle are treated in the ame way a ordinary rectangle The iterative matching automatically take care of the guillotine tructure The iteration top if the contructed meta rectangle are travere Thee travere will be aigned to the tock heet by a rt t decreaing algorithm In every iteration tep all poible alignment of demand and meta rectangle are conidered by making ue of hape function 3 Shape Function A meta rectangle ha everal layout according to the orientation of the horizontal or vertical cut which i ued to plit the meta rectangle into the two buddie it reulted from The layout alo depend on the alignment of the demand rectangle the meta rectangle contain The alignment of the demand rectangle are not xed at any time the algorithm proceed For example, if we match two demand rectangle A and B the reulting meta rectangle AB ha eight eentially dierent layout: A lie above or left from B and both rectangle can be rotated Figure 2 how the eight eentially dierent layout for a meta rectangle coniting of two demand rectangle Figure 2: Layout for a meta rectangle with two demand rectangle A ingle layout of a meta rectangle i repreented by a licing intruction Thi i a four dimenional vector (x; y; o; p), where x and y are the length and width of the meta rectangle in it conidered layout, o 2 fh; vg i the orientation of the horizontal or vertical cut ued to plit it and p i the point (according to o) where the cut tart To decribe all poible layout of a meta rectangle, a hape function i ued Such function are known from oorplanning problem [Sto 1983], [Ott 1983] A hape function R for a rectangle R can be conidered a a decreaing, piecewie linear function The interpretation i that R (w) i an upper bound on the length of a (meta) rectangle that ha the width w So, if a meta rectangle ha been computed which contain a et of demand rectangle and if the dimenion of the tock 4

5 rectangle are known, the maximal needed x-dimenion for a layout of the meta rectangle can be found ubject to the given width of the tock rectangle w l (l,w,h,0)? (w,l,h,0)? f l w - Figure 3: Shape function of a demand rectangle A hape function i conidered a a lit of licing intruction The hape function of a demand rectangle of length l and width w > l conit of two licing intruction ((l; w; h; 0); (w; l; h; 0)) The cut orientation and it tarting point i of no ue here, but later we will need thi information for the cutting of meta rectangle unequal to demand rectangle In Figure 3 an example of a hape function for a demand rectangle i given The hape function of a meta rectangle i calculated by the hape function of the two demand or meta rectangle it conit of, with the help of the compoition procedure: Let f1 and f2 be two hape function and i = (x i ; y i ; o i ; p i ) a licing intruction of f i (i = 1; 2) The compoed licing intruction for a horizontal rep vertical cut i: horizontal cut: comp(hor; 1 ; 2 ) := (max(x 1 ; x 2 ); y 1 + y 2 ; h; min(y 1 ; y 2 )) vertical cut: comp(ver; 1 ; 2 ) := (x 1 + x 2 ; max(y 1 ; y 2 ); v; min(x 1 ; x 2 )) By compoing all licing intruction of f1 with thoe of f2, two hape function f hor and f ver are computed which contain all poible horizontal rep vertical licing intruction Figure 4 how a horizontal and vertical hape function, reulting from the compoition of the hape function of two demand rectangle Three licing intruction per cut orientation are found Thi lead to the fact that a layout of the reulting meta rectangle i not included in the lit of licing intruction, if another layout of the ame width but a maller length or the ame length but a maller width exit Hence, only feaible licing intruction are tored in the hape function For example, layout three and four in the upper row and layout one and two in the lower row of Figure 2 are treated in thi way 5

6 f1 f2 f ver f hor f2 f1 - - a) horizontal Figure 4: Shape function with given cut orientation b) vertical To calculate a ingle hape function f for a meta rectangle, we take the minimum about f hor and f ver by merging thi two function and deleting the dominated licing intruction For example, a licing intruction of f ver i dominated by one of f hor, if f hor (w) < f ver (w) for a given width w If two licing intruction of f hor and f ver have the ame width and the ame length then we will determine here, that the horizontal cut i choen The maximal number of licing intruction that have to be tored in a hape function f which i calculated by compoing two other hape function f1 and f2 i 2(jf1j+jf2j?1), where jf i j denote the number of licing intruction in function f i (ee [Mue 1991]) f hor f ver f Figure 5: Minimum hape function Figure 5 how the minimum hape function of the function f hor and f ver hown in Figure 4 In thi example of calculating the minimum function, no dominated licing intruction have to be deleted - 4 Slicing Tree To tore the actual pair in every iteration tep we ue binary licing tree Every leaf of a tree repreent a demand rectangle, an interior node of a tree repreent a meta rectangle Two node have the ame father if the rectangle repreented by thee node are paired In every node the hape function of the correponding meta rectangle i tored

7 A licing tree i built bottom up At the beginning of the algorithm all licing tree conit of only one node (the root) repreenting the demand rectangle In every iteration tep, ome of the tree are combined by a common father, which i the root of a new licing tree repreenting the new meta rectangle The iteration top, if no more uitable pair can be matched A detailed decription of thi top criteria i given in ection ix Since all node contain a hape function and every hape function decribe all poible layout of a meta rectangle, the licing tree contain an exponential number of poible layout for the tock rectangle When the algorithm ha ended, we can detect how the demand rectangle hould be cut out of the tock rectangle Therefore, a tree i travered in top down order Since every node tore a hape function we can detect how the rt cut in the tock rectangle ha to be made by conidering the hape function in the root of a tree and earching for the rt licing intruction whoe y-value i maller or equal to the width of the tock rectangle Due to the property of the hape function mentioned above, we automatically get the licing intruction which decribe the layout with the minimal length according to the made pairing If the rt cut ha been carried out, the tock rectangle i plit into two part The hape function in the on of the root decribe how thee two part eciently have to be plit further Thi proceeding i hown in Figure by an example By the traveral of all licing tree (one licing tree for every layout) the demand rectangle can be produced (,(000,3200,v,1400),) (,(1400,3200,h,1050),) Figure : Layout and correponding licing tree 5 Maximum Weight Matching To nd out in every iteration tep, which of the available rectangle have to be paired, a maximum weight matching in a complete undirected weighted graph i ued The node of the graph repreent the available (meta) rectangle, the edge repreent the poible pair and the weight give information about the benet of a pair The higher the weight, the better the pair The weight are found by a o called benet function, which evaluate the poible pairing of two rectangle by conidering the correponding hape function For example, a poible benet function for the pairing of two rectangle i to calculate the average wate of all poible layout of the 7

8 reulting meta rectangle and to ubtract thi value from 100 Thi encourage the formation of meta rectangle having many denely packed layout If all weight of the edge are calculated, the maximum weight matching nd the et of edge, which maximize the um of all weight under the contraint that no two edge are adjacent Thi problem can be formulated a a LP-problem a follow: Maximize ubject to nx nx i;j=1 c i;j x i;j x i;j 1 for i = 1; : : : ; n j=1 x i;i = 0 for i = 1; : : : ; n x i;j = x j;i for i; j = 1; : : : ; n x i;j 2 f0; 1g for 1 i j n Here, n i the number of available meta rectangle and c i the cot matrix The value of x i;j i 1 if the rectangle i and j are matched and x i;j = 0 if not Edmond howed, that the maximum weight matching problem can be olved in time O(n 3 ) The algorithm ued in thi application wa uggeted by Gabow [Gab 1973] Figure 7 how the matching algorithm for even rectangle in three iteration tep The edge with zero benet are not hown {1,2} 90 {3,4} {5,7} Iteration {1,2,5,7} {3,4,} Iteration 1 Iteration 3 Figure 7: Maximum weight matching in three iteration 8

9 Iterated Matching Up to now, it ha not been mentioned at which time the iteration top and what a uitable pair of rectangle i The top criteria i choen a follow: One retriction when cutting gla heet i, that the tock rectangle mut be plit into travere at the beginning of the cutting; the rt cut mut be parallel to the horter ide of the tock rectangle, o that the tock rectangle i divided into ection with the ame width a and a maller length than the original heet So the iterated matching top if meta rectangle have been generated with at leat one layout in which they repreent a travere By computing uch meta rectangle the problem of nding the layout for the tock rectangle i reduced to a one dimenional bin packing problem: the travere, all of nearly the ame width a that of the tock rectangle, have to be aigned to the latter in uch a way, that the tock heet are lled nearly complete in length To nd only uitable meta rectangle the algorithm ha to take care of the further retriction a it i the bound of the travere length, the cut ditance and the obervance of the guillotine tructure The lat i preerved ince the iterated matching automatically take care of the guillotine tructure a een above To prevent cut with ditance of le than the given minimum, licing intruction which do not fulll thi ubjection are not tored during the computation of hape function To reduce the number of the licing intruction of a hape function alo thoe intruction are not tored which have a x- or y-value bigger than the larger ide of the tock rectangle The benet function etimate a potential pair with zero value if the reulting meta rectangle doe not have a layout in which it t into the given travere hape All thee obervation are made in any iteration tep The developed algorithm conit of four tage The rt two tage are ued to contruct travere In the third tage the travere will be improved The fourth tage i needed to align the travere to the tock rectangle In the rt tage univeral rectangle are calculated by an iterative ue of the maximum weight matching To call a meta rectangle univeral, it mut comply with two requirement: 1 The average wate of all layout of the meta rectangle mut be maller than ve percent and 2 There mut exit another meta rectangle, o that the meta rectangle reulting from the pairing of thee two i a travere A travere hould have a wate area of le than 4 percent In every iteration tep of tage one a matching graph i formed which node repreent the available non univeral rectangle The benet function in tage one deliver a high value for a poible pair if the average percentage wate of all layout of the potentially reulting meta rectangle i maller than ve percent The benet i zero if all layout of the reulting meta rectangle violate the retriction like travere length and cut ditance After the calculation of all weight in the matching graph the maximum weight matching chooe the bet marriage Stage one end if no more univeral rectangle can be contructed without violating the retriction The econd tage conit of only one iteration tep Here the n contructed univeral rectangle are matched to n=2 travere For thi alo the maximum weight matching i ued A an indicator for the deciion whether a meta rectangle (reulted from the pairing of two univeral rectangle) i a travere or not, the percentage of the ued area of it \relevant" layout i computed To nd the relevant layout, the rt licing intruction of the hape function i choen with a y-value maller than or equal to the width of the tock rectangle By multiplying the x- and the y-value 9

10 of thi licing intruction the area which i claimed by the correponding layout i calculated To compute the percentage of the ued area, only the area of all contained demand rectangle have to be added, multiplied with 100 and divided by the claimed area The higher the percentage of the ued area, the better the travere Thi ha to be repected by the benet function of tage two, alo enuring that none of the other retriction i violated Since the iterated matching i a greedy trategy, the third tage i needed to improve thoe travere with a lot of wate area The trategy of improvement i to chooe the wore travere, for example thoe with le than 9 percent ued area, and improve them by interchanging ubtree of the correponding licing tree To nd out whether an interchangement hould be made or not, the um of all x-dimenion of the travere i calculated before and after the interchangement The idea i that the travere are improved, if the um of all travere length after interchanging i maller than before it, becaue then all demand rectangle have been packed on a maller domain Hence le wate will be produced Figure 8 how thi proceeding by an example R 3 R 4 R 5 R 4 R 1 R 2 R 5 R 1 R 2 R 3 R 1 R 2 R 3 R 4 R 5 R 1 R 2 R 5 R 3 R 4 Figure 8: Improvement of travere by interchanging ubtree Since the travere all have nearly the ame width a the tock rectangle, the problem of packing the rectangle into the tock heet i reduced to a one dimenional bin packing problem Therefore, in the fourth tage the travere are aigned to the tock rectangle Thi could be done by a rt t decreaing algorithm: lit the travere ordered by decreaing x-dimenion and let the tock heet be indexed by increaing number Aign the rt travere of the lit to the tock rectangle with the lowet index it t in Then aign the econd travere by the ame rule If a travere t in none of the already ued tock heet, a new tock rectangle ha to be ued The rt t decreaing nd olution for thi one dimenional bin packing which are at mot 22 percent wore than the optimal olution, and it can unfortunately be thi bad [Gar 1979] Therefore a modied rt t decreaing algorithm i ued The modication i, that after the normal rt t decreaing i completed, a ytematic interchange of travere between the ued tock rectangle i made with the aim to ll the tock rectangle a full a poible After the aignment ha been computed, all travere of one tock heet are paired and the correponding licing tree and hape function are computed So, at the end of tage four for every ued tock rectangle a licing tree i computed 10

11 which give information about how to cut the gla heet The good reult of the tock cutting by iterated matching are hown in the table of the following ection 7 Computational Reult The algorithm, programmed in C, ha been teted with everal intance taken from real demand of a gla factory, containing between 52 and 11 rectangle A PC-48 (33 MHz) wa ued to meaure the time requirement of the algorithm The time give information about the real computation time, ie requirement for input and output are not conidered The tock rectangle are of dimenion about 000 mm x 3200 mm for all intance The maximal travere length i 2300 mm and the minimal cut ditance i 20 mm for all intance except intance three Here the minimal cut ditance i 40 mm ince the thickne of the required tock heet i mm and not 4 mm a it i for the other intance Reult of the Iterated Matching Algorithm intance n area time in ec average wate in % LNS RNS in mm 2 tage 1+2 tage 3 match travere tock :58 4:9 5:4 7:313 7: :11 5:18 5:34 :98 : :05 8:35 9:73 3:39 3: :33 4:59 4:0 5:559 5: :4 4:80 4:84 4:277 4: :73 3:87 3:97 8:75 8: :03 4:82 :40 :137 :241 Table 1: Reult of the four tage algorithm Table 1 how the data of the intance and the computational reult of the four tage algorithm dicued in ection ix The number n of the demand rectangle of the intance i given in the econd column and the um of all demand rectangle area i lited in the third column The time column give information about the time requirement in econd of tage one, two and three Stage one and two are conidered together, becaue they both how how fat the iterated matching work The time needed to run the modied rt t decreaing i not lited, becaue it only take le than one econd The match wate i the average wate in percent of all travere after tage two, the travere wate i the wate after tage three and the tock wate i the average wate of all computed layout at the end of the algorithm The number of the at leat needed tock rectangle (LNS) and of the real needed tock rectangle (RNS) give information about how good the modied rt t decreaing aign the travere to the tock rectangle LNS i a lower bound for the number of required tock rectangle, RNS i the fractional number of really needed tock rectangle LNS and RNS are calculated a follow: Let k be the number of travere after tage three, t i the length of travere i according to the 11

12 given width W and length L of the tock heet Then kx LNS = 1 L i=1 i a lower bound for the number of needed tock heet Notice, that it i not guaranteed that thi bound can really be reached To calculate RNS, let 2 NI be the number of ued tock rectangle R i,i = 1; : : : ; Aume that R i the tock heet which i lled at leat in length and let l denote the ued x-dimenion of R Then RNS = (? 1) + l L i the fractional number of really needed tock rectangle The iterated matching produce travere with little average wate (about 8 percent) in a very hort time Exception are the intance two and three Here the greedy behaviour of the matching trategy ha to be adjuted Thi i done in tage three The improvement of the travere in tage three take in mot cae only a few econd (no more than 95 econd), except for intance two If the third tage for olving intance two i conidered in detail it can be een that mot of the improvement i done in the rt thirty econd Hence, a time retriction for tage three would be ueful Notice, that it i not the number of demand rectangle, that make the algorithm impracticable for intance two Intance ix involve nearly the ame number of demand rectangle but it ha the mallet wate of all computed intance and it take only two minute and 42 econd to compute thi olution The modied rt t decreaing nd good olution for the aignment of the travere to the tock rectangle Thi can be een in the very mall dierence between LNS and RNS Analyzing Table 1, it can be een that the average wate of all computed layout i le than ix percent and that the time requirement, with exception of intance two, i le than three minute Four of the even intance really take le than 112 econd for computation Hence, we can conclude that the iterated matching i a new ecient method for olving the two dimenional cutting tock problem ubject to the retriction of the indutry which i applicable for intance with a large number of rectangle t i Reference [Can 1979] [Cof 1980] [Cof 1990] [Gab 1973] P de Cani, \Packing problem in theory and practice", Department of Engineering Production, Univerity of Birmingham, March 1979 E G Coman, M R Garey, D S Johnon and R E Tarjan, \Performance Bound for Level-Oriented Two-Dimenional Packing Algorithm", SIAM Journal on Computing 9, 4 (1980), pp 808{82 E G Coman and P W Shor, \Average-Cae Analyi of Cutting and Packing in Two Dimenion", European Journal of Operational Reearch 44, 2 (1990), pp 134{ 145 H Gabow, "Implementation of Algorithm for Maximum Matching on Nonbipartite Graph", PhD Thei, Stanford Univerity,

13 [Gar 1979] [Gar 1981] [Gil 191] [Gil 195] [Her 1972] [Mue 1991] [Ott 1983] [Sto 1983] [Whi 1977] MR Garey und DS Johnon, \Computer and Intractability", Freeman, San Francico, 1979, pp 12{127 M R Garey and D S Johnon, \Approximation Algorithm for Bin Packing Problem: A Survey", in Analyi and Deign of Algorithm in Combinatorial Optimization, Vol 2, G Auiello and N Lucertini, ed, Springer Verlag, Berlin, 1981, pp 147{172 P C Gilmore and R E Gomory, \A Linear Programming Approach to the Cutting- Stock Problem", Operation Reearch, Vol 9 (191), pp 849{859 P C Gilmore and R E Gomory, \Multitage cutting tock problem of two and more dimenion", Operation Reearch, Vol 13 (195), pp 94{120 J C Herz, \Recurive Computational Procedure for Two-Dimenional Stock Cutting", IBM Journal of Reearch and Development 1 (1972), pp 42{49 R Muller, \Hierarchiche Floorplanning mit integrierter globaler Verdrahtung", Report No 81, Fachbereich Mathematik{Informatik Univeritat{GH{Paderborn, February 1991, pp 3{4 RHJM Otten, \Ecient Floorplan Optimization", In Proceeding of the International Conference on Computer Deign: VLSI in Computer, IEEE, 1983, pp 499{ 502 L Stockmeyer, \Optimal Orientation of Cell in Slicing Floorplan Deign", Information and Control, No 57 (1983), pp 91{101 C Whitlock and N Chritode, \An Algorithm for Two-Dimenional Cutting Problem", Operation Reearch, Vol 25, Nr 1, January-February 1977, pp 30{44 13