A COMPUTER SlMULATl‘ON OF A PACKAGiNG LINE Thesis for the Degree of M. S. MlCHIGAN STATE UNWERSITY JAMES RANDOLPH STONEMAN 1975 III II III III II III II III III II III L 1293101403909 LIBRARY Michigm State University This‘is to certifythat the thesis entitled A COMPUTER SIMULATION OF A PACKAGING LINE presented by JAMES RANDOLPH STON EMAN has been accepted towards fulfillment of the requirements for M.S. degree in PACKAGING Dr. Wayne H. Clifford Major professor Date February 25, I976 0-7 639 5?.» a? not» '5’! v: 5 2,391? 21.1%» New L ’ 010610 DECIBZUOS ABSTRACT A COMPUTER SIMULATION OF A PACKAGING LINE By - James Randolph Stoneman This thesis and its computer program are designed to simulate the operation of a high speed beverage canning line such as a beer or soft drink line. The packaging line chosen for this study is composed of five machines and a connecting conveyor system. Through detailed time studies, accurate data has been collected on the speeds and relevant downtimes of these machines and programmed into the simulation procedure. Treating these as con- stants, the lengths of the interconnecting conveyor lines can be altered and their effect on total line production can be examined. An optimum conveyor length, matchingi capital investment against production, can be arrived at without having to spend the money or take the risk of actually making a physical change in the line. A COMPUTER SIMULATION OF A PACKAGING LINE BY James Randolph Stoneman A THESIS Submitted to Michigan State University in partial fulfillment of the requirements ' for the degree of MASTER OF SCIENCE School of Packaging 1975 ACKNOWLEDGMENTS. I would like to thank Dr. Wayne Clifford without whose gracious help and assistance this thesis would never have been written. Sincere appreciation is also extended to Mr. Bill Weatherston for his help in completing the basic research required for this paper. Last, but not least, I would also like to thank my wife, Marsha, for her patience and understanding. . ii TABLE OF CONTENTS Page LIST OF TABLES . . . . . . . . . . . . iv LIST OF FIGURES . . . . . . . . . . . . v Chapter- I. INTRODUCTION . . . . . . . . . . . 1 II. PROBLEM STATEMENT . . . . . . . . . 8 III. STIMULATION MODELING . . . . . . . . 11 IV. MODEL VALIDATION (DOWNTIME SUBROUTINE) . . 18 V. COMPUTER PROGRAMMING . . . . . . . . 31 VI. PROOF OF THE COMPUTER.PROGRAM. . . . . . 40 VII. SUMMARY . . . . . . . . . . . . 69; APPENDIX . . . . . . . . . . . . . . 71 BIBLIOGRAPHY . . . . . . . . . . . . . 78 Table > >- D> aoo 953nm _ _ weapons» . _ _ _ . . .. _. ...._.. ._.._..__r-.-...1 13 KLP(I): the production for machine I. NH(I): the number of packages to leave conveyor I to enter machine I. It is the smaller of the machine capacity and the conveyor content. NC(I): the number of packages to leave machine I to enter conveyor I+l. It is the smaller of the conveyor capacity and the machine content. The simulation program can be divided into two parts: (Figure 3.2): I. the exiting packages from M(I) II. the entering packages to M(I) I. The exiting packages A. A check is run on the station machine M(I) to see if it is down. 1. If yes: no units will exit, output = O 2. If no: the machine is Operating. The number of packages (units) in this machine from the previous cycle will exit on this cycle if the exit conveyor C is open. B.- A check is made to see if the empty space on the conveyor C(I+l) is capable of holding all the packages exiting from the maching M(I)' 1. If no: the number on the conveyor is incremented by the number of packages it can accept (denoted by NC) and the number in the machine is lessened by a similar amount, or ICP = ICP + NC NMP = NMP - NC (I) (I) 14 2. If yes: the number on the conveyor is incremented by the entire quantity in the machine and the machine is left empty. Equations = ICP = ICP + NMP(I) NC = NMP NMP(I) = 0 Show this operation. Machine I can now accept packages from conveyor 1. ”II. The entering packages A. A check is run to see if the empty Space in the machine M(I) is capable of holding all the packages exiting from the conveyor C(I)‘ 1. If no: the number in the machine is incre- mented by the number of packages it can accept (denoted by NH) and the number on the conveyor is lessened by a similar amount, or, ICP( ICP - NH I) NMP NMP(I) + NH (I) 2. If yes: the number in the machine is incremented by the entire quantity on the conveyor and the conveyor is left empty. Equations NMP(I) =(I) + ICP(I) NH = ICP ICP10" 43" 12" 12" I' 4" s" 18" 55" 54" 2' 54" 55" 1. 48" 36" 32" 13" 1' 3" 2' l6" 1' 14" 1' 49" 3' s" 34" 44" 28" grip grip grip grip grip grip grip fingers fingers fingers fingers fingers fingers fingers 20 missed missed missed missed missed missed missed chipboard jam chipboard jam chipboard jam chipboard jam chipboard jam chipboard jam sriP' grip grip grip grip grip grip grip grip grip grip grip fingers fingers fingers fingers fingers fingers fingers fingers fingers fingers fingers fingers missed missed missed missed missed misSed missed missed missed missed missed missed chipboard chipboard chipboard chipboard chipboard chipboard chipboard chipboard chipboard chipboard chipboard chipboard chipboard chipboard chipboard chipboard' chipboard chipboard chipboard 21 'l“!'|llll'|‘l T 2 5 T "E'ii‘ll I'll oil-LIT X X X X X |l|l|||||llllllllllr H I. X X ‘ 3 v“ ice—b X '||‘||-|II'III "90 _X X .5 6. s 2 7. x _X X 37 -"o‘l 13“..- DOT X X If X vi. 1 —IX_X X X X v. x < 9 x x x x x x 3.6 ‘ X X X X _X X X v. X 1.1 _ x x _ x x x x x at .— 7. .6 .5 .4 1. o. ooeauauuo o>uuo~o¢ Length of Downtime (Minutes) Figure‘4.l.-Depa11etizer Downtime. z. '45" 4.. 26" 30" so" 1. 40" 3. 15" 41" 12" 4.. 7' 24" 10" 24" 2. 19" 41" 2' 24" 7.. 57" 16" 28" 30" 22 chip jam in hopper suction cup malfunction jam at disc outfeed loose chip - not lower jam.at disc outfeed inspector jam at disc output Filler Downtime Data jam in exiting tunnel jam in exiting tunnel jam in exiting tunnel high can switch malfunction high can switch malfunction low level oil tipped can at filler output jam at filler infeed jam in tunnel bent can jam.in tunnel bent can in worm gear output-one can gap jam in tunnel can partially filled can partially filled can partially filled 1' 12" 1' 2" 22" 2' 34" 1' 10" 6' 20" 4.. 1. 1' 2" s" 7' z." 2. 2' 12" 1' so" 54" 1' 35" so" 42" 1. 1' 26" 56" so" 40" 2. 1. 1. 23 can partially filled can partially filled no grease - one hour-light no C0 light 2 clean filler with.water can partially filled jam.in tunnel bent can at high can limit jam in tunnel no C02 gas low oil level jam in tunnel jam in tunnel jam.in tunnel jam in tunnel jam in tunnel jam.in tunnel jam in tunnel jam in tunnel jam.in tunnel jam.in tunnel no grease one hour-light no grease one hour-light can jam.at infeed no grease one minute-light bent can at high can limit 24 -‘lll'-"l-' l: m 1 - 7 In'll'-“ ‘III I. In“: X 1 6 -Ill'l Ill-'0 IIIIIIo—z o 5 -‘i'l'! 'l'lII-z 1 4 X l!" ll. I" I"! a 3 . {3 X X l‘l""'l'-llx—z X X 2 5 X X |u~ X 'llll'lll'l'- I4": X X .ib X X X X - - 1 _ x _ x — x _ x x x x x x 3... — x x x x x x x 5 a. .m: X X X X X - .is —X X X X X X X 0 o. no a: .w .3 .n a. 9‘ i. eucouauuo o>uuono¢ Tine (Minutes) Figure 4.2.-Filler Downtime. 6 l||||'al‘|ll"llulr 1 5 X l’lll'lll" In? 2 X o. X "lllll‘ll'll‘l'll lo? 1 3 X "ll‘lllll'- l-z 2 2 3.. X X 'l'l|'|l-|'[lelr 24 TTTT . __._. Id 32 _.._._._.._.._H_._. ooceuaooc o>uuonoz Tine (Minutes) rigure 4.3.o-Seaner Downtime. 1. 4.. 1. 1. 1. 3. 32" 3' 33" 1' 32" 1. 3.. 30" 11" 1' 17" 2' 42" 4" . 4.. so" 1. 1. 2. 5. 1. a" 25 no grease one minute-light jam in tunnel no grease one minute-light; bent can bent can jam at rinser Seamer Downtime Data jam at discharge rail bent lid in seamer head bent can in seamer can jam.in seamer output seamer missed lid seamer missed lid seamer missed lid jam.seamer output seamer outputdmachinist missed lid missed lid seamer output jam seamer input jam seamer output jam can jam transfer chain.jammed seamer-output jam seamer missed lid 26 1 ,6 ml 1 S '|l" ‘ all" t ,. ,1“ 2 m Ilnl'e'|lll'- 1.2 2 3 "lll I'II' 1—2 X 2 5 la; X "l.""' 1“" X s "e X X 1 X X X Au .45 3 X X X X X "2 X X X X X X if X X X X X X X 0 8 76 5432 ooceusouo o>uun~os Time (Minutes) Figure 6.4.-Pscker Downtime. -............i.:.-fl ...... .. II ....F 20 - I“ T -llll"l| EDT _I.. a. - .3 E" I'll 'In . I.“ E. I I 2 Ir “ X X X X X 1 X X X X X I.» l...‘ X —X X X X X X 0 n. .1 .6 t, .u .4 or euceuaooa o>uue~o¢ Time (Minutes) Figure 4.S.-Trsymeker Downtime. 27 22" jam seamer output 5" seamer missed lid 10" seamer missed lid 4" seamer missed lid 24" seamer missed lid 1' 34" jam at output 4' 6" output conveyor - air nozzle 1' 10" output conveyor - air nozzle 24" seamer input jam 15" seamer input jam 1' 30" seamer input jam l' 10" seamer input jam 1' bent can at seamer In order to use this data in.a subroutine and make it applicable to this simulation program.the number of occurrences of downtime in each minute time frame were counted and rounded to the nearest whole minute. The size of the breakdowns range from.ane to seven minutes. Each minute for all five machines is represented by one data card for a total of seven cards in all. The card representing the first minute is shown: DATA (NX(1,K), K=1,5) / 49,35,24,30,20/ This translates to mean: 49 - one 35 - one 24 - one 30 - one 20 - one The depalletizer down as follows: minute minute minute minute minute minute \lmUI-L‘UJNH minute 28 minute breakdowns minute breakdowns minute breakdowns minute breakdowns minute breakdowns for the for the for the for the for the depalletizer filler seamer packer ' traymaker data for all seven minutes is broken - 49 occurrences occurrences occurrences occurrence occurrences occurrences I O O N H «P 0‘ occurrences total time 49 12 12 4 10 0 0 minutes minutes minutes minutes minutes minutes minutes 87 minutes It was decided to operate this line at approxi- mately 80% efficiency, therefore 87 minutes or 20% of 'the total running time, the depalletizer is down. This I would signify a total simulation time of 87 e .20 = 435 minutes. The total Simulation run to insure 80% effi- ciency for all five machines is shown in an eighth data card: 29 DATA NTOT / 435,355,220,325,440 / 435 minutes is the total simulation run for the depalletizer 355 minutes is the total simulation run for the filler 220 minutes is the total simulation run for the seamer 325 minutes is the total simulation run for the packer 440 minutes is the total simulation run for the traymaker At what point in time a machine breaks down and how long it remains down is determined by a random number generator (RANF(X)) subroutine in the computer that generates a number sequence between 0.0 and 0.99...9. This sequence is found to be randomly and uniformly distributed based on chi-squared testing. An example of this downtime generation is shown for the depalletizer. Equation: M = NTOT(I() * RANF(0.5) + 1.0 = 435 * (0.130...) + 1.0 = 57.6 A different random number value for M is determined for the depalletizer as well as for each of the other machines at the beginning of each one minute time period. This value is then run through the first column of the seven data cards. Referring to the Subroutine DOWN in Appendix A, it is seen that 57.6 is greater than 55 but. smaller than 59, thus denoting the depalletizer will go 30 down for a period of three minutes. Should M be larger than 62, as it will be 80% of the time, the machine would remain in operation. ”CHAPTER VI COMPUTER PROGRAMMING 5.1 Flow Chart The computation process consists of a main program and two subroutines. The main program, PKGLINE, is concerned with simulating the movement of a high speed assembly line. Each station (machine) is handled individually with its entering and exiting conveyor systems. Checks are run on each station to see if it is operating and to see how many packages each machine is capable of accepting from the entering conveyor. Likewise a check is made to see how many packages the exiting conveyor is capable of accepting from its machine. The subroutine DOWN analyzes and delegates down- time to the various line machines in accordance with observed data gathered from actual high speed assembly line observations and estimates. The subroutine PLINE used in Proof Model 4 uses fields of symbols to present a unique visual "picture" of the units as they are processed through the line. It was written by Dr. wayne Clifford, Assistant Professor 31 32 at the School of Packaging for presentation at a brewery packaging technology seminar at Kellogg Center. A complete simulation run is eight hours. The time increments are one minute each, totaling 480.. The program computes the line status each six seconds and prints out the current standing of each machine and corresponding entering conveyor at the end of each minute. The subroutine PLINE prints its visual every six seconds. Flow charts have been drawn to show both the main program.PKGLINE, and the subroutine DOWN: Figure 5.1.1 for PKGLINE and Figure 5.1.2 for DOWN. The variable names used in the main program: ICP: number of packages on each conveyor. ICM: maximum capacity of each conveyor. NMP: number of packages being processed at each station. NMM: maximum capacity of each station. MD: machine down = 0 if machine is running. 8 1,2,...7 if machine is down. KLP: production of each machine. KPT: distinguishes the individual machines (depalletizer,'filler, seamer, packer traymaker) UPD: indicates if machine is operating or not by printing UP or DN. 33 {Dimension ICP(10). ICM(11), NMPUO), mum, 100.0), I KLP(10), KPT(20), UPP(10), NHXUO) f [ Data NMP, ICP, ICM, .. , MD, RPT, UP, ON I Initialization Subroutine Ranset Random Number Generator l Subroutine in Computer DELT . 1.0 T-O N35 LP-Ol \ Initial ization l l Subroutine Pline Call to Set Up — Subroutine r---—---—--——— JI-lour-l,2 > I I Print Titles, I I I JTime-1,60> .. —- —— -— - - -—-[5—ubroutine Down j KLPU) II 0 T-‘l' + Delt I I I I I I I ® Figure 6.1.l.-Program PKGLINE. Part I 3li " ________ 1 Time - 1,10::>. LP I LP + ICP(N+1) ICP(N+1) - 0 rcrtr) - Icn(I) I I I N+1 - 12 NC I ICH(I+1) - ICP(I+1) >NMP(T) ayes KLP(I) - KLP(I) + NC ICP(I+1) - ICP(I+1) + NC mu) - NMP“) — NC NH - NHH(I) - NMP(I) .L. . mm yes yes -—--—-——-————-—-——-fi NHX(I) I NH 941“" - rcp(r)l NHP(I) I NMP(I) + NH ICP(I) I ICP(I) - NH I I I I I I I I I | l I I l I I l I l I I I I I I I II L _______ Figure 6.1.1.-Program PKGLINE, Part II 135 I UPD(J) - as L -------- Q [Itint ‘1', ICP“), NHPU), UPD(I). 210(1). ‘1'?qu I I I I I I , I I I II I I : I ' I | I I I I I I I I I Figure 6.1.1.—Progran PKGLINE, Part III r—----I I I I I I I I I I I I I I I sun 36 [Dimension HD(5) , MDOWINS) ,NTO‘NS) .NXfl'é Eat. N'X(1,K),(2,K) ...(7,K), NTOT] MIMIC) - MDIK) + 1 Q MINTOTGQ * Ranf(0.5) + 1.0 Figure 6.1.2.-—Subroutin¢ Donn. Part I ’°’ II> sumo 37 ”Figure 5.1.2.-—-Flow Chart for Subroutine DOW, Part II 38 NHX: array necessary for PLINE subroutine DELT & T: time in minutes N: number of machines in the line (5) LP: total line production NC: number of packages to exit machine and enter conveyor NH: number of packages to exit conveyor and enter machine The variable names used in DOWN: DOWN: stores the time in minutes the machine is down NTOT: running time from which random.se1ector picks downtime NX: two dimensional array used in matching length of downtime to machine In the computer numerous computations occur in a simulation run, the number of which depends on the length of the run. In this case when simulating a beverage line for a greater period of time than one eight hour shift, 20 minutes of downtime must be allowed for the entire line for lubrication. Assumptions used in the program: 1. Perpetual infeed to the conveyor entering the depalletizer. . ' 2. Infinite storage space available to accept all the output from the traymaker. 3. All downtime is attributable to the five stations. The conveyors are trouble free. 39 In the actual downtime data studies, large downtimes not characteristic with normal downtime problems were ignored. The incremental revenue on a single unit is assumed to be $0.03. To simplify matters in this paper, the entire amount will be ap- plied to the capital investment in the con- veying mechanisms. CHAPTER VI PROOF OF THE COMPUTER PROGRAM Four models are examined: 1. A model with infitied (500,000 units) conveyor lengths between the stations. 2. A model with very small (140,140,120,140, 160,160) conveyor lengths between the stations. These lengths correspond to the larger machine capacities at either end of each conveyor. 3. A model with different size conveyor lengths between the stations. It will try to optimize output production at the ‘lowest cost of conveying mechanisms. 4. Similar to model three but with the addition of the subroutine PLINE. In each model the cost versus production aspect will be examined. The optimum.conveyor capacities could be found with a saphisticated trail and error subroutine, such as Rosenbrock's HCLMB.11 However, the routine requires numerous multiple runs and its cost makes it prohibitive for this exercise. If one were designing a real assembly 11H.H. Rosenbrock and C. Storey, Computational Techniques for Chemical Engineers (New York: Pergamon Press, 19667, pp. 64-63} 40 41 line for a company this subroutine would be vital to pin- pointing the optimum conveyor lengths. 6.1 Proof Model 1 This model forms a coupling of the five machines in a series as shown in Figure 6.1. It is designed with Very large (500,000 unit) conveyor lengths or reservoirs between the machines. Each reservoir is half full of units at the beginning of the simulation run. Because of the very large reservoirs each machine operates independently of line capacities and downtimes of the others. 'There is no idling time due to the in- ability of the entering conveyor to transport units (cans) or the existing conveyor to accept them. Therefore, each machine is expected to yield its full catalog efficiency of operation and the line efficiency will equal that of the last machine. The simulation result of the proof model is shown in Table 6.1. Each machine turns over its stock of units every six seconds or ten times every minute. At the end of each minute there is a summary of the current status of each entering conveyor and corresponding machine. Example: TIME Cl Ml D1 Pl ........ C5 M5 D5 P5 TOTAL TIME: the time in minutes. 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Example: ‘ UP 0 indicates the machine is operating. DN 1,2,...7 signifies the number of minutes the machine has been down. P1...P5: indicates the total production of the station at the end of that particular minute. TOTAL: total line production. Only the first two pages of the simulation are presented here. It is felt these will show the results equally as well as the entire printout and much less - space will be required. Comments about Table 6.1: 1. The efficiency of the line is assumed to be approximately 80%. 2. The efficiency of each machine, excluding idling time can be found by adding the downtimes of each over the eight hour simulation run, subtracting each from eight hours and figuring the result as a percentage of the total shift. Examples: A. Depalletizer Total Downtime = 85 minutes 480 min - 85 min _ 480 min _ X 100 _' 82.29270 B. Filler Total Downtime = 101 minutes 480 - 101 _180 x 100 = 78.9581 46 C. Seamer Total Downtime 94 minutes 480a5094 x 100 = 81 875% D. Packer Total Downtime = 87 minutes 48045087 x 100 = 81.875% E. Traymaker Total Downtime 118 minutes 480 - 118 480 x 100 = 75.417% The overall line efficiency, based on machine downtimes is 82.222 j 78.958 + .5 .......... . = 79.792%, This figure is very close to the expected 80%. ' Another way of showing line efficiency is based on the number of units produced over time. For this first 'model, efficiency is gauged on the last machine, the traymaker. Total Production: Length of Run: Last Machine: 185,280 units 150 minutes 1600 cpm (cans per minute) 185,280 units 150 min x 1600 cpm x 10 % = 77.2% 3. Filler idling time is as small as possible. Idle time as a percent of total running time is: 19 min 150 min x 100% - 12.7% This idle time is due to the 6 second start-up time step of the filler while it is being loaded with cans. 47 4. Conveyor cost versus production: Conveyor Capacity: 500,000 units/conveyor Each Unit (can): 2.5 inches in diameter Cost of Conveyor: $100.00 per running foot Capital Cost: SGIIDOIIfits 2. ° . conveyor x 5 conveyors x {71% x EsmxflgE-QQ = $52,083,333 Total Output: 185,280 units (cans) Incremental Revenue per unit of Output: $0.03 Length of Run: ' 150 minutes Incremental Profit: 480 min $.03 185,280 units X m1n X Ell—1? = $16,786.88/eight hour shift Necessary time to return investment with 3, eight hour shifts per day working five days per week: $52,083,333 x 1 shift x l_year $17,786.88 3 shifts/day 250working days = 3.9 N 4 yrs. 6.2 Proof Model 2 This proof model is the same as Mbdel 1 except for the length of the transfer mechanisms between the machines. These very small conveyor lengths (140,140,120,l40,160) correspond to the larger machine capacities at either end of each conveyor (Figure 6.2). The model run is started with each conveyor half full. 48 .N Hovoz uooumuu.~.o ouswwm hufiUdauo awn: cod no . huflouauo vac: ova no Announce use: osfl so suaoaauo ans: oNH no hufluuamo pan: 04H «0 huaounao awn: oca do naho>uou " magnum» « owQuOum . uwauoum ouaaamau . ouaawmau . 6 ueaeaeaoao. Il- .I‘II I III 1 II" I. III 49 Interaction between the machines occurs almost 100% of the time which means every breakdown of a machine affects the operation of all the others. It is expected that the efficiency of each machine and the entire line will decrease markedly. The line efficiency for this model is based on the slowest machine, the filler. Table 7.2 shows the computer printout. Some comments follow: 1. The efficiency of each machine, excluding idling time is the same as that discussed in comment (2) of Proof Model 1. The line efficiency based on the . number of units produced is very low. Total Production: 65,760 cans Length of Run: 150 minutes Slowest Machine: M(2)-Filler 1200 cpm 65,760 cans _ 150 min x 1200 cpm x 100 " 35-53% 2. Machine idling time is large. Idle time as a percent of total running time for the filler is W x 1007. = 45.337. This compares to zero idle time for Proof Model 1. Idle time is defined as the number of minutes the filler did not produce its full machine capacity during the simula- tion run. 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O 0 o o o '00—. 0.. ”was:eu—NngmcscoOu~M:m¢~¢conun4mc~o¢cu~nercuc3can"amcsocou~n4mo~co°«~n:mcs¢¢o bthhhscccoccoo0:00CCOOGcreoooooooccouu—uuuuuuuun~~~~~~~~rnnnnnnnnn:J:J:J:J:Jm -_4._4' 4 A A.‘._4 ._4._4 . .4 -‘_‘ A ‘_4_4 hm v. v vvvw W TABLE 7.2.--Proof Model 2, Part II 52 3. .The assumption that no units will enter or exit a station ifits corresponding entering or exiting conveyors are full is in evidence in this model. It can be seen that the capacities and downtimes of each machine materially affect the production of the machines before and after them. A maximUm of 1400 and 1600 units per minute can be carried on the entering conveyors to machines 4 and 5, respectively. However, this quantity never rises this high when both machines arerunning because of the line bottleneck, the filler and seamer machines. Although the packer and traymaker are capable . of producing 1400 and 1600 units respectively, it is of no consequence if the cans are not available. 4. From a cost versus production point of view: Conveyor Capacity: 140+l40+120+l40+l60 = 700 units Each Unit (Can): 2.5 inches diameter Cost of Conveyor: $100.00 per running foot Capital Cost: 2.5 in. 1 ft. x $100.0 700 units x ————T— 1 unit x m. “fir.— = “4583-33 Incremental Revenue: Total Output: ‘ 65,760 Incremental Revenue per Unit of Output: $0.03 Length of Run: 150 minutes . 480 ’ 0.03 . . 65,760 units x ISU—gi%'x §GHIE'= $6,312.96/e1ght hour Shlft 53 Necessary time to recoup investment working 3, eight hour shifts per day, five days per week: $14,583.33 1 shift = , . $6.3I2756'x 3 shifts/day 77 “Orklng daYS- Mbdels l and 2 were examined more to prove that the simulation model functions realistically as a high speed assembly line, given the stated assumptions in section six, rather than as a practical cost versus production example. However, there are important aspects to be noticed in this area. Mbdel I produced more units but lost its profit in excessive capital investment in. conveyor equipment. Mbdel 2 spent too little on capital equipment and suffered from an inefficient line, loss of production and lost market position. 6.3 Proof MOdel 3 Proof Models 3 and 4 are designed in an effort to., optimize production at the lowest capital investment in conveyor systems. Mbdel 3 utilizes conveyor lengths of 10,000, 20,000, 15,000, 15,000, and 30,000 units (Figure 6.3). All four models are similar in that they are arranged in a coupling of five machines in a series and each simulation run is started with all conveyors half full. 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The efficiency of each machine, excluding idling time is the same as that discussed in comment (2) of Proof Mbdel 1. The line efficiency based on the filler is Total Production: 160,200 cans Length of Run: 150 minutes Slowest Machine: M(2)-Filler 1200 cpm 160,200 cans x 100% _ 89% I50 min x 1200 cpm 2. Filler idling time is as small as possible. Idle time as a percent of total running time is: 19 min 100% = 12.7% m.n x This idle time is due to the 6 second start-up time step of the filler while it is being loaded with cans. 3. In analyzing the operation of this model it is found that the 20,000 can capacity of the conveyor entering the filler never falls below 5,000 cans. This section of conveyor forms a 'permanent' storage area for cans. If the size of this conveyor is reduced by g 5,000 cans it is logical to expect no loss in efficiency from this machine. It was noted previously that the filler is the heart of the high speed beverage line. It is of paramount importance to keep this machine at optimum efficiency. 58 The conveyor entering the seamer can handle 15,000 units per minute. The seamer itself produces 1200 cpm. The only time its conveyor runs below this 1200 can machine capacity is the 77.0 to 83.0 minute range. 'This is due to a six minute downtime in the filler. The con- veyor never accumulates more than 11,000 cans at any time in the run, therefore a decrease in capacity to :_ll,000 cans is warranted. The packer can handle 1400 cpm. Its conveyor in this particular model can accommodate 15,000. Only in the 83.0 to 86.0 minute time frame does it fill up. This is due to a six minute jam in the packer. In the 123.0 to 130.0 minute range the machine shows idle capacity due to lack of cans. A larger conveyor seems warranted in this case. p The traymaker is the fastest machine on the line, handling 1600 cpm. Its feeding conveyor was set at a 30,000 unit capacity. The most cans it accumulates (except at the startup) is 9,220 in the 141.0 minute of operation. Although it has idle capacity for a total of 43 minutes because of no cans, little can be done other than to in- crease the size of conveyor C4 entering the packer. 59 4. Conveyor Cost and production figures: Total Conveyor Capacity: 90,000 cans Each Unit: 2.5 inches Cost of Conveyor: $100.00 per running foot ' ' Capital Cost: 90,000 cans x 24—549- x 1 ft x Egg-€99 = $1,875,000 can I2 in Total Output: 160,200 cans Incremental Revenue per Unit of Output: $0.03 Length of Run: 150 minutes Incremental Revenue: 480 . 160,200 cans x ISO—31173 x figzgi -= $15,379.20/eight hour shift Necessary time to return investment working 3, eight hour shifts per day, 5 days a week: $1,875,000 1 shift 8 ' $I5737gTZU-x 3 shifts/day 40.7 m 41 days 6.4 Proof Mbdel 4 Proof MOdel 4 utilizes conveyor lengths of 10,000, 17,000, 11,000, 17,000 and 15,000 units (Figure 6.4). An analysis of Table 6.4 follows: 1. The efficiency of each machine according to accumulated downtime over an eight hour simulation run is again the same as comment (2) of Proof Mbdel l. 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W I?) m amuse-009'.» ac: no ”Jean-ar-UUJQOQQQUaa wt: ”Woo-39%;; 9 2 .u-amaonmfleflaeflmva-m O uourcccctcr-Jc acocccu'n Ltft‘tNat-”OCWJOLGJUU‘wouu-Ddém uou-¢¢o¢uo¢¢¢~caco. «canton-cccoracgootttuocccccrooamcwocmutocuouotoctthuuocccc can creature”. 003°?J‘OU‘E996‘49U3‘ 1:39—090’3‘5‘0'090 ‘3‘: unU‘J‘uJururu-I'NG‘U’S‘UJ‘waJ 0930‘0 U‘C‘OOU‘30ON"”: m on a on «on '0‘ on. do... on. all." c. In «and h V vw v—v vw—v xz.o:.'o:‘:o....t..-..o.o..oT-177:7.o77.:—o...'.. 0:07....o-ovooxvo:onoo'o;00000‘;o.'. can-.99.: EMU-3.1:.csr9uu9fl3mctvscc‘o—0N'h awc~oa~aum:m¢mc a: u—«vn :mdzscz-o—M—Ic (Kc-coauncfiobooc btKKKK“..-o...¢OOO9OOWOOmoow-aowuuuuudung-rmmm~~~~ngrnnmnnaa 10:10 AJOAO TABLE 7.4.-—Proof Model 4, Part II l'.l|l I III! I ‘lll'll In"!!! I'll “ A In alllll 41." 53 152,700 cans _ 150 min x 1200 cpm " 1°07" " 34-3 "' 85% 2. Filler idling time is the same as in Proof Model 3. 3. The conveyor CZ. entering the filler has been reduced to a 17,000 can capacity. The filler still pro- duces at its peak output but the cost of its conveyor has been reduced by $62,500. Conveyor C entering the seamer has been lowered to 11,000 cans as :ecommended in.Mode1 3. The six minute idletime in Mbdel 3 has increased to 20 minutes because the simulation run started with only 4500 cans versus 7500 in Model 3. However, the most units C3 ever accumu- lates is 8,980. A judgment is needed by management at this point concerning conveyor investment as opposed to machine idle time. The packer has an entering conveyor capacity of 17,000 as compared to 15,000 in the last model. Only in the 86.0 minute does it completely fill up. It shows idle capacity in the 123.0 to 131.0 and 145.0 to 150.0 time frames. A.sma11 increase in this conveyor based on changes in conveyor C3 should be experimented with. Conveyor CS accumulates 53 minutes of idle time as opposed to 43 in the previous model but its length has been cut by one half. 64 4. Conveyor cost and production figures: Total Conveyor Capacity: 70,000 cans Each Unit: 2.5 inches Cost of Conveyor: $100.00 per running foot Capital Cost: 70,000 cans x 2'5 in 1 ft $100&00 _E§fi__.x 12 in x "—f__— = $1,458,333 This is a savings of $416,667 over Mbdel 3. Total Output: 152,700 cans Incremental Revenue per Unit of Output: . $0.03 Length of Run: 150 minutes Incremental Revenue: 480 min $0.03 152,700 cans x min x can = $14,659.20/eight hour shift Necessary time to return investment working 3, eight hour shifts per day, 5 days a week: $1,458,333.00 1 shift = . x 3 Shifts/day 33°2'”33 days 5. Table 6.5 shows the simulation program.with the addition of the subroutine PLINE. The regular program calculates the position of the assembly line each six seconds or ten times each minute and prints a numeric summary of the status of the line at the end of each minute of operation. 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A '— ‘ .4; .A fit :fio’owh‘-a~;” ‘;M~ _. 'tk G CC O CH (:0 v :wk 7' :' cc: b‘h-OoO-Q-mfl ~---- 3. :- _. ‘E;-I~v_-W OS A“ ‘ OD “ v. ' fl :- W“ gram-us... aunt-act MHW‘Ib‘m‘ Mb.” a'unngm :2 an... vwssr‘o—UI “finntmg. .4- 4‘ 1 A A k A _4 .4 A 'v v v 00303000 00003000 __ A A .. C. - v - w 7.- TABLE 7.5._--Visual to Proof Model 4, Part II 67 pages are included for a total of 12 minutes. The print- out for an eight hour shift would be lengthy but trends in the line can be spotted very easily by virtually anyone. Four symbols are used in this picture: 1. $ 2. O (alphabetic symbol) 3. S 4. 0 (numeric symbol) 1. Each $ represents 1300 units moving on a conveyor to a station. ' 2. Each 0 (alphabetic symbol) represents 1300 units on a conveyor that are stationary. This is due to a breakdown in one or more of the succeeding machines. 3. Each 5 represents 20 units (cans) that are moving through the machine. 4. Each 0 (numeric symbol) represents 20 units that are stationary in the machine. This machine is run- ning but its exiting conveyor is full. The packages are not moving due to_a breakdown in one or more of the suc- ceeding machines. When a machine goes down all the units it contains are emptied into the exit conveyor if this conveyor has the space available to accept them. In this model 1300 and 20 units are represented. It is very easy to change the amounts to suit the user. Each machine is distinguishable because of the letter that follows its symbols: 68 A - depalletizer B - filler C - seamer D - packer E - traymaker In analyzing the line during the-first minute of operation machines M1, M2, ananz have gone down. fMachine‘Ms is producing cans that move to conveyor 04 fand machine M5 produces units that are counted in the total line production. Looking over the entire 12 minutes conveyor C2 entering the filler tends to decrease its can supply. Conveyor C4 has become well stocked with cans due to the six minutes of downtime in the packer and the traymaker-(M5) soon exhausts its supply which never rises above 140 cans in the 12 minutes. CHAPTER VII SUMMARY A high speed beverage line is a complex system .with many interacting components. It involves a variety of expensive line machines and transfer equipment. Because of this complexity and the variable nature of the machines in terms of breakdowns and idle activity, analytic techniques are inadequate in selecting or designing the optimum.assembly line. Simulation models then, are necessary in trying to predict the conse- quences of changes in conditions, or methods without having to spend the money or take the risk of actually making the change in real life. MOdels are not perfect, they cannot replace the real world, but they can reduce a complex system, in this case a high speed beverage line, to manageable proportions. It can be seen from the Proof Models that there is a trade off between conveyor equipment investment and efficient line production. A larger capital investment, however, does not necessarily mean a more efficient line and vice versa. This balance is different for each comr pany depending on the objectives and attitudes of its higher‘management. 69 70 It was found, while talking to managers of two different high speed brewing lines, that there is a ready market for as much product as can be produced. In this case then, the manufacturer would want to make as efficient use of his packaging line as possible, meaning increased conveyor investment. ‘Without the use of a simulation program, such as this, a company could easily invest more money than is necessary for effi- cient production, thus tying up capital needlessly which could be used to purchase other assets that would better serve its goals in the competitive business arena. ‘Working closely with higher management and running several test models similar to those shown in Proof Models 3 and 4, it is relatively easy to narrow down the ranges of optimum.conveyor length to get the most for the capital invested. A computer program.for the simulation model has been developed and, given the assumptions in section six, has been proven to work successfully. If a high speed packaging line can be described by Figure 3.1 or any combined form of it, this simulation program will be found useful in designing a more cost efficient line. APPENDIX 71 in I' ‘Hi‘ ‘I. . ll I!!! III. I lrk l I I l I I]. ll!‘ ll I'll “ .Il'lilli'llll APPENDIX Figures Al, A2 and A3 show the actual computer program" The main program, PKGLINE is a general repre- sentation of a packaging line. As mentioned previously the substitution of different numbers, sppeds, and capacities of the machines as well as the conveyors is relatively simple. The subroutine DOWN is more specific and is accurate only for a high speed beverage packaging line. The subroutine PLINE is a visual aid to under- standing the workings of PKGLINE and DOWN, 72 f l! A." ml! I! 73 PROGRAM PKGLINE(INPUT.OUTPUT’ ccccc ccccccccccccc E HILL EXIT 0N YHIS HINE T415 15 0(II'(CHANGE IN T) t I! IS VHS ES 9.53.3323“ ‘End'luu UT. .Huob 3N HE I.L~ ; :— £0” YCLE (TH S 3 ) HA3 V IN HINUTES V13 OPE THE I I ,4 I 3:1 or O?SW3.HV1YI YPII : TIME E ”5 H39 99.; . Vu.)9\F Saw «a or... u. 01E... c...Yn..tno .. couvsvoa raon pgsvraus CYCLE. a‘ 33x§3 .._4.u.T~.vuV...vO PFIPAUPS 9. CT. (UNIT... .tCEtu9 NECIIHUV- nuv o On. JCGAPLSC IRHNoruNCur—nzr S QHOUUHY. LCL RT C. Y;3\ T ’ON "9.. 8551 HSUQDIY... O ”*VX lJITnuol‘...P..DI U ”Alan Y. 0.45:7. JNNC P. .22 11.31026: can. IC: T