FLEXIBLE ASSEMBLY SYSTEMS

Transcription

1 FLEXIBLE ASSEMBLY SYSTEMS Job Shop and Flexible Assembly Job Shop Each job has an unique identity Make to order, low volume environment Possibly complicated route through system Very difficult Flexible Assembly Limited number of product types Given quantity of each type Mass production High degree of automation Even more difficult! 144 1

2 Flexible Assembly Systems 1. Sequencing Unpaced Assembly Systems Simple flow line with finite buffers Application: assembly of copiers 2. Sequencing Paced Assembly Systems Conveyor belt moves at a fixed speed Application: automobile assembly 3. Scheduling Flexible Flow Systems Flow lines with finite buffers and bypass Application: producing printed circuit board 145 Objectives Multiple objectives usual Meet due dates w j T j Setting for job j on machine i Maximize throughput s h a, a ijk Minimize work-in-process (WIP) i ik ij 146 2

3 Flexible Flow Shop Stage 1 Stage 2 Stage 3 Stage Unpaced systems Machines can spend as much time as needed on any job Machines have limited buffers Manufacturing systems with: Large items (TV sets, copiers) Similar product with different options requiring different production time 148 3

4 Unpaced system model Material handling system: When a job finishes moves to next station No bypassing (First In First Out) Blocks job if buffer is full Can model any finite buffer situation A buffer is a machine with zero processing time Equivalent to model with no buffers 149 Unpaced system scheduling: cyclic Different jobs are scheduled in a certain way, and this schedule is repeated over and over Contains all product types May contain multiple jobs of same type Practical if insignificant setup time Low inventory holding costs Easy to implement Cyclic schedules often maximizes throughput, but there are no guarantees

5 Defining a cyclic scheduling 1. Find out the Minimum Part Set (MPS) 2. Apply Profile Fitting (PF) heuristic 151 Minimum Part Set l different product types N k jobs of product type k z is the greatest common divisor of N 1,, N l Then N N N Nl,,..., z z z * 1 2 is the smallest set with same proportions as the long range production set. It is called the Minimum Part Set (MPS) 152 5

6 Cyclic schedule The jobs in the MPS are n jobs where 1 l l k z k 1 k 1 n N N Let p ij be the processing time as before A cyclic schedule is determined by sequencing jobs in the MPS Maximizing throughput is equivalent to minimizing cycle time in a steady-state. * k 153 Example Number of products: l = 4 Jobs per product: N k = (12, 3, 12, 18) = 45 jobs Greatest common divisor, GCD: z = 3 Minimum Part Set, MPS: N * = (4, 1, 4, 6) Jobs in MPS: n = 15 cycle MPS MPS MPS 154 6

7 Profile Fitting heuristic Used to solve F m block C max Remember: F m flow shop block limited buffer size. When the buffer is full the upstream machine cannot release jobs C max minimize makespan 155 Profile Fitting heuristic Minimizes MPS cycle time 1. Select one job to go first Arbitrarily or According to some other rule, e.g. Longest Processing Time (LPT) first 2. Generates profile. Departure of job 1 from machine i: X i p i, j1 h, j1 h

8 Profile Fitting heuristic 3. Determine which job goes next Try all other jobs in MPS For job c X X X 1, j2 i, j2 m, j2 max( X max( X X m1, j2 1, j1 i1, j2 p p, X mc 1c ic 2, j1 p, X i1, j1 ) i 2,, m 1 Choose job with smallest total non productive time m Xi, j Xi, j pic 2 1 i1 ) 157 Example Jobs p 1 j p 2 j p 3 j p 4 j buffer * N 1,1,

9 Example Sequence: 1, 2, 3 (ii) (iii) (i) (i) (ii) (ii) (ii) (iii) (iii) (i) (i) (i) (ii) (ii) (ii) (iii) (i) (i) (ii) (ii) cycle time = Example Sequence: 1, 3, 2 (i) (ii) (ii) (iii) (iii) (i) (i) (ii) (ii) (iii) (iii) (i) (i) (ii) (ii) (iii) (iii) (i) (i) (ii) (ii) (iii) cycle time =

10 Discussion: PF Heuristic PF heuristic performs well in practice Refinement: Nonproductive time is not equally bad on all machines Bottleneck machines are more important. Use weight in the sum: Weighted Profile Fitting heuristic (see Example 6.2.3) 161 Additional complications The material handling system does not wait for a job to be complete Paced assembly systems There may be multiple machines at each station and/or there may be bypass Flexible flow systems with bypass

11 Paced Assembly Systems Conveyor moves jobs from one workstation to another at fixed speeds Fixed distance between jobs Spacing proportional to processing time No bypass Unit cycle time time between two successive jobs line balancing Example: automotive industry 163 Paced system model painting sunroof windshield wheels

12 Paced system model Example: Automotive industry Line runs at a constant rate Size of work center proportional to duration of operation It takes longer to install a sunroof than a windshield so the workstation is longer Once the setup is done, the only thing to control is the sequence of cars 165 Paced systems scheduling: setup costs n cars m paint colors Color of each car is known Every time color is changed it has a cost (for cleaning, discarding paint, etc.) Setup or transition cost Problem: Find a sequence that minimizes setup cost

13 Car sequencing with options n cars Each car has different options Sunroof, air conditioning, spoiler, etc. The work center for each option is capacity constrained Can only install sunroofs in 1 out of every 3 cars 167 Car sequencing with options Can only install sunroofs in 1 out of every 3 cars painting sunroof Not enough time to install next sunroof!

14 Car sequencing with options Can only install sunroofs in 1 out of every 3 cars But it gets worse Every car requires a set of options Each option has a capacity constraint 169 Objectives in paced systems Minimize total setup cost Example: paint shop a job with color j is followed by color k has a setup cost of c jk. Meet due dates for make-to-order jobs Total weighted tardiness Spacing of capacity constrained operations i () = penalty for working on two jobs positions apart in i th workstation. Decreases with (as installing sunroofs) Keep rate of material consumption as regular as possible at all workstations

15 Paced system scheduling: heuristic Strategy is grouping jobs by similar features to decrease setup time Strategy has to space out over the sequence jobs with special and long operations 171 Grouping and Spacing (GP) heuristic 1. Determine the total number of jobs to be scheduled 2. Group jobs with high setup cost operations 3. Order each subgroup accounting for shipping dates 4. Space jobs within subgroups accounting for capacity constrained operations

16 Grouping and Spacing heuristic Total number of jobs High number allows a sequence with lower cost. However, probability of disruption is high. Number of jobs is normally between one day and one week. Grouping jobs Ex: determining run lengths of different colors. Grey can go up to 50, and purple may be only 1 or 2. Other examples: options, destination of cars. Similar to computing an Economic Order Quantity (see Lot Scheduling) 173 Grouping and Spacing heuristic Ordering different subgroups Determined by the urgency with which the jobs in a group have to be shipped. Urgency is determined by committed shipping dates of Make To Order jobs and current inventory levels of Make To Stock jobs. Internal sequencing within subgroups Consider capacity constrained operations. Consider most critical operation and space jobs as uniformly as possible. It proceeds similarly to the other operations

17 Grouping and Spacing heuristic Determine # jobs / scheduling horizon Group jobs based on setup cost 175 Grouping and Spacing heuristic Group jobs based on setup cost Order groups w.r.t. shipping/holding cost

18 Grouping and Spacing heuristic Sequence jobs within groups to account capacity constraints 177 Example Single machine, 10 jobs with two attributes. p j = 1, j a j1 : color; a j2 =1: with sunroof. If a j1 a k1 the setup cost is: c a a jk j1 k1 If a j2 = a k2 = 1 and spaced jobs apart the penalty cost is: 2( ) max(3,0) Tardiness w j T j is considered for job j with due date d j. Find sequence that minimize the total cost

19 Data for example Job a 1 j a 2 j d j 2 6 w j Grouping Group A: Jobs 1, 2 and 3 Group B: Jobs 4, 5 and 6 Group C: Jobs 7, 8, 9 and 10 Best order according to setup cost: A B C

20 Grouped jobs A B C Job a j a j d 2 6 j w j Due date Order A C B 181 Capacity constrained operations A C B Job a j a j d 2 6 j w j Total cost: 6 (setup cost) + 3 (capacity constraint) =

21 Flexible Flow Systems At each stage, there are machines in parallel A job is often equivalent to a batch of similar products At each stage, any machine will do Machines might not be identical A job may have to be assigned to a specific machine If a job does not need processing at a certain stage, bypass is allowed If buffer is full, handling system must stop unless recirculation is possible 183 Paced systems model Main goal is to obtain a cyclic scheduling Stage 1 Stage

22 Flexible Flow Line Loading (FFLL) Objectives Maximize throughput Minimize Work-In-Process (WIP) Minimizes the makespan of a day s mix Actually minimization of cycle time for Minimum Part Set (MPS) Reduces blocking probabilities 185 Flexible Flow Line Loading algorithm Three phases: 1. Machine allocation phase assigns each job to a specific machine at each stage 2. Sequencing phase orders in which jobs are released Dynamic Balancing heuristic 3. Time release phase minimize MPS cycle time on bottlenecks

23 FFLL algorithm: machine allocation Parallel machines model Which machine for which job? Basic idea: workload balancing Use LPT dispatching rule Output: allocation of jobs to machines, but not the sequencing of jobs or the timing of processing. 187 FFLL algorithm: sequencing (1) Sequencing the MPS by spreading out jobs sent to the same machine Dynamic Balancing heuristic (1) Workload in a MPS assigned to machine i W i p ij j1 Total workload of MPS n m W Wi i

24 Dynamic Balancing heuristic (2) J k set of jobs loaded into the system Total workload of machine i that entered the system by the time job k is loaded: pij ik [0,1] jjw k i Ideal balanced target: m n m * k ij ij j jj i1 j1 i1 jj k p p p / W k 189 Dynamic Balancing heuristic (3) Overload of machine i due to job k Wi oik pik pk W Cumulative overload of all jobs, including job k O o p W * ik ij ij k i jjk jjk Objective: minimize overload n m k1 j1 max O, 0 ik

25 FFLL algorithm: time release Releasing time by minimizing MPS cycle time Bottleneck cycle time Algorithm Step 1: Release all jobs as soon as possible Step 2: Delay all jobs upstream from bottleneck as much as possible Step 3: Move up all jobs downstream from the bottleneck as much as possible 191 Example

26 Data ' p hj : processing time of job j at stage h Jobs ' p1j ' p2 j ' p3j FFLL: machine allocation Using LPT: Jobs p 1 j p 2 j p 3 j p 4 j p j

27 FFLL: sequencing Workload: from the previous table we obtain each row W1 9 W 9 W W W W 51 each column p 13 p p p p FFLL: sequencing Overload: with job 1 as first job of sequence o o o o o / / / / /

28 FFLL: sequencing Overload matrix: With jobs 1 to 5 as first job of sequence, we can compute matrix o ik 3,71-1,76-1,41 1,41-1,94-2,29 1,24-0,41-1,59 3,06 0,20-0,16-0,73 1,06-0,37 0,94 2,65-1,88 0,88-2,59-2,55-1,96 4,43-1,76 1, FFLL: sequencing Dynamic Balancing heuristic: 3,71 0,00 0,00 1,41 0,00 0,00 1,24 0,00 0,00 3,06 0,20 0,00 0,00 1,06 0,00 0,94 2,65 0,00 0,88 0,00 0,00 0,00 4,43 0,00 1,84 4,84 3,88 4,43 3,35 4,90 First job

29 FFLL: sequencing Dynamic Balancing: selecting 2 nd job Calculate the cumulative overload where O p W * 11 1j 1 1 jj1 * 1 p j{4,1} 1j (36) jj k p / W p /51 j j{4,1} (9 13) / j 199 FFLL: sequencing Dynamic Balancing: selecting 2 nd job Cumulative overload: O O O O i1 i2 i3 i5 ( 5.11, 3.88, 1.26, 1.82, 4.32) ( 0.35, 0.36, 0.90, 3.52, 3.72) ( 0.00, 2.00, 0.33, 1.00, 2.67) ( 0.53, 1.47, 0.69, 1.71, 0.08) Selected next

30 FFLL: sequencing and time release Schedule jobs 4, 5, 1, 3, 2 Initial schedule: 201 Final schedule Machine 4 is the bottleneck

31 Assembly sequence at Toyota Everything operates on Just-in-Time Works to minimize WIP and tardiness Most important objective is to keep the part consumption regular The quantity of a given part consumed per hour should be constant Solution: the next car to build is chosen to minimize error from the desired rate 203 Example Model 1 requires 1 unit Model 2 requires 2 units Model 3 requires 3 units Production needed: 10 cars of model 1 15 cars of model 2 10 cars of model

32 Example So, it is needed (10 1) + (15 2) + (10 3) = 70 If the consumption of unit is to be constant, the consumption is 70/35 = 2 per car Choose car j to minimize: (units used/cars built 2) Toyota problem A car has around 20,000 number of parts. Parts are represented by their respective subassembly Number of subassemblies in Toyota is around 20. Toyota developed the Goal Chasing Method to solve the mixed model assembly, which minimizes the sum of squared error over all subassemblies. See p

33 Flexible Manufacturing Systems Flexible Manufacturing Systems (FMS) Numerically controlled machines Automated Material Handling system Produces a variety of product/part types Scheduling Routing of jobs Sequencing on machines Setup of tools Similar features but more complicated