PRODUCTION ACTIVITY CONTROL (PAC)
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1 PRODUCTION ACTIVITY CONTROL (PAC) Concerns execution of material plans Contains shop floor control (SFC), and vendor scheduling and follow-up SFC encompasses detailed scheduling and control of individual jobs at work centres on the shop floor Vendor scheduling and follow-up largely considered as part of production activity control, but not completely Procurement is seen as a professional activity where information networks, relationships, terms, and conditions are established with vendor companies outside of PAC Release of individual orders and follow-up are part of PAC Objective of PAC is managing the material flow to meet MPC plans Payoffs Ensure meeting the company s customer service goal Reduced work-in-process and lead- Improved vendor performance Effective feedback on shop and supplier s performance against plans Capacity Planning and SFC Insufficient capacity No SFC system will be able to decrease backlogs No SFC system will be able Improve delivery performance or improve output Enough capacity exists to meet peak loads Almost any SFC system will achieve material flow objective Effective utilisation of bottleneck areas is important A good capacity planning provides sufficient capacity with relatively level loading means SFC is straightforward If peaks and valleys in capacity requirements are passed down to the SFC, execution becomes more complex and difficult The Linkage between MRP and PAC MRP appraises the PAC system of all changes in material plan Feedback from PAC to MRP status information and warning signals Production Activity Control March 202
2 Status information includes where things are, notification of operational completion, count verification, order closeout and disposition, and accounting data Warning signals help flag inadequacies in material and capacity plan Basic Data Required for Shop Floor Control Essential inputs are routing and lead of each item Routing specifies each operation to be performed to make the part and which work centre will perform the operation Lead s are typically made up of four elements: Run (Operation or machine run per piece lot size) Set-up ( to prepare the work centre) Move (From one work centre to the next) Queue ( spent waiting to be processed at a work centre, which depends on workload and schedule) The data requirement is described with an example product structure for end item A End item A Lead = Part B Lead = 3 Part C Lead = 2 Part D Lead = 2 Part E Lead = Part P Lead = Part Q Lead = 2 Item P and Q are purchased parts Fig. Product structure of Product A Production Activity Control 2 March 202
3 Table: Routing and lead- data for part D Operation Work centre Run Set-up Move Queue Total Rounded Total lead (days) = 0.0 Table: Routing and lead- data for part E Operation Work centre Run Set-up Move Queue Total Rounded Total lead (days) = 5.0 Queue (the critical element) frequently accounts for 80 percent or more of total lead- Reduced queue means shorter lead and, therefore, reduced work-in-process inventory Reduction requires better scheduling Queue and Move Times in Scheduling Schedule may or may not use lead s that included queue and move s The more usual practice is to prepare the detailed work centre schedule without move and queue s Production Activity Control Techniques The SFC approaches are: Gantt Chart Gantt Charts, Finite Loading, and Priority Sequencing Rules Developed by Henry L. Gantt A simple visual technique used Shows planned work activities versus actual accomplishments on the same scale Production Activity Control 3 March 202
4 National Institute of Technology Calicut Shows a schedule It also shows progress of jobs. So it can be called a progress chart In addition to operation, maintenance also may be represented on it Or shows the sequencing of jobs on machines Shows a schedule In addition to operation, maintenance also may be represented on it Or as a load chart Example of Gantt chart progress chart Fig. Planned work activities Vs Actual accomplishments Production Activity Control 4 March 202
5 A Gantt chart example problem Consider certain orders, which are to be processed in a numerical order of job numbers. All the jobs have same sequence of operations (flow shop). The jobs are processed first in machine A and then in Machine B. The jobs and their requirements in days (set-up and run ) are given below. It is assumed that there is no maintenance involved till all the jobs in the list are processed Table: Requirement in days Job Machine A B Machines A B Time (days) 2 Fig. Gantt chart for the above problem This Gantt chart shows the start and finish for jobs in each machine for the given priority Production Activity Control 5 March 202
6 Find the sequence of jobs using Johnson s rule and develop a Gantt chart for that sequence. Can you notice any difference in completion of all jobs (Makspan) Johnson s Rule: Identify the shortest processing of jobs. If this is in machine put this job first in the sequence else put it in last in the sequence. Remove this job from further consideration. Repeat this process till all jobs are assigned. A three machine n Job problem: A shop has eight shop orders that must be processed sequentially through three work centres. Each job must be finished in the same sequence in which it was started. Time (in hours) required at various work centres are as shown below. Use Johnson s rule to develop the job sequence that will minimize the completion over all shop orders. Job Code A B C D E F G H Work Centre Work Centre Work Centre Note: Johnson s rule is applicable to 3 machine n job problem if the largest processing in 2 nd machine is less than or equal to the smallest processing in machine or machine 3. (The flow shop machine sequence for the jobs is Machine first then Machine 2 and Machine 3.) If this condition is satisfied, convert the three machine problem into 2 machine n job problem. To convert the problem into 2 machine n job problem, consider two fictitious machines. The processing of jobs in first fictitious machine is the sum of the processing of jobs in machines and 2. The processing of jobs in second fictitious machines is the sum of the processing of jobs in machines 2 and 3. Now, consider the problem as a two machine n job problem, determine the sequence of jobs to be processed. This sequence is to be used for preparing the schedule of jobs in machines, 2 and 3. A Gantt chart example Load chart The following hours are required to complete six jobs, which are routed through four work centres. The hours available at the work centres are 40 hrs at W, 32 hrs at X, 40 hrs at Y and 30 hrs at Z. Develop a load chart. Table: Processing s in various work centres Job Hours Required at work centres No. W X Y Z A A A B B B Production Activity Control 6 March 202
7 National Institute of Technology Calicut Week of Work Centre Job No. A2 A22 A23 B4 B5 B6 W Hours Reqd Unused X Avai Hours Reqd Unused Y Avai Hours Reqd Unused Z Avai Hours Reqd Avai Unused 3 Fig. Load Chart - Tabular Form Fig. Load chart (Gantt chart) Production Activity Control 7 March 202
8 Shop Loading Loading determines the work centres to receive the jobs Loading assigns jobs to the work centres, but it does not necessarily specify the order in which jobs will be performed An Infinite Loading Problem MRP computation gives planned order releases for items A through F which are summarised below: Planned order releases (by weeks) Items A 0 5 B 5 27 C 7 3 D E F 3 2 In addition, the following information shows the work centres required to make A F, their operation sequence at the work centres (X, Y, and Z), and setup and machine required per unit: Item Route A X Z B Y C Z D X Y X E Z X F X Item Work centre Operation sequence number Machine setup s (hours) Operation per item (hours) A X A Z B Y C Z D X D Y D X E Z E X F X (a) Develop an infinite-load chart which shows the load in each of the eight weeks for work centres X, Y, and Z. Production Activity Control 8 March 202
9 (b) If the factory is operating at a 40-hour per week capacity (but with ten hours of over per week authorised), how can the work loaded in centres X, Y, and Z be rescheduled to utilize available capacity more evenly? Finite Loading Systems (FLS) FLS simulate actual job order starting and stopping to produce a detailed schedule for each shop order and each work centre Load all jobs in all necessary work centres for the length of the planning horizon Result is a set of start and finish dates for each operation at each work centre The schedule is based on the finite capacity limits at each work centre Employ a material requirements plan from MRP and then schedule work centres based on work centre capacities and job priorities The detailed capacity planning in the engine side of the MPC system uses infinite loading assigns jobs to work centres without considering capacity There is a connection from the detailed capacity planning to the SFC This indicates that the data similar to those used for capacity requirements planning are used to schedule detailed shop operations with finite loading techniques Finite Loading Process Assume that we want to run the finite loading system on Monday morning Assume that at a given machine, a job is partly finished, there is queue of jobs waiting for processing, and some new jobs have arrived at the shop The finite loading system consider the remaining processing of partially processed jobs and all other jobs including the new ones Sequence (priorities) the jobs so that available capacity of the machines is not exceeded The output is the is a set of start s and completion s for each job at each work centre The system often has the capability of look ahead to jobs that have yet to arrive at the work centre So the system some allows a machine to idle for short awaiting an important job to arrive Some systems also allows lot streaming The use of different sized transfer batches has been called lot streaming (i.e., transfer batch size should not necessarily equal the production batch size) Commonly used types of loading are horizontal and vertical loading Production Activity Control 9 March 202
10 Horizontal Loading The entire shop order or job is loaded for all its operations and then takes the next job Process Begins with the highest priority job and schedules it through all work centres Then next highest priority job is scheduled through all work centres This process repeats until all jobs have been scheduled If a lower priority job can be scheduled ahead of a higher priority job at some machine, without delaying the start of higher priority job, the system will do so At s a machine is to sit idle, even if a job is available, because a more important job is coming Vertical Loading Fills a work centre job-by-job It is consistent with how most job shop scheduling research is conducted, as well as with the priority scheduling viewpoint The system looks at a work centre and decide which set of jobs to load next An Example Table: Set-up and Processing Times for Finite Loading Example Job Priority Machines A B C Routes Second A-B-C 2 First A-B-C 3 Third B-C A Job 2 Job B Job 3 Job 2 Job C Job 2 Job Job Fig. An Example of Horizontal Loading (Forward scheduling) Production Activity Control 0 March 202
11 For vertical loading the priority is SPT A Job Job 2 B Job 3 Job Job 2 C Job 3 Job Job Fig. An Example of Vertical loading (Forward scheduling) Horizontal loading is similar to blocked- approach Vertical loading is like event-driven approach Event-driven approach The system reschedules each an event occurs Events are job completion at work centres, arrival of new jobs, and so on The event-driven approach attempts to never idle a machine if any jobs are ready for processing Event-driven schedules performs better in terms of flow, flow variability, and mean tardiness Flow It is the amount of a job spends from the moment it is ready for processing until its completion, and includes any waiting prior to processing Makespan is the total for all jobs to finish processing Average WIP is directly related to the jobs spend in the shop Tardiness Tardiness is related to lateness and earliness and they are the performance measures related to each job s due date Lateness is the amount of a job is past its due date Earliness is the amount of prior to its due date at job processing is complete Tardiness equals lateness if the job is late, or zero if it is on or early Production Activity Control March 202
12 Anther distinction is that of forward (front) scheduling versus backward (back) scheduling Forward (Front) Scheduling Scheduling starts from the current and build the schedule forward in Forward schedule can tell the earliest date that an order can be completed Backward (Back) Scheduling Scheduling starts from the due date of each job and schedules backward in It tells when are order must be started in order to be done by a specific date PRIORITY SEQUENCING RULES Sequence jobs in priority order, without regard to capacity They are generally designed to require very little computational effort and, even for a large number of jobs Most rules sort jobs based on one parameter FCFS (First-Come-First-Served) Choose the jobs in the order in which they arrive. It is viewing as being fair and many service operations use it for that reason However, it is far from optimal for many objectives SPT (Shortest Processing Time) Choose job that has the shortest processing first Processing include set-up also SPT schedules are optimal for number of objectives (performance measures) such as minimising total flow, mean flow, mean waiting, mean tardiness and total lateness SPT is the best rule for performance measures (mean flow, mean lateness and WIP) related to shop congestion SPT, however, may cause long jobs to experience excessive delays because short jobs keep arriving and move to the front of queue Therefore, the variance of job flows may be quite high with SPT One solution would be to use Truncated SPT (TSPT) TSPT imposes a limit on jobs in the queue Any jobs exceeding the limit are sequenced according to FCFS If no jobs exceed the limit use SPT Production Activity Control 2 March 202
13 TSPT reduces flow variance but WIP and mean flow increases compared to SPT Another solution is to use Relief SPT (RSPT) Use FCFS until the queue length hits a value Q Then switch the rule to SPT Results are roughly the same as with TSPT SWPT (Shortest Weighted Processing Time) Explicitly accounts for the value of jobs The weight are expressed as costs relevant to the performance measure of interest delay cost, holding cost or other costs SWPT sequencing minimises total weighted flow n F Total weighted flow w j SWPT sequencing is w Example problem: Truncated SPT (TSPT) P w P 2 w 2 j F Production Activity Control 3 March 202 j P n w n The following list of jobs available today in a critical department includes estimates of their processing s and these jobs are released to the department at various points in : Job Processing (days) Job released date A 8 5 th week st day B 3 5 th week 2 nd day C 7 4 th week 5 th day D 0 3 rd week 4 th day E 6 5 th week 3 rd day F 5 4 th week st day G 4 4 th week 5 th day Today in production calendar is the third day morning of the 5 th week. Assume that the jobs are released in the given day morning and the organisation work for 5 days in a week. If the job waiting exceeds 5 days then, the organisation uses truncated shortest processing method for scheduling the jobs. Prepare the job sequence based on this method and determine the mean flow. The scheduler decided that this schedule is not going to change and further jobs will not be considered for scheduling until all these jobs are completed. Solution: The jobs are taken for scheduling one after another.
14 Job Processing (days) Job released date Waiting until now A 8 5 th week st day 2 B 3 5 th week 2 nd day C 7 4 th week 5 th day 3 D 0 3 rd week 4 th day 9 E 6 5 th week 3 rd day 0 F 5 4 th week st day 7 G 4 4 th week 5 th day 3 Sequence: D F B G E C A Job Processing Completion from Flow Remarks (days) now D FCFS rule is F used B G E SPT is used C A Mean flow = ( )/7 = 28 days EDD (Earliest Due Date) EDD sequences are extremely easy to develop so long as each job has a due date attached EDD minimises maximum lateness and maximum tardiness CR (Critical Ratio) This rule rely on the due date as well as processing CR is the ratio of (Due date Current ) to the processing remaining It is a dynamic rule because the ratio changes as progresses An Example: Single machine static problem Job Processing, P j, in days Due date, d j, (day) Production Activity Control 4 March 202
15 Calculate for each job flow (F j ), Lateness (L j ), Earliness (E j ) and Tardiness (T j ) for the priority rules such as FCFS, SPT, EDD and CR. Compare the rules performance based WIP, mean flow, mean lateness, mean tardiness, maximum flow, maximum lateness and maximum tardiness Note on WIP Calculation WIP is the weighted average of the number of jobs in the system where the weights are the production s for the jobs. Calculation procedure Consider the flowshop example given in the Gantt chart problem. The flow of the jobs for the schedule given in the Gantt chart is Job Flow The average number of jobs in the system is determined by: Four jobs are in the system when job is completed, 3 jobs are in system from first job completion to second job completion, from second job completion to third job completion there are two jobs and one job in the system from third job completion to last job finishing. Therefore, WIP = ( )/2 =62/2 This is also equal to cumulative flow divided by maximum flow. Minimum Slack Time (MST) Slack is the difference between the until the due date and the remaining processing. Choose the job with minimum slack Operation Due Date (ODD) Create operation due dates that serve as intermediate deadlines prior to the real due date Operation due date may be spaced evenly between the job arrival and the final due date, by dividing the interval by the number of operations to performed Another way of setting operation due date is to use proportion of the job s total work Research has shown that later method performs better than former Slack Time per Operation (S/OPN) The job with the smallest ratio between slack and the number of operations remaining is given highest priority Allowance per Operation (A/OPN) Remaining allowance is the between the current date and the due date. Production Activity Control 5 March 202
16 The job with the smallest ratio between remaining allowance and the number of remaining operations is given highest priority Modified Operation Due Date (MOD) Modified due date of an operation is the larger of its original operation due date and its earliest possible finish Minimize the number of Tardy Jobs The algorithm consists of two sets; first set (V) consists of early jobs in EDD order and the second set (U) consists of late jobs The algorithm due to Moore, begins with all jobs in set V, and then successively assigns jobs to set U Steps. Sequence the jobs according to the EDD rule and initially put all jobs in set V 2. Find the first tardy job in set V. Suppose this is the k th job in sequence. If there are no tardy jobs in set V, stop; the sequence is optimal 3. Find the longest job among the first k jobs in sequence. Place this job in set U, and go to step 2. Jobs in Set U are tardy and will occupy a position in sequence after all nontardy jobs Some General Comments on Sequencing Rules Rules may be Local or Global Local rules require information only about the queue at the current machine. E.g. SPT Global rules require information about conditions elsewhere in the shop as well. E.g.- ODD Rules may be Static or Dynamic Static rules have priorities that do not change over E.g. EDD. SPT, on the other hand, is static at a particular machine, but a job s priority can change as the job moves through the shop Dynamic rules generate priorities that can change over. E.g. MST Rules are generally divided into two groups: Rules for relieving shop congestion E.g. SPT Rules for meeting due date E.g. EDD, ODD, MST Production Activity Control 6 March 202
17 Two-Machine Job Shop Problem (Makespan Criterion) A static deterministic job shop problem having two-machines and n jobs with maximum two operations, an optimum solution (Jackson s algorithm) exists. The algorithm minimizes the makespan. Jackson s Algorithm Partition the set of jobs into 4 subset Set {} contains the jobs, which consists of only one operation, to be performed on machine M. Set {2} contains the jobs, which consists of only one operation, to be performed on machine M2. Set {, 2} contains the jobs, which consists of two operations, of which first is to be performed on machine M, and second on M2. Set {2, } contains the jobs, which consists of two operations, of which first is to be performed on machine M2, and second on M. Scheduling Phase First, use Johnson s algorithm to sequence the jobs in set {,2}, disregarding all other jobs. Use the same procedure to sequence the jobs in set {2,} Job in set {} may be scheduled arbitrarily; similarly, job in {2} Solution Phase Preserving the ordering as in the scheduling phase arrange the four set of jobs as follows On M schedule jobs in {,2} first, then jobs in {}, and at the end jobs in {2,} On M2 schedule jobs in {2,} first, followed by jobs in {2}, followed by jobs in {,2} Production Activity Control 7 March 202
18 Machine {, 2} {} {2, } 2 {2, } {2} {, 2} An example Time Fig. Principle of Jackson s Algorithm Machine Processing of jobs in days Job Operation Sequence Solution A A B B B A 5 A B Set (A, B) = {2, 5}, Set (B, A) = {4}, Set (A) = {}, Set (B) = {3} Sequence of jobs in machines Machine A Machine B Production Activity Control 8 March 202
19 Schedule Job Start Machine A Finish Job Start and Finish Time of Jobs Start Machine B Finish Production Activity Control 9 March 202
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