Operations

Operations Scheduling What ' - Setting up of operations processing times so that, jobs will be completed by the time they are due. Purpose – To convert MPS into Day – to – day activities. Instructions about utilisation of Men, M/Cs. Allocation of Prod. Capacities to meet Demands in time (e.g. - Classroom Time-Table, Comp. Lab Timings, Marriage) BILLS OF MATERIALS CHECK INVENTORY CALCULATION OF SCHEDULE CAPACITY PLAN PURCHASE ORDERS PRODUCTION ORDERS INVENTORY FOR FORECAST DEMAND MATERIAL PLAN FORECAST DEMAND PRODUCTION OF STOCK 4 Types of Industrial Controls - 1) Administrative Control 2) Material Control 3) Production Control 4) Financial Control or Budgetary Control PRODUCTION PLANNING & CONTROL (PPC) (Ref. Book - Hazara Chowdhari, Roy) PRODUCTION DEPT. PPC Staff Production Staff Planning Control Section I S II S III AIM OF PPC :- 1. Efficient utilization of Material Resources, People & Facilities, thro’ planning, co-ordination and control of the production activities. 2. PPC issues directives to Production Staff -What to make, how many, when & by what means' to achieve the specified Target Date. 3. To relieve actual Production Staff from non-operating functions. DECISION MAKING IN PRODUCTION PLANNING 1. Production quantity of each component in specific time period. 2. The method that meets quality specs. of the components. 3. To determine capacity of all Mfg. Depts. If imbalance - offload to other firms (out-sourcing). 4. Convert Sales Orders into section-wise Production Orders. Scheduling of Opns. and deciding delivery, fixing buffer stocks 5. Decisions regarding MIS flow of production achievement. PRODUCTION CONTROL ACTIVITIES 1. Check if the resource requirements and availability. 2. Issue directives and instruction sheets to Production staff. 3. Systematic co-ordination and control of various mfg. activities. 4. Analyse production problem and recommend remedial action. 5. Prepare reports on the status of orders. 6. Corrective action for the items running behind schedule. Nature & Importance of Work Centre - Work Centre: Area wherein productive resources are organized and work is completed. It may be - a single M/C or group of M/Cs or Work Area Work Centres can be organized according to * Function in a job-shop system * Product in a flow system * Assy. Line * Group Technology Cell (GT-Cell) Scheduling Process can be M/C Limited or Labour Limited In Machine Limited Process, equipment is critical In Labour Limited Process, people are key resources. Objectives of Work Scheduling 1. To meet Due -Dates 2. To minimize Lead Time and Set-up time 3. To minimize WIP Inventory 4. To maximize M/C or Labour Utilization. Scheduling Methods – * Forward Scheduling * Backward Scheduling Forward Scheduling : Actual Production activity starts as soon as the job order is received. Backward Scheduling: Start backward from the due date. Problem 1 - National Machines Limited has 2 job orders - X and Y. Both orders have to be processed on 2 machines M1 & M2. The route sheets for the jobs are given in the table below. Both the jobs should be ready in the next eight hours and both the machines are to start processing from now onwards. Develop schedules using forward and backward scheduling. Job X Route Sheet | Job Y Route Sheet | Routingsequence | Machine | ProcessingTime(Hrs) | RoutingSequence | Machine | ProcessingTime(Hrs) | 1 | 1 | 2 | 1 | 1 | 2 | 2 | 2 | 3 | 2 | 2 | 3 | 3 | 1 | 1 | | | | Total | 6 | Total | 5 | Scheduling Activities – 1. Decide what to produce & how much ' ( MPS ) 2. Break into component details ( BoM ) 3. Make –or – Buy 4. for Buy – send Purchase Orders, for Make – prepare Routing Sheets (sequence of op. with Work Centre, Tooling, Op. Time) 5. Loading – ( which job assigned to which m/c, when ) 6. Dispatching – Job Orders despatched according to planed sequence. Gantt Chart / Bar Chart - Work Load Chart (shows work-load levels on Eqpts., Departments, Work Centres ) Gaps to be used for Routine Maintenance - Scheduling Charts – To track the movement and progress of jobs as they pass thro’ work centres. Routing Sheets : Explains sequence of Operations – which operation to be done on which machine and in which time period. Loading : Assigning specific job to each work centre. Despatching: Final act of releasing job-ordered to the workers to go ahead. Priority Rules 1. First Come First Served (FCFS) 2. Shortest Processing Time (SPT ) 3. Earliest Due date. (EDD) 4. Least Slack Rule (LSR) Scheduling Problem Q. Five jobs are in a queue into a work under centre. Their required processing times and due dates are given in below. They are listed in the order of their arrival at the work centre. Job | Required. Time (days) | Due date(day hence) | A | 3 | 5 | B | 6 | 8 | C | 2 | 6 | D | 4 | 4 | E | 1 | 2 | Determine the proper job sequence using each of the following prioritizing rules: 1. FCFS, 2. SPT, 3. EDD and 4. LS. For each solution, compute * the average throughput time, * the average number of jobs in the work centre, * and the average late time. Explanatory Notes - * Average Throughput Time : Average time taken by each job i.e. Av. Completion Time * Average No. of jobs in the Work Centre : At any time how many jobs are pending in the work centre = Total Completion Dates Total Reqd. Time for all jobs together * Average Late Time : Total Lt Time / No of Jobs ANSWER: Sequencing of Jobs is set by the rules as below A) First Come First Served (FCFS) JobSeque-nce | Reqd. time (days) | Due date (day hence) | Processing period (Start + reqd.) | Complet-ion date | Day late | A | 3 | 5 | 0 + 3 | 3 | 0 | B | 6 | 8 | 3 + 6 | 9 | 1 | C | 2 | 6 | 9 + 2 | 11 | 5 | D | 4 | 4 | 11 + 4 | 15 | 11 | E | 1 | 2 | 15 + 1 | 16 | 14 | Total | 16 | | | 54 | 31 | Average through put time = 54 / 5 = 10.8 days Average no. of jobs in the work centre = 54 / 16 = 3.38 jobs Average no. of day late = 31 / 5 = 6.2 day. B) Shortest Processing Time ( SPT ) Rule: JobSeq. | Reqd. time (days) | Due date (day hence) | Processing period (Start +reqd.) | Completion date | Day late | E | 1 | 2 | 0 + 1 | 1 | 0 | C | 2 | 6 | 1 + 2 | 3 | 0 | A | 3 | 5 | 3 + 3 | 6 | 1 | D | 4 | 4 | 6 + 4 | 10 | 6 | B | 6 | 8 | 10 + 6 | 16 | 8 | Total | 16 | | | 36 | 15 | Average through put time = 36 / 5 = 7.2 days Average no. of jobs in the WC = 36 / 16 = 2.25 jobs Average no. of day late = 15 / 5 = 3 days. C) Earliest Due Date (EDD) Rule JobSeq. | Reqd. time (days) | Due date (day hence) | Processing period (Start +reqd.) | Completion date | Day late | E | 1 | 2 | 0 + 1 | 1 | 0 | D | 4 | 4 | 1 + 4 | 5 | 1 | A | 3 | 5 | 5 + 3 | 8 | 3 | C | 2 | 6 | 8 + 2 | 10 | 4 | B | 6 | 8 | 10 + 6 | 16 | 8 | Total | 16 | | | 40 | 16 | Average through put time = 40 / 5 = 8 days Average no. of jobs in the work centre = 40 / 16 = 2.5 jobs Average no. of day late = 16 / 5 = 3.2 days. D) Least Slack Rule JobSeq. | Reqd. time (days) | Due Dt. (day hence) | Slack (days) | Processing period (Start +reqd.) | Completion date | Day late | D | 4 | 4 | 0 | 0 + 4 | 4 | 0 | E | 1 | 2 | 1 | 4 + 1 | 5 | 3 | A | 3 | 5 | 2 | 5 + 3 | 8 | 3 | B | 6 | 8 | 2 | 8 + 6 | 14 | 6 | C | 2 | 6 | 4 | 14 + 2 | 16 | 10 | Total | 16 | | | | 47 | 22 | Average through put time = 47 / 5 = 9.4 days Average no. of jobs in the WC = 47 / 16 = 2.94 jobs Average no. of day late = 22 / 5 = 4.4 days. Some more strategies of Scheduling – 5. Slack time remaining per operation (STR/OP) 6. Critical ratio (CR) 7. Last come, first served (LCFS) 8. Random order Johnson’s Job Sequencing Rule (If N jobs are to be performed, involving 2 machines) – Problem - A firm produces 6 types of fan blades. Any fan blade mfg. requires processing on 2 machines- A and B. The processing times (in hrs.) for each blade is given below – Blade type | 1 | 2 | 3 | 4 | 5 | 6 | M/C A | 30 | 100 | 50 | 20 | 90 | 100 | M/C B | 70 | 95 | 90 | 60 | 30 | 15 | Determine an optimum sequence so that the total elapsed time is minimum. Also calculate the total idle time on m/cs. A and B. Solution: Step 1: Identify Least Processing Time among A & B operations. Step 2 : Job with (Least of A ops.) to be given 1st priority : Job with (Least of B ops.) to be given last priority Step 3 : In case of Tie – assign any one job as top & another as last priority 4 | | | | | 6 | 4 | 1 | 3 | 2 | 5 | 6 | Job sequence | Machine A | Machine B | | Time in | Processing time | Time out | Time in | Processing time | Time out | 4 | 0 | 20 | 20 | 20 | 60 | 80 | 1 | 20 | 30 | 50 | 80 | 70 | 150 | 3 | 50 | 50 | 100 | 150 | 90 | 240 | 2 | 100 | 100 | 200 | 240 | 95 | 335 | 5 | 200 | 90 | 290 | 335 | 30 | 365 | 6 | 290 | 100 | 390 | 390 | 15 | 405 | Therefore the total elapsed time is 405 hours. The idle time for machine A is 15 hrs (from 390 hrs to 405 hrs ) and Idle time for m/c B is 20 +25 = 45 hrs (from 0 – 20 and 365-390 hrs.) Jhonson’s rule for N Jobs & 3 M/cs) Condition ( Any 1of the below 2) – * The smallest processing time on machine A should be greater than or equal to the largest processing time on machine B * The smallest processing time on machine C should be greater than or equal to the largest processing time on machine B We assume 2 fictitious m/cs G & H and the corresponding times of Gi and Hi are obtained as – Gi = Ai +Bi Hi = Bi + Ci Now we solve as if - N jobs and 2 machines G & H. Problem : 5 types of jobs to be processed on three machines A, B & C in the order ABC, as - Job | Processing times | | Ai | Bi | Ci | 1 | 18 | 10 | 8 | 2 | 19 | 12 | 18 | 3 | 12 | 5 | 16 | 4 | 16 | 6 | 14 | 5 | 21 | 9 | 10 | Determine the sequence that minimizes the total operation time. Also find the idle time of each machine; A, B & C Solution – Job | Processing Times | | Gi = Ai + Bi | Hi = Bi + Ci | 1 | 28 | 18 | 2 | 31 | 30 | 3 | 17 | 21 | 4 | 22 | 20 | 5 | 30 | 19 | The optimum job sequence is – 3 | 4 | 2 | 5 | 1 | Job seq. | Machine A | Machine B | Machine C | | Time in | Procesg time | Time out | Time in | Procesg time | Time out | Time in | Procesg time | Time out | 3 | 0 | 12 | 12 | 12 | 5 | 17 | 17 | 16 | 33 | 4 | 12 | 16 | 28 | 28 | 6 | 34 | 34 | 14 | 48 | 2 | 28 | 19 | 47 | 47 | 12 | 59 | 59 | 18 | 77 | 5 | 47 | 21 | 68 | 68 | 9 | 77 | 77 | 10 | 87 | 1 | 68 | 18 | 86 | 86 | 10 | 96 | 96 | 8 | 104 | Idle for M/c B = (28- 17) + (47 – 34) + (68 – 59) + (86- 77) = 11 + 13 + 9 + 9 = 42 hrs Idle for M/c C = (34-33) + (59-48) + ( 96- 87) = 1 +11 + 9 = 21 Hrs.