Operations Scheduling Chapter 8

The Hierarchy of Production Decisions The logical sequence of operations in factory planning corresponds to the sequencing of chapters in a production management text book. All planning starts with the demand forecast. Demand forecasts are the basis for the top level (aggregate) planning. The Master Production Schedule (MPS) is the result of disaggregating aggregate plans down to the individual item level. Based on the MPS, MRP is used to determine the size and timing of component and subassembly production. Detailed shop floor schedules are required to meet production plans resulting from the MRP.

Hierarchy of Production Decisions

Scheduling Scheduling: Establishes the timing of the use of equipment, facilities and human activities in an organization Effective scheduling can yield Cost savings Increases in productivity Improved customer satisfaction

Scheduling Techniques Scheduling techniques are designed to disaggregate the master production schedule into time-phased daily or hourly activities. A detailed production schedule must include when and where each activity must take place in order to meet the master schedule.

Scheduling Activities Scheduling involves the following major activities: Routing (determining where the work is going to be done). Short-run capacity planning. Short-run machine, manpower and production scheduling. Determining the sequence of activities (determining when the work is to be done). Dispatching (issuing the order to begin work). Controlling the progress of orders and monitoring the process to determine that operations are running according to plan. Revising the schedule based on changes in order status of jobs, material and/or capacity availability and various other reasons. Expediting (speeding the progress of the work order) late, critical jobs.

Elements of Scheduling Elements of Scheduling Problems: 1. Job arrival patterns (static vs. dynamic). Dynamic arrival pattern means that more jobs will arrive in the system during the time those currently in the system are being processed. In a static system, all jobs that will ever enter the system are known. Most job shops are dynamic. 2. Ratio of workers to machines (machine limited vs. labor limited environment). 3. Priority rules for scheduling. 4. Flow patterns of jobs through the plant. a. Flow shop: All jobs follow the same pattern of flow through the system. In a flow shop, routing is not typically a problem. b. Job shop: Each job follows its own specified pattern. Job shop is more difficult to analyze than the flow shop.

Goals of Production Scheduling High Customer Service: on-time delivery Low Inventory Levels: WIP and FGI High Utilization: of machines and labor

Meeting Due Dates – Measures Service Level: Used typically in make to order systems. Fraction of orders which are filled on before their due dates. Fill Rate: Used typically in make to stock systems. Fraction of demands met from stock. Lateness: Used in shop floor control. Difference between order due date and completion date. Average lateness has little meaning. Better measure is lateness variance. Tardiness: Is equal to the lateness of a job if it is late and zero, otherwise. Average tardiness is meaningful but unintuitive.

Basic Definitions Throughput (TH): for a line, throughput is the average quantity of good (non-defective) parts produced per unit time. Work in Process (WIP): inventory between the start and endpoints of a product routing. Raw Material Inventory (RMI): material stocked at beginning of routing. Finished Goods Inventory (FGI): material held in inventory prior to shipping to the customer. Cycle Time (CT): time between release of the job at the beginning of the routing until it reaches an inventory point at the end of the routing. Makespan: The total amount of time to process a fixed number of jobs. Little’s Law: TH = WIP/CT  WIP=TH*CT  (L=λw) where λ=TH and w=CT

Reducing WIP and Cycle Time Less WIP Equals Shorter Cycle Times (Little’s Law) Shorter cycle time means: Less WIP Better responsiveness All of which reduce costs and improve sales revenue

Classic Scheduling – Assumptions MRP/ERP: Benefits – Simple paradigm, hierarchical approach. Problems – MRP assumes that lead times are an attribute of the part, independent of the status of the shop. MRP uses pessimistic lead time estimates.

Classic Scheduling – Assumptions (cont.) Classic Scheduling: (only classic in academia) Benefits – “Optimal” schedules Problems – Bad assumptions. All jobs available at the start of the problem. Deterministic processing times. No setups. No machine breakdowns. No preemption. No cancellation.

Objectives in Job Shop Scheduling Meet due dates Minimize work-in-process (WIP) inventory Minimize average flow time Maximize machine/worker utilization Reduce set-up times for changeovers Minimize direct production and labor costs (note: that these objectives can be conflicting)

Measures to Evaluate Performance of a Scheduling Method Service Level: Fraction of orders filled on before their due dates (used in make-to-order systems) Fill Rate: Fraction of demand that are met from inventory without backorder (used in make-to-stock systems) Job Flow Time: Time elapsed from the release of a job until it is completed. Lateness: Difference between completion time and due date of a job (may be negative). Tardiness: The positive difference between the completion time and the due date of a job. Makespan: Flow time of the job completed last.

Measures to Evaluate Performance of a Scheduling Method Production Rate Utilization Keep in mind that high utilization means high return on investment. This is good provided that the equipment is utilized to increase revenue. Otherwise, high utilization only helps to increase inventory, not profits.

Terminology Flow shop: n jobs processed through m machines in the same sequence Job shop: the sequencing of jobs through machines may be different, and there may be multiple operations on some machines. Parallel processing vs. sequential processing: parallel processing means that the machines are identical, any job can be processed on any machine.

Common Sequencing Rules FCFS. First Come First Served. Jobs processed in the order they come to the shop. SPT. Shortest Processing Time. Jobs with the shortest processing time are scheduled first. EDD. Earliest Due Date. Jobs are sequenced according to their due dates. CR. Critical Ratio. Compute the ratio of processing time of the job and remaining time until the due date. Schedule the job with the largest CR value next.

Scheduling Service Operations Vs Manufacturing Operations Scheduling service systems presents certain problems not generally encountered in manufacturing systems. This is primarily due to: The inability to store services The random nature of customer requests To avoid problems such as long delays, unsatisfied customers, service systems rely on appoinment systems and reservation systems.

High-Volume Systems Flow system: High-volume system with Standardized equipment and activities Flow-shop scheduling: Scheduling for high-volume flow system Work Center #1 Work Center #2 Output

High-Volume Systems Examples of high-volume products include autos, personal computers, televisons. In process industries, examples include petroleum refining, sugar refining. A major issue in design of high-volume (flow) systems is line balancing.

Success Factors in High-Volume Systems Process and product design Preventive maintenance Rapid repair when breakdown occurs Minimization of quality problems Reliability and timing of supplies

Intermediate-Volume Systems Outputs are between standardized high-volume systems and made-to-order job shops The volume of output is not large enough to justify continuous production. Examples include canned foods, paint and cosmetics.

Intermediate-Volume Systems The three basic issues in these systems are: 1.Run size, 2.Timing, and 3.Sequence of jobs Economic run size: h’ is defined as h’= h(1- λ/P)

Scheduling Low-Volume Systems Loading - assignment of jobs to process centers Sequencing - determining the order in which jobs will be processed Job-shop scheduling Scheduling for low-volume systems with many variations in requirements

Gantt Load Chart Figure 15.2 Gantt chart - used as a visual aid for loading and scheduling

Loading Infinite loading: Assigning jobs to work centers without considering the capacity of work center. Finite loading: Takes into acccount the capacity of work center. Forward scheduling: Scheduling ahead from some point in time. Backward scheduling. Scheduling backwards from due dates.

Loading

LOADING (The Assignment Problem) In many business situations, management needs to assign - personnel to jobs, - jobs to machines, - machines to job locations, or - salespersons to territories. Consider the situation of assigning n jobs to n machines. When a job i (=1,2,....,n) is assigned to machine j (=1,2, .....n) that incurs a cost Cij. The objective is to assign the jobs to machines at the least possible total cost.

The Assignment Problem This situation is a special case of the Transportation Model and it is known as the assignment problem. Here, jobs represent “sources” and machines represent “destinations.” The supply available at each source is 1 unit And demand at each destination is 1 unit.

The Assignment Problem The assignment model can be expressed mathematically as follows: Xij= 0, if the job j is not assigned to machine i 1, if the job j is assigned to machine i

The Assignment Problem

The Assignment Problem Example Ballston Electronics manufactures small electrical devices. Products are manufactured on five different assembly lines (1,2,3,4,5). When manufacturing is finished, products are transported from the assembly lines to one of the five different inspection areas (A,B,C,D,E). Transporting products from five assembly lines to five inspection areas requires different times (in minutes)

The Assignment Problem Example Ballston Electronics manufactures small electrical devices. Products are manufactured on five different assembly lines (1,2,3,4,5). When manufacturing is finished, products are transported from the assembly lines to one of the five different inspection areas (A,B,C,D,E). Transporting products from five assembly lines to five inspection areas requires different times (in minutes)

The Assignment Problem Example Under current arrangement, assignment of inspection areas to the assembly lines are 1 to A, 2 to B, 3 to C, 4 to D, and 5 to E. This arrangement requires 10+7+12+17+19 = 65 man minutes.

The Assignment Problem Example Management would like to determine whether some other assignment of production lines to inspection areas may result in less cost. This is a typical assignment problem. n = 5 And each assembly line is assigned to each inspection area. It would be easy to solve such a problem when n is 5, but when n is large all possible alternative solutions are n!, this becomes a hard problem.

The Assignment Problem Example Assignment problem can be either formulated as a linear programming model, or it can be formulated as a transportation model. However, An algorithm known as Hungarian Method has proven to be a quick and efficient way to solve such problems. This technique is programmed into many computer modules such as the one in WINQSB.

The Assignment Problem Example WINQSB solution for this problem is as follows:

Hungarian Method Example Step 1: Select the smallest value in each row. Subtract this value from each value in that row Step 2: Do the same for the columns that do not have any zero value.

Hungarian Method Example If not finished, continue with other columns.

Hungarian Method Example Step 3: Assignments are made at zero values. Therefore, we assign job 1 to machine 1; job 2 to machine 3, and job 3 to machine 2. Total cost is 5+12+13 = 30. It is not always possible to obtain a feasible assignment as in here.

Hungarian Method Example 2

Hungarian Method Example 2 A feasible assignment is not possible at this moment. In such a case, The procedure is to draw a minimum number of lines through some of the rows and columns, Such that all zero values are crossed out.

Hungarian Method Example 2 The next step is to select the smallest uncrossed out element. This element is subtracted from every uncrossed out element and added to every element at the intersection of two lines.

Hungarian Method Example 2 We can now easily assign to the zero values. Solution is to assign (1 to 1), (2 to 3), (3 to 2) and (4 to 4). If drawing lines do not provide an easy solution, then we should perform the task of drawing lines one more time. Actually, we should continue drawing lines until a feasible assignment is possible.

Sequencing Sequencing: Determine the order in which jobs at a work center will be processed. Workstation: An area where one person works, usually with special equipment, on a specialized job.

Sequencing n jobs on a Single Machine Priority rules: Simple heuristics such as FCFS, SPT, DD, CR are used to select the order in which jobs will be processed. CR= (Due Date – Current Time)/ Processing Time Job time: Time needed for setup and processing of a job.

Example: Sequencing Rules Use the FCFS, SPT, and Critical Ratio rules to sequence the five jobs below. Evaluate the rules on the bases of average flow time, average number of jobs in the system, and average job lateness. (Due Date) Job Processing Time Time to Promised Completion A 6 hours 10 hours B 12 16 C 9 8 D 14 14 E 8 7

Example: Sequencing Rules FCFS Rule A > B > C > D > E Processing Due Flow Job Time Date Time Lateness A 6 10 6 0 B 12 16 18 2 C 9 8 27 19 D 14 14 41 27 E 8 7 49 42 49 141 90

Example: Sequencing Rules FCFS Rule Performance Average flow time: 141/5 = 28.2 hours Average number of jobs in the system: 141/49 = 2.88 jobs Average job lateness: 90/5 = 18.0 hours

Example: Sequencing Rules SPT Rule A > E > C > B > D Processing Due Flow Job Time Date Time Lateness A 6 10 6 0 E 8 7 14 7 C 9 8 23 15 B 12 16 35 19 D 14 14 49 35 49 127 76

Example: Sequencing Rules SPT Rule Performance Average flow time: 127/5 = 25.4 hours Average number of jobs in the system: 127/49 = 2.59 jobs Average job lateness: 76/5 = 15.2 hours

Example: Sequencing Rules Critical Ratio Rule E > C > D > B > A Processing Promised Flow Job Time Completion Time Lateness E (.875) 8 7 8 1 C (.889) 9 8 17 9 D (1.00) 14 14 31 17 B (1.33) 12 16 43 27 A (1.67) 6 10 49 39 49 148 93

Example: Sequencing Rules Critical Ratio Rule Performance Average flow time: 148/5 = 29.6 hours Average number of jobs in the system: 148/49 = 3.02 jobs Average job lateness: 93/5 = 18.6 hours

Example: Sequencing Rules Comparison of Rule Performance Average Average Average Flow Number of Jobs Job Rule Time in System Lateness FCFS 28.2 2.88 18.0 SPT 25.4 2.59 15.2 CR 29.6 3.02 18.6 SPT rule was superior for all 3 performance criteria.

Sequencing n jobs on two machines Johnson’s Rule: technique for minimizing completion time for a group of n jobs to be processed on two machines or at two work centers. Minimizes total idle time Johnson’s Rule requires satisfying the following conditions:

Johnson’s Rule Conditions Job time must be known and constant Job times must be independent of sequence Jobs must follow same two-step sequence Job priorities cannot be used All units must be completed at the first work center before moving to second

Johnson’s Rule Optimum Sequence List the jobs and their times at each work center Find the smallest processing time. If it belongs to the first operation of a job schedule that job next, otherwise schedule that job last. Eliminate the job from further consideration Repeat steps 2 and 3 until all jobs have been scheduled

Johnson’s Algorithm Example Data: Iteration 1: min time is 4 (job 1 on M1); place this job first and remove from lists:

Johnson’s Algorithm Example (cont.) Iteration 2: min time is 5 (job 3 on M2); place this job last and remove from lists: Iteration 3: only job left is job 2; place in remaining position (middle). Final Sequence: 1-2-3 Makespan: 28

Gantt Chart for Johnson’s Algorithm Example Short task on M2 to “clear out” quickly. Short task on M1 to “load up” quickly.

Example A group of six jobs is to be processed through a two-machine flow shop. The first operation involves cleaning and the second involves painting. Determine a sequence that will minimize the total completion time for this group of jobs. Processing times are as follows: http://www.baskent.edu.tr/~kilter

The remaining jobs and their times are Select the job with the shortest processing time. It is job D, with a time of two hours. Since the time is at the first center, schedule job D first. Eliminate job D from further consideration. Job B has the next shortest time. Since it is at the second work center, schedule it last and eliminate job B from further consideration. We now have The remaining jobs and their times are http://www.baskent.edu.tr/~kilter

The shortest remaining time is six hours for job E at work center 1 The shortest remaining time is six hours for job E at work center 1. Thus, schedule that job toward the beginning of the sequence (after job D). Thus, Job C has the shortest time of the remaining two jobs. Since it is for the first work center, place it third in the sequence. Finally, assign the remaining job (F) to the fourth position and the result is http://www.baskent.edu.tr/~kilter

Sequencing Jobs When Setup Times Are Sequence-Dependent http://www.baskent.edu.tr/~kilter

Scheduling Difficulties Randomness in job arrival times Variability in Setup times Processing times Interruptions Changes in the set of jobs No method for identifying optimal schedule Scheduling is not an exact science Ongoing task for a manager

Classic Dispatching Results Optimal Schedules: Impossible to find for most real problems. Dispatching: sorts jobs as they arrive at a machine. Dispatching rules: FIFO – simplest, seems “fair”. SPT – Actually works quite well with tight due dates. EDD – Works well when jobs are mostly the same size. Many (100?) others. Problems with Dispatching: Cannot be optimal (can be bad). Tends to be myopic.

The Difficulty of Scheduling Problems Dilemma: Too hard for optimal solutions. Need something anyway. Classifying “Hardness”: Class P: has a polynomial solution. Class NP: has no polynomial solution. Example: Sequencing problems grow as n!. Compare en/10000 and 10000n10. At n = 40, en/10000 = 2.4  1013, 10000n10 = 1.0  1020 At n = 80, en/10000 = 5.5  1030, 10000n10 = 1.1  1023 3! = 6, 4! = 24, 5! = 120, 6! = 720, … 10! =3,628,800, while 13! = 6,227,020,800 25!= 15,511,210,043,330,985,984,000,000 en/10000 10000n10

The Difficulty of Scheduling Problems NP stands for non polynomial, meaning that the time required to solve such problems is an exponential function of the number of jobs rather than a polynomial function. The problems for which total enumeration is hopeless are known in mathematics as NP hard.

Computation Times Current situation: computer can examine 1,000,000 sequences per second and we wish to build a scheduling system that has response time of no longer than one minute. How many jobs can we sequence optimally?

Effect of Faster Computers Future Situation: New computer is 1,000 times faster, i.e. it can do 1 billion comparisons per second. How many jobs can we sequence optimally now?

Implications for Real Problems Computation: NP algorithms are slow to use. No Technology Fix: Faster computers don’t help on NP algorithm. Scheduling is Hard: Real scheduling problems tend to be NP Hard. Scheduling is Big: Real scheduling problems also tend to be quite large; impossible to solve optimally.

Implications for Real Problems (cont.) Robustness? NP hard problems have many solutions, and presumably many “good” ones. Our task is to find one of these. Role of Heuristics: Polynomial algorithms can be used to obtain “good” solutions. Example heuristics include: Simulated Annealing Tabu Search Genetic Algorithms

The Bad News Violation of Assumptions: Most “real-world” scheduling problems violate the assumptions made in the classic literature: There are always more than two machines. Process times are not deterministic. All jobs are not ready at the beginning of the problem. Process time are sequence dependent. Problem Difficulty: Most “real-world” production scheduling problems are NP-hard. We cannot hope to find optimal solutions of realistic sized scheduling problems. Polynomial approaches, like dispatching, may not work well.

The Good News Due Dates: We can set the due dates. Job Splitting: We can get smaller jobs by splitting larger ones. Single machine SPT results imply small jobs “clear out” more quickly than larger jobs. Mechanics of Johnson’s algorithm implies we should start with a small job and end with a small job. Small jobs make for small “move” batches and can be combined to form larger “process” batches.

The Good News (cont.) Feasible Schedules: We do not need to find an optimal schedule, only a good feasible one. Focus on Bottleneck: We can often concentrate on scheduling the bottleneck process, which simplifies problem closer to single machine case. Capacity: Capacity can be adjusted dynamically (overtime, floating workers, use of vendors, etc.) to adapt facility (somewhat) to schedule.

Minimizing Scheduling Difficulties Set realistic due dates Focus on bottleneck operations Consider lot splitting of large jobs