5.1 - Scheduling - Técnico Lisboa - Autenticação€Machine Scheduling • Minimizing Set‐up Times – SST = Shortest Set‐up Time Heuristic – A metal products manufacturer

5 – Scheduling

Operations Planning and Control

Some Background

• Machines (resources) are– Machines process jobs (molding machine, x‐ray machine, server

in a restaurant, computer…)– Machine Environment

• Single Machine• Parallel Machines (identical vs. different)• Flow Shops: different machines (e.g. assembly lines)

– Each job must be processed by each machine exactly once– All jobs have the same routing– A job cannot begin processing on the second machine until it has completed processing on the first

• Job Shops– Each job may have its own routing

• Open Shops (e.g. car repair shop)– Jobs have no specific routing

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Scheduling Algorithms

• Instance: particular set of data for the model

• Exact algorithm: Optimum solution for every

instance

• Heuristic algorithm: a good solution, we

hope, optimal or close to optimal for every

instance

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why

Measures

• Completion time of job i, Ci

• Flow time of Job i: Fi=Ci‐ri, ri as release date

• Lateness of job i: Li=Ci‐di, di as due date

• Tardiness of job i: Ti = max {0;Li}, Li>0

• Earliness of job i: Ei=max{0,‐Li}

• Number of tardy jobs: Ni

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Some Application

• Gantt Charts – Developed by Henry Gantt (1911)– Consider the following four‐job, three‐machine job‐shop scheduling problem

– Assume the following sequences:• 2‐1‐4‐3 on M1• 2‐4‐3‐1 on M2• 3‐4‐2‐1 on M3

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Some Application

• Gantt Charts (cont.): Example– The Gantt Chart oriented towards machines is:

• Last operation of job 1 is on machine 3 and is completed at time 14. So, completion time, Ci :

– C1 = 14 (on machine 3)– Also: C2 = 11 (on machine 3); C3 = 13 (on machine 1) and C4 = 10 (on machine 1)

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Some Background

• Gantt Charts (cont.): Example– The makespan is:

– Since Fi= Ci‐ri but ri = 0 in this example for all i, then flowtime and completion time are the same.

– Total flowtime is:

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Min

Some Background

• Gantt Charts (cont.): Example– The lateness and tardiness of a job

– The total lateness is ‐ Number of tardy jobs is (δi= 1)

– The total tardiness is

– The maximum tardiness is

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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. Schedule the next job with the shortest CR value.– Compute the ratio of remaining time until the due date and processing time of the job

Common Sequencing Rules

– A machining center in a job shop for a localfabrication company has 5 unprocessed jobs (1 to5).

• Given the processing times and due dates, apply thesequencing rules to determine scheduling options

• Mean Flow time, Average Tardiness and # tardy jobs

Job Number Processing Time Due Date

1 11 61

2 29 45

3 31 31

4 1 33

5 2 32

Single‐Machine Scheduling

• Introduction– Applicability

• Single‐Machine• Aggregated Machines• Bottleneck Machines

– Methods• Simple Methods• Target on performance measures• Optimization procedures (heuristic, optimal)• Sequence‐dependent setup times• Static vs. Dynamic scheduling

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• Minimizing Flow time– What if your in‐process inventory costs dominate?

• Minimize total flow time tends to minimize total holding costs

– Example

• Proposed sequence: 1‐2‐3‐4‐5• Total flowtime (F)= ?• F = p1+ (p1+p2) + (p1+p2+p3)+...+(p1+p2+...+pn)• F = n p1+ (n‐1) p2 + … + pn• For this problem F = 45

12High value P and delivers asap


• Minimizing Flow time– Shortest Processing Time (SPT)

• Sequence of jobs ordered from the smallest to largest processing times

• Is this optimal?• Theorem. SPT sequencing minimizes total flowtime on a single machine with zero release times.

• Proof. We assume an optimal schedule is not an SPT sequence.– S is optimal sequence (assumed) with i and then j– S’ is a schedule where j comes before I– The set of jobs B comes before i/j or j/I and A comes after– pi > pj– TF(S) = TF(B) + (t+pi) + (t+pi+pj) + TF(A)– TF(S‘) = TF(B) + (t+ pj) + (t+ pj+pi ) + TF(A)– TF(S)‐TF(S‘)= pi ‐pj > 0

» t is the completion of the last job in B, TF(A) and TF(B) are total flowtimes of jobs in A and B

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• Minimizing Flow time– Example (cont.)

• Optimal Schedule is 2‐4‐3‐1‐5• Completion times

– C1 = 11, C2 = 2, C3 = 7, C4 = 4 and C5 =15• Total Flow time = Total Completion Time = 39

– Remarks on the SPT Rule• Minimizes total time jobs spend in the system (because all release times are 0)

• Minimizes the mean number of jobs waiting to be processed (mean work in progress)

• Also minimizes Total Lateness (Li=Ci‐di). Why?

14Professor penaltyParkinson Law


• Maximal Tardiness and Maximal Lateness– Due date oriented measure– Earliest due date sequence (EDD)

• Minimizes the Maximal Tardiness (Tmax), Tmax =max{0;Li}

• Minimizes the Maximal Lateness (Lmax), Li=Ci‐di– Example

• EDD sequence is 5‐3‐4‐2‐1• Tardiness of the jobs is (0, 0, 2, 1, 0)

15Customer satisfaction Minmax


• Number of Tardy Jobs• EDD may have several jobs somewhat tardy• If the fixed cost component of jobs being tardy dominates we wish to have the most of them on time

– Moore’s Algorithm• Step1. Compute the tardiness for each job in the EDD sequence. Set NT=0, and let k be the first position containing a tardy job. If no job is tardy go to step 4.

• Step 2. Find the job with the largest processing time in positions 1 to k.

• Step 3. Remove job j* from the sequence, set NT=NT+1, and repeat Step1.

• Step 4. Place the removed NT jobs in any order at the end of the sequence.

• This sequence minimizes the number of tardy jobs

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• Number of Tardy Jobs– Example

• EDD‐sequence 5‐3‐4‐2‐1• Step 1: The tardiness is (0, 0, 2, 1, 0) ⇒Job 4 in the third position is the first tardy job;

• Step 2: The processing times for jobs 5, 3 and 4 are 4, 3, 2, respectively;

– largest processing time for job 5• Step 3: Remove job 5, go to step 1• Step 1: EDD‐sequence is 3‐4‐2‐1; completion times (3, 5, 7, 11) and tardiness (0, 0, 0, 0) ⇒Go to step 4 ….

• Step 4: schedule that minimizes the number of tardy jobs is 3‐4‐2‐1‐5/3‐4‐2‐5‐1 and has 1 tardy job: Jobs 5

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• Precedence Constraints: Lawler’s Algorithm– Objective function–– gi is a non‐decreasing function of the flow time Fi

– Examples• ‐ = minimizing maximum lateness• ; 0 minimizing maximum tardiness

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• Precedence Constraints: Lawler’s Algorithm– Concept

• Back scheduling

• At each step determine the set of jobs V not required to precede any other

• Among V choose job k that satisfies– ∈– ∑ and corresponds to the processing time of the current sequence

– Job k is scheduled last• Determine again V and is reduced by

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Lawler´s Algorithm Example

An automotive painting and repair has 6 cars waiting to repairs.Three (1,2,3) from car rental and he agreed to finish these carsbased on the due dates. Cars 4,5,6 from a retailer dealer, he agreedthat car 4 be completed first (customer is waiting) .The processing times and due dates are available for each job.

How should be the schedule to minimize the maximum tardiness?

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Job 1 2 3 4 5 6

Pi 2 3 4 3 2 1

Due Date 3 6 9 7 11 7


• Minimizing Set‐up Times– Sequence‐dependent set‐up times– The time to change from one product to another may be significant and may depend on the previous part produced

– pij = time to process job j if it immediately follows job i– Examples:

• Electronics industry• Paint shops• Injection molding

– Minimizes makespan, since it also considers set‐up – Problem is equivalent to the traveling salesman problem (TSP)

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Plastic Tops


• Minimizing Set‐up Times– SST = Shortest Set‐up Time Heuristic

– A metal products manufacturer has contracted to ship metal braces each day for four customers. Each brace requires a different set‐up on the rolling mill:

*Job C cannot follow job D, because of quality problems– SST‐heuristic:

» Step 1 starting arbitrarily by choosing one Job: A» Step 2 B has the smallest set‐up time following A; ⇒ A‐B» Step 3 C has the smallest set‐up time of all the remaining jobs following B; ⇒ A‐B‐C

» Step 4 D is the last remaining job; ⇒ A‐B‐C‐D‐A with a makespan of 3 + 4 + 2 + 5 =14

22Starting point


• Minimizing Set‐up Times– A regret based algorithm

• A regret is a penalty for a decision that was not made– Each job must be included once: at least one element from each row

– Pick the smallest element in each row and their sum is the lower bound on makespan

• Reduced matrix– Row reduction– Column reduction– Sum of reduced coef = lower bound

• Find the reduced matrix!– Has a 0 in each column and each row

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• Minimizing Set‐up Times– A regret based algorithm (cont.)

• The reduced matrix is

– If job B does not follow (come after) job A, some other job must follow A A‐B , A‐?; ?‐B;

» C “adds no” set‐up time (is already in the lower bound)– Some job must precede B: D has 0 set‐up

» Thus, we have 0 regret not to chose B to follow A

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• The reduced matrix is (cont.)

– Zero cell C‐D» If C does not precede D then we must select other job to precede D ( ? ‐D)

» A can precede D : A‐D with a regret of 1 time unit» C‐? : C‐A with “zero” regret time unit» Select the jobs pair based on the highest regret value

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• Find the cycle sequence that minimizes the set‐up time• Data

• pij element of the set‐up time matrix (even if reduced)• Rij regret for element ij, where pij = 0• Cmax makespan of the partial sequence• L iteration• n jobs

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• Step 0: C(max) = 0 and L = 1• Step 1: Reduce the Matrix

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3



• Step 2: Calculate the regret

• Step 3 Choose the largest regret : 17• Step 4 Assign a job pair: Job 2 immediately follows job 5 (5‐2)

– L = 1+1;– We prohibit 2‐5

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3



• New Matrix

• Step 1: reduce the matrix– C(max) = 19 + 4+ 1 = 24

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• Step 2 Calculate the regret

• Step 3 Chose the largest regret: 9• Step 4 Assign a job pair: 3‐1

– Prohibit 1‐3

• Step 1 Reduce the matrix: not possible

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• Step 2 Calculate regret

• Step 3 Choose the largest regret: 3• Step 4 Assign job pair : 1‐4;

– Partial sequence: 5‐2, 3‐1‐4– Prohibit 4‐1 and 4‐3 (to keep 3‐1‐4‐3 from being chosen)

• Final Matrix– Choose 2‐3 and 4‐5‐> sequence 3‐1‐4‐5‐2– The total set‐up time is 24

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Parallel Machines

• Introduction– A job can be processed in any of the machines– The time to process a job is the same on any machine

– A job consists of a single operation– Decision

• Which machine processes the job?• In what order?

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Parallel Machines

• Introduction– List Schedule

• Sequence of all jobs• Assign the next job on the list to the machine with the smallest amount of work assigned

• Procedure– Step 0. Let Hi=0, i=1,2,...,m be the assigned workload on machine i, L=([1],[2],...,[n]) the ordered list sequence, Cj=0, j=1,2,...,n, and k=1

– Step 1. Let j*= Lk and Hi*=mini=1,m{Hi};Assign job j* to be processed on machine i*, Cj*=Hi*+pj*,Hi*=Hi*+pj*

– Step 2. Set k=k+1, if k>n, stop. Otherwise go to step 1.

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Parallel Machines

• Flow Time• Consider a facility with 3 identical machines and 15 jobs that need to be done as soon as possible;

• Processing times(after SPT):

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Parallel Machines

• Makespan• Use the longest processing time list (LPT)• Assign the next job on the list to the machine with the least total processing time assigned (heuristic)

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Truck

Flow Shops

• Introduction• Jobs processed sequentially on multiple machines• All jobs processed in the same order

• Makespan on a Two‐Machine Flow Shop– Johnson’s Algorithm

• Example

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MachinesJobs Total

Time1 2 3 4

1 5 4 3 2 14

2 2 5 2 6 15

Flow Shops


• Example (cont.)– Natural schedule for example has a total makespan of 22 h– Sequence 4‐2‐3‐1: makespan of 17

» Is this optimal?• The makespan must be as large as the sum of the processing times on either machine

– Makespan must account for unavoidable idle times» For each machine, add the minimum processing time of a job in the other machine

» Example: the bound becomes 17, so 4‐2‐3‐1 is optimal

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Flow Shops


– Step 1» Select the job with the lowest processing time on each machine

from the schedulable job list» If the list is empty, the procedure is finished» If the processing time of the job selected is from machine 1, go to

step 2, otherwise, go to step 3– Step 2

» Schedule the job in the earliest position of the sequence and remove it from the schedulable job list

» Return to step 1– Step 3

» Schedule the job in the latest position of the sequence and remove it from the schedulable job list

» Return to step 1• Johnson’s algorithm provides the optimal solution

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Flow Shops

• Heuristics– CDS Heuristic

• Convert a m‐machine problem into a two machine problem. How?

• Procedure– Start with: k=1 and l=m; then k=2 and l=m‐1; until: k=m‐1 and l=2

– m‐1 schedules are generated– Use the best of these m‐1 schedules

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Flow Shops


• Data

• Use the CDS to solve the problem• First use the Johnson’s algorithm for machines 1 and 4

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Flow Shops


• Second combine M1 with M2 to pseudomachine 1 and M3 with M4 to pseudomachine 2

• Finally combine M1+M2+M3 into pseudomachine 1 and M2+M3+M4 into pseudomachine 2

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Flow Shops


• Gantt Chart for CDS schedule

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Job Shops

• Introduction– Different routings for different jobs– Difficult to schedule

• precedence constraints• (n!)m possible schedules

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Job Shops

• Two Machine Job Shops– Jackson (1956) – adapted the Johnson’s algorithm to minimize makespan

• Job Sets with Machines A and B– Machine A: {AB}, {A}, {BA}– Machine B: {BA}, {B}, {AB}– Why in this order?– The order of jobs within the set is to be determined

• Procedure– Machine A: {AB} jobs ordered by Johnson’s Algorithm, then {A} in any Shortest Processing Time {BA} jobs in reverse Johnsons’ order

– Machine B: {BA} jobs reverse Johnsons’ order, then {B} in SPT {AB} jobs in ordered by Johnson’s Algorithm

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Job Shops


• Example

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Job Shops


• Example

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Stochastic Scheduling: Static Case

• Single machine case. Suppose that processing times are random variables. If the objective is to minimize average weighted flow time, jobs are sequenced according to expected weighted SPT. That is, if job times are t1, t2, . . ., and the respective weights are u1, u2, . . . then job iprecedes job i+1 if E(ti)/ui < E(ti+1)/ui+1.

Stochastic Scheduling: Static Case (continued)

• Multiple Machines. Requires the assumption that the distribution of job times is exponential, (memoryless property). Assume two parallel machines processing n jobs. Then the optimal sequence is to schedule the jobs according to LEPT (longest expected processing time first).

• Johnsons algorithm for scheduling n jobs on two machines (flow shop) in the deterministic case has a natural extension to the stochastic case as long as the job times are exponentially distributed.

Documents

5.1 - Scheduling - Técnico Lisboa - Autenticação€Machine Scheduling • Minimizing Set‐up Times – SST = Shortest Set‐up Time Heuristic – A metal products manufacturer