shanghai jiao tong university dept. of industrial engineering production planning and control...

Shanghai Jiao Tong UniversityShanghai Jiao Tong University Dept. of Industrial EngineeringDept. of Industrial Engineering

Production Planning and Control

Professor JIANG Zhibin

Department of Industrial Engineering & Management

Shanghai Jiao Tong University

Chapter 5 Operations SchedulingChapter 5 Operations Scheduling

Chapter 5 Chapter 5 Operations SchedulingOperations Scheduling

ContentsContentsIntroduction

Job Shop Scheduling Terminology

Sequencing Rules

Sequencing Theory for a Single Machine

Sequencing Theory for Multiple Machines

Assembly Line Balancing

Advanced Topics for Operations scheduling

Introduction-What is Operations Scheduling ?

Implement the production orders generated in MRP under given objectives ;

Allocate production resources (machine, workers et al.) to production orders (jobs or tasks and their due dates) in an optimized manners;

The results are time allocations of production resources to different jobs (job sequences on each production resources);

All the orders can be completed while all production resources are utilized with their loads

being balanced.

Forecast of future demand

Aggregate plan

Master production schedule (MPS)Schedule of production quantities by

products and time period

Material Requirement Planning (MRP)Generate Production orders and

purchase order

Operations SchedulingTo meet quantities and time

requirements for MRP

Introduction-Objectives of Job Shop Scheduling

Objectives of operations scheduling1) Meet due date;2) Minimize WIP inventory;3) Minimize the average flow time through the systems;4) Provide for high machine/worker (time) utilization (minimize idle

time);5) Reduce setup cost;6) Minimize production and worker costs

Discussion1) and 3) aim at providing a high level of costumer service; 2), 4), 5) and 6) are to provide a high level of plant efficiency; Impossible to optimize all above objectives simultaneously; Proper trade off between cost and quality is one of the most challenging strategic issues facing a firm today;

Introduction-Objectives of operations Scheduling

Discussion (Cont.)Some of these objectives conflicts, e.g.

Reduce WIP inventory Worker idle time may increase or machine utilization may decrease;Reasons: differences in the throughput rate from one part of the system to the another may force the faster operations to wait, if there is no buffer for WIP between 1 and 2 .

Fig 8-3 A Process Composed of Two Operations in Series

Introduction-Functions of Scheduling and Control is SF

The following functions must be performed in scheduling and control a shop floorAllocating orders, equipments, and personnel to work centers or other specified location-Short term capacity planning;Determining the sequence of orders (i. e. job priorities);Initializing performance of the scheduled work, commonly termed the dispatching of jobs;Shop-floor control, involving

Reviewing the status and controlling the progress of orders as they are being worked on;Expediting the late and critical orders;

Revising the schedules in light of changes in order status.

Introduction-Elements of the Shop Floor Scheduling Problems

The classic approaches to shop floor scheduling focuses on the following six elementsJob arrival patterns: static or dynamic

Static: jobs arrive in batch;Dynamic: jobs arrive over time interval according to some statistical distribution.

Numbers and variety of machines in the shop floorIf there is only one machine or if a group of machines can be treated as one machine, the scheduling problem is much more simplified;As number of variety of machines increase, the more complex the scheduling problems is likely to become.

Introduction-Elements of the Shop Floor Scheduling Problems

The classic approaches to job shop scheduling focuses on the following six elements (cont.)Ratio of workers to machines

Machine limited system: more workers than machine or equal number workers and machines;Labor-limited system: more machines than worker.

Flow pattern of jobs: flow shop or job shopFlow shop: all jobs follow the same paths from one machine to the next;Job shop: no similar pattern of movement of jobs from one machine to the next.

Introduction-Elements of the Job Shop Scheduling ProblemsIntroduction-Elements of the Job Shop Scheduling Problems

An assembly line is a classic example of flow shop Every cars go through all the stations one by one in the same sequences; Same tasks are performed on each car in each station; Its operations scheduling is simplified as assembly line balancing; An assembly balancing problem is to determine the number of stations and to

allocate tasks to each station.

Flow shop: Each of the n jobs

must be processed through the m machines in the same order.

Each job is processed exactly once on each machine.

Introduction-Job Shop

Turnning CenterDrilling Center

Milling CenterGrinding Center

A job shop is organized by machines which are grouped according to their functions.

Introduction-Job Shop

Work Center 1Work Center 2

Work Center 3Work Center 4

Not all jobs are assumed to require exactly the same number of operations, and some jobs may require multiple operations on a single machine (Reentrant system, Job B twice in work center 3 ).

Each job may have a different required sequencing of operations. No all-purpose solution algorithms for solving general job shop problems ; Operations scheduling of shop floor usually means job shop scheduling;

Job A

Job B

Introduction-Elements of the Shop Floor Scheduling ProblemsIntroduction-Elements of the Shop Floor Scheduling Problems

The classic approaches to shop floor scheduling focuses on the following six elementsJob sequencing

Sequencing or priority sequencing: the process of determining which job is started first on some machines or work center by priority rule;Priority rule: the rule used in obtaining a job sequencing;

Priority rule evaluation criteria To meet corresponding objectives of scheduling;Common standard measures

Meeting due date of customers or downstream operations;Minimizing flow time (the time a job spends in the shop flow);Minimizing WIP;Minimizing idle time of machines and workers (Maximizing utilization).


1. Parallel processing versus sequential processing Sequencing Processing: the m machines are distinguishable, and

different operations are performed by different machines. Parallel processing: The machines are identical, and any job can be

processed on any machine.

M1 M2

M3 M4

Job A

Job B

M1, M2, M3, and M4 are different; Job A has 2 operations which should

be processed on different Machines: M1and M2;

Job B has 3 operations which should be processed on different Machines: M3, M2 and M4;

M1 M2

M3 M4

Job A

Job B

M1, M2, M3, and M4 are identical;

Jobs A and B can be processed on any one of the 4 machines


2 Flow time The flow time of job i is the time that elapses from the initiation of

the that job on the first machine to the completion of job i. The mean flow time, which is a common measure of system

performance, is the arithmetic average of the flow times for all n jobs

Job 1 Job 2

Job 1

Job 3

Job 2 Job 3

Machines

M1

M2

TimeF1: FT of Job 1

F2: FT of Job 2

F3: FT of Job 3

Mean Flow Time=(F1+F2+F3)/3


2. Make-span The make-span is the time required to complete a group of jobs (all n

jobs). Minimizing the make-span is a common objective in multiple-machine

sequencing problems.

Job 1 Job 2

Job 1

Job 3

Job 2 Job 3

Machines

M1

M2

TimeF1: FT of Job 1

F2: FT of Job 2

F3: FT of Job 3

Make-span of the 3 jobs


3. Tardiness and lateness Tardiness is the positive difference between the completion time and the

due date of a job. Lateness refers to the difference between the job completion time and its

due date and differs from tardiness in that lateness can be either positive or negative.

If lateness is positive, it is tardiness; when it is negative, it is earliness

Due date of Job i

Completion time of Job i

Tardiness of Job iDue date

of Job i

Completion time of Job i

When the completion of Job is earlier than due date, the tardiness is 0

Lateness>0---Tardiness

Lateness<0---Earliness

Sequencing Rules

FCFS (first come-first served) Jobs are processed in the sequence in which they entered the shop; The simplest and nature way of sequencing as in queuing of a bank

SPT (shortest processing time) Jobs are sequenced in increasing order of their processing time; The job with shortest processing time is first, the one with the next

shortest processing time is second, and so on;

EDD (earliest due date)

Jobs are sequenced in increasing order of their due dates;

The job with earliest due date is first, the one with the next earliest due date is second, and so on;

Sequencing Rules

CR (Critical ratio) Critical ration is the remaining time until due date divided by processing

time; Scheduling the job with the largest CR next;

Processing time of Job i

Due date of Job iCurrent time Remaining time of Job i

CRi=Remaining time of Job i/Processing time of Job i

=(Due date of Job i-current time)/Processing time of Job i

CR provides the balance between SPT and EDD, such that the task with shorter remaining time and longer processing time takes higher priority;CR will become smaller as the current time approaches due date, and more priority will given to one with longer processing time..For a job, if the numerator of its CR is negative ( the job has been already later), it is naturally scheduled next;If more jobs are later, higher priority is given to one that has shorter processing time (SPT).

Sequencing Rules

Example 5.1 A machine center in a job shop for a local fabrication company has five unprocessed jobs remaining at a particular point in time. The jobs are labeled 1, 2, 3, 4, and 5 in the order that they entered the shop. The respective processing times and due date are given in the table below.Sequence the 5 jobs by above 4 rules and compare results based on mean flow time, average tardiness, and number of tardy jobs

Job number Processing Time Due Date

1

2

3

4

5

11

29

31

1

2

61

45

31

33

32

Sequencing Rules——FCFS


1

2

3

4

5

11

29

31

1

2

61

45

31

33

32

Job Completion Time Due Date Tardiness

1 11 61 0

2 40 45 0

3 71 31 40

4 72 33 39

5 74 32 42

Totals 268 121

Mean Flow time=268/5=53.6Average tardiness=121/5=24.2No. of tardy jobs=3.

Sequencing Rules——SPT


1

2

3

4

5

11

29

31

1

2

61

45

31

33

32

Job Processing Time Completion Time Due Date Tardiness

4 1 1 33 0

5 2 3 32 0

1 11 14 61 0

2 29 43 45 0

3 31 74 31 43

Totals 135 43


Sequencing Rules——EDD


1

2

3

4

5

11

29

31

1

2

61

45

31

33

32

Job Processing Time Completion Time Due Date Tardiness

3 31 31 31 0

5 2 33 32 1

4 1 34 33 1

2 29 63 45 18

1 11 74 61 13

Totals 235 33


Sequencing Rules——CR

Current time: t=0

Job number Processing Time Due Date Critical Ratio

12345

11293112

6145313332

61/11(5.545)45/29(1.552)31/31(1.000)33/1 (33.00)32/2 (16.00)

Current time: t=31

Job number Processing Time Due Date-Current Time Critical Ratio

1245

112912

30 14 2 1

30/11(2.727)14/29(0.483) 2/1 (2.000)1/2 (0.500)

Current time should be reset after scheduling one job

Sequencing Rules——CR

Current time=60

Job number Processing Time Due Date-Current Time

Critical Ratio

1

4

5

11

1

2

1

-27

-28

1/11(0.0909)

-27/1<0

-28/2<0

Job number Processing Time Completion Time Tardiness

3

2

4

5

1

31

29

1

2

11

31

60

61

63

74

0

15

28

31

13Totals 289 87


Both Jobs 4 and 5 are later, however Job 4 has shorter processing time and thus is scheduled first;

Sequencing Rules——Summary

Rule Mean Flow Time Average Tardiness

Number of

Tardy Jobs

FCFS

SPT

EDD

CR

53.6

27.0

47.0

57.8

24.2

8.6

6.6

17.4

3

1

4

4

Discussions SPT results in smallest mean flow time; EDD yields the minimum maximums tardiness (42, 43, 18, and 31 for the

4 different rules); Always true? Yes!

Sequencing Theory for A Single Machines

Assuming that n jobs are to be processed through one machine. For each job i, define the following quantities: ti=Processing time for job i, constant for job i;

di=Due date for job i, constant for job i;

Wi=Waiting time for job i, the amount of time that the job must wait before its processing can begin.

When all the jobs are processed continuously, Wi is the sum of the processing times for all of the preceding jobs;

t1 t2 t3 t4

W4=t1+t2+t3

Fi=Flow time for job i, the waiting time plus the processing time: Fi= Wi+ ti;

Li=Lateness of job i , Li= Fi- di, either positive or negative;

Ti=Tardiness of job i, the positive part of Li, Ti=max[Li,0] ;

Ei=Earliness of job i, the negative part of Li, Ei =max[- Li,0]

F4=W4+t4


max 1 2max{ , ,..., }

nT T T T Maximum Tardiness

n

iiFF n 1

' 1 Mean Flow Time

For only a single machine, every schedule can be represented by a permutation (ordering) of the integers 1, 2, 3, …, n.

There are totally n! (the factorial of n) different permutations.

Suppose that 4 jobs J1, J2, J3, J4 need to be scheduled

For example a schedule is J3-J2-J1-J4

Cindered as a permutation of integers 1, 2, 3, 4: 3, 2, 1, 4.

A permutation of integers 1, 2, , n is expressed by [1], [2], , [n], which represents a schedule;

In case of a schedule 3, 2, 1, 4, [1]=3, [2]=2, [3]=1, and [4]=4;


1. Shortest-Processing-Time Scheduling Theorem 5.1 The scheduling rule that minimizes the mean

flow time F’ is SPT

The mean flow time of all jobs on the schedule is given by

n

i

n

k

k

iik tFF nn 1 1 1][][

' 11

k

iik tF

1][][

Suppose a schedule is [1], [2], [k], [k+1], [n], the flow time of the job that is scheduled in position k is given by, say job in position 3:

F[3]=t[1]+t[2]=t2+t1

t[1] (t2) t[2] (t1) t[3] (t4) t[4] (t3)


1. Shortest-Processing-Time SchedulingTheorem 5.1 The scheduling rule that minimizes the mean flow time F’ is

SPT

The mean flow time is given by

n

i

n

k

k

iik tFF nn 1 1 1][][

' 11

The double summation term may be written in a different form. Expanding the double summation, we obtaink=1:t[1] ; k=2:t[1]+ t[2]; …;k=n:t[1]+ t[2 +…t[n]

SPT sequencing rule: the job with shortest processing time t is set first

By summing down the column rather than across the row, we may rewrite F’ in the form

nt[1]+(n-1)t[2]+…+t[n]

ttt n][]2[]1[...

Clearly, it is minimized by setting


1. Shortest-Processing-Time Scheduling (Cont.)

Corollary 5.1 The following measures are equivalent: Mean flow time Mean waiting time Mean lateness

SPT minimizes mean flow time, mean waiting time, and mean lateness for single machine sequencing.

2. Earliest-Due-Date Scheduling: If the objective is to minimize the maximum lateness, then the jobs should be sequenced according to their due dates. That is, d[1] d[2]… d[n].


3. Minimizing the number of Tardy Jobs: An algorithm from Moore(1968) that minimizes the number of tardy jobs for the single machine problem.

Step1. Sequence the jobs according to the earliest due date to obtain the

initial solution. That is d[1] d[2],…, d[n]; Step2. Find the first tardy job in the current sequence, say job [i]. If none

exists go to step 4. Step3. Consider jobs [1], [2], …, [i]. Reject the job with the largest

processing time. Return to step2. (Why ?)

Reason: It has the largest effect on the tardiness of the Job[i].

Step4. Form an optimal sequence by taking the current sequence and appending to it the rejected jobs. (Can be appended in any order?) Yes, because we only consider the number of tardiness jobs rather than

tardiness.


Job 1 2 3 4 5 6

Due date 15 6 9 23 20 30

Processing time 10 3 4 8 10 6

Example 8.3

Solution

Job 2 3 1 5 4 6

Due date 6 9 15 20 23 30


Completion time

3 7 17 27 35 41

Longest processing time



Job 2 3 5 4 6

Due date 6 9 20 23 30

Processing time 3 4 10 8 6

Completion time 3 7 17 25 31

Example 8.3 :Solution (Cont.) Longest processing time

Job 2 3 4 6

Due date 6 9 23 30

Processing time 3 4 8 6

Completion time 3 7 15 21

The optimal sequence: 2, 3, 4, 6, 5, 1 or 2, 3, 4, 6, 1, 5. In each case the number of tardy jobs is exactly 2.



Precedence constraints: Lawler’s Algorithm

gi is any non-decreasing function of the flow time Fi

iiiii LdFFg )(

)0,max()( iiii dFFg

Minimizing maximum lateness

Minimizing maximum tardiness

The AlgorithmFirst schedules the job to be completed last, then the job to be completed next to last, and so on. At each stage one determines the set of jobs not required to precede any other. Call this set V. among the set V, choose the job k that satisfies

n

i i

ivi

k

t

gg

1

))((min)(

The processing time of the current sequence

)(maxmin1

iini

Fg

Objective Function


The Algorithm (Cont.)

Consider the remaining jobs and again determine the set of jobs that are not required to precede any other remaining job.

The value of τ is reduced by tk and the job scheduled next to last is now determined.

The process is continued until all jobs are scheduled.

Note: As jobs are scheduled, some of the precedence constraints may be relaxed, so the set V is likely to change at each iteration.



Example 8.4

Job 1 2 3 4 5 6

Processing time

2 3 4 3 2 1

Due date 3 6 9 7 11 7


Example 8.4

Job 1 2 3 4 5 6

Processing time

2 3 4 3 2 1

Due date 3 6 9 7 11 7

Step1: find the job scheduled last(sixth)Not predecessor

τ =2+3+4+3+2+1=15 3 5 6

Tardiness

15-9=6 15-11=4 15-7=8

Job 1 2 3 4 6

Processing time

2 3 4 3 1

Due date 3 6 9 7 7

Step2: find the job scheduled fifth

τ =15-2=133 6

Tardiness

13-9=4 13-7=6

Not predecessor


Example 8.4

Job 1 2 4 6

Processing time

2 3 3 1

Due date 3 6 7 7

Step3: find the job scheduled fourthNot predecessor

τ =13-4=9 2 6

Tardiness

9-6=3 9-7=2

Job 1 2 4

Processing time

2 3 3

Due date 3 6 7

Step4: find the job scheduled third

τ =9-1=82 4

Tardiness

8-6=2 8-7=1

Not predecessor

Because job3 is no longer on the list, job 2 now because a candidate.

Because job6 has been scheduled, so job 4 now because a candidate along with job 2.


Example 8.4

Job 1 2

Processing time

2 3

Due date 3 6

Step5: find the job scheduled secondNot predecessor

The optimal sequence: 1-2-4-6-3-5

Job Processing time

Flow time

Due date Tardiness

1

2

4

6

3

5

2

3

3

1

4

2

2

5

8

9

13

15

3

6

7

7

9

11

0

0

1

2

4

4Maximum tardiness


Assume that n jobs are to be processed through m machines. The number of possible schedules is staggering, even for moderate values of both n and m.For each machine, there is n! different ordering of the jobs; if the jobs may be processed on the machines in any order, there are totally (n!)m possible schedules. (n=5, m=5, 25 billion possible schedules)Even with the availability of inexpensive computing today, enumerating all feasible schedules for even moderate-sized problems is impossible or, at best, impractical.



Machine 1 Machine 2

Job I 4 1

Job J 1 4

Gantt chart Suppose that two jobs, I and J, are to be scheduled on two machines, 1 and 2, the processing times are

Assume that both jobs must be processed first on machine 1 and then on machine 2. There are four possible schedules.



Schedule Total flow time Mean flow time Mean idle time

1 9 (5+9)/2=7 (4+4)/2=4

2 6 5.5 1

3 10 8 5

4 10 9.5 5



1. Scheduling n Jobs on Two Machines Theorem 8.2 The optimal solution for scheduling n jobs on two machines

is always a permutation schedule.

A very efficient algorithm for solving the two-machine problem was discovered by Johnson(1954). Denote the machines by A and B The jobs must be processed first on machine A and then on machine B. Define

Ai=Processing time of job i on machine A

Bi=Processing time of job i on machine B

Rule: Job i precedes job i+1 if min(Ai, Bi-1)<min(Ai+1,Bi)

List the values of Ai and Bi in two columns. Find the smallest remaining element in the two columns. If it

appears in column A, then schedule that job next. If it appears in column B, then schedule that job last.

Cross off the jobs as they are scheduled. Stop when all jobs have been scheduled.


Example 8.5Job Machine A Machine B

1 5 2

2 1 6

3 9 7

4 3 8

5 10 4

Optimal sequence : 2 4 3 5 1



2. Extension to Three Machines The three-machine problem can be reduced to a two-machine problem if

the following condition is satisfied min Aimax Bi or min Cimax Bi

It is only necessary that either one of these conditions be satisfied. If that is the case, then the problem is reduced to a two-machine problem

Define Ai’=Ai+Bi, Bi’=Bi+Ci

Solve the problem using the rules described above for two-machines, treating Ai’ and Bi’ as the processing times.

The resulting permutation schedule will be optimal for the three-machine problem.

If the condition are not satisfied, this method will usually give reasonable, but possibly sub-optimal results.


3. The Two-Job Flow Shop Problem: assume that two jobs are to be processed through m machines. Each job must be processed by the machines in a particular order, but the sequences for the two jobs need not be the same.

Draw a Cartesian coordinate system with the processing times corresponding to the first job on the horizontal axis and the processing times corresponding to the second job on the vertical axis.

Block out areas corresponding to each machine at the intersection of the intervals marked for that machine on the two axes.

Determine a path from the origin to the end of the final block that does not intersect any of the blocks and that minimizes the vertical movement. Movement is allowed only in three directions: horizontal, vertical, and 45-degree diagonal. The path with minimum vertical distance corresponds to the optimal solution.


Example 8.7

A regional manufacturing firm produces a variety of household products. One is a wooden desk lamp. Prior to packing, the lamps must be sanded, lacquered, and polished. Each operation requires a different machine. There are currently shipments of two models awaiting processing. The times required for the three operations for each of the two shipments are

Job 1 Job2

Operation Time Operation Time

Sanding(A) 3 A 2

Lacquering(B) 4 B 5

Polishing( C ) 5 C 3



Minimizing the flow time is the same as maximizing the time that both jobs are being processed.that is equivalent to finding the path from the origin to the end of block C that maximizes the diagonal movement and therefore minimizes either the horizontal or the vertical movement.

or 10+(3+2)=15

or 10+6=16


The problem of balancing an assembly line is a classic industrial engineering problem. The problem is characterized by a set of n distinct tasks that must be

completed on each item The time required to complete task i is a known constant ti.

The goal is to organize the tasks into groups, with each group of tasks being performed at a single workstation

In most cases, the amount of time allotted to each workstation is determined in advance, based on the desired rate of production of the assembly line.


Assembly line balancing is traditionally thought of as a facilities design and layout problem.

There are a variety of factors that contribute to the difficulty of the problem. Precedence constrains: some tasks may have to be completed in a particular

sequence. Zoning restriction: Some tasks cannot be performed at the same

workstation. Let t1, t2, …, tn be the time required to complete the respective tasks. The total work content (time) associated with the production of an item, say T, is

given by

n

iitT

1

For a cycle time of C, the minimum number of workstations possible is [T/C], where the brackets indicate that the value of T/C is to be rounded to the next larger integer.

Ranked positional weight technique: the method places a weight on each task based on the total time required by all of the succeeding tasks. Tasks are assigned sequentially to stations based on these weights.


Example 8.11 The Final assembly of Noname personal computers, a generic mail-order PC clone, requires a total of 12 tasks. The assembly is done at the Lubbock, Texas, plant using various components imported from the Far East. The network representation of this particular problem is given in the following figure.



Task Immediate Predecessors Time1 _ 12

2 1 6

3 2 6

4 2 2

5 2 2

6 2 12

7 3, 4 7

8 7 5

9 5 1

10 9, 6 4

11 8, 10 6

12 11 7

PreconditionThe job times and precedence relationships for this problem are summarized in the table below.


ti=70, and the production rate is a unit/15 minutes;The minimum number of workstations = [70/15]=5


Task Positional Weight1 70

2 58

3 31

4 27

5 20

6 29

7 25

8 18

9 18

10 17

11 13

12 7

The solution precedence requires determining the positional weight of each task. The positional weight of task i is defined as the time required to perform task i plus the times required to perform all tasks having task i as a predecessor.


t3+t7+t8+t11+t12=31

The ranking

1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12


Profile 1 C=15


Station 1 2 3 4 5 6

Tasks 1 2, 3, 4 5, 6, 9 7, 8 10, 11 12


Idle time 3 1 0 3 5 8

The ranking

1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12

Task Immediate Predecessors

Time

1 _ 12

2 1 6

3 2 6

4 2 2

5 2 2

6 2 12

7 3, 4 7

8 7 5

9 5 1

10 9, 6 4

11 8, 10 6

12 11 7


T2=6

Profile 1 C=15


Station 1 2 3 4 5 6

Tasks 1 2,3,4 5,6,9 7,8 10,11 12


Idle time 3 1 0 3 5 8

Cycle Time=15

T1=12

T2=6 T3=6 T4=2

T5=2 T6=12 T9=1

T5=2

T8=5T7=7 T10=4

T10=4 T11=6 T12=7

T12=7

15 Evaluate the balancing results by the efficiency ti/NC;

The efficiencies for Profiles 1 ~ 3 are 77.7%, 87.5%, and 89.7%. Thus the profile 3 is the best one.


Profile 1: Increasing cycle time from 15 to 16

Station 1 2 3 4 5

Tasks 1 2,3,4,5 6,9 7,8,10 11,12

Idle time 4 0 3 0 3

Increasing the cycle time from 15 to 16, the total idle time has been cut down from 20 min/units to 10; resulting in a substantial improvement in balancing rate. However, the production rate has to be reduced from one unit/15 minutes to one unit/16minute;


Alternative 1: Change cycle time to ensure 5 station balance


Profile 2 C=13

Station 1 2 3 4 5 6

Tasks 1 2,3 6 4,5,7,9 8,10 11,12

Idle time 1 1 1 1 4 0


Alternative 2: Staying with 6 stations, see if a six-station balance could be obtained by cycle time less that 15 minutes

13 minutes appear to be the minimum cycle time with six station balance.

Increasing the number of stations from 5 to 6 results in a great improvement in production rate;

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