Scheduling

Production Planning & ControlProduction Planning & Control
Operations SchedulingOperations Scheduling

Operations SchedulingOperations Scheduling
ContentsContents
Introduction
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 ?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
product and time period
Material Requirement Planning (MRP)
Generate production orders and
purchase order
Operations Scheduling
To meet quantities and time
requirements for MRP

SchedulingScheduling
 Scheduling: The allocation of resources over time to accomplish
specific tasks.
 Demand scheduling: A type of scheduling whereby customers
are assigned to a definite time for order fulfillment.
 Workforce scheduling: A type of scheduling that determines
when employees work.
 Operations scheduling: A type of scheduling in which jobs are
assigned to workstations or employees are assigned to jobs for
specified time periods.

IntroductionIntroduction
Operations scheduling is critical to the success of an
organization; however, it can be a very complicated task.
Effective schedules are needed to meet promised customer
delivery dates or inventory targets.
It covers the following areas in particular:
- assign job to a particular work center/ machine
- time of assignment of job and completion
- allocation of resources like manpower and materials
- time sequence of operations
- feedback and control function to take care of deviations

Objectives of OperationsObjectives of Operations
 Making efficient use of the labour.
 Making best possible use of the equipments that
are available for the use.
 Increasing the profit.
 Increasing the output.
 Improving the service level.

Objectives of OperationsObjectives of Operations
 Maximizing the delivery performance i.e. meeting
the delivery dates.
 Minimizing the inventory.
 Reducing the manufacturing time.
 Minimizing the production costs.
 Minimizing the worker costs.

Functions of OperationsFunctions of Operations
 Allocation of the resources.
 Shop floor control.
 Making maximum use of the plant at minimum
possible cost.
 Ensure that the needs of the manpower are
optimum.
 Determination of the sequence of the jobs.

Functions of OperationsFunctions of Operations
 Specifying the start and the end time for each job
(actively scheduled).
 Getting quick feedback from the shops regarding
the delays and the various interruptions.
 Possess up – to – date information for the
availability of the materials, expected delivery
dates etc.
 Possess up – to – date data on the machine
regarding its breakdown, servicing etc.

Types of SchedulingTypes of Scheduling
Types of Operations Scheduling are as follows:
1. Forward operations scheduling –
 Classified on the basis of the time.
 All the activities are scheduled from the date of the planned
order release.
 First task of the job is scheduled.
 Its subsequent task is scheduled on the scheduled
completion of the first task.
 Like this, accordingly all the tasks of the job are scheduled.

Types of SchedulingTypes of Scheduling
2. Backward operations scheduling –
 Also classified on the basis of the time.
 Activities are scheduled from the date or the planned receipt
date.
 The last activity is scheduled first.
 Time of the start of the last task is considered as the time for
the start of the previous activity.

Shop Floor Control (SFC)Shop Floor Control (SFC)
Schedule and monitor day-to-day job shop production • Also
called production control and production activity control
(PAC) • Performed by production control department
• Loading - check availability of material, machines, and labor
• Sequencing - release work orders to shop and issue dispatch
lists for individual machines
• Monitoring - maintain progress reports on each job until it is
complete

Sequencing JobsSequencing Jobs
 Operations schedules are short-term
plans designed to implement the sales and
operations plan
 An operation with divergent flows is often
called a job shop
 Low-to medium-volume production
 Utilizes job or batch processes
 The front office would be the equivalent for a
service provider
 Difficult to schedule because of the variability in
job routings and the continual introduction of
new jobs to be processed

Sequencing JobsSequencing Jobs
• An operation with line flow is often
called a flow shop
– Medium- to high-volume production
– Utilizes line or continuous flow processes
– The back office would be the equivalent for a
service provider
– Tasks are easier to schedule because the jobs
have a common flow pattern through the
system

Introduction-Job ShopIntroduction-Job Shop
A job shop is organized by machines which are grouped
according to their functions.

Introduction-Job ShopIntroduction-Job Shop
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

Job Shop Scheduling TerminologyJob Shop Scheduling Terminology
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
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

3. 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

4. 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

SequencingSequencing RulesRules
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 RulesSequencing Rules
CR (Critical ratio)
 Critical ratio is the remaining time until due date divided by processing
time;
 Scheduling the job with the smallest 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 than one jobs are later, higher priority is given to one that has shorter
processing time (SPT).

Sequencing RulesSequencing 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 dates 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 RulesSequencing Rules————FCFSFCFS
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=53.6
Average tardiness=121/5=24.2
No. of tardy jobs=3.

Sequencing RulesSequencing Rules————SPTSPT
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 RulesSequencing Rules————EDDEDD
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 RulesSequencing Rules————CRCR
Current time: t=0
Job number Processing Time Due Date Critical Ratio
1
2
3
4
5
11
29
31
1
2
61
45
31
33
32
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
1
2
4
5
11
29
1
2
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 RulesSequencing Rules————CRCR
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
13
Totals 289 87
Both Jobs 4 and 5 are later, however Job 4 has shorter processing time
and thus is scheduled first; Finally, job 1 is scheduled last.

Sequencing RulesSequencing Rules————SummarySummary
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 maximum tardiness (42, 43, 18, and 31 for the 4
different rules);
 Always true? Yes!

Sequencing Theory for A Single MachinesSequencing 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] ;

F4=W4+t4

max 1 2
max{ , ,..., }nT T T T= Maximum Tardiness
∑=
=
n
i
iFF 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
 Considered 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;
 [i] denotes the integer that put in the ith
place in the permutation;
 In case of a schedule 3, 2, 1, 4, [1]=3, [2]=2, [3]=1, and [4]=4;

1.Shortest-Processing-Time Scheduling
Theorem 8.1 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
i
ik tFF nn 1 1 1
][][
' 11
∑=
=
k
i
ik 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[2]=t[1]+t[2]=t2+t1
t[1] (t2) t[2] (t1) t[3] (t4) t[4] (t3)

1.Shortest-Processing-Time Scheduling
Theorem 8.1 The scheduling rule that minimizes the mean flow
time F’ is SPT
The mean flow time is given by ∑ ∑∑= = =
==
n
i
n
k
k
i
ik tFF nn 1 1 1
][][
' 11
 The double summation term may
be written in a different form.
Expanding the double summation,
we obtain
k=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]
[1] [2] [ ]
... nt t t≤ ≤ ≤
Clearly, it is minimized by setting

1. Shortest-Processing-Time Scheduling (Cont.)
Corollary 8.1 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].

Sequencing Theory for A Single MachineSequencing Theory for A Single Machine
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 Algorithm
First 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
i
vi
k
t
gg
1
))((min)(
τ
ττ
The processing time of the current sequence
)(maxmin 1
ii
ni
Fg
≤≤
Objective
Function
e.g.: the job among V that has smallest tardiness,
if arranged on position [n].

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 then 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
Due date 3 6 9 7 11 7

Example 8.4
Job 1 2 3 4 5 6
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=13
3 6
Tardines 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 fourth
Not 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=8
2 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, 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 second
Not 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
4 Maximum tardiness

Assume that n jobs are to be processed through m
machines. The number of possible schedules is
astonishing, 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.
Sequencing Theory for Multiple MachinesSequencing Theory for Multiple Machines

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

Example 8.5
Job 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 Ai≥max Bi or min Ci≥max 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

Assembly Line BalancingAssembly Line Balancing
 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
i
itT
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:
Places a weight on each task based on the total
time required by this task and all of the
succeeding tasks;
Tasks are assigned sequentially to stations
based on these weights-the bigger the weight is,
the higher the priority is.

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 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
Precondition
The 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 Weight
1 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 is 77.7%.
The ranking
1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11,
12

Profile 2: 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;
The efficiencies for profile 1~ 3 are 77.7%,
87.5%, and 89.7%. Thus the profile 3 is the
best one.

Scheduling

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