Production Planning & Control

Operations Scheduling

Operations Scheduling

Contents
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 ?

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

Introduction-Objectives of Job Shop Scheduling

Objectives of operations scheduling

1)
2)
3)
4)

Meet due date;

Minimize WIP inventory;
Minimize the average flow time through the systems;
Provide for high machine/worker (time) utilization (minimize idle
time);
5) Reduce setup cost;
6) Minimize production and worker costs
Discussion
1)

and 3) aim at providing a high level of costumer service;

2), 4), 5) and 6) are to provide a high level of workshop 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 another may force the faster operations to
wait.
As an example, if there is no buffer for WIP
between 1 and 2, what happens?

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

Introduction-Functions of Scheduling and Control

The following functions must be performed in scheduling

and controlling a shop floor:
Allocating 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 elements:
Job 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 floor
If 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

(Continued)
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 shop
Flow 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 Shop Floor Scheduling Problems

(Continued)
Job 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 for 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).

Introduction-Elements of the Job Shop Scheduling Problems

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.

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.

Introduction-Job Shop

A job shop is organized by machines which are grouped

according to their functions.

Introduction-Job Shop

Job A
Job B

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 Shop Scheduling Terminology

1. Parallel processing versus sequential processing

Job A
Job B

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

Job A

M1

M2

M3

M4

Job B

M3

M4

, 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;
1

M1, M2, M3, and M4 are

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

Job Shop Scheduling Terminology

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

Machines
M1
M2

Job 1

Job 2

Job 3

Job 1

Job 2

F1: FT of Job 1
F2: FT of Job 2
F3: FT of Job 3

Job 3
Time

Job Shop Scheduling Terminology

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.

Machines
M1
M2

Job 1

Job 2

Job 3

Job 1

Job 2

F1: FT of Job 1
F2: FT of Job 2
F3: FT of Job 3
Make-span of the 3 jobs

Job 3
Time

Job Shop Scheduling Terminology

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 i

Due date
of Job i

Completion
time of Job i

Lateness>0--Tardiness
Lateness<0--Earliness

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

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 ratio is the remaining time until due date divided by processing
time;
Scheduling the job with the smallest CR next;
Current time

Remaining time of Job i

Due date of Job i

Processing 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 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 RulesFCFS

Job

Mean Flow time=268/5=53.6

Average tardiness=121/5=24.2
No. of tardy jobs=3.

Job number

Processing Time

Due Date

1
2
3
4
5

11
29
31
1
2

61
45
31
33
32

Completion Time

1
2
3
4
5

11
40
71
72
74

Totals

268

Due Date

Tardiness

61
45
31
33
32

0
0
40
39
42
121

Sequencing RulesSPT

Mean Flow time=135/5=27.0

Average tardiness=43/5=8.6
No. of tardy jobs=1.

Job number

Processing Time

Due Date

1
2
3
4
5

11
29
31
1
2

61
45
31
33
32

Job
4
5
1
2
3
Totals

Processing Time
1
2
11
29
31

Completion Time Due Date

1
3
14
43
74
135

33
32
61
45
31

Tardiness
0
0
0
0
43
43

Sequencing RulesEDD

Mean Flow time=235/5=47.0

Average tardiness=33/5=6.6
No. of tardy jobs=4.

Job number

Processing Time

Due Date

1
2
3
4
5

11
29
31
1
2

61
45
31
33
32

Job Processing Time

3
5
4
2
1
Totals

31
2
1
29
11

Completion Time
31
33
34
63
74
235

Due Date
31
32
33
45
61

Tardiness
0
1
1
18
13
33

Sequencing RulesCR
Current time: t=0
Job number
Processing Time
1
11
2
29
3
31
4
1
5
2

Due Date
61
45
31
33
32

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

Current time should be reset after scheduling one job

Current time: t=31
Job number Processing Time Due Date-Current Time
1
11
30
2
29
14
4
1
2
5
2
1

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

Sequencing RulesCR
Current time=60
Job number
Processing Time
1
4
5

11
1
2

Mean Flow time=289/5=57.8

Average tardiness=87/5=17.4
No. of tardy jobs=4.
Due DateCurrent Time
1
-27
-28

Critical Ratio
1/11(0.0909)
-27/1<0
-28/2<0

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.
Job number
Processing Time
Completion Time Tardiness
3
31
31
0
2
29
60
15
4
1
61
28
5
2
63
31
1
11
74
13
Totals

289

87

Sequencing RulesSummary
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 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, W is the sum of the
i
processing times for all of the preceding jobs;
t1
t2
t3
t4

W4=t1+t2+t3

F4=W4+t4
F =Flow time for job i, the waiting time plus the processing time: F = W + t ;
i
i
i
i

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] ;

Sequencing Theory for A Single Machines

Maximum Tardiness
Mean Flow Time

T
F

max
'

max{

T ,T
1

,...,T n}

F
i 1

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.

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.
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;

Sequencing Theory for A Single Machines

1.Shortest-Processing-Time Scheduling
Theorem 8.1 The scheduling rule that minimizes the mean flow
time F is SPT.
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:
t[1] (t2)

t[2] (t1)

t[3] (t4)

[k ]

t
i 1

t[4] (t3)

F[2]=t[1]+t[2]=t2+t1
The mean flow time of all jobs on
the schedule is given by

'

F
i 1

[k ]

t
k 1 i 1

[i ]

[i ]

Sequencing Theory for A Single Machines

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

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]

By summing down the column rather

than across the row, we may rewrite F
in the form

i 1

[k ]

k 1 i 1

[i ]

nt[1]+(n-1)t[2]++t[n]
Clearly, it is minimized by setting

[1]

t [2] ... t [ n ]

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

Sequencing Theory for A Single Machines

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 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.

Sequencing Theory for A Single Machine

Example 8.3
Job

Due date

15

23

20

30

Processing time

10

10

Longest processing time

Solution
Job

Due date

15

20

23

30

Processing time

10

10

Completion
time

17

27

35

41

Sequencing Theory for A Single Machine

Longest processing time

Example 8.3 :Solution (Cont.)

Job

Due date

20

23

30

Processing time

10

Completion time

17

25

31

Job

Due date

23

30

Processing time

Completion time

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.

Sequencing Theory for A Single Machines

Precedence constraints: Lawlers Algorithm
g i ( Fi ) Fi d i Li

Objective
Function

min max g ( F )
1i n

Minimizing maximum
lateness

g i ( Fi ) max( Fi d i ,0) Minimizing maximum

tardiness

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

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
g k ( ) min ( g i ( )) e.g.: the job among V that has smallest tardiness,
iv
if arranged on position [n].

i 1 t i
n

The processing time of the current sequence

Sequencing Theory for A Single Machines

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 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.

Sequencing Theory for A Single Machines

Example 8.4

Job

Processing time

Due date

11

Sequencing Theory for A Single Machines

Example 8.4

Not predecessor

Step1: find the job scheduled last(sixth)

Job

Processing time

Due date

11

=2+3+4+3+2+1=15

Tardiness

15-9=6

15-11=4

15-7=8

Step2: find the job scheduled fifth

Job

Processing time

Due date

=15-2=13

Tardines

Not predecessor

13-9=4

13-7=6

Sequencing Theory for A Single Machines

Example 8.4

Not predecessor
Step3: find the job scheduled fourth
Job

Processing time

Due date

=13-4=9

Tardiness

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

9-6=3

9-7=2

Step4: find the job scheduled third

Job

Processing time

Due date

=9-1=8

Tardiness

Not predecessor
Because job6 has been
scheduled, Job 4 now
because a candidate along
with Job 2.
2

8-6=2

8-7=1

Sequencing Theory for A Single Machines

Example 8.4

Not predecessor
Step5: find the job scheduled second

Job
1
2
4
6
3
5

Job

Processing
time

Due date

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

Processing
time

Flow
time

Due date

Tardiness

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

Sequencing Theory for Multiple Machines

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 Machines

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

Machine 1

Machine 2

Job I

Job J

Assume that both jobs must be processed first on machine 1

and then on machine 2. There are four possible schedules.

Sequencing Theory for Multiple Machines

Schedule

Total flow time

Mean flow time

Mean idle time

(5+9)/2=7

(4+4)/2=4

5.5

10

10

9.5

Sequencing Theory for Multiple Machines

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

Sequencing Theory for Multiple Machines

Job

Example 8.5

Machine A

1
2
3
4
5

Optimal sequence : 2

Machine B

5
1
9
3
10

2
6
7
8
4

Sequencing Theory for Multiple Machines

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.

Sequencing Theory for Multiple Machines

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.

Sequencing Theory for Multiple Machines

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)

Lacquering (B)

Polishing( C )

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+6=16

or 10+(3+2)=15

Assembly 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

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
T

t
i 1

Assembly Line Balancing

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.

Assembly Line Balancing

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.

Assembly Line Balancing

ti=70, and the production rate is a unit /15 minutes;
Precondition
The minimum number of workstations = [70/15]=5
The job times and precedence relationships for this problem are summarized
in the table below.
Task

Immediate Predecessors

Time

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

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

12
6
6
2
2
12
7
5
1
4
6
7

Assembly Line Balancing

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

Task

Positional Weight

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

70
58
31
27
20
29
25
18
18
17
13
7

Assembly Line Balancing

Profile 1 C=15
Station

Tasks

2, 3, 4

5, 6, 9

7, 8

10, 11

12

Processing time

12

14

15

12

10

Idle time

Task

Immediate
Predecessors

Time

12

2
3
4
5
6
7
8
9
10
11
12

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

6
6
2
2
12
7
5
1
4
6
7

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

Assembly Line Balancing

Profile 1 C=15
Station

Tasks

2,3,4

5,6,9

7,8

10,11

12

Processing time

12

14

15

10

Idle time

15
Cycle Time=15
T1=12
T2=6
T5=2
T7=7
T10=4
T12=7

T3=6
T6=12
T8=5
T11=6

The ranking

1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12
T2=6
T4=2
T5=2
T9=1
T10=4
T12=7

Evaluate the
balancing results by
the efficiency
ti/NC;
The efficiencies for
Profiles 1 is 77.7%.

Assembly Line Balancing

Alternative 1: Change cycle time to ensure 5 station balance
Profile 2: Increasing cycle time from 15 to 16
Station

Tasks

2,3,4,5

6,9

7,8,10

11,12

Idle time

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;

Assembly Line Balancing

Alternative 2: Staying with 6 stations, see if a six-station
balance could be obtained by cycle time less that 15 minutes
The efficiencies for profile 1~ 3 are 77.7%,
87.5%, and 89.7%. Thus the profile 3 is the
best one.

Profile 2 C=13
Station

Tasks

2,3

4,5,7,9

8,10

11,12

Idle time

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 End !