Transcript Chapter 5 Operations Scheduling Production Planning and Control Professor JIANG Zhibin

Production Planning and Control
Chapter 5 Operations Scheduling
Professor JIANG Zhibin
Department of Industrial Engineering &
Management
Shanghai Jiao Tong University
Shanghai Jiao Tong University
Dept. of Industrial Engineering
Chapter 5 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
products 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 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 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
 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 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 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
Turnning Center
Grinding Center
Drilling Center
Milling Center
A job shop is organized by machines which are grouped
according to their functions.
Introduction-Job Shop
Work Center 1
Work Center 2
Job A
Job B
Work Center 4
Work Center 3
 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;
Introduction-Elements of the Shop Floor Scheduling Problems
 The classic approaches to shop floor scheduling focuses on
the following six elements
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 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).
Job 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.
Job A
Job B
M1
M2
Job A
M3
M4
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
 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
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
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
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.
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
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 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 ration is the remaining time until due date divided by processing
time;
 Scheduling the job with the largest 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 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
Job
Job number
Processing Time
1
2
3
4
5
11
29
31
1
2
Completion Time
1
2
3
4
5
11
40
71
72
74
Totals
268
Due Date
61
45
31
Mean Flow time=268/5=53.6
33
Average tardiness=121/5=24.2
No. of tardy jobs=3. 32
Due Date
Tardiness
61
45
31
33
32
0
0
40
39
42
121
Sequencing Rules——SPT
Job number
Processing Time
1
2
3
4
5
11
29
31
1
2
Job
4
5
1
2
3
Totals
Processing Time
1
2
11
29
31
Due Date
61
45
31
Mean Flow time=135/5=27.0
33
Average tardiness=43/5=8.6
No. of tardy jobs=1.
32
Completion Time Due Date
1
3
14
43
74
135
33
32
61
45
31
Tardiness
0
0
0
0
43
43
Sequencing Rules——EDD
Job number
Processing Time
1
2
3
4
5
11
29
31
1
2
Job Processing Time
3
5
4
2
1
Totals
31
2
1
29
11
Due Date
61
45
31
Mean Flow time=235/5=47.0
33
Average tardiness=33/5=6.6
No. of tardy jobs=4.
32
Completion Time
31
33
34
63
74
235
Due Date
31
32
33
45
61
Tardiness
0
1
1
18
13
33
Sequencing Rules——CR
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
1
11
2
29
4
1
5
2
Due Date-Current Time
30
14
2
1
Critical Ratio
30/11(2.727)
14/29(0.483)
2/1 (2.000)
1/2 (0.500)
Sequencing Rules——CR
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
Due DateCritical Ratio
No. of tardy jobs=4.
Current Time
1
1/11(0.0909)
-27
-27/1<0
-28
-28/2<0
Both Jobs 4 and 5 are later, however Job 4 has shorter processing time and
thus is scheduled first;
Job number
3
2
4
5
1
Totals
Processing Time
31
29
1
2
11
Completion Time
31
60
61
63
74
289
Tardiness
0
15
28
31
13
87
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
F4=W4+t4
 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]
Sequencing Theory for A Single Machines
 Maximum Tardiness
 Mean Flow Time
T
F
max
'
 max{
1

n
T ,T
,...,T n}
1
2
n
F
i 1
i
 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.
 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;

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 5.1 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)
F
[k ]

k
t
i 1
[i ]
t[4] (t3)
F[3]=t[1]+t[2]=t2+t1
The mean flow time of all jobs on
the schedule is given by
F
'

1
n
n
F
i 1
[k ]

1
n
n
k
 t
k 1 i 1
[i ]
Sequencing Theory for A Single Machines
1. Shortest-Processing-Time Scheduling
Theorem 5.1 The scheduling rule that minimizes the mean flow time F’ is SPT
'

1
n
n
F

1
n
n
k
 t
The mean flow time is given by
F
 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
t
[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 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].
Sequencing Theory for A Single Machines
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];
3.



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 Machines
Example 8.3
Job
1
2
3
4
5
6
Due date
15
6
9
23
20
30
Processing time
10
3
4
8
10
6
Longest processing time
Solution
Job
2
3
1
5
4
6
Due date
6
9
15
20
23
30
Processing time
3
4
10
10
8
6
Completion
time
3
7
17
27
35
41
Shanghai Jiao Tong University
Dept. of Industrial Engineering
Sequencing Theory for A Single Machines
Longest processing time
Example 8.3 :Solution (Cont.)
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
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.
Shanghai Jiao Tong University
Dept. of Industrial Engineering
Sequencing Theory for A Single Machines
Precedence constraints: Lawler’s Algorithm
g i ( Fi )  Fi  d i  Li
Objective
Function
Minimizing maximum
lateness
min max g ( F )
1i  n
i
i
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 ( ))
iv
  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 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.
Shanghai Jiao Tong University
Dept. of Industrial Engineering
Sequencing Theory for A Single Machines
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
Sequencing Theory for A Single Machines
Example 8.4
Step1: find the job scheduled last(sixth)
Not predecessor
Job
1
2
3
4
5
6
Processing
time
2
3
4
3
2
1
Due date
3
6
9
7
11
7
τ =2+3+4+3+2+1=15
Step2: find the job scheduled fifth
Tardines
s
3
5
6
15-9=6
15-11=4
15-7=8
Not predecessor
Job
1
2
3
4
6
Processing
time
2
3
4
3
1
Due date
3
6
9
7
7
τ =15-2=13
Tardines
3
6
13-9=4
13-7=6
Sequencing Theory for A Single Machines
Example 8.4
Not predecessor
Step3: find the job scheduled fourth
Job
1
2
4
6
Processing
time
2
3
3
1
Due date
3
6
7
7
τ =13-4=9
Step4: find the job scheduled third
Tardines
s
Job
1
2
4
Processing
time
2
3
3
Due date
3
6
7
τ =9-1=8
Tardines
Because job3 is no
longer on the list,
job 2 now because
a candidate.
2
6
9-6=3
9-7=2
Not predecessor
Because job6 has
been scheduled, so job 4
now because a candidate
along with job 2.
2
4
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
1
2
Processing
time
2
3
Due date
3
6
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
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.
Shanghai Jiao Tong University
Dept. of Industrial Engineering
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
4
1
Job J
1
4
Assume that both jobs must be processed first on
machine 1 and then on machine 2. There are four possible
schedules.
Shanghai Jiao Tong University
Dept. of Industrial Engineering
Sequencing Theory for Multiple Machines
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
Shanghai Jiao Tong University
Dept. of Industrial Engineering
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
been scheduled.
Sequencing Theory for Multiple Machines
Job
Example 8.5
Machine A
1
2
3
4
5
Optimal sequence : 2
Shanghai Jiao Tong University
Machine B
5
1
9
3
10
4
3
2
6
7
8
4
5
1
Dept. of Industrial Engineering
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)
3
A
2
Lacquering(B)
4
B
5
Polishing( C )
5
C
3
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Dept. of Industrial Engineering
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
Shanghai Jiao Tong University
Dept. of Industrial Engineering
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  ti
i 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.
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.
Shanghai Jiao Tong University
Dept. of Industrial Engineering
Assembly Line Balancing
Precondition
The job times and precedence relationships for this problem are summarized
in the table below.
Task
1
2
3
4
5
6
7
8
9
10
11
12
Immediate Predecessors
Time
_
12
1
6
2
6
2
2
2
2
2
12
3, 4
7
7
5
5
1
9, 6
4
ti=70, and the production rate is a unit/15 minutes;
8, 10
6
The minimum
number
of
workstations
= [70/15]=5
11
7
Shanghai Jiao Tong University
Dept. of Industrial Engineering
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
Shanghai Jiao Tong University
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
Dept. of Industrial Engineering
Assembly Line Balancing
Profile 1 C=15
Station
1
2
3
4
5
6
Tasks
1
2, 3, 4
5, 6, 9
7, 8
10, 11
12
Processing time
12
14
15
12
10
7
Idle time
3
1
0
3
5
8
Task
Immediate
Predecessors
Time
1
_
12
2
3
4
5
6
7
8
9
10
11
Shanghai
12
1
2
2
2
2
3, 4
7
5
9, 6
8, 10
Jiao Tong
11
6
6
2
2
12
7
5
1
4
6
University
7
The ranking
1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12
Dept. of Industrial Engineering
Assembly Line Balancing
Profile 1 C=15
Station
1
2
3
4
5
6
Tasks
1
2,3,4
5,6,9
7,8
10,11
12
Processing time
12
14
15
12
10
7
Idle time
3
1
0
3
5
8
15
Cycle Time=15
T1=12
T2=6
T5=2
T7=7
T10=4
T12=7
T3=6
T6=12
T8=5
T11=6
Shanghai Jiao Tong University
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 ~ 3 are
77.7%, 87.5%, and
89.7%. Thus the
profile 3 is the best
one.
Dept. of Industrial Engineering
Assembly Line Balancing
Alternative 1: Change cycle time to ensure 5 station balance
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;
Shanghai Jiao Tong University
Dept. of Industrial Engineering
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
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
 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;
Shanghai Jiao Tong University
Dept. of Industrial Engineering