Title: Production Planning and Control
1Production Planning and Control
Chapter 5 Operations Scheduling
Professor JIANG Zhibin Department of Industrial
Engineering Management Shanghai Jiao Tong
University
2Chapter 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
3Introduction-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.
4Introduction-Objectives of Job Shop Scheduling
- Objectives of operations scheduling
- Meet due date
- Minimize WIP inventory
- Minimize the average flow time through the
systems - Provide for high machine/worker (time)
utilization (minimize idle time) - Reduce setup cost
- 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
5Introduction-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 .
6Introduction-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.
7Introduction-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.
8Introduction-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.
9Introduction-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.
10Introduction-Job Shop
A job shop is organized by machines which
are grouped according to their functions.
11Introduction-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
12Introduction-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).
13Job Shop Scheduling Terminology
- 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, 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, and M4 are identical
- Jobs A and B can be processed on any one of the 4
machines
14Job 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(F1F2F3)/3
15Job 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.
Make-span of the 3 jobs
16Job 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
Latenessgt0---Tardiness
Latenesslt0---Earliness
When the completion of Job is earlier than due
date, the tardiness is 0
17Sequencing 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
18Sequencing 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
CRiRemaining 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).
19Sequencing 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
20Sequencing RulesFCFS
Job number Processing Time Due Date
1 2 3 4 5 11 29 31 1 2 61 45 31 33 32
Mean Flow time268/553.6 Average
tardiness121/524.2 No. of tardy jobs3.
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
21Sequencing RulesSPT
Job number Processing Time Due Date
1 2 3 4 5 11 29 31 1 2 61 45 31 33 32
Mean Flow time135/527.0 Average
tardiness43/58.6 No. of tardy jobs1.
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
22Sequencing RulesEDD
Job number Processing Time Due Date
1 2 3 4 5 11 29 31 1 2 61 45 31 33 32
Mean Flow time235/547.0 Average
tardiness33/56.6 No. of tardy jobs4.
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
23Sequencing RulesCR
Current time t0 Current time t0 Current time t0 Current time t0
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 should be reset after scheduling one
job
Current time t31 Current time t31 Current time t31 Current time t31
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)
24Sequencing RulesCR
Mean Flow time289/557.8 Average
tardiness87/517.4 No. of tardy jobs4.
Current time60 Current time60 Current time60 Current time60
Job number Processing Time Due Date-Current Time Critical Ratio
1 4 5 11 1 2 1 -27 -28 1/11(0.0909) -27/1lt0 -28/2lt0
Both Jobs 4 and 5 are later, however Job 4 has
shorter processing time and thus is scheduled
first
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
25Sequencing 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 maximums tardiness (42,
43, 18, and 31 for the 4 different rules) - Always true? Yes!
26Sequencing Theory for A Single Machines
Assuming that n jobs are to be processed through
one machine. For each job i, define the following
quantities
- tiProcessing time for job i, constant for job
i - diDue date for job i, constant for job i
- WiWaiting 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
- FiFlow time for job i, the waiting time plus
the processing time Fi Wi ti - LiLateness of job i , Li Fi- di, either
positive or negative
- TiTardiness of job i, the positive part of Li,
TimaxLi,0 - EiEarliness of job i, the negative part of Li,
Ei max- Li,0
27Sequencing Theory for A Single Machines
- 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, 13, 22,
31, and 44
28Sequencing Theory for A Single Machines
- 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
29Sequencing Theory for A Single Machines
- Shortest-Processing-Time Scheduling
- Theorem 5.1 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 - k1t1
- k2t1 t2
-
- knt1 t2 tn
- By summing down the column rather than across the
row, we may rewrite F in the form - nt1(n-1)t2tn
SPT sequencing rule the job with shortest
processing time t is set first
30Sequencing Theory for A Single Machines
- 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. - 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, d1? d2? dn.
31Sequencing Theory for A Single Machines
- 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 d1? d2?,, ? dn - 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 Jobi. - 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.
32Sequencing 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
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
33Sequencing Theory for A Single Machines
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.
34Sequencing Theory for A Single Machines
Precedence constraints Lawlers Algorithm
Minimizing maximum lateness
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
35Sequencing 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 t 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.
36Sequencing 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
37Sequencing Theory for A Single Machines
Example 8.4
Step1 find the job scheduled last(sixth)
Job 1 2 3 4 5 6
Processing time 2 3 4 3 2 1
Due date 3 6 9 7 11 7
3 5 6
Tardiness 15-96 15-114 15-78
t 23432115
Step2 find the job scheduled fifth
Job 1 2 3 4 6
Processing time 2 3 4 3 1
Due date 3 6 9 7 7
3 6
Tardiness 13-94 13-76
t 15-213
38Sequencing Theory for A Single Machines
Example 8.4
Step3 find the job scheduled fourth
Job 1 2 4 6
Processing time 2 3 3 1
Due date 3 6 7 7
Because job3 is no longer on the list, job 2 now
because a candidate.
2 6
Tardiness 9-63 9-72
t 13-49
Step4 find the job scheduled third
Because job6 has been scheduled, so job
4 now because a candidate along with job 2.
Job 1 2 4
Processing time 2 3 3
Due date 3 6 7
2 4
Tardiness 8-62 8-71
t 9-18
39Sequencing Theory for A Single Machines
Example 8.4
Step5 find the job scheduled second
Job 1 2
Processing time 2 3
Due date 3 6
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
40Sequencing 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. (n5, m5, 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.
41Sequencing 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.
42Sequencing Theory for Multiple Machines
Schedule Total flow time Mean flow time Mean idle time
1 9 (59)/27 (44)/24
2 6 5.5 1
3 10 8 5
4 10 9.5 5
43Sequencing Theory for Multiple Machines
- 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
- AiProcessing time of job i on machine A
- BiProcessing time of job i on machine B
- Rule Job i precedes job i1 if min(Ai,
Bi-1)ltmin(Ai1,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.
44Sequencing Theory for Multiple Machines
Job Machine A Machine B
1 5 2
2 1 6
3 9 7
4 3 8
5 10 4
Example 8.5
Optimal sequence 2 4 3 5
1
45Sequencing Theory for Multiple Machines
- 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 AiAiBi, BiBiCi
- 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.
46Sequencing Theory for Multiple Machines
- 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.
47Sequencing 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 Job 1 Job2 Job2
Operation Time Operation Time
Sanding(A) 3 A 2
Lacquering(B) 4 B 5
Polishing( C ) 5 C 3
48 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 10616
or 10(32)15
49Assembly 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.
50Assembly 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
- 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.
51Assembly 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.
52Assembly Line Balancing
Precondition The job times and precedence
relationships for this problem are summarized in
the table below.
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
- ?ti70, and the production rate is a unit/15
minutes - The minimum number of workstations 70/155
53Assembly 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.
t3t7t8t11t1231
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 ranking 1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12
54Assembly Line Balancing
Profile 1 C15
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 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
The ranking 1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12
55Assembly Line Balancing
Profile 1 C15
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
- 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.
Cycle Time15
T26
T112
T26
T36
T42
T52
T52
T612
T91
T85
T77
T104
T104
T116
T127
T127
56Assembly 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
57Assembly 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 C13
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