GROUP MEMBERS: HAMIZAH BINTI HAMZAHB050910232 NORAINA BINTI MOHD
YUSOFB050910140 NUR ADILA BINTI RAMLIB050810 SHCEDULLING (PART II)
Slide 2 1. FCFS 2. SPT 3. EDD 4. CR 5. ALGORITHM - Moore 1968 -
Lawlers SEQUENCING THEORY FOR SINGLE MACHINE Slide 3 Example:FCFS
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 Example. 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 Slide 4
Mean Flow time=268/5=53.6 Average tardiness=121/5=24.2 No. of tardy
jobs=3 Slide 5 Example:SPT 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 Mean Flow time=135/5=27.0
Average tardiness=43/5=8.6 No. of tardy jobs=1 Slide 6 Example: EDD
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 Mean Flow
time=235/5=47.0 Average tardiness=33/5=6.6 No. of tardy jobs=4
Slide 7 Current time: t=0 Job numberProcessing TimeDue DateCritical
Ratio 1234512345 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) Example:
Scheduling CR CR (Critical ratio) Critical ratio is the remaining
time until due date divided by processing time; Scheduling the job
with the smallest CR next; Slide 8 Current time: t=31 Job
numberProcessing TimeDue Date-Current TimeCritical Ratio 12451245
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
Current time=60 (31+29) Job numberProcessing TimeDue Date- Current
Time Critical Ratio 145145 11 1 2 1 -27 -28 1/11(0.0909) -27/1
Slide 9 Job numberProcessing TimeCompletion TimeTardiness
3245132451 31 29 1 2 11 31 60 61 63 74 0 15 28 31 13 Totals28987
Mean Flow time=289/5=57.8 Average tardiness=87/5=17.4 No. of tardy
jobs=4. 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. Slide 10 Sequencing Rules Summary RuleMean Flow
TimeAverage 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 31443144 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! Slide 11 Sequencing Theory for A Single Machine AN
ALGORITHM (Moore 1968) Minimizing the number of Tardy Jobs 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. Slide 12 Job123456 Due date1569232030
Processing time10348 6 Example Solution Job231546 Due
date6915202330 Processing time3410 86 Completion time 3717273541
Longest processing time Sequencing Theory for A Single Machine
Slide 13 Job23546 Due date69202330 Processing time341086 Completion
time37172531 Example :Solution (Cont.) Longest processing time
Job2346 Due date692330 Processing time3486 Completion time371521
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 Machine Slide 14 Sequencing Theory for A Single Machines
Example Job123456 Processing time234321 Due date3697117 Precedence
constraints: Lawlers Algorithm Slide 15 Example Job123456
Processing time234321 Due date3697117 Step1: find the job scheduled
last (sixth) Not predecessor =2+3+4+3+2+1=15 356
Tardiness15-9=615-11=415-7=8 Job12346 Processing time23431 Due
date36977 Step2: find the job scheduled fifth =15-2=13 36 Tardiness
13-9=413-7=6 Not predecessor Slide 16 Example Job1246 Processing
time2331 Due date3677 Step3: find the job scheduled fourth Not
predecessor =13-4=9 26 Tardiness9-6=39-7=2 Job124 Processing
time233 Due date367 Step4: find the job scheduled third =9-1=8 24
Tardiness8-6=28-7=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. Slide 17
Example 3 Job12 Processing time 23 Due date36 Step5: find the job
scheduled second Not predecessor The optimal sequence: 1-2-4-6-3-5
JobProcessing time Flow time Due dateTardiness 124635124635
233142233142 2 5 8 9 13 15 3 6 7 9 11 001244001244 Maximum
tardiness Slide 18 SEQUENCING THEORY FOR MULTIPLE MACHINE Slide 19
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) Lets consider a case when each
job must be processed in the following order First on machine 1,
then machine 2. Sequencing Theory for Multiple Machines 1. n jobs
are to be process through m machine Slide 20 Suppose that two
jobs,Y and Z, are to be scheduled on two machines, 1 and 2, the
processing times are Assume that both jobs must be processed first
on machine A and then on machine B. There are four possible
schedules. Sequencing Theory for Multiple Machines Machine 1Machine
2 Job Y Job Z Slide 21 ScheduleTotal flow timeMean flow timeMean
idle time 111(6+11)/2=8.5(5+5)/2=5 27 6.51 312 9.56 412 11.56
Sequencing Theory for Multiple Machines Slide 22 Deterministic
Scheduling with Multiple Machines: Johnsons Rule Name Machine 1 =
A, Machine 2 = B, then a i = processing time for job i on A and b i
= processing time for job i on B Johnsons Rule says that job i
precedes job j in the optimal sequence if Algorithm: Step 1: Record
the values of a i and b j in two columns Step 2: Find the smallest
remaining value in two columns. If this value in column a, schedule
this job in the first open position in the sequence; if this value
in column b, schedule this job in the last open position in the
sequence; Step 3: Cross off each job as it is scheduled Slide 23
Example 1: JobMachine AMachine B 152 216 397 438 5104 Optimal
sequence : 2 4 3 5 1 Sequencing Theory for Multiple Machines jobAB
152 216 397 438 5104 Johnsons schedule: 2 > x > x > x >
x 2 > x > x > x > 1 2 > x > x > 5 > 1 2
> 4 > x > 5 > 1 2 > 4 > 3 > 5 > 1 Slide 24
jobAB 152 216 397 438 5104 Optimal sequence : 2 4 3 5 1 Slide 25
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 A i max B i or
min C i max B i 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 A i =A i +B i, B i =B i +C
i Solve the problem using the rules described above for
two-machines, treating A i and B i 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. Slide 26 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. Graphical procedure developed by Akers (1956): 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. Slide 27 Example 2: 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 1Job2
OperationTimeOperationTime Sanding (A)3A2 Lacquering (B)4B5
Polishing( C )5C3 Sequencing Theory for Multiple Machines Slide 28
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 Slide 29 Example
3: Job 1Job2 Order & Operation TimeOrder & Operation Time
B3A2 D4D5 C2B4 A5C3 B D C A A D B C A C D B F 14+4=18 14+2+2=18 7
11 15 18 B A D C B A C D J1 J2 B A D C B A C D J1 J2 Slide 30 LINE
BALANCING Slide 31 WHAT IS LINE BALANCING??? EVERYONE IS DOING THE
SAME AMOUNT OF WORK DOING THE SAME AMOUNT OF WORK TO CUSTOMER
REQUIREMENT VARIATION IS SMOOTHED NO ONE OVERBURDENED NO ONE
WAITING EVERYONE WORKING TOGETHER IN A BALANCED FASHION Slide 32
Slide 33 Slide 34 Example 6 The Final assembly of NANO 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. Slide 35 TaskImmediate PredecessorsTime 1_12 216 326 422
522 62 73, 47 875 951 109, 64 118, 106 12117 t i =70, and the
production rate is a unit /15 minutes; The minimum number of
workstations = [70/15]=5 The job times and precedence relationships
for this problem are summarized in the table below. Slide 36 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. TaskPositional Weight 170 258
331 427 520 629 725 818 9 1017 1113 127 t 3 +t 7 +t 8 +t 11 +t 12
=31 The ranking 1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12 Slide 37
Station123456 Tasks12, 3, 45, 6, 97, 810, 1112 Processing
time12141512107 Idle time310358 PROFILE 1: C=15 TaskImmediate
Predecessors Time 1_12 216 326 422 522 62 73, 47 875 951 109, 64
118, 106 12117 The ranking 1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12
Slide 38 Station123456 Tasks12,3,45,6,97,810,1112 Processing
time1214152107 Idle time310358 PROFILE 1: C=15 Cycle Time=15 T1=12
T2=6T3=6T4=2 T5=2T6=12T9=1 T8=5T7=7T10=4 T11=6T12=7 The ranking 1,
2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12 15 Evaluate the balancing
results by the efficiency t i /NC; The efficiencies for Profiles 1
is 77.7%. Slide 39 Profile 2: Increasing cycle time from 15 to 16
Alternative 1: Change cycle time to ensure 5 station balance
Station12345 Tasks12,3,4,56,97,8,1011,12 Idle time40303 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; Slide 40 Alternative 2:
Staying with 6 stations, see if a six-station balance could be
obtained by cycle time less that 15 minutes Profile 3: C=13
Station123456 Tasks12,364,5,7,98,1011,12 Idle time111140 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.
