SHCEDULLING
(PART II)
GROUP MEMBERS:
HAMIZAH BINTI HAMZAH
B050910232
NORAINA BINTI MOHD YUSOF B050910140
NUR ADILA BINTI RAMLI
B050810
SEQUENCING THEORY FOR SINGLE
MACHINE
1.
2.
3.
4.
5.
FCFS
SPT
EDD
CR
ALGORITHM
- Moore 1968
- Lawler’s
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
Mean Flow time=268/5=53.6
Average tardiness=121/5=24.2
No. of tardy jobs=3
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
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
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;
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
2
4
5
11
29
1
2
Current time=60 (31+29)
Job number
Processing Time
1
4
5
11
1
2
Due Date-Current Time Critical Ratio
30
14
2
1
Due DateCurrent Time
1
-27
-28
30/11(2.727)
14/29(0.483)
2/1 (2.000)
1/2 (0.500)
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
3
2
4
5
1
Totals
Processing Time
31
29
1
2
11
Mean Flow time=289/5=57.8
Average tardiness=87/5=17.4
No. of tardy jobs=4.
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 maximum tardiness (42, 43, 18, and 31 for the 4 different
rules);
 Always true? Yes!
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.
Sequencing Theory for A Single Machine
Example
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
3
7
17
27
35
41
time
Sequencing Theory for A Single Machine
Example :Solution (Cont.)
Longest processing time
Job
2
3
5
4
6
Due date
6
9
20
23
30
Processing time
3
4
10
8
6
Completion time
3
7
17
25
31
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.
Sequencing Theory for A Single Machines
Precedence constraints: Lawler’s Algorithm
Example
Job
1
2
3
4
5
6
Processing time
2
3
4
3
2
1
Due date
3
6
9
7
11
7
Example
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
Tardiness
3
5
6
15-9=6
15-11=4
15-7=8
Step2: find the job scheduled fifth
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
Tardiness
3
6
13-9=4
13-7=6
Example
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
Tardiness
Because job3 is no
longer on the list, Job
2 now because a
candidate.
2
6
9-6=3
9-7=2
Step4: find the job scheduled third
Not predecessor
Job
1
2
4
Processing time
2
3
3
Due date
3
6
7
τ =9-1=8
Tardiness
Because job6 has been
scheduled, Job 4 now
because a candidate along
with Job 2.
2
4
8-6=2
8-7=1
Example 3
Step5: find the job scheduled second
Job
1
2
4
6
3
5
Job
1
2
Processing
time
2
3
Due date
3
6
Not predecessor
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
MACHINE
Sequencing Theory for Multiple Machines
1. n jobs are to be process through m machine
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)
Let’s
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
Suppose that two jobs,Y and Z, are to be scheduled on two
machines, 1 and 2, the processing times are
Machine 1
Machine 2
Job Y
Job Y
Job Y
Job Z
Job Z
Job Z
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
Schedule
Total flow time
Mean flow time
Mean idle time
1
11
(6+11)/2=8.5
(5+5)/2=5
2
7
6.5
1
3
12
9.5
6
4
12
11.5
6
Deterministic Scheduling with Multiple Machines:
Johnson’s Rule
 Name Machine 1 = A, Machine 2 = B,
then ai = processing time for job i on A
and bi = processing time for job i on B
 Johnson’s Rule says that job i precedes job j in the optimal sequence if
min ai , b j   min a j , bi 
Algorithm:
 Step 1: Record the values of ai and bj 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
Sequencing Theory for Multiple Machines
Job
Example 1:
Machine A
1
2
3
4
5
job
A
B
1
5
2
2
1
6
3
9
7
4
3
8
5
10
4
Machine B
5
1
9
3
10
2
6
7
8
4
Johnson’s 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
Optimal sequence : 2
4
3
5
1
job
A
B
1
5
2
2
1
6
3
9
7
4
3
8
5
10
4
Optimal sequence : 2
4
3
5
1
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.
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.
Sequencing Theory for Multiple Machines
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 1
Job2
Operation
Time
Operation
Time
Sanding (A)
3
A
2
Lacquering (B)
4
B
5
Polishing( C )
5
C
3
Minimizing the flow time is the same as maximizing the time that both jobs are being
processed. That is equivalent to finding the path from the origin to the end of block C that
maximizes the diagonal movement and therefore minimizes either the horizontal or the
vertical movement.
or 10+6=16
or 10+(3+2)=15
15
14+2+2=18
14
C
13
C
12
Example 3:
F
11
10
B
14+4=18
B
9
Job 1
8
7
6
D
5
D
4
3
2
A
A
1
0
0
1
2
B
J1 B
J2 A
3
4
D5
6
7
C8
D
10 11 12 13 14 15 16 17
A
D
A
C
D
B
7
J1 B
J2 A
9
11
D
B
C
15
A
C
C
18
Job2
Order &
Operation
B
Time
Time
3
Order &
Operation
A
D
4
D
5
C
2
B
4
A
5
C
3
2
LINE BALANCING
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
Example 6
The Final assembly of NANO personal computers, a generic mailorder 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.
ti=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.
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
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
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
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
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
2
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
T4=2
T9=1
T10=4
T12=7
The ranking
1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11,
12
Evaluate the balancing
results by the efficiency
ti/NC;
The efficiencies for
Profiles 1 is 77.7%.
Profile 2: Increasing cycle time from 15 to 16
Alternative 1: Change cycle time to ensure 5 station balance
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;
Profile 3: C=13
Alternative 2: Staying with 6 stations, see if a six-station balance could be
obtained by cycle time less that 15 minutes
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
The efficiencies for profile 1~ 3 are 77.7%,
87.5%, and 89.7%. Thus the profile 3 is the best
one.
 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;