1
4.4 Sequencing Theory for Single Machine
1. Shortest Processing Time (SPT)
 Minimizes the mean flow time
 Minimizes waiting time
 Minimizes lateness
for single-machine sequencing.
2. Earliest Due Date Scheduling
 Minimizes the maximum lateness
2
4.4 Sequencing Theory for Single Machine
3. Moore’s algorithm
 Minimizes number of tardy jobs
Step 1:
Sequence the jobs according to the earliest due date to obtain
the initial solution
d[1] d[2],…,  d[n]
Step 2:
Find the first tardy job in the current sequence, say job [i]. If
none exists go to step 4.
3
4.4 Sequencing Theory for Single Machine
Step 3:
Consider jobs [1], [2], …, [i]. Reject the job with the largest
processing time. Return to step2.
Reason: Largest effect on the tardiness of the Job[i]
Step 4:
Form an optimal sequence by taking the current sequence
and appending to it the rejected jobs. The job appended to
the current sequence scheduled in any order.
Reason: Consider number of tardiness jobs rather than tardiness
4
4.4 Sequencing Theory for Single Machine
EXAMPLE 1:
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
Longest processing time
5
4.4 Sequencing Theory for Single Machine
EXAMPLE 1: 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
Optimal sequence: 2, 3, 4, 6, 5, 1 or 2, 3, 4, 6, 1, 5
Number of tardy jobs is two in either case.
No
lateness
6
Exercise 4.4.1:
Seven jobs are to be processed through a single machine. The
processing times and due dates are given below.
Job
1
2
3
4
5
6
7
Processing time
3
6
8
4
2
1
7
Due date
4
8
12
15
11
25
21
Determine sequence of the jobs in order to minimize
a) Mean flow time
b) Number of tardy jobs
c) Maximum lateness
7
4.4 Sequencing Theory for Single Machine
4. Lawler’s Algorithm
 Minimize maximum lateness
 Minimize maximum tardiness
Subject to precedence constraint ( certain job
must be completed before other jobs can begin)
g i ( Fi )  Fi  d i  Li
min max g ( F )
1i  n
i
i
g i ( Fi )  max( Fi  d i ,0)
8
4.4 Sequencing Theory for Single Machine
Rule:
Schedule job in reverse order.
Step 1:
At each stage, determine the set of jobs (named
V) not require to precedes any other. Among set
V, choose job k that satisfies:
g k ( )  min ( g i ( ))
iv
  i 1 ti
Example: Job among V that has smallest tardiness,
if arranged on position [n].
n
Processing time of current sequence.
9
4.4 Sequencing Theory for Single Machine
Step 2:
Now, Job k scheduled last. Consider remaining jobs and
again determine set of jobs that not require to precede
any other remaining job.
Step 3:
After scheduling Job k, τ reduced by tk and job
scheduled next to last is now determined.
Step 4:
Process is continued until all jobs are scheduled.
10
4.4 Sequencing Theory for Single Machine
EXAMPLE 2:
1
2
4
5
3
6
Job
1
2
3
4
5
6
Processing time
2
3
4
3
2
1
Due date
3
6
9
7
11
7
11
4.4 Sequencing Theory for Single Machine
Solution:
Step 1: 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
τ = Total processing time
= 2+3+4+3+2+1
= 15
Job
Tardiness
3
5
6
15-9=6
15-11=4
15-7=8
Min value
Hence, Job 5 scheduled last
12
4.4 Sequencing Theory for Single Machine
Solution: Cont.
Step 2: Find the job scheduled fifth.
1
2
4
5
3
6
Job
1
2
3
4
6
Processing time
2
3
4
3
1
Due date
3
6
9
7
7
τ = New total processing time
= 15 – 2
= 13
Job
Tardiness
Hence, Job 6 scheduled fifth.
3
6
13-9=4
13-7=6
Min value
13
4.4 Sequencing Theory for Single Machine
Solution: Cont.
Step 3: Find the job scheduled fourth.
1
2
4
5
3
6
Job
1
2
4
6
Processing time
2
3
3
1
Due date
3
6
7
7
τ = New total processing time
= 13 – 4
=9
Job
Tardiness
Hence, Job 6 scheduled fourth.
2
6
9-6=3
9-7=2
Min value
14
4.4 Sequencing Theory for Single Machine
Solution: Cont.
Step 4: Find the job scheduled third.
1
2
4
5
3
6
Job
1
2
4
Processing time
2
3
3
Due date
3
6
7
τ = New total processing time
=9–1
=8
Tardiness
Hence, Job 4 scheduled third.
2
4
8-6=2
8-7=1
Min value
15
4.4 Sequencing Theory for Single Machine
Solution: Cont.
Step 5: Find the job scheduled second.
Job
1
2
4
6
3
5
1
2
4
3
Job
1
2
5
Processing time
2
3
6
Due date
3
6
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
The optimal sequence: 1-2-4-6-3-5
Maximum
tardiness
16
Exercise 4.4.2:
Eight jobs are to be processed through a single machine. The
processing times and due dates are given below.
Job
1
2
3
4
5
6
7
8
Processing Time
2
3
2
1
4
3
2
2
Due date
5
4
13
6
12
10
15
19
Furthermore, assume the following precedence relationships
must be satisfied:
2
6
3
1
4
7
8
Determine the sequence in which jobs should be done in order to
minimize maximum lateness subject to precedence restriction.
17
4.5 Sequencing Theory for Multiple Machines
• Analysis of the previous section which several jobs
must be proceed on more than one machine.
•The optimal solution for scheduling n jobs on two
machines is always a permutation schedule (that is,
jobs are done in the same order on both machines).
•Permutation schedules provide better performances
in term of both total and average flow time
• Minimization of the mean idle time in the system
18
4.5 Sequencing Theory for Multiple Machines
1. Johnson’s algorithm : Scheduling n jobs on two
machines
• How to implement this rule:
1. List value of A and B in to columns
2. Find the smallest element in the 2 columns. If it appears in
column A, then schedule that at front of the sequence. If
appears in column B, schedule that at the back of sequence
3. Find the remaining element in the two columns. If it
appears on column A, then schedule that next job. If
appears in column B, then schedule that job last.
4. Cross off the jobs as they are scheduled.
19
4.5 Sequencing Theory for Multiple Machines
EXAMPLE:
• Five jobs are to be scheduled on two machines. The processing
time are:
Job
1
2
3
4
5
Machine A
5
1
9
3
10
Machine A
A2
A4
Machine B
Machine B
2
6
7
8
4
A3
B2
0
1
A5
B4
4
7
A1
B3
13
15
2-4-3-5-1
B5
22 23
B1
27 28
20 30
Exercise 4.5.1:
Six job are to be schedule on two machine. The processing time
are
Job
1
2
3
4
5
6
Machine A
20
16
43
60
35
42
Machine B
27
30
51
12
28
24
21
4.5 Sequencing Theory for Multiple Machines
2. Extension to three machine
• The 3 machine problem can be reduced to a 2
machine problem if the satisfied the following
condition:
min A ≥ max B
or
min C ≥ max B
• only either one of these conditions be satisfied.
• Then reduced to 2 machine problem in the
following way:
A’ = A + B and B’ = B + C
22
4.5 Sequencing Theory for Multiple Machines
EXAMPLE:
Machine
Machine
Job
A
B
C
Job
A'
B'
1
4
5
8
1
9
13
2
9
6
10
2
15
16
3
4
5
8
6
5
2
3
4
6
7
11
3
10
8
4
9
10
5
9
15
Min A = 4
Max B = 6 Min C = 6
Machine A
A1
A4
Machine B
1-4-5-2-3
A5
B1
0
9
A2
B4
18 22
A3
B5
27 32
B2
42
47
B3
52 63
23
71
Exercise 4.5.2:
The following four job must be processed through a threemachine flow shop
Job
1
2
3
4
A
4
2
6
3
Machine
B
2
3
5
4
C
6
7
6
8
24
4.5 Sequencing Theory for Multiple Machines
3. Two job flow shop problem
• Two jobs are processed through m machines.
• Present graphical procedure for solving this problem. The step
are:
1. Draw a cartesian coordinate system with the processing times
corresponding the first job on the horizontal axis and the
processing time corresponding to the second job on the vertical
axis.
2. Block out areas corresponding to each machine at the
intersection of the intervals marked for the machine on the two
axes.
3. Determine the part from the origin to the end of the final block
that does not intersect any of the blocks and that minimize the
vertical movement.
25
4.5 Sequencing Theory for Multiple Machines
EXAMPLE:
• A regional manufacturing firm produces a variety of
household products. One is a wooden desk lamp. Prior to
packing, the lamp 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 2
Operation
Time
Operation
Time
Sanding (A)
3
A
2
Lacquering (B)
4
B
5
Polishing ( C)
5
C
3
26
4.5 Sequencing Theory for Multiple Machines
10
9
8
7
6
5
Total time = 12 + (2 + 2)
= 16
C
Total time = 12 + 3 = 15
B
4
3
2
1
A
1
2 3
Gantt Chart
Solution
4 5 6 7 8 9 10 11 12
A
A1
A2
B1
B
C
2
4
6
B2
C1
8
10
C2
12
14
27
4.5 Sequencing Theory for Multiple Machines
10
9
8
7
6
5
C
Total time = 10 + 6 = 16
Total time = 10 + (3 + 2)
= 15
B
4
3
2
1
A
1
2 3
Gantt Chart
Solution
4 5 6 7 8 9 10 11 12
A1
A2
A
B1
B
C
2
4
6
B2
C1
8
10
C2
12
14
28
Exercise 4.5.3:
Two jobs must be proceeds through four machines in same
order. The processing time in the required sequence are
Job 1
Machine
A
B
C
D
Time
5
4
6
3
Job 2
Machine
Time
A
2
B
4
C
3
D
5
Determine how the two job should be schedule in order to
minimize the total makespan and draw the gantt chart indicating
the optimal schedule.
29
4.6 ASSEMBLY LINE BALANCING
Line Balancing
Line balancing: the process of assigning tasks to
workstations in such a way that the workstations
have approximately equal time requirements.
Cycle Time
Cycle time: maximum time allowed at each workstation to
complete its set of tasks on a unit.
 amount of time allotted to each workstation determined in
advance, based on the desired rate of production of assembly
line.
Maximum cycle time = summation of the task times.
Minimum cycle time = the longest task times.
30
4.6 ASSEMBLY LINE BALANCING
Factors that contribute to the
difficulty of the problem:
- There are precedence
constraints
- Some tasks cannot be
performed at the same
workstation
OT  operating time per day
D = desired output rate
OT
Output rate, D =
CT
OT
CT = cycle time =
D
31
4.6 ASSEMBLY LINE BALANCING
Line Balancing Rules
Some Heuristic (intuitive) Rules:
• Assign tasks in order of most following
tasks.
–
Count the number of tasks that follow
• Assign tasks in order of greatest positional
weight.
–
Positional weight is the sum of each task’s time
and the times of all following tasks.
32
4.6 ASSEMBLY LINE BALANCING
Precedence Diagram
Precedence diagram: Tool used in line balancing to display
elemental tasks and sequence requirements
0.1 min.
1.0 min.
a
b
c
0.7 min.
A Simple Precedence
Diagram
d
0.5 min.
e
0.2 min.
Figure 4.6.1
33
4.6 ASSEMBLY LINE BALANCING
Determine the Minimum Number
of Workstations Required
N=
( t)
CT
 t = sum of task time
What is the minimum number of workstations for the previous
precedence diagram? (assume minimum cycle time)
2.5 mins
N=
 2.5  3
1 min
 t = 2.5 mins
34
4.6 ASSEMBLY LINE BALANCING
EXAMPLE:
Arrange tasks shown in Figure 4.6.1 into 3 workstations.
 Use
a cycle time of 1.0 minute
 Assign tasks in order of the most number of followers
Solution
Eligible
Assign
Task
1.0
0.9
0.2
a, c
c
none
a
c
–
0.9
0.2
2
1.0
b
b
0.0
3
1.0
0.5
0.3
d
e
–
d
e
–
0.5
0.3
Workstation
1
Time
Remaining
Revised
Time
Remaining
Station
Idle Time
0.2
Idle time per cycle
0.0
0.3
0.5
35
4.6 ASSEMBLY LINE BALANCING
Calculate Percent Idle Time
Idle time per cycle
% idletime =
* 100%
(N)(CT)
Efficiency = 100% – Percent idle time
What’s the % idle time and efficiency for the above example?
0.5
Percent idle time =
*100%  16.7%
(3)(1.0)
Efficiency = 100% - 16.7% = 83.3%
36
4.6 ASSEMBLY LINE BALANCING
EXAMPLE:
A manager wants to assign workstations in such a manner that
hourly output is 4 units. Working time is 56 minutes per hour.
What is the cycle time?
Cycle time = operating time/output rate
= 14 min
37
Arrange tasks in
decreasing order of
positional weight
Task
Task time
F
5
D
7
G
6
A
3
B
2
C
4
E
4
H
9
I
5
Assign the tasks above to workstations in the order of greatest positional
weight.
Steps:
1) Arrange the task in the decreasing order of positional weights.
2) Find out the number of workstations
Number of workstations = sum of task time s = 45 = 3.2 =4
cycle time
14
38
Station
14
I
Cycle Time=14 min
F=5
III
IV
D=7
G=6
II
percent idle time =
C=4
H=9
A=3
E=4
B=2
11
*100%
(4)(14)
 19.64%
I=5
39
4.6 ASSEMBLY LINE BALANCING
EXERCISE 4.6.1
A shop wants an hourly output of 33.5 units per hour. The
working time is 60 minutes per hour. Assign the tasks using
the rules:
a) In the order of most following task.
b) In the order of greatest positional weight.
40
4.6 ASSEMBLY LINE BALANCING
EXAMPLE:
41
The job times & precedence relationship for this problem:
Task
Immediate
Predecessor
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
42
Suppose that the company is willing to hire enough workers to produce one assembled
machine every 15 minutes.
Sum of task time = 70 minutes
Minimum no. Workstation = 70/15 = 4.67  5
Positional weight of task i: the time required to perform task i plus the times required to
perform all task having task i ask a predecessor.
TASK
POSITIONAL
WEIGHT
TASK
POSITIONAL
WEIGHT
1
70
1
70
2
58
2
58
3
31
3
31
4
27
6
29
5
20
4
27
6
29
7
25
7
25
5
20
8
18
8
18
9
18
9
18
10
17
10
17
11
13
11
13
12
7
12
7
Arrange tasks in
decreasing order of
positional weight
43
Rank the step in the order of decreasing positional weight:
Assume C = 15 min
Station
1
2
3
4
5
6
Tasks
1
2,3,4
5,6,9
7,8
10,11
12
Total idle
time
Idle time
3
1
0
3
5
8
20
But, minimum possible No. of workstation = 5!!!
This method is heuristic  possible that there is a solution with 5 stations
Use C = 16 min
Station
1
2
3
4
5
Tasks
1
2,3,4,5
6,9
7,8,10
11,12
Total idle
time
Idle time
4
0
3
0
3
10
No. of stations decreases  16 %, cycle time increases  7%
Assume production day = 7 hrs
C = 15  daily production level = 28 units/assembly operation
44
C = 16  daily production level = 26.25 units/assembly operation
Management have to determine whether the decline in the production rate of 1.75
units/day/operation is justified by the savings realised with 5 rather than 6 stations.
Alternative choice:
Stay with 6 stations, but use C = 13 min
Station
1
2
3
4
5
6
Tasks
1
2,3
6
4,5,7,9
8,10
11,12
Total idle
time
Idle time
1
1
1
1
4
0
8
13 min: minimum cycle time with 6 stations. Why not 12 min???!!
Production rate = 32.3 units/day/operation
Increasing No. of stations from 5 to 6  substantial improvement in the throughput rate.
45
4.6 ASSEMBLY LINE BALANCING
EXERCISE 4.6.2
Consider the assembly line balancing problem represented by the figure
above. Determine a balance for
a. C = 20
46
4.7 Stochastic Scheduling
4.7.1 Static Analysis
1. Single machine
 Uncertainty of processing times
 Exact completion time of one or more jobs may not be
predictable
Objective: Minimize expected average weighted flow
time.
 Job i precedes job i+1 if
where job times are t1, t2, …, tn and weights are u1, u2, …,
ui.
47
4.7.1 Stochastic Scheduling: Static Analysis
EXAMPLE:
Jobi
1
2
3
4
5
6
7
8
9
10
Processing Time, ti
1
10
5
2
8
7
8
4
3
6
Importance weight, ui
3
2
1
1
4
2
3
3
2
4
Due date, di
10
20
15
10
10
25
15
25
10
20
Find the optimal sequence that minimize expected
average weighted flow time.
48
4.7.1 Stochastic Scheduling: Static Analysis
Solution:
Jobi
ti
ui
ti/ui
di
Ci
Ti
1
1
3
0.3
10
1
0
8
4
3
1.3
25
5
0
9
3
2
1.5
10
8
0
10
6
4
1.5
20
14
0
4
2
1
2
10
16
6
5
8
4
2
10
24
14
7
8
3
2.7
15
32
17
6
7
2
3.5
25
39
14
3
5
1
5
15
44
29
2
10
2
5
20
54
34
∑Ci =
237
∑Ti=
114
Expected average
weighted flow time
= 237/10
= 23.7
49
4.7.1 Stochastic Scheduling: Static Analysis
2. Multiple Machine
• n jobs are to processed through two identical parallel
machines. Each jobs needs to be processed only once
on either machine.
• The objective is to minimize the expected makespan.
• Parallel is different from flow shop problem.
• In flow shop, jobs are processed first one machine 1
then on machine 2.
• Then the optimal sequence is to schedule the jobs
according to LEPT (longest expected processing time
first) opposite with the SPT rule
50
4.7.2 Stochastic Scheduling: Dynamic Analysis
Dynamic: Jobs arrive randomly over time, and
decisions must be made on an ongoing basis as
how to schedule those jobs.
When jobs arrive shop dynamically over time,
queuing theory provides a means of analyzing
the results.
 Standard M/M/1 queue applies to case of
purely random arrivals to single machine with
random processing times.
51
4.7.2 Stochastic Scheduling: Dynamic Analysis
Selection independent of job processing times
 Same mean flow times, but differ variance
flow times
Selection dependent of job processing times
 Job times realized when job joins the queue
rather than when job enters service
 SPT results lowest expected flow time.
52
4.8 ADVANCED TOPIC FOR OPERATIONS SCHEDULING
FCS - jobs are being scheduled through a number of work centers, each
with one or more machines.
• jobs can pass each other or change their order as they are processed,
depending on their priority.
• a job can be split into two or more parts if this will facilitate scheduling
Advantage:
-The addition of capacity  improves completion dates of jobs & reduces
waiting times.
Bottleneck: a work centre whose capacity is less than the demand placed
on it & less than the capacities of all other resources.
53
54
Batch operation:
Job shop  “shop”, “job”, “work center”
Job  “customer”, “patient”, “client”, “paperwork”
Work centre  “room”, “office”, “facility”, “skill specialty”
In a job shop, batch corresponds to what customer orders
– can include one or several parts or items.
Each part or job is scheduled through the various
machines & work centres according to the equipment &
labour needed to process the job.
55
Batch scheduling:
• each batch flowing through a batch process typically
moves along with many starts & stops, not smoothly – due
to layout of batch process  jobs or customers wait in line
as each batch is transferred from one work centre to the
next.
• batch scheduling problem: network or queues.
•Jobs or customers spend most of their time waiting in line
– amount of time waiting varies with the load of the
process
56
57
Solution:
Jobi
ti
di
Ci
Ti
6
1
25
1
0
5
2
11
3
0
1
3
4
6
2
4
4
15
10
0
2
6
8
16
8
7
7
21
23
2
3
8
12
31
19
4.4.1a) SPT sequence rule,
Mean flow time
= 90/7
= 12.86
∑Ci = 90 ∑Ti= 31
58
Job
1
2
5
3
4
7
6
Due date
4
8
11
12
15
21
25
Processing time
3
6
2
8
4
7
1
Completion time
3
9
11
19
23
30
31
Job
1
5
3
4
7
6
Due date
4
11
12
15
21
25
Processing time
3
2
8
4
7
1
Completion time
3
5
13
17
24
25
Job
1
5
4
7
6
Due date
4
11
15
21
25
Processing time
3
2
4
7
1
Completion time
3
5
9
16
17
4.4.1b) Moore ‘s algorithm.
Optimal sequence: 1, 5, 4, 7, 6, 3, 2 or 1, 5, 4, 7, 6, 2, 3
Number of tardy jobs is two in either case.
59
Jobi
ti
di
Ci
Ti
1
3
4
3
0
2
6
8
9
1
5
2
11
11
0
3
8
12
19
7
4
4
15
23
8
7
7
21
30
9
6
1
25
31
6
∑Ci =
126
∑Ti= 31
4.4.1c) EDD sequence
rule, Maximum lateness
=9
Summary:
Rule
Sequence
Mean flow time
Number of
tardy job
Maximum
lateness
SPT
6-5-1-4-2-7-3
12.86
4
19
Moore’s
1-5-4-7-6-3-2
14.86
2
19
EDD
1-2-5-3-4-7-6
18
5
9
60
Solution:
2
6
3
1
4
7
8
Job
1
2
3
4
5
6
7
8
Processing Time
2
3
2
1
4
3
2
2
Due date
5
4
13
6
12
10
15
19
τ = 2+3+2+1+4+3+2+2
Job
= 19
Tardiness
3
8
19-13=6
19- 19=0
2
6
3
1
4
7
8
Job
1
2
3
4
5
6
7
Processing Time
2
3
2
1
4
3
2
Due date
5
4
13
6
12
10
15
τ = 19- 2= 17
Job
Tardiness
3
7
17-13=4
17- 15=2
61
2
6
3
1
4
7
8
Job
1
2
3
4
5
6
Processing Time
2
3
2
1
4
3
Due date
5
4
13
6
12
10
Job
τ = 17- 2 = 15
Tardiness
2
6
3
1
4
7
3
4
15-13=3
15- 6=9
8
Job
1
2
4
5
6
Processing Time
2
3
1
4
3
Due date
5
4
6
12
10
τ = 15-2 = 13
Job
Tardiness
4
6
13-6=7
13- 10=3
62
2
6
3
1
4
7
8
Job
1
2
4
5
Processing Time
2
3
1
4
Due date
5
4
6
12
Job
τ = 13-3 = 10
Tardiness
2
6
3
1
4
7
2
4
10-4=6
10-6=4
8
Job
1
2
5
Processing Time
2
3
4
Due date
5
4
12
τ = 10- 1= 9
Job
Tardiness
1
2
9-5=4
9-4=5
63
Job
Processing Flow time
time
Due date
Tardiness
2
3
3
4
0
1
2
5
5
0
4
1
6
6
0
6
3
9
10
0
3
2
11
13
0
7
2
13
15
0
8
2
15
19
0
5
4
19
12
7
4.4.2) Maximum tardiness: 7
Optimal sequence: 2-1-4-6-3-7-8-5
64
4.5.1)
A2
Job
1
2
3
4
5
6
A1
B2
0
16
Machine A
20
16
43
60
35
42
A3
B1
36 46
Machine B
27
30
51
12
28
24
A5
B3
73 79
2-1-3-5-6-4
A6
A4
B5
114 130
B6
156 158
B4
182 216 228
4.5.2)
Job
1
2
3
4
A
4
2
6
3
Machine
B
2
3
5
4
Job
1
2
3
4
C
6
7
6
8
min A = 2 max B = 5 min C = 6
min C > max B = 6 > 5
A2
A1
0
5
A3
B1
11
B’
8
10
11
12
2-1-4-3
A4
B2
A’
6
5
11
7
15
18
B4
B3
26
29 33
44
4.5.3)
Job 1
Machine
A
B
C
D
Job 2
Machine
Time
A
2
B
4
C
3
D
5
Time
5
4
6
3
14
12
D
10
8
C
6
B
4
2
A
2
4
6
8
10 12 14 16 18
Total time = 18 + 5 = 23
Total time = 18 + 2
= 20
A1
A2
0
B1
B2
2
C2
6
7
C1
D2
D2
10 11
16 17
20
Sequence for Multiple Machines
Different sequence
14
12
Total time = 16 + 4 = 20
C
10
Total time = 14 + (3 + 4)
= 23
B
8
6
D
4
A
2
2
4
6
8
10
A
B
C
D
12
14
16
A1
A2
B1
B2
C2
C1
D2
5
10
D1
15
20
(a)
CT 
OT
60 min/ hr

 1.80 min/ unit
D 33.5units / hr
t  1.4  0.5  0.6  0.7  0.8  0.5 1.0  0.5  6
t

N

CT
6 min
 3.33  4
1.8 min
Idle time
per cycle =
1.2
Workstation
Time
remaining
Eligible task
Assigned
Task
Revised Time
Remaining
Station Idle
Time
1
1.8
A
A
0.4
0.4
2
1.8
B
E
B
E
1.3
0.5
0.5
D
G
D
G
1.1
0.1
0.1
C
F
G
C
F
G
1.2
0.7
0.2
0.2
3
4
1.8
1.8
Percent idle time =
x 100%
1.2
100%
4(1.8)
 16.7%

Efficiency  100%  %
idle
time
 100%  16.7%
 83.3%
(b)
Task
Positional Weight
Task
Positional Weight
A
6
A
1.4
B
4.6
B
0.5
C
1.6
E
0.8
D
2.2
D
0.7
E
2.3
C
0.6
F
1.0
G
1.0
G
1.5
F
0.5
H
0.5
H
0.5
Arrange tasks in
decreasing order of
positional weight
Station
N
I
II
III
IV
t 
CT
6
 3.33  4
1.8
1.8
Idle time
Cycle Time=1.8 min
A=1.4
E=0.8
D=0.7
C=0.6
G=1.0
B=0.5
H=0.5
F=0.5
B=0.5
0.4
0.3
0.2
0.3
Idle time per
cycle = 0.4 + 0.3
+ 0.2 + 0.3 = 1.2
C  20
Task
Immediate
predecessor
Time
Positional
Weight
1
-
12
68
2
1
5
32
3
1
7
31
4
2
8
27
5
2,3
5
24
6
3
6
25
7
4,5,6
4
19
8
7
3
15
9
8
4
12
10
8
6
14
11
9,10
8
8
 t  68
Arrange task in decreasing order of
positional weight:
Task
Positional
Weight
1
68
2
32
3
31
4
27
6
25
5
24
7
19
8
15
10
14
9
12
11
8
Station
I
II
III
IV
20
Idle time
Cycle Time=20 min
Task 1=12
Task 2=5
Task 3=7
Task 4=8
Task 5=5 Task 7=4 Task 8=3
Task 11=8
Task 9=4
Task 6=6
Task 10=6
3
-1
2
8
Station
Swap Task 6 & Task 5:
I
II
III
IV
20
Idle time
Cycle Time=20 min
Task 1=12
Task 2=5
Task 3=7
Task 4=8
Task 6=6 Task 7=4 Task 8=3
Task 11=8
Task 9=4
Task 5=5
Task 10=6
3
0
1
8