Here are the sequences that would result from applying each priority rule to the jobs: FCFS: A, B, C SPT: A, C, B LPT: B, A, C EDD: B, A, C TSPT: A, C, B LS: A, B, C COVERT: B, A, C The FCFS, EDD, LS and COVERT rules all produce the same sequence of B, A, C since job B has the earliest due date. The SPT and TSPT rules sequence A first since it has the shortest processing time. Only the LPT rule sequences the jobs in the opposite order of B

1. Production Planning
& Control
Chapter 4
Shop Floor Planning
& Control
Chapter4 1

2. Shop Floor Planning
& Control
Once the Broadest Strategic decisions concerning
manufacturing company viz.,
a)What business it is in &
b)What territory it will serve
Are made the operations managers use the forecasts of
expected demand to select most appropriate type of factory
The choice may be one of following types
1.Job Shop
2.Flow Shop
3.Continuous production
4.Project organization for manufacturing or service
Chapter4 2

3. Shop Floor Planning & Control
The Production Environment
In f o r m a tio n E x c h a n g e B e t w e e n L e v e l 3 & 4
Chapter4 3

4. Shop Floor Planning
& Control
B u s in e s s F u n c tio n s ( S 9 5 ) Business Functions
E n t e r p r is e P e r fo r m a n c e R e p o r tin g
Level 4
D em and C ost H u m a n R e s o u rce
P la n n in g
S u p p ly C h a i n P la n n in g
A c c o u n t in g M anagem ent B u s in e s s
P la n n in g &
S a le s & M a te r ia l P ro d u c tio n W a re h o u s e M a in t e n a n c e
D is t r ib u ti o n M anagem ent P la n n in g M anagem ent M anagem ent
L o g is t ic s
D e t a il P ro c e ss R e c ip e P e r fo rm a n c e P ro c e ss
S c h e d u l in g O p t im iz a tio n M anagem ent M anagem ent A n a ly s is Level 3
M a n u f a c tu r in g
R e s o u rce P ro d u c tio n W IP M a t e r ia l P ro d u c tio n Q u a lit y
M anagem ent E x e c u tio n M anagem ent H is to r y M anagem ent
O p e r a tio n s
R e a l- T im e R e a l- T im e
E x e c u tio n SPC Level 2
P la n t F lo o r
R e a l- T im e R e a l- T im e C o n tro l
C o n tro l M o n it o r in g
Chapter4 4

5. Shop Floor Planning
& Control
Several generic tasks characterize production, the process
through which parts and materials are transformed into final
products.
These tasks include, among others, the receipt and
acknowledgment of orders, the acquisition of materials, the
performance of shop floor operations, and the generation of
information needed to support continuous improvement.
Together, these tasks (when properly done) constitute a
qualified production process.
Qualifying a production process is a demanding and important
task that requires people trained and physically qualified for a
given job, machines and
Chapter4 5

6. Shop Floor Planning
& Control
process instruments that can be guaranteed to operate
within specifications, production capacity that can match
the order demand, and the availability of production
capacity in the desired time frames.
The information-processing view of a production facility
is in essence the same as that for an individual work cell
within the facility. Both factories and work cells process
orders and turn out products.
For a factory, the order usually comes from a customer
outside the factory; for a work cell, the order comes from
inside the factory.
Chapter4 6

7. Production/shop floor
Activity Control
The MRP System specifies what products or components
are needed &When they are required.The production
activity control(PAC)directs how,when,where the
product/components should be made in order to ensure
delivery of goods as per schedule.
Priority Production activity Capacity
Control Control Control
Priority Control ensures that production activities are
carried out as per plan
Chapter4 7

8. Objectives of Production
Activity Control
To know The current status of job
To determine what should be next job to be
processed & which work center
To ensure correct quantity of materials are in
right place
To minimize work in process inventory
To minimize setup costs
To maintain control of operations by monitoring
job status
To maximize operational efficiency
Chapter4 8

9. Operations Planning
& Scheduling
In this context the scheduling process
centers around:
1. Time to do the work,
2. The department which will perform the
work,
3. The resources to be applied,
4. Statusing work progress versus work
scheduled, and
5. Monitoring and reporting
Chapter4 9

10. Operations Planning
& Scheduling
Scheduling is an important tool for manufacturing and
engineering, where it can have a major impact on the
productivity of a process.
In manufacturing, the purpose of scheduling is to
minimize the production time and costs, by telling a
production facility what to make, when, with which staff,
and on which equipment. Production scheduling aims to
maximize the efficiency of the operation and reduce costs.
Production scheduling tools greatly outperform older
manual scheduling methods. This provides the production
scheduler with powerful graphical interfaces which can be
used to visually optimize real-time work loads in various
stages of the production,
Chapter4 10

11. Operations Planning
& Scheduling
and pattern recognition allows the software to
automatically create scheduling opportunities which might
not be apparent without this view into the data.
For example, an airline might wish to minimize the
number of airport gates required for its aircraft, in order to
reduce costs, and scheduling software can allow the
planners to see how this can be done, by analyzing time
tables, aircraft usage, or the flow of passengers .
Companies use backward and forward scheduling to
allocate plant and machinery resources, plan human
resources, plan production processes and purchase
materials.
Chapter4 11

12. Operations Planning
& Scheduling
Forward scheduling is planning the tasks from the date
resources become available to determine the shipping date
or the due date.
Forward scheduling is also done if no product is
available on the material availability date calculated by
backward scheduling. The system does an availability
check to determine the first possible date when product
will be available. This new material availability date
forms the starting point for scheduling the remaining
activities. The loading time, pick/pack time, transit time,
and transportation lead time are added to the new material
availability date to calculate the confirmed delivery
Chapter4 12

13. Operations Planning
& Scheduling
Backward scheduling is planning the tasks from the
due date or required-by date to determine the start date
and/or any changes in capacity required.
Backward scheduling is the calculation of deadline
dates: the arrival time at the customer site is calculated as
the earliest possible goods receipt time at the customers
unloading point on the requested delivery date.
All four of the delivery and transportation scheduling lead
times are subtracted from the customer’s requested
delivery date to determine if this date can be met.
The transit time, loading time, and pick/pack time are
subtracted from the customer’s requested delivery date to
calculate the required Chapter4
material availability date. 13

14. Operations Planning
& Scheduling
The system calculates backward scheduling as follows:
Requested delivery date minus transit time = Goods issue
date
Goods issue date minus loading time = Loading date
Loading date minus transportation lead time =
Transportation scheduling date
Loading date minus pick/pack time = Material availability
date
By default, the system will calculate delivery dates the
closest day, taking into consideration the working days of
the shipping point and a rounding profile. In this case the
system assumes a 24 hour work day and lead times can be
entered in days up to 2 decimal points. This is referred to as
Chapter4 14
daily scheduling.

15. Operations Planning
& Scheduling
Precise scheduling calculated down to the day,
hour and minute is supported. This allows the
scheduling of a delivery within a single day. It is
activated by maintaining the working hours for a
particular shipping point.
Backward scheduling is always carried out
first. If the material availability date or
transportation scheduling date is calculated to be in
the past, the system must then use forward
scheduling.
Chapter4 15

16. Operations Planning
& Scheduling
Forward scheduling is also done if no product is
available on the material availability date calculated by
backward scheduling.
The system does an availability check to determine the
first possible date when product will be available. This new
material availability date forms the starting point for
scheduling the remaining activities.
The loading time, pick/pack time, transit time, and
transportation lead time are added to the new material
availability date to calculate the confirmed delivery
date.
Chapter4 16

17. Operations Planning
& Scheduling
Order
Release Dept.X Dept.Y Dept.Z
Date L.T =4WK L.T =5WK L.T =3WK
Operation A Operation B Operation C
Start Finish Start Finish Start Finish Completion
Date
Total manufacturing lead time
Example of forward scheduling
Chapter4 17

18. Operations Planning
& Scheduling
Dept.X Dept.Y Dept.Z
L.T =4WK L.T =5WK L.T =3WK
Operation A Operation B Operation C
Today’s Start Finish Start Finish Start Finish Completion
Date Date
Total manufacturing lead time
Example of backward scheduling
Chapter4 18

19. Stages in Scheduling
Scheduling is performed in two stages
1.Loading: is the process of determine which
work center receives which job.It involves
assessing a jog or task,machine/worker, raw
material availability
2.Dispatching:is sequencing and selecting the
jobs waiting at work center(determining which
job to be done next).It lists all jobs waiting at
work center & arrange as per priority
Chapter4 19

20. Stages in Scheduling
Finite loading :Start with specific Capacity for each
work centers and a list of jobs processed at the work
centers(sequencing)
The work center capacity is allocated to job s by
simulating job starting times and completion times
100Hr/Week
Load in
Standard 50
Hours
0 2 4 Chapter4
6 8 10 20
Period Weeks

21. Stages in Scheduling
Infinite loading :The process of loading work centers
with all jobs without regard to the actual capacity
available at he work center
200
Work center
Capacity 100Hr/Week
100
Load in
Standard
Hours 50
0 2 4 Chapter4
6 8 10 21
Period Weeks

22. Operations Planning
& Scheduling
The benefits of production scheduling include:
Process change-over reduction
Inventory reduction, leveling
Reduced scheduling effort
Increased production efficiency ,Labor load
leveling
Accurate delivery date quotes ,Real time
information
Chapter4 22

23. Gantt Chart
Planning and scheduling complex projects
Gantt Charts are useful tools for analyzing and
planning complex projects. They:
•Help you to plan out the tasks that need to be
completed
•Give you a basis for scheduling when these tasks
will be carried out
•Allow you to plan the allocation of resources
needed to complete the project, and
Help you to work out the critical path for a project
where you must complete it by a particular date.
Chapter4 23

24. Gantt Chart
When a project is under way, Gantt Charts help you to
monitor whether the project is on schedule. If it is not, it
allows you to pinpoint the remedial action necessary to put it
back on schedule.
Sequential and parallel activities:
An essential concept behind project planning (and Critical
Path Analysis) is that some activities are dependent on other
activities being completed first. As a shallow example, it is
not a good idea to start building a bridge before you have
designed it!
These dependent activities need to be completed in a
sequence, with each stage being more-or-less completed
before the next activity can begin. We can call dependent
activities 'sequential' or 'linear'.
Chapter4 24

25. Gantt Chart
Other activities are not dependent on completion of
any other tasks. These may be done at any time
before or after a particular stage is reached. These
are nondependent or 'parallel' tasks.
Drawing a Gantt Chart
To draw up a Gantt diagram (Gant diagram), follow
these steps:
1. List all activities in the plan
For each task, show the earliest start date,
estimated length of time it will take and whether it
is parallel or sequential. If tasks are sequential,
show which stages they depend on.
Chapter4 25

26. Gantt Chart
Gantt Chart
Chapter4 26

27. Gantt Chart
Gantt Chart
Chapter4 27

28. What Is Sequencing?
•Predictable, consistent ordering and delivery of
learning activities, in an instructionally meaningful
manner, regardless of delivery environment
•Designers/authors specify sequencing behaviors
at design/authoring time.
•Activities are sequenced at time of delivery
depending on specified behaviors and the learner’s
actions.
•Sequencing behaviors are external from the
content to enable greater degree of granularity and
reuse
Chapter4 28

29. Priority Sequencing
Priority sequencing in a real world job shop
An engineering industry job shop that
manufactures 19 types of products against orders is
described, with the objective of evaluating a new
sequencing rule based on the monetary value of the
job and its processing time, the highest value time
ratio rule.
Analysis of this rule has not appeared in
literature on job shop scheduling. For purposes of
comparison, based on the performances in past
studies, the following four sequencing rules are
included: Chapter4 29

30. Priority Sequencing
(These rules were evaluated on the following
criteria: work-in- process inventory in monetary
terms, and delivery performance, and a
combination of the two measures.
Results of the study show that the highest value
time ratio rule results in minimum work-in-process
inventory. The shortest processing time rule results
in maximum delivery performance; the same rule
is superior on combined measure of performance.
Chapter4 30

31. Priority Sequencing(Rules)
1) Shortest processing time (SPT) :
The jobs are having the shortest processing time is
given the highest priority.
2) Least Slack per operation (LS): the highest
priority given to job which has least slack
3) First-Cone, first-Served (FCFS):Jobs are selected
in the same sequence as they come
4) Earliest Due date (EDD):Jobs are processed
according to due dates
Chapter4 31

32. Priority Sequencing(Rules)
5) Longest Processing Time(LPT):the highest priority given
to job which has longest processing time
6) Shortest Processing Time(LPT) :Jobs are selected as per
Shortest Processing Time .
7) Random Selection: Jobs are selected randomly
8) Cost over time(COVERT):Uses the ratio of expected
delay cost over the processing cost
9) Preferred Customer order(PCO):As per customer order
10 Least Change Over cost ((LCOC):The sequencing of
jobs is done by analyzing the total cost of making
all the machine changeover between jobs
Chapter4 32

33. Illustration for Sequencing(Rules)
XYZ company has received the following jobs at a
work center to be processed.
The processing time (in days),arrival date and due
date are given Determine the sequence in which jobs
should be processed according to priority rules
FCFS,SPT,LPT,EDD,TSPT,LS & COVERT
Assume today is is day 100 & jobs can not be
delayed more than 60 days
Assume expected cost of delay is Rs 10/day
Chapter4 33

34. Illustration for Sequencing(Rules)
Job Processing Arrival Due Date
Time(Days) Date
A 15 95 185
B 20 110 200
C 10 112 175
D 30 125 235
E 25 125 180
F 18 130 220
Chapter4 34

35. Illustration for Sequencing(Rules)
Priority Rules: FCFS Sequence A B C D E F
Priority Rules: SPT Sequence C A FB E D
Priority Rules: LPT Sequence D E B F C A
Priority Rules: EDD Sequence C E A B F D
For TSTP rule ,it is specified that the jobs can not
be delayed by more than 60 days if we apply SPT
Rule .If none of the jobs violets the constraints ,the
sequence will be identical for SPT and TSPT
rule.To examine this ,we should know the wait time
for the jobs as per SPT rule ,which is determined as
per next slide
Chapter4 35

36. Illustration for Sequencing(Rules)
SPT Rule
Job Processing Arrival Start Date Wait Time
Time(Days) Date Days Days days
C 10 112 112 Nil
A 15 95 122 27
F 18 130 135 7
B 20 110 155 45
E 25 125 175 50
D 30 125 200 75
Chapter4 36

37. Illustration for Sequencing(Rules)
Least Slack Rule
Job Processing Available Slack Days Sequence
(Rank)
Time(Days) time Days
A 15 90(185-95) 75 5
B 20 90 70 3
C 10 63 53 2
D 30 110 80 6
E 25 55 30 1
F 18 90 72 4
JOB sequence E C B F A D
Chapter4 37

38. Dynamic Sequencing Rules
Dynamic slack(DS) rule:When the least slack rule is
used repeatedly at each machine/work center for
sequencing the jobs ,it is known as dynamic slack rule.
Dynamic slack per remaining operation(DS/RO) rule:
In this rule ,the ratio of total slack time available for the
job to the number of operations remaining including the
current operation is obtained.
Total Slack time
DS/RO ratio =
Total number of operations remaining
(including the current operation is obtained.)
Job with Smallest DS/RO ratio is scheduled first
Chapter4 38

39. Dynamic Sequencing Rules
Critical Ratio rule:
The critical ratio rule is designed to give priority to
jobs that have most urgently needed work to meet the
shipping schedule.
Due date-Date now
C/R ratio =
Days required to complete the
= D.D-D.N
L.T.R
Chapter4 39

40. Dynamic Sequencing Rules Illustration
ABC Company has 6 jobs arriving at random at several work
stations & passing through them,requiring different processing
time.For particular work station the data is given below.
Job Arrival Processi Due No of Time for Due
Time ng Time subsequent subsequent time
Time operations operations Hrs
Hrs Hrs
1 12 2 4,00PM 2 12 20
2 2 PM 2 5PM Nil Nil 12
3 3,30PM 4 8PM 4 30 50
4 3,30PM 3 7PM 3 10 25
5 4,30PM 6 12PM 2 20 45
6 6PM 4 8Am 4 25 38
Next day
Chapter4 40

41. Dynamic Sequencing Rules Illustration
Ds /Ro Rule
Job Arrival Processing Due Time Dynamic Slack
Time Time Hrs Available time-Total
Hrs operation time (Hrs)
1 12 2+12 20 20-14=6
2 14 2+nil 12 12-2=10
3 15.5 4+30 50 50-35=16
4 15.5 3=10 25 25-13=12
5 16.5 6+20 45 45-2619
6 18 4+25 38 38-29=9
Chapter4 41

42. Dynamic Sequencing Rules Illustration
Calculation of Ds /Ro Ratio
Job Dynamic Remaining Operations(RO) DS/RO Ratio
Slack(DS)
1 6 2+1=3 6/3=2
2 10 Nil+1=1 10/1=10
3 16 4+1=5 16/5=3.33
4 12 3+1=4 12/4=3
5 19 2+1=3 19/3=6.33
6 9 4+1=5 9/5=1.8
Sequence As per Ds/Ro rule: 6,1,4,3,5,2
Chapter4 42

43. Dynamic Sequencing Rules Illustration
Critical Ratio Rule
Job Processing Time Available time Critical ratio
Hrs (Hrs.)
1 2 4 4/2=2
2 2 3 3/2=1.5
3 4 4.5 4.5/4=1.125
4 3 3.5 3.5/3=1.166
5 6 7.5 7.5/6=1.25
6 4 14 14/4=3.5
Sequence As per CR rule: 3,4,5,2,1,6
Chapter4 43

44. Johnson’s Rule or algorithm
FLOW SHOP SCHEDULING
n JOBS (n JOBS, m MACHINES)
BANK OF m MACHINES (SERIES)
3
1
2 M1 M2 Mm
4 n
Chapter4 44

45. FLOW SHOPS
PRODUCTION SYSTEMS FOR WHICH:
A NUMBER OF OPERATIONS HAVE TO BE DONE ON
EVERY JOB.
THESE OPERATIONS HAVE TO BE DONE ON ALL JOBS IN
THE SAME ORDER, i.e., THE JOBS HAVE TO FOLLOW
THESAME ROUTE.
THE MACHINES ARE ASSUMED TO BE SET UP IN SERIES.
COMMON ASSUMPTIONS:
UNLIMITED STORAGE OR BUFFER CAPACITIES IN
BETWEEN SUCCESIVE MACHINES (NO BLOCKING).
A JOB HAS TO BE PROCCESSED AT EACH STAGE ON
ONLY ONE OF THE MACHINES (NO PARALLEL
MACHINES).
Chapter4 45

46. PERMUTATION FLOW SHOPS
FLOW SHOPS IN WHICH THE SAME SEQUENCE OR
PERMUTATION OF JOBS IS MAINTAINED THROUGHOUT:
THEY DO NOT ALLOW SEQUENCE CHANGES BETWEEN
MACHINES.
PRINCIPLE FOR Fm||Cmax:
THERE ALWAYS EXISTS AN OPTIMAL SCHEDULE
WITHOUT SEQUENCE CHANGES BETWEEN THE FIRST
TWO MACHINES AND BETWEEN THE LAST TWO
MACHINES.
THERE ARE OPTIMAL SCHEDULES FOR F2||Cmax AND
F3||Cmax THAT DO NOT REQUIRE SEQUENCE CHANGES
BETWEEN MACHINES.
Chapter4 46

47. JOHNSON’S F2||Cmax PROBLEM
FLOW SHOP WITH TWO MACHINES IN SERIES WITH
UNLIMITED STORAGE IN BETWEEN THE TWO
MACHINES.
THERE ARE n JOBS AND THE PROCESSING TIME OF JOB j
ON MACHINE 1 IS p1j AND THE PROCESSING TIME ON
MACHINE 2 IS p2j.
THE RULE THAT MINIMIZES THE MAKESPAN IS
COMMONLY REFERRED TO AS JOHNSON’S RULE.
Chapter4 47

48. JOHNSON’S PRINCIPLE
ANY SPT(1)-LPT(2) SCHEDULE IS OPTIMAL FOR
Fm||Cmax.
(THE SPT(1)-LPT(2) SCHEDULES ARE NOT THE ONLY
SCHEDULES THAT ARE OPTIMAL. THE CLASS OF
OPTIMAL SCHEDULES APPEARS TO BE HARD TO
CHARACTERIZE AND DATA DEPENDENT).
Chapter4 48

49. DESCRIPTION OF JOHNSON’S ALGORITHM
1. IDENTIFY THE JOB WITH THE SMALLEST
PROCESSING TIME (ON EITHER MACHINE).
2. IF THE SMALLEST PROCESSING TIME INVOLVES:
• MACHINE 1, SCHEDULE THE JOB AT THE BEGINNING
OF THE SCHEDULE.
• MACHINE 2, SCHEDULE THE JOB TOWARD THE END
OF THE SCHEDULE.
3. IF THERE IS SOME UNSCHEDULED JOB, GO TO 1.
OTHERWISE STOP.
Chapter4 49

50. EXAMPLE
CONSIDER THE FOLLOWING INSTANCE OF THE
JOHNSON’S (Fm||Cmax) PROBLEM:
JOB 1 2 3 4 5
p1j 4 4 10 6 2
p2j 5 1 4 10 3
SEQUENCE:
Chapter4 50

51. SEQUENCE:
EXAMPLE: SCHEDULE 51432
JOB 1 2 3 4 5
p1j 4 4 10 6 2
p2j 5 1 4 10 3
M1
M2
t
Chapter4 51

52. A BOUND ON THE MAKESPAN
FOR JOHNSON’S PROBLEM:
⎧⎛ n ⎞⎛ n ⎞⎫
⎪⎜ ⎟⎜ ⎟⎪
Cmax (OPT ) ≥ max ⎨⎜ min p 2 j + ∑ p1 j ⎟, ⎜ min p1 j + ∑ p 2 j ⎟ ⎬
⎪⎜ j=1,..,n
⎩⎝ j=1
⎟ ⎜ j=1,..,n
⎠⎝ j=1
⎟⎪
⎠⎭
Chapter4 52

53. JOHNSON’S ALGORITHM
LET U = {1, 2,..., n} BE THE SET OF UNSCHEDULED
JOBS.
k =1,
l = n,
Ji = 0, i IDENTIFICATION OF SMALLEST PROCESSING TIME
STEP 1:
= 1, 2, ..., n.
IF U = ∅, GO TO STEP 4.
LET
⎧ ⎫
p i* j* = min ⎨ min p1 j, min p 2 j ⎬
⎩ j=1,..,n j=1,..,n ⎭
IF i* = 1 GO TO STEP 2; OTHERWISE GO TO STEP 3.
Chapter4 53

54. JOHNSON’S ALGORITHM
(CONTINUED)
STEP 2: SCHEDULING A JOB ON EARLIEST POSITION
• SCHEDULE JOB j* IN THE EARLIEST AVAILABLE POSITION: Jk
= j*.
• UPDATE k: k = k + 1.
• REMOVE THE JOB FROM THE SCHEDULABLE SET, U = U – {j*}.
• GO TO STEP 1.
STEP 3: SCHEDULING A JOB ON LATEST POSITION
• SCHEDULE JOB j* IN THE EARLIEST AVAILABLE POSITION: Jl
= j*.
• UPDATE l: l = l - 1.
• REMOVE THE JOB FROM THE SCHEDULABLE SET, U = U – {j*}.
• GO TO STEP 1.
Chapter4 54

55. JOHNSON’S ALGORITHM
(CONTINUED)
STEP 4: SEQUENCE OF JOBS
THE SEQUENCE OF JOBS IS GIVEN BY Ji,
WITH J1 THE FIRST JOB, AND SO FORTH.
Chapter4 55

56. Fm||Cmax
Fm||Cmax IS A STRONGLY NP-HARD
PROBLEM.
AN EXTENSION OF JOHNSON’S ALGORITHM
YIELDS AN OPTIMAL SOLUTION FOR THE
F3||Cmax PROBLEM WHEN THE MIDDLE
MACHINE IS DOMINATED BY EITHER THE
FIRST OR THIRD MACHINE.
Chapter4 56

57. MACHINE DOMINANCE: F3||Cmax
A MACHINE IS DOMINATED WHEN ITS LARGEST
PROCESSING TIME IS NO LARGER THAN THE SMALLEST
PROCESSING TIME ON ANOTHER MACHINE.
FOR F3||Cmax PROBLEM:
⎧ ⎫
p 2 j ≤ max ⎨min p1 j , min p 3 j ⎬
⎩ j ⎭
WHICH IMPLIES THAT MACHINE 2 (DOMINATED MACHINE)
CAN NEVER CAUSE A DELAY IN THE SCHEDULE.
Chapter4 57

58. JOHNSON’S ALGORITHM FOR 3 MACHINES
FOR F3||Cmax, WHENEVER MACHINE 2 IS DOMINATED, i.e.,
min{p1 j } ≥ max{p 2 j } OR min{p 3 j } ≥ max{p 2 j }
j j j j
SOLVING AN EQUIVALENT TWO-MACHINE PROBLEM WITH
PROCESSING TIMES:
p’1j = p1j + p2j AND p’2j = p2j + p3j
GIVES THE OPTIMAL MAKESPAN SEQUENCE TO THE
DOMINATED THREE-MACHINE PROBLEM.
Chapter4 58

59. EXAMPLE: F3||Cmax
CONSIDER F3||ΣCmax WITH THE FOLLOWING JOBS:
JOB 1 2 3 4 5
p1j 4 9 8 6 5
p2j 5 6 2 3 4
p3j 8 10 6 7 11
min{p1 j } = max{p 2 j } = min{p 3 j } =
j j j
Chapter4 59

60. EXAMPLE: PROCESSING TIMES, DUMMY
MACHINES
JOB 1 2 3 4 5
p1j 4 9 8 6 5
p2j 5 6 2 3 4
p3j 8 10 6 7 11
p'1j
p'2j
SEQUENCE:
Chapter4 60

61. SEQUENCE:
EXAMPLE: SCHEDULE 1 4 5 2 3
JOB 1 2 3 4 5
p1j 4 9 8 6 5
p2j 5 6 2 3 4
p3j 8 10 6 7 11
M1
M2
M3
t
Chapter4 61

62. Illustration for Johnson's rule
Estimated processing time (Hours)
Job Work Center Work Center
WC1 WC2
A 2 1
B 4 2.25
C 0.75 2.5
D 1.5 3
E 2 4
F 2 3.5
Chapter4 62

63. Illustration for Johnson's rule
Applying Johnson's rule Job C has the shortest
processing time in either work center (I.e.75hour),
assign job C as as the first job in the sequence.
Next Job A has Shortest time (I.e 1 Hr) assign
Job A as last job in sequence and cross out time
for Job
The Next smallest time is 1.5 hours for job D on
WC1
The next smallest time is 2 hr for job E and F on
WC1,Since there is tie in between jobs E,F choose
the job with smaller subscript I.e job E
Chapter4 63

64. Illustration for Johnson's rule
Assign the job E as the next job after job D
from beginning of the sequence.
The next Job after job E is job F The last job B
has to be accommodated in the gap between job F
and the last job in the sequence i.e Job A
Hence the sequence is C,D,E,F,B,A
Chapter4 64

65. Batch Scheduling
Batch production falls between job shop
production and continuous.
In batch production system, the output is can be
stored as inventory for further processing or as
finished products and can be produced in
substantial volume, even-though the volume may
not justify continuous production.
In these situations, it is necessary to determine
the lot-size for a batch to be produced at one time
in addition to scheduling the batch on the facilities.
Chapter4 65

66. Batch Scheduling
Examples of such production are production of
pharmaceutical products, paints etc., Decision to be taken
by Operations manager are
(i)the lot size; and
(ii)The scheduling decision regarding when to begin the
processing of the batch.
A key-off in the determination of the lot size for an item
is between set-up costs and inventory carrying costs.
Another important consideration is the requirement to
produce a feasible schedule that meets the demand for all
items.
Chapter4 66

67. Batch Scheduling
For example, if set-up costs are low as
compared to inventory carrying costs, it may be
advantageous to go for small lot sizes
But it may not be possible to produce the
required quantities of all items within the
specified time period if these small lot sizes are
employed. This will happen if much of the time is
consumed for machine set-ups thereby reducing
the available production time. To overcome this
problem, larger lot sizes may have to be employed
which will result in higher inventory carrying
costs. Chapter4 67

68. Batch Scheduling
Hence, it is necessary to compute economic lot
sizes while maintaining feasibility in scheduling
batches of such lot sizes for the items to be
produced.
Two types of costs associated with lot
manufacture are:
(a)Set up costs i.e. costs/unit which decrease
with batch size.
(b)Inventory carrying cost which increases with
batch size.
Chapter4 68

69. Batch Scheduling
Set up cost includes:
(i)Cost of releasing work orders, shop orders,
stores requisitions, tool requisitions etc.
(ii)Cost of first off inspection, cost of rejections
till machine set up is ready for production run.
(iii)Machine set-up cost for mounting
accessories, tools, jigs and fixtures on the
machine. Chapter4 69

70. Batch Scheduling
Inventory carrying costs include:
(i)Cost of working capital tied up in average
inventory.
(ii)Cost of handling and storing materials (i.e.
parts produced)
(iii)Insurance charges and taxes.
(iv)Cost of spoilage and obsolescence etc.
Chapter4 70

71. Case1 Batch Scheduling
Instantaneous Supply with no simultaneous consumption
Let A= Annual demand for an Item
S= Set up cost per setup
I =Inventory Carrying cost
C=Cost per unit of item produced
Q=Economical batch Quantity
The Solution:
No of batches per year =A/Q
Setup cost per year =A/Q* S
Average inventory held=Q/2
Chapter4 71

72. Batch Scheduling
Instantaneous Supply with no simultaneous consumption
Inventory carrying cost per year=Q/2*C*I
Total Cost per year =Set up cost per year +
inventory carrying cost per year.
T.C=A/Q*S+Q/2CI
For minimizing the total cost
d T.C = -(AS/Q*Q)+(CI/2)
dQ
For T.C to be minimum d T.C =0
dQ
Or –AS/Q8Q+CI/2=0
Or CI/2=AS/Q*Q, Hence Q= 2AS
Chapter4 ٧ CI 72

73. Batch Scheduling
Instantaneous Supply with no simultaneous consumption
Maximum Inventory
Quantity
Average Inventory
ERQ
Q
Time
Chapter4 73

74. Case 2 Batch Scheduling
Instantaneous Supply with simultaneous consumption
Let A= Annual demand for an Item
S= Set up cost per setup
I =Inventory Carrying cost
C=Cost per unit of item produced
Q1=Economical batch Quantity
d=demand or consumption rate
p=Production rate
The Solution:
No of Set ups per year =A/Q1
Setup cost per year =A/Q1* S
Average inventory held = Maximum inventory/2
Chapter4 74

75. Case 2 Batch Scheduling
Instantaneous Supply with simultaneous consumption
To calculate the inventory built up let us assume
that production period is t1( weeks )and
consumption only period is t2 (say weeks)
Quantity produced in time t1=Q1(at the rate of
“p”per week)
Q1=p*t or t1=Q1/p
Consumption during period t1=dt1
(at the rate of “d”per week)=d*Q1/p or =Q1*d/p
Chapter4 75

76. Case 2 Batch Scheduling
Instantaneous Supply with simultaneous consumption
(Maximum inventory built up =Qty produced-Qty Consumed
During period t1)
=Q1-Q1*d/P =Q1(1-d/P)
Average inventory = Q1/2(1-d/P)
Inventory carrying cost per year= Q1/2(1-d/P)*CI
Total cost per year = Set cost per year+Inventory carrying cost
per year
Tc=A/q1*s+Q1/2(1-d/P)*CI
Chapter4 76

77. Case 2 Batch Scheduling
Instantaneous Supply with simultaneous consumption
For the total cost Tc to be minimum dTc
=0
dQ1
I.e –A.S/Q1*Q1+ (1-d/P)*CI = 0
22
Solving for Q1
2As
Q1=
CI(1-d/p)
Chapter4 77

78. Batch Scheduling
Instantaneous Supply with simultaneous consumption
Consumption only
Production period (t1)
period (t2)
Maximum Inventory
Quantity
Average Inventory
ERQ
Q
Time
Chapter4 78

79. Run out or Run out time method
This method attempts to use the total production
capacity available in each time period to produce just
enough of each product variety so that if production is
stopped ,the finished goods inventory for each product
would be depleted or would run out at the same point
of time.
The run out time is expressed as the ratio of the
current inventory to demand forecast for the period.
Run out Time = Current inventory of Item X
Demand per period for item X
Chapter4 79

80. Aggregate Run out or Run out time method
In case of above , run out time method,it is observed
that ,there is shortage of required capacity .If it is
necessary to produce all items required for a product
,shortage of any single item should not occur In such
case s the aggregate run out method is used.
(Machine hours inventory +(Total Available
for all items) Machine hours)
AROT =
Machine Hours requirements forecasted for all the items
Chapter4 80

81. Scheduling & Controlling Production For
Delivery Schedules - Line of Balance method
Line of balance technique has been used in
production scheduling and control to determine ,at a
view date,not only how many( quantity )of item
should have been completed by that date,but also how
many should have passed through previous operation
stages by that time so as to ensure the completion of
the required delivery schedule.
LOB is a charting and computational technique for
monitoring and controlling products and services that
are made to meet specific delivery schedule
Chapter4 81

82. Line of Balance method -Example
XYZ company has received orders to deliver a a
product for which the operations program and
delivery schedule s are given below
Week No Qty of end product to be delivered
1 5
2 10
3 10
4 10
5 15
Chapter4 82

83. Line of Balance method -Example
Purchased part Fabrication
Assembly Delivery of End product
Develop a LOB chart & determine the Quantities that
should have passed through the upstream processing steps
during the review point at the end of 2nd week
1 2 3 Item A 5
Item B 4
10 9 8 7 6 5Chapter4 4 3 2 1 83

84. Line of Balance method Solution
Method: the five steps required to be followed in LOB
techniques:
Preparation of operation program or assembly chart
Preparation of Cumulative completion/delivery
schedule
Construction of LOB chart
Construction of program progress
Analysis of progress and corrective action
Chapter4 84

85. Line of Balance method -Solution
Preparation of operation program or assembly chart
The operation program shows the “lead time”for each
operation.The lead time is shown as length of time prior to
completion of final Completion operation.
Assembly Delivery of End product
Fabrication
Purchased part
1 2 3 Item A 5
Item B 4
10 9 8 7 6 5Chapter4 4 3 2 1 85

86. Line of Balance method Solution
Preparation of operation program or assembly chart
The delivery date for the finished product (end
item) is zero & the time scale indicating “lead
time”runs from left to right.
The operation program indicates that Purchased part
A must be combined with B in operation stage 4 three
days before completion of end item.
Item B ,prior to combination has undergone a
conversion operation which has to be completed five
days before the completion of end item,The purchased
part for item B must be available ten days prior to
delivery date for end item which means longest lead
time is ten days Chapter4 86

87. Line of Balance method Solution
Stage 2 : Preparation of Cumulative completion
/delivery schedule
The quantities of end item to be completed week by
week and cumulatively indicated in table as shown
below
Week No Qty of end item to be Cumulative Qty to be
completed Nos. completed Nos.
1 5 5
2 10 15
3 10 25
4 10 35
5 15 50
Chapter4 87

88. Line of Balance method Solution
Stage 3 : Construction of line of balance chart
The line of balance shows the quantity of item
that should have completed at each operation
stage in a particular week at which progress will
be reviewed.
So as to meet the delivery schedule
The line of balance chart can be constructed as
shown in next slide
Chapter4 88

89. Line of Balance method Solution
50
45 Line of Balance
40
35 35 Nos.
30 25 Nos.
25 21 Nos.
20 5 Nos.
15
10
5
1 2 3 4 5 1 2 3 4 5
Chapter4 89

90. Line of Balance method Solution
Stage 4 : Construction of progress chart
50 The progress chart for review week is shown
45 below.
40 Excess production
35 35 Nos.
30 Shortage production
25
21 Nos. Exact production
20 15 Nos.
15
10
5
1 2 3Chapter4 4 5 90

91. Line of Balance method Solution
Stage 5 : Analysis of progress & corrective action
delivery schedule
By referring to programme process chart which is
prepared every week the difference between the
desired production (as indicated by line of
balance)for the review week can be compared with
the actual production achieved at the end of the
review week.
The excess production or shortage production can
be found out & appropriate actions taken to bring
production in line
Chapter4 91

92. Line of Balancing Methods
The various Line balancing methods are:
Heuristic method
Linear programming
Dynamic programming
Computer based sampling technique
A heuristic is a method for helping in solving of a problem,
commonly informal. It is particularly used for a method that
often rapidly leads to a solution that is usually reasonably
close to the best possible answer. Heuristics are "rules of
thumb, educated guesses, intuitive judgments or simply
common sense. In more precise terms, heuristics stand for
strategies using readily accessible though loosely applicable
information to control problem-solving in human beings and
machine
Chapter4 92

93. Line of Balancing Methods
Heuristic method (Illustration)
The table below shows the number of work stations
(N),Cycle time( C) & daily production for a product
No of Cycle Time Daily production
Workstations N C Sec (8 Hrs shift)
1 120 240
2 60 480
3 40 720
4 30 960
5 24 1200
6 20 1440
Chapter4 93

94. Line of Balancing Methods
Heuristic method (Illustration)
It is desired to have two assembly lines each
producing 720 units per day,the cycle time will be
40 seconds & there will be 3 work stations in each
assembly line
The precedence diagram is shown in next slide.
The activity time in seconds are given in bracket
for each of twenty operations involved
Assuming that activities may be combined within
given zone ,without violating the precedence
relationship ,assign the activities into three
workstations.This can be done by trial & error basis
by adding activity time
Chapter4 94

95. Line of Balancing Methods
12
4 3 10
5 10
8
1 5 18
5 2
6 11 3 13
15 19
2
7 12
7
5
14 4 4 5
8
2 13 16 20
9 7
6 6
17 9
14
3
Chapter4 95

96. Line of Balancing Methods
Heuristic method (Illustration)
Activities Total time Sec
Work station 1 1,7,2,8,9,11 5+5+2+14+5+7+2=40
Work station 2 4,5,10,12,13,3 3+8+12+7+4+6=40
Work station 3 14,15,16,17,18, 6+3+4+9+10+3+5=40
19,20
A perfect balance is obtained since all work stations have
exact the same workload of 40 sec.
Chapter4 96

97. Scheduling Services
Services are those economic activities in which the
primary product is nether a product nor a
construction.
Some of the examples of services are
Transportation
Utilities
Communication
Wholesale trade
Retail trade
Real estate
Hotel & restaurant
Hospital services
Chapter4 97

98. Scheduling Services
Four approaches that are usually used are
Use of waiting lines
Use of appointment schedules
Personal schedules
Emergency services
Chapter4 98

99. Shop Floor Planning
& Control
End Of
Chapter 4
Chapter4 99

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