unit-4-4Operations Planning and Control-Scheduling.ppt

Operations management

1. Unit –4
-4.4 Operations Scheduling

2. Scheduling
Scheduling deals with the timing of
operations
It is establishing the timing of the use
of equipment, facilities and human
activities in an organization
2

3. Scheduling
It is the last
stage of planning
before
production
occurs
Capacity Planning
(long term; years)
Changes in facilities
Changes in equipment etc
Aggregate Planning
(intermediate term; quarterly or monthly)
Facility utilization
Personnel changes
Subcontracting etc
Master Schedule
(intermediate term; weekly)
Material Requirement Planning-MRP
Disaggregating aggregate plan
Short term Scheduling
(short term; days, hours, minutes)
Work center loading
Job sequencing
3

4. Goals of Scheduling
Efficient utilization of
staff
equipment
facilities
Minimization of
customer waiting time
inventories
processing time
4

5. Scheduling Operations
Companies differ based on product volume
and product variety which affects how
companies organizes their operations
Each kind of company operation needs
different scheduling techniques
Scheduling has specific definitions for routing,
bottleneck, due date, slack and queue
5

6. Scheduling Definitions
Routing: The operations to be performed, their
sequence, the work centers, & the time standards
Bottleneck: A resource whose capacity is less
than the demand placed on it
Due date: When the job is supposed to be
finished
Slack: The time that a job can be delayed & still
finish by its due date
Queue: A waiting line
6

7. Importance of Scheduling
Scheduling executes a company’s strategic business plan
Scheduling affects functional areas
Accounting relies on schedule information and completion of
customer orders to develop revenue projections
Marketing uses schedule effectiveness measurement to determine
whether the company is using lead times for competitive advantage
Operations uses the schedule to maintain its priorities and to
provide customer service by finishing jobs on time
7

8. Type of Scheduling
Forward scheduling
Scheduling ahead, from some point
in time
Forward scheduling starts as soon
as the requirements are known or
when a job is received
Frequently results in buildup of
work-in-process inventory
Backward scheduling
Scheduling by working
backwards from the due
date
begin scheduling the job’s
last activity so that the job
is finished on due date
Due
Date
Now
Due
Date
Now
8

9. Scheduling Operations
Scheduling tasks are largely a function of the
volume of system output
Different kinds of operations need different
scheduling techniques
Scheduling in High-Volume Operations
Scheduling in Intermediate-Volume Operations
Scheduling in Low-Volume Operations
9

10. High-Volume Operations
Also known as flow operations (flow systems)
Scheduling encompasses allocating workloads to
specific work centers and determining the
sequence in which operations are to be performed
Characterized by standardized equipment and
activities that provide high-volume standard items
Designed for high efficiency and high utilization of
labor and equipment
10

11. High-Volume Operations
Bottlenecks are easily identified
Because of the highly repetitive nature of
operations , many of the loading and sequence
decisions are determined during the design of the
operations system.
Scheduling in the high-volume operations is
typically done through line balancing
allocating the required tasks to workstations so that they satisfy technical
(sequencing) constraints and are balanced with respect to equal work
times among work stations
11

12. High-Volume Operations
The success of High-volume operations depends on
the following factors
Process and product design -(interms of cost and manufacturability)
Preventive maintenance
Rapid repair when breakdown occurs
Optimal product mixes
Minimization of quality problems
Reliability and timing of supplies
12

13. Intermediate-Volume Operations
Outputs are between standardized high-volume systems and made-
to-order job shops
Typically produce relatively low-volume standard outputs of similar
products using intermittent process
Work centers periodically shift from one job to another
The run (batch) sizeof jobs, the timingof the job, and the
sequencingof jobs are of significant concern to schedulers
The larger the run size, the fewer the number of runs needed and,
hence, the lower the annual set up costs and set up times
Set up costs are costs required to prepare equipments for job, such as
cleaning, adjusting, and changing tools and fixtures-with every
production run there are set up costs.
13

14. Low-Volume Operations
Low-volume, job shop operations, are designed for
flexibility
Use more general purpose equipment
Customized products with higher margins
The variable work-flow paths and processing time
generates queues, work-in-process inventories, and
capacity utilization concerns that can require more day-to-
day attention than in the high-or intermediate-volume
systems
Scheduling in a low-volume operations typically involves
the use of priority rules
14

15. Job-shop Scheduling
Job-shops schedulingis scheduling for low-volume
operations with many variations in requirements
Two basic issues in job shop processing
Loading
Sequencing
15

16. Loading
Assignment of jobs to processing
(work) centers
Loading techniques included
1.Infinite loading
Assigning specific jobs to work centers
without regard to the capacity of the
work center
2.Finite loading
Jobs are assigned to work center
taking into account the work
center capacity and job processing
times
loads jobs up to a predetermined
capacity level
Loading can be done using forward or backward scheduling
16

17. Gantt Charts
Gantt charts:-used as a visual aid for loading
and scheduling purpose
Load chart shows the loading and idle times of
departments, machines, or facilities
Displays relative workloads over time
Schedule chart monitors jobs in process
All Gantt charts need to be updated frequently
to account for changes
17

18. Load chart
Day
Monday Tuesday Wednesday Thursday Friday
Work
Center
Metal works
Mechanical
Electronics
Painting
Job 349
Job 349
Job 349
Job 408
Job 408
Job 408
Processing Unscheduled Center not available
Job 350
Job 349
Job 295
18

19. Schedule chart
Job
Day
1
Day
2
Day
3
Day
4
Day
5
Day
6
Day
7
Day
8
A
B
C
Now
Maintenance
Start of an
activity
End of an
activity
Scheduled
activity time
allowed
Actual work
progress
Nonproduction
time
Point in time
when chart is
reviewed
19

20. Schedule chart
Illustrate the planned schedule compared to actual performance
Brackets show when activity is scheduled to be finished. Note:
design & pilot run both finish late; feedback has not started yet.
20

21. Input/Output Control
I/O control is a capacity-control technique used to monitor
work flow at individual work centers
Monitors how well available capacity is used and provides
insight into process problems
Identifies overloading and under loading conditions
Prompts managerial action to resolve scheduling problems
Options available to operations personnel include
1.Correcting performances
2.Increasing capacity
3.Increasing or reducing input to the work center
21

22. Input/Output
Control
The backlog for each period is
determined by subtracting the
’’actual output’’ from the
’’actual input’’ and adjusting
the backlog from the previous
period by that amount
Input/output report for a work center
Input Information (in hours) Period
4 5 6 7 8
Planned Input 800 750 800 820 800
Actual Input 750 780 780 810 810
Deviation -50 30 -20 -10 10
Cumulative deviation 0 -50 -20 -40 -50 -40
Output information (in hours) Period
4 5 6 7 8
Planned output 800 800 800 800 800
Actual output 800 750 780 850 825
Deviation 0 -50 -20 50 25
Cumulative deviation 0 0 -50 -70 -20 5
Backlog (in hours) 100 50 80 80 40 25
22

23. Assignment Method
A special class of linear programming
models that assign tasks or jobs to
resources
Objective is to minimize cost or time
Only one job (or worker) is assigned
to one machine (or project)
Hungarian method is the method of
assigning jobs by a one for one matching
to identify the lowest cost solution
23

24. Assignment Method
Determine the optimal assignment of jobs to Machines for
the following data
The numbers in the body of the table represent the value
or cost associated with each job-machine combination
Machines
Job A B C
R 11 14 6
S 8 10 11
T 9 12 7
24

25. Job Assignment-Hungarian
method
1.Row Reduction:-subtract the smallest number in each row
from every number in the row. Enter the results in a new
table
2.Column Reduction:-subtract the smallest number in each
column of the new table from every number in the column.
3.Draw the minimum number of vertical and horizontal lines
necessary to cover all zeros in the table.
•If the number of lines equals either the number of rows or the number of
columns, an optimum assignment is possible. In this case proceed to
step 5
•If the number of lines is lessthan the number of rows or the number of
columns, proceed to step 4
25

26. Job Assignment-Hungarian
method
4.Subtract the smallest number not covered by a line from
all other uncovered numbers. Add the same number to
any number at the intersection of two lines. Numbers
crossed out but not at intersections covering lines carry
over unchangedto the next table. Return to step 3
5.Make the assignments. Begin with rows or columns with
only one zero. Match items that have zeros, using only one
match for each row and each column. Cross out both the
row and column for each row
26

27. Job Assignment-Hungarian
method
A B C
Job
R 11 14 6
S 8 10 11
T 9 12 7
Machines
A B C
Job
R 5 8 0
S 0 2 3
T 2 5 0
Machines
Step 1-Rows
A B C
Job
R 5 6 0
S 0 0 3
T 2 3 0
Machines
Step 2 -Columns
27

28. Assignment Example
Step 3 -Lines
A B C
Job
R 5 6 0
S 0 0 3
T 2 3 0
Machines
Because only two lines are
needed to cover all the zeros, the
solution is not optimal
i.ethe number of lines is less
than the number of rows or the
number of columns
Step 4 –Subtraction and addition
A B C
Job
R 3 4 0
S 0 0 5
T 0 1 0
Machines
The smallest uncovered number is 2
so this is subtracted from all other
uncovered numbers and added to
numbers at the intersection of lines
28

29. Assignment Example
Because three lines are needed,
the solution is optimal and
assignments can be made
i.ethe number of lines is equal to
the number of rows or the
number of columns
Step 3 -Lines
A B C
Job
R 3 4 0
S 0 0 5
T 0 1 0
Machines Begin with rows or columns with
only one zero.
-assign job R to Machine C as this
is the only possible assignment for
the job.Job T must go to Machine
A as Machine C is already assigned.
This leaves job S for Machine B.
Step 5 -Assignments
A B C
Job
R 3 4 0
S 0 0 5
T 0 1 0
Machines
29

30. Assignment Example
Step 5 -Assignments
A B C
Job
R 3 4 0
S 0 0 5
T 0 1 0
Machines
A B C
Job
R 11 14 6
S 8 10 11
T 9 12 7
Machines
From the original cost table
Minimum cost = $6 + $10 + $9 = $25
30

31. Hungarian Method Example2
Determine the optimal assignment of jobs to
machines for the following data
MACHINE
JOB A B C D
1 8 6 2 4
2 6 7 11 10
3 3 5 7 6
4 5 10 12 9
31

32. Hungarian Method Example2
MACHINE
JOB A B C D ROWMIN
1 8 6 2 4 2
2 6 7 11 10 6
3 3 5 7 6 3
4 5 10 12 9 5
Step-1 Row reduction
32

33. Hungarian Method Example2
MACHINE
JOB A B C D
1 6 4 0 2
2 0 1 5 4
3 0 2 4 3
4 0 5 7 4
COLMIN 0 1 0 2
Subtract the smallest number in each row to form a new table and select
column minimum
33

34. Hungarian Method Example2
MACHINE
JOB A B C D
1 6 3 0 0
2 0 0 5 2
3 0 1 4 1
4 0 4 7 2
Step-2 Subtract the smallest number in each column & Enter the
results to form a new table
34

35. Hungarian Method Example2
Step-3 Determine the minimum number of lines needed to cross
Out all zeros. Here we have three lines only and rows are 4, so the
solution is not optimal. Note that the smallest uncovered value is 1
MACHINE
JOB A B C D
1 6 3 0 0
2 0 0 5 2
3 0 1 4 1
4 0 4 7 2
35

36. Hungarian Method Example2
MACHINE
JOB A B C D
1 6+1=7 3 0 0
2 0+1=1 0 5 2
3 0 0 3 0
4 0 3 6 1
Step-4 Subtract the smallest value that has not been crossed out from every
number that has not been crossed out (1 here) and add this to numbers that
are at intersectionsof covering lines
36

37. . Hungarian Method Example2
MACHINE
JOB A B C D
1 7 3 0 0
2 1 0 5 2
3 0 0 3 0
4 0 3 6 1
Step-3 Determine the minimum number of lines needed to cross Out all 0 ( 4),
since this equals the number of rows , we obtain the optimum assignment
37

38. Hungarian Method
Example2
MACHINE
JOB A B C D
1 7 3 0 0
2 1 0 5 2
3 0 0 3 0
4 0 3 6 1
Assignments Cost(From the originalcost table)
1-C $ 2
2-B 7
3-D 6
4-A 5
$20
Step-5 Make the assignments,
start with rows and columns
with Only one 0 (Job 2 to B).
Match jobs with machines that
have 0 costs
38

39. Sequencing
Determining the order in which jobs at
a work center will be processed
A work center is an area in a business
in which productive resources are
organized and work is completed
Job time: Time needed for setup and
processing of a job
39

40. How to Sequence Jobs
Which of several jobs should be scheduled first?
Techniques are available to do short-term planning of jobs
based on available capacity & priorities
Priority rules:
Simple heuristics (Commonsense rules) used to select the order in
which jobs will be processed.
Decision rules to allocate the relative priority of jobs at a work center
Local priority rules: determines priority based only on jobs at that
workstation -(pertaining to single workstation)
Global priority rules: also considers the remaining workstations a job
must pass through -(pertaining to multiple workstation)
40

41. Commonly Used Priorities Rules
First come, first served (FCFS)
Last come, first served (LCFS)
Earliest due date (EDD)-earliest due date first
Shortest processing time (SPT)-shortest job first
Longest processing time (LPT)
Slack per remaining Operations (S/RO)
Slack /(number of remaining operations)
41

42. Commonly Used Priorities
Rules
An index number found by dividing the time remaining
until the due date by the work time remaining on the job
Jobs with lowcritical ratios are scheduled aheadof jobs
with higher critical ratios
Performs well on average job lateness criteria
CR = =
Due date -Today’s date
Processing (lead) time remaining
Processing Time remaining
Workdays remaining
42

43. Assumptions of Priority Rules
1.The set of jobs is known, no new jobs arrive after
processing begins and no jobs are canceled
2.Setup time is deterministic
3.Processing times are deterministic rather than
variables
4.There will be no interruptions in processing such
as machine breakdowns , accidents or worker
illnesses
43

44. How to Use Priority Rules
1.Decide which priority rule to use
2.List all jobs waiting to be processed
with their job time
3.Using priority rule determine which
job has highest priority then second,
third and so on
44

45. Sequencing Example
Job
Job Work
(Processing) Time
(Days)
Job Due
Date
(Days)
A 6 8
B 2 6
C 8 18
D 3 15
E 9 23
45

46. Sequencing Example
Job
Sequence
Job Work
(Processing)
Time
Flow
Time
Job Due
Date
Job
Lateness
A 6 6 8 0
B 2 8 6 2
C 8 16 18 0
D 3 19 15 4
E 9 28 23 5
28 77 11
FCFS: Sequence A-B-C-D-E
46

47. Sequencing Example
Job
Sequence
Job Work
(Processing)
Time
Flow
Time
Job Due
Date
Job
Lateness
A 6 6 8 0
B 2 8 6 2
C 8 16 18 0
D 3 19 15 4
E 9 28 23 5
28 77 11
FCFS: Sequence A-B-C-D-E
Average completion time = = 77/5 = 15.4 days
Sum of total flow time
Number of jobs
Utilization = = 28/77 = 36.4%
Total job work time
Sum of total flow time
Average number of
jobs in the system
= = 77/28 = 2.75 jobs
Sum of total flow time
Total job work time
Average job lateness = = 11/5 = 2.2 days
Total late days
Number of jobs
47

48. Sequencing Example
Job
Sequence
Job Work
(Processing)
Time
Flow
Time
Job Due
Date
Job
Lateness
B 2 2 6 0
D 3 5 15 0
A 6 11 8 3
C 8 19 18 1
E 9 28 23 5
28 65 9
SPT: Sequence B-D-A-C-E
48

49. Sequencing Example
Job
Sequence
Job Work
(Processing)
Time
Flow
Time
Job Due
Date
Job
Lateness
B 2 2 6 0
D 3 5 15 0
A 6 11 8 3
C 8 19 18 1
E 9 28 23 5
28 65 9
SPT: Sequence B-D-A-C-E
Average completion time = = 65/5 = 13 days
Sum of total flow time
Number of jobs
Utilization = = 28/65 = 43.1%
Total job work time
Sum of total flow time
Average number of
jobs in the system
= = 65/28 = 2.32 jobs
Sum of total flow time
Total job work time
Average job lateness = = 9/5 = 1.8 days
Total late days
Number of jobs
49

50. Sequencing Example
Job
Sequence
Job Work
(Processing)
Time
Flow
Time
Job Due
Date
Job
Lateness
B 2 2 6 0
A 6 8 8 0
D 3 11 15 0
C 8 19 18 1
E 9 28 23 5
28 68 6
EDD: Sequence B-A-D-C-E
50

51. Sequencing Example
Job
Sequence
Job Work
(Processing)
Time
Flow
Time
Job Due
Date
Job
Lateness
B 2 2 6 0
A 6 8 8 0
D 3 11 15 0
C 8 19 18 1
E 9 28 23 5
28 68 6
EDD: Sequence B-A-D-C-E
Average completion time = = 68/5 = 13.6 days
Sum of total flow time
Number of jobs
Utilization = = 28/68 = 41.2%
Total job work time
Sum of total flow time
Average number of
jobs in the system
= = 68/28 = 2.43 jobs
Sum of total flow time
Total job work time
Average job lateness = = 6/5 = 1.2 days
Total late days
Number of jobs
51

52. Sequencing Example
Job
Sequence
Job Work
(Processing)
Time
Flow
Time
Job Due
Date
Job
Lateness
E 9 9 23 0
C 8 17 18 0
A 6 23 8 15
D 3 26 15 11
B 2 28 6 22
28 103 48
LPT: Sequence E-C-A-D-B
52

53. Sequencing Example
Job
Sequence
Job Work
(Processing)
Time
Flow
Time
Job Due
Date
Job
Lateness
E 9 9 23 0
C 8 17 18 0
A 6 23 8 15
D 3 26 15 11
B 2 28 6 22
28 103 48
LPT: Sequence E-C-A-D-B
Average completion time = = 103/5 = 20.6 days
Sum of total flow time
Number of jobs
Utilization = = 28/103 = 27.2%
Total job work time
Sum of total flow time
Average number of
jobs in the system
= = 103/28 = 3.68 jobs
Sum of total flow time
Total job work time
Average job lateness = = 48/5 = 9.6 days
Total late days
Number of jobs
53

54. Sequencing Example
Rule
Average
Completion
Time (Days)
Utilization
(%)
Average Number
of Jobs in
System
Average
Lateness
(Days)
FCFS 15.4 36.4 2.75 2.2
SPT 13.0 43.1 2.32 1.8
EDD 13.6 41.2 2.43 1.2
LPT 20.6 27.2 3.68 9.6
Summary of Rules
54

55. Comparison of
Sequencing Rules
No one sequencing rule excels on all criteria
SPT does well on minimizing flow time and number of jobs in the
system
But SPT moves long jobs to
the end which may result
in dissatisfied customers
FCFS does not do especially
well (or poorly) on any
criteria but is perceived
as fair by customers
EDD minimizes lateness
55

56. Performance Calculations con’t
Lateness and Tardiness are both measures
related to customer service
Average tardinessis a more relevant Customer
Servicemeasurement as illustrated belowExample 15-5 Calculating job lateness and job tardiness
Completion
Job Date Due DateLatenessTardiness
A 10 15 -5 0
B 13 15 -2 0
C 17 10 7 7
D 20 20 0 0
Average 0 1.75
56

57. Comparing SPT and S/RO
E done at
end of day 2
A end of
day 5
D at end
of day 9
F at end of
day 14
C at end of
day 20
B done at end
of day 27Performance Measures using SPT
Job Time at
Work Center SPT
301 Due date Completion Lateness TardinessScheduling
Job (days) (days from now) Date (days) (days) Sequence
A 3 15 5 -10 0 2
B 7 20 27 7 7 6
C 6 30 20 -10 0 5
D 4 20 9 -11 0 3
E 2 22 2 -20 0 1
F 5 20 14 -6 0 4
Total 27 Avg. Job Flow 12.83 -8.3 1.2
Total Job Flow Time 77
Makespan 27
Avg. # Jobs 2.85
57

58. Comparing SPT and S/ROPerformance Measures Using S/RO
Job Time Remaining
at Work Remaining Number
Center Job Time at Slack of Operations
301 Other Work Due date Time After Work Scheduling Completion LatenessTardiness
Job (days)Center (days)(days from now)(days)Center 301S/RO Sequence Date (days)(days)
A 3 6 15 6 2 2 2 10 -5 0
B 7 8 20 5 4 1 1 7 -13 0
C 6 5 30 19 3 4.75 6 27 -3 0
D 4 3 20 13 2 4.33 5 21 1 1
E 2 7 22 13 3 3.25 4 17 -5 0
F 5 5 20 10 3 2.5 3 15 -5 0
Total 27 Avg. Job Flow 16.17 -5.0 0.167
Total Job Flow Time 97
Makespan 27
Avg. # Jobs 3.59
B done at
end of day 7
A at end
of day 10
F at end of
day 15
E at end of
day 17
D at end of
day 21
C done at end
of day 27
58

59. Critical Ratio (CR)
An index number found by dividing the time
remaining until the due date by the work
time remaining on the job
Jobs with low critical ratios are scheduled
ahead of jobs with higher critical ratios
Performs well on average job lateness
criteria
CR = =
Due date -Today’s date
Work (lead) time remaining
Time remaining
Workdays remaining
59

60. Critical Ratio Example
Job
Due
Date
Workdays
Remaining Critical Ratio
Priority
Order
A30 4 (30 -25)/4 = 1.253
B28 5 (28 -25)/5 = .601
C27 2 (27 -25)/2 = 1.002
Currently Day 25
With CR < 1, Job B is late. Job C is just on schedule and Job A has some
slack time.
60

61. Critical Ratio Technique
1.Helps determine the status of specific jobs
2.Establishes relative priorities among jobs
on a common basis
3.Relates both stock and make-to-order jobs
on a common basis
4.Adjusts priorities automatically for changes
in both demand and job progress
5.Dynamically tracks job progress
61

62. Sequencing N Jobs on Two
Machines: Johnson’s Rule
Works with two or more jobs that pass
through the same two machines or
work centers
Minimizes total production time and
idle time
62

63. Johnson’s Rule
1.List all jobs and times for each work center
2.Choose the job with the shortest activity
time. If that time is in the first work center,
schedule the job first. If it is in the second
work center, schedule the job last.
3.Once a job is scheduled, it is eliminated from
the list
4.Repeat steps 2 and 3 working toward the
center of the sequence
63

64. Johnson’s Rule Example
Job
Work Center 1
(Drill Press)
Work Center 2
(Lathe)
A 5 2
B 3 6
C 8 4
D 10 7
E 7 12
64

65. BEDCA
Johnson’s Rule Example
Job
Work Center 1
(Drill Press)
Work Center 2
(Lathe)
A 5 2
B 3 6
C 8 4
D 10 7
E 7 12
65

66. Johnson’s Rule Example
Job
Work Center 1
(Drill Press)
Work Center 2
(Lathe)
A 5 2
B 3 6
C 8 4
D 10 7
E 7 12
Time0 3 10 20 28 33
B ACDE
WC
1
WC
2
B ACDE
66

67. Johnson’s Rule Example
Job
Work Center 1
(Drill Press)
Work Center 2
(Lathe)
A 5 2
B 3 6
C 8 4
D 10 7
E 7 12
Time0 3 10 20 28 33
Time01357910111213171921 22 23252729313335
B ACDE
B ACDE
WC
1
WC
2
B E D CA
B ACDE
67

68. Johnson’s Rule Example:Vicki’s Office Cleanersdoes the annual
major cleaning of university buildings. The job requires mopping
(1
st
activity) and waxing (2
nd
activity) of each building. Vicki wants
to minimize the time it takes her crews to finish cleaning (minimize
makespan) the five buildings. She needs to finish in 20 days.Activity 1 Activity 2 Johnson's Activity 1 Activity 2
Hall Mopping (days)Waxing (days) Sequence Mopping (days)Waxing (days)
Adams Hall 1 2 Adams Hall (A) 1 2
Bryce Building 3 5 Chemistry Building (C) 2 4
Chemistry Building 2 4 Bryce Building (B) 3 5
Drake Union 5 4 Drake Union (D) 5 4
Evans Center 4 2 Evans Center (E) 4 2 Activity123456789101112131415161718
MoppingACCBBBDDDDDEEEE
Waxing AACCCCBBBBBDDDDEE
68

69. Scheduling Bottlenecks
In the 1970’s Eli Goldratt introduced optimized
production technology (OPT)
OPT focused on bottlenecks for scheduling &
capacity planning
Definitions:
Throughput:quantity of finished goods that can be sold
Transfer batch:quantity of items moved at the same
time from one resource to the next
Process batch:quantity produced at a resource before
switching to another product
69

70. Theory of Constraints
TOC is an extension of OPT –theory is that a
system’s output is determined by its constraints
1.Identify the bottleneck(s) in the process
2.Exploit (fully utilize) the bottleneck(s)
3.Subordinate all other decisions to Step 2 -
Schedule non-bottlenecks to support maximum
use of bottleneck activities
4.Elevate the Bottleneck(s)
5.Do not let inertia set in
70

71. Scheduling for Service Organizations
Demand management:
Appointments & reservations
Discounts or other promotional schemes
Delayed services or backlogs (queues)
When demand management is not feasible, managing capacity
through staffing flexibility may be used
Scheduling Employees:
Staff for peak demand (if cost isn’t prohibitive)
Floating employees or employees on call
Temporary, seasonal, or part-time employees
71

72. Scheduling Service Employees
Objective
-is to meet staffing requirements with the minimum number of workers
steps
1.Determine the staffing requirements
2.Identify two consecutive days with the lowest total requirements and
assign these as days off
3.Make a new set of requirements subtracting the days worked by the
first employee
4.Apply step 2 to the new row
5.Repeat steps 3 and 4 until all requirements have been met
72

73. Cyclical Scheduling Example
This example shows how service
personnel can be scheduled for seven
day operation giving each employee
two consecutive days off
Day of the week MTWThFSaSu
Number of staff needed5565433
73

74. M T W T F S S
Employee 1 5 5 6 5 4 3 3
Capacity (Employees)
Excess Capacity
Cyclical Scheduling Example
74

75. M T W T F S S
Employee 1 5 5 6 5 4 3 3
Employee 2 4 4 5 4 3 3 3
Capacity (Employees)
Excess Capacity
Cyclical Scheduling Example
75

76. M T W T F S S
Employee 1 5 5 6 5 4 3 3
Employee 2 4 4 5 4 3 3 3
Employee 3 3 3 4 3 2 3 3
Capacity (Employees)
Excess Capacity
Cyclical Scheduling Example
76

77. M T W T F S S
Employee 1 5 5 6 5 4 3 3
Employee 2 4 4 5 4 3 3 3
Employee 3 3 3 4 3 2 3 3
Employee 4 2 2 3 2 2 3 2
Capacity (Employees)
Excess Capacity
Cyclical Scheduling Example
77

78. M T W T F S S
Employee 1 5 5 6 5 4 3 3
Employee 2 4 4 5 4 3 3 3
Employee 3 3 3 4 3 2 3 3
Employee 4 2 2 3 2 2 3 2
Employee 5 1 1 2 2 2 2 1
Capacity (Employees)
Excess Capacity
Cyclical Scheduling Example
78

79. M T W T F S S
Employee 1 5 5 6 5 4 3 3
Employee 2 4 4 5 4 3 3 3
Employee 3 3 3 4 3 2 3 3
Employee 4 2 2 3 2 2 3 2
Employee 5 1 1 2 2 2 2 1
Employee 6 1 1 1 1 1 1 0
Capacity (Employees)
Excess Capacity
Cyclical Scheduling Example
79

80. M T W T F S S
Employee 1 5 5 6 5 4 3 3
Employee 2 4 4 5 4 3 3 3
Employee 3 3 3 4 3 2 3 3
Employee 4 2 2 3 2 2 3 2
Employee 5 1 1 2 2 2 2 1
Employee 6 1 1 1 1 1 1 0
Employee 7 1
Capacity (Employees)
Excess Capacity
Cyclical Scheduling Example
80

81. Cyclical Scheduling Example
M T W T F S S
Employee 1 5 5 6 5 4 3 3
Employee 2 4 4 5 4 3 3 3
Employee 3 3 3 4 3 2 3 3
Employee 4 2 2 3 2 2 3 2
Employee 5 1 1 2 2 2 2 1
Employee 6 1 1 1 1 1 1 0
Employee 7 1
Capacity (Employees) 5 5 6 5 4 3 3
Excess Capacity 0 0 0 0 0 1 0
81

82. Developing a Workforce Schedule:This example shows how a
staff of six people can be scheduled for seven day operation giving
each employee two consecutive days of
Step 1–Find out the minimum number of employees
needed for each day of the week
Step 2–Given the above requirements, calculate the
number of employees needed for each pair of consecutive
days
Step 3-Find the pair of days with the lowest total needed(1) Day of the week MTWThFSaSu
Number of staff needed4553523 (1) Pair of Consecutive DaysTotal of Staff needed
Monday & Tuesday 9 employees
Tuesday & Wednesday 10 employees
Wednesday & Thursday 8 employees
Thursday & Friday 8 employees
Friday & Saturday 7 employees
Saturday & Sunday 5 employees
82

83. Workforce Scheduling con’t
Step 4–Update the number of employees you still need to
schedule for each day
Step 5–Using the updated staffing needs, repeat steps 2
through 4 until you have satisfied all needs (2) Day of the week MTWThFSaSu
Number of staff needed3442423 (2) Pair of Consecutive DaysTotal of Staff needed
Monday & Tuesday 7 employees
Tuesday & Wednesday 8 employees
Wednesday & Thursday 6 employees
Thursday & Friday 6 employees
Friday & Saturday 6 employees
Saturday & Sunday 5 employees
83

84. Scheduling con’t(3) Pair of Consecutive DaysTotal of Staff needed
Monday & Tuesday 5 employees
Tuesday & Wednesday 6 employees
Wednesday & Thursday 4 employees
Thursday & Friday 4 employees
Friday & Saturday 5 employees
Saturday & Sunday 5 employees (3) Day of the week MTWThFSaSu
Number of staff needed2331323 (4) Day of the week MTWThFSaSu
Number of staff needed1231212
(4) Pair of Consecutive DaysTotal of Staff needed
Monday & Tuesday 3 employees
Tuesday & Wednesday 5 employees
Wednesday & Thursday 4 employees
Thursday & Friday 3 employees
Friday & Saturday 3 employees
Saturday & Sunday
3 employees
84

85. Schedule con’t(5) Day of the week MTWThFSaSu
Number of staff needed0120112 (6) Pair of Consecutive DaysTotal of Staff needed
Monday & Tuesday 1 employees
Tuesday & Wednesday 2 employees
Wednesday & Thursday 1 employees
Thursday & Friday 0 employees
Friday & Saturday 0 employees
Saturday & Sunday 1 employees (5) Pair of Consecutive DaysTotal of Staff needed
Monday & Tuesday 1 employees
Tuesday & Wednesday 3 employees
Wednesday & Thursday 2 employees
Thursday & Friday 1 employees
Friday & Saturday 2 employees
Saturday & Sunday 3 employees (6) Day of the week MTWThFSaSu
Number of staff needed0110001
85

86. Final Schedule(7) Day of the week MTWThFSaSu
Number of staff needed0000000 Employees M TWThFSaSu
1 x x x x xoffoff
2 x x x x xoffoff
3 x xoffoffx x x
4 x x x x xoffoff
5 offoffx x x x x
6 x x x xoffoffx
This technique gives a work
schedule for each employee
to satisfy minimum daily
staffing requirements
Next step is to replace
numbers with employee
names
Manager can give senior
employees first choice and
proceed until all employees
have a schedule
86

87. End
87