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Asignatura: Direccio Operacions, Profesor: ju ju, Carrera: Administració i Direcció d'Empreses, Universidad: UB
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Operations scheduling 11
Operations Management
Economy & Business Organization Department
2
Operations scheduling 2
Index
Assignment method of job allocation
o Index methodology
Order sequencing
o Johnson’s rule for N orders and two work centres.
o Johnson’s rule for N orders and three work centres.
o Jackson’s rule
Operations scheduling 33
Activities dedicated to Activities dedicated to program, monitorprogram, monitor
and evaluate
and evaluate short term operation tasks
short term operation tasks
in order to achieve the Master Production
in order to achieve the Master Production
Schedule efficiently taking into account Schedule efficiently taking into account
the available capacity. the available capacity.
Short term scheduling and control
4
Operations scheduling 4
Evaluation and controlEvaluation and control of the orders pending to beof the orders pending to be
produced according to the Materials Planning.produced according to the Materials Planning.
Setting of prioritiesSetting of priorities among the orders that should beamong the orders that should be
manufactured: Operations Scheduling.manufactured: Operations Scheduling.
Monitoring
Monitoring the evolution of
the evolution of the orders in progress
the orders in progress in
in
the different work centres.
the different work centres.
Controlling
Controlling the operations development in the work
the operations development in the work
centres.
centres.
ControllingControlling the capacity of each work centre.the capacity of each work centre.
ProvidingProviding feedbackfeedback to the Planning and Capacityto the Planning and Capacity
Control System.Control System.
Basic functions
Operations scheduling 77
PossiblePossible productiveproductive configurationsconfigurations::
Scheduling according to the configuration
8
Operations scheduling 8
This configuration is used to execute projects (products
This configuration is used to execute projects (products
or unique services, which are complex. The activities to or unique services, which are complex. The activities to
be developed are different and specific for each
be developed are different and specific for each
particular case). particular case).
The planning is unique for each project.The planning is unique for each project.
Planning and control techniques such as : PERT
Planning and control techniques such as : PERT
Project Evaluation and Review
Project Evaluation and Review Tecnique
Tecnique ), CPM (
Critical
Critical
Path Method Path Method ),), PERTPERT--CostCost oror ROYROY can be applied.can be applied.
Characteristics of Productive Configurations
Operations scheduling 99
Normally the production of this configuration is carried out toNormally the production of this configuration is carried out to bebe
stored in the warehouse. The same facilities are used to obtainstored in the warehouse. The same facilities are used to obtain thethe
same product.same product. The machinery is located as a chain.The machinery is located as a chain.
The operation that performs each machinery is always the same.The operation that performs each machinery is always the same. AA
standardised product is obtainedstandardised product is obtained..
The adequate use of the facility depends on the quality of theThe adequate use of the facility depends on the quality of the
process design. If itprocess design. If it’’s correct, the inventory in progress will bes correct, the inventory in progress will be
minimum.minimum.
This system is very
This system is very efficient but NOT flexible
efficient but NOT flexible .
.
Continuous configuration characteristics
A B C D
10
Operations scheduling 10
Lot configuration:
Lot configuration:
The same facilities (work centres) are used toThe same facilities (work centres) are used to
manufacture different articles. manufacture different articles.
Consequently, after a lot production, the work centre
Consequently, after a lot production, the work centre
stops and prepares the next production lot. stops and prepares the next production lot.
In this configuration, we can find two types ofIn this configuration, we can find two types of
processes:
processes:
Production line
Production line
JobJob--shopshop
Lot configuration characteristics
Operations scheduling 1313
In addition, the manufacturing process of a lot canIn addition, the manufacturing process of a lot can
differ greatly in terms of necessary materials, process
differ greatly in terms of necessary materials, process
timing in each work centre, preparation timing, timing in each work centre, preparation timing,……
Characteristics of job-shop configuration
A
B
C
D
E
F
14
Operations scheduling 14
Assignment of job allocation or assignment of machinery
Assignment of job allocation or assignment of machinery
loadsloads: Assignment of the orders that must be produced in: Assignment of the orders that must be produced in
the work centres, indicating which operations must bethe work centres, indicating which operations must be
performed in each one.
performed in each one.
SequencingSequencing: determination of priorities of the orders in the: determination of priorities of the orders in the
different work centres so that the planned delivery dates
different work centres so that the planned delivery dates
are accomplished with the minimum resources andare accomplished with the minimum resources and
inventories.
inventories.
Detailed schedulingDetailed scheduling: determination of initial and final: determination of initial and final
moments of the different activities that must be performed
moments of the different activities that must be performed
in each work centre.in each work centre.
Despatch
Despatch :
Physical delivery by production planning &
Physical delivery by production planning &
control of a work order to the operations department.control of a work order to the operations department.
Fundamental activities
Operations scheduling 1515
Techniques to assign job allocation
17
Operations scheduling 17
Process timing, in hours, for 5 operations in 3 machines
Operation nº Machine 1 Machine 2 Machine 3
22C 100 150 125
27A 200 100 220
44G 25 50 20
32B 40 30 ---
51E 60 50 70
Available hours 160 110 150
The lowest execution times appear in red.
Indices method to assign job allocation
Operations scheduling 2020
Determine the best delivery term (in days) that could be
Determine the best delivery term (in days) that could be
offered to a customer that places a 300 units order. Take
offered to a customer that places a 300 units order. Take
into account that it
into account that it ’
s possible to obtain 4 units/hour in
s possible to obtain 4 units/hour in
machine M1, 6 units/hour in M2 and 3 units/hour in M3.
machine M1, 6 units/hour in M2 and 3 units/hour in M3.
Consider that the machines work 10 hours daily.
Consider that the machines work 10 hours daily.
The availability hours in each work centre are: 280 hoursThe availability hours in each work centre are: 280 hours
in M1, 260 hours in M2 and 290 hours in M3. in M1, 260 hours in M2 and 290 hours in M3.
Operation M1 M2 M
A 200 300 250
B 400 200 440
C 50 100 40
D 80 60 90
E 120 100 140
Indices methodology: Example
21
Operations scheduling 21
M1 M2 M
Operation Hours Indices Hours Indices Hours Indices
A (200) 1 300 1,5 250 1,
B 400 2
(200) 1 440 2,
C 50 1,25 100 2,5 (40) 1
D 80 1,33 (60) 1 90 1,
E 120 1,2 100 1 (140) 1,
Available hours 280 260 290
Solution a)
Assigned hours 200 200 + 60 = 260 40 + 140 = 180
Pending assignment 80 0 110
Solution b)
Assigned hours 200 + 80 = 280 200 40 + 140 = 180
Pending assignment 0 60 110
Indices method: Example 2
Operations scheduling 2222
The present work load expressed in days: The present work load expressed in days:
Solution a) Solution a) Solution b)Solution b)
20 days in M1 20 days in M1 28 days in M128 days in M
26 days in M2 26 days in M2 20 days in M220 days in M
18 days in M
18 days in M 18 days in M
18 days in M
Taking into account the new order:
Taking into account the new order:
In M1: 300 / 4 = 75 hours In M1: 300 / 4 = 75 hours 7,5 days7,5 days
In M2: 300 / 6 = 50 hours
In M2: 300 / 6 = 50 hours
5 days
5 days
In M3: 300 / 3 = 100 hours
In M3: 300 / 3 = 100 hours
10 days
10 days
The best delivery term is obtained with solution b)The best delivery term is obtained with solution b)
assigning the order to M2. Then, delivery term is 25 daysassigning the order to M2. Then, delivery term is 25 days
Indices method
23
Operations scheduling 23
Sequencing
Sequencing establishes the priority of
the orders in the work centres in
order to comply delivery terms with
least inventories and resources.
Operations scheduling 2626
Johnson’s rule for N orders and 2 machines
27
Operations scheduling 27
Sequencing
P1 P2 P3 P4 P
M1 5 --- 9 3 10
M2 2 --- 7 8 4
Jonhson’s rule for N orders and 2 machines
Operations scheduling 2828
that belongs to M2. Consequently, P1 must be that belongs to M2. Consequently, P1 must be
programmed after all the orders pending of
programmed after all the orders pending of
assignment. Operations sequence is as follows:
assignment. Operations sequence is as follows:
Pending sequence :
P1 P2 P3 P4 P
M1 --- --- 9 3 10
M2 --- --- 7 8 4
Johnson’s rule for N orders and 2 machines
29
Operations scheduling 29
M1. Consequently, P4 must be programmed before M1. Consequently, P4 must be programmed before
the rest (BUT after P2). the rest (BUT after P2).
Pending sequence:
P1 P2 P3 P4 P
M1 --- --- 9 --- 10
M2 --- --- 7 --- 4
Johnson’s rule for N orders and 2 machines
Operations scheduling 3232
Definitive tests should be programmed to five patients and alsoDefinitive tests should be programmed to five patients and also
corresponding checkups with the doctor.corresponding checkups with the doctor.
The mobile laboratory will be used from 8 am and itThe mobile laboratory will be used from 8 am and it’’s desirables desirable
to finish these visits as soon as possible.to finish these visits as soon as possible.
The nurse has estimated the following times for each patientThe nurse has estimated the following times for each patient
(time in minutes):
(time in minutes):
Determine the laboratory and doctor visits programming. Is it
Determine the laboratory and doctor visits programming. Is it
possible to finish by 13h?possible to finish by 13h?
Patient A B C D E
Laboratory 120 30 20 40 60
Doctor visit 10 60 120 30 60
Jonhson’s rule for service industries
Hospital – N orders & 2 machines
33
Operations scheduling 33
A
B
C B E D A
C B E D A
8:00 8:20 8:50 9:50 10:30 12:
8:
12:
10:20 11:20 12:
13:
Jonhson’s rule for service industries
Operations scheduling 3434
The 3 machines are called M1, M2 & M3. The index indicates theThe 3 machines are called M1, M2 & M3. The index indicates the
order during the production process.order during the production process.
The algorithm is based in the generation of 2 fictitious machineThe algorithm is based in the generation of 2 fictitious machiness
M4 & M5,M4 & M5,
In M4, the execution time for order i equals the sum of timings
in M1 and M2.
in M1 and M2.
of timings in machines M2 & M3.of timings in machines M2 & M3.
Once these timings are calculated, we apply the JohnsonOnce these timings are calculated, we apply the Johnson’’s rule tos rule to
both fictitious machines considering that work orders pass to M4both fictitious machines considering that work orders pass to M
first and then continue in M5.first and then continue in M5.
It is necessary to verify that
It is necessary to verify that the sum of least times of orders in
the sum of least times of orders in
first and last machines are lower that maximum time in the
first and last machines are lower that maximum time in the
intermediate machine.
intermediate machine.
Johnson’s rule for N orders and 3 machines
35
Operations scheduling 35
Five orders must be processed in 3 machines or workFive orders must be processed in 3 machines or work
centres M1, M2 & M3. The processing times are shown incentres M1, M2 & M3. The processing times are shown in
the table below:the table below:
Johnson’s rule for N orders and 3 machines
Determine the production sequencing
M1 M2 M
P1 5 2 1
P2 1 6 3
P3 9 7 4
P4 3 8 7
P5 10 4 6
Operations scheduling 3838
Jackson algorithm
Sequencing criteria that establishes the sequences of orders along the
work centres.
The programming perspective is:
n / m / F o G / variable to optimize
n = number of pieces (finite or infinite according to queue theory)
m = number of machines of the transformation process of a piece.
F = Production flux A B C D
G = Production flux with reversals A B C D
Variable to optimize = delivery terms, minimum stock, machine
saturation
Johnson’s rule : n / 2 / F / delivery term
Jackson algorithm : n / 2 / G / delivery term
39
Operations scheduling 39
Machine A
Order a 1 3
2 5
4 --
-- 3
3 3
4 2
5 1
3 6
Machine B
Order b
Order c
Order d
Order e
Order f
Order g
Order h
The order is firstly processed
in machine A and afterwards
in machine B.
The order is firstly processed
in machine B and after that in
machine A.
Jackson algorithm for N orders and 2 machines
Consider the orders a, b, c, d, e, f, g, h, that should be processed in
machines A & B. Times and production flow are shown below:
Operations scheduling 4040
Jackson algorithm for N orders and 2 machines
Sequencing criteria:
Machine A: 1 º - Pieces AB sorted by Johnson
2 º - Pieces A
3 º - Pieces BA sorted by Johnson
Machine B: 1 º - Pieces BA sorted by Johnson
2 º - Pieces B
3 º - Pieces AB sorted by Johnson
Machine A
Order a 1 3
2 5
4 --
-- 3
3 3
4 2
5 1
3 6
Machine B
Order b
Order c
Order d
Order e
Order f
Order g
Order h
AB
BA
A
B
AB
BA
AB
BA
Sequence
41
Operations scheduling 41
Jackson algorithm for N orders and 2 machines
Johnson sequencing for pieces AB:
Machine A
Order a 1 3
3 3
5 1
Machine B
Order e
Order g
a – e – g
f – h – b
Johnson sequencing for pieces BA:
Machine A
Order b 5 2
2 4
6 3
Machine B
Order f
Order h
Sequence:
Machine A : a – e – g – c – f – h – b
Machine B : f – h – b – d – a – e – g
AB
AB
BA
BA
A
B
