Sequencing and shedulding problems for Operations management

MODULE V
SEQUENCING & SCHEDULING

Gantt chart
• HL Gantt has developed a
simplified graph which
represents/displays the planned
starting and finishing time of
each task on a time scale. But it
does not show the
interrelationship among the tasks.
• On the left of the chart is a list of
the activities and along the top is
a suitable time scale

Single Machine Scheduling
• Scheduling is the allocation of start and finish time to each particular
order.
• The single machine scheduling problem consists of n jobs with the
same single operation on each of the jobs, while the flow shop
scheduling problem consists of n jobs with m operations on each of
the jobs.
• In this problem, all the jobs will have the same process sequences.

CONCEPT OF SINGLE MACHINE SCHEDULING
1. A set of independent, single-operation jobs is available for
processing at time zero.
2. Set-up time of each job is independent of its position in jobs
sequence. So, the set-up time of each job can be included in its
processing time.
3. One machine is continuously available and is never kept idle when
work is waiting.
4. Each job is processed till its completion without break.

Terminologies
• Processing time (tj) :It is the time required to process job j. The processing time,
tj will normally include both actual processing time and set-up time.
• Ready time (rj): It is the time at which job j is available for processing.
• Due date (dj): It is the time at which the job j is to be completed.
• Completion time (Cj). It is the time at which the job j is completed in a sequence.
• Flow time (Fj):It is the amount of time job j spends in the system. Flow time is a
measure which indicates the waiting time of jobs in a system. This in turn gives
some idea about in-process inventory due to a schedule.

• Lateness (Lj): It is the amount of time by which the completion time
of job j differs from the due date (Lj = Cj – dj).
Lateness can be either positive lateness or negative lateness. Positive lateness
of a job means that the job is completed after its due date. Negative lateness of
a job means that the job is completed before its due date. The positive lateness
is a measure of poor service. The negative lateness is a measure of better
service.
• Tardiness (Tj) : Tardiness is the lateness of job j if it fails to meet its
due date, or zero, otherwise
Tj = max {0, Cj – dj}

SHORTEST PROCESSING TIME (SPT) RULE
• In single machine scheduling problem, sequencing the jobs in
increasing order of processing time is known as the shortest
processing time (SPT) sequencing.
• This method achieves the following objectives :
1. minimizing mean waiting time,
2. minimizing mean flow time,
3. minimizing mean lateness,
4. minimizing the mean number of jobs waiting as in-process
inventory.

Weighted shortest processing time(WSPT)
•In case importance of the jobs varies, a weight wi is assigned to each job, a
larger value indicating greater importance. Processing times are divided by the
weights and jobs sequenced in the order of increasing

EXAMPLE: Eight jobs 1, 2, ..., 8 are to be processed on a single machine. The processing times, due dates and
importance weights of the jobs are represented in table.
• Assuming that no new jobs arrive thereafter, determine using SPT rule and WSPT rule
(i) optimal sequence,
(ii) completion time of the jobs,
(iii) mean flow time as well as weighted mean flow time,
(iv) average in-process inventory,
(v) lateness, mean lateness and maximum lateness,
(vi) number of jobs actually late.

Solution :SPT METHOD
Arrange jobs in the increasing order of processing time (ti)
Job
sequence
4 8 1 3 7 2 5 6
Processingti
me(ti)
3 3 5 6 7 8 10 14
Number of
jobs in the
system
8 7 6 5 4 3 2 1
Completion
time(Ci)
3 6 11 17 24 32 42 56
Due date(di) 25 50 15 15 45 10 20 20
Lateness(Ci-
di)
-22 -44 -4 2 -21 22 22 16

Mean flow time =
Average in-process inventory =
• Mean lateness = = 3.625 minutes
• Maximum lateness = 22 minutes
• Number of jobs actually late = 4 jobs

Solution :WSPT METHOD
Arrange jobs in the increasing order of weighted processing time(ti /Wi)
Job
sequence
3 4 8 7 2 6 1 5
Processing
time(ti)
6 3 3 7 8 14 5 10
Completio
n time(Ci)
6 9 12 19 27 41 46 56
Number of
jobs in the
system
8 7 6 5 4 3 2 1
Due
date(di)
15 25 50 45 10 40 15 20
Lateness(Ci
-di)
-9 -16 -38 -26 17 1 31 36

Mean flow time =
• Mean lateness = = -0.5 minutes

Earliest Due Date (EDD) Rule
• According to this rule jobs are sequenced in the order of non-
decreasing due dates. This rule minimizes the maximum job lateness
as well as maximum job tardiness.
• However, this rule tends to make more jobs tardy and increases the
mean tardiness.

Solve the previous example by applying EDD rule. Compare the results with those obtained from
SPT rule.
• Solution (EDD METHOD):
Job
sequence
2 1 3 5 4 6 7 8
Processing
time(ti)
8 5 6 10 3 14 7 3
Due date(di) 10 15 15 20 25 40 45 50
Number of
jobs in the
system
8 7 6 5 4 3 2 1
Completion
time(Ci)
8 13 19 29 32 46 53 56
Lateness(Ci-
di)
-2 -2 4 9 7 6 8 6

Mean flow time =
• Mean lateness = = 4.5 minutes
• Thus EDD rule has reduced the maximum lateness from 22 minutes to 9
minutes. However, it has increased the mean lateness from 3.625 minutes to 4.5
minutes. Also the number of late jobs has increased from 4 to 6

PARALLEL PROCESSORS UNDER SINGLE
MACHINE SCHEDULING
• In most of the single machine scheduling problems, it is possible to have more than
one machine of the same specification used to process a given set of single operation
jobs. Under this situation, there will be a scope to minimize makespan measure which
is considered to be very important among many measures.
• McNaughton has proposed a method to schedule such jobs on m parallel machines
when pre-emption of jobs is allowed(Interruption). If pre-emption of a scheduled job
is allowed, the processing of the job which is scheduled beyond mean makespan on a
machine is rescheduled to the next machine.
• The mean makespan (M) is computed by the following formula:
M = max {1/m max (tj)}
• where tj is the processing time of the job j, m is the number of parallel machines
(processors) and n is the number of independent single operation jobs.

McNaughton’s Algorithm
• Step 1: Select any one job and schedule it to machine-1 starting at
time zero.
• Step 2: Choose any unscheduled job and schedule it on the same
machine immediately after the previously scheduled job. Repeat this
step until the processing of the current job goes beyond M or until all
the jobs are scheduled.
• Step 3: Reschedule the processing of the job which is scheduled
beyond M to the next machine, starting at time zero. Then, go to step
2.

EXAMPLE : Sequence the following set of jobs with three parallel
processors (m = 3) to minimize makespan. Consider as pre-
emption is possible
Job 1 2 3 4 5 6 7
Duration
(tj)
1 2 3 4 5 6 6
Solution:
Mean makespan ,M = max {1/m, max (tj)}
M= max{*(1+2+3+4+5+6+6),6}= max{9,6} =9

Solve the previous problem if pre-
emption(Interruption) is not allowed
• Step 1. Form Longest Processing
Time(LPT) ordering of the jobs.
• Step 2. Select the job as per LPT
ordering and schedule it to the
machine with the least amount of
utilization (least amount of
processing already assigned).
• Step 3. Repeat Step 2 until all the
jobs are scheduled.

PROCESSING n JOBS THROUGH TWO MACHINES
• There are n different jobs to be processed on two machines and it is desired to
determine the optimal sequence of jobs that minimizes T, the total elapsed
time from the start of the first job on first machine to the completion of the last
job on second machine. The total elapsed time includes the idle time, if any
• The following conditions are assumed :
(i) Only two machines are involved, A and B.
(ii)Each job is processed in the order AB i.e., whichever job is processed first on
machine A must also be processed first on machine B and so on; no passing
being allowed.
(iii)Set-up times of machines A and B are independent of the sequence in which
the jobs are taken up.
(iv)Order of completion of jobs has no significance i.e., no job is required more
urgently than the other.

Johnson’s rule for 2 machine problem
•Step 1: Examine the columns of processing times on machines A and B and find the smallest
value.
• Step 2 : If this smallest value falls in machine A, schedule this job first on machine A. If this
value falls in column B, schedule this job last on machine A (because of the given order AB). If
there are equal minimal values (there is tie) one in each column, schedule the one in the first
column first on machine A; and the one in the second column, last on machine A. If both equal
values are in the first column (A), select the one with lowest entry in column B first. If the equal
values are in the second column (B), select the one with the lowest entry in column A first.
• Step 3 : Cross out the job assigned and continue the process (repeat steps 1 and 2), placing
the jobs next to first or next to last till all the jobs are scheduled. The resulting sequence will
minimize T.

EXAMPLE :
A machine operator has to perform two operations, turning and threading, on a number of different
jobs. The time required to perform these operations (in minutes) for each job is known. Determine
the order in which the jobs should be processed in order to minimize the total time required to turn
out all the jobs.
Also find the total processing time and idle times for turning and threading operations.
Job Time for turning
(minutes)
Time for threading
(minutes)
1 3 8
2 12 10
3 5 9
4 2 6
5 9 3
6 11 1

Sequencing and shedulding problems for Operations management

EXTENSION OF JOHNSON’S RULE (n JOBS THROUGH
THREE MACHINES)
• Extension of Johnson’s rule for
three machine problem can be
used only if any of the following
condition is satisfied.
Minimum time on machine A is ≥
Maximum time on Machine B
OR
Minimum time on machine C is ≥
Maximum time on Machine B

• Create a hypothetical problem with two machines and n jobs as shown in Table
• Then find the optimal sequence using Johnson’s rule
• From the optimal sequence construct the elapsed time table.

Min Ai=2,Max Bi=8,Since Min Ci(8)= Max Bi(8) we can apply
Johnson’s extended rule
Optimal sequence is

PALMER’S HEURISTIC
• If Johnson’s rule cannot be applied for a n Job Three machine problem, solve
using Palmer’s heuristic.
• Slope index ,
Sj = (m – 1) tjm + (m – 3) tj,m–1 + (m – 5) tj,m–2 + … – (m – 3) tj2 – (m – 1) tj1
• Procedure:
Step 1. Compute slope for each job.
Step 2. Arrange the jobs as per the decreasing order of slopes.

Slope calculation (m=3 here)
Job Processing time
Machine 1 Machine 2 Machine 3 Slope index
1 3 4 10 (3-1)*10+(3-3)*4+(3-5)*3=14
2 11 1 5 (3-1)*5+(3-3)*1+(3-5)*11=-12
3 7 9 13 (3-1)*13+(3-3)*9+(3-5)*7=12
4 10 12 2 (3-1)*2+(3-3)*12+(3-5)*10=-16
Optimal sequence= 1-3-2-4

Sequencing and shedulding problems for Operations management

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