Independent Tasks
Scheduling Idpt Tasks
Optimal Schedules
Reducing FinTime
Spring 2015
Mathematics in
Management Science
Independent Tasks
Tasks are independent when there
are no order requirements.
In this case, LPA simply says:
schedule next task on next proc.
Also, for project with independent
tasks, DTA=CPA; this is because
DTPLs are CTPLs.
3
Example
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2
2
Suppose have six independent tasks
with indicated task times.
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Scheduling 2 procs with DTA get
schedule with FinTime=7. Optimal?
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2
3
2
2
3
2
3
2
2
NO! Easy to find schedule with FinTime=6.
Optimal Finishing Time
A schedule is optimal if it has the
shortest possible completion time.
OptFinTime is smallest possible time
to finish overall project; can’t beat.
Always have
OptFinTime ≥ ECT ;
here ECT is the length of a critical
path (longest path from Start to End).
Optimal Finishing Time
Different PLs give different FinTimes.
OptFinTime is given by some PL.
Too many PLs to check every one.
How to tell if a schedule is optimal?
Or at least not too far away?
Generally not easy to tell, but there are
constraints on completion times that
can help.
Optimal Finishing Time
No schedule will take longer than the
total task time TTT=sum all task times.
No schedule can complete in less time
than ECT=length of a critical path.
No schedule can complete in less time
than total task time divided by the
number of processors.
Optimal Finishing Time
No schedule can complete in less time
than total task time divided by the
number of processors. That is, always
have
FinTime ≥ TTT / N
where TTT = total task time
and
N = # of processors .
Example
Have 150 exams to grade. Each exam
takes 1 hour to grade. So, ECT=1.
Only get if have 150 TAs to do grading!
50 TAs  FinT=3 30 TAs  FinT=5
40 TAs  FinT=4 (40+40+40+30=150)
Best FinT is TTT divided by # TAs
where TTT=150 (rounded up ).
Optimal Finishing Time
Always have lower bounds
FinTime ≥ ECT
FinTime ≥ TTT / N
where
ECT = earliest completion time,
TTT = total task time,
N = # of processors .
Optimal Finishing Time
Always have lower bounds
FinTime ≥ ECT
FinTime ≥ TTT / N.
So, if get schedule with FinTime either
ECT or TTT/N, then it is optimal!
Optimal Schedules
A schedule is optimal if
FinTime = CritTime
or
FinTime = TTT / N .
These not necessary conditions!
Example
CritTime=12
TTT=25, so
PL E, D, C, B, A
TTT/2=[12.5]=13
hence schedule
is optimal
P1
P2
Reducing FinTime
Suppose we try to decrease FinTime
by reducing individual task times.
How much effect will this have?
Which task times are the most
important ones to reduce?
[7]
Example
Start
CritPath SCEFE
ECT=14
[5]
[9]
0
[1]
[6]
End
0
[14]
If decrease task time for E by 2, … new ECT=13
If decrease task time for C by 2, … new ECT=12