Transcript Powerpoint slides late jobs

Getting rid of stochasticity
(applicable sometimes)
Han Hoogeveen
Universiteit Utrecht
Joint work with
Marjan van den Akker
Outline of the talk
• Problem description
• How to solve the deterministic problem
• Stochastic processing times
– consequences
– four classes of instances
• Stochastic machine
• Processing times and machine stochastic
Problem description
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•
•
•
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1 machine
n jobs become available at time 0
Known processing time pj
Known due date dj
Known reward wj for timely completion (currently 1)
Decision to make at time 0: accept or reject job j
minimize number of tardy jobs
on a single machine
Moore-Hodgson
1. Number the jobs in EDD order
2. Let S denote the EDD schedule
3. Find the first job not on time in S (suppose
this is job j)
4. Remove from S the largest available job
from jobs 1,…,j
5. Continue with Step 3 for this new schedule S
until all jobs are on time
Solving the problem from scratch
Observations
• First the on time jobs
• On time jobs in EDD order
• Forget about the late jobs (once rejected =
lost for ever)
Knowing the on time set is sufficient
Dominance rule
• Let E1 and E2 be two subsets of jobs 1,…,j
• All jobs in E1 and E2 are on time (feasible)
• Cardinality of E1 and E2 is equal
• The total processing time of the jobs in E2 is
more than the total processing time of the
jobs in E1
Then subset E2 can be discarded.
Proof (sketch)
Take an optimal schedule starting with E2
(remainder: jobs from j+1, …, n)
E2
remainder
time
0
E1
remainder
Apply Dynamic Programming
• Define Ej*(k): feasible subset of jobs 1,…,j
with cardinality k and minimum total
processing time
• Knowledge of the total processing time of
Ej*(k) is sufficient in the next step ⇒ use
state variables fj(k) to represent this, etc.
(details are omitted)
Further remarks
• DP computes more state variables than
necessary
• DP can be used for the weighted case:
– Use fj(W) with W is the total weight of the on time
set (instead of cardinality of the on time set)
• DP can be used for more problems
Stochastic processing times
• Completion times are uncertain
• Decision about accept or reject must
be made before running the schedule
• When do you consider a job on time?
On time stochastically
• Work with a sequence of on time jobs
(instead of a set of completion times)
• Add a job to this sequence and compute the
probability that it is ready on time
• If this probability is large enough (at least
equal to the minimum success probability
msp) then accept it as on time
Classes of processing times
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Gamma distribution
Negative binomial distribution
Equally disturbed processing times pj
Normal distribution
Jobs must be independent
Class 1: Gamma distribution
• Parameters aj and b (common)
• If X1 and X2 follow the gamma distribution
and are independent, then X1+X2 is gamma
distributed with parameters a1+a2 and b (we
call this additive).
More gamma
• Define S as the set containing job j and all its
predecessors in the schedule.
• Define p(S) as the sum of all processing times pj
of jobs in S; a(S) is defined similarly.
• Then Cj=p(S) follows a gamma distribution with
parameters a(S) and b.
Even more gamma
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Denote the msp of job j by yj
Job j is on time if P(Cj  dj)  yj.
The distribution of Cj depends on a(S) only
Given dj and yj, you can compute the
maximum value of a(S) such that
P(Cj  dj)  yj: call this maximum value Dj
Last of Gamma
Important: a(S)  Dj  P(Cj  dj)  yj
(S contains job j and its predecessors)
• Treat Dj as ordinary due dates
• Treat aj as ordinary deterministic processing
times
You can use Moore-Hodgson!
Negative binomial distribution
• Parameters s_j and p (common for all jobs)
• If independent, then C_j=p(S) follows a
negative binomial distribution with
parameters s(S) and p
Solvable like the gamma distribution
Same approach is viable if
• The distribution of pj depends on one specific
parameter aj
• The probability distribution is additive
• P(Cj  dj)  yj does not increase when a(S)
increases, irrespective of yj
The negative binomial distribution
possesses these characteristics
More complicated problems
Normally distributed processing times
pj
Var
dj
mspj
Job 1
12
1
20
0.5
Job 2
8
1
21
0.95
Optimum: first job 2 and then job 1
Necessary for DP: msp values and due
dates are oppositely ordered
Equal disturbances
• On time probability of job j depends on:
– Number of predecessors (on time jobs before j)
– Total processing time of its predecessors
• Dominance rule: given the cardinality of the on time
set, take the one with minimum total processing time
• Use dynamic programming with state variables fj(k)
that indicate the minimum total processing time
possible (as before)
• Hence: Moore-Hodgson’s solves it!
Normal distribution (1)
• Parameters: expected processing time of job j and
variance of job j
• Two parameters ⇒ simple approach fails
• Reminder: expected value and variances of X1+X2 are
equal to the respective sums
• Necessary for computing the on time probability of
job j:
– Total processing time of predecessors
– Total variance of predecessors
Normal distribution (2)
• Dominance rule: if cardinality and total
processing time are equal, then take the set
with minimum total variance (msp > 0.5)
• Use state variables fj(k,P):
– k is cardinality of on time set
– P is total processing time of on time set
– fj(k,P) is minimum variance possible
Normal distribution: details
• Running time pseudo-polynomial
• Problem is NP-hard
• Role of total variance and total processing
time in the dominance rule and in the DP is
interchangeable
Unreliable machine
• Assume deterministic processing times
• Amount of work done by the machine per
period is stochastic
• Define X(t) as the stochastic variable
denoting the amount of work done in [0,t]
• Assume that the probability distribution of
X(t) is known
Solution method
• Use the minimum success probability yj
• Again, use S to denote job j and its
predecessors in the schedule; the total
processing time of S is p(S)
• Job j is stochastically on time if
P(X(dj)  p(S))  yj
Solution method (2)
• Compute Dj as the maximum value of p(S)
such that P(X(dj)  p(S))  yj .
• Treat the Dj values as the traditional due
dates for a reliable machine.
=> Traditional deterministic problem
Moore-Hodgson solves it
Everything unreliable
• Same approach as with the unreliable
machine:
– X(t) denotes the amount of work done in [0,t]
– p(S) is total processing time of the jobs in S; this
is a stochastic variable now
– Job j is stochastically on time if
P(X(dj)  p(S))  yj
Gamma distribution
• Now p(S) is gamma distributed with
parameter a(S)
• Compute Dj as the maximum value of a(S)
such that P(X(dj)  p(S))  yj.
• Treat Dj as ordinary due dates again
• Use aj as processing time of job j
=> Traditional deterministic problem
Moore-Hodgson solves it
Conclusion
• We can sometimes get rid of
stochasticity by
– using minimum success probability
– concept of stochastically on time
• Necessary to compute the probabilities
– either analytically
– or numerically (simulation)
Conclusion (2)
• Unreliable machines are made reliable
again
• Stochastic processing times are
represented by their deterministic
parameter
• Then the deterministic algorithm can be
applied again!