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Getting rid of stochasticity
(applicable sometimes)
Han HoogeveenUniversiteit 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
• 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 order2. Let S denote the EDD schedule3. 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,…,j5. 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
E1 remainder
time0
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.
DP (2)
• Add job j+1• Recurrence relation:
If the number of on time jobs k can be reached without job j+1, then compute fj+1(k)=min{fj(k), fj(k-1)+pj+1}
• Effectively: the longest job of the old setplus job j+1 gets removed.
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
• 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
• 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
pj Var dj mspj
Job 1 12 1 20 0.5
Job 2 8 1 21 0.95
Normally distributed processing times
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)