Transcript Single MC – Independent Jobs 스케줄 이론

스케줄 이론
Single Machine Independent Jobs – part1
Problems without due dates
Single MC – Independent Jobs
스케줄 이론
1. Notational Conventions
 Definition of the problems
 (Method shown in Pinedo's text book 1st Ed.): A/B/C/D
 A) Job arrival pattern (Static=number of jobs, Dynamic=arrival distribution)
 B) Number of machines
 C) Flow pattern (Flow shop, Job shop, General Shop-either Flow shop or Job
shop)
 D) Evaluation Criteria
– (예) Static n job / 2 Machine / Flow shop / Min. Completion Time Problem  n/2/F/Cmin
– (예) n/m/G/Tmin
 I.2.2 Variables
– Problem Variables (Job Descriptors): small letters
– Decision Variables: Capital letters
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Single MC – Independent Jobs
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Jobs and machines
 (Method shown in Pinedo's text book 2st Ed.): ||
 Data (assumed to be given)
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m
number of machines
n
number of jobs
Pij,
tij
processing time of job j at machine i
Pj
tj
processing time of job j at when all machines are identical
rj
release date of job j
dj
due date of job j
wj
weight of job j
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Single MC – Independent Jobs
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Describing a scheduling problem
||
Machine
environment
Objective (to be
minimized)
Process
characteristics and
constraints
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Single MC – Independent Jobs
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Machine environment 
 Single machine and machines in parallel
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1
single machine
Pm
m identical machines in parallel
Qm
m machines in parallel w/different speeds vi
Rm
m unrelated machines in parallel
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Machine environment  (2)
 Machines in series
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Fm
flow shop: all jobs processed in the same order on the machines
FFc
flexible flow shop: same as flow shop but with c stages of parallel
machines
Jm
job shop: each job has its own routing
FJc
flexible job shop: same as job shop but with c stages of parallel
machines
Om
open shop: each job has to be processed on all machines but no
routing restrictions
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Single MC – Independent Jobs
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Processing characteristics and constraints 
could be empty!
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rj
Release dates
sjk
sequence dependent setup times
sijk
sequence and machine dependent setup times
prmp
preemption
prec
precedence constraints
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Processing characteristics and constraints  (2)
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brkdwn
breakdowns
Mj
machine eligibility restrictions
prmu
permutation
block
blocking
nwt
no waiting
recrc
recirculation
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Objectives 
 Performance measures of individual jobs
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Cj
completion time of job j
Lj
lateness = Cj – dj
Tj
tardiness = max(Lj, 0)
Ej
earliness = max(-Lj, 0)
Uj
unit penalty = 1 if Cj > dj and 0 otherwise
hj(Cj)
hj is a non-decreasing cost function
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Objectives  (2)
 Functions to be minimized
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Cmax = max Cj
makespan
Lmax = max Lj
maximum lateness
ΣwjCj
total weighted completion time
Σwj(1-e-rCj)
total weighted discounted Cj
ΣwjTj
total weighted tardiness
ΣwjUj
weighted number of tardy jobs
Σwj' Ej+ Σwj'' Tj
total weighted earliness and tardiness
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Single MC – Independent Jobs
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1. Notational Conventions
 Problem Variables
Jj
j  [1,..., n]
 a. Jobs
sometimes ( J1 , J 2 ,..., J n )
 b. Machines i  [1,..., m]
sometimes (m1 , m2 ,..., mn ) M
i
 c. Processing Time ( t j orPj ): processing time of Job j
( tij or Pij ): i-th operation of j-th Job
 d. Ready Time ( r j ): the earliest time that processing of the first operation of
Job j could begin
 e. Due Date ( d j ): the time by which the processing of the last operation is due to
be completed
 f. Allowed Time (a j ): allowed time in the shop (= d  r )
j
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j
Single MC – Independent Jobs
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1. Notational Conventions
 Decision Variables
 a. Completion Time ( C j ): Absolute time
 b. Flow Time ( F j ): Time spent in the shop = C j  rj
 c. Lateness( L j ) = C j  d j
– Negative Lateness is possible 5 ( C ) - 10 ( d ) = - 5
j
j
– Any Good Points in having Negative L j ?
 d. Tardiness( T j ) = Max{0, L j }
 e. Earliness( E j ) = Max{0,  L j }
– Tardiness and earliness are all positive numbers.
 f. Waiting Time( W j ) = C j  rj  t j
C j  rj  t j  W j
F j  t j  W j  C j  rj
 (Digression)
 We may think of Wij , the waiting time for the i-th operation of Job j
 (?) A schedule is completely described by a set of Wij
 ⇔ 2 Schedules ( ∧ & ~ ) are identical / equivalent (w.r.t. some performance
criteria) iff Wˆ  W element-wise
ij
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ij
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1. Notational Conventions
 Performance Measures
1 n
F   Fi
n i 1
 ↓MFT (Mean Flow Time) =
1
 Tj
n
 ↓Mean Tardiness =
T 
 ↓Max Flow Time =
Fmax  Max1 j n{Fj }
 ↓Max Tardiness =
Tmax  Max1 j n{Tj }
 ↓# of Tardy Jobs =
n
NT    (T j )
j 1
 (T j )  1
if
Tj  0
 (T j )  0 otherwise
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1. Notational Conventions
 (Definition) Regular Measure (Z) of Performance refer to a
performance measure for the following case:
 ① Scheduling objective is to minimize Z.
 ② Performance measure(Z) is a function of completion time. 즉
 ③ Increases only if one of C j ' s increases.
Namely, we have Z   f (C1,..., Cn ) such that
Z  Z   Ci  Ci for at least one i, i=1, ..., n
 (Example) Quantities that are not Regular Measure
 Average Earliness
 Max Earliness
 Difference between the largest to second largest completion times
 We need to consider only Dominant Schedule Sets
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Z  f (C1,..., Cn )
Single MC – Independent Jobs
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1. Notational Conventions
 (Definition) Dominant Set
 A concept to reduce solution space from complete enumeration
 Reasoning Procedures
–
–
–
–
1. Consider an arbitrary schedule
SD
(S is a string consisting of C j ' s)
where D is a set of a certain class of schedules
2. Show that ∃ a schedule S   Dwhere C j '  Cfor
all j.
j
3. For regular measures, above implies
Z   f (Ci )  Z  f (Ci )
4. It is sufficient to consider schedules in D only.
 (Example) Suppose (Mean Tardiness) is the measure of performance in
single m/c scheduling
problem.
Now suppose there exists a job k that
T
n
satisfies d k   t j , then there exists an optimal sequence in which job k is
j 1
assigned the last.
 ⇒ We can consider only n-1 jobs excluding job k.
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2. Introduction
 Single Machine is Not that Restrictive in Real Applications
 Chemical Process Industry: whole facility can be regarded as one M/C.
 Bottleneck Process in Process Industries as well as Machine Industries
(Temporary Bottleneck)
 Single Processor Computing System
 Tape Drive, etc.
 Basic Single-machine Assumptions





1. We have n independent, single-operation jobs.
2. Sequence independent set-up time can be included in each processing time.
3. Job descriptors( t j , rj , d j ) are completely known.
4. No idle time.
5. No interruption once a job is started.
 (Example) Process industry. Is assumption 2 reasonable?
 We need consider only Permutation Schedules (The total
number of possible Schedules: n! )
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2. Introduction
 Permutation Schedule
 Schedules are completely specified by giving processing order (n!).
  So we call these sequencing problems.
  So Performance measures such as “Max flow time”, “Max # of tardy jobs”
are irrelevant.
 Use Bracket to indicate position in sequence
– [5]=2
– d[1] = ?
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2. Introduction
 Theorem 2.1
With above assumptions, schedules without inserted idle time
constitute a dominant set.
 Proof
– Obvious but we need a formal proof.
– (Hint) Consider two schedules S and S'.
 Theorem 2.2
With above assumptions, schedules without preemption constitute
a dominant set.
 Proof
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3. Problems Without Due Dates
 The relationship between FLOW TIME and INVENTORY
 Are they proportional? Let us prove it in two cases (Static and Dynamic)
a. Static Case
– Let J(t) = # of Jobs in System at time t, V(t) = Inventory Level at time t
n
Fmax   t[ j ]   t j
j 1
J
1
[nt[1]  (n  1)t[2]  t[ n ] ]  A / Fmax
Fmax
th
( [i] means the i job in sequence. So [1]=2 , ...)
( A : sum of the areas of rectangles)
1
F  [ F[1]  F[2]  F[ n ] ]  A / n
n
– By rearranging for A, we get
– 즉 minimizing
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F
A  Fmax J  nF  F  J 
is directly proportional to minimizing
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Fmax
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J
Single MC – Independent Jobs
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3. Problems Without Due Dates
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3. Problems Without Due Dates
 b. Dynamic Case (Say, Job arrival is
r1  r2  ...  rn
)
– Case 1. Assume C  C  ...  C . 즉 Jobs are completed in arriving order.
1
2
n
– Also assume everything completed by C .
n
– Now consider
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3. Problems Without Due Dates
– So

Cn
0
n
n
j 1
j 1
J (t )  (C j  rj )   Fj
F

J 
j
Cn
Now, F
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Fj
n
n




F
rn  Fn
n
rn  Fn
F
r

(
) J(
j
n
n

Fn
)
n
n
r
J n
n
 J  (Mean Job Arrival Time)
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3. Problems Without Due Dates
– Case 2. Jobs are Completed In Random Order
– Now consider the case when jobs may finish in random.
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3. Problems Without Due Dates
Let Yj (t)= Fraction of Fj still remaining after time t.

Cn
0
n
J (t ) dt  (1  Y j (Cn )) Fj
j 1
n
n
j 1
j 1
  F j   Y j ( Cn ) F j
Then

J 
Cn
  Fj  Y j (Cn ) Fj
n
Now




Cn
rn  Fn  n
n
Y j (Cn ) Fj
rn Fn

 J(  )  F 
n
n
n
0
Again,
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J (t )dt 
F  J  (Mean Job Arrival Time) in steady state.
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
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3. Problems Without Due Dates
c. Discussion
①
F  J  (Mean Job Arrival Time)
Under steady state assumption (Rate of completion  Rate of arrival)
above holds : 1) Static or Dynamic
2) No matter how we schedule ( FIFO or Whatever )
3) Even with weighted inventory costs
② Also think
F j  C j  rj  C j  d j  d j  r j  L j  a j
F j  C j  rj  C j  rj  t j  t j  W j  t j
Above implies
F  J  (MeanJobArrivalTime)  L  a  W  t
in Steady State
A schedule which minimizes MFT also minimizes inventory (i.e. # of jobs in system), mean lateness,
mean waiting time.
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Proof
 Proof of Theorem 2.1
Let
C j (C j )  Completion time of job j in schedule S (S)
B  Set of Jobs finished before time a
tB  Time when all jobs in B are finished
Then j , such that C j  t B , C j = C j
and j , such that C j > t B , C j  C j
 For regular measures, Z  Z
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Proof
 Proof of Theorem 2.2
Consider a schedule S in which job i started at t B . But before it is completed,
job i is preempted by job j at time t C .
(For simplicity, assume job j is not peempted.)
At some later time t D , work resumes on job i and it is ultimately completed.
S
S'
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Proof
Consider S' where position of job j is interchanged with the first positon of job i.
Simply C j

 C j and nothing changes other than that.
Z  Z for regular measure
Repeating the same for all jobs intervening job i, any preemption can be eliminated
without increasing any regular measure.
 It is Sufficient to consider only schedules without preemption.
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3. Problems Without Due Dates
 Theorem 2.3 SPT (Shortest Processing Time) sequencing minimizes Mean
Flow Time.
 Proof
n
n
n
n
①
F
1
1
1
1
Fi   (Wi  ti )  Wi   ti

n i 1
n i 1
n i 1
n i 1
n
In the last equation, the second term is a constant. So minizing Wi is the same as choosing a
i 1
sequence to make Wi 's as small as possible. But W(1)  0, W(2)  t(1) , W(3)  t(1)  t(2) , ...
So SPT minimzes Mean Flow Time.
② Graphical Proof of Baker's.
③
1 n j
1 n
F
t

n
j 1 k 1
[k ]

(n  j  1)t

n
[ j]
j 1
n
(n-j-1) is a non-increasing sequence. So to make
 (n  j  1)t
j 1
should be in non-decreasing sequence.
④ Formal (?)
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[ j]
as small as possible, t[ j ]
Single MC – Independent Jobs
스케줄 이론
3. Problems Without Due Dates
 (Discussion)
SPT minimizes: Mean Completion Time ( f j  C j - rj ),
Mean Waiting Time (W j  C j - rj - t j ),
Max Waiting Time, ....
LPT Maximizes: ...
 Theorem 2.4 WSPT(Weighted SPT) sequencing minimizes
WMFT (Weighted Mean Flow Time).
WSPT is
t[1]
w[1]

t[2]
w[2]
 .......
t[ n]
w[ n ]
Proof : Follow the reasoning of ④.
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