Spring, 2007 8-1 Scheduling Operations. Spring, 2007 8-2 Scheduling Problems in Operations Job Shop Scheduling. Personnel Scheduling Facilities Scheduling

SpringSpring, 2007, 20078-1

Scheduling Operations


Scheduling Problems in Operations Job Shop Scheduling.

Personnel Scheduling Facilities Scheduling Vehicle Scheduling and Routing Project Management Dynamic versus Static Scheduling


The Hierarchy of Production DecisionsThe logical sequence of operations in factory

planning corresponds to the following sequence All planning starts with the demand forecast. Demand forecasts are the basis for the top level (aggregate)

planning. The Master Production Schedule (MPS) is the result of

disaggregating aggregate plans down to the individual item level.

Based on the MPS, MRP is used to determine the size and timing of component and subassembly production.

Detailed shop floor schedules are required to meet production plans resulting from the MRP.


Hierarchy of Production Decisions










Characteristics of the Job Shop Scheduling Problem

Job Arrival Pattern Number and Variety of Machines Number and skill level of workers Flow Patterns Evaluation of Alternative Rules


Objectives in Job Shop Scheduling

Meet due dates Minimize work-in-process (WIP) inventory Minimize average flow time Maximize machine/worker utilization Reduce set-up times for changeovers Minimize direct production and labor costs

(note: that these objectives can be conflicting)

SpringSpring, 2007, 20078-15Terminology

Flow shop: n jobs processed through m machines in the same sequence

Job shop: the sequencing of jobs through machines may be different, and there may be multiple operations on some machines.

Parallel processing vs. sequential processing: parallel processing means that the machines are identical.

Flow time of job i: Time elapsed from initiation of first job until completion of job i.

Makespan: Flow time of the job completed last. Tardiness: The positive difference between the completion time

and the due date. Lateness: Difference between completion time and due date

(may be negative).


Common Sequencing Rules

FCFS. First Come First Served. Jobs processed in the order they come to the shop.

SPT. Shortest Processing Time. Jobs with the shortest processing time are scheduled first.

EDD. Earliest Due Date. Jobs are sequenced according to their due dates.

CR. Critical Ratio. Compute the ratio of processing time of the job and remaining time until the due date. Schedule the job with the largest CR value next.


Results for Single Machine Sequencing The rule that minimizes the mean flow time of all

jobs is SPT. The following criteria are equivalent:

Mean flow time Mean waiting time. Mean lateness

Moore’s algorithm minimizes number of tardy jobs Lawler’s algorithm minimizes the maximum flow

time subject to precedence constraints.

SpringSpring, 2007, 20078-18Results for Multiple

Machines The optimal solution for scheduling n jobs on two

machines is always a permutation schedule (that is, jobs are done in the same order on both machines). (This is the basis for Johnson’s algorithm.)

For three machines, a permutation schedule is still optimal if we restrict attention to total flow time only. Under rare circumstances, the two machine algorithm can be used to solve the three machine case.

When scheduling two jobs on m machines, the problem can be solved by graphical means.


Stochastic Scheduling: Static Case Single machine case. Suppose that

processing times are random variables. If the objective is to minimize average weighted flow time, jobs are sequenced according to expected weighted SPT. That is, if job times are t1, t2, . . ., and the respective weights are u1, u2, . . . then job i precedes job i+1 if E(ti) / ui < E(ti+1) / ui+1.


Stochastic Scheduling: Static Case (continued)

Multiple Machines. Requires the assumption that the distribution of job times is exponential, (memoryless property). Assume parallel processing of n jobs on two machines. Then the optimal sequence is to to schedule the jobs according to LEPT (longest expected processing time first).

Johnsons algorithm for scheduling n jobs on two machines in the deterministic case has a natural extension to the stochastic case as long as the job times are exponentially distributed.


Stochastic Scheduling: Dynamic Analysis

When jobs arrive to the shop dynamically over time, queueing theory provides a means of analyzing the results. The standard M/M/1 queue applies to the case of purely random arrivals to a single machine with random processing times.

If the selection discipline does not depend on the flow times, the mean flow times are the same, but the variance of the flow times will differ.

If job times are realized when the job joins the queue rather than when the job enters service, SPT generally results in lowest expected flow time.

SpringSpring, 2007, 2007

22 of 52

8-22

Single Machine Deterministic Models

Jobs: J1, J2, ..., Jn

Assumptions:

• The machine is always available throughout the scheduling period.• The machine cannot process more than one job at a time.• Each job must spend on the machine a prescribed length of time.


t

tJktS k

at time processed is job no if0

at time processed is job if)(

1

3

2

J1J2 J3 ...

S(t)

t


Requirements that may restrict the feasibility of schedules:

• precedence constraints• no preemptions• release dates• deadlines

Whether some feasible schedule exist? NP hard

Objective function f is used to compare schedules.

f(S) < f(S') whenever schedule S is considered to be better than S'problem of minimising f(S) over the set of feasible schedules.


1. Completion Time Models

Due date related objectives:

2. Lateness Models

3. Tardiness Models

4. Sequence-Dependent Setup Problems


Completion Time Models

Contents

1. An algorithm which gives an optimal schedulewith the minimum total weighted completion time 1 || wjCj

2. An algorithm which gives an optimal schedule with the minimum total weighted completion time when the jobs are subject to precedence relationshipthat take the form of chains1 | chain | wjCj


Literature:• Scheduling, Theory, Algorithms, and Systems, Michael Pinedo,

Prentice Hall, 1995, or new: Second Addition, 2002, Chapter 3.


1 || wjCj

Theorem. The weighted shortest processing time first rule (WSPT) isoptimal for 1 || wjCj

WSPT: jobs are ordered in decreasing order of wj/pj

The next follows trivially:

The problem 1 || Cj is solved by a sequence S with jobs arranged innondecreasing order of processing times.


Proof. By contradiction.

S is a schedule, not WSPT, that is optimal.

j and k are two adjacent jobs such that

k

k

j

j

p

w

p

w which implies that wj pk < wk pj

t t + pj + pk

j kS:

t t + pj + pk

k jS’

......

... ...

S: (t+pj) wj + (t+pj+pk) wk = t wj + pj wj + t wk + pj wk + pk wk

S’: (t+pk) wk + (t+pk+pj) wj = t wk + pk wk + t wj + pk wj + pj wj

the completion time for S’ < completion time for S contradiction!


1 | chain | wjCj

chain 1: 1 2 ... k chain 2: k+1 k+2 ... n

Lemma. If

n

kjj

n

kjj

k

jj

k

jj

p

w

p

w

1

1

1

1

the chain of jobs 1,...,k precedes the chain of jobs k+1,...,n.

Let l* satisfy

l

jj

l

jj

kll

jj

l

jj

p

w

p

w

1

1

1*

1

*

1max

factor of chain 1,...,kl* is the job that determines the factor of the chain


Lemma. If job l* determines (1,...,k) , then there exists an optimalsequence that processes jobs 1,...,l* one after another withoutinterruption by jobs from other chains.

Algorithm

Whenever the machine is free, select among the remaining chainsthe one with the highest factor. Process this chain up to and includingthe job l* that determines its factor.


Example

chain 1: 1 2 3 4chain 2: 5 6 7

jobs 1 2 3 4 5 6 7wj 6 18 12 8 8 17 18pj 3 6 6 5 4 8 10

factor of chain 1 is determined by job 2: (6+18)/(3+6)=2.67 factor of chain 2 is determined by job 6: (8+17)/(4+8)=2.08chain 1 is selected: jobs 1, 2

factor of the remaining part of chain 1 is determined by job 3:12/6=2 factor of chain 2 is determined by job 6: 2.08chain 2 is selected: jobs 5, 6


factor of the remaining part of chain 1 is determined by job 3: 2 factor of the remaining part of chain 2 is determined by job 7:18/10=1.8chain 1 is selected: job 3

factor of the remaining part of chain 1 is determined by job 4: 8/5=1.6 factor of the remaining part of chain 2 is determined by job 7: 1.8chain 2 is selected: job 7

job 4 is scheduled last

the final schedule: 1, 2, 5, 6, 3, 7, 4


1 | prec | wjCj

Polynomial time algorithms for the more complex precedenceconstraints than the simple chains are developed.

The problems with arbitrary precedence relation are NP hard.

1 | rj, prmp | wjCj preemptive version of the WSPT ruledoes not always lead to an optimal solution,the problem is NP hard

1 | rj, prmp | Cj preemptive version of the SPT rule is optimal

1 | rj | Cj is NP hard


Summary

1 || wjCj WSPT rule

1 | chain | wjCj a polynomial time algorithm is given

1 | prec | wjCj with arbitrary precedence relation is NP hard

1 | rj, prmp | wjCjthe problem is NP hard

1 | rj, prmp | Cj preemptive version of the SPT rule is optimal

1 | rj | Cj is NP hard


36 of 52

8-36

Lateness Models

Contents1. Lawler’s algorithm which gives an optimal schedule

with the minimum cost hmax

when the jobs are subject to precedence relationship 1 | prec | hmax

2. A branch-and-bound algorithm for the scheduling problemswith the objective to minimise lateness 1 | rj | Lmax


Prentice Hall, 1995, Chapter 3.2or new: Second Addition, 2002, Chapter 3.


Lawler’s Algorithm

• Backward algorithm which gives an optimal schedule for1 | prec | hmax hmax = max ( h1(C1), ... ,hn(Cn) )

hj are nondecreasing cost functions

Notation

makespan Cmax = pj completion of the last job

J set of jobs already scheduledthey have to be processed during the time interval

JC complement of set J, set of jobs still to be scheduledJ' JC set of jobs that can be scheduled immediately

before set J (schedulable jobs)

Jjj CpC maxmax ,


Step 1.

J = JC = {1,...,n}k = n

Step 2.

Let j* be such that

CCC Jj

jjJjJj

jj phph min*

Place j* in J in the k-th order positionDelete j* from JC

Step 3.

If JC = then Stopelse k = k - 1

go to Step 2

Lawler’s Algorithm for 1 | | hmax


Example (no precedence relationships between jobs)

jobs 1 2 3pj 2 3 5hj(Cj) 1+C1 1.2C2 10

J = JC={1, 2, 3} jobs still to be scheduledCmax = 10h1(10) = 11h2(10) =12h3(10) =10 Job 3 is scheduled last and has to be processed in [5, 10].

105... 3


J = {3} JC={1, 2} jobs still to be scheduledCmax = 5h1(5) = 6h2(5) = 6Either job 1 or job 2 may be processed before job 3.

105321

105312

or

Two schedules are optimal: 1, 2, 3 and 2, 1, 3


Step 1.

J = , JC = {1,...,n}J' the set of all jobs with no successorsk = n

Step 2.

Let j* be such that

CC Jj

jjJjJj

jj phph'

* min

Place j* in J in the k-th order position Delete j* from JC

Modify J' to represent the set of jobs which can be scheduledimmediately before set J.Step 3.

If JC = then Stopelse k = k - 1

go to Step 2

Lawler’s Algorithm for 1 | prec | hmax


Example. What will happen in the previous example if theprecedence 1 2 has to be taken into account?

J = JC={1, 2, 3} still to be scheduled J'={2, 3} have no successorsCmax = 10h2(10) = 12h3(10) = 10

J = {3}JC={1, 2} still to be scheduled J'={2} can be scheduled immediately before J Cmax = 5h2(5) = 6J = {3, 2} JC={1} J'={1} h1(2) = 3Optimal schedule: 1, 2, 3,hmax = 10

105... 3

105321

10532

2

2


1 || Lmax is the special case of the 1 | prec | hmax

where hj = Cj - dj

algorithm results in the schedule that orders jobs in increasing orderof their due dates - earliest due date first rule (EDD)

1 | rj | Lmax is NP hard , branch-and-bound is used

1 | rj , prec | Lmax similar branch-and-bound


Branch-and-bound algorithm

• Search space can grow very large as the number of variablesin the problem increases!

• Branch-and-bound is a heuristic that works on the idea of successivepartitioning of the search space.

S

S1 S2 Sn

S12 S13

. . .

. . .

. . .

S = S1 S2... Sn

S1 S2... Sn =


S

S1 S2 Sn

S12 S13

. . .

. . .

• We need some means for obtaining a lower bound on the cost forany particular solution (the task is to minimise the cost).

fbound f(x), xS1

fbound f(x), xS2

there is no need to explore S2


Branch-and-bound algorithm

Step 1Initialise P = Si (determine the partitions)Initialise fbound

Step 2Remove best partition Si from PReduce or subdivide Si into Sij

Update fbound

P = P Sij

For all SijP doif lower bound of f(Sij) > fbound then remove Sij from P

Step 3If not termination condition then go to Step 2


*,*,*,*

1,*,*,* 2,*,*,* n,*,*,*

1,2,*,* 1,3,*,*

. . .

. . .

. . .

Branch-and-bound algorithm for 1 | rj | Lmax

Solution space contains n! schedules (n is number of jobs).Total enumeration is not viable !


Branching rule:k-1 level, j1, ... , jk-1 are scheduled,jk need to be considered if no job still to be scheduled can notbe processed before the release time of jk that is: )),(max(min ll

Jlj prtrk

J set of jobs not yet scheduledt is time when jk-1 is completed

*,*,*,*

1,*,*,* 2,*,*,* n,*,*,*

1,2,*,* 1,3,*,*

. . .

. . .

. . .

1st level

2nd level


Lower bound:

• Preemptive earliest due date (EDD) rule is optimal for 1 | rj prmp | Lmax

A preemptive schedule will have a maximum latenessnot greater than a non-preemtive schedule.

• If a preemptive EDD rule gives a nonpreemptive schedule thenall nodes with a larger lower bound can be disregarded.


Example.jobs 1 2pj 4 5rj 3 0dj 4 6

• Non-preemptive schedules

0

1 2

3 7 12

2 1

0 5 9

L1=3L2=6Lmax=6

L1=5L2=-1Lmax=5

• Preemptive schedule obtained using EDD

2 1 2

0 3 7 9

L1=3L2=3Lmax=3

the lowest Lmax !


Example

jobs 1 2 3 4pj 4 2 6 5rj 0 1 3 5dj 8 12 11 10

*,*,*,*

1,*,*,* 2,*,*,* 4,*,*,*

1,2,*,* 1,3,*,*

3,*,*,*

job 2 could be processedbefore job 3

job 1 could be processedbefore job 4

1,3,4,2

L.B. = 5 L.B. = 7

L.B. = 6L.B. = 5


*, *, *, *

1,*,*,*1 [0, 4] L1=-43 [4, 5]4 [5, 10] L4=03 [10, 15] L3=42 [15, 17] L2=5

2,*,*,*2 [1, 3] L2=-91 [3, 7] L1=-14 [7, 12] L4=23 [12, 18] L3=7

1,2,*,*1 [0, 4] L1=-42 [4, 6] L2=-64 [6, 11] L4=13 [11, 17] L3=6

1,3,*,*1 [0, 4] L1=-43 [4, 10] L3=-14 [10, 15] L4=52 [15, 17] L3=5

Schedule: 1, 3, 4, 2,

4,*,*,*either job 1 or 2can be processedbefore 4 !

3,*,*,*job 2can be processedbefore 3 !


Summary

1 | prec | hmax , hmax=max( h1(C1), ... ,hn(Cn) ), Lawler’s algorithm

1 || Lmax EDD rule

1 | rj | Lmax is NP hard , branch-and-bound is used

1 | rj , prec | Lmax similar branch-and-bound

1 | rj, prmp | Lmax preemptive EDD rule


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8-54

Tardiness Models

Contents

1. Moor’s algorithm which gives an optimal schedulewith the minimum number of tardy jobs 1 || Uj

2. An algorithm which gives an optimal schedule with the minimum total tardiness 1 || Tj


Prentice Hall, 1995, Chapters 3.3 and 3.4or new: Second Addition, 2002, Chapter 3.


Optimal schedule has this form jd1,...,jdk

, jt1,...,jtl

NotationJ set of jobs already scheduledJC set of jobs still to be scheduledJd set of jobs already considered for scheduling, but which have

been discarded because they will not meet their due date in theoptimal schedule

meet their due datesEDD rule

do not meet their due dates

Moor’s algorithm for 1 || Uj

SpringSpring, 2007, 20078-56Step 1.

J = Jd = JC = {1,...,n}

Step 2.Let j* be such that j

Jjj dd

Cmin*

Add j* to JDelete j* from JC

Step 3.If *j

dpJj

j

else let k* be such that jJj

k pp

max*

Delete k* from JAdd k* to Jd

Step 4.If Jd = STOPelse go to Step 2.

then go to Step 4.


jobs 1 2 3 4 5pj 7 8 4 6 6dj 9 17 18 19 21

Example

J = , Jd = , JC = {1,...,5}

j*=1 J = {1} , Jd = , JC = {2, 3, 4, 5}, t=7 < 9 = d1

71

71 2

15

71 2

15319

j*=2 J = {1, 2} , Jd = , JC = {3, 4, 5}, t=15 < 17 = d2

j*=3 J = {1, 2, 3} , Jd = , JC = {4, 5}, t=19 > 18 = d3

k*=2 J = {1, 3} , Jd = {2}, t=11


j*=4 J = {1, 3, 4} , Jd = {2}, JC = {5}, t=17 < 19 = d4

j*=5 J = {1, 3, 4, 5} , Jd = {2}, JC = , t=23 > 21 = d5

k*=1 J = {3, 4, 5} , Jd = {2, 1}, t=16 < 21 = d5

optimal schedule 3, 4, 5, 1, 2 Uj = 2

71 3

154

17

71 3

154

175

23


The Total Tardiness

1 || Tj is NP hard

Lemma. If pj < pk and dj < dk then there exists an optimal sequencein which job j is scheduled before job k.

d1 ... dn and pk = max(p1, ... , pn){1, ... ,k-1}any order

k

{k+1, ... ,n}

Lemma.There exists an integer , 0 n-k such that there is anoptimal schedule S in which job k is preceded by jobs j k + andfollowed by jobs j > k + .

k{1, ... ,k-1, k+1, ..., k+ }any order

{k++1, ..., n}any order

kjjk ptC )(

completion timeof job k


PRINCIPLE OF OPTIMALITY, Bellman 1956. An optimal policy has the property that whatever the initial state andthe initial decision are, the remaining decisions must constitute anoptimal policy with regard to the state resulting from the firstdecision.

Algorithm

Dynamic programming procedure: recursively the optimal solution forsome job set J starting at time t is determined from the optimal solutionsto subproblems defined by job subsets of S*S with start times t*t .

J(j, l, k) contains all the jobs in a set {j, j+1, ... , l} with processing time pk

V( J(j, l, k) , t) total tardiness of the subset under an optimal sequenceif this subset starts at time t


Initial conditions:V(, t) = 0V( { j }, t ) = max (0, t + pj - dj)

Recursive conditions:

)))(,)',,1'((

))(,0max(

),)',',(((min),),,((

'

''

k

kk

CklkJV

dC

tkkjJVtkljJV

where k' is such thatpk' = max ( pj' | j' J(j, l, k) )

Optimal value function is obtained forV( { 1,...,n }, 0 )

SpringSpring, 2007, 20078-62Example

jobs 1 2 3 4 5pj 121 79 147 83 130dj 260 266 266 336 337

k'=3, 0 2 dk' = d3 = 266

)560,(294)0),3,5,1((

)430),3,5,5((164)0),3,4,1((

)347),3,5,4((81)0),3,3,1((

min)0}),5,...,2,1({

VJV

JVJV

JVJV

V

V( J(1, 3, 3) , 0) = 0 1, 2 C2 = 200 < 266 = d2

T1+ T2 = 02, 1 C1 = 200 < 260 = d1

T2+ T1 = 0C3(0) - d3 = 121 + 79 + 147 - 266 = 347 - 266 = 81


V( J(4, 5, 3) , 347) 4, 5 T4 = 430 - 336 = 94T5 = 560 - 337 = 229

T4 + T4 = 3175, 4 T5 = 477 - 337 = 140

T4 = 560 - 336 = 224 T5+ T4 = 364

C3(1) - d3 = 347 + 83 - 266 = 430 - 266 = 164C3(2) - d3 = 430 + 130 - 266 = 560 - 266 = 294

V( J(1, 4, 3) , 0)=0 achieved with the sequence 1, 2, 4 and 2, 1, 4V( J(5, 5, 3) , 430)=223V( J(1, 5, 3) , 0)=76 achieved with the sequence 1, 2, 4, 5

and 2, 1, 4, 5 V( , 560)=0

370029476

2231640

317810

min)0},5,...,2,1({v

optimal sequences:1, 2, 4, 5, 3 and2, 1, 4, 5, 3


Summary

1 || Uj forward algorithm

1 || wjUj is NP hard

1 || Tj is NP hard, pseudo-polynomial algorithmbased on dynamic programming


Assembly Line Balancing

Characteristics of the Assembly Line Balancing problem. A collection of n tasks must be completed on each item Tasks are assigned to stations. Tasks must be sequenced

properly, and certain tasks may not be done at the same station.

The objective is to assign tasks to stations to minimize the cycle time, C.

The general problem is difficult to solve optimally, but effective heuristics are available. (the text discusses one known as the ranked positional weight technique.)


Schematic of a Typical Assembly Line

Documents

Spring, 2007 8-1 Scheduling Operations. Spring, 2007 8-2 Scheduling Problems in Operations Job Shop Scheduling. Personnel Scheduling Facilities Scheduling