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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

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

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

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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

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Step 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.

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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

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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

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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

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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

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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 )

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Example

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

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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

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Summary

1 || Uj forward algorithm

1 || wjUj is NP hard

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