Minimizing the makespan for a UET bipartite graph on a single processor with an integer precedence delay

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Minimizing the makespan for a UET bipartite graph on

a single processor with an integer precedence delay

Alix Munier-Kordon

To cite this version:

Alix Munier-Kordon. Minimizing the makespan for a UET bipartite graph on a single processor with

an integer precedence delay. [Research Report] lip6.2001.016, LIP6. 2001. �hal-02545572�

graph on a single processor with an integer

precedence delay

AlixMUNIERKORDON

Lab oratoire LIP6,

4 place Jussieu,75 252 Paris cedex 05

[email protected]

Abstract

We consider aset oftasksof unitexecutiontimes andabipartite precedence delaysgraph with ap ositive precedence delay d : an arc (i;j)ofthisgraph meansthat j canb eexecuted at leastdtimeunits after thecompletiontimeofi. The problemis to sequence the tasks inorderto minimizethemakespan.

Firstly,weprovethattheasso ciateddecisionproblemis NP-comp-lete. Then,weprovideanontrivialp olynomialtimealgorithmifthe degreeofevery tasksfromone ofthetwosets is2. Lastly,we give an

approximationalgorithmwithratio 3 2 .

1 Introduction

Single andmultipro cessorsscheduling problems haveb eenextensively stud-iedin theliterature [16]. Scheduling problems withprecedencedelays arise indep endently in severalimp ortant applicationsand manytheoretical

stud-ies were devotedto these problems : this class of problems was considered forresource-constrained scheduling problem [3,13]. It wasalso studied asa relaxation for thejob-shop problem [1,8]. For computer systems, it

corre-sp ondstothebasic pip elines scheduling problems [15,20].

An instance of a scheduling problem with precedence delays is usually denedbyasetoftasksT =f1;:::;ngwithdurationsp

i

NP-HardProblem Reference 1jchains;d ij =djC max Wikumetal.[23] 1jprec;d ij =d;p i =1jC max Leungetal.[18]

every arc(i;j)2E,taskj can b eexecuted atleastd ij

timeunits after the completion timeof i. The numb er of pro cessors islimited. The problem is tondascheduleminimizing themakespan,orotherregularcriteria. Using

standard notations [16], the minimization of the makespan is denoted by Pjprec;d

ij jC

max .

In this pap er, we supp ose that the graph G is bipartite : T is split

into two setsX and Y and every arc (i;j) 2 E veries i 2 X and j 2 Y. We also consider that there is only one pro cessor, the duration of tasks is one and that the delay is the same for every arc. This problem is noted

1jbipartite;d ij = d;p i =1jC max

. The decision problem asso ciated is called SEQUENCING WITHDELAYS and is dened as:

 Instance: AbipartiteorientedgraphG=(X[Y;E),ap ositivedelay danda deadline D .

 Question : isthere asolution tothesequencing problem withadelay danda makespan smallerthanorequal to D ?

Weprove in section 2that 1jbipartite;d ij =d;p i =1jC max isNP-Hard. The complexity of this problem was a challenging question since several

authorsproved theNP-Hardness of moregeneral instances of this problem as shown in the table 1. In section 3, we prove that if the degree of every task in X is 2, then the problem is p olynomial and we provide a greedy algorithmtosolveit.

Several authorshave adaptedthe classical p olynomial algorithmsform pro cessors and particular graphs structures to a sequencing problem with

a unique delay as shownin the table 2. Note that Bampis [2] proved that Pjbipartite ;p

i

= 1jC max

is NP-Hard, but his transformation do esn't seem tob eeasily extended toourproblem.

Wikum et al. [23] also proved several complexity results, p olynomial

sp ecial cases and approximation algorithms for unusual particular classes of graphs (in fact, sub classes of trees). Munier and Sourd proved that 1jchains;d ij = d;p i = pjC max

Polynomial Problem Reference Comments 1jtree;d ij =d;p i =1jC max

Bruno etal.[6] Based on [14]

1jprec;d ij =1;p i =1jC max

Leung etal. [18] Based on [7]

1jintervalorders;d ij =d;p i =1jC max

Leung etal.[18] Based on [21]

develop ed a p olynomial algorithm forPjtree;d ij D ;p i =1jC max if D is a constant value.

At last,there aresomeapproximation algorithmsforproblems with

de-lays: Graham'slistschedulingalgorithm[11]wasextendedtoPjprec. delays ; d ij =k ;p j =1jC max

togive aworst-casep erformance ratioof 2 1=(m(k+ 1))[15,20]. ThisresultwasextendedbyMunieretal. [19]toPjprec.delays ;

d ij

jC max

. Bernstein and Gerner [5] study the p erformance ratio of the Coman-Graham algorithm for Pjprec. delays;d

ij = d;p i = 1jC max and slightly improve it in [4]. Schuurman [22]develop ed ap olynomial

approxi-mationscheme fora particularclassofprecedence constraints. Weprovein section 4 that the b ound 2 of Graham's list algorithm may b e achieved in

theworstcase for1jbipartite ;d ij

=d;p i

=1jC max

and we developa simple algorithmwith worstcasep erformance ratioequal to 3=2forthis problem.

2 Complexity of the problem

Let us consider a non orientedgraph G =(V;E)and an ordering L of the verticesofG(ie,aone-to-onefunctionL:V !f1;:::;jVjg). Forallinteger

i2f1;:::;jVjg,theset V L

(i)V is:

V L

(i)=fv2V;L(v)iand 9u2V;fv;ug2E and L(u)>ig

VERTEX SEPARATION is thendened as:

 Instance: A nonorientedgraphG=(V;E)and ap ositive integerK.

 Question : Is there an ordering L of the vertices of G such that, for

alli2f1;:::;jVjg,jV L

(i)jK?

This problem is proved to b e NP-complete in [17]. For the following, our pro ofswill b e moreelegant if we consider the converse ordering of the

tasks. Let n=jVj. If weset, 8v2V,L(v)=n L(v),j =n i+1and B

L

0(j)=V L

(i),wegetfor every value j2f1;:::;ng:

B L 0 (j)=fv2V;L 0 (v)>j and 9u2V;fv;ug2E and L 0 (u)jg

So,theequivalent INVERSE VERTEXSEPARATION problemmayb e dened as:

 Instance: A nonorientedgraphG=(V;E)and ap ositive integerK.

 Question : Is there an ordering L of the vertices of G such that, for alli2f1;:::;jVjg,jB

L

(i)jK ?

Weprove thefollowing theorem:

Theorem 2.1. Thereexistsapolynomialtransformationf fromINVERSE

VERTEX SEPARATIONto SEQUENCINGWITHDELAYS.

Proof. Let I b e an instance of INVERSE VERTEX SEPARATION. The

asso ciated instance f(I) is given by a bipartite graph G 0

= (X[Y;E 0

), a delayd anda deadline D dened as:

1. To any vertex v 2 V is asso ciated two elements x v 2 X and y v 2 Y andan arc(x v ;y v )2E 0 .

2. To anyedge fu;vg2E is asso ciated the arcs (x u ;y v ) and (x v ;y u ) in E 0 .

3. Thedelayis d=n 1 K and thedeadline D=2n.

f canb eclearlycomputedinp olynomial time(seeanexample gure1).

Let us supp ose thatL is asolution totheinstance I. Then,webuild a solution tof(I)asfollows :

1. Tasksfrom Y are executedb etweentime nand 2nfollowing L : they areexecuted fromy

L 1 (1) toy L 1 (n) .

2. Letus dene thepartition P i ;i=1:::nof X as: P i =fx L 1 (i) g[fx u ;u2B L (i)g i 1 [ j=1 P j

Tasksfrom X areexecuted b etween 0 and nfollowing P 1

:::P n

a

b

c

d

e

G = (V, E)

K = 2

x

a

x

b

x

c

x

d

x

e

y

e

y

d

y

c

y

b

y

a

d = 2

D = 10

f

Figure 1: Example oftransformationf

x

b

x

a

x

d

x

c

x

e

y

a

y

b

y

c

y

d

y

e

P

₁

P

2

P

4

Figure 2: The schedule asso ciated withL

For example, if we consider the order dened by L(a) = 1, L(b) = 2, L(c) = 3, L(d) = 4 and L(e) = 5, the sets P

i , i = 1:::5, are dened by P 1 =fx a ;x b g,P 2 =fx c ;x d g,P 3 =;,P 4 =fx e gandP 5 =;. Figure2shows thecorresp onding solution forf(I)for ourexample.

Wehavetoprovenowthatthis schedulefulll all theprecedence delays of G

0

. Let us consider the task y L

1 (i)

;i= 1:::n. We must show that all

its predecessors in G 0

are completed attime(n+i 1) d=K+i.

1. We claim that all the predecessors of y L 1 (i) in G 0 are in S i j=1 P j . Indeed,x L 1 (i) 2P j ;jibyconstruction.

The other predecessors of y L 1 (i) are vertices x v with v adjacent to u=L 1

(i)in G. Now, if L(v) <L(u), then x v 2P k with k L(v). Otherwise,v 2B L (i)sox v 2P k withkL(u). 2. Weshowthatj S i j=1 P j

jK+i. Indeed,thissetiscomp osedby: [1]i

tasksx L

1 (j)

,j=1:::i, and[2]tasksx u

withL(u)>i, sou2B L

(i).

G 0

is bipartite, we can exchange the tasks such that tasks from X are all

completedb eforethersttaskfromY. WebuildanorderLfromtasksinY suchthat,8i2f1;:::;ng,L

1

(i)isthetasku2V suchthaty u

isexecuted attimen+i 1. Then, we mustprovethat,8i2f1;:::ng,jB

L

(i)jK.

Letconsideri2f1;:::;ng. Tasksexecutedduringtheinterval[0;K+i) can b e decomp osed into [1]x

L 1 (1) :::x L 1 (i) and [2] A set Q i of K other tasksfrom X[Y. Letb e v2B L

(i). Weclaim thatx v

2Q i

. Indeed, we getthatL(v)>i and there exists u 2V with L(u)i and fu;vg2 E. By denition of G

0 , we have then(x v ;y u )2E,sox v 2Q i . Wededuce thatjB L (i)jjQ i j=K. Corollary 2.2. 1jbipartite;d ij =d;p i =1jC max isNP-Hard.

3 A polynomial special case

Let us consider a non oriented connected graph G =(V;E) without lo ops (i.e. withoutedgesfu;ug,u2V)andanordering Lofthevertices. We set jVj=n. 8i2f1;:::;ng,we dene thesequences E

L

(i)by :

E L

(i)=ffu;vg2E; L(u)ig

E L

(i) is the set of edges adjacent to at least one vertices in fL 1

(1);:::; L

1 (i)g.

We dene the problem MIN ADJACENT SET LINEAR ORDERING by:

 Instance : A non oriented graph G = (V;E) without lo ops and a p ositive integerK.

 Question : Is there an ordering L of the vertices of G such that, for alli2f1;:::;jVjg,jE

L

(i)jK+i?

NoticethattheformulationofthisproblemisquitesimilartoMIN-CUT LINEAR ARRANGEMENT [10], which is NP-complete. In thefollowing,

weconsider thesubproblem ofSEQUENCING WITHDELAYS withthe restrictionthatthe degree ofevery vertexfrom X isexactly 2.

Theorem 3.1. There exists a polynomial transformation from  to MIN

(X[Y;E),a delaydand adeadline D . We build an instancef(I) ofMIN

ADJACENTSET LINEAR ORDERING asfollows :

 G 0

=(Y;E 0

). Foreveryx2Xwith(x;y 1 )and(x;y 2 )2Eisasso ciated anedgee x =fy 1 ;y 2 gin E 0 .  thevalue K =D d jYj 1.

f canb ecomputedinp olynomialtime. Weprovenowthatfisap olynomial

transformation(see gure 3foran example)

f

1

2

3

4

a

b

c

d

d = 2

D = 8

a

b

c

d

K = 1

Figure 3: Example oftransformationf

Let us supp ose that a solution to I is given. Then, without lo osing generality, we can supp ose that the tasks from X are p erformed during

[0;:::;jXj)and tasks from Y during [D jYj;:::;D ). We build a linear ordering L following the sequencing order of tasks Y : 8i 2 f1;:::;jYjg, L(i)istheith taskofY in theschedule.

8i2f1;:::;jYjg,let b et=D jYj+(i 1)=K+i+dthestarting

timeofthetaskL 1

(i)fromY. At timet d=K+i, allthepredecessors of L

1

(1);:::;L 1

(i)must b e completed. Now, forevery edge e x

2 E L

(i)

isasso ciated exactlyone of those predecessors. So, jE L

(i)jK+i.

Conversely, let us supp ose that a solution to f() is given. Then, we p erformtasksfromY following Lduring theinterval [D jYj;:::;D ). We

dene thenthefollowing sequence X i X : 1. X 1 =fx2X ;e x 2E L (1)g, S i

Noticethat,byconstructionthat,8i2f1;:::;ng, i j=1 X i =fe x 2E L (i)g. TasksofXarep erformedduring[0;:::;jX)followingX

1 ;X 2 :::X n . Every task from S i j=1 X i

is then completed at time K +i (see gure 4 for the corresp onding schedule).

X

1

3

1

2

4

c

a

d

b

X

2

X

3

Figure 4: A corresp onding schedule

We mustprove thatthe delays constraints arefullled : let us consider thetask y=(L

1

(i)). Forevery taskx 2 1

(y) is asso ciated e x 2E L (i). So, x 2 S i j=1 X i

and is completed at time K +i. Since y is p erformed at timet=D jYj+i 1,we get:

t (K+i)=D jYj+i 1 (K+i)=d

So,thedelaysare fullled.

Theorem 3.2. Let us consider an instance I of MIN ADJACENT SET LINEAR ORDERING given by a graph G=(V;E)and an integerK >0. A necessaryand suÆcient condition for the existenceof a solution isthat

jEjK+jVj 1

Proof. The conditionis necessary: since thegraphG isconnectedwithout

lo ops, every linear ordering L veries E L

(n 1)=E. So, if L veries the condition,we getthe condition ofthetheorem.

The condition is suÆcient : let us consider a linear ordering L and a

family ofgraph G i

,i=0;:::;ndened such that,

 G 0

=G,

 8i = 1;:::;n, we cho ose a vertex u in the subgraph G i 1 = (V fL 1 (1);:::;L 1

(i 1)g;E)with aminimum degree in G i 1 and we setL(u)=i.  G n =;. We note E i the edges of G i

. Noticethat,8i=1;:::;n, the twosets E L

(i) and E

i

area partitionof E.

We prove by contradiction that the linear ordering L is a solution to

L

in G is greater than or equal to K +2. So, 2jEj jVj(K+2). By

hyp othesis, we get2K+2jVj 2KjVj+2jVj, soK(2 jVj)2.

Since K > 0, we get that jVj < 2, so jVj = 1. In this case, we get

jE L

(1)j=jEj=0,which contradictsjEjK+2.

 Now,letussupp osethat,fori<n 2,8j2f1;:::;ig,jE L (j)jK+j and that jE L (i+1)j(i+1)+K+1. Forevery vertexu 2G i , we setd G i

(u) thedegree ofu in G i

.

Thetotalnumb er of edgesveriesjEj=jE L (i+1)j+jE i+1 j. 1. Byhyp othesis, jE L (i+1)j(i+1)+K+1.

2. Bydenitionofthesequences G i ,jE i+1 j=jE i j d G i (L 1 (i+1)). Sinceu=L 1 (i+1)is thevertexofG i

witha minimum degree,

thenumb er ofarcs ofG i veries 2jE i j(n i)d G i (L 1 (i+1)) So, jE i+1 j 1 2 (n i)d Gi (L 1 (i+1)) d Gi (L 1 (i+1)) We show that d G i (L 1

(i+1))  2. Indeed, let us denote by

e(k )=fL 1 (i+1);L 1 (k )g anedgeofGadjacenttoL 1 (i+1). Then,we geteasily that E

L (i+1) E L (i)=fe(k )2G i g,so d G i (L 1 (i + 1))=jE L (i + 1)j jE L (i)j(i + 1) + K+ 1 (K+ i)=2 We deducethat jE i+1 j n i 2 2 d G i (L 1 (i+1))n i 2

So,thetotalnumb er ofedges of Gveries :

jEj=jE L

(i+1)j+jE i+1

j(i+1)+K+1+n i 2=jVj+K

whichcontradictsthehyp othesis of thetheorem.

Noticethatthis pro ofisconstructive: if thecondition ofthetheorem is

fullled, one can easily implements agreedy p olynomial algorithm tobuild alinear ordering.

Corollary 3.3.  is polynomial.

Inthis section,we considertheanalysis ofthep erformancesof two approx-imation algorithms.

The rst one is the classical Grahamlist scheduling algorithm [12]. At eachtimet,aschedulabletaskischosentob ep erformedwithoutanypriority

rule. For the bipartite graph G = (X[Y;E), it consists on p erforming tasks from X in any order and tasks from Y as so on asp ossible. Several

authorsshowthatthep erformanceratioofthisalgorithmisupp erb ounded asymptotically by2 [15, 20,19]. Weprove here that this b ound isreached forbipartitegraphs:

Theorem 4.1. The performance ratio of a list scheduling for a bipartite

graph tends asymptotically to 2.

Proof. Letus consider avalue d>0 and a bipartitegraph G=(X[Y;E)

withX =fa 1

;:::;a d

g[fbg, Y =fcg and E =f(b;c)g. In the worstcase fortheGrahamlistscheduling algorithm,tasks fa

1 ;:::;a

d

gare p erformed

rst. We getthena schedule of length l 1

=2d+2.

Now,we can geta schedule without idle slotsif wep erformbrst. The length ofthis second schedule is thenl

2

=d+2. The p erformanceratiois thenb oundedby: r=

2d+2 d+2 =2 2 d+2 ! d!1 2.

We present nowa slightly b etter approximation algorithm : let us sup-p ose that G = (X [Y;E) with jXj = n, jYj = m and n  m. In the

opp osite,wemo dify theorientationof theedgesand weconsider thegraph G

0

=(Y [X ;E 0

). We can get a feasible schedule for G by considering the

inverse order ofa schedule forG 0

. Let us consider theset X

1

of tasks from X with a strictly p ositive out-degree (i.e.,X

1

is theset of X with atleast one successorin Y). The idea istoapplya listscheduling algorithmwhich p erformstasksfromX

1 b efore thosefrom X 2 =X X 1 . We denote by C opt (resp. C H

) the makespan of an optimal schedule (resp. a schedule obtained using this algorithm). We set jX

i j=n i ;i=1;2 and p = max(0;d+1 n 2

m). We prove the following upp er b ound on C opt : Lemma 4.2. C opt n+m+p.

Proof. The last task of X 1

is p erformed at time t  n 1

and has at least one successor in Y, so C opt  n 1 +d+1. Now, if p = d+1 n 2 m,

2 1 Otherwise, p=0and wegetobviously C

opt

n+m.

Theorem 4.3. Theperformance ratio of thisalgorithm isbounded by 3 2 .

Proof. We denote by I the idle slots of theschedule obtainedby our algo-rithm. We get,usingthe previous lemma :

C H

=n+m+jIjC opt

+(jIj p)

1. IfjIjp,weget thetheorem.

2. LetusassumenowthatjIj>p. WebuildasubsetI p

Ibyremoving

fromI thepthrstidleslotsinourschedule. Letb eanelementk2I p andt(k )the timeof thisidle slot.

Clearly, by denition of I p

, t(k )p+n. Moreover, there is at least one taskfrom y 2Y p erformedafter t(k ) such thaty is notreadyat timet(k ), sot(k )n 1 +d. We get jIj p=jI p jn 1 +d (p+n) Then, jIj p=jI p jd n 2 max(0;d+1 n 2 m) We deducethat jI p jmin(d n 2 ;m 1) So,jI p jjYj.

Now,the inequalityb etween C H and C opt b ecomes : C H C opt +jI p jC opt +jYj

Since jYjjXj, we get that jYj 1 2 (jXj+jYj)  1 2 C opt and we get thetheorem.

We can prove that the b ound 3 2

is asymptotically tight : indeed, let us consideranintegern>0 andthebipartitegraphG=(X[Y;E)withX = fx 1 ;:::;x n g, Y =fy 1 ;:::;y n

g and the arcs E =f(x i

;y j

);1jing.

We setd=n 1. NotethatjXj=n=jYj. If we p erform task from X such that t(x

i

) = i 1;i = 1;:::;n, then tasks from Y can'tb e p erformed b eforen+d 1. So, we get a makespan

L 1

=3n 2.

Now,ifwep erformtaskfromfromXsuchthatt(x i

)=n i;i=1;:::;n, thenwegeta schedule without idle slotswith makespanL

2 =2n.

Several new questionsarisefrom theresults presentedhere:

 In order to study the b orderline b etween NP-complete and p olyno-mialproblems, thecomplexity of the problem with a bipartite graph wherethedegreeofverticesfromXdo esnotexceed 3isaninteresting

problem.

 Theexistenceofb etterapproximationalgorithmsisalsoaninteresting

question.

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