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Minimizing the volume in scheduling an outtree with
communication delays and duplication
Claire Hanen, Alix Munier
To cite this version:
Claire Hanen, Alix Munier. Minimizing the volume in scheduling an outtree with communication
delays and duplication. [Research Report] lip6.2001.015, LIP6. 2001. �hal-02545560�
outtree with communication delays and duplication Claire Hanen Laboratory LIP6 4, place Jussieu F-75 252 Paris cedex 05
Alix Munier Kordon
IUP Miage, university Paris 12
71, rue Saint-Simon
F-94 017 Creteil Cedex
June 5, 2001
Abstract
Weconsiderin thispap eraschedulingproblemgivenbyntasksof same pro cessing time d, an out-tree, communication delays all equal
toc d and an integer t. The problem is to nd the minimum vol-umeof a feasibleschedulewith makespant. Studying thedominance prop ertiesofsuchschedules,weprovethatthisproblemisp olynomial
using adynamic programmingalgorithm.
1 Introduction
With the recent development of parallel architectures arise a new class of
scheduling problems in which communication delays are considered. The
target machine is a set of parallel pro cessors connected by a network. A
parallel program is mo deled as usual by a directedacyclic graph, the no des
ofwhich aretasks. An arc fromtaski to taskj meansthati computesdata
that is an input for j. If these two tasks are not p erformed by the same
pro cessor, a delay must b e considered b etween the completionof i and the
b eginningof j to dispatchthe data through the network. The aimisto nd
veys [3][9]). Most of them are NP-hard evenwith very strong assumptions.
For example,ifweassume unitarypro cessing timesand unitary
communica-tion delays (UET-UCT task systems),and if in nitely many pro cessors are
available, Picouleau provedin [7] that the makespanminimizationproblem
denoted by Pjpr ec;c jk =1;p j =1jC max isNP-hard.
Taskduplicationmightb eusefultoreducethein uenceofcommunication
delays. Indeed, if a task i has several successors j 1
;:::;j k
, then p erforming
taskionkpro cessorsmayallowtheexecutionofj 1
;:::;j k
onthesepro cessors
just afteri, avoiding communicationthrough the network.
This feature also reduces the problem complexity : if communication
delays are less than or equalto the pro cessing times of the tasks, Colinand
Chretienne[4]provedthatthe makespanminimizationproblemon in nitely
many pro cessors is p olynomial. Unfortunately,if no assumption is made on
the communication delays, the problem Pjdupl ication;c jk
jC max
is NP-hard
[6]. Sothe UET-UCTassumptionseemsto b eab orderlineb etweeneasyand
hard problemsif duplication is allowed.
The main drawback concerning duplication on an unlimited numb er of
pro cessors isthat the numb erof duplicatesmayb e imp ortant. One solution
to reduce it is to x a limited numb er of pro cessors. Veltman [8] proved
that the problem without duplication Pjin tr ee;c jk = 1;p j = 1jC max is
NP-hard. But for this class of graph, duplication is useless since a taskhas
at mostone successor. Noticethat we provedin [5]that a greedy algorithm
with relativep erformanceequalto 2 1
m
may b edevelop edfor theproblem
Pjdupl ication;c jk =1;p j =1jC max .
In this pap er, we consider another way to reduce the numb er of
dupli-cations. We study the minimization of the overall numb er of duplicates (
called the volume of the schedule) for a class of feasible schedules with a
b ounded makespan t. We develop a p olynomial algorithm for tasks with
sameduration d, an out-tree and communicationdelays allequal to cd.
In the second section, we intro duce the notations and we prove several
useful dominance prop erties. In the third section, we prove a recurrence
prop erty on the de nition of the volumes. In the section 4, we prove that
we can computep olynomially these values and an optimal schedule for any
makespan t. In section 5, we present some exp erimental results. The last
Letusconsider asetT of taskswith duration d2IN
indexed from1to jTj
and an out-tree A ro oted by task 1. 8i 2 T, A(i) denotes the sub-tree of
A ro oted by i and, if i 6= 1, p(i) denotes the unique immediate predecessor
of i in A. 8i 2 T, +
(i) is the numb er of immediate successors of i in A.
We consider that the tasks are numb ered such that, 8j 2 +
(i), i < j. We
supp ose that the value of the communicationdelays is c2IN
with cd.
Theduplicationisallowedandthenumb erofpro cessorsisunlimited. For
anyfeasibleschedule, i
()denotesthesetofpro cessors p erformingi2T
and n i
() the numb er of executions of i. v() is the total numb erof tasks
(i.e. , originaland duplicates) of . Clearly,
v()= X i2T n i ()
This value is called the volume of . If there is no confusion, may b e
omitted inthe notations.
8i2 T and 8t2 IN
, we willdenote S i
(t) the set of feasibleschedule of
A(i)with completiontimeb oundedby tand V i
(t)the minimumvolumeofa
scheduleof S i
(t). WewillalsoconsiderthateveryelementfromS i
(t)veri es
the dominanceprop erty 1, 2 and 3proved inthe following. IfS i (t) =;, we setV i (t)=1. Clearly, V i (t)= min 2S i (t) v()
The aim of this pap er is to compute V 1
(t) for any t 2 IN
and a
corre-sp onding schedule.
Property 1 The set of schedules such that :
1. for any task i> 1 with 2 i
()\ p(i)
(), if p(i) is performed by
at time t, then i is performed by at t+d,
2. for any task i>1with 2 i
() p(i)
(), ifthe starting time of the
earliest execution of p(i) ist,then then i isperformedby att+c+p,
is dominant.
Proof
i, then i can b e obviously p erformed by at time t+d. Otherwise,
this execution of i can b e removedfrom and, if needed,executed by
another pro cessor earlier at timet+c.
2. The pro of is similarto the previous one.
Property 2 The set of schedules such that, for any task i 2 T, the
execu-tions of i are all performed at the same time denoted by t i
(), is dominant.
Proof
Let b e a schedule whichveri es the prop erty 1. Leti b e the earliest task
with two executionsp erformed resp ectivelyat timest and t+ with >0.
Ifi=1, thenwecanobviouslyp erformallthe executionsof thero ot at time
0. So, we consideri>1.
All the executions of p(i) are p erformed at time t 0
= t d. Indeed,
if t 0
< t d, then by prop erty 1, the two considered executions of i are
p erformed by other pro cessors. So, they are p erformed simultaneously at
timet 0
+c+p.
So, the rst executionof iisp erformed at the completionof p(i) andthe
second one at timet+c. Since the rst instance ofi is p erformed at timet,
anytasksfrom +
(i)isstartedat mostat timet+d+c,whichisexactlythe
completiontimeofthesecondinstanceofi. So,thisinstancecanb eremoved
without any mo di cationon itsimmediatesuccessors.
Notice that the prop erty 2 is not true anymore if the communication
delays are not all equal to c. For example, let us consider the four tasks
T = f1;2;3;4g and the out-tree pictured by gure 1. For t = 20, the
optimumschedulehas twoexecutionsoftask2withdistinctstarting times.
The followingprop erty is a simpleconsequence ofprop erty 2 :
Property 3 The set of schedules such that,
1. for any task i, if J is the set of immediate successorsof i with 8j2J,
t j ()=t i ()+d, then n i ()=1 J=; + P i2J n j (),
2. the number of executions of any task i 2 T is bounded by the number
1
2
3
4
2
4
4
1
2
3
2
4
20
Figure1: A counterexamplefor prop erty 2
is dominant. In the following, this number of leaves will be denoted by l i
.
Proof
1. By prop erty 2, all the executions of j 2 J are p erformed at t j (), so n i () 1 J=; + P i2J n j
(). Since the overall numb er of duplicates is
minimum,the equalityholds.
2. Weprovebyrecurrenceonthestructureofthetreethat,8i2T;n i
()
l i
.
If task i2T isa leaf,then, by the previous equality,n i ()=1= l i , otherwise, by recurrence,n i ()1 J=; + P j2J l j . Since max(1 J=; ; X j2J l j ) X j2 + (i) l j
this prop erty is true.
3 A recurrence relation on the minimum
vol-umes
Theaimofthispartistoshowarecurrencerelationontheminimumvolumes.
we provethe recurrencerelation.
Letusconsiderthesmallexamplepicturedby gure2. The rstschedule
hasaminimumnumb erofduplicates. Thisexampleleadstothetwofollowing
remarks:
Evenifthe thecompletiontimehasaminimumvalue, thereisno need
to duplicateany path fromthe ro ot to a leafto get a feasibleschedule
as thealgorithmin[4]do es. Itisclear thatthis algorithmpro ducesan
imp ortantnumb erof useless duplicates.
Ifwe decideto not duplicatetask7,weget ascheduleofvolumeequal
to12. So, wemusthereduplicatean internalno dewithout duplicating
its predecessor. This means that the structure of solutions may b e
muchcomplicatedthan in the case where the numb er of duplicates is
not limited.
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
7
8
9
c = d = 1
t = 6
7 is duplicated,
but not its
predecessor.
V1(6)=10
1
2
3
4
5
6
1
2
3
7
8
9
7 is not duplicated.
This schedule is not
optimal.
i
A(i) with completiontime at most t and such that the ro ot i has at most
n executions. Wealso supp ose that S i
(n;t)is reducedto schedules verifying
the 3 previous dominance prop erties.
Clearly, S i
(n;t)S i
(n+1;t). Moreover,by prop erty 3, we get
S i (t)= [ n2IN S i (n;t)=S i (l i ;t) where l i
denotesthe numb er ofleavesof A(i).
8n 2 f1;:::;l i
g, V i
(n;t) is the minimum volume of a schedule from
S i (n;t). Herealso, V i (n;t)V i (n+1;t), so weget V i (t)= min n2f1;:::;l i g V i (n;t)=V i (l i ;t) By convention, if S i (n;t) = ;, we set V i
(n;t) = +1. This leads to the
following lemma:
Lemma 1 If j is aleaf then 8t d, V j
(1;t)=1 and 8t<d, V j
(1;t)=+1
Now,wecan provethe following inequality :
Lemma 2 Let 2 S i
(n;t) and J be the set of immediate successors of i
beginning their executions at the completion time of i. Then,
v()1 J=; + X j2J (n j ()+V j (n j ();t d))+ X j2 + (i) J V j (l j ;t d c) Proof
Tasks (original and duplicates) from can b e partitioned into 3sets :
n i
() executionsof i,
tasksfromA(j),8j2J. The volumeof thesub-scheduleof for A(j)
is then greaterthan V j
(n j
();t d).
Tasks fromA(j),8j2 +
(i) J. Thevolumeof the sub-schedule of
forA(j)isgreaterthan V j (n j ();t d c). Byprop erty 3,n j ()l i , so V j (n j ();t d c)V j (l j ;t d c).
ity: Theorem 1 8t2IN , 8i2T such that + (i)6=;, 8n2f1;:::;l i g, V i (n;t)= min J + (i) (1 J=; + X j2 + (i) J V j (l j ;t d c) + min n j >0; P j 2J n j n X j2J (n j +V j (n j ;t d)) Proof Let 2 S i
(n;t) b e a schedule with a minimumvolume v()= V i
(n;t). By
lemma2, V i
(n;t) is greater than the right termof the equality. If this term
is in nite,also is V i
(n;t).
Conversely,supp ose nowthatthis minimumis nite. Then,wecan build
an optimalschedule of S i
(n;t) by induction by setting :
(B) Ifiisaleaf,thenn=1. Sincethis minimumis nite,t1andp erform
one executionof i at timet.
(I) Now + (i) 6= ;. Let J +
(i) verifying the equality and the
corre-sp onding numb er of executionsn j >0;j 2J . 1. p erform 1 J =; + P j2J n j executionsof i at time0, 2. 8j 2J
, startoneoptimalschedulefromS(n j
;t d)at timedon
the samepro cessors than n j executionsof i. 3. 8j 2 + (i) J
, start one optimal schedule from S(l j
;t d c)
at timed+con new pro cessors.
4 Computation of the volumes
Inthis section,weprovethatthe volumesV i
(n;t)mayb ep olynomially
com-puted usingthe relationexpressed by theorem1. Firstly,we showthat only
a p olynomial numb er of timest 2 IN have to b e considered. Then, we will
intro duce somemiddlesteps for the computation of V i
(n;t). Lastly, we will
b e equal to t ;
= d + c, with ( ; ) 2 IN +
IN. Now, let us consider
t = d+
c the minimallength of a schedule of A without duplication
(this value may p olynomially computed by the algorithm of P.Chretienne
[2]). Weget t
jTjd+(jTj 1)c.
1. 8tt
,A mayb e scheduledwith makespant
without any duplicate,
so V 1 (t)=V 1 (t ). 2. 8tsuch that d t<t
, let the greatest value t ; with t ; t. Then V i (t)=V i (t ; ).
3. If t<d, there is no feasibleschedule,so V i
(t)=1.
So, we will limit the times domain to the values t ; with , jTj 1and d+ ct
. The sizeof this domainisroughly b ounded
by jTj 2
.
ComputationofV i
(n;t) Letusconsiderataski 2T suchthat +
(i)6=;.
Wesupp osethat,foreveryj 2 + (i);8n 0 l j and8t 0 2 wehavecomputed V j (n 0 ;t 0
). The aim ishere to computeV i (n;t), 8nl i and 8t2 . Let us consider + (i)= fj 1 ;:::;j j + (i)j
g the set of immediatesuccessors
ofi. 8k2f1;:::;j +
(i)jg,wewilldenotebyF i
(n;t;k)theminimumvolume
of a schedule of the subgraph A(i) A(j k +1
):::; A(j + (i)
) suchthat :
1. the makespanof isat most t,
2. the numb erof executionsof iin is b ounded by n,
3. there is at least one j 2 fj 1
;:::;j k
g which executions are p erformed
immediatelyat the end of i in.
Any schedule fromS i
(t;n) has eitherno successors of ip erformed at the
completiontimeof i,either at least one. So, we get :
V i (n;t)=min(F i (n;t;j + (i)j);1+ X j2 + (i) V j (l j ;t d c))
The second term of this minimum can b e easily computed. We will
present now how to compute the value F i
(n;t;k) by recurrence on k 2
f1;:::;j +
1 F i (n;t;1)= min 1mmin(n;l j 1 ) (m+V j 1 (m;t d))
Now,letus considerk >1. Bytheorem 1and since J 6=;, we get :
F i (n;t;k+1) = min Jfj 1 ;:::j k +1 g;J6=; ( X j2fj 1 ;:::j k +1 g J V j (l j ;t d c)+ min 0<n j l j ; P j 2J n j n X j2J (n j +V j (n j ;t d)) LetJ
b e an optimalsubset. Three cases may o ccur :
1. If j k +1
62 J
, then there is a communicationdelay b etweeni and
j k +1 and : F i (n;t;k+1) =V j k +1 (l j k +1 ;t d c)+F i (n;t;k) 2. If J =fj k +1 g, then F i (n;t;k+1)= X j2fj 1 ;:::;j k g V j (l j ;t d c) + min 1mmin(n;l j k +1 ) (m+V j k +1 (m;t d)) 3. Otherwise, fj k +1 gJ and F i (n;t;k+1) = min 1m<min(n;lj k +1 ) (m+V j k +1 (m;t d)+F i (n m;t;k)) F i
(n;t;k+1) willb e obtained by gettingthe minimumof these3values.
Complexityof the algorithm Letusconsideri2T,n 2f1;:::;l i
gand
t2 . ThecomplexityofthecomputationofV i (n;t)isO (j + (i)j(j + (i)j+n)).
Wededucethatthe complexityof V i (n;t);n2f1;:::;l i g isO (j + (i)j)l 2 i . So,
the complexity of the computation of every value V i (n;t) is O (jTj 3 :j j) = O (jTj 5 ).
For a xedvalue t,anoptimalscheduleasso ciatedwithV 1
(t)canb ebuilt
using a recursivealgorithmof complexityalso b ounded by O (jTj 5
For these exp eriments, we generate random trees using [1]. We observed
thatexp erimentally,duplicationdo esn'tseemto reducethe makespanofthe
optimalschedulesigni cantly and the resultspresented here are quite p o or.
Ourexplanationisthatthe structureofthe treesrandomlygenerated isvery
irregularandsuchthateveryno dehasafewnumb erofimmediatesuccessors.
For this class oftrees, the duplication do esn't allowto reducethe makespan
of the schedules.
Thefollowingtablesummarizesourexp erimentalresultsforrandomtrees
with 100 tasks.
1 0:75 0:5 0:25 0:1
Percentageof duplication 30% 24% 20% 26% 26%
Averagevalue of improvement 4:5% 3:5% 1:4% 1:1% 0:4%
Averagemaximumvolume 138 140 135 144 140
For every value of the ratio = c
d
, we generate 500 instances of the
schedulingproblem and we computethe following values:
1. Percentage ofduplication,ie. thep ercentageofinstancesb elonging
to the setT forwhichthe duplicationreducesthe minimummakespan
of a schedule.
2. Average value of improvement, ie. if C max
(A) (Resp. C d max
(A))
for A 2 T denotes the minimal value of a schedule of A without
(Resp. with) duplication, we computed the average value of the
ra-tio C max (A) C d max (A) C max (A) .
3. Averagemaximumvolume,ie. theaverageofthevolumeV(C d max
(A))
for everyA2T.
Noticethattheresultsareslightlyb etterfor=1. Now,wecancompare
the result for a completebinary treeof height6 (ie.,with 127 no des).
1 0:75 0:5 0:25 0:1
Averagevalue of improvement 46% 40% 30% 17:6% 7:8%
Averagemaximumvolume 448 448 448 448 448
6 Perspectives
In this pap er, we intro duced a new objective function for scheduling with
communicationdelays inorder to reduceb oth the numb erof duplicatesand
the makespan. We proved that the minimizationof the volume for an
out-treewith equal pro cessing timesd and equal communicationdelays cd is
p olynomial. Many questions arose fromthis result, as :
Is it p ossible to extend these results to other sp ecial classes of
prece-dence graphs ?
Is it p ossible to use this result for minimizing the makespan for a
scheduling problemwith communicationdelays and a limitednumb er
of resources ?
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