ZESZY TY N A U K O W E P O L IT E C H N IK I Ś LĄ SK IEJ Seria: A U T O M A T Y K A z. 134
2002 N r kol. 1554
Jacek B Ł A Ż E W IC Z 1», K laus E C K E R 2», T am as K IS 3), Micha! TA N A Ś1)
1) In stitu te o f C o m p u tin g Science, P oznań U niversity of Technology, P oznań, P oland 2) In s titu te of C o m p u tin g Science, Technical U niversity C lau sth al, G erm any
3) U niversity of Siegen, G erm any
SCHEDULING COUPLED TASKS ON A SINGLE PROCESSOR
S u m m a r y . T h is p a p e r considers a problem of coupled ta sk scheduling o n one processor, w here all processing tim es are equal to 1, th e gap has exact len g th h, precedence co n stra in ts are stric t an d th e criterion is to m inim ize th e schedule length. T h is problem is in tro d u ced e.g. in system s controlling ra d a r operations.
We show th a t th e general problem is N P -h ard . T h is p ap e r also shows a fast ap p ro x im atio n algorithm for chain precedence constraints.
S Z E R E G O W A N IE ZADAŃ S P R Z Ę Ż O N Y C H N A JE D N Y M P R O C E S O R Z E
S tr e s z c z e n ie . W referacie zaprezentow ano problem szeregow ania zad ań sprzężo
nych n a jed n y m procesorze, z jednostkow ym i czasam i w ykonyw ania operacji, sta lą długością przerw y pom iędzy operacjam i, gdzie celem je s t m inim alizacja długości uszeregow ania. P ro b lem te n często w ystępuje w p rak ty ce w system ach sterow ania u rząd zen iam i radarow ym i. W referacie pokazujem y N P -tru d n o ść problem u w p rzy
p ad k u ogólnych ograniczeń kolejnościowych oraz szybki algorytm aproksym acyjny d la ograniczeń kolejnościowych ty p u „łań cu ch ’’.
1. Introduction
A scheduling problem is, in general, a problem answ ering a q uestion of how to allocate some resources over tim e in order to perform a given set of ta sk s [1], In p rac tica l applica
tions resources are processors, money, m anpow er, tools, etc. T asks can be described by a wide range of p ara m ete rs, like read y tim es, due dates, relative urgency factors, precedence constraints a n d m a n y m ore. D ifferent crite ria can b e applied to m easure th e q u ality of a schedule. T h e general form ulation of scheduling problem s an d th e com m only used n o ta
tio n can b e found in books such as [5], [15], [2], [14], [3] and [4]. A survey of th e m ost im p o rta n t resu lts is given in [10].
O ne b ra n c h of scheduling th e o ry is concerned w ith scheduling of coupled tasks. A ta sk is called coupled if it contains two op eratio n s w here th e second has to be processed som e tim e a fte r a com pletion of th e first one. T h is problem , described in [16] an d [17], often a p p e a rs in radar-like devices, w here th e first o p era tio n is th e transm ission of an electro m ag n etic pulse a n d th e second is th e reception of its echo. Several algorithm s designed to solve th e problem of ra d a r pulse scheduling can be found in [8] an d [11].
T h e com plexity of various scheduling problem s w ith coupled ta sk s h a s been stu d ied in [12]. A lth o u g h m ost of th e cases are N P -h ard [12], som e polynom ial algorithm s were found in [13].
A coupled ta sk scheduling problem w ith variable len g th gap is surveyed in [9] a n d [6].
N P -h ard n ess of th is case is proved in [18], w here some in terestin g connections betw een coupled task s an d flow shops are also given.
In th is note, we com plem ent th e above results by presenting th e N P -com pleteness pro o f for th e problem of scheduling coupled task s on a single processor, w ith all processing tim es equal to 1, exact, integer gap length, general stric t precedence co n stra in ts a n d th e o p tim iza tio n c rite ria of m inim izing th e schedule length.
An org an izatio n of th e p ap e r is as follows. T h e problem is form ulated in S ection 2. T h e N P -h ard n ess pro o f is presented in Section 3. T h e ap proxim ation algorithm is presented in S ection 4. W e conclude in Section 5.
2. Problem formulation
We consider th e problem of scheduling n coupled ta sk s on a single m achine, w here each coupled ta sk s Tj consists of tw o op eratio n s T u an d TJ2 an d a gap betw een them . D uring th e gap, a n o th e r ta sk can b e processed. L et p,i and p ,2 den o te th e processing tim es of o p era tio n s an d Ti2, respectively.
T h e gap is ex a ct w hen o p era tio n Tj2 has to s ta r t exactly /q u n its of tim e a fte r th e end of o p e ra tio n T h , w here /i< denotes a le n g th of th e gap. In th is p ap e r, th e only cases considered are those w here all hi are equal, i.e. /q = h, i — 1 , 2 , . . . , n.
Scheduling coupled tasks on a single processor 55
Precedence constraints of coupled task s can be s tric t or weak. Tj -< T j m eans th a t Ti2 -< T ji if precedence co n stra in ts are strict, an d Tj2 -< T j\ A T v -< T ,2 if th e y are weak.
T h e special case of a coupled ta sk problem involves identical tasks. Com monly, such tasks are den o ted by ( p i , h , p 2), w here p\ = p u , p 2 = Pi2, h = hi for all 1 < i < n.
A d ap tin g th e com m only accepted n o ta tio n for scheduling problem s [7], th e scheduling problem considered in th is p ap e r can be den o ted by 1) — coupled, s tr ic t p re c , exact gap\C max, w hich means:
• T h e re is one processor in a system .
• Tasks are coupled an d identical, w ith processing tim es p n = pi2 = 1, Vi<j<„
• G aps are ex a ct an d have uniform length h.
• E very o p e ra tio n h a s a processing tim e equal to 1, a n d th e len g th of
• P recedence co n strain s are strict.
• T h e o p tim iza tio n criterion is to m inim ize th e schedule len g th Cmax — m a x { fj2}, w here t j 2 is a com pletion tim e of T j (its second o peration).
3. NP-hardness of the general case
In case w here precedence co n stra in ts are general th e problem of scheduling coupled tasks on a single processor is N P -h a rd even for u n it processing tim es. W e will prove th is by a series of lem m ae show ing N P -hardness of som e in te rm ed ia te problem s.
L e m m a 1. P ro b le m of Balanced Coloring o f Graphs w ith P artially Ordered Vertices is N P-hard.
P ro o f : T h e p roblem of Balanced Coloring o f Graphs w ith P artially Ordered Vertices (B C G P O V for sh o rt) can be sta te d as follows:
Instance: A d irected, acyclic g rap h G = [V, E ) w here IV^ = q. (It is clear th a t th e arcs define a p a rtia l o rd er in se t V .)
Question: C a n th e vertices of G b e colored w ith I colors such th a t no p air of ad jac en t vertices sh ares th e sam e color and exactly q /l vertices are colored w ith th e sam e color.
(W e will call such a coloring th e balanced coloring.) W ith o u t loss of generality we can assum e th a t q / l € Z +.
F irstly, we prove th a t B C G P O V belongs to class NP. To verify a solution of B C G P O V is enough to verify th a t
1. No p a ir of a d jac en t vertices is m onochrom atic. B ecause each vertex h as no m ore th a n (q — 1) ad jac en t vertices th e com plexity of th is ste p is 0 ( q 2).
2. E x a c tly q /l vertices are colored w ith each color. C om plexity of th is ste p is 0 ( q ).
A so lu tio n of B C G P O V can b e verified in polynom ial tim e, w hich m eans th a t th e problem B C G P O V belongs to class NP.
In o rd er to prove N P-com pleteness of th e problem B C G P O V we will use th e 3 -P a rtitio n problem defined as follows:
Instance: A collection of 3 r item s, bo u n d B e Z + , and size s(a 7) 6 Z + , Vayga su ch th a t B / 4 < s (a j) < B / 2 an d such th a t Y l ajeA s (ai) ~ r ^ -
Q uestion: C a n A be p a rtitio n e d into r disjoint sets A i , . . . , A r such th a t for 1 < i < r, s (ai ) = & (n o te th a t each A,- m ust contain exactly th ree elem ents from A )?
The transform ation: For any instance of th e 3-P artition problem let us define th e corre
sp o n d in g in stan c e of th e B C G P O V problem as follows:
• q = 3 r 2 + r B
• I = r
• For each item a , e A , 1 < j < 3 r let us co n stru c t g rap h G j in th e following way:
1. C o n s tru c t com plete g rap h Kj. on r vertices. D enote one vertex of K { by Vj.
2. C o n s tru c t set of s (a ; ) in d ep en d en t vertices.
3. C re a te all possible edges (vx , v y) such th a t v x € K f vx j* Vj a n d v y e
G ra p h G j is fully defined by a trip le (j , r, s (a j)) so th e c o n stru c tio n h as th e com p lex ity of 0 ( lo g m a x { j, r, s ( a j) } ). I t is clear th a t edges of g rap h G j can b e directed to define a p a rtia l order in th e se t of vertices of G j.
A n ex am ple of such a g rap h is show n in Fig. 1.
Scheduling coupled ta sk s on a single processor 57
Fig. 1. An example of graph G j , where q — 4 , s ( a . j) = 3 Rys. 1. Przyklad grafu G j , dla q = 4 , S^CLj) = 3
T he com plexity of th is tran sfo rm atio n is 0 ( r + log B ) .
It is clear th a t G j ca n b e colored w ith r colors only in th e following way: each vertex from K 3r h as a different color an d all vertices from are colored w ith th e sam e color as vertex Vj.
No tw o vertices of any su b g rap h K } , 1 < j < 3r can sh a re th e sam e color, so after coloring of all K 3, exactly 3 r vertices are colored w ith th e sam e color. All vertices of are con n ected w ith all b u t Vj vertices of K 3, so all vertices of D 3^ y have to be colored w ith th e sam e color as vertex Vj. So, th e se t of vertices of any h a s to be m onochrom atic.
Let A i , . . . , A j be a solution for th e given in stan ce of 3-P artition problem and let Ai = {a<(i),a,(2), a;(3)}. L et us den o te Si = G;(i)UG,-(2) UGi ( 3). G can be colored such th a t sets Di an d D j sh a re th e sam e color if an d only if b o th D , and D j are in th e sam e Si. I t means th a t for all i exactly s(ai(i)) + s(aj(2j) + s(a;(3)) = B vertices of sets D , so exactly 3r + B vertices of g rap h G, are colored w ith each color.
On th e o th e r h an d , le t S i , S2, ■ ■. , S r be a disjoint sets of vertices of g rap h G, such th a t V { G ) = 5Zi=i ■S'ii each v erte x in Si is colored w ith i-th color an d V; |5 ; | = 3 r + B . Let S P b e a su b se t of Si th a t contains only vertices belonging to su b g rap h s. Si has to co n tain 3r vertices belonging to K 3, so |5,D| = B for each i. For each fea-
sible coloring each is m onochrom atic, so Vi<j<3r 3i<<<r D 3s(aj) e ^i- Because Vi<j<3r 5 / 4 < s(a.j) < 5 / 2 each Si has to co n tain exactly 3 su b g rap h s j , j — 1, 2, 3.
So, we can d en o te A i — {a;y) : j = 1 ,2 ,3 } , i = 1 , 2 , . . . , r a n d th is is th e solution of th e 3 -P a rtitio n problem .
O
Now, we will show th a t problem B C G P O V polynom ially transform s to ou r scheduling problem .
L e m m a 2 . T h e p roblem of Balanced Coloring o f Graphs with P artially Ordered Vertices polynom ially tran sfo rm s to 1|(1, h, 1) - coupled, s tr ic t prec, ex a c t gap\Cmax.
P r o o f : L et G = (V , E ) be an instance of problem B C G P O V . L et it co n tain all tr a n sitive arcs. L e t us define a corresponding in stan ce of 1 |(1 ,/i, 1) - coupled, s tr ic t prec, e x a c t gap\C max problem in th e following way:
• n = q
• h = q /l — 1
• For each vertex n,- of g rap h G define th e coupled ta sk T,¡.
• For each arc {vi,V j) of g ra p h G define an arc T { -< 7) of precedence co n stra in ts in th e scheduling problem .
• y ~ Gmax 2n.
L et us assum e th a t a b alanced coloring of G exists. L et S \, b e su b se ts of V {G ) such th a t U |=1 S, — 'F (G ), an d all vertices th a t belong to Si are colored w ith th e i- th color a n d Vi<i<; |S i| = q /l. S ets S t- are p a rtia lly ordered because vertices of G are p a rtia lly ord ered an d G contains all tran sitiv e arcs. Schedule sets Si in accordance to th e p a rtia l o rd er using th e following algorithm :
A lg o rith m 1
b e g i n s := 0
Scheduling coupled task s on a single processor 59
r e p e a t
G et a t a s k T j 6 S i .
Let s be the starting time of the task T j . Si := S J T j
s := s + 1 u n til Si ^ 0 e n d ;
T h e schedule g en erated in such way is show n in Fig. 2.
PS (E ll P 4 (D ) I ; P 3 (C ) i j P 2 ( B ) = 1 p i (A ll I ;
\
l * l |g ||C ||P ,|g . |* , |B ,|C .|P ,|B , |
\
Fig. 2. Schedule of set S{ = {T ^ , Tj,, T c, Trf, Te } where h = 4 Rys. 2. Uszeregowanie zbioru = { T OJ Tj,, T c, Trf, T e } dia h = 4
T his p ro ce d u re g u ara n tee s th a t no precedence co n stra in t will b e violated. T h e schedule contains n o idle intervals, so its le n g th is y.
On th e o th e r h a n d , let us assum e th a t a schedule of le n g th y for th e given in stan c e of the coupled ta sk s p roblem exists. T h e schedule does n o t co n tain idle intervals, so it has to be a sequence o f segm ents as show n in Fig. 2. E ach segm ent co n tain s h + 1 in d ep en d en t tasks, w hich m e an s th a t th e corresponding vertices in G are also in d ep en d en t. So, th e vertices from one segm ent can b e colored w ith th e sam e color, w hich m eans G can be colored w ith n / ( h + 1) colors such th a t exactly h + 1 vertices sh ares th e sam e color.
□
Now we can conclude.
T h e o r e m 1 . P ro b le m l |( l , / i , 1) — co u p led ,stric t prec, exact gap\C max is N P -h ard . P r o o f : Follows im m ediately L em m ae 1 a n d 2.
□
4. The approximation algorithm
In th is section we presen t a fast approxim ation algorithm for th e problem of scheduling coupled ta sk s on a single processor, w ith all processing tim es equal to 1, exact, integer gap of le n g th 2k, general s tr ic t precedence c o n stra in ts an d th e o p tim isa tio n c rite ria of m inim ising th e schedule length.
4.1. T h e alg o rith m
L et c d en o te th e n um ber th e num ber of chains in precedence graph. L e t Sj a n d f j den o te respective sta rtin g an d finishing tim es of chain C j. A n in stan ce of th e problem 1|( 1, 2k, 1) — coupled, s tr ic t ch a in s, ex a ct gap\C max is fully described by given len g th s of th e chains, so it can be encoded using O (lo g n ) bits.
A lgorithm 2.
I n p u t : A set of chains of coupled tasks.
S t e p 1 : W e can consider a chain of coupled ta sk s as one preem ptive task , so ou r problem converts to 2 k + llpm tnlC m ax w ith p reem ptions allowed only in integer p o in ts of tim e.
S t e p 2: S o rt th e p reem ptive ta sk s in descending order of th e ir lengths.
S t e p 3: C o m p u te value:
S t e p 4: Solve th e preem ptive problem using M cN aughton’s rule in tim e w indow o f length D . D en o te a com pletion tim e of th e la st ta sk processed on m achine Pi by
S t e p 5: If chain C) is scheduled in such a way th a t
for any 1 < i < 2 k + 1, i t should b e split in to tw o subchains C j an d C j, w hose processing finishes a n d s ta rts a t tim e 6; respectively. N ote th a t no chain sp lits in to m ore th a n 2 k + 1 subchains.
(
1)
Sj < bi < f j i r j
(
2)
Scheduling coupled ta sk s on a single processor 61
O u t p u t 61, &2i • • • i h k - i an d a set of fours (j , i, s j, / ] ) w here j is th e num ber of chain, i is th e nu m b er of subchain, s'- a n d / j are respective th e s ta r t ancf th e finish tim e of th e subchain in th e preem ptive schedule.
□
T h e to ta l com plexity of th e alg o rith m is
for fixed k.
4.2. S egm ents
Now we have the schedule still in the preemptive form. It should be converted back to the coupled tasks form by choosing the right type of schedule (called segment) in the following way:
• S I: Schedule all subchains Cj such th a t / j < 6afc+i in the way shown in Fig. 2.
The coupled tasks schedule of this p a rt has length of 2(2k + 1)62*4-1 and contains no idle time units.
• S2: Com pute
62* = ^2k — R(.t>2k — 62*4-1, 2k + 1)
where R ( x ,y ) is the remainder of division x by y. Schedule all subchains Cj such th a t s) > 62*4-1 and / j < b2k in the way shown in Fig. 3.
Notice th a t 2fc + 1 such transformations covers entire rectangular area in the preemptive schedule, as shown in Fig. 4. The coupled tasks schedule of this p art has length of 2 • 2k(2k + l)(b2k - 62*4-1) + 2 and contains 2 units of idle tim e regardless of the length of this part.
• S i ': Schedule all subchains Cj such th a t sj > b'2k and f j < b2k in form of segment S i (shown in Fig. 2) w ithout the task E , which renders 1 unit of idle tim e per segment. The couple tasks schedule of this p art has length (2 -2 k + l)(&2jt - b'2k) and contains (62* - b'2k) idle tim e units. Off course (b2k — b'2k) < 2k.
• S3: Schedule all subchains Cj such th a t sj > b2k in the way shown in Fig. 5. The coupled tasks schedule of this part has length of (2k + 2) (61 - &2fc) + 2 • max{z : 2 <
P4 (D) ; : P3 (C) : i P2(B) I I PI (A)| I
ł
ł
^ ■ b , i c , | D|1Ai|a2|B i |c ,|d ,b a 2i
Fig. 3. Conversion to schedule type (segment) S2. Only one time slot is converted (using a task borrowed from the next one) in the figure. In the next time slot the two tasks should be taken from another processor
Rys. 3. Konstrukcja segmentu S2. Wykorzystano jedną jednostkę czasu z uszeregowania dla pro
blemu 2k + l \ p r n t n \ c max i jedno zadanie “pożyczone” z kolejnej jednostki
Fig. 4. A few “time slots” that are used one after another to construct a segment S2. Notice that the shape remains rectangular if its length can be divided by 2k + 1 without remainder Rys. 4. Po utworzeniu 2 fc + 1 kolejnych segmentów S2, wykorzystana część uszeregowania dla
problemu 2fc + l \ p m t n \ C max odzyskuje kształt prostokąta
this p a rt if b ^ - i = b\. If b ^ - i < b\ the number of idle tim e slots is not im portant — the machine
PI
process only one long chain, so shortening of th e schedule will violate precedence constraints.4.3. W o rs t case analysis
There are only three following forms of th e preemptive schedule possible, in general.
• A p a rt S i followed by a p a rt S2 followed by a p art S i'. T he schedule contains no more th a n 0 + 2 + 2fc units of idle time, so
Cm a - C S L < 2 k + 2
Scheduling coupled ta sk s on a single processor 63
P5(E) P4(D) P3(C) P2(B) PI (A)
|A |B b , B C ||a 1^ |b , |b 2|c,[c2|a2|a,ib2|b3|c2B a 31a 4|b^b:
Fig. 5. Conversion to schedule type (segment) S3 Rys. 5. Konstrukcja segmentu S3
A p art 52 followed by a part 5 1 ' followed by a part 53. In this case &2k - i = &i, so the schedule contains no more than 2 + 2k + 2k units of idle time, so
CmaX- C ^ = 4k + 2
A p art 53 only. In this case the instance contains a t least one chain of length 6,, so
^ > ( 2 * + 2 ) 6 ,
The length of schedule generated by this algorithm is
Cmax = (2fc + 2)6, + 2 max{i : 2 < i < 2k - 1 A 6, = 6,}
so
Cm ax- C ^ x < 4 k - 2
To summarise
Cmax - CZX < 4fc + 2
So, the solution generated by this algorithm can be longer than th e optim al one by a constant number of tim e units (the k is fixed) regardless of th e size of the instance.
5. Conclusions
In the paper, scheduling of coupled tasks has been considered. General precedence con
straints resulted in a NP-hardness of the problem, even for unit processing times and equal gap lengths for all the tasks. The other cases, especially where precedence constraints are chain-like or tree-shaped are still open.
In this paper is also shown the approximation algorithm of complexity O (clogc) th a t gen
erates solutions with fixed error comparative to th e optim al ones is given for the strict chains precedence constraints and even length of the gap. T he algorithm can be also applied for in-forest and out-forest precedence constraints.
R E F E R E N C E S
1. B aker K.: Introduction to Sequencing and Scheduling, J. Wiley, New York 1974.
2 . Błażewicz J., C ellary W ., Słowiński R . an d W ęglarz J.: Scheduling u n d er Resource C onstraints: D ete rm in istic Models, J. C. B altzer, Basel 1986.
3. Błażewicz J., E cker K. H., Pesch E ., S chm idt G. a n d W ęglarz J.: Scheduling C om p u te r and M anufacturing Processes (2nd edition), Springer, Berlin, New Y ork 2001.
4. Błażewicz J., Ecker K. H., P la te u B., T ry sta m D.: Handbook o f Parallel and D is
tributed Processing, Springer, Berlin, New York 2000.
5. Conw ay R. W ., M axwell W . L. an d M iller L. W .: Theory o f Scheduling, Addison- Wesley, R eading, M ass. 1967.
6 . G u p ta J. N. D.r Single fa c ility scheduling with two operations per job and time-lags, p re p rin t, 1994.
7. G ra h a m L. R ., Law ler E. L., L e n stra J. K ., R innooy K an A. H. G.: O ptim ization a n d ap p ro x im atio n in d eterm in istic sequencing an d scheduling theory: a survey, Ann.
D iscrete M ath. 5, 1979, 287-326.
8 . F a rin a A., N eri P.: M u ltita rg e t interleaved tracking for p h ased ra d a r array, IE E Proceedings F 2 7 , 1980, 312-318.
9. K ern W ., N aw ijn W . M.: Scheduling m ulti-operations jobs w ith tim e lags on a single m achine, p re p rin t, 1991.
10. Law ler E. L., L e n stra J. K ., R innoy K an A. H. G ., Shm oys D. B: S equencing and scheduling: alg o rith m s a n d com plexity, in S. C. Graves e t al, Handbooks in Operations
Research and M anagem ent Science 4 , N orth-H olland, A m ste rd a m 1993.
11. M ilojevic D. J., Popovic B. M.: Im proved alg o rith m for th e in terleaving of radar pulses, I E E Proceedings F 139, 1992, 98-104.
12. O rm an A. J., P o tts C. N.: O n th e com plexity of coupled task s scheduling, Discrete A pplied M a them atics 7 2 , 1997, 141-154.
Scheduling coupled task s on a single processor 65
13. O rm an A. J., P o tts C. N., S hahani A. K., M oore A. R.: Scheduling for th e control of a m u ltifu n ctio n a l r a d a r system , European Journal o f O perational Research 9 0 , 1996, 13-25.
14. P inedo M.: Scheduling: Theory, A lgorithm s and S y ste m s, P re n tic e H all, Englew ood Cliffs, N. J. 1995.
15. Rinnooy K an A. H. G.: M achine Scheduling Problems: Classification, C om plexity and C om putations, M artinus Nijhoff, T h e H ague 1976.
16. S hapiro R. D.: S cheduling coupled tasks, N aval. Res. Logist. Q uart. 2 7 , 1980, 477-481.
17. O rm an A. J ., S hahani A. K. an d M oore A. R.: M odelling for th e control of a com plex ra d a r system , C om puters Ops Res. 25, 1998, 239-249.
18. Yu W .: T h e T w o-m achine Flow Shop Problem with Delays and the O ne-m achine Total Tardiness Problem , P h . D. T hesis, Technische U niversiteit Eindhoven, 1996.
R ecenzent: Prof. d r hab . Józef G rabow ski
S tre s z c z e n ie
W referacie zaprezentow ano p roblem szeregow ania za d ań , zwanych zadaniam i sprzę
żonymi, sk ład ający ch się z dw óch operacji, przy czym w ykonyw anie drugiej z nich m oże zostać rozpoczęte p o upływ ie określonego czasu od m o m en tu zakończenia pierw szej op
eracji. P ro b lem szeregow ania zad ań sprzężonych je s t często spotykany w p ra k ty c e w systemach sterow ania u rządzeniam i radarow ym i, gdzie pierw szą o p era cją je s t w ysianie impulsu radarow ego, a d ru g ą odbiór pow racającego echa.
W referacie skupiono się n a problem ie szeregow ania zad ań sprzężonych n a je d n y m procesorze, gdzie czasy w ykonyw ania w szystkich operacji są jednostkow e, długość przerw y pomiędzy o p eracjam i je s t sta ła , a celem je s t m inim alizacja długości uszeregow ania.
N ajpierw skupiliśm y się n a p rzy p a d k u ogólnych ograniczeń kolejnościowych i w ykaza
liśmy, że d la ogólnego g rafu ograniczeń kolejnościowych problem szeregow ania za d ań sprzężonych je s t N P -tru d n y , naw et przy stałej długości przerw y i jednostkow ych cza
sach w ykonyw ania operacji. W dowodzie w ykorzystaliśm y jako pośredni e ta p problem sprawiedliwego kolorow ania grafów z częściowo uporządkow anym i w ierzchołkami. N a j
pierw pokazaliśm y pseudow ielom ianow ą transform ację problem u tró jp o d z ia łu zbioru do problemu spraw iedliw ego kolorow ania grafów, a n astęp n ie w ielom ianow ą tran sfo rm ację problemu kolorow ania grafów do danego problem u szeregowania.
W dalszej kołejnos'ci przedstaw iliśm y szybki algorytm zn ajd u ją cy przybliżone roz
w iązanie naszego p roblem u szeregow ania za d ań sprzężonych d la p rzy p a d k u , gdy graf o graniczeń kołejnościow ych m a p o sta ć łańcuchów , a długość przerw y pom iędzy operac
ja m i je s t p a rz y s ta . D okonaliśm y analizy najgorszego p rzy p a d k u i wykazaliśmy, że błąd bezw zględny rozw iązania generowanego przez te n algorytm je s t nie większy o d pewnej stałej niezależnie o d wielkości instancji.
