On the optimality of List Scheduling for online uniform machines scheduling

Transcription

1 On the optimality of Lit Scheduling for online uniform machine cheduling Fangqiu HAN Zhiyi TAN Yang YANG Abtract Thi paper conider claical online cheduling problem on uniform machine We how the tight competitive ratio of LS for any combination of peed of three machine We prove that LS i optimal when 1 = , where 1,, are the peed of three machine On the other h, LS can not be optimal for all combination of machine peed, even retricted to the cae of 1 = 1 = < For m 4 machine, LS remain optimal when one machine ha very large peed, the remaining machine have the ame peed Keyword: Scheduling, Online, ompetitive analyi, Uniform machine 1 Introduction In thi paper we conider claical online cheduling problem on uniform machine We are given a equence J = {J 1, J,, J n } of independent job with poitive ize p 1, p,, p n, which hould be non-preemptively cheduled on m uniform machine M 1, M,, M m The peed of machine M i i i Without lo of generality, we aume 1 = 1 m If a job of ize p j i aigned to machine M i, then p j / i time unit are required to proce thi job If the peed of all the machine are 1, thee machine are called identical machine The job arrive online over lit, ie, each job hould be aigned to a machine before the next job i revealed The goal i to minimize the makepan, which i the maximum completion time among the machine The quality of the performance of an online algorithm i meaured by it competitive ratio For a job equence J an algorithm A, let A (J ) (or briefly A ) denote the makepan produced by A let (J ) (or briefly ) denote the optimal makepan Then the competitive ratio of A i defined a R A = up{ A (J ) J (J ) } The overall competitive ratio of A i the upremum of R A for all poible combination of i, i = 1,, m An online cheduling problem ha a(n) (overall) lower bound ρ if no online algorithm ha a(n) (overall) competitive ratio maller than ρ An online Department of Mathematic, Zhejiang Univerity, Hangzhou 31007, P R hina Supported by the National Innovation Project for Undergraduate of hina orreponding author Department of Mathematic, State Key Lab of AD & G, Zhejiang Univerity, Hangzhou 31007, P R hina Supported by the National Natural Science Foundation of hina ( , ) Fundamental Reearch Fund for the entral Univeritie (010QNA3040) tanzy@zjueducn Department of Mathematic, Zhejiang Univerity, Hangzhou 31007, P R hina Supported by the National Innovation Project for Undergraduate of hina 1

2 algorithm A i called optimal (in the overall ene) if it (overall) competitive ratio matche the (overall) lower bound of the problem An optimal algorithm in the overall ene need not be optimal for any combination of machine peed In 1966, Graham propoed the algorithm Lit Scheduling (LS for hort) proved that the competitive ratio of LS i 1 m for m identical machine [9] Lit Scheduling imply aign each job one by one to the machine where it can be completed earliet Twenty year later, LS wa proved to be the optimal online algorithm for two three identical machine [8] It i remarkable to note that neither LS can be the optimal algorithm, nor another optimal algorithm ha been obtained, even for four identical machine [, 1] If the machine have different peed, ho Sahni proved that the overall competitive ratio of LS i 1+ 5 when m =, + m when m = 3, 4, 5, 6 Whether the upper bound + m i tight or not for m > 6 machine remain open [5] Berman et al deigned the firt online algorithm which ha a contant overall competitive ratio of 588 for arbitrary m They alo proved the overall lower bound i at leat 438 [1] Due to the difficulty in tudying problem with arbitrary machine peed when m i large, often ome aumption are made on the value of i The mot common one i that 1 = 1 = = m 1 < m = For thi pecial cae, ho Sahni proved that the overall competitive ratio of LS im 1 m+1 the bound i tight for any m [5] An improved algorithm with overall competitive ratio trictly le than that of LS wa preented by Li Shi for m 4 machine [11] They alo proved that the overall lower bound for three machine i, thu LS i optimal in the overall ene heng et al preented another algorithm which ha a maller overall competitive ratio of 45 than LS the algorithm given in [11] when m 4 1 A we have een from the identical machine cae, the performance of LS i better when the number of machine i maller In [7], Eptein et al gave a comprehenive tudy on the competitive ratio of LS for two uniform machine They howed that the competitive ratio of LS i }, it matche the lower bound of the problem for any 1, where = 1 i the peed ratio of the two machine Since LS i optimal up to three identical machine, it i a natural quetion whether LS i till optimal for three uniform machine To the author knowledge, neither competitive analyi for problem with arbitrary value of i, i = 1,, 3, ha been conidered, nor improved algorithm have been obtained even for the pecial cae of 1 = 1 = < min{ +1 +1, +1 A trongly related problem i preemptive online cheduling where each job may be cut into piece Thee piece are to be aigned to poibly different machine, in non-overlapping time lot Preemptive cheduling i eaier compared to non-preemptive, ince the optimal chedule can be obtained in polynomial time hen et al deigned an optimal algorithm with competitive ratio for m identical machine [3] An optimal algorithm for two uniform machine with m m m m (m 1) m competitive ratio ( 1 + ) wa given independently by Wen Du [13] Eptein et al [7] Recently, Ebenlendr et al [6] preented an algorithm baed on olution of linear programming, which i optimal for any number of machine any fixed combination of machine peed For three machine, the competitive ratio can be expreed analytically In thi paper, we reviit the optimality of LS for uniform machine cheduling on three machine We firt how the tight competitive ratio of LS for three machine any combination of machine peed 1, Then we prove that LS i optimal when 1 = Note that it contain another pecial cae of i = i 1, which came into attention recently motivated

3 3 by algorithmic game theory [10] On the other h, LS can not be optimal for all combination of machine peed, even retricted to the cae of 1 = 1 = <, becaue we find a new algorithm with competitive ratio trictly le than that of LS when 1 = = < < However, it can be hown that LS remain optimal when the peed of one machine i large, the remaining machine have the ame peed for arbitrary m machine The tructure of the paper i a follow In Section, we preent a bound on the competitive ratio of LS in term of machine peed for m uniform machine, how that thi bound i tight for any combination of machine peed when m = 3 In Section 3, we give both poitive negative reult on the optimality of LS for three machine In Section 4, we generalize the optimality reult to m machine In the ret of the paper, let L i j be the total ize of job cheduled on M i after J j i aigned by an online algorithm, j = 1,,, n, i = 1,,, m For ubet J J, we denote the total ize of job in J by P (J ) ompetitive ratio of LS In thi ection, we preent an upper bound on the competitive ratio of LS in term of machine peed for m uniform machine, how that thi bound i tight for any combination of machine peed when m = 3 The proof of the following theorem ue imilar method a thoe in [9] [5] Theorem 1 The competitive ratio of algorithm LS i at mot { i=1 min } j + (m k) m 1 k m i=k j Proof Without lo of generality, we aume that the lat job J n determine the makepan learly, we have n j=1 p j i=1 i=1 = Li n 1 + p n i i=1, p n, i m By the definition of LS, we have LS = min 1 i m { L i n 1 + p n Therefore, for any k, k = 1,, m, we have ( m ) ( LS 1 = i=k i LS 1 L i n 1 i i=k i=k + p ) n i = i i i=k ( i=1 Li n 1 + p ) n + (m k)pn i=k i i } i=1 i + (m k) m i=k i i=k Li n 1 + (m k + 1)p n i=k i = i=1 i + (m k) m i=k i Hence, LS { min i=1 } i + (m k) m 1 k m i=k i

4 4 Next we prove that the bound in Theorem 1 i tight when m = 3 Theorem The competitive ratio of LS for three uniform machine i where { min, 1 + +, 1 + } + 3 = D 1 = {(, ) 1, }, (, ) D 1, (, ) D, (, ) D 3, D = {(, ) 1, 3 < + +, 3 + > } D 3 = {(, ) 1, } Proof Recall that we aume 1 = 1 The upper bound i a direct corollary of Theorem 1 Next we preent job equence to how that the bound i tight The detailed calculation i contained in the appendix For the area D 1, we ue the equence with job ize p 1 = 1, p = p 3 = For the area D 3, we ue the equence with job ize { + 3, + + 3, , 3 +, } To prove the tightne when (, ) D, we further partition D into three dijoint ubarea D 1, D D 3 a follow (cf Fig 1), D 1 = {(, ) < + 1, 3 + > }, D = {(, ) + 1, }, D 3 = {(, ) > + 1, > 1 + For the area D 1, D D 3, we ue the equence with job ize 5 4 1, 3 < + + } { 3, + 3 +, +, 3 + }, repectively {,, 3, + 3, + 3} { +, + 3 +, 3, 3 + }, Note that the equence ued in the proof of Theorem are nontrivial, epecially for the area where i cloe to Therefore, it may not be eay to anwer the quetion whether the competitive ratio of LS given in Theorem 1 i tight for general m

5 5 Figure 1: The graph of the area 3 Optimality of LS for three machine In thi ection, we concentrate on the optimality of LS for three machine Lemma 31 Let J 0 = {J 1, J,, J k } the ize of J i be i 1, i = 1,, k For any ν 0, there exit two dijoint ubet J 1, J of J 0 uch that J 0 = J 1 J P (J 1 ) k 1 + ν, P (J ) ν k 1 + ν Proof Note that P (J 0 ) = k i=1 i 1 = k 1 Let B = k 1+ν 1 k 1 be a nonnegative integer, b k b k 1 b b 1, b i = 0, 1, i = 1,, k, be the binary repreentation of B, ie, B = k i=1 i 1 b i Let J 1 = {J i b i = 1, i = 1,, k} J = J 0 \J 1, we have k 1 + ν 1 P (J 1) = B = k 1 k 1 + ν 1 + ν, P (J ) = P (J 0 ) P (J 1 ) ( k 1) ( k 1 + ν 1 ) = ν k 1 + ν

6 6 Theorem 31 The competitive ratio of any online algorithm for cheduling on three uniform machine i at leat 1++ when (, ) D 1 Proof We prove the theorem uing an adverary method Note that when (, ) D 1, We further ditinguih two cae ae > 1 (1) We ue a equence with at mot N + 1 job The ize of J i i i 1, i = 1,, N, the lat job ha ize 3 1+ N We firt prove that if algorithm A doe not aign all job except the lat one to M 3, the competitive ratio of A i at leat 1++ In fact, uppoe that the job J k with ize k 1 i the firt job which i not aigned to M 3, 1 k N, then the remaining job in the equence do not arrive we have A k 1 If k = 1, then clearly = 1 A by (1) If k, let J 0 = {J 1, J,, J k } ν = By Lemma 31, there exit two dijoint ubet J 1, J of J 0 uch that J 0 = J 1 J P (J 1 ) k 1 + ν = 3 k 1 +, P (J ) ν k 1 + ν = ( ) k 1 + (When k =, J 0 = J 1 = J = The following analyi i till valid) onider the chedule where job in J 1 are aigned to M 1, J k 1 i aigned to M, job in J {J k } are aigned to M 3 We have max {P (J 1 ), k, P (J ) + k 1 } { 3 k max, k, 3 k } = k, where the lat equality i due to 3 1 Thu, A 1++ by Hence, we aume that algorithm A aign the firt N job with total ize N j=1 j 1 = N 1 to M 3 By (1), = (1 + ) 1 + Hence, { 1 A min N, 1 ( N 1 + )} ( 3 1 N = + 1 ) N after J N+1 i aigned On the other h, let J 0 = {J 1, J,, J N } By Lemma 31, there exit two dijoint ubet J 1, J of J 0 uch that J 0 = J 1 J P (J 1) N, P (J 1 + ) N 1 + onider the chedule where job in J 1 J are aigned to M 1 M, repectively, J N+1 i aigned to M 3 We have { N max, 1 N, 1 N } = N

7 7 A ( ) N 1 1 = N (N ) 1+ N ae < 3 1 Let t =, then t > 1 t 1 = < 1 1 Note that { } t n 1 t i an increaing n t equence lim n 1 n = 1 There exit an integer N, which i the mallet integer uch that t N 1 t N t n > t 1 Therefore, for any k, 1 k < N, t k 1 t 1 tk () We ue a equence with at mot N + job The ize of J i i t i 1, i = 1,, N, the lat two job have ize tn 1 t 1 tn 1 t 1, repectively We firt prove that if algorithm A doe not aign all job except the lat two job to M 3, the competitive ratio of A i at leat 1++ In fact, uppoe job J k with ize t k 1 i the firt job which i not aigned to M 3, 1 k N, then the remaining job in the equence do not arrive we have A tk 1 If k = 1, then clearly = 1 A by (1) If k, there exit a feaible chedule that the firt k job are aigned to M 1, J k 1 J k are aigned to M M 3, repectively Then k { max t j 1, tk, tk 1 t k } = max 1 t 1, tk, tk 1 = tk 1, j=1 where the lat equality i due to t = () Thu A t = by (1) Hence, we aume that algorithm A aign the firt N job with total ize N j=1 tj 1 = tn 1 t 1 to M 3 If J N+1 i aigned to M 1 or M, then A tn 1 t 1 onider the chedule where the firt N 1 job are aigned to M 1, J N J N+1 are aigned to M M 3, repectively Since by the definition of N t, we have N 1 max t j 1, tn 1, tn 1 t 1 j=1 t N 1 > tn = tn 1 tn 1 1 t 1 t 1 { t N 1 1 = max t 1, tn 1, tn 1 t 1 } = tn 1 t 1 Thu, A by (1) Hence we may aume from now on that J N+1 i alo aigned to M 3 by A Then no matter which machine J N+ i aigned to, we have { 1 A t N 1 min t 1, 1 ( t N )} 1 t 1 + tn 1 t 1 + tn 1 t 1 = tn , t 1

8 8 where the lat equality i due to (1) In the optimal chedule, the firt N job are aigned to M 1, J N J N+1 are aigned to M M 3, repectively The completion time of each machine i t N 1 A t 1 Thu, 1++ By Theorem 31, we know that LS i optimal when (, ) D 1 However, LS can not be optimal for any combination of i, i = 1,, 3 We will verify it by preenting another algorithm, Dual, which ha a maller competitive ratio than LS when (, ) L = {(, ) = 1, < < } Recall that LS trie to keep the current completion time of three machine a equal a poible However, the behavior of Dual i oppoite It trie to leave the total ize of job cheduled on M 1 a mall a poible Write = for implicity Let LB j = max{ L1 j 1 +L j 1 +L3 j 1 +p j Algorithm Dual 1 Initially et L i 0 = 0 for i = 1,, 3 let j = 1 +, p j } R() = 3+ + If L3 j 1 +p j < R()LB j, aign J j to M 3 If L3 j 1 +p j R()LB j L j 1 + p j < R()LB j, aign J j to M Ele, aign J j to M 1 3 If no new job arrive, top Otherwie, j = j + 1, return tep Theorem 3 The competitive ratio of Algorithm Dual i le than R() = 3+ + when (, ) L Proof Thi theorem i proved by contradiction Suppoe the theorem i fale, then there exit counterexample we conider the one with the minimum number of job Let J n be the lat job of the equence, then J n determine the makepan by the aumption of minimality We firt how that J n mut be aigned to M 1, the ratio between the completion time of three machine before J n arrive i fixed Then we prove that the lat job aigned to M arrived later than the econd lat job aigned to M 1 Finally, we can derive a contradiction by giving contradicting bound on the proceing time of the lat job aigned to M learly, LB n If J n i aigned to M 3 or M, then Dual = L3 n 1 +pn < R()LB n R() or Dual = L n 1 + p n < R()LB n R(), which i a contradiction Hence, we aume J n i aigned to M 1 thu L 3 n 1 + p n R() = 3 + LB n +, L n 1 + p n R() = 3 + LB n + (3) By the definition of LB n, we have Dual = L 1 n 1 + p n LB n (4) LB n L3 n 1 + L n 1 + L1 n 1 + p n, LB n p n + (5)

9 9 Therefore, ( 1 L3 n 1 + p n + L n 1 + p n + L1 n 1 + p ) n + LB n LB n LB n = L3 n 1 + L n 1 + L1 n 1 + 3p n = L3 n 1 + L n 1 + L1 n 1 + p n + ( + )LB n ( + )LB n ( + )LB n + LB n = 3 + ( + )LB n ( + )LB n + p n ( + )LB n Since the left-h ide of above inequality i identical with the right-h ide, all term are the ame Together with (3), (4) (5), we have LB n = L3 n 1 + L n 1 + L1 n 1 + p n + = p n (L 1 n 1 + p n ) = (L n 1 + p n ) = L 3 n 1 + p n (7) (6) By (6), (7) <, we have L1 n 1 L 3 n 1 = L n 1 L 3 n 1 = + > 0 Without lo of generality, we aume L 1 n 1 = L n 1 = 1 L3 n 1 = Then p + n = + + onider the lat job which i aigned to M 1 We denote it a J r thu L 1 r 1 +p r = L 1 n 1 = 1 Since J r i aigned to M 1, we have L r 1 + p r LB r L1 r 1 + L r 1 + L3 r 1 + p r + L r 1 + p r LB r pr = L r 1 + L3 r 1, (8) + (9) L 3 r 1 + p r LB r 3 + L 1 r 1 + L r 1 + L3 r 1 + p r + + = L r 1 + L3 r 1 (10) + Note that (10) i equivalent to L 3 r 1 Subtitute (11) into (8), we get 3 + ( (L ) r 1 + 1) 4 + p r (11) L r 1 + p r 3 + ( + ) (L r 1 + 1) ( + ) 4 + (L r 1 + 1) 3 + ( + ) ( + ) 4 + p r = L r p r, ie, ( + )L r 1 (6 + 5 )p r (3 + ) (1)

10 10 Since < < p r L 1 r 1 + p r = 1, L r < 1 Hence, there exit J l, r < l < n, uch that J l i the lat job which i aigned to M Thu, L 1 l 1 = 1 L l 1 + p l = L n 1 = 1 Since J l i aigned to M, we have ie L 3 l 1 + p l LB l 3 + L 1 l 1 + L l 1 + L3 l 1 + p l + + p l (3 + ) 4 + ( + ) ( + ) L 3 l 1 = L3 l 1 +, Since L 3 l 1 L3 n 1 =, we have + p l (3 + ) 4 + ( + ) ( + ) + = +, (13) On the other h, (9) i equivalent to Subtituting (14) into (1), we have p r + + L l 1 (14) ( + )L r 1 ( + )(6 + 5 ) + L r 1 (3 + ), i e, Thu, ince <, L r p l = 1 L l 1 1 L r = < +, which contradict (13) Unfortunately, Theorem 3 only exclude the poibility that LS i the optimal algorithm for any combination of machine peed Dual doe not eem to be a ubtantial improvement over LS in it current verion However, it doe give ome hint on how an improved algorithm can poibly be defined In fact, Dual can be viewed a a cla of algorithm with parameter R() We conjecture that the competitive ratio of Dual can be maller if we chooe a maller value of R() than 3+ + But it eem to be not an eay tak to obtain it analytical bound

11 11 4 Generalization to m 4 machine In Section, we have preented a bound on the competitive ratio of LS for m uniform machine From the reult in Section 3, we can conclude that not only the bound i tight, but alo it matche the lower bound for ome combination of machine peed when m = 3 In thi ection, we try to generalize the optimality reult on LS to m 4 uniform machine Though only the pecial cae of 1 = 1 = = m 1 < m = i conidered, it indicate that LS can alo be optimal for ome combination of machine peed Theorem 41 For the pecial cae of 1 = 1 = = m 1 < m =, the competitive ratio of any online algorithm i at leat +m 1 when m Proof We ue a equence with at mot N( m 1) + m + 1 job The job equence, except for the lat one, i partitioned into N + 1 ucceive batche The ith batch ha m 1 job with the ame ize of i 1, i = 1,, N The (N + 1)th batch ha m job with the ame ize of N The lat job ha ize N We firt prove that if A doe not aign all job except the lat one to M m, the competitive ratio of A i at leat +m 1 In fact, uppoe the jth job in the firt batch i the firt job that i aigned to one of the firt m 1 machine Then the remaining job in the equence do not arrive, A = 1 In the optimal chedule, all job are aigned to M m = j Hence, A = j m m m 1 m 1 + m 1 Therefore, we aume that all job in the firt batch are aigned to M m = + m 1 Suppoe the jth job in the ith batch i the firt job that i aigned to one of the firt m 1 machine, i =,, N + 1 Then the remaining job in the equence do not arrive A = i 1 In a feaible chedule, each of the firt m 1 machine procee exactly one job from each of the firt i batche, the completion time of M l i i l=1 l 1 = i 1, l = 1,, m 1 Each job in the (i 1)th batch i aigned to one of the next m 1 machine eparately The completion time of M l i i, l = m,, m ( m 1) All job in the ith batch are aigned to M m, the completion time of M m i ji 1 i Hence, i A Therefore, we aume that A aign all job except for the lat one to M m m i 1 +m 1 When the lat job arrive, no matter which machine it i aigned to, we have { A min N, ( m } 1)(N 1) + m N + N = N ( + m 1) ( m 1) In the optimal chedule, each of the the firt m 1 machine procee exactly one job from each of the firt N batche, the completion time of M l i N 1, l = 1,, m 1 Job in the (N + 1)th batch are aigned eparately to M l, l = m,, m 1 The completion time of each of thee machine i N The lat job i aigned to M m Hence, = N A N ( + m 1) N ( m 1) = + m 1 ( m 1) N + m 1 (N )

12 1 By Theorem1 41, we have the following corollary orollary 41 For the pecial cae of 1 = 1 = = m 1 < m =, LS i the optimal algorithm with competitive ratio +m 1 when m Acknowledgement The author would like to acknowledge two anonymou referee for their careful reading of the paper helpful comment Reference [1] Berman, P, harikar, M, Karpinki, M, On-line load balancing for related machine, Journal of Algorithm, 35, , 000 [] hen, B, van Vliet, A, Woeginger, G J, New lower upper bound for on-line cheduling, Operation Reearch Letter, 16, 1-30, 1994 [3] hen, B, van Vliet, A, Woeginger, G J, An optimal algorithm for preemptive on-line cheduling, Operation Reearch Letter, 18, , 1995 [4] heng, T E, Ng, T, Kotov, V,A new algorithm for online uniform-machine cheduling to minimize the makepan, Information Proceing Letter, 99, , 006 [5] ho, Y, Sahni, S, Bound for lit chedule on uniform proceor, SIAM Journal on omputing, 9, , 1980 [6] Ebenlendr, T, Jawor, W, Sgall, J, Preemptive online cheduling: optimal algorithm for all peed, Algorithmica, 53, 504-5, 009 [7] Eptein, L, Noga, J, Seiden, S, Sgall, J, Woeginger, G J, Romized on-line cheduling on two uniform machine, Journal of Scheduling, 4, 71-9, 001 [8] Faigle, U, Kern, W, Turan, G, On the performance of on-line algorithm for particular problem, Acta ybernetica, 9, , 1989 [9] Graham, R L, Bound for certain multiproceing anomalie Bell Sytem Technical Journal, 45, , 1966 [10] Kovac, A, Tighter approximation bound for LPT cheduling in two pecial cae, Journal of Dicrete Algorithm, 7, , 009 [11] Li, R H, Shi, L J, An online algorithm for ome uniform profeor cheduling, SIAM Journal on omputing, 7, 414-4, 1998 [1] Rudin III, J F, hraekaran, R, Improved bound for the online cheduling problem, SIAM Journal on omputing, 3, , 003 [13] Wen, J, Du, D, Preemptive on-line cheduling for two uniform proceor, Operation Reearch Letter, 3, , 1998

13 13 Appendix: Proof of Theorem learly, LS alway aign the firt job to M 3 Hence, Area D 1 Note that for the area D 1, (15) Let p 1 = 1, p = p 3 = By (16) (15), p 1 + p = = p + 1 (16) p 1 + p + p 3 = = p 3 Hence, J J 3 are both aigned to M 3 by LS, LS = 1++ In the optimal chedule, J i i aigned to M i, i = 1,, 3, = 1 Thu LS = 1++ Area D 3 Hence, Note that for the area D 3, (17) + 8 ( ) + 1 (18) Let p 1 = + 3, p = ++ 3, p 3 = , p 4 = 3 + p 5 = By (17), p 3 0 p By (18), 3 = ( ) ( + 3) = + 3 ( + 1) + 3 = Summing it with 3, we have which i equivalent to , p = = p 1 + p Therefore, J will be aigned to M by LS

14 14 Since by (18), < 0 Thu Therefore, = ( ) + ( ) ( ) + ( ) = ( = 5 ) > 0 p 3 = = p 1 + p 3 On the other h, by (17), Moreover, ( + ) ( + 1) Summing up the above two inequalitie, we have which i equivalent to , p 3 = = p + p 3 Hence, J 3 will be aigned to M 1 by LS By (18), p 1 + p 4 = = p 3 + p = p + p 4 Thu J 4 hould be aigned to M 3 by LS Note that p 1 + p 4 + p 5 = p + p 5 = p 3 + p 5 = No matter which machine J 5 i aigned to, we have LS = In the optimal chedule, J 3 J 4 are aigned to M 1, J 1 J are aigned to M, J 5 i aigned to M 3 Hence = Thu LS = The remaining part of the proof will concentrate on the area D, Thu 3 < + + (19) Area D 1 For the area D 1, 3 + > (0) < + 1 (1) Let p 1 = 3, p = + 3 +, p 3 = +, p 4 = 3 + learly, p 0 by (19)

15 15 Since we have Therefore, J will be aigned to M by LS , p = = p 1 + p By (1), 3 = ( )( + ) < +, which i equivalent to p 1 + p 3 = + 1 < = p + p 3 On the other h, p 1+p 3 = + 1 < + = p 3 Therefore, J 3 will be aigned to M 3 by LS Note that p 1 + p 3 + p 4 = p + p 4 = < 3 + = p 4, where the inequality i due to (0) Hence, LS = In the optimal chedule, J 3 i aigned to M 1, J 1 J are aigned to M, J 4 i aigned to M 3 Hence = + Thu LS = Area D For the area D, () < (3) Let p 1 =, p =, p 3 = 3, p 4 = + 3, p 5 = + 3 learly, p 4 0 by (3) Since J will be aigned to M by LS p = 1 < + = p + p 3, By (), 3 ( + 1) + 1 Thu we have Therefore, J 3 will be aigned to M 3 by LS Since > 3, we have On the other h, p 1 + p 3 = + 1 < 3 = p 3 p 1 + p 3 = = p + p 3 p + p 4 = < + + = p 1 + p 3 + p 4 ( 1) 3 + ( ) + ( + ) 0

16 16 by (3) Thu Therefore, J 4 will be aigned to M by LS Note that p + p 4 = = p 4 p 1 + p 3 + p 5 = p + p 4 + p 5 = = p 5, where the inequality i due to (0) Hence, LS = In the optimal chedule, J 1 J are aigned to M 1, J 3 J 4 are aigned to M, J 5 i aigned to M 3 Hence = + Thu LS = Area D 3 For the area D 3, Moreover, by (19), We firt give a technique claim > 1 + laim 41 For the area D 3, > 0 Proof We ditinguih two cae according the value of ae By (4) (6) direct algebraic calculation, > + 1 (4) (5) < , (6) = 3( ) + + > 3(1 ) + + ( + ) (1 ) + + = (1 ) Therefore, it i neceary to prove (1 ) > 0, which i equivalent to < 0 Define F ( ) = Note that F ( ) = F ( ) = learly, F ( ) > 0 when Thu F ( ) = 0 ha exactly one root ince F (1) = 9 < 0 F ( 9 4 ) = > 0 A F (1) = 9 < 0 F ( 9 4 ) = < 0, we have F () < 0 when ae > 9 4

17 17 By (6), (5) direct algebraic calculation, = 3( ) + + > 3 ( + ) ( Therefore, it i neceary to prove ( ) ( ) ( ) + + ) which i equivalent to > 0 Define G( ) = Note that G ( ) = , G ( ) = , G ( ) = G (4) ( ) = > 0, + + learly, G (4) ( ) > 0 when > 9 4 Moreover, G ( 9 4 ) = 4008 > 0, G ( 9 4 ) = > 0, G ( 9 4 ) = > 0 G( 9 4 ) = > 0 Thu G() > 0 when > 9 4 Let p 1 = +, p = + 3 +, p 3 = 3, p 4 = 3 + learly, p 0 by (19) By laim 41, we have p = < = p 1 + p Therefore, J will be aigned to M by LS By (4), 3 ( + 1) Hence, Therefore, J 3 will be aigned to M 3 by LS Note that p 1 + p 3 = = p + p 3 3 = p 3 p 1 + p 3 + p 4 = p + p 4 = = p 4, where the inequality i due to (0) Hence, LS = In the optimal chedule, J 1 i aigned to M 1, J J 3 are aigned to M, J 4 i aigned to M 3 Hence = + Thu LS = 1++ +