Minimizing Total Completion Time in a Flow-shop Scheduling Problems with a Single Server
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1 Joural of Aled Mathematcs & Boformatcs vol. o SSN: (rt) (ole) Sceress Ltd 0 Mmzg Total omleto Tme a Flow-sho Schedulg Problems wth a Sgle Server Sh lg ad heg xue-guag Abstract We cosder the roblem of two-mache flow-sho schedulg wth a sgle server ad equal rocessg tmes we show that ths roblem s NP -hard the strog sese ad reset a busy schedule for t wth worst-case boud 7 / 6. Mathematcs Subect lassfcato: 90B35 Keywords: flow-sho schedulg roblem total comleto tme worst-case troducto We cosder the two-mache flow-sho schedulg roblem wth mmzg total comleto tme ad equal rocessg tmes that Deartmet of Mathematcs Hube Uversty for Natoaltes shlg59@6.com School of Mathematcs ad Statstcs Wuha Uversty e-mal: chegxueguag60@ms.com. Artcle fo: Receved : Setember 0. Revsed : October 9 0 Publshed ole : December 30 0
2 34 Mmzg Total omleto Tme a Flow-sho... s F. omlexty results for F roblem obtaed by Garey et al [] J.A. Hoogeree [] studed some secal cases for two-mache flow-sho roblems wth mmzg total comleto tmes ad roved that the roblem wth equal rocessg tmes o frst mache that s F s NP -hard the strog sese ad reset a O( log ) aroxmato algorthm for t wth worst-case boud 4 / 3.omlexty results for flow-sho roblems wth a sgle server was obtaed by Brucher et al [3]. ths aer we derve some ew comlexty results for two-mache roblem wth a sgle server troduce a mroved algorthm ad rove that ts worst case s 7 / 6 the boud s tght. omlexty of the F S roblem Let deote the comleto tmes of ob J o mache M. f there are o dle tmes o M ad M we have s s s max{ } s for.... s Theorem The roblem of decdg whether for a gve stace of the F S roblem there exsts a schedule wth cost o more tha a gve threshold value y s NP -hard the strog sese. Proof. Our roof s based uo a reducto from the roblem Numercal Matchg wth Target Sums or short T S whch s kow to be NP -hard the strog sese. T S Gve two multsets X x... x } ad Y y... y } of ostve tegers { { ad a target vector { z... z } where ( x y ) z s there a osto of the set X Y to dsot set Z... Z each cotag exactly oe elemet
3 Sh lg ad heg xue-guag 35 from each of X ad Y such that the sum of the umbers Z equal z for...? () P -obs: s b b; s b x b(... ) () Q -obs: s 0 b; s b y b(... ) (3) R -obs: s 0 b; s b z b(... ) (4)U -obs: s 0 b; s 0 b(... ) (5)V -obs: s 0 b; s 0 b(... ) (6)W -obs: s 0 b; s 0 b(... ) (7) L -obs: s 4 b b; s b b(... ). Observe that all rocessg tmes are equal to b.to rove the theorem we show that ths costructed f the F S roblem a schedule S 0 b 0 satsfyg ( S ) y x ( x y ) (77 3 4) / exsts f ad oly f T S has a soluto. Suose that T S has a soluto. The desred schedule S 0 exsts ad ca be descrbed as follows. No mache has termedate dle tme. M rocess the obs order of the sequece.e. the sequece {... } P Q R U V W L P Q R U V W L Whle M rocess the obs the sequece {... } P Q R U V W L P Q R U V W L as dcated Fgure. The we defe the sequece ad show Fgure. Obvously these sequece ad fulflls( S) ( ) y. oversely assume that the flow-sho schedulg roblem has a soluto ad wth ( S) y. osderg the ath comosed of M oeratos of obs { P Q R U V W } M oeratos of obs
4 36 Mmzg Total omleto Tme a Flow-sho... R U V W... R U V W L } we obta that { L ( S) 3b x 5b x y 7b x y z 8b 9b 0 b... (3 ( )) b x (5 ( )) b x y (7 ( )) b... ( ) b x ( ) (77 3 4) /. x y b y So we have( S) y.. P Q R U V W L P Q R U V W L P Q R U V W L P V Q R U W L Fgure : Gat chart for the F S roblem (a) f S has a artto the there s a schedule wth fsh tmes y. Oe such schedule s show Fgure. (b) f S has o artto the all schedule must have a fsh tmes y. Sce S has o artto the x y z (... ). Let x y z (... ) so S x x y b ( ) ( ) (77 3 4) / y 3 Worst-case for the F S roblem examg worst schedule we restrct ourselves to busy schedule. A busy schedule s a schedule whch at all tmes from start to fsh at least oe server s rocessg a task.
5 Sh lg ad heg xue-guag 37 Theorem The F S roblem let S 0 be a busy schedule for ths roblem S be the otmal soluto for the F S roblem the ( S0)/ ( S ) 7/6 the boud s tght. Proof. For a schedule S let ( S)( ;... ) deote the total dle tmes of ob J o M.osderg the ath comosed of M oeratos of obs... M oerato of ob we obta that () ( s s ) osderg the ath comosed of M oeratos of obs M oerato of ob... we obta that s ( s ) () osderg the ath comosed of M oeratos of obs...l M oerato of ob l... So we have we obta that l ( s s l ) ( ) 0 6 ( S ) (( ( s ) )) (( ( s ) )) ( (( ( s ) ( s (( s ) ( s )) 7 ( S ) ( S0)/ S ) 7/ 6. l To rove the boud s tght troduce the followg examle as follows ad show Fgure ad Fgure 3. () P -obs: s b b s b b( ) ; () Q -obs: s b s 0 b( 34) 0. (3)
6 38 Mmzg Total omleto Tme a Flow-sho... Fgure : ( S0) 35b Fgure 3: ( S ) 30b b So we have ( S )/ ( S ) 35b /30 7/ 6 the boud s tght. 0 Refereces [] M.R.Garey D.S.Johso ad R.Seth. The comlexty of flowsho ad obsho schedulg Math. Oer.Res.m (976) 7-9. [] J.A. Hoogeree T.Kaweguch. Mmzg total comleto tme a two-mache flowsho: aalyss of secal cases Processg of the 5th PO coferece (996) [3] P. Brucker S. KustG.Q. Waget al. omlexty of results for flow-sho roblems wth a sgle server Euroea J. Oer. Res. 65() (005)
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