93189654 Phuong Phap Toi Uu Full
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Trang 1 Giáo trình Phƣơng pháp tối ƣu Đại học Hải Phòng. Giảng viên: Lê Đắc Nhƣờng Tài liệu tham khảo 1. Nguyễn Đức Nghĩa T ối ưu hóa (Quy hoạch tuy ế n tính và r ờ i r ạc) . NXB Giáo dục 1999 2. Bùi Minh Trí Quy ho ạch toán h ọc . NXB Khoa h ọc k ỹ thuật 1999 3. Bùi Minh Trí T ối ưu hóa.Tập 1,2. NXB Khoa h ọc k ỹ thuật 2005 4. Bùi Minh Trí Bài t ập t ối ưu hóa. NXB Khoa h ọc k ỹ thuật 2005 5. Phí Mạnh Ban Quy ho ạch tuy ến tính . NXB Đại học sƣ phạm 2005 6. Phí Mạnh Ban Bài t ập quy ho ạch tuy ến tính. NXB Đại học sƣ phạm 2004 7. Trần Vũ Thiệu Giáo trình T ối ưu tuy ến tính. NXB ĐHQG Hà Nội 2004 8. Phạm Trí Cao Tối ƣu hóa . ĐH K inh tế thành phố Hồ Chí Minh 2005 9. Phạm Trí Cao Bài t ập tối ƣu hóa. ĐH Kinh tế thành phố Hồ Chí Minh 2005 10. PGS. TS Bùi Th ế Tâm Gi ải các bài toán t ối ƣu trên Excel. Phòng tối ƣu và điều khi ển. Vi ện Toán học 11. Hoàn g Tụy Lý thuy ết t ối ưu (Bài giảng l ớ p cao h ọc). Vi ện toán học 2003 12. PGS.TS Nguyễn Nhật l ệ T ối ưu hóa ứ ng d ụng. NXB Khoa học k ỹ thuật 2001 13. Lê Mƣu Dũng Nh ập môn các phương pháp tối ưu. NXB Khoa học k ỹ thuật 1998 14. Phan Q uốc Khánh – Tr ần Huệ Nƣơng Quy Ho ạch Tuy ến Tính. Nhà xuất bản Giáo Dục 15. Đặng Văn Uyên Quy ho ạch tuy ến tính . NXB Giáo dục 1998
Transcript of 93189654 Phuong Phap Toi Uu Full
Trang 1 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Ti liu tham kho 1.Nguyn c Ngha Ti u ha (Quy hoch tuyn tnh v ri rc). NXB Gio dc 1999 2.Bi Minh Tr Quy hoch ton hc. NXB Khoa hc k thut 1999 3.Bi Minh Tr Ti u ha.Tp 1,2. NXB Khoa hc k thut 2005 4.Bi Minh Tr Bi tp ti u ha. NXB Khoa hc k thut 2005 5.Ph Mnh Ban Quy hoch tuyn tnh. NXB i hc s phm 2005 6.Ph Mnh Ban Bi tp quy hoch tuyn tnh. NXB i hc s phm 2004 7.Trn V Thiu Gio trnh Ti u tuyn tnh. NXB HQG H Ni 2004 8.Phm Tr Cao Ti u ha. H Kinh t thnh ph H Ch Minh 2005 9.Phm Tr Cao Bi tp ti u ha. H Kinh t thnh ph H Ch Minh 2005 10. PGS. TS Bi Th Tm Gii cc bi ton ti u trn Excel. Phng ti u v iu khin. Vin Ton hc 11. Hong Ty L thuyt ti u (Bi ging lp cao hc). Vin ton hc 2003 12. PGS.TS Nguyn Nht l Ti u ha ng dng. NXB Khoa hc k thut 2001 13. L Mu Dng Nhp mn cc phng php ti u. NXB Khoa hc k thut 1998 14. Phan Quc Khnh Trn Hu NngQuy Hoch Tuyn Tnh. Nh xut bn Gio Dc 15. ng Vn Uyn Quy hoch tuyn tnh. NXB Gio dc 1998 Trang 2 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Mc lc Chng 1M hnh bi ton ti u 1. PHN LP BI TON .............................................................................................................. 5 1.1. Nghin cu ban u............................................................................................................... 5 1.2. Phn lp bi ton................................................................................................................... 5 1.3. Phn lp bi ton theo phc tp ca thut ton ................................................................. 6 1.3.1 Lp bi ton P, NP. .......................................................................................................... 6 1.3.2 Lp bi ton NP- Hard, NP- Complete. ................................................................................. 6 1.3.2.1.Cc khi nim. ............................................................................................................. 6 1.3.2.2. Bi ton NP- Hard. ........................................................................................................ 7 1.3.2.3. Bi ton NP- Complete. ................................................................................................. 7 2. GII THIU V BI TON TI U ......................................................................................... 7 2.1 Xy dng m hnh ton hc cho mt s vn thc t ............................................................ 8 2.2 Mt s m hnh thc t ........................................................................................................... 9 2.2.1 Bi ton vn u t .......................................................................................................... 9 2.2.2 Bi ton lp k hoch sn xut ........................................................................................ 10 2.2.3 Bi ton vn ti ............................................................................................................... 12 2.2.4 Bi ton ct vt liu ........................................................................................................ 14 3. BI TON TI U DNG CHUN TC, DNG CHNH TC.............................................. 15 3.1 Bi ton ti u dng tng qut............................................................................................... 15 3.1.1 Dng tng qut ............................................................................................................... 15 3.1.2 Phn loi bi ton ti u ................................................................................................. 16 3.2 Bi ton ti u dng chnh tc v chun tc ........................................................................... 16 3.2.1 Bi ton ti u dng chnh tc ......................................................................................... 16 3.2.2 Bi ton ti u dng chun tc ........................................................................................ 17 2.3.3 Bin i bi ton ti u tng qut v dng chnh tc hoc chun tc ................................ 17 Bi tp chng 1 ......................................................................................................................... 20 Chng 2. Tp phng n ca bi ton ti u .................................................................... 22 1. MT S K HIU V NH NGHA ...................................................................................... 22 2. PHNG N C S CHP NHN C ......................................................................... 23 2.1 nh ngha ............................................................................................................................ 23 2.2 S tn ti phng n c s chp nhn c ........................................................................ 24 2.3 Tiu chun ti u .................................................................................................................. 24 3.KHI NIM LI V CC TNH CHT ..................................................................................... 24 3.1 T hp li ............................................................................................................................. 24 3.2. Tp hp li .......................................................................................................................... 25 Trang 3 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 3.3 im cc bin ca mt tp hp li ........................................................................................ 25 3.4 a din li v tp li a din.................................................................................................. 26 3.4.1. a din li ..................................................................................................................... 26 3.4.2. Siu phng - Na khng gian ......................................................................................... 26 3.4.3. Tp li a din ............................................................................................................... 26 4. C IM CA TP PHNG N ...................................................................................... 27 5. PHNG PHP HNH HC .................................................................................................. 28 5.1 Ni dung phng php ......................................................................................................... 28 5.2 V d .................................................................................................................................... 29 Bi tp chng 2 ........................................................................................................................... 32 Chng 3. Phng php n hnh ........................................................................................ 33 1. NG LI CHUNG V C S CA PHNG PHP N HNH ................................ 33 2. THUT TON N HNH DNG BNG ............................................................................... 33 2.1 Bng n hnh ...................................................................................................................... 35 2.2 V d .................................................................................................................................... 36 3. TNH HU HN CA THUT TON N HNH ................................................................. 43 3.1 Tnh hu hn ca thut ton n hnh ................................................................................... 43 3.2 Hin tng xoay vng ........................................................................................................... 44 3.3 Cc bin php chng xoay vng ............................................................................................ 45 3.3.1 Phng php t vng .................................................................................................... 46 3.3.2 Qui tc Bland .................................................................................................................. 48 4. THUT TON N HNH HAI PHA ...................................................................................... 48 4.1 M t thut ton .................................................................................................................... 48 4.2 V d. ................................................................................................................................... 51 5. THUT TON N HNH HAI PHA CI BIN ..................................................................... 52 5.1 M t thut ton .................................................................................................................... 52 5.2 V d .................................................................................................................................... 53 6. PHNG PHP NH THU (M PHNG PHP)........................................................ 54 6.1 M t thut ton .................................................................................................................... 55 6.2 V d .................................................................................................................................... 56 Chng 4. L thuyt i ngu v bi ton ti u i ngu 1. BI TON I NGU............................................................................................................. 61 2. QUI TC CHUYN BI TON TI U TNG QUT SANG BI TON I NGU ......... 61 2.1 Qui tc chuyn i ................................................................................................................ 61 2.2 V d .................................................................................................................................... 63 Trang 4 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 2.3 ngha kinh t ca bi ton i ngu .................................................................................... 64 3. CC NH L I NGU ........................................................................................................ 65 4. THUT TON N HNH I NGU................................................................................... 69 Chng 5. Bi ton vn ti ...................................................................................................... 73 1. PHT BIU BI TON, S TN TI CA NGHIM TI U ............................................. 73 1.1 Pht biu bi ton ................................................................................................................. 73 1.2 S tn ti nghim ti u ........................................................................................................ 74 2. TIU CHUN NHN BIT PHNG N CC BIN .......................................................... 75 2.1 Bng vn ti ......................................................................................................................... 75 2.2 Cc nh ngha v nh l ...................................................................................................... 75 3. CC PHNG PHP TM PHNG N XUT PHT ....................................................... 76 3.1 Phng php gc Ty Bc ................................................................................................... 76 3.2 Phng php cc tiu cc ph ............................................................................................ 77 3.2.1 Phng php cc tiu cc ph theo dng...................................................................... 77 3.2.2 Phng php cc tiu cc ph theo ct......................................................................... 77 3.2.3 Phng php cc tiu cc ph ton bng...................................................................... 78 3.3 Phng php Fghen........................................................................................................ 78 3.4 Phng php Larson R.E ..................................................................................................... 81 4. TIU CHUN TI U V THUT TON TH V ................................................................. 81 4.1 Tiu chun ti u. ................................................................................................................. 81 4.2 Thut ton th v ................................................................................................................... 81 5. TRNG HP KHNG CN BNG THU PHT................................................................. 84 5.1 Tng lng pht ln hn tng lng thu: m ni ji 1 j 1a b= => ......................................................... 84 5.2 Tng lng pht nh hn tng lng thu: m ni ji 1 j 1a b= =< ......................................................... 84 6. MT S V D ........................................................................................................................ 85 Chng 6. Gii bi ton ti u trn my tnh 1. GII BI TON TI U ......................................................................................................... 86 2. GII BI TON VN TI ........................................................................................................ 89 Trang 5 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Chng 1 M hnh bi ton ti u 1. PHN LP BI TON 1.1. Nghin cu ban u * Biu din bi ton:Input: Thng tin u vo Output: Kt qu u ra 1.2. Phn lp bi ton. Ti sao phi phn lp bi ton? Li ch ca vic phn lp? liu sc mnh ! Cc Bi tonBi ton cha c li giiBi ton c li giiBi ton khng gii cBi ton gii cBi ton d giiBi ton kh gii Bi ton chia thnh 2 loi: Bi ton c li gii: Bi ton cha c li gii (Open Problem). Bi ton c li gii c chia thnh 2 loi. Bi ton khng th gii c. Bi ton c th gii c Bi ton c th gii c chia thnh 2 loi. Bi ton thc t gii c: BT tr c, BTd (Easy). Bi ton thc t kh gii c: BT bt tr c (Interactability), BTkh (Hard). Bi ton thc t kh gii c:2 loi. Bi ton thc t kh gii:Kh va phi (Binary Hard). Trang 6 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Bi ton thc t kh gii:Rt kh (Unary Hard). Ch :Cn phn bit giakhng th giivkh gii (bt tr).1.3. Phn lp bi ton theo phc tp ca thut ton1.3.1 Lp bi ton P, NP. 1). Vi mt bi ton, c hai kh nng xy ra: c li gii.Cha c li gii. 2). Vi bi ton c li gii, c haitrng hp xy ra: - Gii c bng thut ton. - Khng gii c bng thut ton. 3). Vi bi ton gii c bi thut ton cng chia thnh hai loi: + Thc t gii c:D gii.c hiu l thut ton c x l trong thi gian nhanh, thc t cho php, l thut ton c phc tp thi gian a thc. + Thc t kh gii:Kh gii. c hiu l thut ton phi x l trong nhiu thi gian, thc t kh chp nhn, l thut ton c phc tp thi gian l trn a thc (hm m). P: l lp bi ton gii c bng thut ton n nh, a thc (Polynomial). NP : l lp bi ton gii c bng thut ton khng n nh, a thc. P c NP. Ch : Hin nay ngi ta cha bitPNP. 1.3.2 Lp bi ton NP- Hard, NP- Complete. 1.3.2.1.Cc khi nim. a.Khi nim "Dn v c". Bi ton Bc gil "Dn v cbi ton Amt cch a thc, khiu:B A. Nu c thut tonn nh athc gii bi ton A th cng c thut tonn nh a thc gii bi ton B. Ngha l:Bi ton A "kh hn" bi ton B, hay B "d hn A. - Bc din t bng ngn ng ca bi ton A. (Tc l: B l trng hp ring caA). - Gii cA Gii cB. Ch :Quan h c tnh cht bc cu, tl:C B v B A C A. b.Khi nim "Kh tng ng". Bi ton A gi l kh tng ng bi ton B, k hiu A ~ B, nu : A B v B A Trang 7 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 1.3.2.2. Bi ton NP- Hard. * Bi tonAc gi l NP - hard (NP- kh)nu L e NP u l L A. * Lp bi tonNP - hardbao gm tt c nhng bi ton NP - hard. Bi tonNP hard c th nmtrong hocngoilp NP. 1.3.2.3. Bi ton NP- Complete. a. Khi nim Bi ton NP- Complete.* Bi ton A c gi l NP - Complete (NP- y )nu A l NP Hard vAe NP. Tm li: Bi tonNP Completelbi tonNP - hard nm trong lp NP. * Lp bi tonNP - Completebao gm tt c nhng bi ton NP - Complete. LpNP Complete l c thc, v Cook vKarp ch ra BT u tin thuc lp ny. l bi tontha c:SATISFYABILITY. b.Chng minh bi ton l NP Hard. Cch 1: Theo nh ngha * Bi tonAc gi lNP - hard(NP- kh)nu L e NPu l L A. +Chngminhtheonhnghagpnhiukhkhnvphichngminh:MibitontrongNP u d hn A. + Theo cch 1, nm 1971Cook vKarp ch ra BT u tin thuc lp NP - hard. l bi ton tho c(Satisfyability). Cch 2 + chng minh bi ton A l NP hard, trong thc t ngi ta thng da vo bi tonB no c bit l NP - Hard v chng minh rngB A. Theo tnh cht bc cu ca quan hdn v, A tho mn nh ngha NP hard. Theo cch hiu trc quan: B khthAcngkh. 2. GII THIU V BI TON TI U BitontiubtnguntnhngnghincucanhtonhcNganiting,Vins KantorovichL.V. trongmtlot cc cng trnhvbitonlpk hoch snxutc cng bnm1938.Nm1947nhtonhcMDantzignghincuvxutphngphp nhnh(SimplexMethod)giibitontiutuyntnh.Nm1952phngphpn hnh c ci t v chy trn my tnh in t M. C th tmnhngha ti u hallnhvc ton hcnghin cu ccbi tonti u m hm mc tiu (vn c quan tm)v cc rngbuc (iu kinca bi ton) u l hm v cc phng trnh hoc bt phng trnh tuyn tnh. y ch l mt nh ngha mh,bi ton quy hoch tuyn tnh s c xc nh r rng hn thng qua cc m hnh v v d. Trang 8 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 2.1 Xy dng m hnh ton hc cho mt s vn thc t Cc bc nghin cu v ng dng mt bi ton quy hoch tuyn tnh (QHTT) in hnh l nh sau: Bc 1: Xc nh vn cn gii quyt, thu thp d liu. Xy dng m hnh nh tnh cho vn t ra, tc l xc nh cc yu t c ngha quan trngnhtvxclpccquilutmchngphituntheo.Thngthngbcnynm ngoi phm vi ca ton hc Bc 2: Lp m hnh ton hc. Xydngmhnhtonhcchovnangxt,tcldintlididngngnng tonhcchomhnhnhtnh.Nhvy,mhnhtonhcltrutnghadidng ngnng tonhc cahin tng thc t, cnphicxydng sao chovic phn tch n chophp tahiucbn cht cahin tng. Mhnhton hc thit lp mi quanh gia cc bin s v cc tham s iu khin hin tng. Trongbcny,mtvicrtquantrnglcnphixcnhhmmctiu,tclmt c trng bng sm gitr cng ln (cngnh) ca n tng ngvi tnhhung cng tt hn i vi ngi cn nhn quyt nh. Bc th 2 bt u i hi nhng kin thc ton hc nht nh.Nh vy, sau hai bc u ta pht biu c bi ton cn gii. Bc 3:Xy dng cc thut ton giibi ton mhnhhobngngn ng thun li cho vic lp trnh cho my tnh. Cc thut ton ti u halmt trong nhng cng c c lc gii quyt cc biton t ra. Cn nhn mnh rng, thng thng cc bi ton thc t c kch thc rt ln, v th, gii chng cn phi s dng n my tnh in t.Bc 4: Tnh ton th v iu chnh m hnh nu cn. Trongbcnycnkimchngliccktqutnhtonthuctrongbc3.Trong bc ny cn phixc lpmc ph hp camhnh l thuytvivnthc t m n m t. thc hin bc ny, c th lm thc nghim hoc p dng phng php phn tch chuyn gia. y c 2 kh nng: Kh nng 1: Cc kt qu tnh ton ph hp vi thc t. Khi c th p dng n vo vic gii quytvn thc tt ra. Trong trnghpmhnh cn c s dng nhiu ln, s xut hinvnxy dngh thng phnmmm bogiao din thun tin giangi s dng v my tnh, khng i hi ngi s dng phi c trnh chuyn mn cao v ton hc. Kh nng 2: Cc kt qu tnh ton khng ph hp vi thc t. Trong trng hp ny cn phi xem xt cc nguyn nhn ca n. Nguyn nhn u tin c th do cc kt qu tnh ton Trang 9 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng trongbc3lchacchnhxccnthit.Khicnphixemliccthutton cng nh cc chng trnh tnh ton trong bc ny.Mtnguynnhnkhcrtcthldomhnhxydngchaphnnhcy hintngthct.Nuvycnphirsotlibc1,trongvicxydngmhnhnh tnh c yu t hoc quy lut no b b st khng? Cui cng cn phi xem xt hoc xy dng limhnh ton hc bc 2. Nh vy, trong trnghpkt qutnh tonkhng ph hp vi thc t chng ta cn phi quayli kim tratt c cc bc thc hin trc ,v rt c th s phi lp i lp li nhiu ln cho n khi kt qu tnh ton ph hp vi thc t. Bc 5: p dng gii cc bi ton thc t. 2.2 Mt s m hnh thc t Mhnhhalmtlnhvcnghinculthuytring,ihitrctinlshiubit nhng kin thc trong lnh vc ca i tng cn m phng. Trongmc ny taxtvim hnh truyn thng ca tiu haminh ha chovicxy dng m hnh ton hc cho cc bi ton c ni dung kinh t, k thut. 2.2.1 Bi ton vn u t Ngi ta cn c mt lng (ti thiu) cht dinh dng i=1,2,..,m do cc thc n j=1,2,...,n cung cp.Gi s : +aij l s lng cht dinh dng loi i c trong 1 n v thc n loi j(i=1,2,...,m) v (j=1,2,..., n) +bi l nhu cu ti thiu v loi dinh dng i +cj l gi mua mt n v thc n loi j Vntralphimuaccloithcnnh thnotngchiphbratnhtm vn p ng c yu cu v dinh dng. Vn c gii quyt theo m hnh sau y: +Gi xj 0 (j= 1,2,...,n) l s lng thc n th j cn mua Tng chi ph cho vic mua thc n l: V chi ph b ra mua thc n phi l thp nht nn yu cu cn c tha mn l: Lng dinh dng i thu c t thc n 1 l : ai1x1 (i=1m) Trang 10 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Lng dinh dng i thu c t thc n 2 l : ai2x2 ......................................................... Lng dinh dng i thu c t thc n n l : ainxn Vy lng dinh dng th i thu c t cc loi thc n l: ai1x1+ai2x2+...+ainxn(i=1m) Khi theo yu cu ca bi ton ta c m hnh ton sau y: 2.2.2 Bi ton lp k hoch sn xut 2.2.2.1 V d Mtcssnxutdnhsnxut2loisnphmAvB.Ccsnphmnyc ch to t ba loi nguyn liu I, II, III. S lng n v d tr ca tng loi nguyn liu v s lngnvtngloinguynliucndngsnxutramtnvsnphmmiloi c cho trong bng di y: Loinguyn liu Nguyn liu d tr S lng n v nguyn liu cn cngcho vic sn xut mt n v sn phm AB I1823 II3054 III2516 Hylp k hoch snxut, tc ltnhxem cn snxut baonhiunv sn phmmi loi tin li thu c l ln nht, bit rng bn mt n v sn phm A thu li 3 trm nghn ng, bn mt n v sn phm B thi li 2 trm nghn ng. Ta xy dng m hnh ton hc cho bi ton trn: Gixvy theo th t l slngnv snphm Av B cn snxuttheokhoch. Khi tin li thu c s l: z = 3x + 2y Do nguyn liu d tr c hn nn x v y phi chu nhng rng buc no , c th l: + s 2x 3y 18 (rng buc v nguyn liu I) Trang 11 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng + s 5x 4y 30 (rng buc v nguyn liu II) + s x 6y 25 (rng buc v nguyn liu III) Ngoiracncccrngbucrttnhinnal> > x 0, y 0 vsnvsnphm khng th m. Bng ngn ng ton hc bi ton trn c th c pht biu nh sau:Tm x v y sao cho ti biu thc z = 3x + 2y t gi tr ln nht vi cc rng buc: + s + s+ s> >2x 3y 185x 4y 30x 6y 25x 0, y 0 2.2.2.2 M hnh ca bi ton lp k hoch sn xut T m loi nguyn liu hin c ngi ta mun sn xut n loi sn phm. Gi s : +aij l lng nguyn liu loi i dng sn xut 1 sn phm loi j (i=1,2,...,m) v (j=1,2,..., n) +bi l s lng nguyn liu loi i hin c +cj l li nhun thu c t vic bn mt n v sn phm loi j Vn t ra l phi sn xut mi loi sn phm l bao nhiu sao cho tng li nhun thu c t vic bn cc sn phm ln nht trong iu kin nguyn liu hin c. Gi xj 0 l s lng sn phm th j s sn xut (j=1,2,...,n) Tng li nhun thu c t vic bn cc sn phm l: V yu cu li nhun thu c cao nht nn ta cn c : + Lng nguyn liu th i=1m dng sn xut sn phm th 1 l ai1x1 + Lng nguyn liu th i=1m dng sn xut sn phm th 2 l i2 2a x+ ...............................................+ Lng nguyn liu th i=1m dng sn xut sn phm th n l ainxn Trang 12 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Vy lng nguyn liu th i dng sn xut l cc sn phm l: ai1x1+ai2x2+...+ainxn V lng nguyn liu th i=1m dng sn xut cc loi sn phm khng th vt qu lng c cung cp l bi nn: ai1x1+ai2x2+...+ainxn bi (i=1,2,...,m) Vy theo yu cu ca bi ton ta c m hnh sau y: 2.2.3 Bi ton vn ti 2.2.3.1 V d C mt loi hng cn c vn chuyn t hai kho (trm pht) P1 v P2 ti ba ni tiu th (trm thu) l T1, T2, T3. Bng diy chobit s lnghnh cnvn chuyn ii mikho vslnghngcnnhnminitiuthvccphvnchuynmtnvhnht mi kho ti ni tiu th tng ng. Trm pht Trm thu Lng pht T1T2T3 P1 52330 P221175 Lng thu352545 Hy lp k hoch vn chuyn tha mn mi yu cu thu pht sao cho chi ph vn chuyn l nh nht. Nu k hiu ijx(I = 1, 2 v j = 1, 2, 3) l lng hnh cn vn chuyn t kho Pi n ni tiu th Tj th m hnh ton hc ca bi ton vn ti s l: Tm cc s ijx(I = 1, 2 v j = 1, 2, 3) sao cho ti biu thc: + + + + + 11 12 13 21 22 235x 2x 3x 2x x x min vi cc rng buc sau: Trang 13 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng + + = + + =+ =+ => =11 12 1321 22 2311 2112 22ijx x x 30x x x 75x x 35x x 25x 0, i 1, 2 v j=1, 2, 3 2.2.3.2 M hnh bi ton vn ti Ngi ta cn vn chuyn hng ho t m kho n n ca hng bn l.+Lng hng ho kho i l si (i=1,2,...,m) +Nhu cu hng ho ca ca hng j l dj (j=1,2,...,n).+Cc vn chuyn mt n v hng ho t kho i n ca hng j l cij 0 ng. Gisrngtnghnghoccckhovtngnhucuhnghocccahngl bng nhau, tc l: = == m ni ji 1 j 1s dBi ton t rallp k hochvn chuyn tin cc l nhnht,viiukin lmi ca hng u nhn hng v mi kho u trao ht hng. Gixij 0 l lng hng hophivn chuynt kho in ca hng j. Ccvn chuyn chuyn hng ho i n tt c cc kho j l: =ni j ijj 1c xCc vn chuyn tt c hng ho n tt c kho s l: = == m ni j iji 1 j 1z c x Theo yu cu ca bi ton ta c m hnh ton sau y: Trang 14 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 2.2.4 Bi ton ct vt liu Trongthct,tathngphictnhngvtliudi(nhthanhthp,ngnc,bng giy) c ti cho trc thnhnhng on ngn hnvi slng nht nh sdng. Nn c nh th no cho tn t vt liu nht? 2.2.4.1 V d Mtphnxngsnxutthpcnhngthanhthpnguyndi3.8mt.Cnctthnh ba loi on ngn hn l 1 2 3T,T ,Tvi di tng ng l 1.8 mt, 1.4 mt v 1.0 mt. C tt c 5 mu ct khc nhau (cho trong bng). Hicnphicttheomimubaonhiuthanhthpnguynvaslngcc on 1 2 3T,T ,Tm phn xng cn sao cho tng phn thp tha l nh nht? Loi on cn Mu ct S on cn c IIIIIIIVV 1Tdi 1.8 mt20101400 2Tdi 1.4 mt02001400 3Tdi 1.0 mt012301300 2.2.4.2 M hnh bi ton ct vt liu Gi ix (j=1,,5)lsthanhthpnguyncncttheomuj.Son 1T thucl 1 3 52x x x + + . Phn xng cn c 400 on loi 1T . V th, cc bin s cn phi tha mn l: 1 3 52x x x 400 + + =Tng t, thu c s on 2 3T ,Tphn xng cn, cc bin s phi tha mn: 2 52x x 400 + =2 3 4x 2x 3x 1300 + + =Tng s thp tha l: 1 4 5f 0.2x 0.8x 0.6x = + +(mt). Bi ton trn c pht biu nh sau:Tm cc bin s 1 2 3 4 5x ,x ,x ,x ,xsao cho: 1 4 5f 0.2x 0.8x 0.6x min = + + Tha mn cc iu kin sau: ( )1 3 52 52 3 4j2x x x 4002x x 400x 2x 3x 1300x 0 j 1..5+ + = + =+ + => = Trang 15 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 3. BI TON TI U DNG CHUN TC, DNG CHNH TC 3.1 Bi ton ti u dng tng qut 3.1.1 Dng tng qut Tngqutnhng bi ton tiu c th trn, mtbi ton ti u lmt mhnh ton tm cc tiu (min) hoc cc i (max) ca hm mc tiu tuyn tnh vi cc rng buc l bt ng thc v ng thc tuyn tnh.Dng tng qut ca mt bi ton ti u l: Trong : (I)Hm mc tiu L mt thp tuyn tnh ca cc bin s,biu th mt i lng no m ta cn phi quan tm ca bi ton. (II)Cc rng buc ca bi ton (cc rng buc cng bc) Lccphngtrnhhocbtphngtrnhtuyntnhnbins,sinhratiu kin ca bi ton (III) Cc cc hn ch v du ca cc bin s (Cc rng buc t nhin) Ngi ta thng trnh by bi ton quy hoch tuyn tnh di dng ma trn nh sau: Gi ai (i=1m) l dng th i ca ma trn A, ta c: Trang 16 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Ngi ta gi: +A l ma trn h s cc rng buc. +c l vect chi ph (cT l chuyn v ca c) +b l vect gii hn cc rng buc. 3.1.2 Phn loi bi ton ti u a. Theo jX+ { }j j j j jX x : h x h = s svi j jh , h = + = Bi ton ti u lin tc. + jXl nhng tp ri rc Bi ton ti u ri rc. + jXl tp s nguyn Bi ton quy hoch nguyn. b. Theo hm f(x) cn ly g(x) +Cc hm f(x),( )ig x l cc hm tuyn tnh Bi ton ti u tuyn tnh +Cchmf(x),( )ig x khnglcchmtuyntnh(phituyn)Bitontiuphi tuyn. +Nuccthamsxcnhf(x),( )ig x lcchngsBitontiuttnh. Ngc li cc tham s l cc i lng ngu nhin Bi ton ti u ngu nhin. +Nu cctham s jX clpvi thigian Bi ton tiutnh. Ngc li jX ph thuc vo thi gian Bi ton ti u ng. Chng ta ch nghin cu lp bi ton ti u tuyn tnh lin tc, tt nhv tnh; lp bi ton ti u ri rc. 3.2 Bi ton ti u dng chnh tc v chun tc 3.2.1 Bi ton ti u dng chnh tc Bi ton ti u chnh tc l bi ton ti u m trong cc rng buc ch c du = v cc bin s u khng m. Tc l: Trang 17 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng
3.2.2 Bi ton ti u dng chun tc Bitonti u chun tcl bi tonti u m trong cc rngbuc ch cdu > v cc bin s u khng m. Tc l: 2.3.3 Bin i bi ton ti u tng qut v dng chnh tc hoc chun tc Ngi ta c th bin i bi ton quy hoch tuyn tnh dng tng qut thnh bi ton quy hoch tuyn tnh dng chnh tc nh cc quy tc sau y: +a rng buc bt ng thc dng s v dng > bng cch nhn 2 v vi -1 n nij j i ij j ij 1 j 1a x b a x b= =s > +a rng buc = v dng >. Khi : nij j inj 1ij j inj 1ij j ij 1a x ba x ba x b===>= > Trang 18 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng +a rng buc dng v dng = th ngi ta tr vo v tri ca rng buc mt bin ph xn+i 0 c du = nnij j n i ij 1ij j ij 1n ia x x ba x bx 0+==+ => > Nu( )1 2 n n ix ,x ,...x ,x +lnghimcahth( )1 2 nx , x ,...x lnghimcabtphng trnh xut pht. +a rng buc dng v dng = th ngi ta cng vo v tri ca rng buc mt bin ph xn+i 0 c du =. nnij j n i ij 1ij j ij 1n ia x x ba x bx 0+==++ =s > +Ccbinphchlnhngilnggiptabinccrngbucdngbtng thcthnhngthc,nphikhngnhhnggnhmmctiunnkhng xut hin trong hm mc tiu. +Nu bin xj 0 th ta t xj = - xj vi xj 0 ri thay vo bi ton. +Nu bin xj l tu (khng c iu kin v du) th ta t c th a v hiu ca hai bin khng m: + = j j jx x xvi + > >j jx 0, x 0 +Trong trng hp trong s cc rng buc c dng m v phi ca dng l gi tr m th i du c hai v c v phi l mt gi tr khng m. +Chuyn i bi ton min v bi ton max nh sau: ( ) { }max f x : x D eTng ng vi bi ton: ( ) { }min f x : x D eNgha l li gii ca bi ton ny cng l li gii ca bi ton kia v ngc li. ( )( )( )( )x X x Xf x min f x f x max f xe e ( = = Trong xl phng n ti u. Trang 19 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Da vo cc php bin i trn m ngi ta c th ni rng bi ton quy hoch tuyn tnh chnh tc l bi ton quy hoch tuyn tnh m trong cc rng buc ch c du = , v phi v cc bin s u khng m. V d: Bin i bi ton quy hoch tuyn tnh sau y v dng chnh tc : Tin hnh cc thay th sau: Ta c: Hay Trang 20 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Bi tp chng 1 1.Mtxnghipcthsdngtia510gimycn,360gimytinv150gimy mi ch ti ba loi snphm A, Bv C. ch tomtnv sn phm A cn 9gi my cn, 5 gi my tin, 3 gi my mi; mt n v sn phm B cn 3 gi my cn, 4 gi my tin; mtnv snphm C cn 5gimy cn, 3 gi my tin, 2gi mymi.Mi sn phm A tr gi 48 nghn ng, mi sn phm B tr gi 16 nghn ng v mi sn phm C tr gi 27 nghn ng. Vn t ra l x nghip cn ch to bao nhiu n v sn phm miloitngsgitrsnphmxnghipthucllnnhtviiukinkhng dng qu s gi hin c ca mi loi my. a)Lp m hnh bi ton ti u tuyn tnh cho vn trn b)a bi ton ti u tuyn tnh thu c v dng chnh tc 2.Mttrichnnuigiasccnmua3loithcntnghp 1 2 3T,T ,T .Theocngthcch bin th: Trong 1 kg T1 c 3 n v dinh dng D1, 1 n v dinh dng D2 Trong 1 kg T2 c 4 n v dinh dng D1, 2 n v dinh dng D2 Trong 1 kg T3 c 2 n v dinh dng D1, 3 n v dinh dng D2 Chobitgimua1kgT1l15nghnng,1kgT2l12nghnng,1kgT3l10nghn ngv mi ba n chogia sc cn ti thiu 160 n v dinhdngD1 v 140 n vdinh dngD2.Vnltmslngkg 1 2 3T,T ,T cnmuachiphmuathcnchomt ba ca gia xc l t nht. a)Lp m hnh bi ton ti u tuyn tnh cho vn trn b)a bi ton ti u tuyn tnh thu c v dng chnh tc 3.Mt nh my cn thp c th snxut 2 loi snphm thp tmv thp cun. Nu ch sn xut mt loi snphm th nh my ch c th snxut 200 tn thp tm hoc 140 tn thp cuntrongmtgi.Linhunthuckhibnmttnthptml25USD,mttnthp cun l 30USD. Nh my lm vic 40 gi trong mt tun v th trng tiu th ti a l 6000 tn thp tm v 4000 tn thp cun. Vn t ra l nh my cn sn xut mi loi sn phm lbao nhiutrongmttuntlinhuncaonht.Hytrnhbybitontiucho vn trn. 4.Mtxng lm ca st c nhng thanh thpdi 12mt, cn ct thnh 8 ondi 4 mt, 5 on di 5 mt v 3 on di 7 mt. C 5 mu ct nh sau: -Mu 1: 3 on 4 mt, khng tha -Mu 2: 1 on 4 mt v 1 on 5 mt, tha 3 mt -Mu 3: 1 on 4 mt v 1 on 7 mt, tha 1 mt -Mu 4: 2 on 5 mt, tha 2 mt Trang 21 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng -Mu 5: 1 on 5 mt v 1 on 7 mt, khng tha Lp bi ton ti u tuyn tnh tm cc mu ct tit kim nht. 5.C 3 ngi cng phi i mt qung ng di 10km m ch c mt chic xe p mt ch ngi. Tc ib ca ngi thnhtl4km/h, ngi thhai l2km/h, ngi thba l 2km/h.Tcixepcangithnhtl16km/h,ngithhail12km/h,ngi th ba l 12km/h. Vn t ra l lm sao thi gian ngi cui cng n ch l ngn nht. Hy trnh by bi ton ti u cho vn trn. 6.Mt nh my sn xut ba loi tht : b, ln v cu vi lng sn xut mi ngy l 480 tn tht b,400 tntht ln,230 tntht cu.Mi loi u c thbnc dngtihoc nu chn. Tng lng cc loi tht c thnu chn bnl 420 tn trong giv250 tn ngoi gi. Linhun thuc tvicbnmttn miloi tht c chotrongbng sau y: Hy trnh by bi ton ti u nh my sn xut t li nhun cao nht. 7.Mt xng mc lm bn v gh. Mt cng nhn lm xong mt ci bn phi mt 2 gi, mt ci gh phi mt 30 pht. Khch hng thng mua nhiu nht l 4 gh km theo 1 bn do tl snxut giaghvbn nhiu nht l 4:1. Gibn mt ci bn l135USD, mt ci gh l 50USD. Hy trnh by bi ton ti u xng mc sn xut t doanh thu cao nht, bit rng xng c 4 cng nhn u lm vic 8 gi mi ngy. 8.Mtnhmysnxuthaikium.Thigianlmramtcimkiuthnhtnhiu gp 2 ln thi gian lm ra mt ci kiu th hai. Nu sn xut ton kiu m th hai th nh my lmc 500 cimi ngy. Hng ngy, th trng tiu th nhiu nht l150 ci m kiu thnhtv 200 cikiuthhai. Tinli khibnmt cim kiu thnhtl8USD, mtcimthhail5USD.Hytrnhbybitontiunhmysnxuttli nhun cao nht. 9.Trong hai tun mt con g mi c 12 trng hoc p c 4 trng n ra g con. Sau 8tunthbnttcgconvtrngvigi0,6USDmtgv0,1USDmttrng.Hy trnhbybitontiubtr100gmitrnghocptrngsaochodoanhthul nhiu nht. Trang 22 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Chng 2 Tp phng n ca bi ton ti u 1. MT S K HIU V NH NGHA Bi ton ti u dng chnh tc: 1, max / min== nj jjcx c x0Ax bx= > Chng ta thng hay s dng cch vit di dng ma trn - vect ca bi ton.K hiu: -c l vect h s hm mc tiu -b l vect iu kin -A l ma trn rng buc-Ax = b gi l h rng buc c bn -> x 0 gi l rng buc du (day rng buc trc tip) ca bi ton ti u chnh tc. -K hiu { }= ej ijA a :i Il vect ct th j ( e j J ) ca ma trn A. -H rng buc c bn c th vit thnh: ==nj jj 1A x b-Vectnchiuxthomnttcccrngbuccabitoncgilphng n chp nhn c (li gii chp nhn c). Tp{ } : Ax = b, x 0 = > D x-Tt c cc phngn chp nhn c cabi tonc gi l min chpnhn c hay min rng buc ca bi ton. -Phng n chp nhn c x* em li gi tr ln nht cho hm mc tiu, tc l:( ) s e f x c' x*, x D c gi l phng n ti u, cn gi tr= f* cx * - gi tr ti u ca bi ton. Trang 23 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 2. PHNG N C S CHP NHN C Khi nimphngn c s chp nhnc gimtvai tr quan trng trong thutton n hnh gii bi ton ti u. 2.1 nh ngha Xt bi ton ti u dng chnh tc, gi thit rng hng ca ma trn A l m (Rank (A) = m), tc l rng buc c bn Ax = b gm m phng trnh c lp tuyn tnh. nh ngha 2.1. Ta gi c s ca ma trn A l mt b gm m vect ct c lp tuyn tnh { }=j1 j2 jmB A , A ,..., Aca n. Gis( ) =BB A I, J ,trong{ } =B 1 mJ j ,..., j lmtcscamatrnA.Khivect ( ) =1 2 nx x ,x ,...,xtho mn: j N Bx 0, j J J/ J = e = Khi : +Cc bin j Bx , j J ec gi l cc bin c s (cc bin c gi tr khc 0)+Cc bin j Nx , j J e- cc bin phi c s (cc bin c gi tr bng 0) Cch xc nh cc bin c s 1.Chn mt c s B ca ma trn A 2.t Nx 0 = . 3.Xc nh Bxt h phng trnh BB.x b =V d: Xt bi ton ti u tuyn tnh sau: 1 2 3 4 5 6 76x 2x 5x x 4x 3x 12x min + + + 1 2 3 41 53 62 3 7jx x x x 4x x 2x x 33x x x 6x 0; j 1..7+ + + = + =+ =+ + => = Xtcs { }4 5, 6 7 4B A ,A A ,A E = = .Phngncstngngvinl ( ) x 0,0,0,4,2,3,6 = . Mt c s khc ca Al { }2 5, 6 7B A , A A , A = viphngn c s tng ng l( ) x 0,4,0,0,2,3, 6 = . Cth nhnthyxl phngn chpnhn c cnxkhng l phng n chp nhn c (v 7x 6 0 = < ) Trang 24 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng nhngha 2.2. Phngn c s cgil phng ncschpnhnc (li gii c s chp nhn c) nu nh n l phng n chp nhn c. Mnh2.1.Gisx lphngncschpnhnccabitontiutuyn tnh tng ng vi c s B. Khi tm c vectc sao cho x l phng n ti u duy nht ca bi ton. { }max/min c x : Ax = b, x 0 >2.2 S tn ti phng n c s chp nhn c Mnh 2.2. Phng n chp nhn c x l phng n c s chp nhn c khi m ch khi tp cc vect ct ca ma trn A ng vi cc thnh phn khc khng ca n to thnh mt h vect c lp tuyn tnh. nhl2.1Gisbitontiutuyntnhdngchnhtccphngnchpnhn c. Khi n c t nht mt phng n c s chp nhn c. 2.3 Tiu chun ti u nh ngha 2.3. Phng n c s chp nhn c x c gi l khng thoi ha (khng suy bin) nu nh tt c cc thnh phn c s ca n l khc khng. 3.KHI NIM LI V CC TNH CHT 3.1 T hp li nhngha3.1Chomim ix trongkhnggianRn .imxcgilthplicacc im ix nu: ni 1 2 ni 1 2 ni 1x x x x ... x== o = o + o + + o Trong : 1 2 n, ,..., 0 o o o >v 1 2 n... 1 o + o + + o = . - Khi x l t hp li ca 2 im 1x , 2x ngi ta thng vit : ( ) ( )1 2x x 1 x 0 1 = + s sNu0 1 < V d: Trong R2 mt na khng gian c xc nh bi phng trnh:1 2 1 2ax bx c hay ax bx c + > + sSiu phng v na khng gian u l cc tp hp li. 3.4.3. Tp li a din Giao ca mt s hu hn cc na khng gian trong nR c gi l tp li a din. Trang 27 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Tp li a din l mt tp hp li. Nu tp li a din khng rng v gii ni th l mt a din li. 4. C IM CA TP PHNG N nh l 4.1: Tp hp cc phng n ca mt bi ton ti u l mt tp li a din. Nu tphpli a din nykhng rngv gii nithlmt a din li, s im cc bin ca n l hu hn. nh l 4.2: Tp hp cc phng n ti u ca mt quy hoch tuyn tnh l mt tp li. Xt bi ton ti u chnh tc: Gi s A=[aij ]m.n c cp m.n, m n, rank(A)=m. Gi jA (j=1,2,...,n) ct th j ca ma trn A, bi ton ti u chnh tc trn c th vit: Gi | |{ }T1 2 n1 2 n 1 2 nS x x , x ,..., x 0/ x A x A ... x A b = = > + + + = ltpccphngncabi ton. T0 0 0 01 2 nx x , x ,..., x S( = e l mt phng n khc 0. nhl4.3.iukincnv 0x lphngnccbin(imccbincaS)l cc ct jA ng vi 0jx>0 l c lp tuyn tnh. H qu 4.1:-Sphngnccbincamtbitontiuchnhtclhuhn.Sthnh phn ln hn 0 ca mt phng n cc bin ti a l bng m. -Khisthnhphnlnhn0camtphngnccbinbngngmth phng n c gi l mt phng n c s. nh l 4.4. Nu tp cc phngn camt biton tiu chnh tc khng rng th bi ton c t nht mt phng n cc bin. B : Nu: xl mt phng n ti u ca quy hoch tuyn tnh. 1 2x , x l cc phng n ca quy hoch tuyn tnh. Trang 28 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng xl t hp li thc s ca 1 2x , xth 1 2x , x cng l phng n ti u ca quy hoch tuyn tnh. nhl4.5.Nubitontiuchnhtccphngntiuththsctnhtmt phng n cc bin l phng n ti u. V d: xt bi ton ti u chnh tc sau:maxz(x) = 1 22 3 x x +1 2 31 21 2 34x 2x x 5x 3x 1x , x , x 0+ + = + => Vi h A1 A2 ta tnh c T113 1x , , 03 10 (= ( Vi h A1 A3 ta tnh c| |T2x 1, 0, 1 =Vi h A2 A3 ta tnh cT31 13x 0, ,3 3 (=( Choln lt ccgi tr 1 2 3x , x , x bng0 ri thaythvo h rngbucta cc ktqu nh trn. V ccthnhphn caphngn ccbin l >0nn tachixt 2x v 3x . Khi thay gi tr cc 2 3x , x vo hm mc tiu ta c: z(2x ) = 2*1 + 3*0 = 2 z(3x ) = 2*0 + 3*1/3 = 1 Vy| |T2x 1, 0, 1 =l mt phng n ti u. nhl4.6.iukincnvmtbitontiucphngntiultpcc phng n khng rng v hm mc tiu b chn. nh l 4.7. Nu tp cc phng n ca mt bi ton ti u khng rng v l mt a din li th bi ton ti u s c t nht mt phng n cc bin l phng n ti u. 5. PHNG PHP HNH HC 5.1 Ni dung phng php Khng gim tng qut, gi s bi ton ti u c dng: ( ) ( )1 1 2 2f x c x c x Min Max = + Cc rng buc: i1 1 i2 2 ij1 1 j2 2 ji ja x a x bD a x a x bx , x ,.. 0+ >= + s> Trang 29 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Nhn xt: +Tp phng n ca bi ton ti u nm trong gc phn t th nht 1 2Ox x+Mi btphng trnh 1 1 2 2 i i iax a x b + > , 1 1 2 2 j j ja x a x b + sucxcnhthngqua ccphngtrnhngthngtngngl 1 1 2 2 i i iax a x b + = v 1 1 2 2 j j ja x a x b + = vi min xc nh nm v mt pha ca ng thng ng vi bt ng thc. +Tpphngnlmtagiclihocmtagictriravtn(agickhng ng). +Qutch ccim( )1 2, x xti hm mc tiu f nhngi trxcnh 1 1 2 2c x c x z + =ng thng nyvunggcvivc t( )1 2, cc cgi l ng mc. Vimi gi tr z thay i ta c h cc ng mc song song. +Khi ta di chuyn ng mc theo mt phng thc no s dn n s thay i gi tr ca hm mc tiu. Thut ton Bc 1: Biu din tp cc phng n trn mt phng ta .Bc 2: Nu tp phng nD| =th kt thc, ngc li sang bc 3. Bc 3:+V ng mc 1 1 2 2c x c x z + =vi mt gi tr z c nh (gi tr z ty ) +Xcnhhngtnghocgimcangmc(datrnphngphp xc nh min du trong mt phng) +Chn( )1 2' ', ' x x x , tnh gi tr 1 1 2 2' ' ' c x c x z + = .+So snh z, z: Nu z < z hng di chuyn lm tng gi tr hm mc tiu Nu z > z hng di chuyn lm gim gi tr hm mc tiu Bc4:Dichuynngmctheohngtnghocgimgpvtrtihnl giaoimcanvingmctaccgitrzlphngntiu.Nhn xt: khi ngmc chia khnggian thnhmt tphaynmvmt pha thta thu c phng n ti u. Nhn xt: Phng php th gii cc bi ton 2 n tin li v c th m rng gii vi bi ton nhiu hn hai n bng cch chuyn cc n cn li v biu din thng qua 2 n. 5.2 V d V d 1: Xt bi ton ti u sau: Trang 30 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng ()1 21 21 21 21 2f x 3x 2x maxx x 4x 2x 145x 2x 30x , x 0= + > + s+ s> Gii:A,B,C,D,O l cc im cc bin. Gi tr hm mc tiu ti l: z(A)=3.6+2.0=18z(B)=3.4+2.5=22 z(C)=3.2+2.6=18z(D)=3.0+2.8=8 z(O)=3.0+2.0=0 Phng n ti u ca bi ton t c ti B: 1 2x 4, x 5 = =V d 2:Gii bi ton ti u sau:1 221 24 ax x x m + 1 21 21 21 23x 3313D=5 8 80, 0xx xx xx x+ s + s+ s> Gii: V th ln lt cc hm s trong mt phngx1Ox2 trong gc phn t th nhtv xc nh cc min khng gian tng ng xc nh D. * V ng thng: 1 23x 33 x + =Chn x1 = 0 x2= 33 im (0, 33) Chn x2 = 0 x1= 11im (11, 0) 1 213 x x + =Chn x1 = 0 x2= 13 im (0, 13) Chn x2 = 0 x1= 13im (13, 0) Trang 31 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 1 25 8 80 x x + =Chn x1 = 0 x2= 10 im (0, 10) Chn x2 = 0 x1= 16im (16, 0) * V ng mc vi gi tr chn z = 168 1 221 24 168 x x + =Chn x1 = 0 x2= 7 im (0, 7) Chn x2 = 0 x1= 8im (8, 0) Phng n ti u: x* = (8, 5) 1 28, 5 x x = = max21.8 24.5 288 f = + = Kim tra bng i s nh sau: x* l giao im ca hai ng thng: 1 21 25 8 8013x xx x+ = + = 2 2 13 15 5, 8 x x x = = = Biu din cc ng thng trn mt phng x1Ox2 Trang 32 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng V d 3( )1 2f x 2x x min = + ( )( )( )()( )1 21 21 212x 2x 2 12x 3x 6 24x 5x 20 3x 0 4x 0 5 + > s+ s> > Hy gii bi ton bng phng php th. Kt qu:( )1 245 8 82x* x , x , , f *11 11 11| |= = = |\ . Bi tp chng 2 Gii nhng bi ton ti u sau y bng phng php hnh hc: Trang 33 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Chng 3 Phng php n hnh 1. NG LI CHUNG V C S CA PHNG PHP N HNH Phng php n hnh c xy dng da trn hai nhn xt sau: Nu bi ton ti u c phng n ti u th c t nht mt nh ca D l phng n ti u. (Vi D l tp cc rng buc t nhin v rng buc bt buc ca bi ton) a din li D c mt s hu hn nh. Thut ton gm 2 giai on: Giai on I: Trc ht tm mt phng n cc bin (tc tm mt nh) Giai on II: Kim tra iu kin ti u i vi phng n : +Nuiukin ti u c thamn th phngn lti u. Nu khng ta chuyn sang tm phng n cc bin mi sao cho ci tin gi tr hm mc tiu t gi tr min/max. +Kim tra iu kin ti u i vi phng n mi. Chng ta thc hin mt dy cc th tc nh vy cho n khi nhn c phng n ti u hoc n tnh hung khng c phng n ti u. Vi mi vct phi c s
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tnh c lng: k jk j kj Jz c ceA = nh l 1.1(Tiu chun ti u):Nuccclngcaphngnccbin( )1 2 nx x ,x ,...,x = thamn0kA > vimi k J eth x l phng n ti u ca bi ton ti u. 2. THUT TON N HNH DNG BNG Khng lm gim tnh tng qut, ta xt bi ton ti u dng chnh tc 1, maxnj jjcx c x== 0Ax bx= > Trong : A l ma trn c kch thc m x n, b l vct c kch thc m x 1. Thut ton ca phng php n hnh c thc hin nh sau: Trang 34 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Bc 1: Tm mt phng n cc bin xut pht x v c s ca n { }jA , j J evi J l tp cc ch s c s. Bc 2: a.Xc nh cc h s jkz bi h:k jk jj JA z Ae= b.i vi mik J e , tnh cc c lng: k jk j kj Jz c ceA = Bc 3: a.Nu( ) 0kk J e A > x l phng n ti u. Dng thut ton b.Ngc li, sang bc 4 Bc 4: a.Nu( ) 0, 0,k jkk J z j J - e A < s e bi ton ti khng c nghim ti u (z khng b chn trn).Dng thut ton. b.i vi mik J esao cho0kA < , tn ti: 0jkj J z e > chn: { } min | 0s k kA = A A = ` ) a vc t rA ra khi c s. Ta c phng n cc bin mi x vi c s{ } { } ' \ J J r s =.Quay tr li bc 2. Trang 35 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Thut ton n hnh c din t theo s khi nh hnh v sau: Hnh 2.1: Lu thut ton n hnh 2.1 Bng n hnh d tnh ton, ngi ta thc hin th tc n hnh theo bng sau gi l bng n hnh: jC C sPhng n 1C2C jC rC mC kC sC 1A2A jA rA mA kA sA 1C1A1x10000 1kz 1sz 2C2A2x 01000 2kz 2sz jCjAjx00100 jkz jsz rCrArx00010 rkz rsz mCmAmx00001 mkz msz 1 Xc nh x, J, kA2.0,kk J A > e3. x ti u 4.0,0,kjkz j J-A ` ) 9In kt qu8Bin i bng 10. Dng Trang 36 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng f00 0 0 0 kA sA Nuttcccstrongdngcui(trf)u>0,nghal0,kk A > khixl phng n ti u. Nu dng cui (tr f) c nhng s m th xem th c ct ct dng cui cng mt s m m mi s trong ct u 0 hay khng? +Nu c th bi ton khng c phng n ti u. +Nu khng th chn ct s sao cho: { } min | 0s k kA = A A = ` ) Ct s gi l ct xoay . Vc t sAc a vo c s. Dng r gi l dng xoay. Vc t rAc a ra khi c s. Phn t0rsz >l giao ca ct xoay v dng xoay gi l phn t trc.Cc phn t,jsz j s = gi l phn t xoay. Tathucbngnhnhmitbngnhnhcbngcchthay,r rc A trongdng xoay bng,s sc A . Sau thc hin php bin i di y: 1.Chia mi phn t dng xoay cho phn t trc (c s 1 v tr trc), kt qu thu c l dng chnh. 2.Lymidngkhctritchcadngchnhnhnviphntxoaytngng (c s 0 mi v tr ca ct xoay). 3.Dng mi = Dng c tng ng Dng chnh x phn t xoay Lu rng sauphpxoayth vtr sAta thu c s0vlcny sAtr thnhvct nh v c s, ngha l ta lm mt i s m nh nht dng cui cng ca bng c. Ton th php bin i trn gi l php xoay quanh trc rsz . Sau khi thc hin php xoay ta c mt phng n mi v mt c s mi. Nu cha t yu cu ngha l cn0kA a.Gii bng phng php hnh hc + Gii phng trnh: - 1 22x 3x 22 + = .Cho 1 222 1x 0 x 77 3= = = ; 2 1x 0 x 11 = =- 1 2x x 10 + = . Cho 1 2x 0 x 10 = = ; 2 1x 0 x 10 = =- 2x 6 =+ Phng trnh ng mc: - 1 2x 4x 4 + = . Cho 1 2x 0 x 1 = = ; 2 1x 6 x 4 = =+ V th. Ta c phng n ti u l: 1 2 maxx 2,x 6 f 2 4* 6 26 = = = + = b.Gii bng phng php n hnh Ta thm vo 3 bin ph:3 4 5x 0, x 0, x 0 > > >Bi ton dng chnh tc l: Trang 38 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng ( )1 2 3 4 51 2 31 2 42 5jf x x 4x 0x 0x 0x max2x 3x x 22x x x 10x x 6x 0; j 1..5= + + + + + + = + + =+ => = Ta c phng n cc bin xut pht l: 1 2x= x= 0 l cc bin phi c s 3 4 5x 22, x 10, x 6 = = =l cc bin c s Vi cc vect c s l:3 4 51 0 0A 0 ; A 1 ; A 00 0 1|| || ||| | |= = =| | || | |\. \. \. Lp bng n hnh: jC C sPhng n 14000 1A2A3A4A5A 0 3A2223100 0 4A10 11010 s= 2 0 5A6 01001 r =5 f=0-1-4000 jC C sPhng n 14000 1A2A3A4A5A 0 3A42010-3 0 4A4 1001-1 s= 1 4 2A6 01001 r =3 f=0-10000 jC C sPhng n 14000 1A2A3A4A5A 1 1A2101/20-3/2 0 4A200-1/212 4 2A6 01001 f=0001/205/2 Phng n ti u l:1 2 4 maxx 2,x 6, x 2 f 2 4* 6 26 = = = = + = Trang 39 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng V d 2: Gii bi ton ti u sau:1 221 24 ax x x m + 1 21 21 21 23x 3313D=5 8 80, 0xx xx xx x+ s + s+ s> Ta a bi ton v dng chnh tc bng cch a vo 3 bin ph 3 4 50, 0, 0 x x x > > > . Bi ton dng chnh tc tng ng l: 1 2 3 4 521 24 0 0 0 ax x x x x x m + + + + 1 2 31 2 41 2 53x 3313D=5 8 800, 1..5jx xx x xx x xx j+ + = + + =+ + => = Ta c phng n cc bin xut pht:3 4 533, 13, 80 x x x = = =l cc bin c s 1 2 3 4 50 0 1 0 00 ; 0 ; 0 ; 1 ; 00 0 0 0 1A A A A A| | | | | | | | | | |||||= = = = = ||||| |||||\ . \ . \ . \ . \ . vi 3 4 51 0 00 ; 1 ; 00 0 1A A A| | | | | | |||= = = ||| |||\ . \ . \ .l cc vc t c s Ta lp bng n hnh sau: jC C sPhng n 2124000 1A2A3A4A5A 0 3A33 31100 0 4A13 11010 s= 2 0 5A80 58001 r =5 f=0-21-24000 Tnh:f = 0*33 + 0*13 + 0*80 = 0 Tnh: k jk j kj Jz c ceA = 3*0 + 1*0+ 5*0 21 = -21;1*0 + 1*0+ 8*0 24 = -24 1*0 + 0*0 + 0*0 0 = 0;1*0 + 0*0 + 0*0 0 = 0 Trang 40 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Tm:{ } min | 0s k kA = A A = ` ) Ta c s = 2, r = 5. Chia dng xoay cho phn t trc jC C sPhng n 2124000 1A2A3A4A5A 0 3A33 31100 0 4A 13 11010 s= 2 0 5A 10 5/81001/8 r =5 f=0-21-24000 Thay A5 bng A2. Cc dng khc thc hin bin i quay quanh phn t trc chnh. jC C sPhng n 2124000 1A2A3A4A5A 0 3A23 19/8010-1/8 0 4A 3 3/8001-1/8 24 2A 10 5/81001/8 f=240-60003 Ct phng n bin i:33 33 10*1 = 23;13 13 10*1 = 3; Ct A1:3 3 5/8 * 1 = 19/8;1 1 5/8 * 1 = 3/8 Ct A2:1 1 1* 1 = 0;1 1 1* 1 = 0 Ct A3:1 1 0* 0 =10 0 0* 0 = 0 Ct A4:.. jC C sPhng n 2124000 1A2A3A4A5A 0 3A33 31100 0 4A 13 11010 s= 2 0 5A 80 58001 r =5 f=0-21-24000 jCC sPhng n 2124000 Trang 41 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 1A2A3A4A5A 0 3A23 19/8010-1/8 0 4A 3 3/8001-1/8 24 2A 10 5/81001/8 f=240-60003 jC C sPhng n 2124000 1A2A3A4A5A 0 3A23 19/8010-1/8 0 4A 3 3/8001-1/8 24 2A 10 5/81001/8 f=240-60003 jC C sPhng n 2124000 1A2A3A4A5A 0 3A4001-19/32/3 21 1A 8 1008/3-1/3 S=2 24 2A 5 010-5/31/3 R=4 f=288000161 Vy phng n ti u l: x1=8, x2 = 5; x3=4; f=288. V d 3: Gii bi ton ti u bng phng php hnh hc v n hnh 1 218 6 ax + x x m1 21 21 21 2- 4x 3 63 15D=4 4, 0+ s + s s>xx xx xx x i.Phng php hnh hc - 1 2- 4x 3 6 + = x . Cho 1 2x 0 x 2 = = ;2 12x 0 x3= = - 1 23 15 + = x x . Cho 1 2x 0 x 5 = = ;2 1x 0 x 15 = = - 1 24 4 = x x . Cho 1 2x 0 x 1 = = ;2 1x 0 x 4 = = Trang 42 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng ng mc: 1 218 6 36 + = x x . Cho 1 2x 0 x 6 = = ;2 1x 0 x 2 = = Hm mc tiu khng b chn Bi ton khng c phng n ti u. ii.Phng php n hnh Ta a bi ton v dng chnh tc bng cch a vo 3 bin ph 3 4 50, 0, 0 x x x > > > . Bi ton dng chnh tc tng ng l: 1 2 3 4 518 6 0 0 0 ax + + + + x x x x x m1 2 31 2 41 2 5- 4x 3 63 15D=4 40; 1..5+ + = + + = + => =jx xx x xx x xx j Ta c phng n cc bin xut pht: 1 2x =x =0 l cc bin phi c s 3 4 5x 6, x 15, x 4 = = =l cc bin c s Vi cc vect c s l: 3 4 51 0 0A 0 ; A 1 ; A 00 0 1|| || ||| | |= = =| | || | |\. \. \. Lp bng n hnh: jC C sPhng n 186000 1A2A3A4A5A 0 3A6-43100 0 4A 15 -13010 s= 1 0 5A 4 1-4001 r =5 f=0-18-6000 Trang 43 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng jC C sPhng n 186000 1A2A3A4A5A 0 3A6-4 -13 104 0 4A 19 -11011 18 1A 4 1-4001 f=00-78000 V 2
l0,nuthnh phn khc khng u tin ca n l dng. Vc t v e Rn gi l t vng m v k hiu l v l0. Gi s v, w e Rn, ta ni v l t vng ln hn w v vit l v >l w, nu v - w >l 0. Tng t nh vy c th a vo khi nim t vng nh hn, t vng bng. Gi s z1,z2, . ..,zkl ccvct Rn. Vc tzs(1 sk) s sc gi l cc tiutvng (cc i t vng) ca cc vc t z1, z2, . . ., zk, nu zslzj ),j =1, 2, . . ., k. Khi ta s vit lzs =lex - min{ zj : j =1, 2, . . ., k } zs = lex - max{ zj : j =1, 2, . . ., k } V d: Cho 4 vc tz1 = (1/4, 1, 1/4, 3/4), z2 = (2, 1, 10, 1), z3 =(1/4, 1, 1/12, 1/6), z4 = (2, 1, 5, -4), Khi zs= lex - min{zj : j =1, 2, 3, 4}, zs= lex - max{zj : j =1, 2, 3, 4}. Qui tc cc tiu t vng chn dng xoay. Ch rng trong thut ton n hnh dng xoay cn chn l dng m ti t gi tr. u0=min{xi /xij0 : xij0 >0, i e JB} Trongtrnghpcnhiuchsdngicngtcctiutrongbiuthctrntachn dng xoay l dng i0 m ti zi/xi0j0= lex - min{ z j / xi0j0 : xi0j0 >0}(1.1) trong zi= (xi, x1, . . . ., xij0, xin),(i = 1, . . ., m) - dng th i ca bng n hnh. Nghi l khi c nhiu dng c th chn lm dng xoay ta s chn dng xoay l dng cc tiu t vng. Qui tc (1.1) c gi l qui tc cc tiu t vng. Chrngvic chndngxoaytheoqui tc cc tiu tvng sl n tr,vnu c hai dng cc tiu t vng th suy ra rank A < m. K hiu zi(i = 1, . . ., m) - dng th i ca bng n hnh : Trang 47 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng zi= (zi, z1, . . . ., zij0, zin) =(xi, x1, . . . ., xij0, xin), (i = 1, . . ., m) cnz0 = (z00, z01, . . ., z0n) =(f(x), A 1, . . ., A n)l dng c lng. nh l 3.1: Gi s bng n hnh xut pht ta c : zi>l0, i =1, 2, . . ., m. v trong thut ton n hnh ta s dng cc qui tc sau y chn dng xoay, ct xoay : a) Ct xoay s c chn ty trong s cc ct c c lng dng (z0s= A s > 0) ; b) Dng xoay i0 = r c chn theo qui tc cc tiu t vng : zr/zrs = lex - min{zi/ zis : zis >0} Khitrongqutrnhthchinthuttonnhnhccdngnyslunltvng dng cn dng c lng z0 s l tvng gimngtv thutton n hnh s dng sau hu hn bc. Chngminh:Trchttachrarngccdngzi,i=1,2,...,m,sltvngdng saumiphpbininhnh.dngxoayrtaczr= 1zrszr,trongzrs>0-phnt xoay. V vy, nu zr >l 0 thzr>l0. By gi, xt i=r. Khi nu zis> 0 thtqui tc t vng suy ra : zi =zi -zzisrszr = zis 1 1zzzzisirsr|\
|.|>l0 cn nu ziss0 th zi =zi -zzisrszr = zi+ zzisrszr>l 0. Vyzilunltvngdngsaumiphpbininhnh.Vctz0sauphpbin i n hnh s tr thnh z0 =zo -zzsrs0zr = z0+ zzsrs0zr , mt khc, do z0s>0 v zr >l0, nn t suy raz0 0 ` (1.2) Khng gim tnh tng qut ta c th gi thit rng bi> 0, i=1, 2, . . ., m(1.3) (nu c bi< 0 ch cn nhn hai v ca phng trnh tng ng vi -1). Theo cc thng s ca bi ton (1.3) ta xy dng bi ton ph thuc sau y exumin,Ax+xu=b,x>0,xu> 0(1.4) trong xu= x(Ju),Ju= { n + 1, . . ., n + m} l m - vt cc bin gi, e=(1, 1, . . ., 1) - m - vc t c cc thnh phn u bng 1. B sau y cho thy mi lin h gia bi ton (1.2) v bi ton (1.4). B 1 : Bi ton (1.2) c phng n chp nhn c khi v ch khi thnh phn x*u trong phng n ti u (x*, x*u) ca bi ton (1.4) l bng khng. Chng minh: Trang 49 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng iu kin cn: Gi s x* l phng n chp nhn c ca bi ton (1.2). Khi r rng (x*, x*u = 0) l phng n chp nhn c ca bi ton (1.4). Mt khc, do ex*u=0sexu , vi mi (x, xu) l phng n chp nhn c ca bi ton (1.27) nn (x*, x*u) l phng n ti u ca n. iu kin : R rng (x*, x*u = 0) l phng n ti u ca bi ton (1.4) th x* l phng n chp nhn c ca bi ton (1.2). B c chng minh. i vi bi ton (1.4) ta c ngay mt phng n c s chp nhn c ca n l x(J) = 0,x(Ju)= b vi c s tng ng B= {An+1, . . ., An+m}, trong khiu Aj,je Julvc t ct trongma trnrngbuc ca bi ton (1.4) tng ng vi bin gi xj, j e Ju. V vy ta c th p dng thut ton n hnh gii bi ton (1.4) bng thut tonn hnhc gi lpha th nht ca thut tonnhnh hai pha gii bi ton (1.2), v bi ton (1.4) cn c gi l bi ton pha th nht. Kt thc phath nht ta sxydng c phngn c s chpnhn c ti u (x*, x*u) vi c s tng ng l B*= {Aj : jeJ*B} ca bi ton (1.4).C th xy ra mt trong 3 kh nng sau: i) x*u=0 ; ii) x*u= 0 v ma trn B* khng cha cc ct ng vi bin gi, tc l n cha ton ct ca ma trn rng buc ca bi ton (1.2): B*= {Aj : jeJ*B}, J*B Ju= f. iii) x*u= 0 v ma trn B* c cha ct ng vi bin gi, tc l B*= {Aj : jeJ*B}, J*B Ju =f. Ta s xt tng trng hp mt. i) x*u=0, th theo b 1, bi ton (1.2) l khng c phng n chp nhn c, thut ton kt thc. ii) Trong trng hp ny x* l phng n c s chp nhn c ca bi ton (1.2) vi c stngnglB*.Btutntacthtinhnhthuttonnhnhgiibiton (1.2). Giai on ny gi l pha th hai ca thut ton n hnh hai pha v ton b th tc va trnh by c gi l thut ton n hnh hai pha gii bi ton ti u (1.2). iii)Khiuxi* ,i*eJ*BJulmtthnhphnbingitrongphngncschp nhn c ti u (x*, x*u) ca bi ton (1.4). Khiuxi*j, jeJJulccphntcadngi*trongbngnhnhtngngvi phng n ti u (x*, x*u) ca bi ton (1.4). Trang 50 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Nu tmc ch s j* eJ \ J*B sao choxi*j*
=0 th thchin mtphp bin i bng nhnhviphntxoaycchnlxi*j* tasacthnhphnbingixi*rakhi c s v thay vo ch ca n l bin xj* Nuxi*j =0,jeJ\J*B,thiucnghilphngtrnhtngngvintrongh phngtrnhtuyntnhAx=blhqucaccphngtrnhcnli.Khitbngn hnhtngngviphngntiu(x*,x*u)cabiton(1.4)tacthxabdngni trn v ng thi xa b lun ct ng vi bin gi xi* . Trong c hai trng hp va nu ta u loi b c bin gi xi*
khi c s. Ln ltimdin tt c cc thnhphn bingi trong c s, tcli n trnghp ii),ngthitrongqutrnhnytacngsloicttcccrngbucphthuctuyn tnh trong h Ax = b. Khi ta c th bt u pha th hai ca thut ton n hnh hai pha. Nh vy, thut ton n hnh hai pha p dng gii bi ton (1.2) s cho php :1) hoc l xc nh c rng bi ton khng c phng n chp nhn c; 2) hoc l xc nh c tnh khng b chn di ca hm mc tiu ca bi ton ; 3) hoc l xc nh c phng n c s ti u ca bi ton ; ng thi trong qu trnh thc hin thut ton ta cng loic ttc cc rng buc ph thuc tuyn tnh. Tktqulmviccathuttonnhnhhaiphatacthchngminhccnhl quan trng sau y:nh l 4.1: Nu bi ton ti u c phng n ti u th cng c phng n ti u c s.Chngminh:Gisbitontiulcphngntiu.Khi,thuttonnhnh haiphapdnggiibitontrachcthktthctnhhung3),tclthuc phng n c s ti u c n. nh l 4.2: iu kin cn v bi tonti u cphng n ti ul hmmc tiu ca n b chn di trn min rng buc khc rng.Chng minh:iu kin cn: Gi sx*l phngn ti u cabi ton. Khi,f(x) > f*(x), x eD, tc l hm mc tiu ca bi ton l b chn di. iukin:Nuhmmctiucabitonlbchnditrnminrngbuckhc rng, thp dng thuttonn hnh haiphagii ta ch c thkt thc thut ton tnh hung 3), tc l tm c phng n ti u ca n. Trang 51 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 4.2 V d.Gii bi ton ti u tuyn tnh sau y bng thut ton n hnh hai pha f(x) =2x1+x2 + x3 min, 1 2 3 4 51 2 3 4 51 23 4 5jx +x+ x+ x+ x=5x +x+2x + 2x+2x=8x +x =2 x+ x+ x=3x 0, j=1, 2, . . ., 5> Bi ton ph tng ng c dng x6 + x7 + x8 + x9 min, x1+x2 + x3 + x4 + x5+ x6=5, x1+x2 + 2x3 + 2x4 + 2x5+ x7=8, x1+x2 + x8 =2, x3 +x4+ x5 +x9 =3, xj>0, j=1, 2, . . ., 9. Phng n c s chp nhn c ca bi ton ph l (x, xu)=(0, 0, 0, 0, 0, 5, 8, 2, 3) vi c s tng ng l B= E4 - ma trn n v cp 4.C s gm ton cc bin gi, ta bt u pha th nht ca thut ton n hnh hai pha. Cc kt qu tnh ton ca pha th nht c ghi vo bng 1.1 sau: Bng 1.1: Bng n hnh pha 1 jCC s Phng n 000001111A1A2A3A4A5A6A7A8A9 1 1 1 1 A6 A7 A8 A9 5 8 2 3 1 1 1 0 1 1 1 0 1 2 0 1* 1 2 0 1 1 2 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 R = 3 S = 9 A = (Zj-Cj) f = 18334*440000 1 1 1 0 A6 A7 A8 A3 2 2 2 3 1* 1 1 0 1 1 1 0 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 -1 -2 0 1 R=1 S=6 Trang 52 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng f = 63*3000000-4 jCC s Phng n 000001111A1A2A3A4A5A6A7A8A9 0 1 1 0 A1 A7 A8 A3 2 0 0 3 1 0 0 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 -1 -1 0 0 1 0 0 0 0 1 0 -1 -1 1 1 f = 000000-300-1 Phathnhtcathuttoncktthctnhhungiii).Xabkhibngnhnh cc dng ng vi bin gi x7, x8 v cc ct ng vi cc bin ny ta tip tc thc hin pha th hai ca thut ton. Cc kt qu tnh ton trong pha th hai c tip tc trong bng 2 Bng 1.2: Bng n hnh pha 2 jCCPhng21100u sn A1A2A3A4A5 2 1 A1 A3 2 3 1 0 1* 0 0 1 0 1 0 1 R=2 S=1 A = (Zj-Cj)f = 701*011 1 1 A2 A3 2 3 1 0 1 0 0 1 0 1* 0 1 R=4 S=3 f = 5-1001*1 1 0 A2 A4 2 3 1 0 1 0 0 1 0 1 0 1 f = 2-10-100 Phng n ti u : x*= ( 0,2,0, 3,0 ).Gi tr ti uf *=2 5. THUT TON N HNH HAI PHA CI BIN 5.1 M t thut ton Thuttonnhnh2phacibinthcchtlthuttonnhnh2phanhngnhm gimsctcamatrntnhtonbanu,cngnh gimkhilngtnhton,thayv chn cc bin c s ban u ton l nhng bin gi, ta s chn ra trong nhng bin ph no lvec-tnvlmmtbincs,vnhvysgimsbingi,thmchcths khng c bin gi. Cho v d rng buc5x1 + 3x2< 1thay v bin i qua 2 giai on: + bin du Ta a bi ton v dng chnh tc nh cc bin ph x4v x5 f(x) =-2x1 + x2 - 2x3 min 1 2 3 41 2 31 2 3 5jx -x+2x + x 82x + 2x - x= 4- x +x+x x = 1x0, j=1..5=> Ma trn cu h s rng buc: 1 2 3 4 5A A A A A1 1 2 1 02 2 1 0 01 1 1 0 1A ( ( (= ( ( A ch cha mt vect n v A4. c ma trn n v hng 3 ta thm vo ma trn A thm 2vct nv: A6 = (0, 1, 0)v A7 = (0, 0,1) bng cch thmvov tri cc rngbuc trn hai bin gi x6 v x7nh sau: 1 2 3 41 2 3 61 2 3 5 7jx -x+2x + x 82x + 2x - xx = 4- x +x+x x x = 1x0, j=1..7=+ +> Hm mc tiu ca bi ton tr thnh: Trang 54 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng f(x) =-2x1 + x2 - 2x3+ x6 + x7 min Pha 1 gm 1 bin ph v 2 bin gi. Bng n hnh pha 1 nh sau: Bng 1.3: Bng n hnh pha 1 cj C s Phng n 0000011 A1A2A3A4A5A6A7 0 1 1 A4 A6 A7 8 4 1 1 2 -1 -1 2 1* 2 -1 1 1 0 0 0 0 -1 0 1 0 0 0 1 R=2 S=7 A = (Zj-Cj)F=513*00-100 0 1 0 A4 A6 A2 9 2 1 0 4* -1 0 0 1 3 -3 1 1 0 0 -1 0 0 -1 2 -1 0 1 0 R=1 S=6 F=24*0-3020-3 0 0 0 A4 A1 A2 9 12 32 0 1 0 0 0 1 3 34 14 1 0 0 -1 12 12 0 14 14 1 12 12 F=000000-1-1 Phathnhtcathuttoncktthctnhhungii).Xabkhibngnhnh cc ct ngvi bin gix5,x6v tip tc thc hin pha thhai ca thut ton. Cc ktqu tnh ton trong pha th hai c tip tc trong bng n hnh pha 2 Bng 1.4: Bng n hnh pha 2 cjCPhng-21-100u snA1A2A3A4A5 0 -2 1 A4 A1 A2 9 1/2 3/2 0 1 0 0 0 1 3* -3/4 1/4 1 0 0 -1 1/2 -1/2 R=3 S=4 A = (Zj-Cj)0011/4*0 -3/2 -1 -2 1 A3 A1 A2 3 11/4 3/4 0 1 0 0 0 1 1 0 0 1/3 1/4 -1/12 -1/3 1/4 -5/12 A = (Zj-Cj)000-11/12-7/2 Phng n ti u : x*= (11/4,3/4, 3 ). Gi tr ti uf*=-31/4 6. PHNG PHP NH THU (M PHNG PHP) a vo cc bin ph trong bi ton ti u dng chun chuyn v dng chnh tc. Trang 55 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 6.1 M t thut ton ()1 1 2 2 n nnij j ij=1jc x c x ... c x maxa x b, i 1..m Px 0, j 1..n+ + + = => = Ta a vo m bin gi to khc vi bin ph: n 1 n 2 n mx ,x ,...,x 0+ + +>V mt s M > 0 rt ln, ln hn bt k s no cn so snh vi n. Ta chuyn bi ton (P) v bi ton (M) sau y: ( )1 1 2 2 n n n 1 n 2 n mi1 1 i2 2 in n n k jjc x c x ... c x Mx Mx ... Mx maxa x a x ... a x x b , i 1..m Mx 0; j 1..m n+ + +++ + + + + + + = => = + Khi a M > 0 rt ln vo chng khc ta nh thu rt nng vo cc bin gi to dng tng ng khin cho trong phng n ti u (nu c) th cc bin gi to bng 0 tt c. Khi bi ton M sau khi chuyn i c gi l bi ton nh thu. Gii bi ton (M) bng phng php n hnh. Tnh hung 1: Bi ton (M) c phng n ti u dng ( )*x , 0, 0,..., 0khi x* l phng n ti u ca bi ton ban u (P) (vi tt c cc bin gi to u bng 0) Tnh hung 2: Bi ton (M) cphngnti u (x,y) trong vecty 0 = cn bin gi to dng. Khi bo ton (P) khng c phng n khng c phng n ti u. Ch1:Vcchshmmctiucabiton(M)phthuctuyntnhvoM,m k jk j k kz c c A = Aphthuctuyntnhvocchshmmctiu kA phthuc tuyn tnh vo M. Phn tch kAthnh hai thnh phn: k k kM A = o +q-Nu kq > 0 kA >0 -Nu kq < 0 kA ij j i ia x b, b 0 thtrchttaphitrvtrichomt bin ph iym khng cng na. Sau thm vo bin gi to nix 0+ >v rng buc tr thnh: nij j i n i ij 1a x y x b+= + = Trn hm mc tiu: i nif f 0.y Mx += + 6.2 V d V d 1. Gii bi ton ti u sau bng phng php n hnh 1 2 3 41 2 3 41 2 3 41 2 3 4j2x x x x maxx x 2x x 22x x 3x x 6x x x x 7x 0, j 1..4+ + = + + =+ + + => = Gii a vo 3 bin gi to khng m 5 6 7x , x , x , ta c bi ton tng ng sau: 1 2 3 4 5 6 71 2 3 4 51 2 3 4 61 2 3 4 7j2x x x x Mx Mx Mx maxx x 2x x x 22x x 3x x x 6x x x x x 7x 0, j 1..7+ + + = + + + =+ + + + => = Trong M>0 l s rt ln.Ta c phng n xut pht:( ) x 0, 0, 0, 0, 2, 6, 7 = Trang 57 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Vi cc bin c s l:5 6 7x 2, x 6, x 7 = = =Cc vect c s xut pht l:5 6 71 0 0A 0 ; A 1 ; A 00 0 1|| || ||| | |= = =| | || | |\. \. \. Trang 58 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Ta c bng n hnh (phn di c chia thnh 2 dng): 1 1 1M 2 4M A = o + q = jC C sPhng n 21-1-1- M- M- M 1A2A3A4A5A 6A7A - M 5A2 1 -12-1100 - M 6A6 21- 31010 - M 7A7 1111001 ko-2-111000 kq- 4-10-1000 jC C sPhng n 21-1-1- M- M- M 1A2A3A4A5A 6A7A 2 1A21-12-1 00 - M 6A2 0 3 - 7310 - M 7A5 02-1201 ko0-35-100 kq0-58-500 jC C sPhng n 21-1-1- M- M- M 1A2A3A4A5A 6A7A 2 1A8/310-1/30 0 1 2A2/3 01- 7/310 - M 7A11/3 00 11/3 01 ko00-220 kq00-11/300 jC C sPhng n 21-1-1- M- M- M 1A2A3A4A5A 6A7A 2 1A31000 1 2A3 0101 - 1 3A1 0010 ko0002 kq0000 f=8 Trang 59 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng V d 2. Gii bi ton ti u sau bng phng php n hnh 1 2 3 41 2 3 41 2 3 41 2 3 4j3x x 3x x minx 2x x x 22x 6x 3x 3x 9x x x x 6x 0, j 1..4 + + + + = + + + = + => = Gii a bi ton trn v bi ton max tng ng: 1 2 3 41 2 3 41 2 3 41 2 3 4j3x x 3x x maxx 2x x x 22x 6x 3x 3x 9x x x x 6x 0, j 1..4 + + + = + + + = + => = a vo 3 bin gi to khng m 5 6 7x , x , x , ta c bi ton tng ng sau: 1 2 3 4 5 6 71 2 3 4 51 2 3 4 61 2 3 4 7j3x x 3x x Mx Mx Mx maxx 2x x x x 22x 6x 3x 3x x 9x x x x x 6x 0, j 1..7 + + + + = + + + + = + + => = Trong M>0 l s rt ln.Ta c phng n xut pht: ( ) x 0, 0, 0, 0, 2, 9, 6 =Vi cc bin c s l:5 6 7x 2, x 9, x 6 = = =Cc vect c s xut pht l:5 6 71 0 0A 0 ; A 1 ; A 00 0 1|| || ||| | |= = =| | || | |\. \. \. Ta c bng n hnh (phn di c chia thnh 2 dng): Trang 60 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng jC C sPhng n 3-1-31- M- M- M 1A2A3A4A5A 6A7A - M 5A212-11100 - M 6A 9 2-633010 - M 7A 6 1-11-1001 ko-3131000 kq- 45-3-3000 jC C sPhng n 3-1-31- M- M- M 1A2A3A4A5A 6A7A 3 1A212-11 00 - M 6A 5 0-105110 - M 7A 4 0-32-201 ko070200 kq013-7100 jC C sPhng n 3-1-31- M- M- M 1A2A3A4A5A 6A7A 3 1A31006/5 0 -3 3A 1 0-211/50 - M 7A 2 010-12/51 ko 0702 0 kq 0-1012/5 0 jC C sPhng n 3-1-31- M- M- M 1A2A3A4A5A 6A7A 3 1A31006/5 -3 3A 5 001-23/5 - 1 2A 2 010-12/5 ko00094/5 kq0000 max minf 3* 3 3* 5 1* 2 8 f 8 = = = Trang 61 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Chng 4 L thuyt i ngu v bi ton ti u i ngu Vi mibi tonti u, ta cth thit lp tngng chonmt bi ton khcgi lbi ton i ngu ca n. Khi nim i ngu l mt trong cc khi nim c bn ca ti u ha. Trong nhiutrnghp ccnhng ktlun chpnhn c chomt trong cc bitontiuthvicnghincubitoningucantrathuntinhn.Hnna, khi phn tch song song mt cp bi ton i ngu ta c th thu c nhng kt lun hay c v ton hc ln kinh t. J. Von Neumann xy dng m hnh bi ton ti u i ngu v mt s nh l i ngu vonm1947 davo ccktqu cal thuyt tr chi, nhng ntnnm1951 cc kt qu ny mi c cng b bi mt s nh ton hc khc nh Gale, Kuhn, Tucker. giiquytbitoningu,nhtonhcC.E.Lemkearaphngphpn hnh i ngu vo nm 1954. 1. BI TON I NGU Xt bi ton ti u dng chun (P): min (P*): max Xt bi ton ti u dng chnh tc (P): min (P*): max 2. QUI TC CHUYN BI TON TI U TNG QUT SANG BI TON I NGU 2.1 Qui tc chuyn i Trong trng hp bi ton ti u tuyn tnh tng qut, nhng quy tc sau y c p dng xy dng bi ton i ngu tng ng Ax bx 0>>A'y cy 0>>Ax = bx 0>A'y c s Trang 62 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Bi ton gcBi ton i ngu -Hm mc tiu MaxMin -Bin i ngu Mi rng bucMt bin i ngu -Cho ph i ngu v gii hn rng buc Chi ph i nguGii hn rng buc -Ma trn rng buc Ma trn rng bucMa trn chuyn v -Chiu ca rng buc v du ca bin Rng buc trong bi ton max c du Bin i ngu trong bi ton minc du (tri chiu) Rng buc trong bi ton max c du =Bin i ngu trong bi ton minc du ty Rng buc trong bi ton max c du >Bin i ngu trong bi ton minc du (tri chiu) Bin ca bi ton maxc du 0 Rng buc i ngu ca bi ton minc du (cng chiu) Bin ca bi ton maxc du ty Rng buc i ngu ca bi ton minc du =Bin ca bi ton maxc du 0 Rng buc i ngu ca bi ton minc du 0 (cng chiu) Xt cc rng buc dng ma trn ca mt bi ton quy hoch tuyn tnh tng qut nh sau : K hiu: l dng th I (i=1..m) l ct th j (j=1..n) Khi , mi lin h gia hai bi ton i ngu c th c trnh by nh sau: s >s>>s sTiAjA Trang 63 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Cch nh:-Bi ton gc min, rng buc chung cng du, rng buc bin tri du -Bi ton gc max, rng buc chung tri du, rng buc bin cng du 2.2 V d V d 1. Bi ton gc (P) Vit di dng ma trn ta c: Cc bin i nh sau: ty Bi ton i ngu tng ng l ()1 2 41 2 3 42 3 41 2 3 41 2 3 4f x x 2x 3x minx 3x 4x x 7x 2x 6x 15x 7x x 8x 2x 0; x 0, x , x ty = + + + + > + + s+ + + = > s() ( ) ( )1 2 3 41234f x 1, 2,0,3 . x , x , x , x minx1 3 4 1 7x0 1 2 6 . 1x5 7 1 8 2x= | |> | | | | | || | s || | ||= |\ . \ .\ .1 2 3 4x 0; x 0,x ,x ty > s1 2 3 4x 3x 4x x 7 + + + > 1y 0 >2 3 4x 2x 6x 1 + + s 2y 0 s1 2 3 45x 7x x 8x 2 + + + = 3y1x 0 > 1 3y 5y 1 + s2x 0 s 1 2 33y y 7y 2 + > 3x ty 1 2 34y 2y y 0 + + =4x ty 1 2 3y 6y 8y 3 + + = Trang 64 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng V d 2.Bi ton gc (P) Bi ton i ngu tng ng l: 2.3 ngha kinh t ca bi ton i ngu Xt bi ton lp k hoch sn xut: Mt x nghip c s lng g (B1) v axits (B2) tng ng l 5000 m3, 90 tn (cc yu t sn xut khc c s lng ln). X nghip c th sn xut ra 3 loi giy A1, A2, A3. Mc tiu hao cc loi nguynliu (yu t snxut) snxut ra1 tn giy thnh phm nh sau: Gi bn 1 tn giy A1, A2, A3 tng ng l 9, 12 v 10 triu ng (gi s cc sn phm sn xut ra u c th tiu th c ht). Lp k hoch sn xut ti u. Gi xj l s tn giy loi Aj cn phi sn xut. Ta c m hnh ton hc sau: Tm sao cho: ()1 2 31 31 2 31 2 31 2 31 2 3f * y 7y y 2y maxy y 13y y 7y 24y 2y y 0y 6y 8y 3y 0;y 0, y ty = + + s + > + + =+ + => s () ( ) ( )1 2 3123f x 1, 1,2 . x , x , x max1 3 5 x 62 0 1 . x 41 4 7 x 2= > | | | || | || |s || | || | = \ . \ .\ .1 2 3x 0; x 0,x ty > s( ) ( ) ( )1 2 3123f * y 6,4, 2 . y ,y ,y min1 2 1 y 13 0 4 . y 15 1 7 y 2= > | | | || | || |s || | || |=\ . \ .\ .1 2 3y 0; y 0,y ty s >( )1 2 3x , x , x Trang 65 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Gisbygicngimunmuatonbslngccyuttsnxutcax nghip. Khi gi bn nn t l bao nhiu? Gi yi l gi bn 1 n v yu t sn xut loi Bi, i=1,2 Gi bn khng th m nn Ta c s tin thu c khi bn cc yu t sn xut dng sn xut ra 1 n v sn phm loi Aj l: Loi A1: y1 + 20y2 Loi A2: 3y1 + 30y2 Loi A3: 2y1 + 24y2 Ta thy c 2 tng sau: -ivingibn:Gibnccyutsnxutchcchpnhnkhistinthu c do bn cc yu t sn xut dng sn xut ra mt n v sn phm loi Aj phi khngthnstinthuckhixnghipsdngccyutsnxutsn xut ra 1 n v sn phm loi Aj. Tc l: -i vingi mua: Ch chp nhn gi tr ccyu t snxutkhi tng s tin dng mua tt c cc yu t sn xut l t nht. Tc l: Tm li, ta c m hnh: Tm (y1,y2) sao cho: 3. CC NH L I NGU Ta thy, bi ton i ngu P* cng l bi ton ti u tuyn tnh. Do gii (P*) c 3 cch. Cch 1: Dng phng php n hnh gii trc tip (P*) ()1 2 31 2 31 2 3jf x 9x 12x 10x maxx 3x 2x 500020x 30x 24x 90000x 0, j 1..3= + + + + s+ + s> =iy 0 >1 2 1 2 1 2y 20y 9; 3y 30y 12; 2y 24y 10 + > + > + >1 2f* 5000y 90000y min = + 1 21 21 21 2if* 5000y 90000y miny 20y 93y 30y 123y 30y 12y 0, i 1,2= + + >+ >+ >> = Trang 66 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Cch 2: Gii bi ton (P) bng thut ton n hnh i ngu. Ta c phng n ti u ca bi ton gc (P), ng thi c lun phng n ti u ca bi ton i ngu (P*) bng cch gii h phng trnh tuyn tnh.Cch 3: Gii bi ton i ngu (P*) bng phng php n hnh. T phng n ti u ca (P*) ta suy ra phng n ti u x* ca bi ton gc (P) Vn t ra l t phng n ti u x* ca (P) lm th no suy ra c phng n ti u y* ca (P*). Vn trn c gii quyt thng qua cc nh l i ngu. Xt cp bi ton i ngu: (P): (P*):Vi X l min rng buc (tp phng n) ca bi ton (P) Y l min rng buc (tp phng n) ca bi ton (P*) nh l 3.1. (nh l i ngu yu): x l phng n ca (P), y l phng n ca (P*) th: H qu 3.1: NuX = C v hm mc tiu f khng b chn di trn min X th Nu v hm mc tiu f* khng b chn trn trn min Y th nh l 3.2. (nh l i mnh): Nu (P) c phng n ti u l x* th (P*) cng c phng n ti u l y* v f(x*)=f*(y*) Nu (P*) c phng n ti u l y* th (P) cng c phng n ti u l x* v f(x*)=f*(y*) H qu 3.2: (P) v (P*) c phng n (P) v (P*) c phng n ti u. V gi tr ti u ca cc hm mc tiu bng nhau. Nhn xt: t nh l ny ta dng kt qu sau kim tra phng n ti u x l phng n ca (P), y l phng n ca (P*) v f(x) = f*(y) x l phng n ti u ca (P) v y l phng n ti u ca (P*). x Xf(x) c, x mine= y Yf * (y) b,y maxe= f(x) =c, x f*(y) =b, y >Y = CY = C X = C Trang 67 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng i vi cp bi ton (P) v (P*) ta c cc kt qu sau: C 2 bi ton cng c phng n th c 2 bi ton cng c phng n ti u v gi tr ti u ca 2 hm mc tiu lun bng nhau. Ch 1 bi ton c phng n th c 2 bi ton cng khng c phng n ti u (gi s (P) c phng n th f(x) khng b chn di, hoc (P*) c phng n th f*(y) khng b chn trn) C 2 bi ton cng khng c phng n th hin nhin chng khng c phng n ti u. Rng buc cht l rng buc xy ra du = Rng buc lng l rng buc xy ra du bt ng thc thc s >, || | ||s |\ . \ .\ .> =maxf 54 =( )1 2 3f y 50y 16y 23x min = + + Trang 68 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Ta c: Giihtrntac: 1 2 323 6y 2,y ,y .5 5= = = Vy 23 6y* (2, , )5 5= .Kimtray*lphng ntiuca(P*):Th 23 6y* (2, , )5 5= vo3rngbuccnli 2 3y 0,y 0 s > ,1 2 35y 3y 4y 2 + + > thythamny*lphngnca(P*).M=54y*l phng n ti u duy nht ca (P*). V d 2.( )1 2 3f x 6x 2x 5x max = + + 123j2 3 1 x 101 0 2 x 81 2 5 x 19x 0; j 1..4s | || || | | | |s | | | | | |s\ .\ .\ .> = a)Gii (P) bng thut ton n hnh b)Vit (P*) v gii (P*) Gii a)pdngphngphpnhnhtacphngntiuca(P)lx=(4,0,2)v maxf 34 = . b) Bi ton i ngu (P*) l ( )1 2 3f * y 10y 8y 19y min = + + 123i2 1 1 y 103 0 2 y 81 2 5 y 19y 0; i 1..3> | | | | | | |||> ||| |||>\ . \ . \ .> = Ta c: 31 1 2 33 1 2 34 2.0 5.2 14 19 y 0x 4 0 2y y y 6x 2 0 y 2y 5y 5+ + = < = = > + + == > + + = 1232 3 15 3 4 2y1 0 0 2y1 1 3 1y6 2 1 4y 0,y 0,y ty | | | |> | | || | ||> | || |> ||\ .\ . \ .s >2 13 1 2 34 1 2 3x 14 0 y 2x 6 0 y y 3y 1x 5 0 6y 2y y 4= > = = > + + == > + + =*max minf f = Trang 69 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Gii h trn ta c 7 4y* , ,03 3| |=|\ .. Kim tra ta c y* l phng n ca (P*) nn l phng n ti u ca (P*) v maxf 34 = . 4. THUT TON N HNH I NGU Thc cht caphngphpn hnh i ngu chnh lp dng phng php nhnh giibitoninguviphngnxutphtlphngnccbincabitoni ngu. Kt qu cui cng ta s thu c l phng n ti u ca bi ton gc. Khng mt tnh tng qut, ta xt cp bi ton ti u i ngu sau: (P): min (P*): max Gistacmtphngnccbiny0 khngthoihacabiton(P*)viccrng bucclptuyntnh: 0j jA ,y c , j J = e .Vmtkhc: 0k jA ,y c , k J < e .Trong J m =v h { }jA j J egi l c s i ngu.Gii h phng trnh:j jj Jkx .A bx 0, k Je== e ta thu c mt gi phng n ca (P) tng ng vi c s i ngu J. Cc bin xj , j J ecngcgilccbincscagiphngn.Nuttcccbincscagi phng n x u khng m th x c gi l phng n ti u ca bi ton (P). Bc 1: Xy dng bng n hnh cho gi phng n x vi c s J: k0, k J A s eBc 2: Kim tra tiu chun ti u cho gi phng n x: a.Nu> ejx 0, j J th x l phng n ti u ca bi ton (P). Thut ton dng. b.Nu- < ejx 0, j J th sang bc 3 Bc 3: a.Nu- ejkz 0, k J Hm mc tiu cu bi ton i ngu (P*) khng b chn trn bi ton (P) khng c phng n chp nhn c. Thut ton dng. b.i vi mi< ejx 0, j J tn tik J esao cho: A'y c s Trang 70 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng - Chn { }e= ejx j J3. x ti u 4.0,0,- ejjkxz j J 5Khng c phng n ti ung ngSai Sai 7 A A u = = < ` )k ks rkrs rsmin | z 0z z 9In kt qu8Bin i bng 10. Dng 6 { }e= < er j jj Jx min x | x 0, j J Trang 71 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng V d 1f(x)= x1 x1-2x4 +2x5 -3x6 min x1+x4 +x5-x6= 2 x2+x4 +x6= 12 x3 +2x4 +4x5 +3x6 = 9 xj> 0 , j=1, 2, 36 Gii Bi ton i ngu:f*(y) = 2y1 +12y2 +9y3 max y1 s 1 y2 s -1 y3 s 0 y1 + y2 + 2y3s - 2 y1 + 4y3s2 -y1 + y2 + 3y3 s -3 Ta d thy mt phng n cc bin: = = = 1 2 3y 1, y 1, y 1 Vi c s 1 2 4A ,A ,A , v rng: ( )| | |= = | |\ .011A , y 1, 0, 0 1 11; ( )| | |= = | |\ .021A , y 0, 1, 0 1 11;( )| | |= = | |\ .041A , y 1, 1, 2 1 21 tm gi phng n ta gii h j jj Jx .A be=, Tc l:11 42 4 2445x2x x 215x x 12 x22x 99x2= + = + = = == Trang 72 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Cn tm cc s jkz t h:e= e = jk j kj Jz .A A , k J, k 3, 5, 6 ( ) ( ) ( ) = = = t t t3 5 6A 0, 0, 1 , A 1, 0, 4 , A 1, 1, 1Kt qu cho trong bng sau: C s Gi phng n 1-10-22-3 1A 2A 3A 4A 5A 6A 1 -1 -2 1A 2A 4A -5/2 15/2 9/2 1 0 0 0 1 0 -1/2 -1/2 1/2 0 0 1 1 -2 2 -5/2 -1/2 3/2 R=1 S=6 -1900-10-5-2 -3 -1 -2 6A 2A 4A 1 8 3 -2/5 -1/5 3/5 0 1 0 1/5 -2/5 1/5 0 0 1 2/5 -9/5 7/5 1 0 0 -17-4/50-3/50-21/50 Ta kt thc v> ejx 0, j J. Vy phng n ti u ca bi ton cho l x*=(0,8,0,9,0,1); f(x*)= -17. 5. Vn phng n cc bin v c s xut pht pdngthuttonnhnhingu,trctintaphixcnhcmtphng n cc bin xut pht cho n. -Nu bi ton dng chnh tc c mt c s gm cc vect { }ie , ta lp bng n hnh ng vi c s ny, nuA s ek0, k Jth ta ly lm c s i ngu xut pht v p dng thut ton. -Nu bit mt phng n cc bin y ca bi ton i ngu ca bi ton dng chnh tc, ta cnxc nh c s cay,tmmatrnh s phn tch theo c s nyvlp bng n hnh tng ng. NuA s ek0, k Jth taly nlm c s i nguxut phtv p dng thut ton. Trang 73 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Chng 5 Bi ton vn ti Bitonvntilmtdngcbitcabitontiu,bivytacthdngcc phngphpgiibitontiugii.Tuynhindotnhcthcan,ngitaxy dng cc phng php gii ring. Ta s xt mt s phng php gii bi ton vn ti. 1. PHT BIU BI TON, S TN TI CA NGHIM TI U 1.1 Pht biu bi ton Cmaim 1 2 mA ,A ,...,A cngsnxutmtloihanghavicclnghngtng ng l 1 2 ma ,a ,...,a . C n ni tiu th loi hng 1 2 nB ,B ,...,B vi cc yu cu tng ng l 1 2 nb ,b ,...,b . n gin ta s giiAl im pht i,= i 1..m jBl im thu j,j 1..n =Hng c th tr t mt im pht bt k (i) n mt im thu bt k (j). K hiu: - ijcl chi ph chuyn tr mt n v hng t im pht (i) n im thu (j) - ijxl lng hng chuyn tr t im pht (i) n im thu (j) Bi ton t ra l: Xc nh nhng i lng ijx cho mi con ng (i, j) sao cho tng chi ph chuyn tr l nh nht vi gi thit l:m ni ji 1 j 1a b= == Tc l lng hnh pht ra bng ng lng hngyu cu cc im thu (iukin cn bng thu pht).Dng ton hc ca bi ton vn ti l: ( )m nij iji 1 j 1c x min 1.1= = ( )nij ij 1x a, i 1..m 1.2== = ( )mij ji 1x b , j 1..n 1.3== = Trang 74 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng ( )m ni, j i ji 1 j 1a b 0, a b 1.4= => = Hrngbuc(1.2),(1.3)cm+nphngtrnh,m*nn,tuynhindo(1.4)nnbtk phng trnh no trong m + n phng trnh cng l h ca cc phng trnh cn li v c th b i. Bi ton vn ti r rng l bi ton ti u dng chnh tc. V th ta c th gii n bng cc thut ton cabi ton ti u ha chnh tc, tuynhinvic lm sdn n nhng chi ph tnh ton khng cn thit. Do tnh cht c th ca bi ton chng ta s s dng mt cu trc c bit c th hn so vi bng n hnh gii. a vo cc k hiu sau: ( ) =T11 12 1n 21 22 2n m1 m2 mnC c ,c ,...,c ,c ,c ,...,c ,...,c ,c ,...,c( ) =T11 12 1n 21 22 2n m1 m2 mnX x ,x ,...,x ,x ,x ,...,x ,...,x ,x ,...,x=1 1 ... 1 0 0 ... 0 ...... 0 0 0 00 0 ... 0 1 1 ... 1 ...... 0 0 0 0... ... ... ... ... ... ... ... ...... ... ... ... ...0 0 ... 0 0 0 ... 0 ...... 1 1 1 1A1 0 ... 0 1 0 ... 0 ...... 1 0 0 00 1 ... 0 0 1 ... 0 ...... 0 1 0 0... ... ... ... 0 0 ... 0 ...... ... ... ...0 0 ... 1 0 0 ... 1 .| | | | | | | | | | | | |\ ......... 0 0 0 1 ( ) =1 2 m 1 2 nB a ,a ,...,a ,b ,b ,...,bTa c th a bi ton vn ti v dng ma trn:( ) = >f X CX minAX=BX 0 nh ngha: Vect ( )=ijX xtha mn tt c cc rng buc ca bi ton vn ti c gi l phng n. C ngha l: Rank A = m + n - 1 1.2 S tn ti nghim ti u nh l 5.1: iu kin cn v bi ton vn ti c phng n ti u l tng tt c cc lng pht phi bng tng tt c cc lng thu, ngha l: m ni ji 1 j 1a b= == Trang 75 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 2. TIU CHUN NHN BIT PHNG N CC BIN 2.1 Bng vn ti Lpmt bng Tgmm hngvn ct. Ti cc(i, j) ta ghi cc stng ng cho trc vo gc v cc lng ijxca phng n X. 1b jb nb1a11c iai1c ijxijc man1c Mt (i, j) m>ijx 0c gi l s dng. Tp hp cc s dng s to thnh dy chuynnu cc cp s dng lin nhauc xp trong mt hng hay trong mt ct. V d: ( ) ( ) ( ) ( ) ( ) ( )+ 1 1 1 2 2 2 2 3 s s s s 1i , j , i , j , i , j , i , j ,..., i , j , i , jHoc:( ) ( ) ( ) ( ) ( ) ( )+ 1 1 2 1 2 2 3 2 s s s 1 si , j , i , j , i , j , i , j ,..., i , j , i , jDy chuyn c gi l khp kn hay l mt chu trnh nu: + += =s 1 1 s 1 1j j hay i i2.2 Cc nh ngha v nh l nhngha:Mttphpcspthtcccabngvnticgilmtchu trnh nu ng n tha mn cc tnh cht sau: i)Hai cnh nhau nm trong cng mt dng hay mt ct; ii)Khng c ba no nm trn cng mt dng hay mt ct; iii) u tin nm trong cng mt dng hay ct vi cui cng. Mt s dng chu trnh: ** o o + + ** ** oo ++ oo++ Trang 76 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Gi G l tp hp cc s dng: ( ) { }= = + ij>0G i, j | x , G m n 1 MtphngnXcabitonvntichocgilkhngthoihanu: = + G m n 1, ngc li thoi ha nu< + G m n 1. nh l 5.2: H thng vect ct ca bi ton vn ti l c lp tuyn tnh khi v ch khi cc tng ng vi cc vect ca h thng khng to thnh chu trnh. Hqu:VectXlphngnccbinkhivchkhitpccsdngtngng khng lp thnh chu trnh. nh l5.3: Gi s Xlmtphng n ca biton vnti v tp cc s dng Glp thnh chu trnh th th bao gi cng c th iu chnh c X chuyn sang mt phng n mi X khng xu hn m tp G khng lp thnh chu trnh. 3. CC PHNG PHP TM PHNG N XUT PHT 3.1 Phng php gc Ty Bc Lp bng vn ti T, qu trnh xy dng phng n xut pht theo phng php gc Ty-Bc c tin hnh nh sau: + Bt ut gctrn bn tri ca bng T tc l(1,1) (nnm v tr gc Tybc ca bng), tin hnh phn phi lng hng cn chuyn vo ny. ( ) =11 1 1x min a , bcc lng thu pht cn li l:= = = i i 1 1 11a ' a, i 1; a ' a x ; = = = j j 1 1 11b ' b , j 1; b ' b x ; + Nu( ) = =11 1 1 1x a min a , bth=1a ' 0 . Khi xa dng th nht ca bng T ta thu c bng T gm m-1 dng v n ct vi cc lng pht v thu tng ng l ia ' , i=1..m; jb ' ,j=2..n. + i vi bng T ta li thc hin th tc phn phi nh l n p dng i vi bng T, tc llibtutgcTybcvphnphilnghngvnchuynvonysaocho hoc lch ht hng im pht, hoc l tha mn ht nhu cu tiu th ca im thu tng ng vi n. R rng saumi ln phnphi, ta sxai c1dng (hay1 ct) ca bng,nn sau ngm+n-1lnphnphi,thtctrnphiktthc(dolnphnphicuicngtaxa ngthicdngvct).Vvy,phngnxydngctheophngphpnysc khng qu m + n -1 thnh phn khc 0. V d 1: Xy dng phng n cho bi ton vn ti theo phng php gc Ty bc vi s liu cho trong bng sau: Trang 77 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng jbia30604625 50 30 4 20 7 12 7 70 5 40 9 30 6 1 41 8 2 16 9 25 1 Gi tr hm mc tiu thu c l: f(X) = 4*30 + 7*20 + 9*40 + 6*30 + 1*25 = 969. 3.2 Phng php cc tiu cc ph Trong phngphpgc Tybc,khi tinhnhphnphi cc lng hngvn chuynta lun chn gc Tybcmkhng chn cc phvn chuyn ca cc. Vvy, c thxutnhngphngphpkhccchnccphvnchuynvihyvngtm cphngnvichiphvnchuynnhhn.Ccphngphpdatrntngtrn gi l phng php cc tiu cc ph. 3.2.1 Phng php cc tiu cc ph theo dng Qutrnhphnphicthchingingnhphngphpgctybc,chkhcl cchnphnphikhngphilgctybcmlcccphnhnhttrong dng u tin ca bng. V d 2:Xydngphngn chobi tonvn titheophngphp cc tiu cc ph theo dng vi s liu cho trong bng sau: jbia30604625 50 30 4 20 7 12 7 70 5 9 45 6 25 1 41 8 40 2 1 9 1 Gi tr hm mc tiu thu c l: f(X) = 4*30 + 7*20 + 1*25 + 6*45 + 9*1+ 2*40 = 644. 3.2.2 Phng php cc tiu cc ph theo ct Qutrnhphnphicthchingingnhphngphpgctybc,chkhcl cchnphnphikhngphilgctybcmlcccphnhnhttrong ct u tin ca bng. Trang 78 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng V d 3:Xydngphngn chobi tonvn titheophngphp cc tiu cc ph theo dng vi s liu cho trong bng (nh v d 2). Tin hnh lm tng t. Gi tr hm mc tiu thu c l: f(X) = 4*30 + 2*41 + 7*19 + 6*46 + 1*24+ 7*1 = 642. jbia30604625 50 30 4 19 7 12 1 7 70 5 9 46 6 24 1 41 8 41 2 9 1 3.2.3 Phng php cc tiu cc ph ton bng Qutrnhphnphivbinibngtngt2phngphptrn,chkhclc chn phn phi l c cc ph nh nht trn ton bng. V d 4:Xydngphngn chobi tonvn titheophngphp cc tiu cc ph trn ton bng vi s liu cho trong bng sau: jbia306046 25 50 30 4 19 7 1 12 7 70 5 9 45 6 25 1 41 8 41 2 9 1 Gi tr hm mc tiu thu c l: f(X) = 1*25 + 2*41 + 4*30 + 6*45 + 7*19+ 1*12 = 652. 3.3 Phng php Fghen Phng php ny cho phng n cc bin kh tt theo ngha kh gn vi phng n ti uvgitrhmmctiuvchcnsaumtstbclpcathuttonthvlcth tm c phng n ti u. Gi s ( )=ijm*nC cl ma trn cc ph ca bi ton vn ti. Ta tin hnh nh sau: i)i vi mi hng v mi ct ca C ta tnh hiu s gia hai gi tr cc ph nh nht trn hng(ct) . Hiu s nybiu thlngpht ti thiu phi chunuta phn sai lng hng vo c cc ph nh nht trn hng (ct) . Trang 79 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng ii)Chnhnghayctchiusnylnnht.Nucnhiuhng(ct)nhthth chn mt hng (ct) bt k trong s . iii)Phn lng hng ti a c th vo c cc ph nh nht trn hng (ct) chn. Gi s l (r, s) . Gim lng cung hng r v lng cu ct s mt s bng lng hng phn phi. Vic ny s tha mn mt rng buc cung hay mt rng buc cu hoc c th l c hai. Loi b (khng cn xt tip) rng buc tha mn bng cch nh du cho vo hng hau ct tng ng ca ma trn cc ph. Nu chai rngbuc cung, cu cng thamnng thi th ch loibmt hng (ct) m thi. Trong trng hp ny c lng cung v cu cn li ca hng (ct) u tr thnh 0. iv)Lp li ccthaotc trn cho ti khi ch cnli mthng haymt ctduy nht.V lng hng c xc nh nh cc lng hng phn trc . V d 5:Xydng phngn cho bi tonvn ti theophng php Fghenvi s liu cho trong bng sau: jbia30604625 50 30 4 19 7 12 1 7 70 5 9 46 6 24 1 41 8 41 2 9 1 Lp bng jbia30604625 Hiu s 50471273 7059614 4182911 Hiu s1 530 Phnlnghng tia cho c cc ph nh nht trn ct2l Min {7, 9, 2} = 2lng hng bng Min{41, 60}=41. Loi dng 3 v phn ht hng.Lp bng mi. jbia30194625 Hiu s 50471273 7059614 Trang 80 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 08291x Hiu s12 6 6 Phn lng hng ti a cho c cc ph nh nht trn ct 3 l min {46, 70} = 46 vi chi ph min {6, 12} = 6. Loi b ct 3. jbia3019025 Hiu s 50 471273 2459614 08291x Hiu s12x6 Phn lng hng ti a cho c cc ph nh nht trn ct 4 l min {24, 25} = 24 vi chi ph min {1, 7} = 1. Loi b dng 2. jbia301901 Hiu s 50 471273 05961x 08291x Hiu s47x7 Phn lng hng ti a cho c cc ph nh nht trn ct 2 l min {19, 50} = 19 vi chi ph 7. Loi b ct 2. jbia30001 Hiu s 31 471273 05961x 08291x Hiu s4xx7 Phn lng hng ti a cho c cc ph nh nht trn ct 4 l min {30, 30} = 30 vi chi ph 4. Loi b dng 1, ct 1 jbia30000 Hiu s 30 471273 05961x Trang 81 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 08291x Hiu s4xxx Kt qu: f(X) = 2*41 + 6*46 + 1*24 + 7*19 + 7*1 + 4*30 = 642. 3.4 Phng php Larson R.E y lphng php ci tin phng phpFghen c ara nm1972 tuy phctp sovitnhtonbngtaynhngcthtnhtonnhanhchngtrnmytnh.Thayvdng cc cc ph ijc cho ta dng cc cc ph c chun ha xc nh nh sau: = == m nij ij pj iqp 1 q 11 1c ' c c cm n iu ny c ngha l mi phn t ijcb tr i mt lng bng trung bnh cc cc ph trn hng v ct ca n. Sau ta p dng phng php Fghen i vi ma trn C. 4. TIU CHUN TI U V THUT TON TH V 4.1 Tiu chun ti u. nh l 5.4: Phngn X cabi ton vn til ti utn ti cc s iu (i=1..n) v jv(j=1..m) sao cho: 1)( ) + s ei j iju v c i, j T 2)+ =i j iju v c nu>ijx 0Ccs iu (i=1..n)v jv (j=1..m)cgilccthvtngngviccimphtv im thu. 4.2 Thut ton th v Bc 1: Xc nh phng n ban u +Kim tra iu kin cn bng thu pht, nu khng thc hin bin i. +Tm phng n xut pht theo mt trong cc phng php trnh by trn. Bc 2: Tm cc th v +NuccsdngGlpthnhchutrnhthtasdngnhl5.3phv chu trnh, chuyn phng n xut pht v phng n cc bin. +Xcnhcchthngthv iu (i=1..n)v jv (j=1..m)theonhl5.4.Vgi thit bi ton khng thoi ha nn tp cc s dng( ) { }= >ijG i, j | x 0c ng m + n -1 , do c m + n -1 phng trnh: + =i j iju v c vi>ijx 0 Trang 82 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng +xcnhm +nn iu(i=1..n)v jv(j=1..m), nh vy s c mt iu hoc mt jvcxc nh tyv m +n -1n cn li sxcnhduynht tm + n -1 phng trnh. Qui tc: u tin cho 0iu = 0 (0i thng l dng u tin hoc ldng chamt s dng). Sau xc nh cc= j ij iv c ucho ct ct dng 0i mt s s dng. Tip xc nh= i ij ju c vcho dng i ct ct phn r mt s s dng. Vi qui tc xc nh tt c cc dng v ct thuc G Bc 3: Tnh cc c lng +Vi mi( )e i, j G ta xc nh cc c lngAij sau y: ( )A = + ij i j iju v c+Nu( ) A s ij0, i, jth phng n c l phng n ti u. +NuA >ij0 vi t nht mt (i,j) th phng n c cha ti u, ta c th iu chnh h gi tr hm mc tiu. Bc 4: iu chnh phng n +Gisviphmtiuchuntiul(i*,j*)tclA >i*j*0 (nucnhiuvi phmtachnngviMax{ A >ij0 }vihyvnghmmctiugimnhanh nht). +( ) e i*, j * G by gi ta thm (i*, j*) vo tp G, khi c thy gm m + n s dng. (i*, j*) s lp vi cc ca G mt chu trnh K duy nht.+Chia K thnh2 phn +K(tp cc chn)v K (tp cc l). Coi (i*, j*) l chn, tc l( )+e i*, j * K . Bc 5: Chuyn sang phng n mi +Xc nh s( ) { }u = e = >s sij i jmin x | i, j K x 0( )( )( )+ + u e= u eeijij ijijx , nui, j Kx ' x , nui, j Kx , nu i,j K Trang 83 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng += u =s s s si j i jx ' x 0 v vy ( )s si ,jb loi,= ui*j*xdo (i*, j*) s tr thnh s dng. +( ) ( )s sG' G\ i , j i*, j * = vngmm+n1sdngvkhngtothnhchu trnh. Quay li bc 2. Taxcnhh thngthvmi ngvimiphngnXv G.Tip tc qu trnh trn n khi no xy ra tnh hung( ) A s ij0, i, jth nhn c phng n ti u. Nu bi ton khng thoi ha th sau mt s hu hn bc bin i s c li gii. Ch:NussdngN3.( ) A s ij0, i, j6 ( ) { }u = e = >s sij i jmin x | i, j K x 0 7Tm X 4. Xopt False True Trang 84 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 5. TRNG HP KHNG CN BNG THU PHT 5.1 Tng lng pht ln hn tng lng thu: m ni ji 1 j 1a b= => Hngthascthmvomtimthuothn+1vilngyucul: m nn 1 i ji 1 j 1b a b+= == vi cc ph l i,n 1c 0, i 1..m+ = = .Ta c bi ton tng ng l: ij ijmij ji 1n 1ij ij 1iji jc xx b , j 1..n 1x a , i 1..mx 0, i 1..m, j 1..n 1a b=+== = += => = = += 5.2 Tng lng pht nh hn tng lng thu: m ni ji 1 j 1a b= =< Hngthascthmvomtimphtothm+1vilnghngbthiul: n mn 1 j ij 1 i 1a b a+= == vi cc ph l m 1,jc 0, j 1..n+= = .Ta c bi ton tng ng l: ij ijm 1ij ji 1nij ij 1iji jc xx b , j 1..nx a , i 1..m 1x 0, i 1..m 1, j 1..na b+=== == = +> = + == Trang 85 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng 6. MT S V D V d 7: Gii bi ton vn ti vi cc s liu cho trong bng sau: jbia185195200310 250126147 3401417713 300161358 F = 195*6 + 55*7 + 185*14 + 155*7 + 45*5 + 255*8 = 7495 V d 8: Gii bi ton vn ti vi cc s liu cho trong bng sau: jbia611285 1089910 1570208 25102515 F = 8*6 + 9*3 + 10*1 + 0*11 + 8*4 + 5*25 = 242 V d 9: Gii bi ton vn ti vi cc s liu cho trong bng sau: jbia20203015 155134 252467 455348 F= 1*5+4*10+2*20+7*5+3*15+4*30=285 V d 10: Gii bi ton vn ti vi cc s liu cho trong bng sau: jbia5152010 102123 256042 151482 F= 2*5 + 1*5 + 0*10 + 5*15 + 8*5 + 2*10 =150 V d 11: Gii bi ton vn ti vi cc s liu cho trong bng sau: jbia25402010 404378 206234 355386 F=340 Trang 86 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Chng 6 Gii bi ton ti u trn my tnh 1. GII BI TON TI UXt bi ton ti u: trongQlmttrongccphptonquanh,,=thtccphptonquanh trong cc rngbucl tu. Nhvybi ton(1) c thlbiton ti u thng thng, ti u nguyn hay ti u boolean. Cch b tr d liu cho trn bng tnh: Hng cui cng l cc gi tr ban u ca cc bin cc cng thc ca Excel hot ng, c th ly gi tr ca tt c cc bin bng 1. Xt bi ton: Cc bc thc hin gii bi ton: Bc 1. Nhp d liu bi ton vo bng tnh di dng sau: Trang 87 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng Phng n ban u X = (1, 1, 1), n c th khng chp nhn c. Bc 2. Tnh gi tr hm mc tiu ti E2 bng cng thc = SUMPRODOCT($B$7:$D$7, B2:D2) HmSumproductchotchvhngcahaidy.CopycngthctE2sangdy cc E3:E6 nhm tnh gi tr v tri ca bn rng buc bi ton (1). Bc 3. Dng lnh Tools / Solver, xut hin hp thoi Solver Parameters. Mc Set Target Cell: chn ch (cha gi tr hm mc tiu), c th nhy vo biu tng ca Excel bn phi hp vn bn xc nh , trong v d chn E2.Mc Equal To: chn Max nu cc i hm mc tiu, chn Min nu cc tiu hm mc tiu, chnValueofvnhpgitrnumunchbngmtgitrnhtnh,trongvdchn Min.Mc By Changing cells: chn cc cha cc bin ca bi ton, ta chn khi B7:D7. Nhy nt Add nhp tt c cc rng buc vo khung Subject to the Constraints (dng utrongkhungngvirngbuckhngmtrnccbin,dngthhaingvihairng buc u bi ton, dng cui ng vi 2 rng buc cui. Khi nhy nt Add, hin hp thoi: chn loi rng buc (>= = = Cch b tr d liu trn bng tnh ca bi ton vn ti Trang 90 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng V d1: Tmphngn ccbin ca bitonvn ti cvectlngphtvlng thu theo th t l a = (90, 100, 110), b = (50, 80, 95, 75) v ma trn cc ph c c = ((((
9 5 12 48 6 10 53 16 2 11 Xt bi ton vn ti c 3 im pht v 4 im thu c nhp vo bng tnh: KhiA2:D4lmatrnchiphvnchuyn,khiA7:D9lphngnvnchuyn(gitr ban u cho tt c bng 1), khi F7:F9 l kh nng ca 3 im pht, khi A11:D11 l nhu cu ca 4 im thu, khi E7:E9 l lng hng pht t mi im pht i theo phng n X chn, khiA10:D10llnghngnhnctimiimthujtheophngnX.Gisrng tng lng hng c trong cc kho bng tng nhu cu ca cc ni thiu th. Qu trnh dng Solver gii bi ton vn ti trn theo cc bc: Bc1.NhpchiphvnchuynvoccA2:D4,nhpkhnngcaccimpht vo F7:F9, nhu cu cc im thu A11:D11, phng n ban u A7:D9. Tnh gi tr hmmc tiutrongF3 theo cng thc = Sumproduct (A2:D4, A7:D9), hm ny tnh tng cc tch ca tng cp phn t trong hai khi . Tnh lng hng pht ca im pht1tiE7theocngthc=SUM(A7:D7),tngttnhcccE8:E9.Tnhlng hng nhn c ca im thu 1 ti A10 theo cng thc = SUM(A7:A9), tng t tnh c cc B10:D10. Bc 2. Dng lnh Tools/ Solver vi cc la chn hm mc tiu v cc rng buc: Bc3.TronghpthoiSolverOptionsphichnAssumeLinearModel.Cuicngta nhnc gi tr ti u hmmc tiu bng1420,phngnvn chuyn ti u:x[1,2]= 80, x[2,3]= 35, x[2,4]= 65, x[3,1]= 50, x[3,3]= 60 trong bng tnh kt qu: Trang 91 Gio trnhPhng php ti u i hc Hi Phng. Ging vin: L c Nhng V d2: Tmphngn ccbin ca bitonvn ti cvectlngphtvlng thu theo th t l a = (105, 60, 85), b = (35, 48, 80, 95), ma trn cc ph c v phng n x c = ((((
12 10 12 205 8 3 119 2 7 4x = ((((
50 35 0 00 45 15 045 0 25 35 Hy xy dng phng n cc bin khng xu hn x Gi tr ti u ca hm mc tiu trong phng n x bng 2075. Tnh gi tr ti u?Bc 1: Nhp d liu v tnh gi tr hm mc tiu, lng hng pht, lng hng thu nh bng tnh Bc 2: Dng lnh Tools/ Solver vi cc la chn hm mc tiu v cc rng buc: Bc 3: Cui cng gi tr ti u hm mc tiu bng 1724, phng n vn chuyn ti u: x[1,1]= 35, x[1,2]= 48, x[1,3]= 22, x[2,4]= 60, x[3,3]= 58, x[3,4]= 27 trong bng tnh kt qu:
