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Trng i hc Nng nghip I
PGS. TS. NGUYN HI THANH
Ti u ha Gio trnh cho ngnh Tin hc
v Cng ngh thng tin
Nh xut bn Bch khoa H Ni
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M s: 920 2006 / CBX / 01 130 / BKHN
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MC LC
M U 6 CHNG I. BI TON TI U TNG QUT V NG DNG 7
1. BI TON TI U TNG QUT V PHN LOI 7 1.1. Bi ton ti u tng qut 7 1.2. Phn loi cc bi ton ti u 8 2. NG DNG BI TON TI U GII QUYT CC VN THC T 9 2.1. Phng php m hnh ha ton hc 9 2.2. Mt s ng dng ca bi ton ti u 10 CHNG II. PHNG PHP N HNH GII BI TON QUY HOCH TUYN TNH
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1. M HNH QUY HOCH TUYN TNH 16 1.1. Pht biu m hnh 16 1.2. Phng php th 17
2. PHNG PHP N HNH 19 2.1. Tm hiu quy trnh tnh ton 19 2.2. Khung thut ton n hnh 23 3. C S TON HC CA PHNG PHP N HNH 23 3.1. Pht biu bi ton quy hoch tuyn tnh dng chnh tc 23 3.2. Cng thc s gia hm mc tiu 25 3.3. Tiu chun ti u 26 3.4. Thut ton n hnh cho bi ton quy hoch tuyn tnh dng chnh tc 27
4. B SUNG THM V PHNG PHP N HNH 29 4.1. a bi ton quy hoch tuyn tnh v dng chnh tc
4.2. Phng php n hnh m rng 4.3. Phng php n hnh hai pha 4.4. Phng php n hnh ci bin
29 31 33 35
BI TP CHNG II 41 CHNG III. BI TON I NGU V MT S NG DNG 44 1. PHT BIU BI TON I NGU 44 1.1. Pht biu bi ton 44 1.2. ngha ca bi ton i ngu 45 1.3. Quy tc vit bi ton i ngu 46 1.4. Cc tnh cht v ngha kinh t ca cp bi ton i ngu 48
2. CHNG MINH MT S TNH CHT CA CP BI TON I NGU 53 2.1. nh l i ngu yu 54 2.2. nh l i ngu mnh 54 2.3. nh l lch b 56
3. THUT TON N HNH I NGU 57
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3.1. Quy trnh tnh ton v pht biu thut ton 3.2. C s ca phng php n hnh i ngu
57 61
4. BI TON VN TI 62 4.1. Pht biu bi ton vn ti
4.2. Cc tnh cht ca bi ton vn ti 4.3. Phng php phn phi gii bi ton vn ti
62 66 68
4.4. Phng php th v gii bi ton vn ti 72 4.5. C s ca phng php phn phi v phng php th v 74 BI TP CHNG III 78 CHNG IV. QUY HOCH NGUYN 81 1. PHNG PHP CT GOMORY GII BI TON QUY HOCH TUYN TNH NGUYN
81
1.1. Pht biu bi ton quy hoch tuyn tnh nguyn 81 1.2. Minh ha phng php Gomory bng th 82 1.3. Gii bi ton quy hoch tuyn tnh nguyn bng bng 84 1.4. Khung thut ton ct Gomory 86 2. PHNG PHP NHNH CN LAND DOIG GII BI TON QUY HOCH TUYN TNH NGUYN
87
2.1. Minh ha phng php nhnh cn bng th 87 2.2. Ni dung c bn ca phng php nhnh cn
2.3. Khung thut ton nhnh cn Land Doig 88 88
3. GII BI TON QUY HOCH TUYN TNH NGUYN BNG QUY HOCH NG
90
3.1. Bi ton ngi du lch 90 3.2. Quy trnh tnh ton tng qut 91 3.3. p dng quy hoch ng gii bi ton quy hoch tuyn tnh nguyn 93 3.4. Bi ton ci ti
3.5. Hp nht ha cc rng buc ca bi ton quy hoch tuyn tnh nguyn 95 100
BI TP CHNG IV 103 CHNG V. MT S PHNG PHP QUY HOCH PHI TUYN 105 1. CC KHI NIM C BN CA BI TON TI U PHI TUYN 105 1.1. Pht biu bi ton ti u phi tuyn 105 1.2. Phn loi cc bi ton ti u phi tuyn ton cc 106 1.3. Bi ton quy hoch li
1.4. Hm nhiu bin kh vi cp mt v cp hai 107 108
2. MT S PHNG PHP GII BI TON QUY HOCH PHI TUYN KHNG RNG BUC
109
2.1. Phng php ng dc nht 109 2.2. Phng php Newton
2.3. Phng php hng lin hp 111 113
3. THIT LP IU KIN TI U KUHN TUCKER CHO CC BI TON QUY HOCH PHI TUYN C RNG BUC
116
3.1. Hm Lagrange 116 3.2. Thit lp iu kin Kuhn Tucker 117 4. MT S PHNG PHP GII QUY HOCH TON PHNG 120 4.1. Bi ton quy hoch ton phng 120 4.2. Pht biu iu kin Kuhn Tucker cho bi ton quy hoch ton phng 121
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4.3. Phng php Wolfe gii bi ton quy hoch ton phng 4.4. Gii bi ton quy hoch ton phng bng bi ton b
121 123
5. QUY HOCH TCH V QUY HOCH HNH HC 126 5.1. Quy hoch tch
5.2. Quy hoch hnh hc 126 129
BI TP CHNG V 133 CHNG VI. MT S VN C S CA L THUYT QUY HOCH LI V QUY HOCH PHI TUYN
136
1. TP HP LI 136 1.1. Bao li 136 1.2. Bao ng v min trong ca tp li 138 1.3. Siu phng tch v siu phng ta ca tp li
1.4. Nn li v nn i cc 139 144
2. NG DNG GII TCH LI VO BI TON QUY HOCH TUYN TNH 145 2.1. im cc bin v hng cc bin 145 2.2. Biu din tp li a din qua im cc bin v hng cc bin
2.3. iu kin ti u trong phng php n hnh gii bi ton quy hoch tuyn tnh
148
150 3. CC TNH CHT CA HM LI 152 3.1. Cc nh ngha v tnh cht c bn 152 3.2. Di vi phn ca hm li 153 3.3. Hm li kh vi 155 3.4. Cc i v cc tiu ca hm li 158 4. CC IU KIN TI U FRITZ JOHN V KUHN TUCKER 162 4.1. Bi ton ti u khng rng buc 162 4.2. Bi ton ti u c rng buc 164 4.3. iu kin ti u Fritz John
4.4. iu kin ti u Kuhn Tucker 166 166
5. MT S PHNG PHP HNG CHP NHN GII BI TON QUY HOCH PHI TUYN
170
5.1. Phng php hng chp nhn 5.2. Thut ton Frank Wolfe gii bi ton quy hoch li c min rng buc l tp li a din
170
172 5.3. Phng php gradient rt gn
5.4. Phng php n hnh li Zangwill 172 174
6. GII THIU PHNG PHP IM TRONG GII BI TON QUY HOCH TUYN TNH
177
6.1. Bi ton ellipsoid xp x 177 6.2. Mt s thut ton im trong 181 BI TP CHNG VI 183 TI LIU THAM KHO 186
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M u Ti u ha, c khi ngun nh mt ngnh ca Ton hc, c rt nhiu ng dng hiu qu
v rng ri trong quy hoch ti nguyn, thit k ch to my, iu khin t ng, qun tr kinh doanh, kin trc th, cng ngh thng tin, trong vic to nn cc h h tr ra quyt nh trong qun l v pht trin cc h thng ln. Chnh v vy, cc lnh vc ca Ti u ha ngy cng tr nn a dng, mang nhiu tn gi khc nhau nh Quy hoch ton hc, iu khin ti u, Vn tr hc, L thuyt tr chi Hin nay, mn hc Ti u ha c a vo ging dy trong nhiu chng trnh o to i hc cho cc ngnh khoa hc c bn, k thut cng ngh, kinh t qun l, sinh hc nng nghip, x hi nhn vn, sinh thi mi trng vi thi lng thng thng t ba cho ti su hc trnh. i vi sinh vin cc ngnh Tin hc, Cng ngh thng tin v Ton Tin ng dng, mn hc Ti u ha l mt mn hc c s khng th thiu. Gio trnh Ti u ha ny c bin son vi mc ch cung cp cho sinh vin nm th hai ngnh Tin hc ca Khoa Cng ngh thng tin, Trng i hc Nng nghip I, mt s kin thc c bn v cc lnh vc quan trng ca Ti u ha. Qua gio trnh ny, sinh vin cn nm c c s l thuyt mt mc nht nh, nm chc cc thut ton ti u c bn p dng trong vic xy dng cc phn mm ti u tnh ton gii cc bi ton kinh t, cng ngh, k thut v qun l.
Chng I gii thiu tng quan v ngn gn bi ton ti u tng qut v phn loi cc bi ton ti u c bn, cng nh gii thiu mt s v d v m hnh ti u pht sinh trong thc t. Phn u trnh by v Quy hoch tuyn tnh bao gm chng II, III v IV. Phn ny nhn mnh vo vic trnh by cc phng php v thut ton c in ca Quy hoch tuyn tnh, nh phng php n hnh (bao gm c phng php hai pha v phng php n hnh ci bin dng ma trn nghch o), phng php n hnh i ngu, phng php th v gii bi ton vn ti, cc phng php ct Gomory v nhnh cn Land Doig cng nh phng php quy hoch ng gii bi ton quy hoch tuyn tnh nguyn. Phn sau ca gio trnh bao gm hai chng v Quy hoch phi tuyn. Chng V trnh by mt s phng php v thut ton ti u phi tuyn khng c rng buc v c rng buc, bao gm phng php ng dc nht, phng php Newton, phng php hng lin hp, cc phng php gii quy hoch ton phng thng dng, phng php quy hoch tch v quy hoch hnh hc. Chng VI gii thiu v c s l thuyt ca quy hoch li v quy hoch phi tuyn. Phn gii thiu v mt lp phng php im trong gii bi ton quy hoch tuyn tnh cui gio trnh mang tnh cht tham kho, c th dnh cho sinh vin nghin cu theo nhm v tho lun. Vic chng minh mt s nh l kh nn sinh vin t nghin cu, khng c tnh bt buc. Khi bin son, chng ti lun c mt nguyn vng l lm sao vic trnh by cc phng php ti u cp ti trong gio trnh cng phi p ng c tiu chun ti u, sinh vin phi hiu c v lm c. Chnh v vy, cc phng php lun c trnh by mt cch c th thng qua cc v d mu t d ti kh, m nhng v d ny c th c s dng nhiu ln tit kim thi gian.
Mt s ti liu ngi hc c th tham kho thm v Quy hoch tuyn tnh l: Nguyn c Ngha, Ti u ha, Nxb. Gio dc, 2002; Phan Quc Khnh Trn Hu Nng, Quy hoch tuyn tnh, Nxb. Gio dc, 2003. V Quy hoch phi tuyn c th c thm mt s chng lin quan trong cc sch tham kho sau: Bazaraa M.S, Shetty C.M, Nonlinear programming: Theory and algorithms, John Wiley and Sons, New York, 1990; Horst R, Hong Ty, Global optimization: Deterministic approaches, Springer Verlag, Berlin, 1993; Bi Th Tm Trn V Thiu, Cc phng php ti u ha, Nxb. Giao thng vn ti, 1998. Ngi c cng c th s dng Internet tm kim cc tp ch v ti liu lin quan.
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Chng I Bi ton ti u tng qut v ng dng
1. Bi ton ti u tng qut v phn loi
1.1. Bi ton ti u tng qut Ti u ha l mt trong nhng lnh vc kinh in ca ton hc c nh hng n hu ht
cc lnh vc khoa hc cng ngh v kinh t x hi. Trong thc t, vic tm gii php ti u cho mt vn no chim mt vai tr ht sc quan trng. Phng n ti u l phng n hp l nht, tt nht, tit kim chi ph, ti nguyn, ngun lc m li cho hiu qu cao.
V d 1. Tm 1x D [ 2,2, 1,8] R = sao cho f(x) = x3 3x + 1 Max. Bi ton ti u trn c dng cc i ho c gii nh sau: Cho f(x) = 3x2 3 = 0, ta c cc
im ti hn l x = 1 v x = +1. Xt gi tr hm s f(x) ti cc im ti hn va tm c v ti cc gi tr x = 2,2 v x = 1,8 (cc im u mt ca on [2,2, 1,8]), ta c f(2,2) = 3,048 , f(1) = 3, f(1) = 1, f(1,8) = 1,432. Vy gi tr x cn tm l x = 1. Kt qu ca bi ton c minh ho trn hnh I.1.
Cho hm s f: D Rn R. Bi ton ti u tng qut c dng: Max (Min) f(x), vi x D Rn. Nh vy, cn tm im x = (x1, x2, ..., xn) D Rn sao cho hm mc tiu f(x) t c gi tr ln nht i vi bi ton Max cc i ho (gi tr b nht i vi bi ton Min cc tiu ho).
y
3,048
1 0 1 1,18
3
x
2,2
1
1,432
Hnh I.1. th hm f(x)
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im x = (x1, x2, ..., xn) D Rn c gi l phng n kh thi (hay phng n chp nhn c hoc phng n, nu ni vn tt) ca bi ton ti u: Max (Min) f(x), vi x D Rn. Min D c gi l min rng buc. Cc to thnh phn ca im x c gi l cc bin quyt nh, cn x cng c gi l vc t quyt nh.
Xt bi ton cc i ho: Max f(x), vi x D Rn. im x* = ( )1 2 nx , x , ..., x Rn c gi l im ti u (hay phng n ti u) ton cc nu x* D v f(x*) f(x), x D. im x Rn c gi l im ti u (hay phng n ti u) a phng nu x D v tn ti mt ln cn N nh ca im x sao cho f( x ) f(x), x N D.
i vi bi ton cc tiu ho Min f(x), vi x D Rn, im x* Rn c gi l im ti u (hay phng n ti u) ton cc nu x* D v f(x*) f(x), x D. im x Rn c gi l im ti u (hay phng n ti u) a phng nu x D v tn ti mt ln cn N nh ca im x sao cho f( x ) f(x), x N D.
D thy, mi phng n ti u ton cc cng l phng n ti u a phng, trong khi mt phng n ti u a phng khng nht thit l phng n ti u ton cc. Trn hnh I.1, im x = 1 ch l phng n ti u a phng khi xt bi ton cc tiu ho.
V d 2. Xt bi ton ti u sau: Max 1 2f (x) 8x 6x= + , vi iu kin rng buc x D = { (x1, x2) R2: 4x1 + 2x2 60; 2x1 + 4x2 48, x1 0, x2 0}.
Bi ton ti u trn y cn c gi l bi ton quy hoch tuyn tnh. Ngi ta chng minh c rng mi phng n ti u a phng ca bi ton quy hoch tuyn tnh cng ng thi l phng n ti u ton cc.
1.2. Phn loi cc bi ton ti u Cc bi ton ti u, cng cn c gi l cc bi ton quy hoch ton hc, c chia ra
thnh cc lp sau: Bi ton quy hoch tuyn tnh (BTQHTT), Bi ton ti u phi tuyn hay cn gi l bi ton quy hoch phi tuyn (BTQHPT), bao
gm c bi ton quy hoch li (BTQHL) v bi ton quy hoch ton phng (BTQHTP), Bi ton ti u ri rc, bi ton ti u nguyn v hn hp nguyn. Bi ton quy hoch ng, Bi ton quy hoch a mc tiu, Bi ton quy hoch ngu nhin / m ... Cc phng php ton hc gii cc lp bi ton ti u tng qut nh nu trn y c gi
l cc phng php ti u ton hc (hay cc phng php quy hoch ton hc). Trong gio trnh ny, trc ht chng ta nghin cu cc phng php gii BTQHTT, bao gm c cc BTQHTT nguyn v hn hp nguyn. Sau , chng ta s xem xt cc phng php gii mt s dng c bit ca BTQHPT. Cc phng php c xem xt ch yu v kha cnh th tc tnh ton thng qua cc v d n gin, nhm gip cho sinh vin ngnh Tin hc, Cng ngh thng tin khi hc gio trnh ny vo nm hc th hai c th lm quen vi t duy lp trnh tnh ton. Phn cui ca gio trnh s cp ti mt s c s l thuyt ca gii tch li v quy hoch phi tuyn, l cc vn c
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tnh cht nn tng i vi nhng sinh vin quan tm v c hng tip tc nghin cu lnh vc Ti u ha.
2. ng dng bi ton ti u gii quyt cc vn thc t
2.1. Phng php m hnh ho ton hc Nhiu vn pht sinh trong thc t c th gii c bng cch p dng cc phng php
ti u ton hc. Tuy nhin, im mu cht y l t bi ton thc t cn xy dng c mt m hnh ti u thch hp da vo cc dng bi ton ti u bit. Sau cn p dng phng php ti u ton hc v quy trnh tnh ton thch hp tm ra li gii cho m hnh t ra.
Cc bc cn thit tin hnh khi p dng phng php m hnh ho ton hc c th c pht biu mt cch khi qut nh sau:
Trc ht phi kho st bi ton thc t v pht hin vn cn gii quyt. Pht biu cc iu kin rng buc v mc tiu ca bi ton di dng nh tnh. Sau
la chn cc bin quyt nh / cc n s v xy dng m hnh nh lng cn gi l m hnh ton hc.
Thu thp d liu v la chn phng php ton hc thch hp gii quyt m hnh trn. Trong trng hp m hnh ton hc l m hnh ti u, cn la chn phng php ti u thch hp gii m hnh.
Xc nh quy trnh gii / thut ton. C th gii m hnh bng cch tnh ton thng thng trn giy. i vi cc m hnh ln, bao gm nhiu bin v nhiu iu kin rng buc cn tin hnh lp trnh v gii m hnh trn my tnh tm ra phng n tha mn m hnh.
nh gi kt qu tnh ton. Trong trng hp pht hin thy c kt qu bt thng, cn xem xt nguyn nhn, kim tra v chnh sa li m hnh hoc d liu u vo hoc quy trnh gii / thut ton / chng trnh my tnh.
Kim chng cc kt qu tnh ton trn thc t. Nu cc kt qu thu c c coi l hp l, ph hp vi thc t hay c cc chuyn gia nh gi l c hiu qu hn so vi cc phng n trc y th cn tm cch trin khai phng n tm c trn thc t.
R rng rng gii quyt cc vn pht sinh t cc bi ton thc t cn c c s hp tc cht ch gia cc chuyn gia trong lnh vc chuyn mn, cc chuyn gia Ton, Ton ng dng v cc chuyn gia Tin hc, k s lp trnh. iu ny l c bit cn thit khi gii quyt cc bi ton cho cc h thng ln. Vic thit lp c mt m hnh hp l, phn nh c bn cht ca bi ton thc t ng thi kh thi v phng din tnh ton lun va mang tnh khoa hc thun ty, va c tnh ngh thut. Cc thut ng sau thng gp khi p dng phng php m hnh ho ton hc:
Ton ng dng (Applied Mathematics). Vn tr hc (Operations Research vit tt l OR). Khoa hc qun l (Management Science vit tt l MS). ng dng my tnh (Computer Applications). M hnh ti u (Optimization Models)
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2.2. Mt s ng dng ca bi ton ti u Nhng nm gn y, nhiu bi ton thc t c gii quyt bng phng php m hnh ha
ton hc rt thnh cng. Trong s cc m hnh ton hc c p dng c nhiu m hnh ti u, c gii quyt thng qua cc bi ton ti u kinh in. Trong trng hp hm mc tiu cng nh tt c cc rng buc u l cc hm tuyn tnh, th bi ton ti u l BTQHTT. BTQHTT c th gii c bng mt s phng php ti u quen bit (nh phng php n hnh, phng php n hnh ci bin hay cc phng php im trong). BTQHTT v ang c s dng rng ri trong quy hoch ti nguyn, qun l s dng t cng nh nhiu lnh vc ca qun l, kinh t v qun tr kinh doanh.
Trong trng hp hoc hm mc tiu hoc mt trong s cc rng buc l phi tuyn, chng ta c BTQHPT. Trong cc m hnh ti u da trn BTQHPT ni chung, v trong cc m hnh ti u trong lnh vc nng nghip ni ring, li gii ti u ton cc c mt ngha quan trng. Chng hn trong thit k my nng nghip, sau khi dng phng php phn tch hi quy nhiu chiu, ta thng thu c hm mc tiu c dng phi tuyn. Cc bi ton ti u ton cc cng c th ny sinh trong quy hoch kinh t sinh thi vng, hay xc nh c cu t canh tc cy trng. Bi ton t ra l phi tm c li gii ti u ton cc. C rt nhiu phng php gii cc lp bi ton ti u phi tuyn ring bit, nhng cha c phng php no t ra hu hiu cho mi bi ton ti u phi tuyn, c bit l cho cc bi ton vi mt s hay tt c cc bin quyt nh nhn cc gi tr nguyn.
Sau y l cc v d minh ho mt s ng dng ca bi ton ti u. V d 3. Bi ton quy hoch s dng t (M hnh ti u tuyn tnh gii bi ton quy
hoch s dng t trn a bn x ng D, huyn Gia Lm, tnh H Ni) Chng ta xt m hnh ti u vi mc tiu cn cc i ho l hiu qu kinh t. thit lp
m hnh, trc ht chn cc bin quyt nh. Da vo kt qu cc d liu thu c, ta chn cc bin quyt nh nh sau: xj vi j = 1, 2, , 18 l din tch cc loi cy trng, n v tnh l ha (theo th t l: la xun, la ma, ng xun, ng ng, ng bao t ng, lc xun, u xanh xun, u tng ng t chuyn mu, u tng ng t ba v, da chut xun, da chut bao t, mp ng xun, rau mi tu, rau gia v, u c ve ng, t xun, c chua xun, c chua ng), x19 l din tch ao h th c, xj vi j = 20, , 23 l s u vt nui trong nm (tru, b, ln, gia cm). Cn x24 l s cng lao ng thu ngoi, x25 l lng tin vn vay ngn hng, n v tnh l nghn ng. Lc chng ta c BTQHTT sau vi 33 rng buc (cha k iu kin khng m ca cc bin).
Hiu qu kinh t cn cc i ha l: f(x) = 4306,14x1 + 4168,73x2 + 3115,21x3 + 3013,11x4 + 4158,68x5 + 4860,91x6 + 4295,31x7 + 3706,11x8 + 3788,25x9 + 12747,31x10 + 12752,96x11 + 12064,81x12 + 79228,88x13 + 35961,31x14 + 10823,91x15 + 7950,16x16 + 7928,06x17 + 5738,46x18 + 11129,50x19 + 429,00x20 + 674,00x21 + 219,50x22 + 11,10x23 15,50x24 0,12x25 Max.
Cc rng buc hay cc iu kin hn ch c nh lng nh sau: x1 80,88; x2 75,78; x3 64,89; x4 64,89; x5 10,50; x6 64,89;
x7 64,89; x8 16,50; x9 45,30; x10 5,50; x11 8,50; x12 6,80; x13 13,70; x14 14,50; x15 4,80; x16 4,50; x17 4,20; x18 10,20; x19 33,11; x20 40,00; x21 180,00; x22 4280; x23 18800;
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x5 + x9 + x11 + x13 + x18 45,30; x3 + x6 + x7 + x10 + x 12 + x16 + x17 64,89; x4 + x8 + x14 + x15 64,89; x1 + x13 80,88; x2 + x13 75,88;
205,5x1 + 150x3 + 75,75x4 + 75x5 + 225,5x6 + 221,5x7 + 102,7x8 + 100,75x9 + 360 x10 + 140x11 + 385x12 + 1833,6x13 + 1446,3x14 + 210,25 x15 + 410,5x16 + 360,5 x17 + 176x18 + 67x19 + 20x20 + 16x21 + 9x22 + 0,3x23 x24 226149,00;
201,5x2 + 150x3 + 75,25x4 + 102,7x8 + 100,75x9 + 140x11 + 2475,4x13 + 1446,3x14 + 210,25x15 + 176x18 + 58x19 + 16x20 + 12x21 + 7x22 + 0,2x23 x24 152190,00;
2871,89x1 + 2691,89x2 + 2243,62x3 + 2243,66x4 + 3630,89x5 + 4780,06x6 + 2229,11x7 + 2401,41x8 + 2326,88x9 + 16440,61x10 + 16058,39x11 + 15960,61x12 + 68494,59x13 + 23146,11x14 + 13676,26x15 + 6061,76x16 + 11083,11x17 + 10391,89x18 + 18058x19 + 1223x20 + 1098,5x21 + 624,5x22 + 12x23 15,5x24 x25 3881500;
3,5x5 + 8x6 + 3,5x7 + 4,1x8 + 3,5x9 + 4,16x10 + 3,5x11 + 4x 12 + 12,1x13 + 14,4x14 + 3,42x15 + 11,58x16 + 8x17 + 7,5x18 3 x20 2x21 0,95x22 0,0052x23 0; 5,1x1 + 4,96x2 + 3,85x3 + 3,8x4 921,25;
Cc bin u phi tha mn iu kin khng m: xj 0, j = 1,25 . Bng cch p dng phng php n hnh gii BTQHTT c th tm c phng n ti
u ca m hnh trn nh sau:
x1 = 67,18; x2 = 62,18; x3 = 25,19; x4 = 45,59; x5 = 10,50; x6 = 18,7; x9 = 2,40; x10 = 5,50; x11 = 8,50; x12 = 6,80; x13 = 13,70; x14 = 14,50; x15 = 4,80; x16 = 4,50; x17 = 4,20; x18 = 10,20; x19 = 33,11; x20 = 40,00; x21 = 180; x22 = 4280; x23 = 18800; x25 = 2368646. Hiu qu kinh t cc i t c l 4325863 (nghn ng).
V d 4. Bi ton cc i ho gi tr sn xut (M hnh ti u phi tuyn gii bi ton cc i ho gi tr sn xut trn mt hc ta nui c ti huyn Vn Giang, tnh Hng Yn)
S dng s liu iu tra 112 h nui c vng ng trong thuc 4 x thuc huyn Vn Giang, Hng Yn, tm phng trnh hi quy m, chng ta nhn c hm gi tr sn xut (dng Cobb Douglas) chnh l hm mc tiu cn cc i ho sau y:
z = f(x) = 19,375 x10,236 x20,104 x30,096 x40,056 x50,056 e0,168 x6 e0,066 x7 Max trong :
z : gi tr sn xut bnh qun 1 ha 1 nm (triu ng / ha),
x1 : chi ph ging bnh qun 1 ha 1 nm (triu ng / ha),
x2 : chi ph thc n bnh qun 1 ha 1 nm (triu ng / ha),
x3 : chi ph lao ng bnh qun 1 ha 1 nm (triu ng / ha),
x4 : chi ph khu hao v thu t bnh qun 1 ha 1 nm (triu ng / ha),
x5 : cc chi ph khc bnh qun 1 ha 1 nm (triu ng / ha),
x6 , x7: cc bin 0 1 gi nh v hnh thc nui,
x6 = 1 i vi nui chuyn canh, x6 = 0 i vi nui tng hp,
x7 = 1 vi hnh thc nui 1 loi c chnh kt hp vi cc loi c khc,
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x7 = 0 vi hnh thc nui 2 loi c chnh kt hp vi cc loi c khc.
t: x1 + x2 + x3 + x4 + x5 = TC, vi TC l mc u t / tng chi ph.
Ty theo tng mc u t / tng chi ph ta c mt trong cc rng buc:
Vi mc u t di 40 triu ng / ha: x1 + x2 + x3 + x4 + x5 < 40,
Vi mc u t 4050 triu ng / ha: 40 x1 + x2 + x3 + x4 + x5 < 50, Vi mc u t 5060 triu ng / ha: 50 x1 + x2 + x3 + x4 + x5 < 60, Vi mc u t 6070 triu ng / ha: 60 x1 + x2 + x3 + x4 + x5< 70, Vi mc u t trn 70 triu ng / ha: x1 + x2 + x3 + x4 + x5 70. Vi hnh thc nui ta c rng buc: x6 + x7 = 1(x6, x7 ch nhn gi tr 0 hoc 1).
Trn y l BTQHPT, vi 5 bin lin tc v 2 bin nguyn dng 0 1. S dng phng php ti u phi tuyn thch hp c tn gi l RST2ANU gii BTQHPT ton cc hn hp nguyn thit lp trn y ta c kt qu trong bng I.1.
Bng I.1. Kt qu c cu u t ti u vng ng
u t (tr/ha) < 40 40 50 50 60 60 70 > 70
x1 35 45% 39 45% 39 45% 35 45% 35 40%
x2 15 20% 17 25% 17 23% 15 20% 18 25%
x3 15 20% 15 20% 15 20% 16 19% 17 23%
x4 10 15% 7 15% 8 15% 9 13% 10 15%
x5 10 15% 10 15% 9 15% 9 15% 10 15%
Gi tr sn xut (tr / ha) < 78,1 78,1 88,3 88,3 97,5 97,5 106 > 106
Thu nhp rng (tr / ha) 38,138,3 38,337,5 37,536
Vic thc hin c cu u t ti u lm gi tr sn xut (GO) cng nh thu nhp rng (NI = GO TC) tng mc u t tng ln r rt so vi thc t sn xut ti a phng. c bit, mc u t 50 triu ng / ha cho ta thu nhp hn hp cao nht l 38,3 triu ng / ha, ln hn 8 triu ng / ha so vi trc khi p dng c cu u t ti u cng nh hnh thc nui thch hp. Ti mc u t ny, c cu u t ti u l x1 t 19,6 21,5 triu ng (chim 39,2 42,2%), x2 t 8,6 9,8 triu ng (17,2 19,6%), x3 t 8,6 9,9 triu ng (17,2 19,8%), x4 t 4,7 6,4 triu ng (9,4 12,8%), x5 t 4,9 6,3 triu ng (9,8 12,6%) vi hnh thc nui chuyn canh (x6 = 1).
Mt cch c th hn, khi p dng phng php ti u thch hp ti mc u t 50 triu ng / ha c th tm c phng n ti u sau: zmax = 88,360733 vi x1 = 21,498072, x2 = 9,528987, x3 = 8,758034, x4= 5,138906, x5 = 5,076000, x6 = 1 v x7 = 0.
V d 5. Bi ton ti u thng s sng phn loi (M hnh ti u phi tuyn gii quyt vn tnh ton mt s thng s hnh hc v ng hc ca c cu sng phn loi dao ng)
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V d ny ch nu vn tt mt ng dng ca m hnh ti u phi tuyn mt mc tiu trong vic tm nghim ca h phng trnh phi tuyn pht sinh trong qu trnh tnh ton mt s thng s hnh hc v ng hc ca c cu sng phn loi dao ng (cn ch rng nhiu phng php tnh ton thng dng khc ca gii tch s t ra khng hiu qu):
r cos1 + v cos2 + //3v cos3 + v4 cos4 xC1 = 0, r sin1 + v sin2 + //3v sin3 + v4 sin4 yC1 = 0, r cos1 + v cos2 + /3v cos(3 ) + v5 cos5 xD1 = 0, r sin1 + v sin2 + /3v sin(3 ) + v5 sin5 yD1 = 0.
Trong h phng trnh phi tuyn trn cc thng s bit l: r = 0,05m; v = 0,30m; //3v = 0,15m;
/3v = 1,075m; v3 = 1,025m; v4 = 0,50m; v5 = 0,40m; xC1 = 0,365m; yC1 =
0,635m; xD1 = 1,365m; yD1 = 0,635m; = /8. gii h phng trnh phi tuyn khi 1 = k/8 (k = 0, , 9), chng ta cn cc tiu ho
hm mc tiu sau:
z = (r cos1 + v cos2 + //3v cos3 + v4cos4 xC1)2 + (r sin1 + v sin2 + //3v sin3 + v4sin4 yC1)2 + (r cos1 + v cos2 + /3v cos(3 ) + v5 cos5 xD1)2 + (r sin1 + v sin2 +
/3v sin(3 ) + v5sin5 yD1)2 min
Kt qu tnh ton c tng hp trong bng I.2 vi zmin = 0.
Bng I.2. Kt qu tnh ton gi tr cc thng s ca sng phn loi
1 [0,2] 2 [0,] 3 [0,] 4 [0,] 5 [0,] 0 0,226128 0,551311 1,783873 1,666775
/18 0,199269 0,550518 1,784628 1,670250 2/18 0,170835 0,550590 1,782751 1,668853 3/18 0,143343 0,550490 1,778826 1,663697 4/18 0,112669 0,552073 1,770032 1,652171 5/18 0,090986 0,551991 1,759350 1,639575 6/18 0,066036 0,553576 1,745374 1,622823 7/18 0,051284 0,554296 1,730174 1,602970 8/18 0,039053 0,555262 1,713242 1,581813 9/18 0,033773 0,556277 1,695605 1,560720
V d 6. Bi ton thit k trc my (M hnh quy hoch phi tuyn a mc tiu gii quyt bi ton thit k trc my)
Trong v d ny chng ta cp ti mt m hnh ti u phi tuyn hai mc tiu.
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Mc tiu 1 l cc tiu ho th tch ca trc my:
f1(x) = 0,785 [x1(6400 x22) + (1000 x1) (1000 x22)] (mm3),
Mc tiu 2 l cc tiu ho nn tnh ca trc:
f2(x) = 3,2981059
317 4 8 4 8 4
2 2 2
1 1 10x4,096 10 x 10 x 10 x
+ (mm/N).
y, x = (x1, x2) l vc t quyt nh, vi x1, x2 l cc bin quyt nh sau: x1 di phn gip ni trc, x2 ng knh trong ca trc. Cc thng s khc c th hin trong cc hm mc tiu f1(x) v f2(x).
Vy cn phi chn cc gi tr cho cc bin quyt nh (cn gi l cc bin thit k) x1, x2 ti u ho ng thi cc mc tiu 1 v 2 trong cc iu kin rng buc sau:
g1(x) = 180 6
17 4
2
9,78 10 x4,096 10 x
0 (1.1)
g2 (x) = 75,2 x2 0 (1.2) g3 (x) = x2 40 0 (1.3) g4 (x) = x1 0 (1.4)
Cc iu kin (1.2), (1.3), (1.4) l d hiu, cn iu kin (1.1) ny sinh l do yu cu: Mt mt, trc my phi chu ng c ti mc ti a lc Fmax = 12000 N. Mt khc, nn kt ni cho php l 180 N/mm.
Vic pht biu bi ton ti u a mc tiu di dng ton hc (chnh l vic lp m hnh ton hc cho vn pht sinh) l mt khu rt quan trng nhm m t tt nht hnh vi ca h thng ang c xem xt, mt khc nhm tm ra c cc phng php ti u ho c hiu qu i ti mt phng n tt v mang li li ch. Sau y, vi mc ch tm hiu bc u, vic p dng phng php tng tc ngi my tnh gii bi ton ti u hai mc tiu c thit lp trn y s c trnh by mt cch vn tt.
Trc ht, hai mc tiu f1(x) v f2(x) c chuyn thnh hai hm thuc m phn nh tho mn ca ngi ra quyt nh i vi tng mc tiu. Cc hm thuc m ny l cc hm tuyn tnh tng khc, c vit di dng gin lc nh sau cho mt s nt ni suy:
0 nu f1 6,594106 = a1 1(f1) = 0,5 nu f1 = 4106 = b1 1 nu f1 2,944106 = c1, 0 nu f2 0,499103 = a2 2(f2) = 0,5 nu f2 = 0,450103 = b2 1 nu f2 0,338103 = c2.
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Lc c th p dng php ni suy tuyn tnh tnh cc gi tr ca 1(f1) hoc 2(f2) ti cc gi tr khc ca f1 hay f2. Cc hm thuc m ny cho php quy cc n v o khc nhau ca f1 v f2 vo cng mt thang bc o, l tha dng ca ngi ra quyt nh / ngi gii bi ton. Phn tch hm thuc m 1, c th thy: ngi ra quyt nh s c tho mn 0 i vi mi phng n x = (x1, x2) lm cho f1 6,594106, tho mn 1 nu f1 2,944106 v tho mn 0,5 nu f1 = 4106. tho mn 0,5 c coi l tho mn ti thiu v mc f1 = 4106 = b1 c gi l mc u tin tng ng i vi mc tiu f1. Tng t chng ta c th phn tch v hm thuc 2 v mc u tin b2.
Chng ta xt hm phi tuyn g(x) = Min {1[f1(x)], 2[f2(x)]} v bi ton maxmin c thit lp cho hai hm mc tiu ring r trn di dng BTQHPT: Max g(x) = MaxMin{1[f1(x)], -2[f2(x)]} vi cc rng buc (1.1), (1.2), (1.3) v (1.4).
Vic gii BTQHPT trn y c thc hin nh mt phng php ti u phi tuyn thch hp, c ci t t ng trn my tnh tm ra cc phng n ti u ca m hnh phi tuyn hai mc tiu ban u. iu chnh thch hp gi tr ca cc mc u tin b1 v b2, c th tm c cc phng n ti u khc nhau. Chng hn, vi b1 = 3,6106, b2 = 0,435103 s nhn c phng n ti u x = (x1, x2) = (235,67; 67,67) vi f1(x) = 3,58106 v f2(x) = 0,433103. y l phng n c cc chuyn gia nh gi l hp l v c la chn trin khai trong vic thit k trc my.
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Chng II Phng php n hnh gii bi ton
quy hoch tuyn tnh
1. M hnh quy hoch tuyn tnh
1.1. Pht biu m hnh Vi mc ch tm hiu bc u, xt m hnh ton hc sau y, cn gi l m hnh quy
hoch tuyn tnh hay bi ton quy hoch tuyn tnh (BTQHTT), m trong chng ta mun ti u ho / cc i ho hay cc tiu ho hm mc tiu:
z = f(x) = c1x1 + c2x2 + .... + cnxn Max (Min), vi cc iu kin rng buc
a11x1 + a12x2 + ... + a1nxn b1 a21x1 + a22x2 + ... + a2nxn b2 ...
am1x1 + am2x2 + ... + amnxn bm x1, x2, ..., xn 0 (iu kin khng m).
V d 1. Xt BTQHTT: Max z = 8x1 + 6x2, vi cc rng buc
4x1 + 2x2 60 2x1 + 4x2 48 x1, x2 0.
Cn tm cc gi tr ca cc bin quyt nh x1, x2 cc rng buc c tho mn v hm mc tiu t gi tr ln nht.
Bi ton ny c ngha kinh t nh sau: Gi s mt x nghip sn xut hai loi sn phm I v II. sn xut ra mt n v sn phm I cn c 4 n v nguyn liu loi A v 2 n v nguyn liu loi B, cc ch tiu cho mt n v sn phm loi II l 2 v 4. Lng nguyn liu d tr loi A v B hin c l 60 v 48 (n v). Hy xc nh phng n sn xut t li nhun ln nht, bit li nhun / n v sn phm bn ra l 8 v 6 (n v tin t) cho cc sn phm loi I v II.
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1.2. Phng php th
Phng php th c ngha minh ha v gip hiu bn cht vn . Bc 1: V min cc phng n kh thi (cn gi l min rng buc) l tp hp cc phng
n kh thi (cc phng n, nu ni mt cch ngn gn). Mi phng n c th hin qua b s (x1, x2), tho mn tt c cc rng buc c k c iu kin khng m ca cc bin (xem hnh II.1).
Trc ht chng ta v ng thng c phng trnh l 4x1 + 2x2 = 60 bng cch xc nh hai im thuc ng thng: (x1 = 0, x2 = 30) v (x1 = 15, x2 = 0).
ng thng ny chia mt phng lm hai na mt phng. Mt phn gm cc im (x1, x2) tho mn: 4x1 + 2x2 60, phn cn li tho mn: 4x1 + 2x2 60. Ta tm c na mt phng tho mn: 4x1 + 2x2 60.
Tng t, c th v ng thng c phng trnh l 2x1 + 4x2 = 48 bng cch xc nh hai im thuc ng thng l (x1 = 0, x2 = 12) v (x1 = 24, x2 = 0). Sau tm na mt phng tho mn: 2x1 + 4x2 48.
Lc ny, giao ca hai na mt phng tm c trn y cho ta tp hp cc im (x1, x2)
tho mn cc rng buc. Tuy nhin, tho mn iu kin khng m ca cc bin, ta ch xt cc im nm trong gc phn t th nht. Vy min cc phng n kh thi (ni vn tt hn, min phng n) l min gii hn bi t gic OABC (cn gi l tp li a din v l min to nn bi giao ca cc na mt phng).
Bc 2: Trong min (OABC) ta tm im (x1, x2) sao cho
z = 8x1 + 6x2 t gi tr ln nht.
Cch 1. Dng ng ng mc. Ty theo gi tr ca x1, x2 m z c nhng mc gi tr khc nhau.
30
4x1 + 2x2 = 60
O
4
8
12
x1
2x1 + 4x2 = 48
x2
6 15 3 24
A
B
C
Hnh II.1. Phng php th gii bi ton quy hoch tuyn tnh
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V ng ng mc: 8x1 + 6x2 = c mc c = 24, (ta c th chn gi tr c bt k, nhng chn c = 24 l bi s chung ca 6 v 8 vic tm ta cc im ct hai trc ta thun li hn). D dng tm c hai im nm trn ng ng mc ny l (x1 = 0, x2 = 4) v (x1 = 3, x2 = 0). Cc im nm trn ng ng mc ny u cho gi tr hm mc tiu z = 24.
Tng t, c th v ng ng mc th hai: 8x1 + 6x2 = 48 i qua hai im (x1 = 0, x2 = 8) v (x2 = 0, x1 = 6). Chng ta nhn thy, nu tnh tin song song ng ng mc ln trn theo hng ca vc t php tuyn nG (8, 6) th gi tr ca hm mc tiu z = 8x1 + 6x2 tng ln.
Vy gi tr z ln nht t c khi ng ng mc i qua im B(12, 6) (tm c x1 = 12, x2 = 6 bng cch gii h phng trnh 4x1 + 2x2 = 60 v 2x1 + 4x2 = 48).
Do , trong cc phng n kh thi th phng n ti u l (x1 = 12, x2 = 6). Ti phng n ny, gi tr hm mc tiu l ln nht zmax = 8 12 + 6 6 = 132.
Nhn xt. Phng n ti u (nu c) ca mt BTQHTT vi min phng n D, l mt tp li a din c nh, lun t c ti t nht mt trong cc nh ca D. Cc nh ny cn c gi l cc im cc bin ca tp li a din D (chnh xc hn, im cc bin l im thuc tp li a din, m khng th tm c mt on thng no cng thuc tp li a din nhn im l im trong). Nhn xt trn y l mt nh l ton hc (xem thm chng VI) c chng minh mt cch tng qut. Ni mt cch hnh nh, mun t c phng n ti u cho cc BTQHTT th cn phi mo him i xt cc im cc bin ca min phng n.
Cch 2. T nhn xt trn, i vi BTQHTT c phng n ti u v c min phng n D l tp li a din c nh, ta c th tm phng n ti u bng cch so snh gi tr ca hm mc tiu ti cc im cc bin ca D. Quay li v d 1, ta c gi tr z ti O(0, 0): z (0, 0) = 0, ti A(0, 12): z(0, 12) = 72, ti C(15, 0): z(15, 0) = 120 v ti B(12, 6): z(12, 6) = 132 (t zmax).
Nhn xt. Xt BTQHTT c phng n ti u v c min phng n D l tp li a din c nh. tm phng n ti u, ta xut pht t mt im cc bin no v tm cch ci thin hm mc tiu bng cch i ti im cc bin k tt hn. Tip tc nh vy cho ti khi tm c phng n ti u. Quy trnh gii ny bao gm hu hn bc do s im cc bin l hu hn.
i vi BTQHTT trong v d 1, quy trnh gii c minh ho nh sau:
O(0, 0) A(0, 12) B(12, 6) dng z = 0 z = 72 z = 132
hoc:
O(0, 0) C(15, 0) B(12, 6) dng z = 0 z = 120 z = 132
Quy trnh gii BTQHTT tng qut c s khi gin lc nh trnh by trn hnh II.2. Trong s trn, v mc ch trnh by vn n gin, chng ta khng cp ti cc trng hp khi BTQHTT c min phng n l tp rng (lc ta khng tm c phng n cc bin xut pht) cng nh khi ta khng tm c im cc bin k tt hn mc d iu kin ti u cha tho mn (lc hm mc tiu z khng b chn).
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S khi
2. Phng php n hnh
2.1. Tm hiu quy trnh tnh ton Phng php n hnh l phng php s gii BTQHTT theo s trn. gii v d
cho, trc ht chng ta cn a BTQHTT v dng chnh tc bng cc bin b khng m x3 v x4 nh sau:
Max z = 8x1 + 6x2 + 0x3 + 0x4
vi cc rng buc
4x1 + 2x2 + x3 = 60
2x1 + 4x2 + x4 = 48
x1, x2, x3, x4 0. Ch . BTQHTT c dng chnh tc l BTQHTT vi cc bin khng m, cc rng buc c
du =, h s v phi ca cc rng buc khng m. Ngoi ra, mi phng trnh bt buc phi c mt bin ng c lp vi h s +1.
Cch lp v bin i cc bng n hnh
Bt u
Nhp d liu
Tm im cc bin xut pht
Tm im cc bin k
tt hn Kim tra iu kin
ti u
In v lu tr kt qu
Dng
ng
Sai
Hnh II.2. S khi gii BTQHTT
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gii BTQHTT dng chnh tc trn y, cn lp mt s bng n hnh nh trong bng II.1. Trc ht, cn in s liu ca bi ton cho vo bng n hnh bc 1:
Ct 1 l ct h s hm mc tiu ng vi cc bin c s chn. Phng n xut pht c th chn l x1 = x2 = 0 (y chnh l im gc to O(0, 0) trn hnh II.1), do x3 = 60, x4 = 48. Nh vy ti bc ny chng ta cha bc vo sn xut, nn trong phng n cha c n v sn phm loi I hay loi II no c sn xut ra (ch sn xut ra cc lng nguyn liu d tha, ta cng ni l cc sn phm loi III v IV), v gi tr hm mc tiu z tm thi bng 0.
Bng II.1. Cc bng n hnh gii BTQHTT
c1 = 8 c2 = 6 c3 = 0 c4 = 0 H s hm mc tiu cj
Bin c s Phng n x1 x2 x3 x4
Bng n hnh bc 1
0 0
x3 x4
60 48
4 2
2 4
1 0
0 1
Hng z z0 = 0 z1 = 0 z2 = 0 z3 = 0 z4 = 0
Hng j = cj zj 1 = 8 2 = 6 3 = 0 4 = 0 Bng n hnh bc 2
8 0
x1 x4
15 18
1 0
1/2 3
1/4 1/2
0 1
Hng z z0 = 120 z1 = 8 z2 = 4 z3 = 2 z4 = 0
Hng j = cj zj 1 = 0 2 = 2 3 = 2 4 = 0 Bng n hnh bc 3
8 6
x1 x2
12 6
1 0
0 1
1/3 1/6
1/6 1/3
Hng z z0 = 132 8 6 5/3 2/3
Hng j = cj zj 0 0 5/3 2/3
Cc bin b c gi tr ln hn 0 c ngha l cc nguyn liu loi tng ng cha c s dng ht. Ta gi cc bin x3 v x4 l cc bin c s v chng c gi tr ln hn 0 cn x1 v x2 l cc bin ngoi c s v chng c gi tr bng 0. Vi bi ton c hai rng buc, ti mi bc ch c hai bin c s.
Ct 2 l ct cc bin c s. Trong ct 3 (ct phng n) cn ghi cc gi tr ca cc bin c s chn.
Cc ct tip theo l cc ct h s trong cc iu kin rng buc tng ng vi cc bin x1, x2, x3 v x4 ca bi ton cho.
Phn tch bng n hnh bc 1 H s ng vi bin x1 trn hng th nht l a11 = 4 c ngha l t l thay th ring gia
mt n v sn phm loi I v mt n v sn phm loi III l 4 (gii thch: xt phng trnh (hay
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rng buc) th nht 4x1 + 2x2 + x3 = 60, x1 tng mt n v th x3 phi gim bn n v nu gi nguyn x2). Tng t ta c th gii thch c ngha ca cc h s aij khc cho trn hng 1 v hng 2 trong bng n hnh bc 1.
Chng ta xt hng z ca bng n hnh. tnh z1, cn p dng cng thc z1 = (ct h s ca hm mc tiu) (ct h s ca bin x1) = 04 + 02 = (gi mt n v sn phm loi III)(t l thay th ring loi I / loi III) + (gi mt n v sn phm loi IV)(t l thay th ring loi I / loi IV) = tng chi ph phi b ra khi a thm mt n v sn phm loi I vo phng n sn xut mi = 0. Cc gi tr zj, vi j = 1, 2, 3, 4, c tnh tng t v chnh l cc chi ph khi a thm mt n v sn phm loi xj vo phng n sn xut mi. Cn z0 l gi tr ca hm mc tiu t c ti phng n ang xt: z0 = (ct h s ca hm mc tiu) (ct phng n) = 060 + 0 48 = 0.
Trn hng j cn ghi cc gi tr j , j = 1, 2, 3, 4, tnh theo cng thc j = cj zj = li nhun / n v sn phm chi ph / n v sn phm. Vy j l "li bin" / mt n v sn phm khi a mt thm mt n v sn phm loi xj vo phng n sn xut mi. Nu j > 0 th hm mc tiu cn tng c khi ta a thm cc sn phm loi j vo phng n sn xut mi. C th chng minh c j chnh l o hm ring jz / x ca hm mc tiu z theo bin xj . Nh vy, x1 tng ln 1 th z tng ln 8 cn x2 tng ln 1 th z tng ln 6 .
Do 1 v 2 u ln hn 0 nn vn cn kh nng ci thin hm mc tiu khi chuyn sang (hay xoay sang) mt phng n cc bin k tt hn (quay li nhn xt mc 1.2, phn gii bi ton bng phng php th: im cc bin k ca im O(0, 0) c th l A(0, 12) hay C(15, 0)).
Th tc xoay (pivotal procedure)
Bc 1: Chn ct xoay l ct bt k c j > 0. Lc bin xj tng ng vi ct xoay c chn lm bin c s mi do xj tng ko theo hm mc tiu tng. y ta chn a x1 vo lm bin c s mi.
Bc 2: Chn hng xoay xc nh a bin no ra khi tp cc bin c s (v ti mi bc s bin c s l khng thay i). chn hng xoay, ta thc hin quy tc t s dng b nht bng cch ly ct phng n (60, 48)T chia tng ng cho ct xoay (4, 2)T chn t s b nht. Mt iu cn ch l ta ch xt cc t s c mu s dng.
V Min {60/4, 48/2} = 60/4 t c ti hng u, nn hng xoay l hng u (hng tng ng vi bin x3). Do cn a x3 ra khi tp cc bin c s.
Bc 3: Chn phn t xoay nm trn giao ca hng xoay v ct xoay.
Bc 4: Xoay sang bng n hnh mi, xc nh cc bin c s mi in vo ct bin c s, ng thi thay cc gi tr trong ct h s hm mc tiu. Sau , tnh li cc phn t ca hng xoay bng cch ly hng xoay c chia cho phn t xoay c hng mi tng ng.
Bc 5: Cc phn t cn li ca bng n hnh mi tnh theo quy tc hnh ch nht:
(1)mi = (1)c (2)c (4)c/(3)c, trong (3) l nh tng ng vi phn t xoay (xem hnh I.3).
22
Gii thch. Cc bc xoay trn y ch l php bin i tng ng h phng trnh
4x1 + 2x2 + x3 = 60 (2.1)
2x1 + 4x2 + x4 = 48 (2.2)
c h
x1 + (1/2)x2 + (1/4)x3 = 15 (2.1)
0x1 + 3x2 (1/2)x3 + x4 = 18 (2.2)
bng cch ly phng trnh (2.1) chia cho 4 (phn t xoay) c (2.1), ri ly (2.2) tr bt 2 (2.1)/4 c (2.2). y chnh l ni dung ca bc 4 v bc 5. Cn vic thc hin bc 3 s m bo rng gi tr ca cc bin c s mi khng m (x1 = 15, x4 = 18).
p dng th tc xoay cho cc phn t nm trn hng 1 v 2 ca bng n hnh bc 1, sau tnh cc gi tr trn hng zj v j tng t nh khi lp bng n hnh bc 1, chng ta s nhn c bng n hnh bc 2.
Phn tch bng n hnh bc 2 Bng bc 2 c th c phn tch tng t nh bng bc 1. Cn ch rng lc ny ta
ang v tr ca im C(15, 0) v x1 = 15 cn x2 = 0 (xem hnh II.1). Ti im ny gi tr ca hm mc tiu l z0 = 120 c ci thin hn so vi bc 1. Ta thy 2 = 2 > 0 nn cn c th ci thin hm mc tiu bng cch a bin x2 vo lm bin c s mi. Thc hin cc bc xoay sang phng n cc bin k tt hn, chng ta s c bng n hnh bc 3.
Phn tch bng n hnh bc 3
Ti bng n hnh bc 3 ta thy iu kin ti u c tho mn (j 0, j =1,4 ) nn khng cn kh nng ci thin phng n. Phng n ti u t c ti x1 = 12, x2 = 6, x3 = 0, x4 = 0, tc l ti im cc bin B(12, 6) vi gi tr zmax = 132 (xem thm hnh II.1).
Mt s ch iu kin ti u cho cc BTQHTT dng Max l j 0, j . i vi cc BTQHTT cn cc tiu ho hm mc tiu th iu kin ti u (hay tiu chun
dng) l j 0, j (nu j* sao cho j* < 0 th cn tip tc ci thin hm mc tiu bng cch chn ct j* lm ct xoay).
(1)
(2) (3)
(4)
Chng hn: nu (1)c = 4,(2)c = 2, (3)c = phn t xoay = 4, (4)c = 2 th (1)mi = 4 22/4 =3
Hnh II.3. Quy tc hnh ch nht
23
Trong thc tin gii cc BTQHTT dng tng qut c th xy ra trng hp khng tm c phng n xut pht (tc l khng c phng n kh thi). Lc ny c th kt lun m hnh thit lp c cc iu kin rng buc qu cht ch, cn xem xt ni lng cc iu kin ny.
Trong trng hp ta tm c ct xoay m khng tm c hng xoay th kt lun hm mc tiu khng b chn trn (i vi cc BTQHTT dng Max) hoc khng b chn di (i vi cc BTQHTT dng Min).
Trong cc trng hp trn cng phi dng li v kt lun m hnh quy hoch tuyn tnh thit lp khng ph hp vi thc t.
2.2. Khung thut ton n hnh Sau y l khung thut ton ca phng php n hnh c pht biu cho BTQHTT cc
i ha dng chnh tc. Bc khi to Tm mt phng n cc bin ban u.
Tnh j = cj zj, j = 1,n , trong n l s bin ca bi ton ang xt. Cc bc lp
Bc 1: Kim tra iu kin ti u. Nu iu kin ti u j = cj zj 0, j = 1,n c tho mn th in / lu tr kt qu ca bi ton v chuyn sang bc kt thc.
Bc 2: Nu tn ti mt ch s j sao cho j > 0 th tin hnh th tc xoay gm nm bc bit, tnh li cc j, j = 1,n v quay li bc 1 (Ch : Trong trng hp ta tm c ct xoay m khng tm c hng xoay th kt lun hm mc tiu khng b chn, in / lu tr kt qu ca bi ton v chuyn sang bc kt thc).
Bc kt thc. Dng.
3. C s ton hc ca phng php n hnh
3.1. Pht biu bi ton quy hoch tuyn tnh dng chnh tc
Xt BTQHTTdng sau y (vi cc rng buc u c du =):
Max (Min) z = c1x1 + c2x2 + ... + cnxn
vi h iu kin rng buc
11 1 12 2 1n n 1
21 1 22 2 2n n 2
m1 1 m2 2 mn n m
j
a x a x ... a x ba x a x + ... a x = ba x a x ... a x b
x 0, j 1,n.
+ + + = + + + + + = =
Chng ta s dng cc k hiu sau (T l k hiu chuyn v): Vc t h s hm mc tiu c = (c1, c2, , cn)T Rn, Vc t quyt nh x = (x1, x2, , xn)T Rn, Vc t h s v phi b = (b1, b2, , bm)T Rm,
24
Ma trn h s cc iu kin rng buc
A =
11 12 1n
21 22 2n
m1 m2 mn
a a ... aa a ... a... ... ... ...
a a ... a
Rmn,
trong aj = (a1j, a2j, ,amj)T l vc t ct j ca ma trn A, j = 1,n . Vi cc k hiu trn, BTQHTT c vit ngn gn l:
Max z = cTx, vi x D = {x Rn: Ax = b, x 0}. (2.3) BTQHTT trn y c gi l BTQHTT dng chun tc nu hng ca A bng m v b 0
(cc ta ca b u khng m). Ngoi ra, nu A c m vc t ct l cc vc t n v c lp tuyn tnh th BTQHTT dng chun tc tr thnh BTQHTT dng chnh tc. Trong trng hp BTQHTT dng chnh tc, khng lm gim tnh tng qut, chng ta lun c th coi m vc t ct aj , j = n m 1,n + l cc vc t n v c lp tuyn tnh,
V d 2. Chng ta xt li v d 1 ca chng ny.
Max z = 8x1 + 6x2 + 0x3 + 0x4
vi cc rng buc
4x1 + 2x2 + x3 = 60
2x1 + 4x2 + x4 = 48
x1, x2, x3, x4 0. y l BTQHTT dng chnh tc. Gi s ma trn A c phn r theo khi di dng A =
[N B] vi B l ma trn kh nghch. Chng ta s s dng cc k hiu sau: J = {1, 2, ..., n} l tp cc ch s, JB = {j: aj l vc t ct ca B} l tp ch s cc bin c s, JN = J \ JB = {j : aj l vc t ct ca N} l tp cc ch s cc bin ngoi c s. Lc , c th vit vc t
quyt nh di dng x = ( )TT TN Bx , x v vc t h s hm mc tiu c = ( )TT TN Bc , c . Trong v d 2, ta c: JN = {1, 2}, JB = {3, 4}. D dng thy, phng n ban u
x = ( )TT TN Bx , x = (0, 0, 60, 48)T, trong xN = (x1, x2)T = (0, 0)T v xB = (x3, x4)T = (60, 48)T. Vc t h s hm mc tiu l c = ( )TT TN Bc , c = (8, 6, 0, 0)T vi cN = (8 6)T, cB = (0 0)T. Cc vc t ct ca ma trn rng buc A l:
a1 = 42
, a2 = 24
, a3 = 10
, a4 = 01
.
Vy A = (a1, a2, a3, a4) = [N B] vi N = 4 22 4
, B = 1 00 1
.
25
Cn ch rng: Ax = b [N B] NB
xx
= b NxN + BxB = b BxB = b xB = B1b.
Phng n cc bin i vi BTQHTT (2.3) dng chnh tc lun c th tm c mt phng n xut pht x =
(0, 0, , 0, b1, b2, , bm)T, trong n m ta u tin u bng 0. y l mt phng n cc bin. Mt cch tng qut, xt mt phn r ty ca ma trn A = [N B] vi B l ma trn vung c to nn t m vc t ct c lp tuyn tnh ca A, N l ma trn c to nn t cc vc t ct cn li. Lc , mt phng n cc bin ca BTQHTT tng ng vi s phn r trn ca A l
mt phng n c dng x = ( )TT TN Bx ,x trong xN = 0, xB 0. Ma trn B c gi l ma trn c s tng ng vi x (c th xem thm v vn phng n cc bin trong chng VI). Nh vy, mt phng n cc bin khng c qu m ta dng. Phng n cc bin c ng m ta dng c gi l phng n cc bin khng suy bin, nu tri li, l phng n cc bin suy bin.
3.2. Cng thc s gia hm mc tiu Xt BTQHTT (2.3) dng chnh tc, gi s x l phng n cc bin tng ng vi phn r
A = [N B], vi B l ma trn c s, cn x l mt phng n khc. t x = x x l vc t s gia cc bin quyt nh. Chng ta tm cch thit lp cng thc s gia hm mc tiu:
cT x cTx = cT( x x) = cTx.
Ta thy ngay A x = Ax = b nn Ax = 0. K hiu x = NB
xx
, ta c Ax = 0 [N
B] NB
xx
= 0 NxN + BxB = 0 BxB = NxN xB = B1NxN.
Vy cTx = T TN B(c ,c ) NB
xx
= TNc xN + TBc xB = TNc xN TBc B1NxN
= ( TNc TBc B
1N)xN = ( TNc TBc B1N)xN + ( TBc TBc B1B)xB
= [ TNc TBc B
1N, TBc TBc B
1B] NB
xx
.
t = [ TNc TBc B1N, TBc TBc B1B] = [N, B], th cTx = x. y chnh l cng thc s gia hm mc tiu cn thit lp.
Quay li v d 2, trong bng n hnh bc 1, chng ta c:
= 1 11 0 4 2 1 0 1 0
(8,6) (0,0) , (0,0) (0,0)0 1 2 4 0 1 0 1
= (8, 6, 0, 0) = (1, 2, 3, 4).
26
Nhn xt. C th chng minh c rng , 1, . = =j jz j nx
Chng hn, tng ng vi
bng n hnh bc 2 ta c: z = z(x + x) z(x) = cT(x + x) cTx = cTx = x = 1x1 + 2x2 + 3x3 + 4x4 = 0x1 + 2x2 + (2)x3 + 0x4. R rng rng, 1
1
z 0x
= = , 22z 2x = = , 33
z 2x = = , 44
z 0x = = .
3.3. Tiu chun ti u
Xt phng n cc bin x ca BTQHTT (2.3) dng chnh tc: x = ( )TT TN Bx , x (tng ng vi phn r A = [N B], vi B l ma trn c s). Lc ny, x D ta c:
cT x cTx cT x cTx 0 cTx 0. Vy tiu chun x l phng n ti u l: cTx 0, x x 0, x
(N, B) NB
xx
= NxN + BxB 0,x NxN 0,x (do B = 0).
nh l 1. Xt BTQHTT (2.3) dng chnh tc. iu kin mt phng n cc bin x =
( )TT TN Bx , x (tng ng vi phn r A = [N B], vi B l ma trn c s) l phng n ti u l N = TNc
TBc B
1N 0. Ngc li, nu x l phng n cc bin ti u khng suy bin th ta cng c N = TNc TBc B1N 0.
Chng minh
iu kin . Nu N 0, th NxN 0, x D, (ch rng xN = 0 lun ng, nn cng lun c xN = x N xN 0). Do B = 0 nn NxN + BxB 0, x hay x 0,x. Vy cT x cTx, x D. Do x l phng n ti u.
iu kin cn. S dng phng php chng minh phn chng, gi s x l phng n cc bin ti u khng suy bin v iu kin N 0 khng c tho mn. Lc tn ti ch s j* JN sao cho j* > 0. Xt phng n x = x + x. Chng ta s ch ra cch xy dng x sao cho x l phng n kh thi tha mn cT x > cTx hay cTx < 0, t suy ra x khng phi l phng n ti u.
Tht vy, chn xN sao cho: xj = 0, j JN, j j* v xj* = > 0. Chn xB sao cho: Ax = 0 [N B] N
B
xx
= 0 NxN + BxB = 0 BxB = NxN
xB = B1NxN xB = B1aj*. Trong v d 2, ta thy: NxN = 1
2
x4 2x2 4
=
4 22 4 0
= 42
= a1 ,
v i j* = 1.
27
x l phng n kh thi, cn phi c x 0. D thy x N 0 theo cch xy dng xN. Cn x B = xB + xB = xB B1aj* . x B 0 phi chn theo quy tc t s dng b nht (nh bit mc 2.1 khi m t th tc xoay).
Trng hp 1: B1aj* 0. Lc ny, khi thc hin quy tc t s dng b nht (ly ct phng n chia cho ct aj*) ta khng nhn c mt t s no c mu s dng. x B 0, chng ta c th chn > 0 v ln tu . Do cTx = j* + khi chn +. iu ny chng t phng n x khng phi l phng n ti u v BTQHTT (2.3) dng chnh tc c hm mc tiu khng b chn trn.
Trng hp 2: Vc t B1aj* c ta dng. cho d hiu, xt li v d 1 v bng n hnh II.1 (bc 2). Do x1 v x4 l cc bin c s v j* = 2 nn:
B = 4 02 1
B1 =
1 / 4 01 / 2 1
B1aj* =
1 / 4 01 / 2 1
24
=1 / 2
3 .
Vy: x B = xB + xB = xB B1aj* = 1518
1 / 23
0 15 (1 / 2) 0
18 3 0.
Chn = Min 15 18,1 / 2 3
= 6 theo quy tc t s dng b nht s m bo x B 0.
Do x l phng n cc bin khng suy bin nn xB > 0 ko theo > 0. Cui cng, ta c cTx = x = NxN + BxB = NxN = j* xj* = j* > 0. Do , phng n x khng th l phng n ti u (pcm).
Nhn xt Nu tn ti ch s i* JB sao cho xi* = 0 (nh bit, phng n cc bin x lc ny c
gi l phng n cc bin suy bin), th t iu kin x B = xB + xB = xB B1aj* 0 c th xy ra trng hp chn c = 0. Do cTx = j* = 0, tc l hai phng n x v x cho cng mt gi tr hm mc tiu. Trong cc trng hp nh vy c th xy ra hin tng xoay vng: Chng hn, khi chuyn t x sang x , ri li chuyn t x sang mt phng n x no m vn cha ci thin c gi tr ca hm mc tiu. Sau , li c th xy ra vic chuyn t x v x. Nh vy qu trnh gii BTQHTT theo thut ton n hnh s b treo ti vng lp x x x x. khc phc hin tng xoay vng c th p dng mt s th tc tnh ton. Cch n gin nht l p dng quy tc t s dng b nht vi s b sung sau: Nu c nhiu ch s ng vi t s dng b nht, th chn ngu nhin mt trong cc ch s xc nh hng xoay tng ng.
Trong qu trnh gii BTQHTT (2.3) dng chnh tc khi xut pht t mt phng n cc bin, bng th tc xoay ta lun chuyn t phng n cc bin ny sang phng n cc bin khc cho ti khi cc du hiu dng c tha mn (tc l khi tiu chun ti u c tha mn hay khi kt lun c BTQHTT cho c hm mc tiu khng b chn trn).
3.4. Thut ton n hnh cho bi ton quy hoch tuyn tnh dng chnh tc Xt BTQHTT dng chnh tc:
Max z = c1x1 + c2x2 + ... + cnxn + cn+1xn+1 + ... + cn+mxn+m
28
vi cc rng buc
a11x1 + a12x2 + ... + a1nxn + xn+1 = b1
a21x1 + a22x2 + ... + a2nxn + xn+2 = b2
...
am1x1 + am2x2 + ... + amnxn + xn+m = bm
x1, x2, ..., xn, xn+1, ..., xn+m 0 Bc khi to Nhp cc h s hm mc tiu c, ma trn rng buc A v cc h s v phi b. t d1 = cn+1, ..., dm = cn+m , tc l cB = (d1, ..., dm)T. t ch s bin c s: r(1) = n + 1, ..., r(m) = n + m.
Gn xr(i) = bi , i = 1,m . t flag = 2.
Cc bc lp Bc 1: Tnh cTx = z = d1xr(1) + ... + dmxr(m).
Tnh zj = m
pj pp 1
a d=
, j = 1,n m+ . Tm = [N, B] = [ TNc TBc B1N, TBc TBc B1B], trong B = 0. Nh vy
j = cj zj, vi zj =m
pj pp 1
a d=
, j N v j = cj zj = 0, j B, (tc l zN = TBc B1N v zB = TBc B
1B).
Bc 2: Nu tn ti ch s j N sao cho j > 0 th thc hin th tc xoay. Xc nh ct xoay: chn ct xoay s ng vi mt ch s j c tnh cht j > 0. Thng
thng chn j ng vi j > 0 ln nht, hoc chn ngu nhin. Xc nh hng xoay q theo quy tc t s dng b nht:
r (q ) r ( i )is
qs is
x xMin , a 0
a a = >
.
Trong trng hp khng tn ti ais > 0, t flag = 0 v chuyn sang bc kt thc. Xc nh phn t xoay aqs. Tnh li ( chuyn sang bng n hnh mi): bq: = bq/aqs, aqj: = aqj/aqs, j. i q tnh li bi
: = bi bqais v aij = aij aqj ais, j. t li ch s cc bin c s: r(q) := s, dq := cs, xr(i) = bi i =1,m . Quay v bc 1. Bc 3: Nu j 0, j N th t flag = 1 v chuyn sang bc kt thc.
29
Bc kt thc Ghi li d liu u vo ca BTQHTT v kt qu cui cng. Nu flag = 0 th kt lun
BTQHTT c hm mc tiu khng b chn trn. Cn nu flag = 1 th kt lun BTQHTT c phng n ti u tm c. Dng.
4. B sung thm v phng php n hnh
Xt BTQHTT dng tng qut:
Max (Min) z = c1x1 + c2x2 + .... + cnxn
vi cc iu kin rng buc
a11x1 + a12x2 + ... + a1nxn b1 a21x1 + a22x2 + ... + a2nxn b2 ...
am1x1 + am2x2 + ... + amnxn bm x1 0, x2 0, ..., xn 0.
Trong k hiu c th hiu l , hoc = i vi cc rng buc. i vi iu kin v du ca cc bin 0 c th hiu l 0, 0 hoc c du tu . Mun gii mt BTQHTT c dng tng qut, trc ht cn a n v dng chnh tc. C th nhc li vn tt, BTQHTT dng chnh tc l bi ton vi cc bin khng m, cc rng buc vi du =, h s v phi ca cc rng buc khng m. Ngoi ra, mi phng trnh bt buc phi c mt bin ng c lp vi h s +1.
4.1. a bi ton quy hoch tuyn tnh v dng chnh tc
V d 3. Xt li v d 1, trng hp cc rng buc u c du . Max z = 8x1 + 6x2, vi cc rng buc
1 2
1 2
1 2
4x 2x 602x 4x 48x ,x 0.
+ +
a BTQHTT v dng chnh tc nh bit bng cch thm hai bin b (slack variables) x3 v x4. Ta c BTQHTT dng chnh tc:
Max z = 8x1 + 6x2 +0x3 + 0x4
1 2 3
1 2 4
1 2 3 4
4x 2x x 602x 4x x 48x ,x ,x ,x 0.
+ + = + + =
Lc ny, trong h hai iu kin rng buc c hai bin ng c lp trong tng phng trnh vi h s +1, nn c th tm c phng n cc bin xut pht bt u qu trnh gii bi ton.
V d 4. Trng hp c iu kin rng buc vi du hoc =.
30
Max z = 8x1 + 6x2, vi cc rng buc
1 2
1 2
1 2
4x 2x 602x 4x 48x ,x 0.
+ +
Ta thm hai bin b x3 (slack variable) mang du +, x4 (surplus variable) mang du c h iu kin rng buc (mang du =)
1 2 3
1 2 4
1 2 3 4
4x 2x x 602x 4x x 48x ,x ,x ,x 0.
+ + = + =
Phi thm bin gi x5 (x5 gi l lng vi phm ca phng trnh th hai) c h iu kin rng buc
1 2 3
1 2 4 5
1 2 3 4 5
4x 2x x 602x 4x x x 48x ,x ,x ,x ,x 0.
+ + = + + =
Lc ny, do c hai bin ng c lp trong tng phng trnh vi h s +1, nn c th tm c phng n cc bin xut pht bt u qu trnh gii bi ton bng phng php n hnh vi hm mc tiu l Max z = 8x1 + 6x2 + 0x3 + 0x4 Mx5 , trong M + v biu thc Mx5 gi l lng pht (nh thu). Bi ton c a v dng chnh tc. Lng vi phm x5 cng ln th hm mc tiu cng gim, gi tr ca hm mc tiu ch c th t Max khi x5 = 0.
V d 5. Trng hp c bin khng dng. Max z = 8x1 6x2
1 2 3
1 2 4
1 2 3 4
4x 2x x 602x 4x x 48x 0,x 0,x 0,x 0.
+ + + =
Lc ny mun gii bi ton bng phng php n hnh ta phi i bin x/2 = x2. Ta c BTQHTT vi cc bin u khng m.
Max z = 8x1 + 6x/2
/1 2 3
/1 2 4
/1 2 3 4
4x 2x x 60
2x 4x x 48
x ,x ,x ,x 0.
+ =
V d 6. Trng hp c bin vi du tu . Max z = 8x1 + 6x2, vi cc rng buc
31
1 2
1 2
1 2
4x 2x 602x 4x 48x 0,x
+ + c du tu .
Lc ny ta vit bin x2 di dng x2 = x/2 x//2 vi
/2 2//2 2
x max{0,x }
x max{0, x }
= = th m bo
/2//2
x 0
x 0.
Cc rng buc s l
/ //1 2 2 3
/ //1 2 2 4
/ //1 2 2 3 4
4x 2x 2x x 60
2x 4x 4x x 48x ,x ,x ,x ,x 0.
+ + = + + =
Bi ton vi hm mc tiu Max z = 8x1 + 6x/2 6x//2 + 0x3 + 0x4 v cc iu kin rng buc
trn l BTQHTT dng chnh tc.
Kt lun. Bao gi cng a c BTQHTT bt k (cc bin c du tu , cc rng buc c
th , hay =) v dng chnh tc. Bin c du 0 c thay bng mt bin c du 0, bin c du tu c thay bi hiu ca hai bin u c du 0. Rng buc c a v = bng cch thm mt bin b (thiu), rng buc a v rng buc = bng cch thm mt bin b (tha) v mt bin gi, mi rng buc c du = c thm mt bin gi. S bin gi ca BTQHTT dng chnh
tc nhn c chnh l tng s cc rng buc v =.
4.2. Phng php n hnh m rng
Phng php n hnh m rng cn gi l phng php nh thu M c p dng gii
BTQHTT c bin gi.
V d 7. Xt BTQHTT: Max z = 8x1 + 6x2, vi cc rng buc
1 2
1 2
1 2
4x 2x 602x 4x 48x ,x 0.
+ + (2.4)
hay: Max z = 8x1 + 6x2 +0x3 + 0x4, vi cc rng buc
1 2 3
1 2 4
1 2 3 4
4x 2x x 602x 4x x 48x ,x ,x ,x 0,
+ + = + = (2.5)
Ta c th a bi ton v dng chnh tc sau gi l bi ton M: Max z = 8x1 + 6x2 +0x3 + 0x4 Mx5 (trong M +)
32
vi cc rng buc
1 2 3
1 2 4 5
1 2 3 4 5
4x 2x x 602x 4x x x 48x ,x ,x ,x ,x 0.
+ + = + + = (2.6)
Cch 1. C th gii BTQHTT vi cc iu kin rng buc (2.4) bng phng php th
nhn c kt qu: phng n ti u l (x1 = 0, x2 = 30) v zmax = 180.
Cch 2. Gii BTQHTT vi cc iu kin rng buc (2.6) bng cch lp bng n hnh nh
thng thng nhng ch h s M + (xem bng II.2). Ti bng n hnh cui cng, ta thy j 0, j, nn phng n ti u t c vi x2 =
30, x4 = 72, x1 = x4 = x5 = 0 v zmax = 180.
Bng II.2. Cc bng n hnh gii bi ton M
8 6 0 0 M H s hm mc tiu
Bin c s Phng n x1 x2 x3 x4 x5
0 M
x3 x5
60 48
4 2
2 4
1 0
0 1
0 +1
Hng z z0 = 48M z1 = 2M z2 = 4M z3 = 0 z4 = M z5 = M
Hng j 1 = 8 + 2M 2 = 6+4M 3 = 0 4 = M 5 = 0 0 6
x3 x2
36 12
3 1/2
0 1
1 0
1/2 1/4
1/2 1/4
Hng z 72 3 6 0 3/2 3/2
Hng j 5 0 0 3/2 M3/2 0 6
x4 x2
72 30
6 2
0 1
2 1/2
1 0
1 0
Hng z 180 12 6 3 0 0
Hng j 4 0 3 0 M
Ch Mt khi mt bin gi c a ra khi c s th khng bao gi quay li na (bn c
hy t chng minh iu ny). Do ta c th xo ct bin gi i khi bng n hnh.
Nu du hiu dng xut hin (j 0, j) nhng vn cn bin gi vi gi tr dng trong s cc bin c s th iu ny chng t bi ton ban u khng th c phng n kh thi (c th chng minh iu ny bng phn chng).
Vi v d trn (xem bng II.2) ta thy qu trnh gii chia lm hai pha: pha 1 nhm gii bi ton M cho ti khi bin gi (x5) c a ra khi tp cc bin c s c c phng n cc bin xut pht cho BTQHTT vi cc rng buc (2.5) v pha 2 nhm tm phng n ti u cho bi ton ny.
33
4.3. Phng php n hnh hai pha T trc ti nay, chng ta lun gi s rng BTQHTT c xem xt lun c phng n v
c th bit c mt phng n (cc bin) ban u ca n khi to qu trnh gii. Trong mc ny chng ta s i xt cc trng hp khi cha bit BTQHTT c phng n hay khng, cng nh cha bit c phng n cc bin ban u. i vi nhng trng hp ny c th s dng phng php n hnh hai pha. Chng ta s trnh by phng php n hnh hai pha thng qua v d sau.
V d 8. Xt li v d 7.
Max z = 8x1 + 6x2, vi cc rng buc
1 2
1 2
1 2
4x 2x 602x 4x 48x ,x 0.
+ +
hay: Max z = 8x1 + 6x2 + 0x3 + 0x4, vi cc rng buc
1 2 3
1 2 4
1 2 3 4
4x 2x x 602x 4x x 48x ,x ,x ,x 0.
+ + = + =
Trc ht cn tr li cu hi BTQHTT dng chun tc trn y c phng n hay khng, nu c th cn tm mt phng n cc bin xut pht ca n.
Pha 1. Tm mt phng n cc bin xut pht bng cch xt BTQHTT sau y:
Min = x5, vi cc rng buc 1 2 3
1 2 4 5
1 2 3 4 5
4x 2x x 602x 4x x x 48x ,x ,x ,x ,x 0.
+ + = + + = (2.7)
Mc ch ca pha 1 l gii BTQHTT vi cc rng buc (2.7) hay cn gi l bi ton . Nu tm c phng n ti u ca bi ton vi cc bin gi u nhn gi tr bng 0 th iu ny chng t BTQHTT vi cc rng buc (2.5) c phng n. Trong trng hp d dng tm c mt phng n cc bin ca n (xem bng II.3).
Ti bng n hnh cui cng, ta thy j 0, j, nn phng n ti u t c vi x2 = 12, x3 = 36, x1 = x4 = x5 = 0 v min = 0. Do chng ta a ra kt lun l BTQHTT vi cc rng buc (2.5) c phng n x1 = 0, x2 = 12, x3 = 36, x4 = 0.
Nhn xt. Mt cch tng qut, c th khng nh c rng, nu bi ton c phng n ti u vi gi tr hm mc tiu l 0 th BTQHTT ban u c phng n, trong trng hp tri li th n khng c phng n. Nu bi ton c gi tr ti u min = 0, th ta c ngay phng n cc bin xut pht cho BTQHTT ban u v chuyn sang pha 2 bng cch b cc ct c bin gi (cng nh cc hng ng vi bin c s l bin gi) v thay li cc h s hm mc tiu.
34
Bng II.3. Cc bng n hnh gii bi ton pha 1
0 0 0 0 1 H s hm mc tiu Bin c s Phng n
x1 x2 x3 x4 x5
0 1
x3 x5
60 48
4 2
2 4
1 0
0 1
0 +1
Hng 0 = 48 1 = 2 2 = 4 3 = 0 4= 1 5 = 1 Hng j 1 = 2 2 = 4 3 = 0 4 = 1 5 = 0
0 0
x3 x2
36 12
3 1/2
0 1
1 0
1/2 1/4
1/2 1/4
Hng 0 = 0 0 0 0 0 0 Hng j 0 0 0 0 1
Pha 2. Gii BTQHTT vi cc rng buc (2.5) cn c phng n cc bin va tm c pha 1 (xem bng II.4): Max z = 8x1 + 6x2 +0x3 + 0x4, vi cc rng buc
1 2 3
1 2 4
1 2 3 4
4x 2x x 602x 4x x 48x ,x ,x ,x 0.
+ + = + =
Nhn xt. Kt qu gii v d trn bng phng php n hnh hai pha cng ging vi kt qu t c khi gii bng phng php n hnh m rng. Tuy nhin, khi s dng phng php n hnh hai pha, chng ta trnh c s phin phc trong vic khai bo gi tr dng ln ca tham s M nh trong phng php n hnh m rng.
Bng II.4. Cc bng n hnh gii bi ton pha 2
8 6 0 0 H s hm mc tiu
Bin c s Phng n x1 x2 x3 x4
0 6
x3 x2
36 12
3 1/2
0 1
1 0
1/2 1/4
Hng z z0 = 72 z1 = 3 z2 = 6 z3 = 0 z4 =3/2
Hng j 1 = 5 2 = 0 3 = 0 4 = 3/2 0 6
x4 x2
72 30
6 2
0 1
1 1/2
1 0
Hng z 180 12 6 3 0
Hng j 4 0 3 0
Ti bng n hnh cui cng, ta thy j 0, j, nn phng n ti u t c vi x2 = 30, x4 = 72, x1 = x3 = 0 v zmax = 180.
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4.4. Phng php n hnh ci bin Bng II.5 l bng n hnh tng qut bc lp th k. hiu bng ny ch cn so snh n
vi cc bng n hnh ngay trong bng II.4 trn y hoc cc bng n hnh ca bng II.1. D dng nhn thy rng cc biu thc tnh ton u xoay quanh ma trn B1, trong B l ma trn c s bc k. chuyn sang bng n hnh bc lp th k+1 tip theo, cn tnh c 1nextB
,
vi Bnext l ma trn c s bc k + 1.
Bng II.5. Bng n hnh dng tng qut
NTc TBc
H s hm mc tiu Bin c s Phng n TNx
TBx
cB xB B1b B1N B1B
Hng z TBc B1N TBc B
1B
Hng j TNc TBc B1N TBc TBc B1B
Ni dung ca phng php n hnh ci bin (hay cn gi l phng php n hnh dng ma trn nghch o) l vic tnh 1nextB
c da vo cc thng tin cn thit v ti thiu nht c
c t B1. V vy cc thng tin cn thit, phi lu tr mi bc tm bng n hnh bc sau, l t hn nhiu so vi phng php n hnh thng thng. Chng ta trnh by ngn gn phng php n hnh qua v d 9.
V d 9. Min z = 3x1 2x2 + 0x3 + 0x4 + 0x5 + 0x6 vi cc iu kin rng buc
x1 + 2x2 + x3 = 6 2x1 + x2 + x4 = 8 x1 + x2 + x5 = 1 x2 + x6 = 2 x1, x2, x3, x4, x5, x6 0.
Xt bng n hnh bc 1 (bng II.6), ta c
B = [a3, a4, a5, a6] =
1 0 0 00 1 0 00 0 1 00 0 0 1
B1 = I, N = [a1, a2] =
=
1 22 1
1 10 1
TNc TBc B1N = [3, 2] [0, 0, 0, 0] I N = [3, 2] = [1, 2].
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Bng II.6. Bng n hnh bc 1
3 2 0 0 0 0 H s hm mc tiu Bin c s Phng n
x1 x2 x3 x4 x5 x6
0
0
0
0
x3
x4
x5
x6
6
8
1
2
1 2
2 1
1 1
0 1
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
Hng z 0 0 0 0 0 0
Hng j 3 2 0 0 0 0
Do 1 = 3, ta c th a bin x1 vo c s (k hiu j0 = 1 l ch s ct ca bin a vo c s). xc nh bin a ra khi c s, ta tnh xB = B1b = Ib =[6, 8, 1, 2]T. Sau tnh ct h
s tng ng vi ct xoay j0 = 1 xc nh c trn l = B1a1 = Ia1 = [1, 2, 1, 0]T = [1, 2, 3, 4]T. p dng quy tc t s dng b nht, xt cc t s 6/1, 8/2, 1/1 v 2/0. T s dng b nht l 8/2, ng vi ta th 2 nn cn a bin x4 ra khi c s. Vy ch s ca hng xoay l i0 = 2 ( y r(2) = 4, xem li thut ton n hnh mc 3.4, nn hng xoay l hng 2) v bin a ra khi c s l x4.
Ta i tm 1nextB . C th nhn thy rng Bnext = [a3, a1, a5, a6] = BV = [a3, a4, a5, a6]V, trong
V = [e1, , e3, e4], vi ei l vc t n v dng ct c ta th i l 1, cn l ct h s ng vi bin a vo c s. Trong v d trn, l ct h s ng vi bin x4. C th kim tra c:
Bnext =
1 1 0 0 1 0 0 0 1 1 0 00 2 0 0 0 1 0 0 0 2 0 00 1 1 0 0 0 1 0 0 1 1 00 0 0 1 0 0 0 1 0 0 0 1
=
V1 c th tm c t V theo quy tc: thay ct = [1, 2, 3, 4]T bi ct [1/2, 1/2, 3/2, 4/2]T = [1/2, 1/2, 1/2, 0]T. D dng kim tra c:
VV1 = 1 1 0 0 1 1 / 2 0 00 2 0 0 0 1 / 2 0 0
I0 1 1 0 0 1 / 2 1 00 0 0 1 0 0 0 1
= .
Mt cch tng qut hn c th kim nghim c rng:
37
VV1 = 1 1 2
2 2
3 3 2
4 4 2
1 0 0 1 / 0 00 0 0 0 1 / 0 0
I0 1 0 0 / 1 00 0 1 0 / 0 1
=
V1
=
1 2
2
3 2
4 2
1 / 0 00 1 / 0 00 / 1 00 / 0 1
.
Ta thy V1 l ma trn thu c t V bng cch thay ct 2 ca V (ct c ch s trng vi ch s ca hng xoay i0 = 2) bi ct mi, thu c bng cch ly tt c cc phn t ca ct 2 nhn vi 1/2, ring ta th i0 = 2 c thay bi 1/2.
T phn tch trn, chng ta nhn c cng thc tnh 1nextB = V1B1, trong
V1 c xc nh theo quy tc nht nh (ch cn tng qut ha quy tc bit). Vi v d 9, trong bng n hnh bc 1 chng ta c Bnext = [a3, a1, a5, a6] v:
1nextB = V1B1=
1 1 / 2 0 00 1 / 2 0 00 1 / 2 1 00 0 0 1
.
Sau y chng ta tm cch tm tt phng php n hnh ci bin di dng bng (xem bng II.7). Trc ht, xt bng n hnh bc 1 (bng II.6). Trong bng ny chng ta b i hng zj, b i cc ct tng ng vi cc bin ngoi c s x1 v x2 th c bng II.7. Cn thm vo mt hng mi TBc B
1 v mt ct mi c cc phn t u bng 0, tr phn t cui bng 1. Ngoi ra,
vit thm vo ct xoay ng vi bin s a vo c s. Lc u, ma trn c s B l ma trn n v nn B1 B. Xt ma trn 1B (c l ma trn B1 bao), thu c t B bng cch thm vo ct mi v cc phn t tng ng ca hng j.
Bng II.7. Bng n hnh ci bin bc 1 1B H s hm mc tiu cB Bin c s Phng n
B1 Ct mi Ct (x1)
0 0 0 0
x3 x4 x5 x6
6 8 1 2
1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
0 0 0 0
1 2
1 0
z = TB Bc x = 0 Hng TBc B
1 0 0 0 0 1 3
tm cc s gia hm mc tiu, ta ly 1 nhn vi hng cui ca ma trn B1 bao, ri li nhn vi cc ct tng ng a1 v a2 trong ma trn A (c l ma trn A m rng), thu c bng cch thm vo ma trn A hng cui l hng cT:
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A =
12
103
2 1 0 01 0 1 01 0 0 11 0 0 02 0 0 0
00010
[1,2] = [0, 0, 0, 0, 1]
12
103
21112
=[3, 2].
Tht vy, do 1 = c1 TBc B1a1 v 2 = c2 TBc B1a2 nn
[1, 2] = [c1, c2] TBc B1[a1, a2] = [ TBc B1, 1] 1 21 2
a ac c
= [3, 2].
Vy ct l ct ng vi bin x1, = B1a1 = Ia1 = [1, 2, 1, 0]T = [1, 2, 3, 4]T c 1 = 3. Vi ct xoay xc nh c, ta tm c hng xoay v phn t xoay theo quy tc thng thng. Sau xoay sang bng n hnh ci bin mi da trn phn t xoay tm c, ma trn B1 bc mi cng c th tm theo cc quy tc ca th tc xoay. Ring hng TBc B
1 c tnh bng cch ly ct cB nhn (theo kiu tch v hng) vi cc ct ca B1(xem bng II.8).
Bng II.8. Bng n hnh ci bin bc 2
1B H s hm mc tiu cB Bin c s Phng n B1 Ct mi
Ct (x2)
0 3 0 0
x3 x1 x5 x6
2 4 5 2
1 1/2 0 0 0 1/2 0 0 0 1/2 1 0 0 0 0 1
0 0 0 0
3/2 1/2 3/2 1
z = TB Bc x = 4/3 Hng TBc B
1 0 3/2 0 0 1 1/2
tm ct trong bng II.8, trc ht cn tm s gia hm mc tiu cho cc bin ngoi c s:
[2, 4] = [c2, c4] TBc B1[a2, a4] = [ TBc B1, 1] 2 42 4
a ac c
= [0, 3/2, 0, 0, 1]
21112
01000
=[1/2, 3/2].
Vy ta xc nh c bin a vo c s l bin x2 v ct in vo bng II.8:
= B1a2 = 1 1 / 2 0 0 2 3 / 20 1 / 2 0 0 1 1 / 20 1 / 2 1 0 1 3 / 20 0 0 1 1 1
= .
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Sau chuyn sang bc tip theo (xem bng II.9).
Bng II.9. Bng n hnh ci bin bc 3
1B H s hm mc tiu cB Bin c s Phng n
B1 Ct mi Ct
2
3
0
0
x2
x1
x5
x6
4/3
10/3
3
2/3
2/3 1/3 0 0
1/3 2/3 0 0
1 1 1 0
2/3 1/3 0 1
0
0
0
0
z = TB Bc x = 38/3 Hng TBc B
1 1/3 4/3 0 0 1
Chng ta i tnh cc s gia hm mc tiu ng vi cc bin ngoi c s:
[3, 4] = [c3, c4] TBc B1[a3, a4] = [ TBc B1, 1] 3 43 4
a ac c
= [1/3, 4/3, 0, 0, 1]
10000
01000
= [1/3, 4/3].
Vy phng n ti u tm c l x1 = 10/3, x2 = 4/3, x3 = 0, x4 = 0, x5 = 3, x6 = 2/3, vi gi tr nh nht ca hm mc tiu l zmin = 38/3.
Ch
Phng php n hnh ci bin cho php tnh ma trn nghch o ca ma trn c s
bc k+1 theo cng thc 1 1 1 1 1 1 1k 1 k k k k 1 1 1B V B ... V V ...V B
+ = = = . Hn na dng ca cc ma trn 1iV
,i cng rt n gin. Do c th thy, phng php n hnh ci bin gim c khi lng tnh ton kh nhiu khi so snh vi phng php n hnh.
C th p dng phng php hai pha cho phng php n hnh ci bin. Lc ny cc du hiu dng khng c g thay i: Nu pha 1 kt thc vi phng n ti u cha bin gi nhn gi tr dng th bi ton khng c phng n. Nu trong khi tin hnh pha 2, ta tm c ct xoay m khng tm c hng xoay th bi ton c hm mc tiu khng b chn. Bi ton s c phng n ti u nu pha 2 kt thc vi du hiu ti u (vi BTQHTT dng Min th du hiu ti
u l j 0, j). trnh by vn n gin, sau y chng ta pht biu thut ton n hnh ci bin mt cch s b cho trng hp bit mt phng n xut pht (BTQHTT dng Min).
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Khung thut ton n hnh ci bin
Bc khi to
Tm mt phng n cc bin ban u.
Xc nh cc bin c s xB, cc h s hm mc tiu tng ng cB. Xc nh ch s ca m bin c s: r(1), r(2), ..., r(m).
Tm ma trn c s B ng vi cc ct vi ch s: r(1), r(2), ..., r(m), ma trn nghch o B
1, ma trn bao 1B vi T 1Bc B l hng cui ca ma trn bao.
Thit lp ma trn m rng A = [ N , B ] v tnh cc s gia hm mc tiu ng vi cc
bin ngoi c s theo cng thc: N = TNc TBc B1N = [ TBc B1, 1] N . t k := 1.
Cc bc lp (bc lp th k)
Bc 1: Kim tra iu kin dng.
Nu N 0 th bi ton c phng n ti u, ghi li kt qu v chuyn sang bc 3. Nu tri li, tn ti j JN sao cho j < 0 th chn xj l bin a vo c s. Thit lp ct = B1aj. Tm hng xoay bng quy tc t s dng b nht. Nu khng
chn c hng xoay (khi 0) th bi ton c hm mc tiu khng b chn di, ghi li kt qu v chuyn sang bc 3.
Bc 2:
Chn c hng i lm hng xoay, a bin xr(i) ra khi c s v tm ch s ca bin c s mi a vo r(i) := j. Xc nh li xB v cB, B v N.
Thc hin th tc xoay tnh li B1, tnh li T 1Bc B v ma trn bao 1B . Tnh cc s gia
hm mc tiu ng vi cc bin ngoi c s theo cng thc N = TNc TBc B1N = [ TBc B1, 1] N . t k := k + 1, sau quay v bc 1.
Bc 3: Dng v in ra kt qu.
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Bi tp chng II
Bi 1. Xt BTQHTT dng Max:
Max z = 6x1 + 4x2
vi cc iu kin rng buc
2x1 + 3x2 100 4x1 + 2x2 120 x1, x2 0.
a. Hy gii bi ton bng phng php th. b. Hy gii bi ton bng phng php n hnh. c. Minh ha ngha kinh t ca bi ton trong mt tnh hung thc t.
Bi 2. Xt BTQHTT dng Min:
Min z = 3x1 x2
vi cc iu kin rng buc
x1 2x2 4 x1 + x2 8 4x1 + 2x2 20 4 x1 8 x2 4 x1, x2 0.
a. Hy gii bi ton bng phng php th. b. Hy gii bi ton bng phng php n hnh.
Bi 3. Xt BTQHTT dng Max:
Max z = 5x1 + x2 + 6x3 + 2x4
vi cc iu kin rng buc
4x1 + 4x2 + 4x3 + x4 44 8x1 + 6x2 + 4x3 + 3x4 36 x1, x2, x3, x4 0.
a. Hy gii bi ton bng phng php n hnh. b. Gii bi ton bng phng php n hnh ci bin. c. Gii bi ton bng phn mm Excel hay phn mm Lingo.
42
Bi 4. Xt BTQHTT dng Min:
Min z = 2x1 + x2 x3 x4
vi cc iu kin rng buc
x1 x2 + 2x3 x4 = 2
2x1 + x2 3x3 + x4 = 6
x1 + x2 + x3 + x4 = 7
x1, x2, x3, x4 0. a. Hy gii bi ton bng phng php n hnh m rng (phng php M). b. Gii bi ton bng phn mm Excel hay phn mm Lingo.
Bi 5. Xt BTQHTT dng Min:
Min z = 3x1 + 2x2 + 8x3
vi cc iu kin rng buc
4x1 3x2 + 12x3 12 x1 + 4x3 6 x2 x3 = 2
x1, x2, x3 0. a. Hy a bi ton v dng chnh tc. b. Hy gii bi ton bng phng php n hnh m rng (phng php M). c. Hy gii bi ton bng phng php n hnh hai pha. d. Gii bi ton bng phng php n hnh ci bin. e. Gii bi ton bng phn mm Excel hay phn mm Lingo.
Bi 6. Xt BTQHTT dng Max:
Max z = x1 + x2
vi cc iu kin rng buc
x1 + x2 + x3 = 1
x1 2x2 + x4 = 0
x1 + 2x2 + x5 = 3
x1, x2, x3, x4, x5 0. Xt phng n (0, 1, 0, 2, 1)T. a. Tm ma trn c s B tng ng vi phng n.
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b. Hy vit cng thc s gia hm mc tiu cho phng n trn v cho bit phng n cho c phi l phng n ti u khng?
c. Nu phng n cho khng phi l phng n ti u, hy thc hin th tc xoay v cho bit ma trn c s bc tip theo. Tm s gia hm mc tiu tng ng.
d. Gii thch ti sao bi ton trn c hm mc tiu khng b chn trn?
Bi 7. Xt BTQHTT dng chnh tc. Gi s chng ta bit mt phng n ti u ca n l x*
v B l ma trn c s tng ng vi x*. Chng minh rng nu tn ti ch s j JN sao cho: cj cBB1aj = 0 th bi ton cho c v s phng n ti u. Hy chn mt v d n gin minh ha trng hp trn.
Bi 8. Hy kim tra li kt qu ca v d 3 chng I (Bi ton quy hoch s dng t trn a bn
x ng D, huyn Gia Lm, tnh H Ni) bng phn mm Excel hay Lingo. Bi 9. Hy lp chng trnh my tnh bng ngn ng Pascal hay ngn ng C gii BTQHTT
dng chnh tc theo thut ton n hnh gii BTQHTT c pht biu ti mc 3.4 ca chng II.
Bi 10. Hy pht biu thut ton hai pha v lp chng trnh my tnh bng ngn ng Pascal hay
ngn ng C gii BTQHTT dng tng qut. Chy kim th chng trnh trn mt s v d bit.
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Chng III Bi ton i ngu v mt s ng dng
1. Pht biu bi ton i ngu
1.1. Pht biu bi ton Tng ng vi mi BTQHTT (cn gi l bi ton gc) c mt bi ton i ngu. Bi ton
i ngu ca BTQHTT cng l mt BTQHTT. Nh vy, bi ton gc v bi ton i ngu ca n lp thnh mt cp BTQHTT, tnh cht ca bi ton ny c th c kho st thng qua bi ton kia. Nhiu quy trnh tnh ton hay phn tch c hon thin khi xem xt cp bi ton trn trong mi lin quan cht ch ca chng, mang li li ch trong vic gii quyt cc vn pht sinh t thc t. Vi mc ch tm hiu bc u, chng ta xt bi ton gc l bi ton quy hoch tuyn tnh (BTQHTT) dng Max vi cc rng buc ch c du v cc bin u tho mn iu kin khng m.
Bi ton gc
Max z = c1x1 + c2x2 + .... + cnxn
vi cc iu kin rng buc
a11x1 + a12x2 + ... + a1nxn b1 a21x1 + a22x2 + ... + a2nxn b2 ...
am1x1 + am2x2 + ... + amnxn bm x1, x2, ..., xn 0.
Lc BTQHTT sau y c gi l bi ton i ngu ca BTQHTT trn.
Bi ton i ngu
Min u = b1y1 + b2y2 + .... + bmym
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vi cc iu kin rng buc:
a11y1 + a21y2 + ... + am1ym c1 a12y1 + a22y2 + ... + am2ym c2 ...
a1ny1 + a2ny2 + ... + amnym cn y1, y2, ..., ym 0.
Cc bin y1, y2, ..., ym c gi l cc bin i ngu. Trong trng hp ny, do bi ton gc c m rng buc, nn bi ton i ngu c m bin i ngu. Bin i ngu yi tng ng vi rng buc th i ca bi ton gc.
1.2. ngha ca bi ton i ngu V d 1. Xt bi ton gc
Max z = 2x1 + 4x2 + 3x3
vi cc rng buc
3x1 + 4x2 + 2x3 60 2x1 + x2 + 2x3 40 x1 + 3x2 + 2x3 80 x1, x2, x3 0.
Cn tm cc gi tr ca cc bin quyt nh x1, x2, x3 cc rng buc c tho mn v hm mc tiu t gi tr ln nht.
Bi ton ny c ngha kinh t nh sau: Gi s mt x nghip sn xut ba loi sn phm I, II v III. sn xut ra mt n v sn phm I cn c 3 n v nguyn liu loi A, 2 n v nguyn liu loi B v 1 n v nguyn liu loi C. Cc ch tiu cho mt n v sn phm loi II l 4, 1 v 3. Cn cho n v sn phm loi III l 2, 2 v 2. Lng nguyn liu d tr loi A v B hin c l 60, 40 v 80 (n v). Hy xc nh phng n sn xut t li nhun ln nht, bit li nhun / n v sn phm bn ra l 2, 4 v 3 (n v tin t) cho cc sn phm loi I, II v III.
Gi s c mt khch hng mun mua li cc n v nguyn liu loi A, B v C. Bi ton t ra l cn nh gi cc n v nguyn liu. R rng rng gi cc nguyn liu c quy nh bi gi tr ca sn phm m chng to nn. Nu cc sn phm ny mang li li nhun ln trn th trng th gi c nh ca cc nguyn liu ny phi cao, cn nu tri li th gi c nh ca chng l thp. Mt khc, li nhun ca cc sn phm thu c trn th trng li ph thuc vo nhiu yu t nh: gi c cc sn phm c bn trn th trng ( c th trng chp nhn), lng d tr nguyn liu hin c, h s chi ph sn xut
Nh vy, gi c nh ca cc nguyn liu A, B v C ph thuc vo: H s hm mc tiu ca bi ton gc: c1 = 8, c2 = 4 v c3 = 63. Ma trn rng buc cc h s chi ph sn xut:
46
A = 3 4 22 1 21 3 2
.
H s d tr cc loi nguyn liu:
b = 604080
.
Tuy nhin, mi ph thuc khng d dng xc nh c. gii quyt vn ny hon ton c th da vo vic phn tch bi ton i ngu. Gi y1 l gi c nh mt n v nguyn liu loi A, y2 l gi c nh mt n v nguyn liu loi B, cn y3 l gi c nh mt n v nguyn liu loi C (y1, y2, y3 0).
Chng ta hy i xt bi ton i ngu:
Min u = 60y1 + 40y2 + 80y3
vi cc iu kin rng buc
3y1 + 2y2 + y3 2 4y1 + y2 + 3y3 4 2y1 + 2y2 + 2y3 3 y1, y2, y3 0.
Tht vy, u = 60y1 + 40y2 + 80y3 chnh l tng chi ph phi b ra nu ngi khch hng mun mua 60 n v nguyn liu loi A, 40 n v nguyn liu loi B v 80 n v nguyn liu loi C. Tt nhin ngi khch hng mun tng chi ph u cng b cng tt.
Xt rng buc th nht. V tri l 3y1 + 2y2 + y3 chnh l s tin khch hng phi b ra mua 3 n v nguyn liu loi A, 2 n v nguyn liu loi B v 1 n v nguyn liu loi C. y l s nguyn liu cn thit sn xut ra mt n v sn phm loi I. R rng rng, ngi khch hng khng th mua c s nguyn liu ny thp hn li nhun m mt n v sn phm loi A mang li khi c bn ra trn th trng (2 n v tin t). iu ny dn n rng buc th nht 3y1 + 2y2 + y3 2. Tng t chng ta c th lp lun c ngha kinh t ca rng buc th hai cng nh rng buc th ba ca bi ton i ngu.
1.3. Quy tc vit bi ton i ngu tng qut
Xt cp bi ton gc v bi ton i ngu trong v d 1 c cho trong bng III.1.
Nhn xt. BTG l bi ton Max BTN l bi ton Min. Cc h s hm mc tiu ca BTG Cc h s v phi ca BTN. Cc h s v phi ca BTG Cc h s hm mc tiu ca BTN. Ma trn h s ca BTG l A Ma trn h s ca BTN l AT. Bin 0 ca BTG (3.2) Rng buc ca BTN (3.2).
47
Rng buc BTG (3.1) Bin 0 ca BTN (3.1). Bng III.1. Cp bi ton gc v bi ton i ngu
Bi ton gc (BTG) Bi ton i ngu (BTN)
Max z = 2x1 + 4x2 + 3x3 vi cc rng buc:
3x1 + 4x2 + 2x3 60 2x1 + x2 + 2x3 40 (3.1) x1 + 3x2 + 2x3 80 x1, x2, x3 0 (3.2)
Min u = 60y1 + 40y2 + 80y3 vi cc rng buc:
3y1 + 2y2 + y3 2 4y1 + y2 + 3y3 4 (3.2) 2y1 + 2y2 + 2y3 3 y1, y2, y3 0 (3.1)
T cc nhn xt trn y, chng ta xem xt cc quy tc vit bi ton i ngu cho mt BTQHTT dng tng qut.
Xt bi ton gc l BTQHTT dng tng qut sau y:
z = c1x1 + c2x2 + .... + cnxn Max vi cc iu kin rng buc:
a11x1 + a12x2 + ... + a1nxn b1 a21x1 + a22x2 + ... + a2nxn b2 ...
am1x1+ am2x2 + ... + amnxn bm x1 0, x2 0, ..., xn 0 .
Trong , k hiu c th hiu l , hoc = i vi cc rng buc. i vi iu kin v du ca cc bin, k hiu 0 c th hiu l 0, 0 hoc c du tu .
Sau y l cc quy tc vit bi ton i ngu tng qut: Quy tc 1: BTG l bi ton Max BTN l bi ton Min. Quy tc 2: Cc h s hm mc tiu ca BTG Cc h s v phi ca BTN. Quy tc 3: Cc h s v phi ca BTG Cc h s hm mc tiu ca BTN. Quy tc 4: Ma trn h s ca BTG l A Ma trn h s ca BTN l AT. Quy tc 5: Bin 0 ca BTG Rng buc ca BTN. Bin 0 ca BTG Rng buc ca BTN. Bin c du tu ca BTG Rng buc = ca BTN. Quy tc 6: Rng buc BTG Bin 0 ca BTN. Rng buc BTG Bin 0 ca BTN. Rng buc = BTG Bin c du tu ca BTN.
48
Ch . Cc quy tc vit bi ton i ngu tng qut trn y c p dng khi bi ton gc cho l BTQHTT dng Max. Trong mc 1.4 (tnh cht 1) ngay tip theo, chng ta s m rng cc quy tc ny cho BTQHTT dng Min. Bng III.2 sau y cho bit cch vit bi ton i ngu trong mt trng hp c th.
Bng III.2. Vit bi ton i ngu cho bi ton gc dng Max
Bi ton gc (BTG) Bi ton i ngu (BTN)
Max z = 2x1 + 4x2 + 3x3 vi cc rng buc:
3x1 + 4x2 + 2x3 60 2x1 + x2 + 2x3 = 40 (3.3)
x1 + 3x2 + 2x3 80 x1 0, x2 0, x3 du tu . (3.4)
Min u = 60y1 + 40y2 + 80y3 vi cc rng buc:
3y1 + 2y2 + y3 2 4y1 + y2 + 3y3 4 (3.4) 2y1 + 2y2 + 2y3 = 3
y1 0, y2 du tu , y3 0. (3.3)
1.4. Cc tnh cht v ngha kinh t ca cp bi ton i ngu Trong phn ny chng ta s nghin cu cc tnh cht ca cp bi ton i ngu c
pht biu mc 1.1 v ngha kinh t ca chng thng qua mt v d n gin. V d 2. Xt li cp bi ton gc v bi ton i ngu trong v d 1 (bng III.1). Tnh cht 1. Bi ton i ngu ca bi ton i ngu li chnh l bi ton gc. Tnh cht ny c th c chng minh mt cch tng qut. Tuy nhin, trnh by vn
n gin, hy xt bi ton gc sau:
Max z = 2x1 + 4x2 + 3x3
vi cc rng buc
3x1 + 4x2 + 2x3 60 2x1 + x2 + 2x3 40 x1 + 3x2 + 2x3 80 x1, x2, x3 0.
Lc , bi ton i ngu l:
Min u = 60y1 + 40y2 + 80y3
vi cc iu kin rng buc:
3y1 + 2y2 + y3 2 4y1 + y2 + 3y3 4 2y1 + 2y2 + 2y3 3 y1, y2, y3 0.
49
hay:
Max t = 60y1 40y2 80y3
vi cc iu kin rng buc
3( y1) + 2( y2 ) + ( y3) 2 4( y1) + ( y2 ) + 3( y3) 4 2( y1) + 2( y2 ) + 2( y3) 3 y1, y2, y3 0.
Chng ta i tm bi ton i ngu cho BTQHTT trn theo cc quy tc bit, vi cc bin i ngu c k hiu l x1, x2 v x3.
Min v = 2x1 4x2 3x3
vi cc rng buc
3x1 4x2 2x3 60 2x1 x2 2x3 40 x1 3x2 2x3 80 x1, x2, x3 0.
t z = v, d thy rng y chnh l bi ton gc cho ban u:
Max z = 2x1 + 4x2 + 3x3
vi cc rng buc:
3x1 + 4x2 + 2x3 60 2x1 + x2 + 2x3 40 x1 + 3x2 + 2x3 80 x1, x2, x3 0.
Bng III.3. Vit bi ton i ngu cho bi ton gc dng Min
Bi ton gc (BTG) Bi ton i ngu (BTN)
z = 60x1 + 40x2 + 80x3 Min vi cc iu kin rng buc:
3x1 + 2x2 + x3 2 4x1 + x2 + 3x3 4 (3.5) 2x1 + 2x2 + 2x3 = 3
x1 0, x2 du tu , x3 0. (3.6)
u = 2y1 + 4y2 + 3y3 Max vi cc rng buc:
3y1 + 4y2 + 2y3 60 2y1 + y2 + 2y3 = 40 (3.6)
y1 + 3y2 + 2y3 80 y1 0, y2 0, y3 du tu . (3.5)
Tnh cht 1 khng nh vai tr bnh ng ca bi ton gc v bi ton i ngu. Bi vy, c
th gi cc BTQHTT ny l cp bi ton i ngu (m khng cn phi phn bit u l bi ton
50
gc, cn u l bi ton i ngu). Hn na, c th b sung vo cc quy tc vit bi ton i ngu
nh trong nhn xt sau y.
Nhn xt. Cc quy tc vit bi ton i ngu tng qut mc 1.3 cng c th c theo
chiu ngc li. Chng hn, quy tc 1 cng c th c hiu l BTG l bi ton Min BTN l bi ton Max. i vi cc quy tc khc cng c iu tng t (v d minh ha trong bng
III.3).
Tnh cht 2. Vi mi phng n x ca bi ton gc (bi ton Max) v vi mi phng n y
ca bi ton i ngu (bi ton Min), ta lun c z(x) u(y). Tip tc xt v d 2 minh ho tnh cht ny. Bng III.4 sau y cho bit phng n ti
u ca bi ton gc (sau khi a bi ton gc v dng chnh tc bng cch s dng 3 bin b
thiu x4, x5 v x6). Cn bng III.5 trnh by kt qu gii bi ton i ngu bng phng php
n hnh m rng (sau khi thm vo ba bin b tha y4, y5, y6 v ba bin gi y7, y8, y9).
Bng III.4. Phng n ti u ca bi ton gc
c1 = 2 c2 = 4 c3 = 3 c4 = 0 c5 = 0 c6 = 0 H s cj Bin c s Phng n
x1 x2 x3 x4 x5 x6
4
3
x2
x3
6 23
16 23
1/3
5/6
1
0
0
1
1/3
1/6
1/3
2/3
0
0
0 x6 26 23 5/3 0 0 2/3 1/3 1
Hng z 76 23 23/6 4 3 5/6 2/3 0
Hng j 11/6 0 0 5/6 2/3 0
Tnh cht 2 c th c minh ho trong hai bng III.4 v III.5. Vi mi phng n x ca bi ton gc v mi phng n y ca bi ton i ngu ta u c z(x) 76 23 u(y).
V mt ngha kinh t, c th lp lun l gii tnh cht ny nh sau: Vi mi phng n nh gi nguyn liu th tng chi ph (pha mun mua) phi b ra mua cc n v nguyn liu khng bao gi thp hn c tng li nhun mang li khi dng cc n v nguyn liu sn xut ra sn phm v tiu th chng trn th trng. Tht vy, z(x) = 60x1 + 40x2 + 80x3 chnh l tng li nhun mang li trong mt phng n sn xut. Cn u(y) = 2y1 + 4y2 + 3y3 l tng gi tr c nh ca ngun d tr nguyn liu c s dng trong cc phng n sn xut. R rng, mt phng n nh gi hp l ngun nguyn liu s phi tho mn u(y) z(x). Trong trng hp tng qut, chng ta c th thay cm t ngun d tr nguyn liu bi cm t ngun d tr ti nguyn c ngha tng qut hn.
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Bng III.5. Phng n ti u ca bi ton i ngu
60 40 80 0 0 0 M M M H s CB
Bin c s
B
Phng n yB
y1 y2 y3 y4 y5 y6 y7 y8 y9
M y7 2 3 2 1 1 0 0 1 0 0
M y8 4 4 1 3 0 1 0 0 1 0
M y9 3 2 2 2 0 0 1 0 0 1
Hng uj 9M 9M 5M 6M M M M M M M
Hng j 60 9M
40 5M
80 6M
M M M 0 0 0
60 y1 2/3 1 2/3 1/3 1/3 0 0 1/3 0 0
M y8 4/3 0 5/3 5/3 4/3 1 0 4/3 1 0
M y9 5/3 0 2/3 4/3 2/3 0 1 2/3 0 1
Hng uj 40+3M 60 40 M
20 +3M
20 +2M
M M 20 2M
M M
Hng j 0 M 60 3M
20 2M
M M 20 +3M
0 0
60 y1 1 1 1/4 3/4 0 1/4 0 0 1/4 0
0 y4 1 0 5/4 5/4 1 3/4 0 1 3/4 0
M y9 1 0 3/2 1/2 0 1/2 1 0 1/2 1
Hng uj 60+M 60 15+ 3M/2
45+ M/2
0 15 +M/2
M 0 15 M/2
M
Hng j 0 25 3M/2
35 M/2
0 15 M/2
M M 15+ 3M/2
0
60 y1 5/6 1 0 2/3 0 1/3 1/6 0 1/3 1/6
0 y4 11/6 0 0 5/3 1 1/3 5/6 1 1/3 5/6
40 y2 2/3 0 1 1/3 0 1/3 2/3 0 1/3 2/3
Hng uj 76 23 60 40 5313 0 6
23 16
23 0 6
23 16
23
Hng j 0 0 26 23 0 6 23 16 23 M M 6 23
M 16 23
Tnh cht 3. Nu tn ti hai phng n x* ca bi ton gc v y* ca bi ton i ngu sao cho z(x*) = u(y*) th x* chnh l phng n ti u ca bi ton gc, cn y* l phng n ti u ca bi ton i ngu. Ngc li, nu x* l phng n ti u ca bi ton gc, cn y* l phng n ti u ca bi ton i ngu th z(x*) = u(y*).
Tnh cht ny c minh ho r trong cc bng III.4 v III.5. Lc ny, z(x*) = u(y*) = 76 23 . V mt ngha kinh t, tnh cht ny ch ra rng tng chi ph thp nht phi b ra nu
52
mun mua cc n v nguyn liu (trong mt phng n nh gi ti u) chnh bng tng li nhun cao nht khi s dng cc n v nguyn liu (trong mt phng n sn xut ti u). Khng th tn ti mt phng n nh gi cho php tng gi c nh nh hn c tng li nhun ln nht.
Mt cch tng qut, gi tr cc ti nguyn ca mt cng ty c c nh da trn trnh t chc sn xut, trnh cng ngh v gi tr th trng ca cc sn phm m cc ti nguyn ny to nn ti thi im hin ti. Quy tc ny t ra c bit cn thit trong vic nh gi ti nguyn / ti sn ca mt cng ty. i vi cc cng ty lm n thua l th gi c nh cc ti nguyn thng kh thp, cn cc cng ty lm n pht t th gi c nh cc ti nguyn thng cao.
Tnh cht 4. Xt cp phng n ti u (x*, y*) ca cp bi ton i ngu. Nu mt iu kin rng buc hay iu kin v du c tho mn khng cht (khng xy ra du =) trong mt bi ton, th iu kin tng ng trong bi ton kia phi c tho mn cht (xy ra du =). Tnh cht ny cn c gi l tnh cht lch b: Nu trong mt iu kin xy ra lch dng th trong iu kin tng ng lch l bng 0.
Trc ht, chng ta hy minh ho tnh cht ny qua v d 2. T bng III.4 ta thy 1x = 0,
2x = 6 2
3, 3x
= 16 23
. Cn bng III.5 cho bit 1y = 5
6, 2y
= 23
, 3y = 0.
i vi bi ton gc ta c
3 1x + 4 2x
+ 2 3x
= 60 (tho mn cht) (3.7)
2 1x + 2x
+ 2 3x
= 40 (tho mn cht) (3.8)
1x + 3 2x
+ 2 3x
< 80 (tho mn khng cht) (3.9)
1x
= 0 (tho mn cht), (3.10)
2x = 6 2
3 > 0 (tho mn khng cht) (3.11)
3x = 16 2
3 > 0 (tho mn khng cht). (3.12)
Cn i vi bi ton i ngu ta c
3 1y + 2y
+ 3y
> 2 (tho mn khng cht) (3.10)
4 1y + 2y
+ 3 3y
= 4 (tho mn cht) (3.11)
2 1y + 2 2y
+ 2 3y
= 3 (tho mn cht) (3.12)
1y = 5
6 > 0 (tho mn khng cht), (3.7)
2y = 2
3 > 0 (tho mn khng cht), (3.8)
3y = 0 (tho mn cht). (3.9)
53
Chng ta i phn tch ngha kinh t ca cc cp iu kin tng ng. Xt cp iu kin tng ng: 1x
+ 3 2x + 2 3x
< 80 (3.9) tho mn khng cht nn 3y = 0 (3.9) tho mn cht.
iu ny c ngha l trong phng n sn xut ti u lng nguyn liu loi C cha c s dng ht. Do gi c nh ca cc n v d tha ra c coi l bng 0. Xt cp iu kin tng ng: 2x
= 6 23 > 0 tho mn khng cht (3.11) nn 4 1y + 2y
+ 3y = 4 tho mn cht
(3.11). iu ny c ngha l nu mt loi sn phm c a vo sn xut trong phng n sn xut ti u th tng gi c nh cc n v ca cc loi nguyn liu to nn mt n v sn phm loi ny chnh bng li nhun m n v sn phm mang li.
Ngc li, xt cp iu kin tng ng: 1y = 56 > 0 (3.7) tho mn khng cht nn 3 1x
+
4 2x + 2 3x
= 60 (3.7) tho mn cht. Nh vy, nu gi c nh ti u cho mi n v nguyn liu loi A l dng th iu ny chng t nguyn liu loi A ang c s dng ht (vt cn) trong mt phng n sn xut ti u. Cn khi xt cp iu kin tng ng: 3 1y
+ 2y + 3y
> 2
(3.10) tho mn khng cht nn 1x = 0 (3.10) tho mn cht. iu ny chng t rng, nu tng
gi c nh cc n v ca cc loi nguyn liu to nn mt n v sn phm loi no cao hn li nhun m mt n v sn phm loi ny mang li th loi sn phm ny khng c sn xut ra trong phng n sn xut ti u.
Tnh cht 5. Phng n ti u ca bi ton i ngu c th tm c trong bng n hnh ti u ca bi ton gc v ngc li.
Xt v d 2. Phng n ti u ca bi ton i ngu 1y = 56 , 2y
= 23 , 3y = 0 c th tm
c trong hng zj ca bng III.4 ng vi cc ct bin b x4, x5 v x6. iu ny c th c gii thch nh sau: Ti tnh hung phng n sn xut ti u, nu chng ta mun sn xut thm mt n v nguyn liu no (xem li chng II, mc 2.1), th phi b ra mt chi ph tng ng cho trong hng zj. chnh l gi c nh (bin) ca mi n v nguyn liu (cn gi l gi bng shadow price).
Tng t, phng n ti u ca bi ton gc 1x = 0, 2x
= 6 23
, 3x = 16 2
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