1. HC VIN CNG NGH BU CHNH VIN THNGTON KINH T (Dng cho sinh vin h o to i hc t xa) Lu hnh ni b H NI - 2007

2. HC VIN CNG NGH BU CHNH VIN THNG TON KINH TBin son : PGS.TS. NGUYN QUNGTS. NGUYN THNG THI 3. LI NI UNhm p ng nhu cu ging dy v hc tp mn hc Ton kinh t dnh cho sinh vin ho to i hc t xa, Hc vin Cng ngh Bu chnh Vin thng (Hc vin) t chc bin sontp Sch hng dn hc tp (Sch HDHT) mn hc Ton kinh t theo ng chng trnh o toC nhn ngnh Qun tr kinh doanh ca Hc vin.Tp sch c bin son trn c s k tha, chn lc b sung tp gio trnh Ton chuynngnh c Nh xut bn Bu in n hnh vo thng 9 nm 2003 v cc bi ging Ton kinht c s dng, ging dy cho chng trnh o to i hc chnh quy ngnh Qun tr Kinhdoanh ti Hc vin.Ni dung tp sch c cu trc gm 7 chng:Chng 1. Cc kin thc m u v phng php ti uChng 2. M hnh ti u tuyn tnhChng 3. Mt s m hnh ti u tuyn tnh khcChng 4. Cc bi ton ti u trn mng.Chng 5. Phng php m hnh ho v m hnh ton kinh t.Chng 6. L thuyt Phc v m ngChng 7. L thuyt qun l d tr. to iu kin thun li cho sinh vin c kh nng t hc, t nghin cu, cc tc gikhng i su vo cc vn l lun v k thut ton hc phc tp, m ch tp trung trnh by, giithiu nhng kin thc c bn ch yu thit thc v cp nht, lm c s cho vic hc tp nghincu phn tch kinh t ni chung v hc tp cc mn chuyn ngnh Qun tr kinh doanh. cuimi chng, sau phn khi qut v tm tt cc vn c bn, ch yu ca l thuyt, cc tc gia ra cc bi tp mu v phn tch cch gii ngi hc c th t gii c nhng bi ton linquan n l lun hc. Phn bi tp cui mi chng cng s gip ngi hc t nghin cu, vndng cc l lun hc vo phn tch, l gii cc ni dung thc tin lin quan.Mc d cc tc gi u t nghin cu chn lc bin son nghim tc p ng yu cuging dy v hc tp ca mn hc, nhng chc tp sch s khng trnh khi nhng thiu st nhtnh. Cc tc gi rt mong nhn c s gp ca bn b ng nghip, bn c v cc bn sinhvin ln xut bn sau c hon thin hn. CC TC GI 4. Chng I: Mt s kin thc m u CHNG I: MT S KIN THC M U1.1. I TNG NGHIN CU CA MN HC1.1.1. Tng quan v ti u ho.Trong hot ng thc tin, nht l trong qu trnh qun l, iu khin h thng kinh t - xhi, chng ta lun mong mun t c kt qu tt nht theo cc tiu chun no . Tt c nhngmong mun thng l li gii ca nhng bi ton ti u no . Mi vn khc nhau cathc t dn n cc bi ton ti u khc nhau. gii cc bi ton , mt lot cc l thuyt tonhc ra i t c s l lun, a ra cc gii php tm li gii, chng minh tnh hi t, tnhkh thi ca cc bi ton thc t v.v. T hnh thnh mt lp cc phng php ton hc gip tatm ra li gii tt nht cho cc bi ton thc t, gi l cc phng php ti u ha. Lp ccphng php ti u ha bao gm nhiu l thuyt ton hc khc nhau, tiu biu l: Qui hoch tonhc, l thuyt tr chi, l thuyt th v.v. Trong qui hoch ton hc, tiu biu l Qui hoch tuyn tnh, Qui hoch phi tuyn, Quihoch ng, Quy hoch tham s, Qui hoch nguyn v.v.Trong l thuyt tr chi, tiu biu l L thuyt la chn quyt nh, Bi ton tr chi chinlc, bi ton tr chi vi phn v.v. Trong L thuyt th c cc bi ton ti u trn mng, biton PERT, Cc bi ton ng i v.v.Cc lp phng php ton hc thuc L thuyt ti u c th biu din bi s sau:L thuyt ti u Cc phng php ti uM hnh ti u .......... M M .....Quy LLM hnhhnh hochthuytthuyt hnhphcqun lton th tr chiton v md tr hckinh tng1 12 3Quy hoch ton hc Quy Quy QuyQuy .....hoch hoch phi hochhochtuyn tuyn ngtham s tnh3 5. Chng I: Mt s kin thc m u3L thuyt tr chi Bi tonBi ton Bi ton..... la chntr chi tr chi quyt chinvi phn nhlc1.1.2. Bi ton ti u tng qut.Bi ton quy hoch ton hc tng qut c pht biu nh sau:Cc i ha (cc tiu ha) hm f (x) max (min)(1.1)Vi cc iu kin: gi (x) (=, ) bi (i = 1, m )(1.2)x X. IRn .(1.3)Hm f (x) cho (1 -1) gi l hm mc tiu.Cc hm gi (x) (i = 1, m ) gi l hm rng buc.Tp hp D = {x X | gi (x) (=, ) bi, i = 1m} ,(1.4)Gi l min rng buc chp nhn c. - Mi mt bt ng thc, ng thc trong (1.2) gi l mt rng buc ca bi ton (1.1) -(1.2) - (1.3)- im x = (x1, x2, ..., xn) D gi l mt phng n ca bi ton (1.1) - (1.2) - (1.3) hay lmt gii php chp nhn c. - Mt phng n x* D lm cc i (cc tiu) hm mc tiu gi l phng n ti u (hayli gii hoc phng n tt nht).Theo nh ngha trn th x* D l phng n ti u khi v ch khif (x*) f (x), x D, (i vi bi ton max) hayf (x*) f(x), x D, (i vi bi ton min).Gi tr f(x*) gi l gi tr ti u (tt nht) ca hm mc tiu, hay l gi tr ti u ca biton (1.1) - (1.2) - (1.3).1.1.3. Phn loi cc bi ton ti u. a - Nu hm mc tiu f(x) v cc rng buc gi (x) l hm tuyn tnh (bc 1) th bi ton (1.1)- (1.2) - (1.3) gi l mt Qui hoch tuyn tnh . (trng hp ring l bi ton vn ti).b - Nu biu thc hm mc tiu f(x) v cc rng buc gi (x) (i = 1, m ) l hm ph thuctham s, th bi ton (1.1) (1.3) gi l qui hoch tham s.4 6. Chng I: Mt s kin thc m uc - Nu bi ton (1.1) (1.3) c xt trong qu trnh nhiu giai on hoc trong qu trnhthay i theo thi gian th gi l Qui hoch ng. d - Nu bi ton (1.1) (1.3) m hm mc tiu f(x) hoc c t nht mt trong cc hm gi(x), (i = 1, m ) l phi tuyn th gi l Qui hoch phi tuyn, trng hp ring l Qui hoch li hocQui hoch lm.Qui hoch li (lm) l Qui hoch ton hc m hm mc tiu f(x) l li (lm) trn tp hpcc rng buc D li (lm). e - Nu bi ton (1.1) (1.3) m min rng buc D l tp ri rc th gi l Qui hoch rirc.g - Nu bi ton(1.1) (1.3) c cc bin xi IR1 l thnh phn i trong vc t x X IRn,ch nhn cc gi tr nguyn, th gi l Qui hoch nguyn. h - Nu bi ton (1.1) (1.3) m cc bin xi IR1 ch nhn cc gi tr O hoc 1, gi l Quihoch Bul (xi l thnh phn i ca vc t x). i - Nu bi ton (1.1) (1.3) m trn min D ta xt ng thi nhiu mc tiu khc nhau,gi l Qui hoch a mc tiu v.v.1.1.4. Ni dung nghin cu ca mn hc. a. Quy hoch tuyn tnh. b. Bi ton vn ti. c. Bi ton ti u trn mng. d. M hnh kinh t v m hnh ton kinh t. e. M hnh phc v m ng. g. M hnh qun l d tr.1.2. C S GII TCH LI.1.2.1. Khng gian tuyn tnh n chiu (Rn). a. Vc t n chiu. Mt h thng c sp , gm n s thc, dng x = (x1 x2, ..., xn), gi l mt vc t n chiu. Th d: x = (4, 0, 5, 10, 15) l mt vc t 5 chiu. Cc s xi, i = 1, n , gi l thnh phn th i ca vc t x. Hai vc t x =(x1, x2, ..., xn) v (y1, y2, ..., yn) gi l bng nhau, nu xi = yi, (i = 1, n ). Khi ta vit x y. Vy x y xi =yi, (i = 1, n ). Cho hai vc t x = (x1, x2, ..., xn) y = (y1, y2, ..., yn) v R1. Ta nh ngha php cng hai vc t x v y l vc t x+y, c xc nh nh sau: x+y= (x1+ y1, x2 + y2, ..., xn + yn)(1.5)1 Php nhn vc t x vi mt s R l vc t x, c xc nh nh sau:5 7. Chng I: Mt s kin thc m u x = (x1, x2, ..., xn) (1.6) - Vc t = (0, 0, ....., 0) gm cc thnh phn ton l s 0, gi l vc t khng. * Cc tnh cht ca php cng vct v nhn vct vi mt s. - Nu x v y l hai vct n chiu th x+y cng l vc t n chiu. - Vi mi vc t n chiu x v y ta u c: x+y =y+x. - Vi mi vc t n chiu x, y v z ta u c: x + (y+z) = (x+y) +z. - Lun tn ti vct n chiu sao cho +x = x+ =x. - Mi vct n chiu x lun tn ti vc t n chiu -x sao cho: x+ (-x)=(-x) +x = - k R v vi mi vc t n chiu x th kx cng l vc t n chiu. - k R v vi mi vc t n chiu x v y ta c: k (x+y) = kx+ky. - l, k R v vi mi vc t n chiu x ta lun c: (k +l ) x = kx +lx. - l, k R v vi mi vc t n chiu x ta lun c: k(lx) = (kl) x. - Mi vc t n chiu ta lun c: 1.x = x. b. Khng gian tuyn tnh n chiu Rn. Tp hp tt c cc vc t n chiu, trong xc lp php ton cng Vc t v nhn vc tvi mt s thc nh (1.5) v (1.6) v tho mn 10 tnh cht nu trn, gi l mt khng gian tuyntnh n chiu. K hiu IRn.1.2.2. Mt s tnh cht i vi vc t trong Rn. a. nh ngha. Cc vc t xi Rn, i = 1, m , gi l c lp tuyn tnh num i =1 i xi = i = 0, i = 1, m .m - Nu tn ti t nht mt s j 0 , 1 j m, sao cho i =1i xi = , th ta ni rng ccvc t x Rn, i = 1, m , l ph thuc tuyn tnh. m - Nu tn ti vc t xi Rn, sao cho: x = i =1 ixi, vi t nht mt i 0, 1 i m, th x gi il t hp tuyn tnh ca cc vc t x , (i = 1, m ). m m - Nu x =i =1 i xi vi i 0, i = 1, m , v i =1i = 1 th x gi l t hp li ca cc vc txi, i = 1, m . - Trong khng gian vc t Rn, h n Vc t c lp tuyn tnh lp thnh c s ca IRn. Gi s C1, C2, ..., Cn l mt c s ca Rn, khi x Rn u c th biu din tuyn tnhmt cch duy nht qua cc Vc t c s. Ci, (i = 1, n ).6 8. Chng I: Mt s kin thc m u b. Cho hai vc t bt k x, y Rn, x = (x1, x2, ... xn) v y = (y1, y2, ...., yn) , ta gi tch vhng ca hai vc t x v y l mt s thc, k hiu l , c xc nh nh sau:m =i =1 xi yi .- di ca Vc t x Rn l s thc, k hiux , c xc nh nh saun x = < x, x > = x i2 i =1- Ch : Tch v hng hai vc t c cc tnh cht sau:b1, < x, y > = < y, x >. (Tnh giao hon) x, y Rn .b2, < x1+x2, y > = < x1, y > + < x2, y >, x1, x2, y Rn .(Tnh phn phi i vi php cng).b3, < x, y > = < x, y > , R1, x, y Rn .b4> < x, x > 0 x Rn, du bng xy ra khi x = .Vi mi x, y Rn, ta nh ngha khong cch gia hai vc t x, y, k hiu (x, y) l sthc, c xc nh nh sau: n ( x, y ) =x y = < x y, x y > = ( xi y i ) 2 .i =1 Ch : Khong cch gia hai vc t x, y Rn, chnh l di ca vc t hiu x+ (-1)y: = x- y. (Hiu ca hai Vc t).1.2.3. Khng gian clt.Mt khng gian tuyn tnh n chiu, trong xc nh php ton tch v hng, do xcnh mt khong cch gia hai vc t, gi l khng gian clt, k hiu IRn.1.2.4. Tp Compact.a. Cc nh ngha.Dy {xk} |Rn, gi l hi t n im xo IRn khi k, nu lim (xk, xo) = 0. Khi ta k ni {x } c gii hn l x khi k , v vit: lim x = x .k o k o k - Mt tp hp S = {xIR (x, a) r, a IRn, r IR1}, gi l mt hnh cu tm a, bn knhn:r trong IRn.- Hnh cu S ni trn, to thnh mt ln cn ca im a, gi l r -ln cn ca a.- Cho tp hp A IRn, im x A c gi l im trong ca A nu - ln cn ca xnm trn trong A.- im x A IRn, c gi l im bin ca A, nu mi ln cn ca x u c cha ccim thuc A v cc im khng thuc A.- Cho tp hp A IRn, ta ni tp hp A l gii ni nu hnh cu cha trn n, ngha l s thc r ln v im a IRn sao cho x A ta u c (x, a) < r.7 9. Chng I: Mt s kin thc m u* Nhn xt. T nh ngha ca dy hi t v tp gii ni, ta suy ra, mt dy {xk} IRn, hit bao gi cng gii ni. - Mt tp hp G IRn c gi l m, nux G, tn ti mt hnh cu tm x cha trntrong G. - Mt tp hp F IRn c gi l ng, nu nh mi dy hi t {xk} F IRn, u hi tn mt im xo F.* Nhn xt. Mt tp hp cha mi im bin ca n l mt tp hp ng.b. Tp Compact. - Tp hp C IRn c gi l tp hp Compct nu t mi dy v hn {xk} C, u cth trch ra mt dy con {xkn} hi t n mt phn t thuc C.- Mt tp C l Compact khi v ch khi C ng v gii ni.- Tp Compact M ca tp ng C cng ng trong C.- Tp con M ng C Compact cng l tp Compact.- Hm f(x) lin tc trn tp Compact C s t gi tr ln nht, nh nht trn C.1.2.5. ng thng, on thng, siu phng.a. nh ngha ng thng v on thng trong IRn.- Cho hai im a, b |Rn. Ta gi ng thng qua a, b l tp hp cc im x IRn cdng:x = a + (1 - )b, IR1- Nu 0 1 th ta c on thng ni hai im a, b, k hiu [a, b]. Ch - Trong khng gian hai chiu IR2, phng trnh bc nht ax + by = c, xc nh mtng thng, mt bt phng trnh ax+by c hoc ax+by c, xc nh na mt phng trong IRn. - Trong khng gian ba chiu IR3, mt phng trnh bc nht ax+by+cz=d xc nh mt mtphng, mt bt phng trnh bc nht ax+by+cz d hoc ax + by + cz d xc nh mt nakhng gian. Ta m rng kt qu trn cho khng gian IRn.b. Siu phng trong IRn . - Siu phng trong khng gian IRn l tp hp tt c cc im x = IRn, thomn phng trnh bc nht:a1 x1 + a2 x2 + ... + an xn = . n n- Mt bt phng trnh bc nht dng ai xi hoc ai xi xc nh mt na khngi =1i =1ngian ng trong IR .1.2.6. Tp hp li .a. nh ngha.Tp hp x IRn c gi l tp hp li nu cng vi vic cha hai im x, y, n cha con thng ni hai im y.iu ny c ngha l X = {z |Rn: z = a + b, a, b IRn, [0, 1]}8 10. Chng I: Mt s kin thc m uV d. C khng gian IRn, na khng gian |Rn, cc a gic trong |Rn, cc khong ,on [a, b] trong IR1... l cc tp hp li. xx yx yA yB CTp A: li Tp B v C: khng li.b. nh l 11.Giao ca hai tp hp li l tp hp li.Chng minh. Ly hai im bt k x, y A B x, y A v x, y B V A li nn [x, y] A.B li nn [x, y] B. => [x, y] A B. Vy A B li.H qu 1. Giao ca mt s bt k tp li l tp li.H qu 2. Tp hp cc nghim ca h bt phng trnh bc nht dng:a11x1 + a12x2 +........ + amxn b1a21x1 + a22x2 +........ + a2nxn b2--------------------------------------am1x1 + am2x2 +........+ amnxn bm,l mt tp hp li, gi l khc li a din, trong |Rn. Ch . Mt khc li a din gii ni gi l a din li, k hiu D. Giao ca cc tp hp licha D ta gi l bao li ca D. K hiu [D].c. im cc bin.nh ca a din li hoc khc li gi l im cc bin. R rng im cc bin x khng th l im trong ca on thng ni hai im no thuc D, ngha l khng th tn ti hai im x1, x2 D sao cho x= x1+(1- )x2, (0, 1).1.2.7. Hm li .a. nh ngha.Mt hm f(x), xc nh trn tp hp li C |Rn, c gi l hm li nu cp im x1,x2 C v s [0, 1] ta lun lun c:9 11. Chng I: Mt s kin thc m uf( x 1 + (1 ) x 2 ) f(x1) + (1 - ) f(x2)(1.7)Nu trong (1.7) xy ra du th hm f(x) gi l hm li cht. Nu trong (1.7) xy ra du th hm f(x) gi l hm lm, xy ra du > th hm f(x) gi lhm lm cht. f(x) f(x2) f(x1) + (1 -) f(x2) f(x1 + (1 - x2))f(x1)0xx x2 xCh . Nu hm f (x) li trn tp C IRn th hm - f (x) lm trn tp C, ngc li nu f (x)lm trn tp li C IRn th hm - f (x) li trn tp hp C.- Ta ni hm f(x) xc nh trn tp li C t cc tiu tuyt i ti x* C nu f(x*) f(x),xC, t cc i tuyt i ti x* c nu f(x*) f(x), x C.- Ta ni hm f (x) xc nh trn tp li C, t cc tiu a phng ti x*C nu ln cn Bca x* sao cho f(x) f(x), x B . - Ta ni hm f (x) xc nh trn tp li C, t cc i a phng ti x*C, nu ln cnB ca x* sao cho f(x) f(x), x B .b. nh l 1.2.Mi im cc tr a phng ca hm li trn tp hp li u l im cc tr tuyt i. Chng minh. Gi s x* l cc tiu a phng nhng khng cc tiu tuyt i trn tp Cli, nh vy x1 C sao cho f (x*) ) f(x1). Xt t hp li ca hai im x* v x1:X = x* + (1 - ) x1, 0 1.Nu = 0 th x x1. Khi o (0, 1) sao cho x B , vi [0, o) ly 1 (0, o)1ta c: x(1)= (1-1) x* + 1 x B .Do f li nn c f ((1-1) x*+1x1) (1-1) f (x*) +1 f(x1). ((1-1) f (x*) +1 f(x*) = f (x*), iu ny mu thun vi hm f (x*) t cc tiu aphng ti x*. T suy ra iu phi chng minh.H qu 1.Mi im cc i a phng ca hm lm trn tp hp li u l cc i tuyt i.- Ta gi o hm theo hng z ca hm f ti x l i lng: f (x + z) f (x)f (x, z) = lim , nu gii hn ny tn ti. 0 10 12. Chng I: Mt s kin thc m u c - B 1.1. Nu hm f (x) l hm li kh vi trn C li. Khi x C v vi mi z sao cho x+z Cth f (x, z) tn ti v nghim ng bt ng thc v ng thc sau: i) f (x, z) f (x +z) - f (x). nf ( x ) ii) f (x, z) = zi = < f(x), z >.i =1 x1 f ( x ) f ( x ) f ( x ) Trong : Vc t f (x) = x , x ,..., x gi l graient ca hm f(x) ti x, 1 2n z = (z1, z2... zn)1.2.8. Mt s tiu chun nhn bit hm li. Cho x, z IRn, t hm s () = f(x+z), [0, 1],(1.8) nh l 1.3. Hm f(x) l li trn IRn khi v ch khi hm s () l li vi [0, 1] v x, z |Rn . nh l 1.4.a. Hm f(x) kh vi trn IRn l li khi v ch khi x, z IRn cho trc, hm () = < f(x+ z), z > khng gim theo . b. Hm f(x) kh vi hai ln trn IRn l li khi v ch khi x, y IRn cho trc, dng tonphng < P(x) z, z > l xc nh khng m. Ch . Mt dng ton phng

l xc nh khng m khi v ch khi

0, z IRn . H qu 1.1 Mt hm bc hai dng f(x) = < c, x > +< Px, x >, trong P = (p ij)nxn l ma trn i2xng cp nxn, l mt hm li khi v ch khi ma trn P l xc nh khng m.Ch . ma trn P l xc nh khng m th iu kin cn v l tt c cc nh thccon chnh ca ma trn ny khng m, ngha l:a 11 a 12 ........ a 1 na 11a 12 a 21 a 22 ....... a 2 n 1 = a11 0 ;2 = 0, ..., n = 0a 21a 22.......... .......... ...a n 1 a n 2 ........ a nnBI TP CHNG I.Bi 1. Mt doanh nghip c 300 n v nguyn liu loi A, 500 n v nguyn liu loi Bv 200 n v nguyn liu loi C sn xut 4 loi sn phm I, II, III, IV. nh mc nguyn liucn thit v tin li ca sn xut cho bi bng 1. Hy lp k hoch sn xut ca x nghip trn saocho thu c li sut ln nht. Bng 111 13. Chng I: Mt s kin thc m uHng hoI II III IV Nguyn liu A: 300 12515 6 B: 500 14879 C: 280 1713 912 Li (n v tin)5 846Bi 2. Cn sn xut t nht 75 sn phm loi A, 58 sn phm loi B v 64 sn phm loi C.Ngi ta c th p dng 3 cch sn xut I, II, III, IV. Trong mt n v thi gian, nng sut v chiph ca tng cch sn xut cho bi bng 2. Bng 2 Cch sn xutIII IIILoi sn phmA 75367B 58593C 64284Chi ph (n v tin) 243 Hy lp k hoch sn xut sao cho chi ph nh nht m vn t c cc yu cu t ra.Bi 3. Mt Cng ty c ba x nghip cng loi: A, B, C c kh nng sn xut c 3 loisn phm: I, II, III. Bit rng nu u t mt n v tin vo x nghip A trong mt nm s snxut c 1200 sn phm loi I, 800 sn phm loi II v 1050 sn phm loi III. u t vo xnghip B mt n v tin, c 1000 sn phm loi I, 740 sn phm loi II, 900 sn phm loi III.u t vo x nghip C mt n v tin th sn xut c 1100 sn phm loi I, 600 sn phm loiII, 1000 sn phm loi III. nh mc tiu hao nguyn liu v lao ng ca mi x nghip trong snxut c cho bng 3. Nguyn liu, lao ng hng nm Cng ty c th cung cp cho sn xut baloi sn phm ny l 390.000 KG v 200.000 gi cng. Theo k hoch phi sn xut t nht l23.000 n v sn phm loi I, 18.000 n v sn phm loi II, v 21.000 n v sn phm loi III.Hy tm mt phng n u t sao cho thu c cc sn phm theo k hoch m vn u t tnht. Bng 3nh mc hao ph ng. liu (Kg/sn phm) v lao ng (g/sn phm) Doanh I II III nghipNg. liuLao ng Ng. liu Lao ng Ng. liu Lao ng A4 21048 4, 5 B 4, 2 3 9 4, 5 7, 8512 14. Chng I: Mt s kin thc m uC4, 5 2, 510, 558, 4 4 Bi 4. Mt x nghip qun i c 4 loi my: A, B, C, D, sn xut ra 6 loi sn phm I, II,III, IV, V, VI. S gi ca mi loi my sn xut mi loi sn phm v gi tin mi loi snphm ghi bng 4. Nng lc sn xut ca cc lmy u c hn, nu dng qu s b hng. Gi strong 1 tun, mi my loi A, B, C, D tng ng lm vic khng qu 850, 700, 100 v 900 gi.Hy lp mt phng n sn xut thu c sn phm mi loi ln nht m vn bo m an toncho my mc v thit b.Bng 4Sn phmLoiLoi LoiLoiLoiS gi snLoi I IIIIIIVV VIxut 1 sp trn my. A0, 01 0, 01 0, 010, 03 0, 03 0, 03 B0, 020, 05 C0, 020, 05 D0, 030, 08 Gi 1 sn phm (/v tin)0, 40 0, 28 0, 320, 72 0, 64 0, 60Bi 5. Mt my bay vn ti qun s c trng ti M. Cn ch n loi thit b bng my bay.Trng lng loi bu kin i, (i = 1, n ) l i, c gi tr i . Hy tm phng n ch mi loi thit bbao nhiu n v ln my bay trng lng tng cng khng vt qu ti trng ca my bay mt c tng gi tr ln nht ? (Bi ton Qui hoch nguyn). 13 15. Chng II: Quy hoch tuyn tnhCHNG II: QUY HOCH TUYN TNH2.1. MT S BI TON THC T DN TI M HNH QUY HOCHTUYN TNH2.1.1. Bi ton lp k hoch sn xut.Gi s mt Cng ty sn xut n loi sn phm v phi s dng m loi nguyn liu khc nhau.Gi xj l sn lng sn phm loi j, (j = 1, n ) m Cng ty s sn xut, cj l tin li (hay gi) mtn v sn phm loi j, aij l chi ph nguyn liu loi i, (i = 1, m ), sn xut ra mt n v snphm loi j, bi l lng nguyn liu loi i ti a c th c.Trong cc iu kin cho, hy xc nh sn lng xj, j = 1, n sao cho tng tin li (haytng gi tr sn lng hng ho) l ln nht vi s nguyn liu hin c.Bi ton thc tin trn, c th m hnh ton hc nh sau:Tm x = (x1, x2, ..., xn) IRn , lm cc i hm mc tiu: nf(x) =j =1cj xj maxvi cc iu kin: nj =1aij xj bi, i = 1, m ,xj 0, j = 1, nBi ton trn l mt bi ton Qui hoch tuyn tnh.2.1.2. Bi ton vn ti.C m kho hng cng cha mt loi hng ho, Ai , i = 1, m (Ai im pht th i). Lng hng kho Ai l ai, (i = 1, m ). C n a im tiu th hng Bj, nhu cu tiu th im Bj l bj, j = 1, n(Bi im thu th i). Bit rng cc ph vn chuyn mt n v hng ho t im pht Ai n imthu Bj l cij. Hy lp k hoch vn chuyn hng ho t cc a im pht n cc a im thuhng sao cho tng chi ph vn chuyn l nh nht.Nu ta k hiu xij l lng hng vn chuyn t im pht Ai, (i = 1, m ) n im thu Bj, vi(j = 1, n ), th ta c th m hnh ton hc bi ton thc t nh sau:Tm vc t x= (x1, x2,..., xn+m) IRnxm ,sao cho:mnF(x) = i =1 j =1 cij xij minvi cc iu kin:14 16. Chng II: Quy hoch tuyn tnh nj =1 xij = ai, i = 1, m mi =1 xij = bi, j = 1, n xij 0, i = 1, m , j = 1, nNgoi ra bi ton phi tho mn iu kin: nmj =1 bj = i =1 ai (cn bng thu v pht).y l mt dng ca bi ton Quy hoch tuyn tnh.2.1.3. Bi ton ngi bn hng (Bi ton ci ti).Mt ca hng cn phi vn chuyn mt lng hng trn mt chuyn nng khng c qub kg. C n loi vt m ca hng cn phi vn chuyn i bn, mi vt loi j, (j = 1, n ), ckhi lng aj kg. V c gi tr l cj . Hy xc nh xem trong mt chuyn hng, ca hng cn aln phng tin vn chuyn cc vt no tng gi tr cc vt thu c l ln nht.Nu ta k hiu xj l s vt loi j s a ln phng tin vn chuyn, ta c m hnh tonhc bi ton nh sau:Tm x = (x1, x2,...,xn) |Rn sao cho:n f(x) = j =1cj xj maxVi iu kin: nj =1 aj xj b xj 0, j = 1, n xj - nguyn, j = 1, ny l bi ton Qui hoch nguyn.2.1.4. Bi ton lp k hoch u t vn cho sn xut. Cn phi u t vn vo m x nghip sn xut ra n loi sn phm. Do trang b k thut -cng ngh v t chc sn xut khc nhau nn hiu qu ca vn u t vo cc x nghip cngkhc nhau. Qua phn tch, ngi ta bit rng khi u t mt n v tin vo x nghip th i, i =1, m , trong mt nm s sn xut ra c bij n v sn phm loi j, j = 1, n . Tng s nguyn liuv lao ng hng nm c th cung cp l A v C (tnh theo gi/cng). Hy xc nh mt k hochu t sao cho m bo sn xut c t nht Bj n v sn phm loi j m tng s vn u t nhnht, bit rng cc nh mc hao ph v nguyn liu v lao ng khi sn xut ra mt n v snphm loi j x nghip i, i = 1, m , tng ng l aij v cij, i = 1, m , j = 1, n .15 17. Chng II: Quy hoch tuyn tnhGi vn u t vo x nghip i l xi n v tin. Khi s lng sn phm loi j sn xut x nghip i l bij xi v s nguyn liu s dng x nghip ny sn xut ra cc sn phm j l aij nbij xi .Vy ton b nguyn liu s dng x nghip i l j =1aij bijxi v tng s nguyn liu s mndng cho k hoch sn xut chung l: i =1 j =1 aij bij xi. mnTng t, ta suy ra tng s lao ng s dng trong k hoch sn xut l: cij bij xi i =1 j =1mTng s vn u t, theo bi ton t ra, l i =1xi v tng s sn phm loi j sn xut c ml i =1 bij xi .Theo mc tiu ca bi ton thc t t ra th bi ton c th m hnh ton hc nh sau:Tm vc t x = (x1, x2 ,..., xn) IRm sao cho:mf(x) =i =1xi minvi iu kin:m n i =1j =1aij bij xi Am n i =1j =1cij bij xi Cmi =1 bij xi Bj, j = 1, nxi 0, i = 1, my l mt dng ca bi ton Qui hoch tuyn tnh.2.2. M HNH BI TON QUY HOCH TUYN TNH.2.2.1. Bi ton quy hoch tuyn tnh tng qutTm x = (x1, x2...xi,...xn) IRn.nSao cho: f(x) = Cj xj max (min) (2.1)j = 1Tha mn iu kin: n aij xj (, = ) bi( i= 1, m )(2.2)j = 1xj 0(j = 1, n )(2.3)16 18. Chng II: Quy hoch tuyn tnh xy dng c s l lun gii bi ton, ch cn xt mt trong hai dng bi ton, chng hnbi ton tm gi tr ln nht (f max ) ca hm mc tiu, cn bi ton tm gi tr b nht (f min ) ca hm mc tiu c th chuyn i nh sau:* Gi nguyn h rng buc ( 2.2 ) v ( 2.3 ) n* a hm mc tiu: f(x) =j =1Cj xj minnv f (x) = - f (x) = ( - Cj ) xj max, ta c m hnh bi ton:j = 1Tm x = ( x1 , x2 , ..., xj ,... xn ) IRnnSao cho: f (x) = (- Cj ) xj max(2.4)j = 1 nTho mn iu kin: aij xj (, =, ) bi ( i = 1, m ) (2.5)j = 1xi 0 ( j = 1, n ) (2.6)B : Nu bi ton (2.4) (2.6) c xopt = x*, th bi ton (2.1) (2.3) vi f (x) min cng c xopt = x* v fmin = - fmaxTht vy, theo gi thit (2.4) (2.6) c xopt = x* vi hm mc tiunf (x) = (-cj ). xj max , th:j = 1f (x) f (x*) ( xD - tp cc phng n ) nn nn (-cj ). xj ( - cj ). x * jcj. x * j cj xjj = 1j = 1 j = 1 j = 1 f (x) f (x*) (x D) x* = xopt ca (2.1)(2.3) vi f(x) min.n n fmin = cj x * = - j (-cj) x * = - f j max( pcm ) j = 1j = 1Nh vy mi bi ton (2.1) - (2.3) vi f(x) min c th chuyn f (x) max.2.2.2. Dng chun tca- Dng y Tm x= (x1 ,.... , xj ,.... xn ) IRnSao cho: f(x) = c1x1 +...+ci xi +...+ cn xn max (2.7)17 19. Chng II: Quy hoch tuyn tnhTho mna11 x1 +...+ a1i xi +...+a1n xn b1a21 x1 +...+ a21xi +...+a2n xn b2------------------------------------ai1 x1 +...+ aii xi +...+ ain xn b1 (2.8)--------------------------------am1 x1+...+ami xi +...+ amn xn bm xi 0( i = 1, n ) (2.9)b. Dng rt gn. n f(x) = cixi maxj = 1 naii xi bi( i= 1, m )j = 1 xi 0 ( = 1, n )Tnh cht ca hm mc tiu (2.7) v dng bt phng trnh ca h rng buc (2.8) xut phtt ngha thc tin ca bi ton t ra. Chng hn nh bi ton lp k hoch sn xut hiu qukinh t tng cng ln nht, khi phi hn ch chi tit nguyn liu s dng. Ngc li, trong bi ton xc nh vn u t cho sn xut phi khai thc ti a trang b kthut - cng ngh sao cho t c yu cu v gi tr sn phm lm ra m vn u t t nht.2.2.3 Dng chnh tca- Dng y n f (x) = cixi max (2.10) i = 1 n aiixi = bi (i = 1, m )(2.11) i = 1 xi 0(i = 1, n )(2.12)b. Dng ma trn:Gi ma trn hng, gm cc phn t l h s cc n trong hm mc tiu l C: C = [ c1 c2...cn]Ma trn ct: b1 x1 x B = b2 , x = 2 bm xn 18 20. Chng II: Quy hoch tuyn tnha11 ......a1n a ......a Ma trn h s cc n (2.11): A = 212n ............... a m1 ......a mn Khi bi ton (2.10) (2.12) c dng ma trn: Tm X sao cho: f(x) = C.X max A.X = BX 0 c. Dng vc t: Gi vc t: c = ( c1 , c2 , ... , cn ) x = ( x1 , x2 , ... , xn ) a1 j Vc t ct lp bi h s cc n (2.2)2: Aj = a 2 j ( j = 1, n )a mj Vc t ct lp bi h s t do (3.5): b1 b2 B = ... b m Khi , bi ton trn c dng vc t: Tm X sao cho: f(x) = maxn Aj xj = Bj = 1 x0Trong : = cjxj - tch v hng ca 2 vc t c v x.Nh vy, bi ton QHTT chnh tc c th vit di dng ma trn hoc vc t.2.3. CC PHP BIN I DNG CA BI TON QUY HOCH TUYNTNH2.3.1. Rng buc: n aij xj bi j = 1 nn C th a v rng buc: ( -aij ) . xj - bi aij xj bi bng cch nhn j = 1 j = 12 v ca (2.7) vi (-1) ri t aij = - aij, bi = -bi19 21. Chng II: Quy hoch tuyn tnh2.3.2. ng thc: n aij xj = bij = 1C th a v 2 rng buc bt ng thc: n naij xj bi aij xj bij = 1 j = 1 n n (- aij ) xj - biaij xj bij = 1 j = 1Vi aij = - aij , bi = - bi .2.3.3. Bin xj t do c th thay bi hiu ca 2 bin khng m, bng cch t:xj = xj - xn + j vi xj 0 , xn+j 0Trong : xj = max { 0 ; xj } xn + j = max { 0 ; - xj}2.3.4. Mt rng buc bt ng thc: naij xj bi c th a v rng buc ng thc, bng cch a vo bin ph (hoc lj = 1bin b) xn + i 0: n aij xj + xn + i = bij = 1 nMt rng buc dng khc: aij . xj bi c th a v rng buc ng thc, bng cchj = 1a vo bin ph xn + i 0: n aij xj - xn + i = bij = 1nVy ta c: aij xj + xn + i = bi j = 1 n aij xj bi xn + i 0j = 1 nn aijxj bi aij xj - xn + i = bij = 1j = 1xn + i 020 22. Chng II: Quy hoch tuyn tnh2.3.5. nh l .nNu vc t x = (1 , 2 , .... n ) nghim ng bt phng trnh: aij.xjbi th vcj = 1 ()t X = (1 , 2 , ... , n , n + 1 ) s nghim ng phng trnh:n aijxj +) x n + i= bi j = 1 ( xn + i 0v ngc li . V d 1: a bi ton QHTT sau v dng chnh tc: f(x) = 2x1 3x4 + x5 + 2x6 max x1 + x2 3x4 + 2x6 = 5 2x2 3x3 + x4 + x54 3x1 x2 + 2x3 2x5 3 xj 0 ( j = 1, 2, 5, 6) Trc ht a h rng buc dng ng thc, bng cch a vo 2 bin ph. x7 0 ; x8 0,ta c: f(x) = 2x1 3x4 + x5 + 2x6 max x1 + x2 3x4 + 2x6 = 5 2x2 3x3 + x4 + x5 + x7 = 4 3x1 x2 + 2x3 2x5 =3 xj 0 ( j: 1, 2, 5, 6, 7, 8)Xt rng buc du cc n, ta thy x3 , x4 khng rng buc v du (khng n), nn t: x3 = x3 x9 Vi x3 0 , x9 0 x4 = x4 x10 Vi x4 0 , x10 0Bi ton c dng: f(x) = 2x1 3 (x4 x10) + x5 + 2x6 max x1 + x2 3 (x4 x10) + 2x6 =5 2x2 3 (x3 x9) + x4+ x5 + x7 = 4 3x1 x2 + 2 (x3 x9) 2x5x8=3 xj 0 ( j = 3 ; 4 ; 9 ; 10 ) , xj 0 ( j = 1 ; 2 ; 5 ; 6 ; 7 ; 8) l bi ton QHTT chnh tc.Tuy nhin, sau khi thc hin cc php bin i, s bin ca bi ton tng ln, song nu sdng linh hot cc php bin i i s, s bin c th gim bt, bi ton rt gn hn.V d 2: a bi ton sau v dng chun tc: f(x) = x1 + x2 + 2x3 + 7x4 + 5x5 max 21 23. Chng II: Quy hoch tuyn tnh x1 + x2 + x4 + 5x5 = 22 (a) x1 + x2 + x3+ 2x4 + 4x5 = 25 (b) x1+x3+x5 = 9 (c) xj 0 ( j = 1,5 ) p dng cc php bin i i s h rng buc: Tr tng v ca (b) cho (a), ta c:x3 + x4 x5=3 Tr tng v ca (b) cho (c), ta c: x2 + 2x4 + 3x5 = 16(d) Tr tng v ca (a) cho (d), ta c: x1 x4 + 2x5 =6 Vy h rng buc: x1 + x2+ x4 + 5x5 = 22 x1 x4 + 2x5 =6 x1 + x2 + x3 + 2x4 + 4x5= 25 x2 + 2x4 +3x5 = 16 x1+ x3+ x5=9 x3 + x4 x5=3 x1 = 6 ( x4 + 2x5 ) 0 x2 = 16 ( 2x4 + 3x5 ) 0 x3 = 3 ( x4 x5 ) 0Thay vo mc tiu, cho kt qu: f(x) = 4x4 + 2x5 + 28Bi ton tng ng vi: f(x) = 4x4 + 2x5 + 28 max - x4 + 2x5 62x4 + 3x5 16x4 + x5 3xj 0 ( j = 4, 5)y l bi ton QHTT dng chun tc, c rt gn hn. Nh vy, mt bi ton QHTT dng chun, bng phng php dng bin ph, ta lun ac v bi ton dng chnh tc. i vi bi ton QHTT dng chnh tc khng gim tnh tngqut, ta gi thit rng:i) H rng buc (2.11) gm m phng trnh c lp: Gi thit ny lun c thc hin, v nu ngc li th trong h c mt hay mt s phngtrnh l t hp tuyn tnh ca cc phng trnh cn li, th loi khi h cc phng trnh ny, c h mi gm cc phng trnh c lp vi nhau.ii) v phi ca h rng buc (2.11): bi 0 ( i = 1, m ) Gi thit ny lun thc hin, v ngc li nu phng trnh th i: bi< 0 th nhn 2 v caphng trnh ny vi -1, ta c phng trnh mi tng ng.n (- aii ) xi = bi vi bi = bi 0 j = 1iii) Trong h rng buc (2.11), s phng trnh m nh thua s n n: m < n22 24. Chng II: Quy hoch tuyn tnhGi thit ny a ra nhm m bo D ca bi ton QHTT, v ngc li m n, th hrng buc (2.11) lun a v h gm n, phng trnh, n n tng ng s c nghim duy nht(h tng thch) hoc h v nghim (h khng tng thch), trong c 2 trng hp , vic giibi ton l v ngha.Vy mi bi ton QHTT lun a v dng chnh tc: nf (x) = cjxj maxj = 1n aij .xj = bi (i = 1, m ) j = 1xj 0(j = 1, n ) Vi: bi 0(i = 1, m )m < n v m phng trnh c lp vi nhau .2.4 - MT S TNH CHT CA BI TON QUY HOCH TUYN TNH.a- nh l 1.Tp D cc phng n ca bi ton QHTT chnh tc (2.10) (2.12) l mt tp li.nh l 2.Nu tp D cc phng n ca bi ton QHTT chng tc (2.10) (2.12) khng rng v bchn, th D l mt a din li.* nh l c th chng minh bng phng php quy np theo s bin ca bi ton. chng minh D l mt a din li, ta ch cn chng t rng, trong D c mt s hu hn cc phngn, m mi phng n thuc D u l t hp li ca cc phng n trong D.nh l 3. Nu bi ton QHTT chnh tc (2.10) (2.12) c li gii v tp D cc phng nca n l mt a din li, th c t nht mt im cc bin ca D l phng n ti u.nh l 4. Nu bi ton QHTT chnh tc (2.10) -(2.12) c li gii, th tn ti t nht 1 imcc bin ca tp D cc phng n l phng n ti u (gi l phng n cc bin ti u).* nh l ny lm c s l lun cho phng php gii bi ton. Nh n ng l phi phitm phng n ti u trong tp v s phng n, ta ch cn tm trong tp hu hn cc phng ncc bin.Tuy vy, khng loi tr c nhng phng n ti u khng phi l im cc bin, th hin nh l sau:nh l 5. Nu bi ton QHTT chnh tc (2.10) - (2.12) c x1 , x2 , ... , xk l nhng phngn cc bin ti u, th mi t hp li ca chng cng l phng n ti u.nh l 6. phng n x = (x1 , x2 ,... ,xn) ca bi ton QHTT chng tc (2.10) (2.12)l phng n cc bin, iu kin cn v l cc vc t ct Aj ca ma trn h s trong (2.12) ngvi cc thnh phn xj > 0 lp thnh h c lp tuyn tnh. * Ta phn tch ngha nh l.Xt bi ton (2.10) (2.12) dng vc t:23 25. Chng II: Quy hoch tuyn tnh f(x) = max nj=1A j xj = BX 0 a1 j a2 j Trong : Aj = ( j = 1, n ) ; c, = ( c1 , c2 , .... , cn ) ... a mj b1 b2 B = ; x = ( x1 , x2 , ... , xn )... b m l cc vc t ct v hng. x = ( x1 , x2 , ... xj , ... xn ) phng n cc bin { Aj: j J , xj > 0 } h vc t c lp tuyn tnh.Vy nh nh l, m vic xem xt mt phng n ca bi ton QHTT c phi l phng ncc bin hay khng , c thay th bng vic xem xt s c lp tuyn tnh hay ph thuc tuynca h vc t ct Aj , j J. Xt v d sau lm r cc khi nim trn: V d. Cho bi ton QHTT: f(x) = 2x1 + x2 - 3x3 + 5x4 max3x1 + 4x2 - x3 + x4= 11 x1 - 3x2- x4 0- 2x1+ x3 + 3x4 = 9xj 0 ( j = 1;4 ) Tm mt phng n cc bin ca bi ton. Chng hn, ta xt h 2 vc t: ( A2 , A4 ) , vi:4 1 A2 = 3 ; A4 = 1 l h c lp tuyn tnh, v ma trn lp bi A2 , A4 :0 3 41 A = 3 1 c hng: rank A = 2 0 3 Vy (A2 , A4 ) - h vc t c s: J = {2 , 4} - tp ch s cc vc t c s x2, x4 - bin c s x1, x3 - bin phi c s, v 1; 3 J v x1 = x3 = 024 26. Chng II: Quy hoch tuyn tnhT: - 2x1 + x3 + 3x4 = 9 x4 = 3 > 0T: 3x1 + 4x2 - x3 + x4 = 11 x2 = 2 > 0Vi x1 = x3 = 0 ; x2 = 4 , x4 = 3 cng tho mn: x1 - 3x2 - x4 0Vy x = ( x1 , x2 , x3 , x4) = ( 0 , 2 , 0 , 3 ) l phng n cc bin.- Nu a bi ton v dng chnh tc, phi a vo bin ph x5 0 , bi ton c dng:f (x) = 2x1 + x2 - 3x3 + 5x4 max3x1 + 4x2 - x3 + x4= 11x1 - 3x2 - x4 + x5 = 0 - 2x1 + x3 + 3x4= 9 xj 0 ( j = 1;5 ) Tng t xt nh trn, h vc t ( A2 , A4 , A5 ) - c lp tuyn tnh l h vc t c s caphng n cc bin (khng suy bin): x = ( 0, 2 , 0 , 3 , 9 )Vy: x = ( 0 , 2 , 0 , 3 , 9 ) c s: ( A2 , A4 , A5 ) duy nht.Ngi ta chng minh c rng:i. i vi mt phng n cc bin suy bin ca bi ton QHTT. C th c nhiu c s khc nhau. Song i vi phng n cc bin khng suy bin ch cmt c s duy nht.V d. Xt bi ton QHTT:f(x) = x1 + 4x2 - 5x3 + 2x4 maxx1 + 5x2 - 3x3 + x4 = 6 3x2 + 3x3- 2x4 = 37x1- 2x3=7 xj 0 ( j = 1;4 )Tm mt phng n cc bin suy bin v c s ca n.Nu ly x3 = x4 = 0 , t h rng buc ta c: x1 = x2 = 1Vyx = ( 1 , 1 , 0 , 0 ) l mt phng nVix1, x2 > 0 xt h vc t (A1,A2) ma trn A lp bi A1, A2: 1 5A = 0 3 c hng A = 2 ( A1 , A2 ) - h vc t c lp tuyn tnh. Do x = ( 1 , 1 , 0 7 0, 0 ) l mt phng n cc bin. Hn na s thnh phn dng trong x l 2 < 3 = m. Vy x = ( 1 ,1 , 0 , 0 ) l phng n cc bin suy bin.Xt h vc t ( A1 , A2 , A3 ) ma trn A lp bi A1 , A2 , A3: 1 5 3A = 0 3 1 det A = 92 0 7 0 225 27. Chng II: Quy hoch tuyn tnhVy h ( A1 , A2 , A3 ) - c lp tuyn tnh, l mt c s ca x.Vy i vi phng n cc bin suy bin x = ( 1 , 1 , 0 , 0 ) c 2 c s khc nhau ( A1 , A2 ,A3 ) v ( A1 , A2 , A4 ).* iu kin tn ti li gii ca bi ton QHTT l.*- Tp cc phng n: D *- Hm mc tiu b chn trn min D.2.5. PHNG PHP HNH HC GII BI TON QUY HOCH TUYNTNH 2 BIN.2.5.1 Biu din hnh hc quy hoch tuyn tnh 2 binXt bi ton QHTT chun tc 2 bin. f(x) = c1x1 + c2 x2 max ai1 x1 + ai2 x2 bi ( i = 1, m ) xj 0 ( j = 1,2 )- T ngha hnh hc ca siu phng trong IR2: H = {x = (x1, x2 ): a1 x1 + a2 x2 = b}chia IR2 thnh 2 na mt phng: D+ = {x = (x1, x2 ): a1 x1 + a2 x2 b} D = {x = (x1, x2 ): a1 x1 + a2 x2 b}th mi bt phng trnh tuyn tnh trong h rng buc ai1 x1 + ai2 x2 bi ( i= 1, m )s xc nh mt na mt phng . Vy min rng buc D, xc nh bi h rng buc l giao ca m na mt phng, s l agic li hay khc li (nu D ) hoc khng tn ti (nu D = ) xc nh na mt phng (2.2)1, trc ht phi xc nh ng thng : (Hi): ai1 x1 + ai2 x2 = bi ( i = 1, m )Sau xc nh vc t php tuyn ca n: ni = {ai1, ai2} ( i = 1, m ) th phn na mt phng ( Di ): ai1 x1 + ai2 x2 bi ( i = 1, m ) s nm v pha ngc hng vi ni , cn na mt phng: (i = 1, m ) s nm v pha cng hng vi ni (i = 1, m ), k c bin ca (Hi).Minh ho hnh hc+ Ch : Ngoi phng php xc nh gia mt phng ( Di) hoc Di nu trn, c th xcnh bng cch: xt im gc ta O (0,0) thuc na mt phng no, nh thay to O (0,0)vo h rng buc hoc ngc li.26 28. Chng II: Quy hoch tuyn tnhx2 + (D i): ai1 x1 + ai2 x2 bi ni = {ai1, ai2} ( Di ): ai1 x1 + ai2 x2 bix1 0(Hi): ai1 x1 + ai2 x2 = bi - ni Hnh 2.1:- T ngha hnh hc, i vi hm mc tiu: f(x) = c1x1 + c2 x2 ta xt phng trnh ngthng:c1x1 + c2 x2 = (*) 1vi IR Ta thy: Khi thay i, (*) s xc nh trn mt phng to 0x1 x2 cc ng thng songsong vi nhau (v cng vung gc vi vc t php tuyn ni = {c1, c2}), gi l cc ng mc(mc gi tr ). Mi im x = ( x 1, x 2) D, s nm trn ng mc vi gi tr: = c1 x 1 + c2 x 2 Xem hnh 2.2. Vy theo ngn ng hnh hc, c th pht biu bi ton (2.10) ( 2.12 ) nhsau: Trong s cc ng mc (2.13) tm ng mc vi gi tr ln nht c th: x2 n ={c1,c2} = xc1 x +c2 2 - c1x1+c2x2 = 0x1 Hnh 2.2 27 29. Chng II: Quy hoch tuyn tnhmax = c1 x1 + c2 x 2 **vi x* = ( x1 , x 2 ) D* *khi x* = ( x1 , x 2 ) = xopt vi f (x) max = max * *2.5.2. Phng php hnh hc gii bi ton QHTT 2 bin .a. Nhn xt:Hm mc tiu: f (x) = c1 x1 + c2 x2 c th biu din bng dng vc t, nh khi nim catnh v hng: f(x) = ( c , x ) vi c = ( c1, c2 ) , x = ( x1, x2 ) IR2Ta thy: f(x) = c1 x1 + c2 x2 = Khi dch chuyn song song cc ng mc (*) theo hng vc t php tuyn c = {c1, c2}, th gi tr ng mc (tc f (x)) s tng.Ngc li, khi dch chuyn theo hng ngc li ca c , (hay cng hng vi vc t i cac l - c ), th gi tr ng mc (hay f(x) s gim ) .Vy gii bi ton trn ta tm x* = ( x1 , x 2 ) D m max = f (x*). * *T , ta c th nu cc bc gii bi ton trn bng phng php hnh hc nh sau:b. Cc bc gii bi ton . gii bi ton trn ta tin hnh:i - Xc nh min rng buc D ca bi ton trong h trc to 0x1x2.ii V th ng mc (*): c1x1 + c2 x2 = vi no ,iii - Xc nh vc t php tuyn c = {c1,c2} v dch chuyn song song cc ng mc (*)theo hng ca vc t c , cho n v tr ti hn (v tr ti hn l v tr m ng mc vn cn ctmin D, nhng nu tip tc dch chuyn th s khng ct min D na) iv - im ( hoc nhiu im ) ca D nm trn giao im ca ng mc v tr ti hn vimin D, l li gii ca bi ton.Minh ho hnh hc (hnh v 2.3) x2 Bc1x1*+c2x2*=max x2A DCc D 0 x1 * - c1x1+c2x2 =Hnh 2.328 30. Chng II: Quy hoch tuyn tnhc. Ch : trn, tin li ta xt bi ton QHTT chun tc, i vi bi ton QHTT bt kcng c th gii c bng phng php hnh hc. C th xy ra cc trng hp:i) - Min D = tc cc na mt phng xc nh bi h rng buc.ai1x1 + ai2x2 bi(hoc ai1x1 + ai2x2 bi)Khng c im chung, th bi ton v nghim. ii) Min D : - Nu D - a gic li, th c duy nht 1 im cc bin l phng n ti u; hoc c v sphng n ti u, khi 2 im cc bin l cc phng n ti u (theo tnh cht bi ton QHTT).- Nu D - khc li (a gic li khng gii ni), th bi ton c mt phng n cc bin tiu, nu min D nm v mt pha ca ng mc (2.4) ct ng mc (2.4) ti 1 im, hoc biton c v s phng n ti u, nu c 2 im cc bin l cc phng n ti u, hoc bi tonkhng c li gii (f(x) khng b chn). C th minh ho bng mt s v d sau: 2.5.3 - V d:V d 1: Gii bi ton sau bng phng php hnh hc:f(x) = 4x1 + 5x2 max 2x1 + x2 8 x1 + 2x2 7x2 3 xj 0 ( j =1 , 2 )Xc nh min rng buc D (hnh v 2.4) l a gic li (tam gic ABC)x28 2x1 + x2 8 n3A B x2 3D2 x1 + 2x2 7C5/2 4 7 04x1 + 5x2=10x1Hnh 2.429 31. Chng II: Quy hoch tuyn tnh Xt ng mc: 4x1 + 5x2 = 10 ( cho = 10, d v) Dch chuyn ng mc song song vi nhau theo hngn = {4; 5}, nh B ( 5 ;3 ) D 2trn ng mc cui cng l im cc bin ti u: 5 xopt = ( ;3 ) 25 max = f(x) max = 4. + 5.3 = 252 V d 2: Gii bi ton sau bng phng php hnh hc: f(x) = 4x1 + 3x2 maxx1 + x2 4 - x1 + 3x2 3 2x1 - x2 2 xj 0 ( j = 1,2 ) Xc nh min rng buc D (hnh v 2.5) Ta thy D = Vy bi ton khng c li gii.V d 3: Gii bi ton sau bng phng php hnh hc: x2 f(x) = 3x1 + 4x2 min x1 + 2x2 4 - x1 + x2 2- x1 + x2 2 2x1 - 4x2 12 xj 0 ( j = 1,2)x1 + x2 44 - x1 + 3x2 3 1x1 0-3 01 4 x1-2Hnh 2.5:30 32. Chng II: Quy hoch tuyn tnhXc nh min rng buc D ca biton: khc li (hnh v 2.6)x2Xt ng mc: x2 0 3x1 + 4x2 = 4Dch chuyn song song cc ng- x1 + x2 2 mc theo hng ngc vi n = {3;4}n = {3;4}Ct D ti im A (0 ;2) l duy nht. 2 Ax1 + 2x2 4Vy A (0;2) l im cc bin tiu: x1 0 1 Xopt = (0;2) BC min = f(x) min =-24/346 x1 = 3.0 + 2.4 = 8 3x1 + 4x2=42x1 - 4x2 1 Hnh 2.6:Ch : Nu bi ton trn, gi nguyn min rng buc, cnf(x) = 3x1 + 4x2 max th bi ton s v nghim, bi v khi dch chuyn song song cc ng mc theo hng n = {3 ; 4},s c v s cc ng thng song song vi ng mc i qua im cc bin C v c khong cchln tu n n, tc f (x) + (f(x) khng b chn trn )V d 4: Gii bi ton sau bng phng php hnh hc: f(x) = 3x1 + 6x2 minx1 + 2x2 2-2x1 + 3x2 60 x1 4Xc nh min rng buc D ca bi ton (hnh 2.7)Xt ng mc: 3x1 + 6x2 = 18 Dch chuyn song song cc ng mc theo hng ngc vi n = {3;6}, c 2 im A (0;1), F(2;0) nm trn ng mc cui cng.Vy c 2 im cc bin ti u bi ton c v s li gii l mi t hp li ca 2 imcc bin ti u trn, tc l v s cc im nm trn on thng [ AF ] cng l phng n ti u(trng hp ng mc cui cng trng vi mt cnh ca min D)31 33. Chng II: Quy hoch tuyn tnh x2x1=4 n = {3;6} 2x1 + 3x2 6C 3 D 2 3x1 + 6x2 6 B 1x2 0x1 + 2x2 22D 0 F4 E 6 x1 Hnh 2.7 Qua phng php hnh hc, ta thy rng: - Nu bi ton QHTT c phng n ti u, th t nht 1 nh ca min D l ti u. - Nu min D gii ni v khc rng, th chc chn c phng n ti u. - Nu min D khng gii ni, nhng hm mc tiu b chn trn min D, th cng chc chnc phng n ti u.2.6. PHNG PHP N HNH2.6.1. C s l lun ca phng php a. ng li chung Xt bi ton quy hoch tuyn tnh chnh tcnf(X) = c x j =1j j min (max)(2.13) n ___ a i x j = b i (i = 1, m) j= (2.14) ___ x j 0( j = 1.n ) ____ Vi gi thit m < n; b1 0 (i = 1, m) v bi ton khng suy bin. Dng vct ca bi ton: f(X) = < C, X> min (max)n = A 1 x j B j =1 ____ x j 0( j = 1, n )32 34. Chng II: Quy hoch tuyn tnhXut pht t mt phng n cc bin x(0) = (x01, x0n vi c s J0, ta tm cch nh gi x(0),nu x(0) cha ti u th tm cch chuyn sang phng n cc bin x(1) tt hn. V s phng ncc bkn l hu hn nn sau mt s hu hn bc lp s tm c phng n cc bin ti u hocpht hin bi ton khng c li gii.b. c lng cc binGi s x(0) l mt phng n cc bin, c s J0. Gi kl c lng ca bin xk theo c s J0 cxc nh bi: ____k = c j z jk c k (k = 1, n)jJ 0Trong zjk l h s phn tch ca vct Ak qua vct c s Aj (j J0) ngha: _____Ak =zjJ 0 A ,jkdo zjk = A-1j Ak (j J0, k = 1, n )Ch rng c lng cc bin c s xj: j = 0 (j J0). V khi :1 nu k = jZjk = 0 nu k jc. Du hiu ti u Nu phng n cc bin x(0), c s J0 ca bi ton (2.13) - (2.15) c k < 0 (> 0). (k J0)th X(0) l phng n ti u duy nht.Nu phng n cc bin X()), c s J0 tho mn du hiu ti u m k = 0. (k J0) th biton c th c phng n ti u khc ngoi X(0).d. nh l c bni vi phng n cc bin X(0), c s J0 ca bi ton (2.13) - (2.15) m k > 0 (< 0),(kJ0) th s sy ra 1 trong 2 trng hp sau:Nu c mt k > 0 (< 0), (k J0) m zjk 0, (j J0) th bi ton khng c li gii. Nu mi k > 0 (< 0), (k J0) u tn ti t nht mt Zjk > 0, (j J0) th chuyn sang cphng n cc bin mi X(1) tt hn X(0); f [X(1)] < f [X(0)] ([X(1)] > f[X(0)]).2.6.2. Thut ton ca phng php n hnh Ton b qu trnh tnh ton c sp xp theo mt trnh t cht ch m bo hiu qu cavic tm li gi ca bi ton QHTT. Trnh t c gi l thut ton.Khng mt tnh tng qut ta xt bi ton QHTT dng chun.nf (x) = c x j =1 j j min 33 35. Chng II: Quy hoch tuyn tnh x 1 + a 1m +1 x m +1 + .... + a 1n x n = b 1 x + a 2 m +1 x m +1 + .... + a 2 n x n = b 2 2 ................................................... x + amim +1 x m +1 + .... + a mim x n = b m m _________ x j 0( j = 1, n ), b i 0(i = 1, m), m < n Bi ton c ngay phng n cc bin x(0) = (x 1 , x 2 ,...., x 0 , 0.... 0) = (b1, b2, ....,bm,0,...,0)0 0 mVi c s J0 gm m vct n v A1, A2, ..., Am; J0 = { 1,2, ...., m}Thut ton n hnh gm cc bc sau:Bc1. Lp bng n hnhTa lp mt bng ghi cc h s phn tch ca vct B v vct Ak, k = (1,2, ...., n) qua c s J0 theo mu di y. Vi phng n ny th Zjk = ajk (j J0).Bng ny gi l bng n hnh vi phng n cc bin x(0), c s J0.HCPhng c1 c2... cJ ... cm cm+1 ... c8... cnssn A1 A2.. AJ...Am Am+1 ... A8... An c1 A1X1 01 0 ... 0.... 0 Z1m+1...Z18....Z1n c2 A2X0 20 1 .... 0.... 0 Z2m+1...Z28....Z2n .... ......... .................................................... cr ArX0r 0 0 .... 1.... 0 Zrm+1... Zr8 ....Zrn .............................................................. cm AmX0m 0 0 .... 0 ... 1 Zm m+1... Zm.8....Zmnf(X (())) 0 0.... 0 .... 0 m+1... 8 .... nHng cui bng gi l hng clng k, (k = 1, 2, ...., n)f[x(0)] = ; c xjJ n j 0 j k = c ZjJ nj jk - ck (k J0); k = 0 (k J0 )Bc 2. Kim tra tnh ti u ca phng n cc bin x(0).Nu k 0, (k J0) th x(0) = x0pt, fmin = f [x(0)] thut ton kt thc.Nu tn ti k > 0, (k J0) m Zjk 0, (j J0) th bi ton khng c li gii v f[x] - , thutton kt thc.Nu mi k > 0, (k J0) m Zjk > 0, (j J0) th chuyn sang bc 3.Bc 3. Tm vct a vo c s v vct loi khi c s.Gi s: max { k > 0, (k J0) } = , vct As c a vo c s, tnh: x0 jx00 = min zjt > J0; j J 0 , gi s 0 = r , vct Ar loi khi c s. Z js Z rs34 36. Chng II: Quy hoch tuyn tnh Nh vy, J1 = {J0{r}{S}}. H s phn tch Zrs nm ti giao ca hng r v ct s gi l phnt trc ca php bin i.Bc 4. Bin i bng, xy dng phng n cc bin mi x(1) vi c s J1. Trong bng nhnh tng ng vi phng n cc bin x(1), c s J1 ta thay cs, As vo v tr ca cr, Ar cc cj, Aj (j r, jJ0) c gi nguyn. Cc thnh phn ca x(1) c tnh theo cng thc: nu j J0 0 x ( 0) x2 = r j nu j J0, j = r Z rs x (r 0 ) x j (0) Z js nu j J0, j r Z rsC s J1 = {J0{r}{S}: {J1} = {J0} = mH s phn tch Z (jk) (j J1, k J1) c xc nh bi cng thc:1 x (rk )0nu j J0, j = rZ Z (jk) 1= rsZ Z rk nu j J0, j rZ jk Z rs jsCng thc ny ng cho c cc thnh phn ct phng n v hng clng 1.Ta c: f[x(1)] = f[x(0)] - , hay f [x(1)] = c xjJ 1j(1 )j 1 = k - , hay 1 = k k c jJ i c Z 1jk c k (k J1)Xem X(1) ng vai tr nh X(0), lp li qu trnh trn t bc hai tr i sau mt s hu hnln hoc pht hin bi ton khng c li gii hoc tm c phng n ti u ca bi ton.Ch khi thc hin thut ton:i vi bi ton f(x) max c th gii trc tip bng thut ton n hnh vi vct A, ca vo c s c s = min {k < 0,k J0} cn xc nh vct loi khi c s Ar v cc thnh phnx (j1) , Z (jk) (j J1, k J1) c tnh tng t, hoc cng c th chuyn thnh bi ton: 1 g(x) = - f(x) min nhng ch rng fmax = - gmin.2.6.3. Phng php tm phng n cc bin xut pht - bi ton MKhi dng thut ton n hnh gii bi ton QHTT chnh tc vi gi thit bit mt phngn cc bin vi c s n v tng ng. Nhng nhn chung gi thit ny khng phi bao gi cngc ngay. V vy p dng c thut ton n hnh cn phi c phng php tng qut chophp tm c mt phng n cc bin m khng ph thuc vo cu trc ring bit ca bi ton.Mt trong nhng phng php nh vy l phng php bin gi hay phng php tm phng ncc bin xut pht.a. Ni dung phng php 35 37. Chng II: Quy hoch tuyn tnh Xy dng bi ton mi l bi ton bin gi hay bi ton M t bi ton ang xt. Bi tonM c ngay phng n cc bin xut pht v c iu kin p dng thut ton n hnh gii, ng thi t kt qu ca bi ton M a ra c kt lun cho bi ton ang xt.b. Xy dng bi ton MXt bi ton chnh tc:nf(x) =c xj =1 j j min (2.16) n c j x j = bi (i 1,m) (2.17) j=1 x 0( j = 1,n) j (2.18)____Bi ton (2.16) (2.18) gi l bi ton u. Gi thit bi 0 (i = 1, m) v ma trn cc h strong h rng buc (2.17): A = (aij)mxn khng cha vct n v no.Bi ton M c xy dng nh sau:____Thm vo v tri ca phng trnh th i (i = 1, m) trong h rng buc (2.17) mt bin gi____xn+ i 0 (i = 1, m) . H s ca cc bin gi xn+i trn hm mc tiu u bng M, vi m l s dngln tu (M > 0 ), bi ton M c dng: ___ nmf( X ) =c xj =1 j j +M Xi =1 n +i min n a j x j + x n + i = bi (i 1, m) j=1 x 0( j = 1, n) jBi ton M c ngay phng n cc bin xut pht:__x (0) = ( 0,0,.....,0, b1, b2,...., bm) vi c s J0 l: Em = (An+1, An+2, ..., An+m ); n sJ0={n+1, n+2,...,n + m}Do vy p dng c thut ton n hnh gii bi ton M.T cch xy dng bi ton M nh trn ta thy:__Nu x = x 1 , x 2 ,....., x n , 0,0,....,0 l phng n ca bi ton M th x = (x1, x2, ..., xn) n s __l phng n ca bi ton ban u v ngc li, ng thi f(x) = f( x ) .c. Mi quan h gia bi ton M v bi ton ban uNu bi ton M c:36 38. Chng II: Quy hoch tuyn tnh * X opt = X = x 1 , x * ,....x * , 0,0,....,0 th bi ton ban u c Xopt = x* = (x 1 , x * ,...., x * ) v____* *2n 2n m s__*f( x ) = f(x*). x opt = x = (x 1 , x * , ,....., x * + m ) trong t nht mt__ __**Nu bi ton M c 2 n ____x * + i > 0(i = 1, m) th bi ton ban u khng c phng n no (khng gii c).nNu bi ton M v nghim th bi ton ban u cng v nghim.d. Ch khi gii bi ton MNu bi ton ban u c nghim Xopt th nghim ny ch c th nhn c sau t nht m + 1bng n hnh khi gii bi ton M. Nu trong ma trn h s trong h rng buc (1.17): A =(ajj)mxn cha m1 vct n v khc nhau (m1 < m) th khi xy dng bi ton M ch cn thmm -m1 bin gi.__V hm mc tiu f ( x ) l tuyn tnh i vi M nn cc s c lng K ca bi ton Mc dng:Do :k = Mk + k (, R; k J) k > hk > h k = h ; k > h mi bc ci tin phng n cc bin, nu mt bin gi b a ra khi c s th n khngth quay li c s c na. Do bng n hnh ng vi phng n cc bin mi ta khngcn tnh ton vi ct ng vi vct tng ng vi bin gi .Vi bi ton f(x) max bi ton M tng ng c hm mc tiu l:__n mf( x ) = c j x j M x n + i max (M >> 0)j=1i =1Cc iu kin cn li tng t bi ton f(x) min2.6.4. Cc bi tp muBi 1. Gii bi ton sau bng phng php n hnh.f(x) = 2x3 + x4 minx 1 + 2 x 3 + x 4 = 6x 2 + x 3 2 x 4 = 2 ____x j 0( j = 1,4 )Bi ton c dng chun, phng n cc bin xut pht:X(0) = (6,2,0,0) vi c s { A1, A2} = E2Lp bng n hnh: 37 39. Chng II: Quy hoch tuyn tnh CjAjPhng 00 2-1c sn A1 A2 A3A4 0 A1 61 0 21X(0) 0 A2 20 1 1200 0-21X(1) -1A4 61 0 2 10A2142 1 5 0 -6 -1 0-40T bng n hnh tng ng vi phng n cc nin x(1) ta thy phng n x(1) c k 0, (k =___1,4) . c bit k 0 (k J1). Do bi ton c nghim duy nht:Xopt = X(1) = (0,14,0,6) v fmin = f [X(1) ] = - 6.Bi 2. Gii bi ton sau bng phng php n hnhf(x) = - x1+ 4x2 + 3x3 max 2 x 1 + x 2 2 x 3 + x 4 = 16 4 x + 2 x + x = 8 1 3 5 x 1 + 2 x 2 x 3 + x 6 = 12 ____ x 0( j = 1,6 ) jBi ton c ngay phng n cc bin xut pht:x(0) = (0,0,0,16,8,12) vi c s {A4, A5, A6 }Lp bng n hnh:C1A1 Phng - 1 4 30 00 C s C s nA1A2A3A4 A5 A60 A4 16 2 1-21 00x (0) 0 A58-4 0 20 100 A6 12 1 2-10 010 1-4-30 000 A4 10 3/2 0 -3/2 1 0-1/2 x(1) 0 A58-4 0 30 104 A26 1/2 1 -1/2 0 0 1/2 243 0 -50 020 A4 16 -3/20 013/4 -1/2x(2) 3 A34-2 0 101/2 04 A28 -1/21 001/4 1/2 44-7 0 005/2 2Ta thy trong bng n hnh ng vi phng n cc bin x(2), c s J2 c 1 = - 7 < 0 ngthi Zj1 < 0 j J2 nn bi ton khng c phng n ti u v tr s hm mc tiu tng v hn trntp phng n.Bi 3. Gii bi ton sau bng phng php n hnhf(x) = - 5x1 - 2x2 - 10/3x3 min38 40. Chng II: Quy hoch tuyn tnh2 x 1 + 4 x 2 + 3x 3 464 x + 2 x + 3x 38 1353x 1 + x 3 21____x 0( j = 1,3 ) ja bi ton v dng chnh tc cng l bi ton dng chun: f(x) = - 5x1 - 2x2 - 10/3x3 min2 x 1 + 4 x 2 + 3x 3 + x 4 = 464 x + 2 x + 3x + x = 38 135 53x 1 + x 3 + x 6 = 21____x 0( j = 1,6 ) j Bi ton dng chun c ngay phng n cc bin xut pht:x(0) = (0,0,0,46,38,21) vi c s n v {A4, A5, A6} = E3. Lp bng n hnh. C1A1 Phng -5- 2 -10/3 0 0 0 CC n A1A2 A3 A4 A5 A6 ss0A4462 4 3 1 00x (0) 0A5384 2 3 0 100A6213 0 1 0 01 0 5 210/3 0 000A4320 47/31 0-2/3 x(1) 0A5100 25/30 1-4/3 -5A1 71 01/30 0 1/3-350 25/30 0-5/30A4120 0-1 1-22x (2 -2A2 50 15/60 1/2-4/6 -5A1 71 01/30 0 1/3-450 0 0 0-1-1/3 0 A4180 -6/50 1 -7/56/5 x(3 -10/3A5 606/51 03/5 -4/5-5 A1 51 -2/50 0 -1/53/5-450 0 0 0-1-1/3____ Trong bng n hnh ng vi phng n cc bin x(2) c k 0 (k = 1,6 ) nn:Xopt = X(2) = (7,5,0,12,0,0) Do bi ton ban u c: x 1 = (7,5,0) v fmin = -450pt 39 41. Chng II: Quy hoch tuyn tnh Tuy nhin trong bng n hnh vi x 2 c 3 = 0 (3 J0) v Z23 = 5/6 > 0; Z33 = 1/3 > 0. 0ptTa a A3 vo c s s c phng n Xopt khc. X0pt = X(3) = (5,0,6,18,0,0) nn bi ton cthm phng n ti u: x 2 = (5,0,6). opt Vy bi ton c v s phng n ti u, tp phng n ti u c dng: {Xopt )={ x 1 opt + (1 ) x 2 ;0 1 } v fmin= -45opt Bi 4. Gii bi ton bng phng php n hnh f(x) = - x1 +x2 max x 1 + x 3 + x 5 = 5 3x x + 5x 10 1 4 5 x 2 x 5 = 1 ____ x 0( j = 1,5 ) j a bi ton v dng chnh tc vi bin ph x6 0 f (x) = - x1 + x2 max x 1 + x 3 + x 5 = 5 3x = x + 5x + x = 10 1 4 56 x 2 x 5 = 1 ___ x 0( j = 1,6) j Bi ton c phng n cc bin xut pht: X(0) = (0,1,5,0,0,10) vi c s {A3, A6, A2} = E3 Lp bng n hnh C1 C A1 Phng-1 100 0 0s C s n A1 A2 A3 A4A5A6 0A15 -1 01 0 10x (0) 0A6 103 00-1 51A210 10 0-101 10 00-1 0 0A33 -8/4 0 1 1/50 -1/5x(1) 0A523/5 0 0-1/511/5 1A233/5 1 0-1/501/5 38/50 0-1/5 01/5 0A4 15 -8 0 5 1 0-1 x (2 0A55 -1 0 1 0 1 0 1A26 -1 1 1 0 0 060 0 1 0 0 040 42. Chng II: Quy hoch tuyn tnh___Trong bng n hnh ng vi phng n x(2) c k 0 (k = 1,6) nn x(2) l phng n ccbin ti u, trong 1 = 6 = 0 (1,6 J2) nhng Zj1 < 0, Zj6 0 (j J2) do : xopt = x(2) =(0,6,0,15,5) l phng n cc bin ti u duy nht ca bi ton ban u v fmax = f[x(2)] = 6.Bi 5. Cho bi tonf (x) = - 2x1 - x2 + x3 + x4 - 4x5 max x 1 x 3 + 4 x 5 2 x 5 4 3x + 2 x x + x 24 1 2 34 5x 1 + 3x 2 + x 3 + 2 x 4 x 5 = 46 ___ x 0( j = 1,5) ja. Chng t x(0) = (0,2,0,20,0) l phng n cc bin, li dng x(0) gii bi ton bng phngphp n hnh.b. Da vo phng n x(0) tm li gii ca bi ton khi hm mc tiu c dng:f (x) = 5x1 + 2x2 + 4x3 + 4x4 - 2x5 min Gii.a. V x(0) R5 tho mn mi rng buc ca bi ton, tho mn cht cc rng buc 2,3 v 3rng buc du: x 10 ) = x 30 ) = x 50 ) = 0 . H 5 rng buc cht ny c lp tuyn tnh v nh thc ca (((cc matrn to bi cc vct ny:3 2 -11 05 312 -11 000 0 =-100 010 00 000 1 Nn x(0) l phng n cc bin ca bi ton. gii bi ton ny bng phng php nhnh, trc ht ta a bi ton v dng chnh tc:f (x) = -2x1 - x2 + x3 + x4 + x5 min x 1 x 3 + 4 x 5 2 x 5 + x 6 = 4 3x + 2 x x + x + x = 24 1 23 4 7 5x 1 + 3x 2 + x 3 + 2 x 4 x 5 = 46 ___ x 0( j = 1,7) jT phng n cc bin X(0) ca bi ton cho suy ra phng n cc bin ca bi tonchnh tc l x-(0) = (0,2,0,20,0,2,0) vi c s {A2, A4, A6} v J0 = {2,4,6}. Ta tm matrn h s ____phn tch ca vct Ak (k = 1,7 ) v vct B qua c s J0 bng phng php bin i s cp trncc hng ca ma trn.41 43. Chng II: Quy hoch tuyn tnh 1 1 4 0 2 1 0 4 h1h1h+ 2h 3 2 1 1 0 0 1 24 h 3 22h2 3 h 5 3 1 2 - 1 0 0 46 1 1 4 0 2 1 0 4 2 1h+ +1h1h h h 3 2 h 3 2 1 1 0 0 1 24 h 3 h2h2 3 2 1 1 3 0 1 0 2 2 2 0 1 0 1 1 2 2 1 0 5 1 2 0 3 20 1 1 3 01 02 2 Lp bng n hnh ng vi phng n cc bin x(0) v p dng thut ton n hnh.Ton b qu trnh tnh ton bng C1A1 -2 -1 3 1-4 0 0Phng CCnA 1 A2 A3A4 A5 A6A7 ss (0)x 0A62-20-1 01 1 -21A4 20 10 5 1 -2 0 -3 -1A22 11-3 01 0218 2050 1 0-5 0 A66 -9/5 0 01/5 3/5 1 -13/5 3 A541/5 0101/5 -2/5 0 -3/5x(1) -1 A2 148/5 13/5 -1/5 0 1/5-2-10 0-1 3 0-2-4 A5 10-30 01/3 1 5/3 -13/3 3 A58-10 11/3 0 2/3 -7/3x(2) -1 A2 16 11 02/3 0 1/3 -2/3 -32-10 0-20 -5 11Trong bng n hnh ng vi phng n cc bin X(2) c 7 = 11 > 0 nhng Zj7 < 0 (jJ2)nn bi ton khng c li gii v f(x) - trn tp phng n.b. Trong trng hp f (X) = 5X1 + 2X2 + 4X3 + 4X4 - 2X5 minTa lp bng n hnh ng vi cc ck mi v vi x(0) = (0,2,0,20,0,2,0).42 44. Chng II: Quy hoch tuyn tnhC1 A15 24 4 -2 0 0PhngC CnA1A2 A3A4 A5 A6 A7s sx(0) 2A62-2 0 -1 011-2 4 A4 20 1 05 1 -2 0 -3 2 A22 1 1 -3 010 284 2 0 10 0 -4 0 -8 0 A66 -9/5001/5 3/5 1 -13/5x(1) 4 A541/5011/5 -2/5 0 -3/5 2 A2 148/5103/5 -1/5 0 1/5-2-1 00-130-2-2 A5 10-3 001/3 1 5/3 -13/3 (2)x 4 A58-1 011/3 0 2/3 -7/3 2 A2 16 1 102/3 0 1/3 -2/344-1 00-200-2____Bng n hnh tng ng vi phng n cc bin X(1) c k 0 (k = 1,7 ) nn X (1) = X(1) opt= (0,14,4,0,0,6,0) v fmin = 44. Nhng trong c 5 = 0 (5 J1) do a A5 vo c s ta cphng n ti u khc: X ( 2 ) = X(2) = (0,16,8,0,10,0,0) vi c s {A5, A3, A2}. optVy khi bi ton c v s phng n ti u: {Xopt}={ X (1) +(1-) X ( 2 ) ; 0 1 } v fmin = 44 optoptBi 6. Gii bi ton bng phng php n hnhf (x) = 2x1 + x2 - x3 - x4 minx 1 x 2 + 2 x 5 2 x 4 = 23x + x 3x + x = 6 123 4x 1 + x 2 + x 3 + x 4 =7 ___x 0 ( j = 1,4) jV trong matrn h s ca cc n trong h rng buc khng cha vct n v no, ta lp biton M vi cc n gi x5, x6, x7 0 c dng: __f ( x ) = 2x1 + x2 - x3 - x4 + M (x5 + x6 +x7) minx 1 x 2 + 2 x 5 x 4 + x 5 = 22 x + x 3x + x x=6 12 3 4 6x 1 + x 2 + x 3 + x 4 + x 7 = 7 ___x 0 ( j = 1,7); M >> 0 jBi ton M c phng n cc bin xut pht:43 45. Chng II: Quy hoch tuyn tnh___()) x = (0,0,0,0,2,6,7) vi c s n v: {A5,A6,A7} = E3Lp bng n hnhC1A121 -1-1 M M M PhngCCA1 n A2 A3A4 A5 A6 A7ssM A5 21-1 2-1 1 0 0__( 0 ) x MA6 62 1-3 1 0 1 0M A7 71 1 1 1 0 0 115M 4M-2M-11 M+10 0 0__(1) 2A1 21-1 2 -10 0 x MA6 20 3-731 0 MA7 50 2-120 1__( 2 )7M+4 0 5M-3-8M+5 5M-1x 2A1 8/31 0 -1/3 00-1A4 2/30 1 -7/3 10M A3 11/3 0 0 11/3 01 11M+14 0-2 11M+8 0 0__( 3 ) 3 3x 2A13 1000-1A43 0101-1A31 00102 0-2 00___Bng n hnh ng vi phng n cc bin x-(3) c k 0 (k = 1,7) nn bi ton M c____( 3 )phng n ti u: x opt = x = (3,0,1,3,0,0,0) trong cc n gi x5 = x6 = x7 = 0. Vy bi tonban u c phng n ti u l:__ xopt = (3,0,1,3) v fmin = f( x opt) = 2Bi 7. Gii bng phng php n hnh bi ton sauf(x)t = x1 - x2 +3x3 min 2 x 1 + x 2 + 3 x 5 = 2 x 1 + x 2 + 2 x 3 =1 ___ x j 0 ( j = 1,3) Gii.a v bi ton M c 2 n gi x4, x5 0, c dng:___f( x ) = x1 - x2 +3x3 + M (x4 + x5) min 2 x 1 + x 2 + 3 x 5 + x 4 = 2 x 1 + x 2 + 2 x 3 + x 5 =1 ___ x j 0 ( j = 1,5); M >> 0 44 46. Chng II: Quy hoch tuyn tnh__( 0 )Bi ton M c phng n cc bin xut pht: x = (0,0,0,2,1) vi c s n v {A4, A5 }= E2.Lp bng n hnh:C1 A 11 -12 M M PhngC C nA1 A2A3A4 A5s s__( 0 )x M A42-21310 MA51 11201__(1) x 3M-M-12M+1 5M-2 0 0 MA41/2- 7/2 -1/2 0 1 MA31/2 1/2 1/2 1 0M + 2 7M M+4 0 0 2 2 2__((1) ___Bng n hnh ng vi phng n cc bin x c k 0 (k = 1,5) nn bi ton M c:__ __(1)1 1x opt = x = 0,0, , ,0) . Nhng c n gi x4 = 1/2 > 0 nn bi ton u khng c phng n ti2 2u (v nghim)Bi 8. Gii bng phng php n hnh bi ton sau. f(x) = x1 - 2x2 - 4x3 max2 x 1 + 3x 2 4 x 5 2 3 x 5 x + x-3123x 1 + x 2 x 4 = 6 ___x 0 ( j = 1,4)jGii. a bi ton v dng chnh tc vi hai bin ph x5, x6 0 ta c: f(x) = x1 - 2x2 - 4x3 max 2 x 1 + 3x 2 4 x 5 + x 5=5 3 x 5 x + x + x =-3 123 6 x 1 + x 2 x 4 = 6 ___ x 0 ( j = 1,6) j Trong ma trn h s ca n h rng buc ca bi ton chnh tc c mt vct n vA5. a v bi ton M vi hai bin gi x7, x8 0 ta c:45 47. Chng II: Quy hoch tuyn tnh ___ f( x ) = x1 - 2x2 =- 4x3 - M (x7 + x8) max2 x 1 + 3x 2 4 x 5 + x 5=5 3 x + 5 x + x + x + x = 3123 6 7x 1 + x 2 x 4 + x 8 =6 ___x 0 ( j = 1,8) jBi ton M c phng n cc bin xut pht:__( 0 )x = (0,0,0,0,5,0,3,6) vi c s: {A5, A7, A8} = E3Lp bng n hnhC1 A12 A11-1-1MMMPhngC Cn A1A2A3 A4 A5 A6 A7 A7 ss__( 0 )0 A55 2 3 -4010 00 x -MA73 3 5 -100 -1 10-M A86 1 10 -100 01 -9M 04M-4 -6M+2 M0 M00__(1) 0A5 16/5 1/5 0-17/5 0 1 3/5 0 x -2A23/5 3/5 1 -1/5 0 0-1/5 0-M A8 27/5 2/5 01/5 -10 1/5 1-27+6 -2M+10 -M-2 M 0-M-2 0 555 5__( 2 ) 0 A5 3 0 -1/3-50/15 01 2/3x 1 A1 1 15/3 -1/300-1/3 M A8 5 0 -2/31/3 -10 1/3-5M+1 02M+11M - 1 -M+10-M-1 3330A553 0-70 -1014__( 3 ) 1A1 6 1 10-100x 0A315 0-21-3016 03 0 0000A6 53/40 -7/4 0 -10/4 1/4 1__( 4 ) 1A1 6 11 0-100 x 0A37/40 -1/4 1 -2/4 -1/4 06 03 00 00 __( 3 ) ___Bng n hnh ng vi phng nx ca bi ton M c k 0 (k = 1,8) nn__ __( 3 ) x opt = x = (6,0,15,0,53,0,0,0) v cc n gi x7 = x8 = 0. Do bi ton ban u c: __( 3 )x (opt = (6,0,15,0) v fmax = 6. Nhng trong bng n hnh ng vi x c 6 = 0 (6 J3) v Z36 =1)1 > 0 do a A6 vo c s ta c phng n ti u khc:46 48. Chng II: Quy hoch tuyn tnhx (opt) = (6,0,1 / 4,0,0,53 / 4,0,0) cc n gi x7 = x8 = 0 nn bi ton u c4x (opt) = (6,0,7 / 4,0) .2Vy khi bi ton c v s phng n ti u:{Xopt} = { x (1) + (1 ) x ( 2 ) ;0 1optopt} v fmax = 6Bi 9. Gii v bin lun bi ton sau theo tham s f(x) = 5x1 + 2x2 + 5x3 min 3 ( x 1 1) x 2 + 4 x 5 2 - 3 x 1 + x 2 + x 33 ___ x j 0 ( j = 1,3); 0 Gii. a bi ton v dng chnh tc vi 2 bin ph x4, x5 0 sau a v bi ton M__vi n gi x6 0: f( x ) = 5x1 + 2x2 + 5x3 + Mx6 min3 ( x 1 1) x 2 + 4 x 5 + x 6 = 2 - 3x 1 + x 2 + x 3 + x 5=3 ___x j 0 ( j = 1,6); 0; M >> 0__ ( 0 )Bi ton M c ngay phng n cc bin xut pht: x = (0,0,0,0,3,2) vi c s n v:{A6, A5 } = E2.Lp bng n hnh C1 A1 Phng 52500 M C C n A1A2A3A4 A5 A6 s s__( 3 )MA62 3-14 -10 1x 0 A53 1101 0__( 4 ) 2M 3M-5 -M-2 4M-5 - M 00x 5 A3 1/2 3/4 -1/41-1/4 00 A5 5/2/4 5/4 0 1/4 15/2 15-20 -13/40-5/4 0__( 2 ) 4x 5 A1 2/(3)1 -1/(3) 4/(3) -1/(3) 00 A5 7/3 04/3 -1/31/3 1 10/(3) 0-5- 6 20-15 -503 3 3 (0)Trong bng n hnh ng vi xtn ti: 1 = 3M -5 v 3 = 4M -5 > 0. 4 (1 )a. Nu 0 1 v 3 > 0. a vct A3 vo c s ta c phng n x = 3(0,0,1/2,0,5/2,0) n gi x6 = 0 nn bi ton ban u c xopt = (0,0,1/2) v fmin = 5/2.47 49. Chng II: Quy hoch tuyn tnh__( 0 )__(1) 4b. Nu th 1 3 > 0 a vct A1 vo c s ci tin x x ta c phng n 3__(1) 10 xopt= (2/(3),0,0,0,7/3,0) n gi x6 = 0 nn bi ton ban u c xopt = (2/(3),0,0)v fmin = 34Kt lun: nu 0 < bi ton c xopt = (0,0,1/2) v fmin = 5/2.34 10Nu bi ton c xopt =(2/(3),0,0) v fmin=3 3Bi 10. Gii v bin lun bi ton sau theo tham s a: f(x) = -2x1 - 5x2 + x3 max ax 1 2 x 2 1 x 1 + ax 2 + x 3 = 5 ___ x j 0 ( j = 1,3) Gii.Nu a 0 th rng buc 1 khng tho mn nn bi ton khng c phng n no (tp cc phngn D = ).Nu a > 0, bi ton v dng chnh tc vi bin ph x4 0, sau a v bi ton M vibin gi x5 0.__f( x ) = -2x1 - 5x2 + x3 - Mx5 maxax 1 2 x 2 - x 4 + x 5 = 1x 1 + ax 2 + x 3 = 5 ___x j 0 ( j = 1,5); M >> 0 ___( 0 )Bi ton M c phng n cc bin xut pht x = (0,0,5,0,1 vi c s n v: {A5, A3 } = E2.Lp bng n hnhA1 Phng - 2 C1 -5 10 -MC nC A1A2A3 A4A5ss__( 0 ) A5 -M1a -20 -10x A5 5 11a1 0 1__(1) -M+5 - aM+3 2M + a + 50 M 0 x - M A5 1 - 5a0 - 2 - a2-a - 10- 2 A1 51a1 0 1 __(1)0 (2+a2)M-2a+5 aM-3 M 0x -2 A1/2 1 - 2/a 0 -1/a1 1 A5 5- 1/a0a + 2/a1 1/a 5- 3/a 0 a + 6/a+5 0 3/a48 50. Chng II: Quy hoch tuyn tnh (0)Trong bng n hnh ng vi phng n x: tn ti 1 = - aM + 3 < 0, nn vct A1 ca vo c s, v chn:1 5 1 5 = min . nu > 0 0 (a < 1/5) nn bi ton u khng cphng n ti u.1 1(1 ) Nu 5 a , vct A5 c a ra khi c s, ta c phng n x opt = (1/a,0,5-a 51/a,0,0) n gi x5 =0, nn bi ton ban u c xopt = (1/a,0,5-1/a) v fmin = 5-3/a. Kt lun: Nu a < 1/5 bi ton ban u v nghim 1 Nu a bi ton c: xopt = (1/a,0,5-1/a) v fmin=5-3/a. 52.7. I NGU CA BI TON QUY HOCH TUYN TNH.2.7.1 Cc dng bi ton i ngu. a, Bi ton i ngu khng i xng Xt bi ton qui hoch tuyn tnh dng chnh tc. n f(X) = c xj= 1 j j min(2.19) n aij x j = bi (i = 1, m)(2.20) j =1 x 0( j = 1, n) (2.21) j Tng ng vi n ta c bi ton i ngu khng i xng: n(Y) = b yj= 1 i i max (2.22)m aij x j c j ( j = 1, n)(2.23) i =1y R (i = 1, m) (2.24) j b. Bi ton i ngu i xng. Xt bi ton qui hoch tuyn tnh dng chun tc.nf(X) = c x j= 1 j j min (2.25) n aij x j bi (i = 1, m) (2.26) j =1x 0( j = 1, n)(2.27) j49 51. Chng II: Quy hoch tuyn tnhTng ng vi n ta c bi ton i ngu i xng.n(Y) = b y j= 1 i i max (2.28)m aij x j c j ( j = 1, n) (2.29) i =1y 0 (i = 1, m) (2.30) jc. Bi ton i ngu tng qutXt bi ton qui hoch tuyn tnh dng tng qut sau:nf(X) = c x j= 1 j j min n aij x j = bi(i I1 )(2.31) j =1 n aij x j bi (i I 2 )(2.32) j =1 n a x b ij j i (i I 3 )(2.33) j =1 xj R ( j J1 )(2.34) x 0( j J2 )(2.35) j xj 0 ( j J3 )(2.36) Trong : Ii , Ji (i = 1,3 ) - Tp ch s cc phng trnh, bt phng trnh v cc n. I1+I2+I3= m , J1+ J2+J3= n Vi Ii,Ji- lc lng ca tp Ii, Ji (i = 1,3 ).Tng ng vi n ta c bi ton i ngu tng qut sau: n (Y) = b yj= 1 i i max (2.37)m aij y j = c j ( j J1 ) (2.20) i =1m aij y j c j( j J2 ) (2.38) i =1m a y c( j J3 )(2.39) i =1 ij jjyj R(i I1 ) (2.40)y 0 (i I 2 )(2.41) jyj 0(i I 3 )(2.42)Ch : Khi thnh lp bi ton i ngu cn ch ma trn h s cc n trong h rng bucca hai bi ton l hai ma trn chuyn v ca nhau.50 52. Chng II: Quy hoch tuyn tnh2.7.2. Cp rng buc i ngu Gi hai cp rng buc (k c rng buc v du) trong hai bi ton i ngu cng tng ngvi mt ch s l mt cp rng buc i ngu.Chng hn, vi cp bi ton i ngu i xng c tt c m+n cp rng buc i ngu sau:n a x j=1 ij j bi ,(i = 1, m) yi 0,(i = 1, m)n a x i=1 ij j bi ,(i = 1, m) yi R,(i = 1,m) m xj 0,( j = 1,n) a ij ci ,( j = 1,n) i=12.7.3. Cc tnh cht ca bi ton i ngu.1. i vi hai phng n bt k X,Y ca mt cp bi ton i ngu f(X) min ta lun c:f(X) (Y)2. Nu i vi hai phng n X* , Y* ca mt cp bi ton i ngu m f(X*) = (Y*)th X* , Y* l hai phng n ti u.2.7.4. Quan h ca cp bi ton i ngunh l: i vi mt cp bi ton i ngu, bao gi cng ch xy ra 1 trong 3 trng hp sau: C hai bi ton cng khng c phng n, hin nhin c hai bi ton u khng gii c. C hai bi ton c phng n th c hai bi ton u gii c. Khi vi mi cpphng n ti u X* Y* ta lun c:f(X*) = (Y*) . Mt trong hai bi ton khng c phng n th bi ton cn li nu c phng n th cngkhng c phng n ti u.2.7.5. Cc bi tp muBi 1. Hy thit lp bi ton i ngu ca bi ton sau:a. f(X) = x1-3x2+ 2x3 min2 x1 + x 2 =3 x + 3x x=2 1 2 35x+ 7x3 = 14 1x j 0( j = 1,3)b. f(X) = -3 x1-5x2 +4x3+x4 maxx1 + x 2 + x 3 x 4 6 3x + 2 x + x 5 2 342 x1 + x 2 3x 4 4x j 0( j = 1,4 )51 53. Chng II: Quy hoch tuyn tnhGii a. Ta thy bi ton qui hoch tuyn tnh dng chnh tc c hm mc tiu f(X) min, nn hmmc tiu trong bi ton i ngu l (Y) max v h rng buc bi ton s c dng " ".Bi ton cho c 3 rng buc (khng k rng buc v du) vi du "=" nn trong bi toni ngu s c 3 bin: y1,y2, y3 v cc bin ny s nhn cc gi tr tu .Theo nh ngha bi ton i ngu, ta c bi ton i ngu khng i xng sau:(Y) = 3y1 + 2y2 +14y3 max 2 y1 y 2 + 5y 3 1 y + 3y 3 1 2 y + 7 y 2 23 y i R , ( i = 1,3) b. y l bi ton qui hoch tuyn tnh dng chun tc c hm mc tiuf(X)max, nn hm mc tiu trong bi ton i ngu l (Y) min v h rng buc s cdu " ".Bi ton cho c 3 rng buc vi du " " (khng k rng buc du) nn bi ton ingu s c 3 bin y1, y2, y3 0.Theo nh ngha bi ton i ngu, ta c bi ton i ngu i xng sau y:(Y) = 6y1-5y2+4y3 miny1 + 2 y 2 3y1 3y 2 + y 3 5y1 + 2 y 2 4 y + y 3y 1 1 23y i 0 ( i = 1,3)Bi 2. Xt bi ton: f(X) = -2x1+x2-4x3 min3x1 6x 2 + 2 x 4 = 82 x + 3x + x 9 12 34 x 2 + 7 x 3 + 5x 4 10x1 , x 3 0, x 4 0, x 2 RVit bi ton i ngu v ch ra cc cp rng buc i ngu ca hai bi ton.Gii Bi ton i ngu.(Y) = -8y1+9y2+10y3 max. 3y1 - 2y2-2- 6y1 + 3y2 - 4y3 = 1 y2 + 7y3 - 4 2y1 + 5y3 0y1 R,, y2 0, y3 052 54. Chng II: Quy hoch tuyn tnhCc cp rng buc i ngu. 3x1-6x2+2x4 = 8 y1 R -2x1+3x2+x3 9 y2 0 -4x2+7x3+5x4 10 y3 0 x1 0 3y1-2y2 -2 x2 R -6y1+3y2-4y3 = 1 x3 0 y2 +7y3 -4 x4 0 2y1+5y3 0Bi 3. Cho bi ton: f(X)= 7x1+6x2-12x3+x4 max.2x1 - 2x2 - 3x3 + 2x4 = 83x2 + 2x3 - 2x4 - 12x1 -3x3 + x4 = 10x3 0 (j = 1,4)Lp bi ton i ngu ca bi ton trn v xc nh cc cp rng buc i ngu.Xt 2 vct X0 = (0,6,0,10), Y0 = (-3,0,7). Chng t rng X0, Y0 l phng n ti u ca cp biton i ngu.GiiBi ton i ngu c dng:(Y) = 8y1- y2 + 10y3 min 2 y1 + 2 y 3 7 2 y + 2 y 6 1 3 3y1 + 2 y 2 3y 3 12 2 y 2 y + y 1 123 y1 R , y 2 0, y 3 R Cc cp rng buc i ngu:2x1-2x2-3x3+2x4 = 8 y1 R3x2 + 2x3 - 2x4 -1 y2 02x1-3x3+x4 = 10 y3 Rx1 0 y1 +2y3 7x2 0 -2y1 +3y2 6x3 0 -3y1 +2y2 -3y3 -12x4 0 2y1-2y2+y3 -1Theo tnh cht 2 i ngu, chng minh X0, Y0 l phng n ti u ta phi chng minh:X0, Y0 l phng n ca cp bi ton i ngu v f(X0)= (Y0). 53 55. Chng II: Quy hoch tuyn tnhTht vy, thay X0 = (0,6,0,10) vo tt c cc rng buc ca bi ton gc ta thy X0 tho mnmi rng buc ca bi ton gc.Suy ra X0 l phng n ca bi ton gc. Thay Y0 = (-3,7,0) vo cc rng buc ca bi ton i ngu thy tho mn suy ra Y0 lphng n ca bi ton i ngu.Mt khc ta li c:f(X0) = 7.0+6.6-12.0+10 = 46(Y0) = 8.(-3)-0+10.7= 460 0 Nnf(X ) = (Y ). 00Vy X , Y l phng n ti u ca cp bi ton i ngu (tnh cht 2).Bi 4. Cho bi ton. f(X) = -x1-5x2+4x3-2x4 min x1 4 x2 + x3 6 x4 13 x + 2x + 3 x4 9 1 2 3 x x x + 2 x 8 1 234 x j 0 ( j = 1; n)Dng nh l i ngu chng t bi ton cho gii cGiiBi ton i ngu:(Y) = 13y1+9y2+8y3 maxy1 + y 3 3y 3 1 4 y 1 + 2 y 2 y 3 5 y1 y 3 46y + 3y + 2 y 2 1 2 3y i 0 ( i = 1,3) chng t bi ton cho gii c ta cn chng minh:- Bi ton gc c phng n.- Bi ton i ngu c phng n. Tht vyVi bi ton gc, xt vct X0 = (0,0,0,0). Thay vo h rng buc ca bi ton ta c: x1-4x2+x3-6x4 = 0 0.Gi s X l phng n ti u, theo nh l v lch b mi phng n ti u Y ca biton i ngu phi tho mn h phng trnh: y1 + 4 y 2 y 3 = 5 y1 + 4 y 2 y 3 = 5 3y1 y 3 = 9 y + 2 y = 76y 2 y 3 = 12 1 21y1 = 3 + 3 y 3 1 y 2 = 2 + y 3 6y 3 RVy nghim ca h l:1 1 Y = (y1,y2,y3) = ( 3 + y 3 , 2 + y 3 , y 3 ), y3 R3 6Thay Y vo cc rng buc cn li ca bi ton ta c:11 -y1+y2+y3 15 3- y3 +2 + y3 +y3 15 y3 12.(1)361 1 y1-y1-2y3 6 -3 + y3 -2 - y3 -2y3 6 y3 -6 (2)3 61 y1 0 -3+ y3 0 y3 9 (3)3 y3 0 (4)T (1), (2), (3), (4) Y mun l phng n th y3 phi tho mn iu kin: 0 y3 9Kt lun: X l phng n ti u ca bi ton gc v Y vi 0 y3 9 l phng n ti uca bi ton i ngu (theo nh l lch b).Bi 2. Mt nh my c 3 phng php khc nhau sn xut ra 3 mt hng 1,2,3. s lnghng loi i phi sn xut ti thiu l bi chic. Nu p dng cch sn xut th j trong mt n vthi gian th chi ph l cj v thu c ai j n v hng loi i. Cc s liu c cho trong bng sau:58 60. Chng II: Quy hoch tuyn tnh j1 2 3 bji 14 4 1 16 25 3 1 20 31 1 1 3cj108 2Yu cu i vi nh my l cn s dng phng php sn xut sao cho m bo nhu cu vsn phm ng thi chi ph l t nht. a. Lp dng ton hc ca bi ton trn? Vit bi ton i ngu ca n (ni r ni dung kinht cc bin trong bi ton i ngu).b. Dng l thuyt i ngu, chng t vct X = (4,0,0) l phng n ti u ca bi tongc. Phn tch ngha kinh t trong mi quan h gia hai bi ton trnc s hai phng n ti uca chng.Giia. Gi xj l thi gian p dng phng php sn xut th j (j= 1,3 ). Dng ton hc ca biton ny l:Tm vct X = (x1,x2,x3) R 3 :f(X) = 10x1+8x2+2x3 min 4 x1 + 4 x 2 + 2 x 3 16 5x + 3x + x 20 12 3 x + x + x 3 1 2 3 x j 0 ( j = 1,3) Bi ton i ngu:Gi yi l gi tr ca 1 n v hng loi i (i= 1,3 ). Ta c:(Y) = 16y1+20y2+3y3 max 4 y1 + 5y 2 + y 3 10 4 y + 3y + y 8 12 3 2 y + y + y 2 12 3 y i 0 ( i = 1,3) Vy ni dung kinh t th f(X) chnh l tng chi ph, (Y) l tng gi tr sn phm sn xutc.59 61. Chng II: Quy hoch tuyn tnhb. Gi s X = (4,0,0) l phng n ti u v X tho mn lng cc rng buc (3) v x1 = 4 >0 nn theo nh l v lch b mi phng n ti u Y =(y1,y2,y3) ca bi ton i ngu phi y 3 = 0tho mn h phng trnh: 4 y1 + 5y 2 + y 3 104 4y1+5y2 = 10 y2 = 2-y1 , y1 R5 4Vy nghim ca h l: Y= (y1, 2- y1,0 ) , y1 R 5Thay vo cc rng buc cn li ta c:45 4y1+3y2+y3 8 4y1+3(2- y1) 8 y1 (7)54 4 2y1+y2+y3 2 2y1 + 2-y1 2 y1 0(8) 5 55 y2 0 2- y1 0 y 1 (9) 42 y1 0 (10)T (7), (8), (9), (10) y1 = 0Vy Y = (0,2,0) l phng n ca bi ton i ngu, nn theo nh l v lch b th X =(4,0,0), Y = (0,2,0) tng ng l phng n ti u ca bi ton gc v bi ton i ngu.Phn tch ngha kinh t: Ta c x1 = 4 >0 chng t cch sn xut th nht c s dng, ch cn 4 n v thi giancho cch sn xut th nht l p ng c nhu cu v sn phm v tng cc gi tr cc sn phmsn xut c l (Y) =16.0 +20.2+3.0 = 40. Tng ng trong nh gi ti u rng buc (4) thomn cht, iu ny ni nn vi phng php sn xut mi hao ph b ra u c chuyn hothnh gi tr sn phm. y l cch sn xut ti u cn c p dng. Vi cch sn xut ny nuta tng thm 1 n v thi gian th c th xy dng c phng n ti u mi vi tng gi tr snphm cao hn trc l 10.Thay phng n ti u Y vo cc rng buc ca bi ton i ngu ta c rng buc (5)tho mn lng, chng t vi phng php sn xut th hai mi hao ph b ra khng c chuynho ht thnh gi tr sn phm nn khng s dng phng php sn xut ny (x2 = 0). Trong phng n ti u Y ta c y2 = 2>0, iu ny ni nn sn phm hng loi 2 c nh gic gi tr v chnh n c vai tr lm tng tng gi tr sn phm. Tng ng trong phng n ti u Xrng buc (2) tho mn cht, ngha l sn phm hng loi 2 sn xut va nhu cu ti thiu, snphm n u tiu thu n y. V vy trong phng n sn xut theo hng sn phm hng loi 2phi tng thm v s lng. y rng buc (6) ca bi ton i ngu cng tho mn cht, iu ny ni nn c th sdng cch sn xut th 3 thay th cch sn xut th nht trong trng hp cch sn xut th nhtgp kh khn v trang thit b, nhin liu.... Rng buc (3) ca bi ton gc tho mn lng nn60 62. Chng II: Quy hoch tuyn tnhtrong phng n nh gi ti u Y, sn phm ny c xem l khng c gi tr, ngha l trongphng n sn xut tip theo khng nn ch trng sn xut sn phm ny.2.9. PHNG PHP GII BI TON I NGU I XNG2.9.1. t vn .i vi cp bi ton i ngu i xng , li gii ca bi ton ny c th thng qua li giica bi ton kia (bng phng php n hnh) m khng cn gii trc tip n.Xt cp bi ton i ngu i xng sau: n(P): f(X) = c x j =1j j max n a ijx j b i (i = 1, m) j=1 x 0 ( j = 1, n) jn(Q): (Y) =b yj =1i i minm a ijx j c j ( j = 1, n) i =1y 0 (i = 1, m) jGi s Y* =(y1*, y2*, ...,ym*) l phng n ti u ca bi ton (Q). tm Y* ta gii bi ton (P).a bi ton (P) v dng chnh tc, ta dc bi ton (P): n(P): f(X) =c xj =1j j maxn a ij x j + x n + i = b i (i = 1, m) j=1x 0( j = 1, n + m) jGii bi ton (P) bng phng php n hnh: Nu bi ton (P) v nghim th bi ton (Q) cng v nghim. Nu bi ton (P) c nghim tc c phng n ti u th b ton (Q) cng c phng nti u v t dng cui cng ca bng n hnh ng vi phng n ti u ca bi ton (P) ta suyra c phng n ti u ca bi ton (Q).Cng thc suy nghim: Y* =(y1*, y2*, ...,ym*) = (*n+1, *n+2,....., *n+m).Trong *n+i , (i= 1, m ) l cc s kim tra ca cc bin ph xn+i 0,(i= 1, m ) trong bng n hnh ng vi phng n ti u ca bi ton (P)). 61 63. Chng II: Quy hoch tuyn tnhGi s x* =(x1*, x2*, ...,xn*) l phng n ti u ca bi ton (P) th nghim X* ca biton cng c tm thng qua cc s kim tra tng ng ca bi ton i ngu (Q) trong bngn hnh ng vi phng n ti u. x* =(x1*, x2*, ...,xn*) = (-*m+1,- *m+2,....., -*m+n).Trong *m+j , (j= 1, n ) l cc s kim tra ca cc bin ph ym+j 0 (j= 1, n ) trong bngn hnh ng vi phng n ti u ca bi ton (Q)).i vi cp bi ton i ngu i xng , vai tr ca bi ton gc v bi ton i ngu c ththay th cho nhau.2.9.2. Cc bi tp muBi 1. Cho bi ton: f(X) = x1+3x2 +4x3+x4 min x1 2 x 2 + 2 x 4 4 3x x + 4 x 9 23 4 3x + x + 2 x x 10 12 3 4 x j 0 ( j = 1,4) - Vit bi ton i ngu.- Gii bi ton i ngu bng phng php n hnh, t suy ra nghim cho bi ton trn.Gii(Y) = 4y1- 9y2+10y3 max. y1 3y 3 1 2 y1 + 3y 2 + y 3 3 y2 + 2 y3 4 2 y + 4 y y 1 1 2 3 y i 0 (i = 1,3) a bi ton i ngu v dng chnh tc bng cch thm vo bi ton cc bin ph yi 0(i= 1,3 ) . Ta c:(Y) = 4y1- 9y2+10y3 max. y1 3y 3 + y 4 = 1 2 y1 + 3y 2 + y 3 + y5 = 3 y2 + 2 y3 + y 6 = 4 2 y + 4 y y + y = 1 1 2 3 7 y i 0 (i = 1,7) T bi ton dng chnh tc ny ta c ngay phng n cc bin xut pht Y0 = (0,0,0,1,3,4,1) vi c s n v E4 = {A 4 , A 5 , A 6 , A 7 } trong R4 .Lp bng n hnh gii bi ton chnh tc vi phng n cc bin xut pht Y0 .62 64. Chng II: Quy hoch tuyn tnh Bng 3 c j 0 j= 1,7 . Phng n ti u ca bi ton chnh tc l:311Yopt = ( ,0,2,,4,0,0) , max = 26.223 Suy ra phng n ti u ca bi ton i ngu l: Yopt = ( ,0,2) , max = 26.2 T bng 3 ta suy ra phng n ti u ca bi ton gc l:Xopt = (x1,x2,x3,x4) = ($, 5, 6, 7) = (0,0,6,2) v fmin = max = 26. Bi 2. Cho bi ton:f(X) = 2x1- x2 + 4x3 max x1 + x 2 2 x 3 5 x 2 x 4 12 x + x2 23 x j 0 ( j = 1,3) Vit bi ton i ngu. Tm nghim ca bi ton trn thng qua bi ton i ngu ca n. Gii Bi ton i ngu:(Y) = 5y1+4y2+2y3 min y1 + y 2 2 y 2 y + y 1 12 3 2 y + y 4 13 y i 0 (i = 1,3) (Y) = 5y1+4y2+2y3 min y1 + y 2 2 y + 2 y y 1 12 3 2 y + y 4 13y i 0 (i = 1,3) ( Y ) = 5y1+4y2+2y3 min y1 + y 2 y 4 =2 y + 2 y y + y = 1 12 35 2 y + y y6 = 4 13 y i 0 (i = 1,6) ( Y) = 5y1+4y2+2y3 +My7+My8 min 63 65. Chng II: Quy hoch tuyn tnh y1 + y 2 y 4 + y 7 = 2 y + 2 y y + y = 1 123 5 2 y + y y 6 + y8 = 4 13 y i 0 (i = 1,8) T bi ton ny ta c ngay phng n cc bin xut pht: Y0 = (0,0,0,0,1,0,2,4) vi c s n v E3 = {A 7 , A 5 , A 8 } trong R3. Lp bng n hnh gii bi ton vi phng n cc bin xut pht Y0 bit. Bng 3 c: j 0 j = 1,6 . Phng n ti u ca bi ton (M) l: Yopt = (0,2,4,0,1,0,0,0) , opt = 16. Phng n ti u ca bi ton chnh tc l: Yopt = (0,2,4,0,1,0) , min = 16. Phng n ti u ca bi ton i ngu l: Yopt = (0,2,4) , min = 16. T bng 3, suy ra phng n ti u ca bi ton gc l: Xopt = (x1,x2,x3) = (-4 ,- 5 ,- 6) = (4,0,2) v fmax = min = 16.BI TP CHNG 2Gii bng phng php n hnh 1. f(x) = 2x1 + 3x3 maxx 1 + x 2 + x3 =5x + 3x 2 + x 4 =9 1x 1 x5 =4x + 2x 2 + x 6 =8 1___x j 0 ( j = 1,6) S: xopt = (3,2,0,0,1,1); fmax = 12 2. f(x) = 2x1 +x2 + 2x3 + 5x4 - 5x5 - 5x6 max 2 x 1 + 3 x 2 + x 3 + x 6 =1x + 4 x + x + x=4 12 3 5- x 1 3x 2 + x 3 + x 4=4 ___x 0 ( j = 1,6) j S: bi ton khng c li gii (f(x) + ) 3. f(x) = 7x1 -21x2 + 14x3 min64 66. Chng II: Quy hoch tuyn tnh 2 x 1 3x 2 + x 3 6 x + 7 x 3x1 1 2 3 - 3x 1 x 2 + x 3 = 3 ___ x 0 ( j = 1,3) j S: xopt = (0,1/7,0,45/7,0,20/7); fmin = -3 4. f(x) = x1 +2x2 - 2x4 - x5 - 3x6 min x 1 + x 2 + x 4 - x 6 = 2 x + x + x = 12 45 6 4x 1 + x 3 + 2x 4 + 3x 6 = 9 ___ x 0 ( j = 1,6) j 5. f(x) = 2x1 +x2 - x3 -x4 max x 1 + x 2 + 2x 3 - x 4 5 x + 2x 2 + x 47 1 x 1 + x 3 + 2x 43 ___ x 0 ( j = 1,4) j S: xopt = (3,2,0,0); fmax = 8 6.f(x) = -5x1 +5x2 - 9x3 + 3x4 min 3 x 1 + 2 x 3 x 4 13 4 x 3x + 8x 25 1 34 x 2 2 x 3 + 2x 4 =4 - 2x + x 3x - 13 1 34 ___ x j 0 ( j = 1,4) S: bi ton khng c li gii (f(x) - )9 7.f(x) = 2x1 +7x2 - 5x3 +x 2 min2 x 1 2 x 2 x 3 + 3x 4 = 14 x 4 x + x 8 2 34 x 2 2 x 3 + 3x 4=4 - 2x + x 3x - 20 134 ___ x j 0 ( j = 1,4) S: xopt = (0,0,34,16): fmin = -98 8. Cho bi ton quy hoch tuyn tnh65 67. Chng II: Quy hoch tuyn tnh f(x) = 6x2 +4x4 + x5 + x6 minx 1 x 2 + 3 x 3 - 7x 4 - 2x 5 + x 6= 14 4 x + 6 x 9 x + 21x + 5x - 2x= - 451 23 45 6- 2x 1 + 4 x 3 x 3 + 2x 4 - x 5 + x 6= - 15___x 0 ( j = 1,6) jChng t x(0) = (12,0,0,0,7,16) l mt phng n cc bin ca bi ton. Li dng x(0) giibi ton bng phng php n hnh.Hng dn v p s: Chng t x(0) tho mn cht 6 rng buc trong h rng buc ca bi ton v 6 vct tngng c lp tuyn tnh. gii bng phng php n hnh cn bin i matrn m rng ca h phng trnhtrong h iu kin a cc vct A1, A5, A6 v cc vct n v. Ta c bi ton tng ng.f(x) = 6x2 + 4x4 + x5 + x6 minx 1 x 2 + x 3 - 2x 4 = 144 x + x + x = 16 24 62x 2 x 3 + 3x 4 + x 5 = 7 ___x 0 ( j = 1,6) jPhng n cc bin x(0) = (12,0,0,0,7,16) ng vi c s n v:{A1, A5, A6, } = E3Bi ton c v s phng n ti u, tp cc phng n ti u:{xopt} = {0 x (opt) + 1 x (opt + 2 x (opt) 0 1) 2} vi 0 0, 1, 2 1 v 0 + 1 2 = 1Trong : x ( 0 ) = (50/3,0,0,7/3,41/3). v fmin = 23opt9. Cho bi ton:f(x) = x1 +2x2 + x3 +3x4 - 2x5 max3x 2 + 7 x 2 x 3 + 7x 4 - 2x 5 =9x + 8x + 9x + 7x= 25 24 5x + x + x=3 345 ___ x j 0 ( j = 1,5) Chng t x(0) = (3,0,0,2,1) l mt phng n cc bin ca bi ton. Gii bi ton viphng n cc bin xut pht . H.D: ch ra x(0) tho mn cht 5 rng buc c lp tuyn tnh. Gi tng t bi tp 9. xopt =(5,1,3,0,0); fmax = 10.10. f(x) = 2x1 +10x2 + 4x3 +8x4 + 8x5 + 3x6 max66 68. Chng II: Quy hoch tuyn tnhx 1 + x 5=5 3 x 1 + x 5 + 2x 6 = 11 536 x1 + x 2 - x 6 = 455___x j 0 ( j = 1,6) .S: xopt = (5,7,0,0,4,5); fmax = 127 11. f(x) = 5x1 + 4x2 + 5x3 + 2x4 + x5 + 3x6 min2x 1 + x 2 + 3x 3 + 4x 5 = 464 x + 3x x + 2x= 38 1 3 4 53x 1 + x 3 + x 6 = 21 ___x 0 ( j = 1,6) j S:x (1) = (7,12,0,5,0,0); x ( 2 ) = (5,0,0,9,0,6)opt opt{ xopt} = { x (opt + ( 1- ) x (opt) ; 0 1} v fmin = 791) 2 12. f(x) = 2x1 + x2 - x3 - 4x4 min2x 2 2 x 3 + 3x 4 122 x 3x + x= 10 12 32x 1 3x 2 2x 3 - 4x 4 =6___x 0 ( j = 1,4)j S: xopt = (5,0,0,1); fmin = 6 13. f(x) = - x1 + x2 - 3x3 max- 2x 1 + x 2 + 3x 3=2x 1 + x 2 + 2 x 3=1 ___x j 0 ( j = 1,3) S: bi ton khng c phng n 14. f(x) = - x1 - 2x2 - x4+5 max 2x 1 + 3x 2 4x 3 5 3 x 5 x + x -3 1 23 x 1 + x 2 x 4 =6 ___ x 0 ( j = 1,4) j S: Bi ton c v s phng n ti u. Tp phng n ti u c dng: {xopt} = { x (opt +(1 ) x (opt) + 0 1 1)2} 67 69. Chng II: Quy hoch tuyn tnh (1 ) Trong : x opt= (6,0,15,0); x ( 2 ) ;= (6,0,7/4,0);fmax= 11opt 15. f(x) = 6x1 + x2 - 2x3 min 9x 1 + x 2 + x 3 18 15x + x 2 x = 20 1 23 3x 1 + x 3 =3 ___ x 0 ( j = 1,3) j S: x0pt = (1,5,0), fmin = 11 16. f(x) = x1 + 2x2 - 4x3 + 3x4 min2x 1 x 2 + x 3 + x 4=4 6 x + 3x + 3x + 2x= 181 2 3 4- x 1 + x 2 - x 3 + x 4 = 10 ___x 0 ( j = 1,4)j S: xopt = (2,6,0,6), fmin = 32 17. f(x) = x1 - 2x2 +x3 -8x4 + x5 + x6 max x 1 + 4 x 2 - x 3 x 4 + x 6= 12 x 2 x =6 3 4 - 2x 2 + 3x 3 + 6 x 4 2x 5 - x 6 = 12 ___ x 0 ( j = 1,6) j Nu c4 = -11 th c kt lun g ? S: tr s ca f(x) khng b chn trn tp phng n (f(x) + ) Vi c4 - 11 bi ton c phng n cc bin ti u: xopt = (11,0,6,0,3,0) v fmax = 20 18. f(x) = -2x1 - 3x2 +6x3 -4x4 min 2x 1 + 3 x 2 - x 3 + x 4 =1 4 x 1 + 6 x 2 + 3x 3 - 2x 4 =3 ___ x j 0 ( j = 1,4) S: bi ton c tp phng n ti u: {xopt} = { x (opt +(1 ) x (opt) + 0 1 1)2} Trong : x (1) = (0,2/5,1/5,0); x ( 2 ) = (3/5,0,1/5,0); fmin = 0 optopt 19. f (x) = 3x1 + 4x2 - 6x4 max68 70. Chng II: Quy hoch tuyn tnh x 1 + 4 x 2 - 2x 3 + x 4 = 10 2 x x + x-2 2 3 5 7 x 2 + x 2 + 4 x 3 22 4x + 2 x 2 x 4 x = 20 23 45 ___ x j 0 ( j = 1,5) S: Bi ton khng c phng n ti u(f (x) + ). 20.f (x) = 4x1 - 2x2 + x3 - x4 min 2x 1 + 2x 2 + x 4 = 10 x 1 + 5 x 2 + 3x 3 + 3 x 4 32 2 2 x 1 2 x 2 + 2 x 3 + x 4 = 16 ___ x j 0 ( j = 1,4) S: xopt = (8,3,0,6); fmin = 20 21. f (x) = x1 - 2x2 - 2x3 + x6 min 4x 1 + 2x 2 x 4 + x 5 + 2x 6 = 5 x 3x + 2 x - x 10 2 3 46 3x 1 x 2 5x 3 + 3x 4 - 4x 63 x + x + x 2 x + x=4 1 2 345 ___ x j 0 ( j = 1,6) S: Tr s: f (x) - trn tp phng n. Bi ton khng c phng n ti u Gii v bin lun cc bi ton sau y theo tham s m 22. . Cho quy hoch tuyn tnh: f (x) = - 2x1 + x2 max - x 1 + ax 21 x 1 a (3 - x 2 ) = 10 - 3a ___ x j 0 ( j = 1,2); a tham s S: Nu a 0 bi ton khng c phng n ti u (f (x) + ) 11 Nu 0 < a < 1/3; xopt = (9/2; 11/2a); fmax = 9 2a 1 Nu a ; xopt = (10,0); fmax = - 20 3 23. f (x) = 3x1 + 4x2 min69 71. Chng II: Quy hoch tuyn tnhax 1 2x 2 2x 1 + a ( x 2 - 2) + x 3 = 2 (2 - a) ___x j 0 ( j = 1,3); a tham s1 S: Nu a 1; xopt = (, 0,0,); fmin = 2a 2a 25.f (x) = 4x1 + 2x2 max ax 1 + x 2 6 ax 3 1 - 2ax 1 x 2 - 10 ___ x 0 ( j = 1,2); a tham s j S: Nu a 0 bi ton v nghim12 Nu 0 < a 2; xopt = (3/a,3); fmax = +6 a Nu a > 2; xopt = (0,6); fmax = 12 26. f (x) = x1 + x2 + 2x3 - 3x4 min x 1 + x 2 + ax 3 = 2 x + 2ax + x = 3 134 4x 1 + ax 3=1 ___ x 0 ( j = 1,4); a tham s j4 1 4 11 9 S: Nu a < ; xopt = ( , ,0, ); f min . 7 4 7 423 27. f (x) = ax1 + x2 + x3 + 3x4 - x5 min270 72. Chng II: Quy hoch tuyn tnh - 2x 1 + 3x 2 + x 3 + 2x 4 - 6x 5 = 12 5 x 1 + 7 x 2 + 3x 4 - 5x 5 22 2 1 4x 1 x 2 x 4 + 2 x 5 38 2 ___ x j 0 ( j = 1,5) Gii bi ton vi a - 2. Tm gi tr ca a bi ton c v s phng n ti u. Tm mt phng n ti u c x2 = 32 S: x (1) = (0,12,0,0,4); fmin = 8 opt 36 68 Vi a = - 2 tm c phng n ti u na: x ( 2 ) = (opt ,36,0,0, );5 5 Nn tp phng n ti u l: {xopt} = { x (opt + (1- ) x (opt) ; 0, 1 }1)2 Da vo yu cu c x2 = 32 suy ra x (1) + (1 - ) x ( 2 ) = 32 opt opt1 Gii ra c = suy ra phng n ti u cn tm l:6xopt = (6,32,0,0,12) Gii cc bi ton qui hoch tuyn tnh thng qua bi ton i ngu: 28.f(x) = x1 +5 x2 + x3 min 3x1 + 2 x 2 + x 3 4 2 x + x x 2 12 3 x 3x 2 x 1 12 3 x j 0 ( j = 1,3) 44 S: xopt = ( , 0, 0) , fmin =33 29. f(x) = x1+5 x2 + 4x3 - 6x4 max 2 x1 + 3 x2 4 x3 5 x4 1 5 x + 6 x x + x 2 1 2 34 4 x + x 2 x + 3x 2 12 3 4 x 0 ( j = 1, 4) j S: Bi ton khng c phng n. 30. f(x) = -x1+2 x2 + 2x3 +x4 +5x5 max 71 73. Chng II: Quy hoch tuyn tnh2523x1 + 5x 2 x 4 33 2 x 3x + x + 5 x 4 x 13 3 12 33 4 3 5 3x1 + 7x 2 x 3 3x 4 7x5 5x 0 ( j = 1,5) j S: Xopt= (6,0,5,2,0) , fmin = 6. 31. f(x) = 3x1+ x2 + 2x3 -x4 -2x5 max5x1 + x 2 x 3 2 x 4 + 8x 5 5 23 x + 3x 2 x 6x + 20x 202 1 23 34517 3 3 15 x1 + x 2 x 3 x 4 + 6x5 44 222 1 3 x1 + 2 x 33x j 0 ( j = 1,5) S:Xopt = (0,15,6,2,0) , fmax = 25. 32. Gii bi ton qui hoch tuyn tnh sau thng qua bi ton i ngu ca n: f(X) = 2x1+ x2 -x3 min x1 2 x 2 + 2 x 3 + x 4 1x + 3x 2 x 2 x 3 1 23 4 2 x 1 + x 2 2 x 3 x 4 b 32 x + x 3x 3x 21 2 34x1 , x 2 , x 4 0, x 3 0 - Nu trnh t gii bi ton trn. - Vi b3 nhn gi tr dng, tm kt qu ca bi ton.- Tm b3 bi ton c phng n ti u. - Thng qua vic tnh ton gii bi ton i ngu, hy xc nh xem c du hiu no chngt bi ton gc cho c nhiu phng n ti u? Nu c, hy tm mt phng n ti u khc cabi ton.33. Cho bi toan:f(X) = (C,X) max n A j x j = B j= 1 X 0 72 74. Chng II: Quy hoch tuyn tnh Chng minh rng nu Aj= Ak , j k v cj < ck th iu kin cn phng n X* ti u l x*j= 0. HD: Xt hai rng buc j v k ca bi ton i ngu. 34. Chng minh rng bi ton:f(X) = (C,X) max AX B , (A 0 , B 0) X 0 Gii c nu vi mi ch s j u c t nht mt ai j >0.HD: Xt bi ton i ngu nhn thy v tri ca mi rng buc u c t nht mt ai j >0.Suy ra bi ton c phng n. 35. Cho bi ton: n f(x) = c xj= 1 j j maxn a ijx j b i ( i = 1, m) j= 1x 0 ( j = 1. n ) jDng bi ton i ngu chng minh bi ton trn khng gii c nu bi < 0 i vi : a ij >0 j. 36. Vit bi ton i ngu v ch ra cp rng buc i ngu ca bi ton sau: a. f(X) = x1+3x2-x3+2x4 min x1 + 5x 2 + 3x 3 = 3 x + 2 x x = 2 2 3 43x + 2 x + x = 5 124x j 0 ( j = 1,4 ) b. f(X) = 2x1-x2+5x3 max x1 3x 2 + x 3 + 2 x 4 8 3x + x 2 x 4 123 5x + x 12 2 4 x j 0 ( j = 1,4 ) c.f(x) = x1+2x2+3x3+4x4 + 5x5 min73 75. Chng II: Quy hoch tuyn tnh 7 x1 + 3x2 x3 + 2 x 4 1 4 x1 + 2 x3 x4 + 3x5 2 2 x2 + 3x3 + 5 x4 x5 4 x + 3x x + 2 x 3 12 3 5 x j 0 ( j = 1,5) d. f(x) = -4x1+2x2-5x3-x4+x5 max2 x1 + x 3 x 4 + 3x5 12x + 3x x + x = 23 12 3 54 x x + 3x x 4 12 4 5x j 0 ( j = 1,5)e. f(x) = 2x1-3x2+4x3+5x4 -x5 min x1 + 3 x 2 5 x 3 + x 5 x 6 = 4 2 x1 x 2 + x 4 + 2 x 6 83 x + 7 x 2 x 6 13 5 5 x 2 x 3 + 3 x 4 = 12 2 x + x + 3 x 10 1 46 x6 1 x 2 , x 4 0 ; x 3 , x 5 0; x1 , x 6 R 37. Dng l thuyt i ngu chng t cc bi ton sau gii c:a. f(X) = 2x1 +5x3 +3x4 maxx1 2 x 3 + 2 x 4 = 53x + 4 x x = 9 2345x 3 + 3x 4 + 2 x5 = 14x1 , x 2 , x 3 0; x 2 , x5 R b.f(X) = 3x1 +2x3 +4x5 min2 x1 x 3 + 3x5 164 x 3x + x 9 3 45 x 2 + x 3 2 x5 = 11x1 , x 3 , x5 0; x 2 , x 4 Rc. f(X) = x1+2x2 +3x3 +...+nxn min74 76. Chng II: Quy hoch tuyn tnhx1 1x + x 2 12x1 + x 2 + x 3 3.............................x1 + x 2 +..............+ x n nx j 0 ( j = 1, n )38. Dng l thuyt i ngu, chng minh rng cc bi ton sau khng gii c:a. f(X) = -2x1 +x2+4x3+3x4 min3x1 + x 2 2 x 3 + 5x 4 = 302 x + x 2 x = 20 2 342 x + 2 x + x 2 x = 12 1 23 4x j 0 ( j = 1,4 )b. f(X) = 2x1 +3x3 -4x3+x4+2x5 max 2 x 2 + 5x 3 x 4 3x5 7 x + 2 x x 11 2 35 x1 + 4 x 2 x 3 + 6x 4 x5 0 3x + 2 x = 24 34 x 4 , x5 0; x1 , x 2 , x 3 R c. f(X) = -x1 +2x2+4x3-x4-3x5+5x6 min3x1 2 x 3 + x5 = 12 4 x 2 + x 6 = 82 x1 + x 2 + x 3 + 3x 4 2 x5 + 4 x 6 15 x + 5x + 4 x 2 x 7 1 356x j 0 ( j = 1,3) , x j R ( j = 4 ,6)39. Cho bi ton:f(X) = -2x1-6x2+5x3-x4-4x5 maxx1 4 x 2 + 2 x 3 5x 4 + 9 x5 = 3x 3x + 4 x 5x = 6 2 34 5x x + x x = 1 2 3 45x j 0 ( j = 1,5)p dng l thuyt i ngu, chng minh rng X* = (0,0,16,31,14) l phng n ti u ca biton cho.75 77. Chng II: Quy hoch tuyn tnh HD: Vit bi ton i ngu v s dng nh l lch b. 40. Bit rng X* = (0,5,0,3) l phng n ti u ca bi ton: f(X) = 10x1+5x2+13x3+16x4 min 2 x1 + 3x 2 + x 3 + x 4 16 x + 2 x + 3x + 4 x 22 12 34 3x + x + 4 x + 5x 20 12 34 x j 0 ( j = 1,4 ) Tm phng n ti u ca bi ton i ngu:3 S: Yopt = Y = (0, ,2).2 41. Cho bi ton: f(X) = x1+3x2+2x3+3x4+5x5 min x1 + 2 x 2 x 3 + x 4 x5 = 3 x x + 2 x + 4 x 18 2345 x + 3x + 2 x 10 235 x j 0 ( j = 1,5) 1 4Tm phng n ti u ca bi ton cho bit rng Y* = ( ,, 0) l phng n ti u 3 3ca bi ton i ngu. S: Xopt = X = (0,0,0,5,2). 42. Kim tra tnh ti u ca phng n X = (5,-6,1,-4,0) i vi bi ton: f(x) = x1-2x2+x3-x4+x5 min x1 2 x 2 + x 3 + 3x 4 + x5 = 6 2 x + 3x 2 x x + x 4 1 23 45 x1 + 3x 3 4 x 4 8 x1 , x 3 , x5 0; x 2 , x 4 R HD: S dng nh l lch b. 43. Cho bi ton f(x) = 3 x1+9x2-2x3+x4-4x5 min x1 + 5x 2 3x 3 + x 4 2 x5 = 6 3x 4 x + 2 x x + x = 4 123 4 5 4 x x + 2 x 3x 2 134 5 x j 0 ( j = 1,5) V vct X0 = (2,0,0,8,6). Vit bi ton i ngu v ch ra cp rng buc i ngu ca hai bi ton.76 78. Chng II: Quy hoch tuyn tnhPhn tch tnh cht ca vct X0 i vi bi ton cho. Xc nh tp phng n ti u vcc phng n cc bin ti u ca 2 bi ton (nu c).1 5S: Xopt = X = (-1+ x5,0,0,-7+ x5,x5) , x5 02 2 311X* = ( ,0,0, ,5) 2 244. Cho bi ton: f(X) = x1- x2 +6x3+5x4+c5 x5 +c6x6 minx1 x 2 x 4 + 3x5 3x 2 x x + 2 x + 3x 5 2 3 456 x + 2 x + x + 3x x + 3x = 10 1 2 3456x j 0 ( j = 2 ,6), x1 RV phng n X = (0,2,3,1,0,0).Vit bi ton i ngu. X0 c phi l phng n cc bin hay khng?Tm iu kin i vi c5, c6 X0 l phng n ti u, khi xc nh phng n ti u ca biton i ngu. S: X0 - khng l phng n cc bin.c5 3 , c 6 2Y opt = (3,-2,2).45. Xt bi ton:f(X) = x1+7x2 +3x3+5x4- x5 -5x6 min 2 x1 3x 2 + x 3 + x 4 x5 = b1 x + 2 x 5x x + 3x b 123562 6x1 + 7 x 2 + 3x 3 x 4 + 4 x5 + x 6 = b 3 x1 , x 3 , x 4 0; x 6 0; x 2 , x5 R Dng l thuyt i ngu chng minh bi ton trn gii c vi mi vct B = (b1, b2 , b3).Xc nh phng n ti u ca bi ton i ngu theo vct B.Tm phng n ti u ca cp bi ton i ngu khi B = (6,6,-4).Cho b2 = 4, b3 = 5. Xc nh b1 mi phng n ca bi ton i ngu u ti u.S: 3b1 + b2 + b3 >0 Y = (8,5,3). 3b1+ b2 + b3 0, cn gi l s dng, khoanh trn xij li.+ loi: L khng c hng, tc xij=0, ta trng .+ Dy chuyn: l mt on thng hay mt dy lin tip cc on thng gp khc m hai umt l hai ch nm trn cng mt hng hoc mt ct vi mt chn khc thuc dy truyn cabng vn ti.+ Chu trnh: L dy chuyn khp kn Nh vy mt hng hoc mt ct m chu trnh i qua th ch i qua hai v s . Do s t nht ca mt chu trnh l 4.+ Ma trn X = (xi j)m.n tha mn h (2) - (4) c gi l mt phng n ca bi ton.+ Phng n: X = (xi j)m.n tha mn c gi l phng n cc bin ca bi ton vn tinu tp hp cc tng ng vi cc thnh phn dng ca n khng to thnh chu trnh. + Phng n X = (xi j)m.n c gi l phng n cc bin khng suy bin nu s chn can ng bng m+n+1.+ Phng n X = (xi j)m.n c gi l phng n cc bin suy bin nu s chn ca n nhthua m+n+1. + Mt phng n tho mn yu cu (1) c gi l phng n ti u (nghim) ca bi ton, khiu l Xopt.3.1.2. Tnh cht chung ca bi ton vn ti ng - Bi ton vn ti ng l mt trng hp ring ca bi ton QHTT dng chnh tc. Trongh (m+n) phng trnh rng buc ch c (m+n-1) phng trnh c lp tuyn tnh do mtphng n cc bin c ti a (m+n-1) thnh phn dng. 81 83. Chng III: Cc m rng ca quy hoch tuyn tnh- Cc vc t iu kin Aj tng ng vi bin xij c thnh phn i v thnh phn (m+j) bng 1,cc thnh phn cn li u bng 0.- Bi ton lun lun c li gii.3.1.3. Phng php th v gii bi ton vn ti dng nga. Phng php tm phng n cc bin xut pht. xy dng mt phng n cc bin xut pht ngi ta thng s dng mt trong baphng php sau:- Phng php gc Ty-Bc.Phng php gc ty bc gm nhng bc sau:Bc 1. Chn nm dng 1, ct 1 ca bng vn ti.Bc 2. Phn lng hng h = min{a1, b1} vo (1,1)Bc 3. nh du hng (ct), theo lng hng trm pht (trm thu) tng ng ht( ).Bc 4. Quay tr v bc 1 thc hin cng vic nhng cn li.- Phng php min- cc.Ni dung phng php min cc gm cc bc sau y.Bc 1. Chn c cc ph thp nht phn hng gi s l (i,j).Bc 2. Phn lng hng h = min {ai, bj} vo (i,j)Bc 3. nh du cc thuc hng i, hoc ct j nu trm pht Ai pht ht hng, hoctrm thu Bj nhn hng.Bc 4. Quay tr li bc 1 thc hin cng vic nhng cn li.- Phng php xp x Phoghel.* nh ngha. lch ca hng (ct) l hiu s gia c cc ph thp th nh tr i ccc ph thp th nht ca hng (ct) .Ni dung ca phng php Phoghen gm cc bc sau:Bc 1. Chn hng hoc ct c chnh lch ln nhtBc 2. Chn c cc ph thp nht thuc hng (ct) c chnh lch ln nht, gi s(i,j).Bc 3. Phn lng hng h = min{ai, bj} vo (i,j)Bc 4. nh du cc thuc hng (ct), theo trm pht Ai pht ht hng hoc trmthu Bj nhn hng, quay tr v t bc 1 tip tc thc hin thut ton.b. Tiu chun ti uPhng n cc bin khng suy bin X=(xij)m.n c gi l phng n ti u khi v ch khitn ti cc s ui (i= 1.m ) cho cc hng v cc s vj (j= 1.n ) cho cc ct ca bng vn ti sao cho: ui + v j = cij , (i. j ) : xij > 0 (1) ui + v j cij , (i. j ) : xij = 0 (2) Phng trnh (1) ng vi (i.j) l chn.82 84. Chng III: Cc m rng ca quy hoch tuyn tnh Phng trnh (2) ng vi (i.j) l loi. Cc s ui ,vj c gi l h thng th v, trong ui (i= 1.m ) c gi l th v hng, vj(j= 1.n ) gi l th v ct. c. Thut ton th vBc 1: Tm phng n cc bin xut pht X0= (xij )m..xnS dng mt trong 3 phng php trnh by trn tm phng n cc bin xut pht(nu phng n tm c l phng n suy bin th ta phi b xung chn khng cphng n khng suy bin, chn c vai tr nh cc chn khc). Bc 2: Kim tra tnh ti u ca phng n. + Xy dng h thng th v. H (1) l h phng trnh c (n+ m) n v (n + m -1) phng trnh c lp tuyn tnh nn h(1) c v s nghim. Nu cho ui (i= 1.m ) hoc vj (j= 1.n ) mt gi tr a tu th mi gi tr khcu xc nh c mt cch duy nht theo (1). + Tnh cc s kim tra.Da vo (2) ta t ij= ui+ vj-cij (i= 1.m , j= 1.n ) gi l c lng kim tra v tnh ij ngvi cc loi. C hai kh nng xy ra: - Nu mi ij 0 (i = 1.m , j = 1.n ) th phng n ang xt l ti u (thut ton kt thc). - Nu tn ti ij >0 (i = 1.m , j = 1.n ) th phng n ang xt cha ti u, chuyn sangbc 3. Bc 3: Xy dng phng n mi + Chn iu chnh: (r,s) gi l iu chnh nu: rs = max {ij > 0 (i = 1.m , j = 1.n )}.+ Tm chu trnh iu chnh: L chu trnh vi xut pht l iu chnh, cc cn li l chn. Gi V l tp hp cc thuc chu trnh iu chnh.+ nh du cc ca chu trnh, bt u t iu chnh nh du (+) ri xen k nhau nh du (-),(+)... cho n ht chu trnh. K hiu V+ l tp hp cc c du (+), V- l tp hp cc c du (-). Khi :V= V+ V-. + Xc nh lng hng iu chnh: q = min{xij: (i,j) V- }, q > 0. + iu chnh sang phng n mi: X1 = (x 1 ) mxn vi:ij x ij, nu (i, j) Vxij = x ij + q , nu (i, j) V + 1 x ij - q , nu (i, j) V- Gi X1 ng vai tr nh X0 ri quay li bc 2 cho n khi tm c phng n ti u. Ch :83 85. Chng III: Cc m rng ca quy hoch tuyn tnh - Nu iu chnh khng duy nht (tc c hai hay nhiu c ij = rs) th ta chn theonguyn tc: Xt theo hng t trn xung di trong bng vn ti gp no u tin th ta chnlm iu chnh.- Nu c th th ta chn (i0,j0) no m:q .i0,j0 = max {q. ij }>0th gi tr hm mc tiu gim nhanh hn, thut ton c rt ngn.3.1.4. Cc dng bi tp muBi 1: Gii bi ton vn ti vi s liu cho trong bng sau:Bng 1bj30 253540ai 45 9 12 7 50 5 46 2 35 5 61 3Gii:Cch 1: Tm phng n cc bin bng phng php gc Ty Bc.Bc 1: Tm phng n xut pht X0 bng phng php gc Ty Bc.S dng phng php gc Ty Bc ta c phng n cho bng 2.Bng 2: X0 bj30 25 35 40 ui aj459121 7-801 - 30 2 + 155054623 3 - 10 4355+ 53558 6 -1 16 34+6 - 35vj 9 1 3 -1+ Ta thy X0 l phng n cc bin khng suy bin.Bc 2: Kim tra tnh ti u ca X0.+ Xy dng h thng th v: Cho u1 = 0 khi ta tnh c cc s ui v vj cn li nh trongbng 1.84 86. Chng III: Cc m rng ca quy hoch tuyn tnh+ Tnh s kim tra: S kim tra ij= ui+ vj-cij (i = 1.m , j = 1.n ) c ghi gc trn bnphi nh trong bng 1. Trong bng 1 ta thy X0 cha ti u v tn ti mt s c lng kim trakhng m. Khi ta chuyn sang bc 3.Bc 3: Xy dng phng n mi X1.+ Chn iu chnh. iu chnh l (1,3) v 13 = 8 = max{7,8,1,6 }.+ Tm chu trnh iu chnh.Chu trnh iu chnh nh trong bng 1.+ nh du chn l xen k nh trong bng 1. (Bt u t iu chnh nh du (+))+ Xc nh lng hng iu chnh.Lng hng iu chnh q = min{30,10,35}=10.+ iu chnh sang phng n mi X1 = (x 1 )ijDa vo cng thc: x ij nu (i, j) V x /ij = x ij + q nu (i, j) V+ x ij - q nu (i, j) V -ta c bng 3 sau:Bng 3: X1bj 3025 35 40 uiaj 459 129 70 0 - 2025 + 505-1 4 -8 62-5- 35 + 15 355 6 -9 16 3-4 + 10- 25 vj9 1117 85 87. Chng III: Cc m rng ca quy hoch tuyn tnhBng 4: X2 bj30253540ui aj459-1 297 - 0 2520505-4 6 2 4- 15+ 35355 6-9 163 5 30 + - 5vj 01 2 -2 bng 3 ta coi X1 nh X0 ri quay li bc 2. C tip tc nh vy ta tm c phng nX2 , X3 v phng n ti u cho bng 6.Bng 5: X3 bj30253540ui aj459-1 2 7 - 225 205055416 2 6+ - 10 40355 6 1 3 0 1- 30+ 5vj 4-10 -4Bng 6: X4 aj30253540ui bj459-3 1 2 7 -4025 20505 4-4 6-5 2 -1 1040355 6-6 1 3-1 -1 2015vj 61 2 386 88. Chng III: Cc m rng ca quy hoch tuyn tnh Kt lun: 0 25 20 0 Xopt =X4 = 10 0 0 40 ; fmin = f(Xopt )= 310 (vcp). 20 0 15 0 Cch 2: Bc 1: Tm phng n xut pht X0 bng phng php Min- Cc. Bng 1: X0aj 30 2535 40 uibj 4591 27 -1 0 - 20 25+ 0 5054160 2-4 10 40 3553 6-6 132 -1 +- 35 vj912 6Khi s dng dng phng php Min-cc ta c phng n cho bng 1 nh trn.Phng n X0 bng 1 l phng n suy bin, ta thy s chn l m+n-1=5 nn ta phi b sung chn 0 vo (1,3). Bc 2: Kim tra tnh ti u ca X0. + Xy dng h thng th v. Cho u1 =0 khi ta tnh c cc ui ,vj cn