TRNG I HC NNG NGHIP IPGS.TS. NGUYN HI THANH

TON NG DNG(Gio trnh Sau i hc)

NH XUT BN I HC S PHM

M s: 01.01.1/121. H 2005

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Mc lc

M u CHNG I. MT S M HNH V PHNG PHP TI U 1. M hnh quy hoch tuyn tnh 1.1. Cc bc cn thit khi p dng phng php m hnh ho 1.2. M hnh quy hoch tuyn tnh 1.3. Phng php n hnh 1.4. Gii m hnh quy hoch tuyn tnh bng cc phn mm tnh ton 1.5. Mt s ng dng ca phng php n hnh 2. B sung thm v phng php n hnh 2.1. a BTQHTT v dng chnh tc 2.2. Phng php n hnh m rng 3. M hnh quy hoch tuyn tnh a mc tiu 3.1. Cc khi nim c bn 3.2. Mt s phng php gii BTQHTT a mc tiu 3.3. Phng php tho dng m tng tc gii BTQHTT a mc tiu 4. M hnh ti u phi tuyn n v a mc tiu 4.1. Mt s khi nim c bn 4.2. Mt s phng php v phn mm gii bi ton ti u phi tuyn n mc tiu 4.3. Mt s phng php gii bi ton ti u phi tuyn a mc tiu CHNG II. CC M HNH MNG 1. M hnh mng vn ti 1.1. Pht biu bi ton vn ti 1.2. To phng n vn ti xut pht 1.3. Phng php phn phi gii bi ton vn ti 1.4. Phng php phn phi ci bin gii bi ton vn ti 2. M hnh mng PERT 2.1. Cc khi nim c bn v PERT 2.2. S PERT vi s liu ngu nhin 2.3. iu chnh d n khi k hoch mt s hot ng b ph v 2.4. Tnh thi gian rt gn ti u bng phng php n hnh 2.5. p dng mng PERT trong phn tch chi ph v qun l ti chnh d n 3. Mt s m hnh mng khc 3.1. Bi ton cy khung ti thiu 3.2. Bi ton tm ng i ngn nht v quy hoch ng 3.3. p dng quy hoch ng cho mt s bi ton ngnh in

5 7 7 7 7 11 14 16 17 17 19 21 21 23 25 29 29 31 37 41 41 41 42 44 48 51 51 56 57 59 59 62 62 64 67

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CHNG III. GII THIU L THUYT M PHNG V M HNH HNG CH 1. Mc ch v cc cng c ca m phng 1.1. Khi nim v m phng ngu nhin 1.2. Cc cng c ch yu ca m phng 1.3. M phng mt s phn phi xc sut 2. p dng m phng ngu nhin 2.1. Vai tr ca phng php m phng 2.2. Cc bc cn tin hnh khi p dng m phng 2.3. Mt s v d v p dng phng php m phng 3. Mt s vn v m hnh hng ch 3.1. Mt s yu t c bn ca h thng hng ch 3.2. Cc ch s cn kho st 3.3. Tnh ton cc ch s 3.4. p dng m phng cho mt s h thng hng ch CHNG IV. PHN TCH MARKOV V NG DNG 1. Cc khi nim c bn v xch Markov 1.1. Mt s nh ngha 1.2. Ma trn xc sut chuyn trng thi v phn phi dng 1.3. Cc tnh cht v nh l 2. Mt s ng dng ca phn tch Markov 2.1. Tm cn bng th phn 2.2. Chnh sch thay th vt t thit b 2.3. Phn tch Markov trong d bo tht thu cho cc hp ng thc hin trc 2.4. Tm phn phi gii hn cho mt h thng k thut 2.5. Mt ng dng ca qu trnh sinht cho h thng hng ch 3. M phng xch Markov 3.1. M phng xch Markov thi gian ri rc 3.2. M phng xch Markov thi gian lin tc Phn bi tp Phn ph lc Ti liu tham kho

73 73 73 73 74 78 78 78 79 88 88 92 92 94 105 105 105 106 111 111 112 112 113 116 120 123 123 124 126 137 145

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M u

Trong mt vi nm gn y, cc mn hc Ton Tin ng dng c a vo chng trnh o to Sau i hc cho mt s chuyn ngnh kinh t k thut nh Qun tr kinh doanh, Qun l t ai, Cng ngh thng tin, Sinh hc ti mt s trng i hc trong nc. Cc mn hc ny, tuy s n v hc trnh cha nhiu nhng gip cho hc vin cao hc cng nh cc nghin cu sinh c nhng kin thc c s v nng cao v Ton hc v Tin hc, c bit v cc phng php tnh ton khoa hc (Scientific Computing Methods), l cc vn ht sc cn thit cho cc ti nghin cu khoa hc ca h. iu ny cng ph hp vi xu th chung trong o to Sau i hc ti cc trng i hc nc ngoi, vi cc mn hc v Ton Tin nng cao cho hc vin cao hc, thng chim thi lng kh ln ti khong 200 n 250 tit bao gm nhiu ni dung phong ph v cp thit. Xut pht t nhng l do trn v da trn cc kinh nghim tch lu c trong qu trnh dy mt s mn hc cho chng trnh Cao hc Qun l t ai v Cao hc in (Trng i hc Nng nghip I), Cao hc Ton Tin ng dng (Trng i hc Bch khoa H Ni), Cao hc Qun tr kinh doanh (ti mt s trng i hc khc), chng ti bin son gio trnh ny vi mong mun vic ng dng cc phng php ton hc, cc phng php vn tr hc c trin khai rng ri hn v mang li cc hiu qu thit thc hn. Gio trnh vi thi lng t 45 ti 60 tit, trc ht, dnh cho hc vin cao hc ngnh in, vi cc ni dung c Khoa C in v Khoa Sau i hc, Trng i hc Nng nghip I, thng qua. Cc ch trong gio trnh bao gm: mt s m hnh v phng php ti u, cc bi ton v mng, gii thiu v quy hoch ng, mt s ng dng ca l thuyt hng ch (Waiting Line Theory) v m phng ngu nhin (Stochastic Simulation), cc khi nim c bn v ng dng ca qu trnh ngu nhin Markov. y l cc ch chnh v Ton ng dng v Vn tr hc m hc vin cao hc ca nhiu chuyn ngnh kinh t k thut ti cc trng i hc nc ngoi bt buc phi hc. Cc ch ny c th gip ch khng ch cho vn qun l s dng in m cn cho vn thit k v xy dng cc h thng k thut in. Gio trnh cng c th c ly lm ti liu tham kho v cc phng php ton ng dng hay m hnh ho cho chng trnh Cao hc cc chuyn ngnh nh: Qun l t ai, Kinh t nng nghip v mt s chuyn ngnh kinh t k thut khc. Khi bin son gio trnh, chng ti lun ch nhn mnh kha cnh ng dng cc phng php ton hc v kha cnh tnh ton khoa hc vi cc v d minh ho chn lc, nhm gip cho hc vin hiu r nn p dng cc phng php vo cc vn nghin cu no v p dng chng nh th no cho mt s trng hp c th. Do thi lng ca mn hc, gio trnh khng i su vo vn chng minh ton hc ca cc phng php ny cng nh cc ng dng tng qut ca chng trong cc h thng ln.

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Hi vng rng, nhng hc vin cao hc quan tm ti cc phng php ton hc c trnh by trong gio trnh c th t mnh tip tc c nhng nghin cu chuyn su hn sau ny. Chng hn, vi kin thc v quy hoch ng v cc phng php ti u phi tuyn m gio trnh cung cp, ngi c c th tip tc nghin cu v cc phng php quy hoch ng nhm p dng vo cc h iu khin ti u trong t ng ho. Cn vi mt s ch v xch Markov v ng dng cng nh m phng xch Markov, ngi c c th tip tc nghin cu v cc m hnh ngu nhin nh qu trnh sinht hay qu trnh hi phc c nhiu ng dng rng ri trong ngnh in, in t v Vin thng hay Cng ngh thng tin. y l mt trong s khng nhiu cc gio trnh v Ton ng dng dnh cho chng trnh Sau i hc cc chuyn ngnh kinh t k thut ti cc trng i hc trong nc, nn mc d chng ti ht sc c gng trong qu trnh bin son, nhng chc chn gio trnh khng trnh khi cn tn ti nhng im hn ch. Chng ti rt mong nhn c cc kin ng gp ca cc nh khoa hc, cc thy gio, c gio, cc hc vin cao hc, tin s gio trnh c hon chnh, chnh xc v sinh ng hn. Cui cng, tc gi xin chn thnh cm n Khoa Sau i hc v Khoa C in, Trng i hc Nng nghip I v nhng gip qu bu trong qu trnh bin son; cm n B mn Ton, ging vin ng Xun H v B mn Tin hc, cc hc vin cao hc chuyn ngnh in kho 10 v 11, Trng i hc Nng nghip I; k s Phan Vn Tin v cc hc vin cao hc chuyn ngnh Ton Tin ng dng, kho 1 v 2, Trng i hc Bch Khoa H Ni, dnh kin ng gp v tham gia hon chnh mt s ni dung ca gio trnh ny. Tc gi cng xin chn thnh cm n cc kin phn bin qu bu ca cc ng Trng b mn Ton, Trng i hc Nng nghip I v Trng khoa Ton Tin ng dng, Trng i hc Bch khoa H Ni. H Ni, ngy 19 thng 5 nm 2005 PGS.TS. Nguyn Hi Thanh

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Chng I MT S M HNH V PHNG PHP TI U1.1.1.

M hnh quy hoch tuyn tnhCc bc cn thit khi p dng phng php m hnh ho

Trc ht phi kho st, pht hin vn cn gii quyt. Pht biu cc iu kin rng buc, 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 s liu, xc nh phng php gii quyt. nh ra quy trnh gii / thut gii. C th gii m hnh bng cch tnh ton thng thng. i vi cc m hnh ln, gm nhiu bin v nhiu iu kin rng buc cn lp trnh v gii m hnh trn my tnh. nh gi kt qu. Trong trng hp pht hin thy c kt qu bt thng hoc kt qu khng ph hp vi thc t, cn kim tra v chnh sa li quy trnh gii hoc m hnh. Trin khai cc phng n tm c trn thc t. Cc thut ng sau thng gp khi p dng phng php m hnh ho: ng dng ton / Ton ng dng (Mathematical Applications hay Applied Mathematics). Vn tr hc (Operations Research vit tt l OR). Khoa hc qun l (Management Science vit tt l MS)1.2. M hnh quy hoch tuyn tnh

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 = c1x1 + c2x2 + cnxn Max (Min) vi cc iu kin rng buc: a11x1 + a12x2 +... +a1nxn a21x1 + a22x2 +... +a2nxn ... am1x1 + am2x2 +... +amnxn

b1 b2 bm

x1, x2,..., xn 0 (iu kin khng m)

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V d: z = 8x1 + 6x2 Max vi cc rng buc: 4x1 + 2x2 60 2x1 + 4x2 48 x 1 , 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 trn mi n v sn phm bn ra l 8 v 6 (n v tin t) cho cc sn phm loi I v II. Phng php th Phng php th c ngha minh ho v gip hiu bn cht vn . Bc 1: V min rng buc / min cc phng n kh thi, 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) cn gi l vc t nghim, tho mn tt c cc rng buc c (xem hnh I.1). Trc ht chng ta v th 4x1 + 2x2 = 60 bng cch xc nh hai im trn th: (x1 = 0, x2 = 30) v (x2 = 0, x1 = 15).

x2 30 4x1 + 2x2 = 6012

A B 2x1 + 4x2 = 48 C

8 4 O 3 6

15

24

x1

Hnh I.1. Phng php th gii bi ton quy hoch tuyn tnh

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th trn l mt ng thng chia mt phng lm hai na mt phng: mt phn gm cc im (x1, x2) tho mn 4x1 + 2x2 60; mt phn tho mn 4x1 + 2x2 60. Ta tm c na mt phng tho mn 4x1 + 2x2 60. Tng t, c th v th 2x1 + 4x2 = 48 bng cch xc nh hai im thuc th (x1 = 0, x2 = 12) v (x2 = 0, x1 = 24). Sau tm na mt phng tho mn 2x1 + 4x2 48. Lc ny, giao ca hai na mt phng tm c trn cho ta tp hp cc im (x1, x2) tho mn hai rng buc u tin. 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 l min gii hn bi t gic OABC (cn gi l n hnh 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. 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 to cc im ct hai trc to thun li hn). D dng tm c hai im nm trn ng ng mc ny l (x1 = 0, x2 = 4) v (x2 = 0, x1 = 3). 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 n (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).Kt lun: 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 ca bi ton trn (hay cc BTQHTT khc, nu c) lun t c ti mt trong cc nh ca n hnh hay cn gi l cc im cc bin ca n hnh (chnh xc hn, im cc bin l im thuc n hnh, m khng th tm c mt on thng no cng thuc n hnh nhn im l im trong). Nhn xt trn y l mt nh l ton hc 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, tm phng n ti u ta ch cn so snh gi tr ca hm mc tiu ti cc im cc bin ca min phng n.

Tnh 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; ti B(12, 6): z(12, 6) = 132 = Max{z(O), z(A), z(B), z(C)}. Vy zmax = 132.

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Nhn xt: Mun tm phng n ti u ca BTQHTT ta xut pht t mt im cc bin no , tm cch ci thin hm mc tiu bng cch i ti im cc bin k n. Tip tc nh vy cho ti khi tm c phng n ti u. Trong trng hp BTQHTT c phng n ti u th quy trnh gii ny bao gm hu hn bc (do s im cc bin l hu hn).

i vi BTQHTT ang xt, quy trnh gii c minh ho nh sau: O(0, 0) z=0 hoc: O(0, 0) z=0S khi

A(0,12) z = 72

B(12,6) dng z = 132

C(15, 0) z = 120

B(12, 6) dng z = 132

Bt u

Nhp d liu

Tm im cc bin xut pht

Kim tra iu kin ti u ng In v lu tr kt qu

Sai

Tm im cc bin k tt hn

DngHnh I.2. S khi gii BTQHTT

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Quy trnh gii BTQHTT tng qut c s khi gin lc nh trnh by trn hnh I.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 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 tp cc gi tr hm mc tiu z khng b chn).1.3. Phng php n hnh

y 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 cch thm vo cc bin b khng m x3 v x4 nh sau: z = 8x1 + 6x2 + 0x3 + 0x4 Max vi cc rng buc: 4x1 + 2x2 + x3 2x1 + 4x2 = 60

+ x4 = 48

x 1 , x2 , x 3 , x4 0Cch lp v bin i cc bng n hnh

gii BTQHTT dng chnh tc trn y, cn lp mt s bng n hnh nh trnh by trong bng I.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)), 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 II 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. 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.

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Bng I.1. Cc bng n hnh gii BTQHTT

H s hm mc tiu cj 0 0 Hng z

Bin c s x3 x4

Phng n 60 48 z0 = 0

c1 = 8 x1 4 2 z1 = 01 = 8

c2 = 6 x2 2 4 z2 = 02 = 6

c3 = 0 x3 1 0 z3 = 03 = 0

c4 = 0 x4 0 1 z4 = 04 = 0

Hng j = cj zj 8 0 Hng z Hng j = cj zj 8 6 Hng z Hng j = cj zjPhn tch bng n hnh bc 1

x1 x4

15 18 z0 = 120

1 0 z1 = 81 = 0

1/2 3 z2 = 42 = 2

1/4 1/2 z3 = 23 = 2

0 1 z4 = 04 = 0 1/6 1/3

x1 x2

12 6 z0 = 132

1 0 8 0

0 1 6 0

1/3 1/6 5/35/3

2/32/3

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 / 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 mt 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 + 048 = 0. Trn hng j cn ghi cc gi tr j, j = 1, 2, 3, 4, tnh theo cng thc j = cj zj = li nhun trn mt n v sn phm chi ph trn mt n v sn phm. Vy j l "li bin"/mt n v sn phm khi a thm mt n v sn phm loi j vo phng n sn xut mi. Nu j > 0 th hm mc tiu cn tng c khi ta a thm cc n v sn phm loi j vo phng n sn xut mi. C th chng minh c j chnh l o hm ring z/xj 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.

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Do 1 v 2 u dng 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 phn gii bi ton bng phng php th: im cc bin k ca im (0, 0) c th l A(0, 12) hay C(15, 0)).Th tc xoay (pivotal procedure) Bc 1: Chn ct xoay l ct c j > 0 tc l chn bin xj lm bin c s mi do xj tng ko theo hm mc tiu tng. y ta chn a x1 vo (nh du ct 1). Bc 2: Chn hng xoay xc nh a bin no ra khi s 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 ta nh du vo hng xoay l hng u (hng tng ng vi bin x3). Do cn a x3 ra khi 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 c 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).(2) (3)

Chng hn: (1)c = 4, 2(c) = 2 (3)c = phn t xoay = 4, (4)c = 2 2 (1)mi = 4 2 = 3. 4(1) (4)Hnh I.3. Quy tc hnh ch nht

Gii thch: Cc bc xoay trn y ch l php bin i tng ng h phng trnh

4x1 + 2x2 + x3 2x1 + 4x2 c h

= 60 (a)

+ x4 = 48 (b)

x1 + (1/2)x2 + (1/4)x3 0x1 + 3x2

= 15 (a)

(1/2)x3 + x4 = 18 (b)

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bng cch ly phng trnh (a) chia cho 4 (phn t xoay) c (a), ri ly (b) tr bt 2 (a)/4 c (b). y chnh l ni dung ca bc 4 v bc 5. Cn 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; 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 chn bin x2 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, 2, 3, 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.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 tn ti j m j 0 th cn tip tc ci thin hm mc tiu bng cch chn ct j lm ct xoay...). 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, xem thm mc 1.2). 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). Khi dng qu trnh gii v kt lun m hnh quy hoch tuyn tnh thit lp khng ph hp vi thc t.1.4. Gii m hnh quy hoch tuyn tnh bng cc phn mm tnh ton

Hin nay c nhiu phn mm tnh ton gii BTQHTT kh hiu qu nh Excel, Lingo. Nhng phn mm ny rt thn thin vi ngi dng. Tuy nhin cn nhn mnh rng, vic pht biu c m hnh bi ton v phn tch, nh gi c kt qu mi chnh l nhng khu quan trng nht trong phng php m hnh ho. Sau y, chng ta dng phn mm Lingo gii v d xt trn. z = 8x1 + 6x2 Max

14

vi cc rng buc: 4x1 + 2x2 60 2x1 + 4x2 48 x 1 , x2 0.

gii bi ton ny, chng ta cn ci t Lingo vo trong my tnh. Nhn vo biu tng Lingo trn mn hnh vo ca s Lingo. Sau thc hin cc lnh Lingo: Menu > New > v g vo cc d liu ca bi ton nh hnh I.4.

Hnh I.4. Nhp d liu ca bi ton quy hoch tuyn tnh trong Lingo

Tip theo, cn nhy chut vo nt LINGO v gii bi ton thu c kt qu chi tit nh trn hnh I.5.

Hnh I.5. Kt qu gii bi ton quy hoch tuyn tnh trong Lingo

Kt qu chi tit cho ta bit gi tr cc i ca hm mc tiu l 132 vi phng n ti u l: x1 = 12, x2 = 6. Cc gi tr ti u ca cc bin i ngu l y1 = 5/3 v y2 = 2/3 (cn gi l cc gi c nh hay gi bng Shadow Prices).

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1.5.

Mt s ng dng ca phng php n hnh

(Gii cc bi ton quy hoch sn xut trong lnh vc c kh v in lc)Bi ton phn phi in nng

C ba h ph ti cn c cung cp in nng t hai ngun in nm cch xa nhau. Gi thnh truyn ti mt n v in nng t ngun i n h tiu th j l cij. Kh nng cung cp in nng ca mi ngun b gii hn bi tr lng hin c ca chng l A1 v A2. Nhu cu tiu dng ca cc h tiu th l B1, B2 v B3. Gi xij l lng in nng c a t ngun i ti h tiu th j. Cn phi xc nh cc xij sao cho tng chi ph l nh nht. Nh vy ta c BTQHTT sau: z=

c xi =1 j=1 ij

2

3

ij

Min

vi cc iu kin rng buc l: x11 + x12 + x13 A1, x21 + x22 + x23 A2, x11 + x21 = B1, x12 + x22 = B2, x13 + x23 = B3, xij 0, i = 1, 2 v j = 1, 2, 3. Bi ton trn y (hoc dng tng qut hn) c th gii c bng phng php n hnh bit hay phng php phn phi s c nghin cu mc 1.3, chng II.Bi ton phn ti cho my

Mt x nghip c hai loi my M1 v M2. Cc loi my ny c th sn xut c ba loi sn phm P1, P2 v P3 vi cc nng sut l aij, chng hn my M1 sn xut sn phm P2 vi nng sut a12. Mi n v sn phm mang li li sut cj vi j = 1, 2, 3. Mi thng x nghip phi sn xut sn phm loi j khng t hn bj n v v khng vt qu dj n v, j = 1, 2, 3. Hy lp k hoch phn ti cho cc my sao cho t tng li nhun ln nht. D thy bi ton ny dn ti BTQHTT sau: z=3 2

c j a ijx ij Maxj=1 i =1

vi cc iu kin rng buc:

16

a11x11 + a21x21 b1, a12x12 + a22x22 b2, a13x13 + a23x23 b3, a11x11 + a21x21 d1, a12x12 + a22x22 d2, a13x13 + a23x23 d3, x11 + x12 + x13 m1, x21 + x22 + x23 m2, xij 0, i = 1, 2 v j = 1, 2, 3. (trong m1 v m2 l tng thi gian chy my M1 v M2). Bi ton trn y cn c th pht biu mt cch tng qut hn v vn gii c bng phng php n hnh. Hn na, trong lnh vc quy hoch sn xut hay qun l kinh doanh, ni ring trong ngnh c kh v in lc, BTQHTT c ng dng rt rng ri v mang li hiu qu cn thit.2.2.1.

B sung thm v phng php n hnha BTQHTT v dng chnh tc

V d 1: (Trng hp cc rng buc u c du )

z = 8x1 + 6x2 Max vi cc rng buc:4x1 + 2x 2 60 2x1 + 4x 2 48 x , x 0 1 2

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 l: z = 8x1 + 6x2 + 0x3 + 0x4 Max4x1 + 2x 2 + x 3 = 60 2x1 + 4x 2 + x 4 = 48 x , x , x , x 0 1 2 3 4

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. Mt cch tng qut, BTQHTT dng chnh tc l bi ton vi cc

17

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. V d 2: (Trng hp c iu kin rng buc vi du )

z = 8x1 + 6x2 Max vi cc rng buc:4x1 + 2x 2 60 2x1 + 4x 2 48 x , x 0 1 2

Ta thm cc bin b x3 (slack variable) mang du +, x4 (surplus variable) mang du c h iu kin rng buc sau:4x1 + 2x 2 + x 3 = 60 2x1 + 4x 2 x 4 = 48 x , x , x , x 0 1 2 3 4

Phi thm bin gi x5 (x5 gi l lng vi phm ca phng trnh th hai) c h iu kin rng buc4 x 1 + 2 x 2 + x 3 = 60 2 x 1 + 4 x 2 x 4 + x 5 = 48 x , x , x , x , x 0 1 2 3 4 5

Lc ny, 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 z = 8x1 + 6x2 + 0x3 + 0x4 Mx5 Max, 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 3: (Trng hp c bin khng dng)

z = 8x1 6x2 Max vi cc rng buc:4x1 + 2x 2 + x 3 60 2x1 + 4x 2 x 4 = 48 x 0, x 0, x 0, x 0 2 3 4 1

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.

18

z = 8x1 + 6x'2 Max vi cc rng buc: 4x1 2x '2 + x 3 60 2x1 4x '2 x 4 = 48 x , x ' , x , x 0 1 2 3 4

V d 4: (Trng hp c bin vi du tu )

z = 8x1 + 6x2 Max vi cc rng buc:4x1 + 2x 2 60 2x1 + 4x 2 48 x 0, x du tu 2 1

Lc ny ta vit bin x2 di dng x2 = x'2 x''2 vi x '2 = max[0, x 2 ] x ' 0 th m bo 2 x ''2 = max[0, x 2 ] x ''2 0

Cc rng buc s l4x1 + 2x '2 2x ''2 + x 3 = 60 2x1 + 4x '2 4x '2 + x 4 = 48 x , x ' , x '' , x , x 0 1 2 2 3 4

Bi ton vi hm mc tiu l: 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 , , =) v dng chnh tc.2.2. 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:

z = 8x1 + 6x2 Max vi cc rng buc:

19

4x1 + 2x 2 60 (a) 2x1 + 4x 2 48 x , x 0 1 2

hay: z = 8x1 + 6x2 +0x3 + 0x4 Max vi cc rng buc4x1 + 2x 2 + x 3 = 60 (b) 2x1 + 4x 2 x 4 = 48 x , x , x , x 0 1 2 3 4

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 +) vi cc rng buc4x1 + 2x 2 + x 3 = 60 (c) 2x1 + 4x 2 x 4 + x 5 = 48 x , x , x , x , x 0 1 2 3 4 5

Cch 1: C th gii BTQHTT vi cc iu kin rng buc (a) 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 (c) bng cch lp bng n hnh nh thng thng nhng ch h s M + (xem bng I.2).Bng I.2. Cc bng n hnh gii bi ton M

H s hm mc tiu 0 M

Bin c s x3 x5

Phng n 60 48 z0 = 48M

8 x1 4 2 z1 = 2M

6 x2 2 4 z2 = 4M

0 x3 1 0 z3 = 0

0 x4 0 1 z4 = M

Mx5 0 +1 z5 = M

Hng z Hng j 0 6 Hng z Hng j 0 6 Hng z Hng j x4 x2 x3 x2

1 = 8+2M36 12 72 3 1/2 3 5 72 30 180 6 2 12

2 = 6+4M0 1 6 0 0 1 6 0

3 = 01 0 0 0 2 1/2 3

4 = M1/2

5 = 0 1/2 1/43/2

1/4 3/23/2 1 0 0 0

M 3/2 1 00

4

3

M

20

Ti bng n hnh cui cng, ta thy j 0 j nn phng n ti u t c vi x2 = 30, x4 = 72, cc xj khc = 0 v zMax = 180.Lu Khi mt bin gi c a ra khi c s th khng bao gi quay li na. Do ta c th xo ct bin gi 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 bng phn chng). Vi v d trn (xem bng I.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 s bin c s (lc ny c phng n cc bin xut pht cho bi ton (b)) v pha 2 nhm tm phng n ti u cho bi ton (b). Phn mm tnh ton Lingo c th gii c tt c cc BTQHTT khng i hi ngi dng phi a chng v dng chnh tc. 3.3.1.

M hnh quy hoch tuyn tnh a mc tiuCc khi nim c bn

Pht biu m hnh

Trong cc bi ton k thut, cng ngh, qun l, kinh t nng nghip v.v... ny sinh t thc t, chng ta thng phi xem xt ti u ho ng thi mt lc nhiu mc tiu. Cc mc tiu ny thng l khc v th nguyn, tc l chng c o bi cc n v khc nhau. Nhng tnh hung nh vy to ra cc bi ton ti u a mc tiu. Nh vy, chng ta cn phi ti u ho (cc i ho hoc cc tiu ho tu theo tnh hung thc t) khng phi l ch mt mc tiu no , m l ng thi tt c cc mc tiu t ra. Bi ton ti u a mc tiu m trong min rng buc D l tp li a din v cc mc tiu zi = fi(X), vi i = 1, 2,, p, l cc hm tuyn tnh xc nh trn D, c gi l bi ton quy hoch tuyn tnh a mc tiu. Khi , ta c m hnh ton hc sau y c gi l m hnh quy hoch tuyn tnh a mc tiu :Max CX vi rng buc X D, trong : C l ma trn cp p n D = { X Rn: AX B}

vi A l ma trn cp m n v B Rm.V d: BTQHTT vi hai mc tiu

f1(X) = x1 + 2x2 Min hay z1 = f1 (X) = x1 2x2 Max z2 = f2(X) = 2x2, Max

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vi cc rng buc x1 + x 2 3 x1 + x 2 3 x , x 0 1 2

Ta c th vit bi ton ny di dng ma trn nh sau: Max CX vi rng buc X D = {X R2 : AX B}, trong X = (x1, x2)T, B = (3, 3, 0, 0)T, cn 1 C= 0 2 , 2 1 1 1 1 . A= 1 0 0 1

C th ni, BTQHTT a mc tiu l BTQHTT m trong chng ta phi ti u ho cng mt lc nhiu mc tiu. Tuy nhin, cc mc tiu ny thng i chi cnh tranh vi nhau. Vic lm tt hn mc tiu ny thng dn ti vic lm xu i mt s mc tiu khc. V vy vic gii cc bi ton ti u a mc tiu, tc l tm ra mt phng n kh thi tt nht theo mt ngha no , thc cht chnh l mt bi ton ra quyt nh. C th thy li y mt ln na khng nh "Ti u ho chnh l cng c nh lng ch yu nht ca qu trnh ra quyt nh". Hin ti cc ti liu, sch chuyn kho, tp ch cp nht v lnh vc lin ngnh Ton Tin, Khoa hc qun l, Cng ngh, Kinh t, in, C kh nng nghip,... cp rt nhiu ti bi ton ti u a mc tiu. Vn nghin cu c s l thuyt, thut ton, lp m hnh, xy dng h my tnh tr gip quyt nh, v p dng cc m hnh ti u a mc tiu cho cc qu trnh cng ngh, qun l,... l mt vn lin ngnh c rt nhiu nh khoa hc v k s thc hnh quan tm.Phng n ti u Pareto

Khi nim then cht trong ti u ho a mc tiu l khi nim phng n ti u Pareto.nh ngha: Mt phng n ti u Pareto X* c tnh cht sau y: Trc ht n phi thuc vo min cc phng n kh thi ca bi ton, tc l phi tho mn tt c cc rng buc: X* D. Vi mi phng n kh thi khc X D m c mt mc tiu no tt hn (fi(X) tt hn fi(X*)) th cng phi c t nht mt mc tiu khc xu hn (fj(X) xu hn fj(X*), j i).

Ni mt cch khc, khng tn ti mt phng n kh thi no X D c th tri hn X trn tng th.*

minh ho nh ngha trn, ta xt v d cho.

22

V d:

Xt BTQHTT vi hai mc tiu. f1 (X) = x1 + 2x2 Min f2 (X) = Vi cc rng buc x1 + x 2 3 x1 + x 2 3 x ,x 0 1 2y 3 2 d A D

2x2

Max

n2o -3 B 3 x

n1Hnh I.6. Minh ho th BTQHTT hai mc tiu

Min cc phng n kh thi D (min gii hn bi on AB v cc tia Ad, Bx) c biu th trn hnh I.6. n1 (1, 2) l hng gim ca mc tiu 1, cn n 2 (0, 2) l hng tng ca mc tiu 2. Lc ny A(0, 3) cng nh B(3, 0) l hai phng n ti u Pareto ca bi ton trn. D thy tp hp P tt c cc phng n ti u Pareto bao gm cc im nm trn on AB v Ad.3.2. Mt s phng php gii BTQHTT a mc tiu

nh ngha 1

Gii bi ton ti u ton cc a mc tiu l chn ra t tp hp P cc phng n ti u Pareto ca bi ton mt (hoc mt s) phng n tt nht (tho mn nht) theo mt ngha no da trn c cu u tin ca ngi ra quyt nh. Trong v d trn, tu theo c cu u tin ca ngi ra quyt nh, chng ta c th

23

chn ra mt hoc mt s im ti u Pareto nm trn AB hoc tia Ad lm phng n ti u ca bi ton.

Cch 1: Bng mt phng php ti u ton hc thch hp tm ra tp hp P tt c cc phng n ti u Pareto. Ngi ra quyt nh s ra c cu u tin ca mnh i vi tp P nhm tm ra phng n ti u Pareto tho mn nht cho bi ton a mc tiu ban u. Cch 2: Vic tm tp hp P trong trng hp cc bi ton nhiu bin l kh kh v mt nhiu thi gian. V vy, so vi cch 1, cch 2 s tin hnh theo trnh t ngc li. Trc ht ngi ra quyt nh s ra c cu u tin ca mnh. Da vo c cu u tin , cc mc tiu s c t hp vo mt mc tiu duy nht, tiu biu cho hm tng tin ch ca bi ton. Bi ton ti u vi hm mc tiu t hp ny s c gii bng mt phng php ti u ton hc thch hp, tm ra mt (hoc mt s) phng n ti u Pareto. Lc ny, ngi ra quyt nh s chn ra trong s cc phng n ti u Pareto mt phng n tt nht.Chng ta s tip tc phn tch cch th 2. R rng, ngi ra quyt nh khng th ra c cu u tin ca mnh mt cch chnh xc ngay t u. Trong qu trnh gii bi ton, trong mi bc lp, sau khi xem xt li c cu u tin ra, cng nh phng n trung gian va tm c, ngi ra quyt nh c th da vo cc thng tin thay i li c cu u tin ca mnh. Sau , qu trnh gii li c tip tc, cho ti khi mt phng n ti u cui cng c a ra.nh ngha 2

Phng php gii bi ton ti u a mc tiu da trn s tr gip ca h my tnh, nhm gip ngi ra quyt nh tng bc thay i cc quyt nh trung gian mt cch thch hp i ti mt phng n ti u Pareto tho mn nht, c gi l phng php tng tc ngi my tnh. Phng php tng tc ngi my tnh gii bi ton ti u a mc tiu c cc yu t cu thnh sau: C cu u tin ca ngi ra quyt nh v hm t hp tng ng. Kiu tng tc ngi my tnh: cho bit cc thng tin no my tnh phi a ra li trong cc bc lp trung gian, v cch thay i cc thng s ca c cu u tin t pha ngi ra quyt nh. K thut ti u ton hc c xy dng da trn l thuyt ti u ho nhm tm ra cc phng n ti u Pareto cho cc bi ton cn gii trong cc bc lp trung gian. Cho ti thi im hin nay, hng chc phng php gii BTQHTT a mc tiu c cp ti trong cc tp ch chuyn ngnh, m a s chng u c nhng ng dng rt thnh cng trong nhiu lnh vc, nh: phng php tham s, phng php nn php tuyn, phng php vc t cc i, phng php trng s tng tc ca Chebysev, phng php tho dng m tng tc ca Nguyn Hi Thanh.

24

3.3.

Phng php tho dng m tng tc gii BTQHTT a mc tiu

Thut gii a. Bc khi to

Nhp s liu cho cc hm mc tiu tuyn tnh zi (i = 1, 2,..., p) v m iu kin rng buc. Gii BTQHTT cho tng mc tiu zi (i = 1, 2,..., p) vi m rng buc ban u, thu c cc phng n ti u X1, X2,..., Xp (nu vi mt mc tiu no bi ton khng cho phng n ti u th cn xem xt chnh sa li cc iu kin rng buc ban u). Tnh gi tr hm mc tiu ti p phng n X1, X2,..., Xp. Lp bng payoff. Xc nh gi tr cn trn ziB v gi tr cn di z iw ca mc tiu zi (i=1, 2,..., p). Xc nh cc hm tho dng m 1(z1), 2(z2),..., p(zp) cho tng mc tiu da vo thng tin t bng payoff theo cng thc: i (z i ) = t k: = 1.b. Cc bc lp (xt bc lp th k)

z i z iw , i = 1, 2,..., p. z iB z iw

Bc 1: Xy dng hm mc tiu t hp t cc hm tho dng trn:w11(z1) + w 22(z2) +... + w pp(zp) Max Trong : w1, w2,..., wp l cc trng s phn nh tm quan trng ca tng hm tho dng trong thnh phn hm t hp, vi w1 + w 2 +... + w p = 1 v 0 w 1, w 2,..., w p 1.

Bc 2: Gii BTQHTT vi hm mc tiu t hp v m rng buc ban u tm c phng n ti u ca bc lp th k l X(k) v gi tr ca cc hm mc tiu zi cng nh ca cc hm tho dng i(zi) (vi i =1, 2,..., p). Nu ngi ra quyt nh cm thy cha tho mn vi cc gi tr t c ca cc hm mc tiu cng nh ca cc hm tho dng th phng n thu c X(k) cha phi l phng n ti u tho mn nht. t k:= k + 1, quay v bc 1. Nu ngi ra quyt nh cm thy tho mn th phng n thu c l X(k). Chuyn sang bc 3.

Bc 3: Kt thc.

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V d: Gii BTQHTT hai mc tiu.

z1 = 8x1+ 6x2 Max z2 = x1 + 3x2 Max vi cc rng buc:(D) 4x1 + 3x2 60 2x1 + 4x2 48 x1, x2 0

a. Bc khi to

Gii BTQHTT cho tng mc tiu trong v d trn ta c hai bi ton: Max z1 = 8x1 + 6x2 Max vi iu kin rng buc (D) cho phng n ti u X1(12, 6) v Max z1 = 132; z2 = x1 + 3x2 Max cho phng n ti u X2(0, 12) v Max z2 = 36. Nh vy min phng n ti u Pareto chnh l mi phng n thuc AB (xem hnh I.1), vi A(0, 12) v B(12, 6). n1 (8, 6) l hng tng ca mc tiu 1, cn n 2 (1, 3) l hng tng ca mc tiu 2. Do , khi chn phng n ti u Pareto dch dn t B v A th z1 gim, z2 tng. Cn tm phng n ti u Pareto "tho mn nht" thuc AB bng cch thng lng gia z1 v z2. Lp bng payoff cho cc mc tiuPhng n Xi X1(12, 6) X2(0, 12) z1 132 72 z2 30 36

W B W Da trn thng tin ca bng payoff, ta c z1 = 72, z1 = 132; cn z 2 = 30,

z B = 36. Do , on bin thin cn xt cho z1 l [72, 132] v cho z2 l [30, 36]. T 2 chng ta c th thit lp cc hm tho dng m ng vi hai mc tiu cho nh sau:W 72 z1 z1 z1 z 72 z = 1 = = 1 1,2 1 ( z1 ) = B W 60 60 132 72 60 z1 z1

Hm tho dng m trn y ph thuc vo z1, nn ph thuc vo (x1, x2). Khi c mt phng n kh thi (x1, x2) ta tnh c tho dng 1 ( z1 ) i vi mc tiu z1. Tng t i vi z2 ta c hm tho dng m:W z2 z2 z 30 z 2 = 2 = 5. B W 36 30 6 z2 z2

2 (z2 ) =

Lp hm tho dng t hp u = w1 1 (z1 ) + w2 2 (z 2 ) , trong w1, w2 l cc trng s tho mn 0 w1, w2 1 v w1 + w2 = 1.

26

b. Cc bc lpz1 z z z 1,2) + 0,5 ( 2 5) = ( 1 + 2 ) + 1,9. 60 6 120 12 z z cc i ho hm tho dng t hp, ta ch cn tm Max 1 + 2 . Vy chng ta cn 120 12

Xt w1 = 0,5 v w2 = 0,5, th c u = 0,5 (

gii bi ton: Max u =

z1 z + 2 vi cc rng buc (D), hay bi ton tng ng: 120 12 z = 120u/18 = x1 + 2x2 Max vi cc rng buc (D). Gii BTQHTT ny ta s c kt qu (0, 12). Nu thay i gi tr ca b trng s (w1, w2) th s thu c cc phng n Pareto khc nhau, nhng u nm trn on AB.

Ch : Ch rng, vi mi b trng s (w1, w2) ta tm c mt phng n ti u Pareto (x1, x2) bng cch lm cc i ho hm tho dng t hp u thit lp c. C th chn cc b trng s w1, w2 khc nhau ta c cc phng n ti u Pareto khc nhau. T cc phng n , chng ta chn c phng n ti u Pareto tt nht.Phn mm MULTIOPT phin bn 1.0 c xy dng da trn phng php tho dng m tng tc nhm gii BTQHTT a mc tiu. Chng trnh c xy dng bng ngn ng Visual Basic, c giao din n gin v d dng. Cc chc nng chnh ca phn mm bao gm: Chc nng nhp d liu cho BTQHTT a mc tiu mt cch trc quan: nhp d liu qua bn phm hoc nhp d liu t tp. Xut bi ton c nhp ra tp. Gii bi ton: theo di bng payoff v gii theo phng php trng s. Xut kt qu ra tp: xut kt qu trung gian ra tp / xut kt qu cui cng ra tp. Sau khi ci t chng trnh, mt lin kt n chng trnh s c to ti menu Start > Programs ca Windows. khi ng chng trnh, kch hot menu Start > Programs > Ti u a mc tiu. Sau , cn phi chn bi ton bng cch vo menu Chng trnh, chn Bi ton tuyn tnh. Ngi s dng c th nhp d liu theo mt trong hai cch sau:

Nhp d liu t bn phmChn menu X l bi ton > Nhp d liu > Nhp t bn phm. Cc s liu u vo ca chng trnh bao gm: n s bin; m s cc rng buc; m1 s cc rng buc = ;

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p s cc mc tiu; ep s dng nh; gz s dng ln.

Nhp d liu t tpThc hin lnh X l bi ton>Nhp d liu>Nhp t tp, sau chn tp d liu.V d:

Gii bi ton quy hoch tuyn tnh ba mc tiu. z1 = x1 2x2 + 3x4 Max z2 = 3x1 + 6x2 9x4 Max z3 = 5x1 + x2 + 2x3 Max vi cc rng buc: 2x1 + x3 6; x1 + x2 + 5x3 6; 6x1 + 5x2 3x3 6; x3 + 4x4 8; x1, x2, x3, x4 0. Nh vy, y chng ta mun cc i ho ng thi ba hm mc tiu trn min D cc phng n kh thi tho mn ng thi bn rng buc v iu kin khng m ca cc bin.

Hnh 1.7. Nhp d liu t bn phm

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Trong bc khi to, ta nhp d liu t bn phm nh trn hnh I.7. Sau khi nhp d liu, thc hin tnh cc bc trung gian (tnh cc gi tr trong bng payoff, tnh cc gi tr cho cc hm tho dng) bng cch chn X l bi ton>Tnh ton>Tnh trung gian. Thc hin X l bi ton>Tnh ton>Phng php trng s. Trn mn hnh xut hin form nhp cc gi tr trng s (xem hnh I.7: cc gi tr ca trng s c nhp sao cho c tng l 1). Sau khi nhp cc gi tr trng s nhn vo nt Gii bi ton gii bi ton (xem hnh I.8).

Hnh I.8. Nhp cc gi tr trng s v gii bi ton

Kt qu trn hnh I.8 l x1 = 1,45, x2 = 0, x3 = 0,91, x4 = 0, 1 (z1 ) = 0,42, 2 (z 2 ) = 0,58, 3 (z 3 ) = 1; z1 = 1,45, z2 = 4,36 v z3 = 9,09 ng vi b trng s w1 = 0,2, w2 = 0,3 v w3 = 0,5.4. M hnh ti u phi tuyn n v a mc tiu M hnh ti u tng qut

4.1. Mt s khi nim c bn

M hnh ti u tng qut, hay bi ton ti u tng qut, c dng:

F(X) Min (Max) vi X D Rn.

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y F(X) c th l mt hm v hng hay hm vc t, tuyn tnh hay phi tuyn. Trong trng hp F(X) l hm v hng th ta c m hnh ti u n mc tiu, cn nu F l hm vc t th c m hnh ti u a mc tiu. D c gi l min rng buc hay min phng n kh thi, thng c biu din bi cc ng thc v/hoc cc bt ng thc.M hnh ti u phi tuyn n mc tiu

Dng chnh tc ca bi ton ti u mt mc tiu c biu din nh sau:

f(X) Min (Max), vi: (i) gj(X) 0, (ii) gj(X) = 0,

X = (x1, x2, , xn) Rn, j = 1, 2, , k, j = k+1, k+2, , m,

Trong cc bi ton thc t c th b sung cc rng buc (iii) ai xi bi, i = 1, 2, , n.

Trong trng hp hoc hm mc tiu f(X) hoc c t nht mt trong cc hm rng buc gj(X), j = 1, 2, , m, l hm phi tuyn, chng ta c bi ton ti u phi tuyn. Khi tt c cc to xi u bt buc nhn cc gi tr nguyn, i = 1, 2, , n, th ta c bi ton ti u nguyn. Cn nu ch c mt s to (nhng khng phi tt c cc to ) bt buc nhn gi tr nguyn th ta c bi ton ti u hn hp nguyn. K hiu D l min cc phng n (min rng buc) cho bi cc rng buc (i), (ii) v/hoc (iii) th bi ton ti u trn y c th vit gn hn nh sau:

f(X) Min (Max) vi X D.Lc ny, i vi bi ton cc tiu ho, X* D c gi l phng n ti u ton cc nu X D ta lun c: f(X*) f(X). Trong trng hp f(X*) f(X) ch ng vi X D trong mt ln cn no ca X* th X* c gi l phng n ti u a phng. Mt cch tng t, ta c th nh ngha khi nim phng n ti u ton cc hoc a phng cho bi ton cc i ho. Nu chng ta ch quan tm ti vic tm kim phng n ti u ton cc th ta c bi ton ti u ton cc. Trong cc bi ton ti u phi tuyn ng dng ni chung, trong lnh vc c kh in lc ni ring, phng n 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 f(X) c dng phi tuyn. Bi ton t ra l phi tm c phng n ti u ton cc. C rt nhiu phng php gii cc lp bi ton ti u phi tuyn, nhng cha c phng php no t ra hu hiu cho mi bi ton ti u phi tuyn, c bit l cc bi ton ti u nguyn v hn hp nguyn.

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M hnh ti u phi tuyn a mc tiu

M hnh ti u a mc tiu c dng:

zj = fj(X) Min (Max), X = (x1, x2, , xn), j = 1, 2,, p (p 2) vi: (i) gj(X) 0, (ii) gj(X) = 0, (iii) ai xi bi, j = 1, 2, , k, j = k+1, k+2, , m, i = 1, 2, , n.

Trong cc bi ton thc t c th b sung cc rng buc

Trong m hnh ny, ta c p mc tiu cn ti u ho, cc h s ca cc hm mc tiu v rng buc ni chung c gi s l cc gi tr thc xc nh. Trong trng hp c t nht mt trong cc hm mc tiu hay cc hm rng buc l hm phi tuyn, chng ta c bi ton ti u phi tuyn a mc tiu. i vi bi ton ti u phi tuyn a mc tiu chng ta cng c khi nim phng n ti u Pareto nh trnh by trong mc 3.1 v 3.2 i vi BTQHTT a mc tiu. Cng nh i vi cc BTQHTT a mc tiu, phng php gii bi ton ti u phi tuyn a mc tiu da trn s tr gip ca h my tnh, nhm gip ngi ra quyt nh tng bc thay i cc quyt nh trung gian mt cch thch hp i ti mt phng n ti u Pareto tho mn nht, c gi l phng php tng tc ngimy tnh.4.2. Mt s phng php v phn mm gii bi ton ti u phi tuyn n mc tiu

Cc phng php gii bi ton ti u ton cc

Cc phng php gii bi ton ti u ton cc phi tuyn n mc tiu c phn ra thnh hai lp: phng php tt nh (deterministic methods) v phng php ngu nhin (stochastic methods). Phng php tt nh s dng cc tnh cht gii tch ca hm mc tiu v cc hm rng buc. Mt s dng bi ton ti u ton cc vi nhng tnh cht gii tch nht nh ca hm mc tiu v cc hm rng buc c th gii c bng cc phng php tt nh thch hp, chng hn nh phng php quy hoch ton phng, quy hoch tch, quy hoch li, quy hoch d.c Trong cc trng hp phng n ti u ton cc c th tm c sau mt s hu hn bc tnh ton vi chnh xc chn trc. Tuy nhin, i vi nhiu lp bi ton ti u ton cc phng php tt nh t ra khng c hiu qu. Trong khi , cc phng php ngu nhin nh: phng php a khi to (multistart), m phng ti (simulated annealing), thut gii di truyn (genetic algorithm), c th p dng gii cc bi ton ti u ton cc dng bt k, khng i hi cc tnh cht c bit ca hm mc tiu hay cc hm rng buc. Cc phng php ngu nhin c bit t ra c hiu qu i vi cc bi ton ti u phi tuyn nguyn v hn hp nguyn. Tuy nhin, cc phng php ny thng ch cho phng n gn ti u kh tt sau mt s hu hn bc m khng kim sot c chnh xc ca phng n tm c.

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Phn mm Lingo gii bi ton quy hoch ton phng V d: Gii bi ton ti u phi tuyn dng ton phng

z = 8x12 + 6x22 Max, vi cc rng buc: 4x1 + 2x2 60 2x1 + 4x2 48 x 1 , x2 0.

gii bi ton trn, chng ta nhn vo biu tng Lingo trn mn hnh vo ca s Lingo. Sau thc hin cc lnh Lingo: Menu > New > v g vo cc d liu ca bi ton (tng t nh khi gii BTQHTT bng phn mm Lingo, xem li mc 1.4, hnh I.4).

Hnh I.9. Kt qu bi ton quy hoch ton phng trong Lingo.

Tip theo, cn nhy chut vo nt LINGO v gii bi ton thu c kt qu chi tit nh trn hnh I.9. Kt qu trn cho ta bit gi tr cc i ca hm mc tiu l 180 vi phng n ti u l: x1 = 15, x2 = 0. Cc gi tr ti u ca cc bin i ngu l y1 = 5/3 v y2 = y3 = y4 = 0.Gii bi ton ti u phi tuyn bng phn mm RST2ANU

Phn mm RST2ANU 1.0 c s dng gii cc bi ton ti u ton cc phi tuyn dng tng qut vi cc bin lin tc, cc bin nguyn v cho cc bi ton hn hp nguyn. Qu trnh xy dng phng php tnh ton ti u, thut gii, ci t trn ngn

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ng C v sau ny l ngn ng Visual C++ 6.0 cng nh chy th nghim ko di gn tm nm. Ngoi u im gii c cc bi ton hn hp nguyn, phn mm c tin cy rt cao trong vic tm ra cc phng n ti u ton cc v c giao din thn thin i vi ngi s dng. Phn mm c ng gi trnh sao chp v c th dng gii cc bi ton ln khi c ci t trn h my tnh mnh.

Thut giiThut gii ngu nhin RST2AN (hay RST2ANU), c a ra bi C. Mohan v Nguyn Hi Thanh. Thut gii ny l thut gii tm kim ngu nhin c iu khin, c kt hp thut ton m phng ti (SA). Thut gii RST2AN l thut gii lp, bao gm hai pha: pha cc b v pha ton cc. Sau y l thut gii RST2AN c pht biu mt cch ngn gn cho bi ton ti u chnh tc dng cc tiu ho.

Trong pha ton cc, mt s lng thch hp ln cc phng n kh thi c c pht sinh ra mt cch ngu nhin v lu tr trong mng c tn A. nh du hai im c gi tr hm mc tiu ln nht v nh nht tng ng l M v L. Trong pha cc b, cc phng n c x l nhm thu c gi tr tt hn ca hm mc tiu. Trong pha ny, thut gii xc nh X l im c ni suy bc hai da trn phng n L v hai phng n khc c chn ngu nhin trong mng A. Nu nh X l phng n kh thi th vi f(X) f(M), M s c thay th bi X trong mng A; cn vi f(X) > f(M), M s c thay th bi X vi xc sut p= exp((f(X)f(M))/(f(X)f(L))), trong > 0 l tham s c la chn thch hp. Nu X khng phi l phng n kh thi, b qua X v chn hai phng n khc trong A mt cch ngu nhin ri cng vi L tip tc sinh ra phng n mi. Qu trnh c th tip din nh vy cho ti khi tp hp cc phng n trong A s c xu hng co cm li xung quanh mt phng n ti u ton cc.V d: Gii bi ton ti u phi tuyn hn hp nguyn.

z = x10,6 + x20,6 + x30,4 + 2x4 + 5x5 4x3 x6, Min vi cc rng buc: x2 3x1 3x4 = 0; x3 2x2 2x5 = 0; 4x4 x6 x1 + 2x4 x 2 + x5 x3 + x6= 0;

4; 4; 6;

x1 3; x2 4; x3 4; x4 1; x5 2; x6 6; x1, x2, x3, x4, x5, x6 0; x4, x5, x6 l cc bin nguyn.

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Hng dn s dngChng trnh c gi gn trong mt file chy duy nht mang tn rst2anu1.0.exe. Khi bt u khi ng chng trnh, ngi dng s c hi m ng k s dng chng trnh. Mi ngi dng s c cp mt m ng k v phi c m ng k mi s dng c chng trnh, do chng trnh khng th b sao chp. Sau khi nhp m ng k, ngi dng c th nhp bi ton mt cch d dng (xem hnh I.10) vi: NX l s bin ca bi ton. XINT xc nh bin nguyn v bin khng nguyn. Nh trong hnh trn, XINT = 0,0,0,1,1,1 cho bit ba bin u l bin lin tc, ba bin sau l bin nguyn.

Hnh I.10. Mn hnh giao din sau khi nhp xong d liu

FX l xu xc nh hm rng buc, c nhp theo c php ca EvaluateExpression. Cc bin c vit bng k hiu X c km theo ch s. V d, X1 l bin th nht, X5 l bin th 5. Nu bi ton ti u l bi ton tm cc tiu th la chn MIN v ngc li chn MAX vi bi ton tm cc i. Feas xu cho bit cc hm rng buc, c nhp cch nhau bi du chm phy hoc xung dng. Cc xu ny cng tun theo c php ca EvaluateExpression. Rules l cc xu ch ra cc lut. y, mt lut c th coi nh l mt lnh gn gi tr ca mt bin bi gi tr ca mt biu thc cc bin khc. MINX l mng xc nh cn di cho cc bin, cc gi tr vit cch nhau bi du phy (,).

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MAXX l mng xc nh cn trn cho cc bin, cc gi tr vit cch nhau bi du phy (,). NA l kch thc ca mng A (c th chn tu , ti thiu l 2(n + 1) vi n l s bin ca bi ton). MAX RANDOM l s ln c gng ti a tm mt phng n chp nhn c bng phng php ngu nhin. ITERLAST, ISLAST, IFLAST l cc gii hn v s vng lp, s ln tht bi trong vic ci thin gi tr hm mc tiu, s ln tht bi trong vic ni suy phng n mi chp nhn c. Epsilon1, epsilon2 l cc s dng nh nhm xc nh tiu chun co cm ca mng A theo thut gii. Beta l hng s s dng trong cng thc tnh xc xut thay th mt phng n tt hn trong mng A bi mt phng n ti hn. Prob file v Res file l cc tp u vo v tp kt qu. C th son sn tp bi ton u vo ri np bi ton. Cng c th lu mt bi ton nhp ra tp. Chy chng trnh Sau khi nhp bi ton hay np bi ton t tp, c th chy chng trnh bng cch kch chut vo nt RUN. Trong khi chy chng trnh, trng thi pha trn nt RUN s xut hin dng ch SEARCHING. Khi thut gii chy xong th trng thi s tr v READY cho bit sn sng cho cc bi ton tip theo. Mi thng tin v phn mm v cch s dng s c bit nu kch chut vo nt ABOUT. Sau khi chy xong chng trnh, kt qu chy s c xem trc tip khi kch chut vo nt RESULTS v c th lu ra file vn bn, bao gm phng n ti u, gi tr hm mc tiu, mng A, c cu trc nh trn hnh I.11.

Hnh I.11. Cu trc file kt qu

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Nh vy, bi ton c gii xong, vi kt qu: x1 = 2/3, x2 = 2, x3 = 4, x4 = 0, x5 = 0, x6 = 0, v gi tr ti u ca hm mc tiu l 11,95913.Bi ton ti u thng s sng phn loi Chng ta c th s dng phn mm RST2ANU tm nghim ca h phng trnh phi tuyn sau pht sinh trong vic 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 + lcos2 + l3cos3 + l4cos4 xC1 = 0; r sin1 + lsin2 + l3sin3 + l4sin4 yC1 = 0; r cos1 + lcos2 + l3cos(3 ) + l5cos5 xD1 = 0; r sin1 + lsin2 + l3sin(3 ) + l5sin5 yD1 = 0;Trong h phi tuyn trn cc thng s bit l: r = 0,05m; l = 0,30m; l3 = 0,15m; l3 = 1,075m; l3 = 1,025m; l4 = 0,50m; l5 = 0,40m; xC1 = 0,365m; yC1 = 0,635m; xD1 = 1,365m; yD1 = 0,635m; = /8. s dng phn mm RST2ANU gii h phng trnh phi tuyn cho 1 = k/8 (k = 0, , 9), trc ht chng ta cn thit lp hm mc tiu sau: z = (rcos1 + lcos2 + l3cos3 + l4cos4 xC1)2 + (rsin1 + lsin2 + l3 sin3 + l4sin4 yC1)2 + (rcos1 + lcos2 + l3cos(3 ) + l5cos5 xD1)2 + (rsin1 + lsin2 + l3sin(3 ) + l5sin5 yD1)2 Min. Kt qu c cho trong bng I.3 vi zmin = 0.Bng I.3. Kt qu tnh ton gi tr cc thng s ca sng phn loi

1 [0,2]0

2 [0,]0,226128 0,199269 0,170835 0,143343 0,112669 0,090986 0,066036 0,051284 0,039053 0,033773

3 [0,]0,551311 0,550518 0,550590 0,550490 0,552073 0,551991 0,553576 0,554296 0,555262 0,556277

4 [0,]1,783873 1,784628 1,782751 1,778826 1,770032 1,759350 1,745374 1,730174 1,713242 1,695605

5 [0,]1,666775 1,670250 1,668853 1,663697 1,652171 1,639575 1,622823 1,602970 1,581813 1,560720

/182/18 3/18 4/18 5/18 6/18 7/18 8/18 9/18

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4.3. Mt s phng php gii bi ton ti u phi tuyn a mc tiu

Phng php tng tc ngimy tnh

Phng php PRELIME (PREference Level Interactive Method) hay cn gi l phng php tng tc da trn mc u tin do C. Mohan v Nguyn Hi Thanh xut. Cn phng php trng s quy chun l do Andrezj Osyczka xut. Cc phng php ny u thuc lp phng php tng tc ngimy tnh gii bi ton ti u a mc tiu vi cc yu t cu thnh sau: C cu u tin ca ngi ra quyt nh v hm t hp tng ng. Kiu tng tc ngi my tnh: cc thng tin no my tnh phi a ra trong cc bc lp trung gian, v cch thay i cc thng s ca c cu u tin t pha ngi ra quyt nh. K thut ti u ton hc c xy dng da trn l thuyt ti u ho nhm tm ra cc phng n ti u Pareto cho cc bi ton cn gii trong cc bc lp trung gian.Bi ton thit k trc my

Bi ton c hai mc tiu sau: 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) = 3,298105 [( 1 1 10 9 ) x13 + 8 ] (mm/N) 8 4 4 4 4,096 10 7 x2 10 x2 10 x2

Trong , X= (x1, x2) l vc t quyt nh hay vc t phng n, 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). Chng ta cn 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 9,78 10 6 x1 0, 4 4,096 10 7 x 2 (1) (2) (3) (4)

g2 (X) = 75,2 x2 0, g3 (X) = x2 40 0, g4 (X) = x1 0,

trong cc iu kin (2), (3), (4) l d hiu (ngoi ra, ta gi thit x1 1000), cn iu kin (1) ny sinh do yu cu sau: Mt mt, trc my phi chu ng c ti mc ti a lc Fmax = 12000N. Mt khc, nn kt ni cho php l 180N/mm.

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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 t thc t) 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, chng ta hy phn tch vn tt hai phng php gii bi ton thit k trc my nu ra trn.Phng php trng s quy chun

Trong yu t cu thnh th nht, hm t hp cc mc tiu cho bi f(X) = 1f1(X) + 2 f2(X), trong 1, 2 l cc trng s khng m ng vi cc hm f1(X) v f2(X), 1 + 2 = 1. Do gi tr ca hm f1(X) thng ln gp rt nhiu ln gi tr ca hm f2(X), 1 v 2 c quy chun nh sau: f(X) = 1'f1(X) + 2'f2(X), vi 1' = 1.106/2,961 ; 2' = 2.10+3/0,338. yu t cu thnh th hai, trong cc bc lp trung gian, ngi ra quyt nh thay i ln lt cc cp trng s (1, 2) vi cc gi tr l (0,2; 0,8), (0,8; 0,2), (0,6; 0,4) v (0,4; 0,6). Cp trng s cui cng cho phng n ti u Pareto tho mn nht l x1 = 237,1 v x2 = 68,2, vi f1(X) = 3,529 106 ; f2(X) = 0,437 103. Cn yu t cu thnh th ba, tc gi Andrezj Osyczka s dng thut ton ti u d tm ngu nhin.Phng php tng tc da trn mc u tin PRELIME

Trc ht, yu t cu thnh th nht, hai mc tiu f1(X) v f2(X) c chuyn thnh hai hm (lin) 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 im ni suy: 0 1 = 0,5 1 nu f1 6,594 10 = a1 nu f1 = 4 106 = b1 nu f1 2,944 106 = c1O c1 b1 a16

1 0,5

1

f1

0 2 = 0,5 1

nu f2 0,499 10 = a2 nu f2 = 0,450 103 = b2 nu f2 0,338 103 = c2

3

1 0,5

2

O

c2

2 0

a2

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th ca cc hm thuc m cho cc hnh v trn. Phn tch hm thuc m 1, ta thy: ngi ra quyt nh s c tho mn 0 i vi mi phng n lm cho f1 6,594 106; tho mn 1 nu f1 2,944 106 ; v tho mn 0,5 nu f1 = 4106. tho mn 0,5 c coi l tho mn ti thiu v mc f1 = 4 106 = 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. Sau , hm tho dng t hp dng MaxMin c thit lp cho hai hm mc tiu ring r trn di dng: Max{Min[1, 2]} nhm tm ra phng n tho dng (x1, x2) trong min rng buc ca bi ton. i vi yu t cu thnh th hai, ngi ra quyt nh s cn c vo thng tin do my tnh a ra iu chnh cc mc u tin b1 v b2. Thay i cc cp mc u tin (b1, b2) t (4106; 0,45103) sang (3,6106; 0,435103), s nhn c phng n sau (x1, x2) = (235,67 ; 67,67) vi (f1, f2) = (3,58106; 0,433103). Trong yu t cu thnh th ba, cc tc gi dng thut ton tm kim ngu nhin c iu khin RST2ANU kt hp vi thut ton m phng ti (SA) tm ra cc phng n ti u Pareto cho cc bi ton trung gian thng qua vic gii bi ton ti u phi tuyn n mc tiu dng Max{Min[1, 2]}.

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40

Chng II

CC M HNH MNG1.1.1.

M hnh mng vn tiPht biu bi ton vn ti

Bi ton vn ti c p dng rt rng ri trong lnh vc lp k hoch phn b sn phm hng ho (dch v) t mt s a im cung / cp pht ti mt s a im cu / tiu th. Thng thng, ti mi a im cung (ni i) ch c mt s lng gii hn hng, mi a im cu (ni n) ch cn mt s lng nht nh hng. Vi cc cung ng vn chuyn hng a dng, vi cc ph vn ti khc nhau, mc tiu t ra l xc nh phng n vn ti ti u. Ni cch khc, vn t ra l cn xc nh nn vn chuyn t mi a im cung ti mi a im cu bao nhiu n v hng nhm tho mn nhu cu ca tng ni n ng thi t tng chi ph vn ti l nh nht.V d: Ta c 3 im cung cp hng A, B, C v 4 im cu S, T, U v V vi lng hng cung v cu ti mi im cng nh cc ph vn ti trn mt n v hng cho mi cung ng nh trong bng II.1.Bng II.1. Cc d liu ca bi ton vn ti

im cung A B C Tng

Lng hng 5000 6000 2500 13500

im cu S T U V Tng

Lng hng 6000 4000 2000 1500 13500

Ni i A B C

Cc ph vn chuyn/n v hng cij (USD) n) S 3 7 2 T 2 5 5 U 7 2 4 V 6 3 5

T im cung i n im cu j ta c cc ph vn ti / mt n v hng l cij bit, chng hn nh c11 l 3USD / mt n v hng. Cn thit lp phng n vn ti hng p ng c cung cu v tng chi ph vn ti l nh nht. Ch rng bi ton vn ti ang xt c tng cung bng tng cu, nn c gi l bi ton vn ti cn bng thu pht. y l dng n gin nht trong cc dng bi ton vn ti.

41

1.2.

To phng n vn ti xut pht

Khi nim bng vn ti

Bng vn ti c m hng, n ct gm m n , m l s im cung, n l s im cu vi cc ph cij c ghi trong (i, j) cho cung ng (i, j). Khi m =3, n = 4 nh trong v d trn, ta c bng vn ti II.2.Bng II.2. Bng vn ti

3 7 2

2 5 5

7 2 4

6 3 5

Cung 1: 5000

Cung 2: 6000 Cung 3: 2500 Tng: 13500

Cu1: 6000

Cu 2: 4000

Cu 3: 2000

Cu 4: 1500

Ta cn tm phng n phn hng vo cc (i, j) sao cho tng theo hng hay ct u khp vi cc lng cung, cu v tng chi ph vn ti l nh nht. Mi (i, j) biu din mt cung ng vn chuyn hng t im cung i v im cu j.Cc phng php to phng n xut pht

C mt s phng php to phng n xut pht. Ta nghin cu hai phng php sau y.a. Phng php "gc ty bc"

Phng php ny c pht biu nh sau:

Phn pht hng ti a vo gc ty bc ca bng vn ti. Sau khi (hng) cung hoc (ct) cu tho mn th ta thu gn bng vn ti bng cch b bt hng cung hoc ct cu i (ch b mt trong hai th hoc hng hoc ct, y l ton t hoc loi tr, OR exlusive). Tip tc lp li hai bc trn y cho ti khi hng c phn phi ht vo cc (cc c phn hng c gi l s dng).Bng phng php gc ty bc ta to c phng n A trong bng II.3 vi su s dng (1, 1), (2, 1), (2, 2), (2, 3), (3, 3) v (3, 4).Bng II.3. Phng n xut pht vi phng php gc ty bc

3 5000 7 1000 2

2 5 4000 5

7 2 1000 4 1000

6 3 5 1500

42

Tng chi ph vn ti:

CPVT = (3 5 + 7 1 + 5 4 +2 1 + 4 1 + 5 1,5) 1000 = 55500.b. Phng php cc ph ti thiu

Phng php ny c pht biu tng t phng php "gc ty bc" nhng u tin phn pht hng vo c cc ph b nht (nu c nhiu nh vy th chn bt k trong s ). Lc ny ta c phng n xut pht l phng n B cho trong bng II.4.Bng II.4. Phng n xut pht vi phng php cc ph ti thiu

3 1000 7 2500 2 2500 Tng chi ph vn ti:

2 4000 5 5

7 2 2000 4

6 3 1500 5

CPVT = (3 1 +2 4 + 7 2,5 + 2 2 + 3 1,5 + 2 2,5) 1000 = 42000Nhn xt

Phng php cc ph ti thiu thng cho phng n xut pht tt hn phng php gc ty bc. Bng vn ti c s s dng l 3 + 4 1 = 7 1 = 6. Mt cch tng qut bng vn ti m hng, n ct c s s dng l m + n 1. Bi ton vn ti cng l BTQHTT. Trong v d ang xt, nu k hiu xij l lng hng vn chuyn trn cung ng (i, j) th chng ta BTQHTT sau:z = c11x11 + c12 x12 +... + c34x34 Min vi cc rng buc: x11 + x12 + x13 + x14 = 5000 x + x + x + x = 6000 22 23 24 21 x 31 + x 32 + x 33 + x 34 = 2500 x11 + x 21 + x 31 = 6000 x12 + x 22 + x 32 = 4000 x13 + x 23 + x 33 = 2000 x14 + x 24 + x 34 = 1500 x 0 i = 1, 2,3; j = 1, 2,3, 4 ij

43

H cc rng buc c 12 bin vi 7 phng trnh. Nu ly tng 3 phng trnh u tr i tng 3 phng trnh tip theo th c phng trnh cui. C th kim nghim d dng, s phng trnh c lp tuyn tnh ca h l 7 1 = 6. Mi phng n xut pht A hay B tm c ca bi ton vn ti chnh l mt phng n cc bin xut pht khi gii BTQHTT. Bi ton vn ti c th hon ton gii c bng phng php n hnh. Tuy nhin do cu trc c bit ca mnh, bi ton vn ti c th gii bng phng php c bit vi thut ton chuyn dng.1.3. Phng php phn phi gii bi ton vn ti

Chng ta p dng phng php ln (tm dch t Stepping Stone Method), hay chnh thc hn cn gi l phng php phn phi (Distribution Method) gii bi ton vn ti. Phng php ln l mt quy trnh tnh ton nhm tng bc ci thin phng n vn ti c cui cng tm c phng n vn ti ti u.Xc nh hiu sut ca cc cha s dng

Quay li bng vn ti II.3 vi phng n xut pht tm c theo phng php gc ty bc. Trong bng ch c mt s s dng, ta coi chng nh cc hn ni ln trong mt ci ao. Xt mt (i, j) bt k cha s dng trong phng n c. Ta cn tnh hiu sut ca , k hiu l eij, (e l vit tt ca t effect) theo cc bc sau: u tin ta cn tm mt ng i c tnh cht: i qua mt (i, j) cha s dng ( xut pht) v mt s s dng khc, mi bc phi i theo hng hoc theo ct xen k nhau (khng c i lin hai bc trn mt hng hay mt ct) cui cng quay v (i, j). iu ny ging nh ang trn thuyn, mun ra khi thuyn m khng t ta phi nhy qua cc hn ni ln trong ao cui cng li quay v thuyn (v vy phng php c tn l phng php ln). Mt iu th v na l con ng nhy trn cc hn nh vy l duy nht. Tm li xut pht t (1, 2) chng hn, ta s c ng i nh sau: (1, 2) (2, 2) (2, 1) (1, 1) (1, 2), trn ng i ny ch duy nht c mt cha s dng (xem bng II.5).Bng II.5. Tnh hiu sut cc cha s dng

3 5000 7 2 1000 (7) 6000

2 5 5 4000 (2) 4000

7 2 1000 4 1000 2000

6 3 5 1500 1500

5000 6000 2500

nh du cng tr xen k ti cc nh trn ng i m trong cha s dng c nh du +. Gi s ta cn lun chuyn mt n v hng theo ng i xc

44

nh m vn tho mn c cung cu (tc l cc mang du +: (1, 2) v (2, 1) c thm mt n v hng, cc mang du : (2, 2) v (1, 1) rt bt i mt n v hng). Lc ny tng chi ph s thay i mt lng tin l: e12 = +c12 c22 + c21 c11= 2 5 + 7 3 = +1. Ni cch khc, tng chi ph vn ti s tng thm ln 1USD cho mi mt n v hng lun chuyn theo ng i trn. Ta gi e12 l hiu sut ca (1, 2). Tng t: e13 = 7 2 + 7 3 = +9, e14 = 6 5 + 4 2 + 7 3 = +7, e24 = 3 5 + 4 2 = 0, e31 = 2 7 + 2 4 = 7, e32 = 5 5 + 2 4 = 2. Ch c hai vi hiu sut m l (3, 1) v (3, 2) (xem bng II.5) c th la chn a vo s dng trong phng n mi. Ta quyt nh trong phng n mi s chn (3, 2) a vo s dng, mi n v hng a vo s dng ti (3, 2) s lm tng chi ph gim 2USD. K hiu e = e32.

Ch : C th chng minh c eij = ij vi ij l gi tr trn hng ng vi ct xij nu gii bi ton vn ti bng phng php n hnh.Xc nh lng hng a vo chn

Nh trn phn tch, mt n v hng a vo (3, 2) lm gim tng chi ph vn ti 2 USD. Ta cn tm q, lng hng ti a c th a vo (3, 2). ng i qua (3, 2) v mt s c s dng l: (3, 2) (2, 2) (2, 3) (3, 3) (3, 2), vi cc c nh du cng tr xen k ( (3, 2) mang du +). Lng hng q c tnh theo quy tc: q = gi tr nh nht ca cc lng hng ti cc mang du () = Min {lng hng ti (2, 2), lng hng ti (3, 3)} = Min {4000, 1000} = 1000. Vy trong phng n mi, lng hng ti cc mang du + (cc (3, 2), (2, 3)) c tng thm 1000 n v, cn ti cc mang du (cc (2, 2) v (3, 3)) lng hng gim i 1000 n v (xem bng II.6). Phng n mi gm 6 s dng ( (3, 3) ng vi q =1000 b loi ra).Bng II.6. Phng n vn ti sau hai bc

3 5000 7 2 1000 (5) 6000

2 5 3000 5 1000 4000

7 2 2000 4 2000

6 3 5 1500 1500

5000 6000 2500

Tng chi ph vn ti:

CPVT = (3 5 + 7 1 + 5 3 + 2 2 + 5 1 + 5 1,5) 1000 = 53500;

45

hoc

CPVTmi = CPVT c e q = 55500 2 1000 = 53500.iu kin ti u

Thc hin theo quy trnh trn cho ti khi tt c cc hiu sut eij 0 (i, j) l cc cha s dng. y chnh l iu kin ti u hay iu kin dng. iu kin ny thc cht l iu kin ij 0 vi mi bin ngoi c s xij nu gii bi ton bng phng php n hnh. gii tip bi ton, cn tnh cc hiu sut cho cc cha s dng trong phng n mi: e12 = 2 5 + 7 3 = +1; e14 = 6 5 + 5 5 + 7 3 = +5; e31 = 2 7 + 5 5 = 5; e13 = 7 2 + 7 3 = +19; e24 = 3 5 + 5 5 = 2; e33 = 4 5 + 5 2 = +2.

Ta quyt nh s dng chn (3, 1) trong phng n mi v e31 = 5. Tm c q = 1000 theo quy tc bit. C hai ng vi q tm c, chng ta ch b i (2, 1) cn phi gi li (3, 2) a vo s dng. Phng n sau bc th ba cho trong bng II.7.Bng II.7. Phng n vn ti sau ba bc

3 5000 7 2 1000 6000

2 5 4000 5 0 4000

7 2 2000 4 2000

6 3 5 1500 1500 (2)

5000 6000 2500

Tng chi ph vn ti:

CPVT = 53500 5 1000 = 48500.Tip tc tnh cc hiu sut: e12 = +1; e14 = 6 5 + 2 3 = 0; e24 = 3 + 5 + 5 5 = 2; e13 = 7 2 + 5 5 + 4 = 9; e21 = 7 2 + 5 5; e33 = 4 5 + 5 2 = 2.

Chn (2, 4) a vo s dng v tnh q = 1500. T c phng n mi sau bn bc nh trong bng II.8.

46

Bng II.8. Phng n vn ti sau bn bc

3 5000 7 2 1000 6000

2 5

(4)

7 2

6 5000 3 2000 1500 5 2000 1500 6000 2500

2500 5 1500 4000 4

Tng chi ph vn ti:

CPVT = 48500 2 1500 = 45500.Tip tc tnh cc hiu sut: e12 = 2 5 + 2 3 = 4; e14 = 6 3 + 5 5 + 2 3=2; e33 = 4 5 + 5 2 = 2; e13 = 7 2 + 5 5 + 2 3 = 4; e21 = 7 2 + 5 5 = 5; e34 = 5 5 + 5 2 = 3;

Ta c e12 = 4 v chn (1, 2) lm chn vi q = 1500 v chuyn sang phng n mi nh trong bng II.9.Bng II.9. Phng n vn ti sau nm bc

3 3500 7 2 2500 6000

2 1500 5 2500 5 4000

7 2 2000 4 2000

6 3 1500 5 1500

5000 6000 2500

Tng chi ph vn ti:

CPVT = 45500 41500 = 39500.Lc ny eij 0 vi mi (i, j) cha s dng. iu kin ti u c tho mn. Phng n vn ti ti u cho trong bng II.9 vi tng chi ph nh nht l 39500.Bi ton vn ti khng cn bng thu pht

Trng hp tng lng cung ln hn tng lng cu, cn b tr thm mt im cu gi m mi chi ph vn ti n u c coi bng 0. Tng t, nu cu vt cung th cn b tr mt im cung gi v coi mi chi ph vn chuyn t i u bng 0.

47

1.4.

Phng php phn phi ci bin gii bi ton vn ti

Phng php ln hay phng php phn phi c mt nhc im l vic tnh hiu sut ca cc kh di dng. V vy, ta s nghin cu phng php phn phi ci bin nhm tnh cc hiu sut eij ngn gn hn. Xt phng n xut pht tm c bng phng php cc ph cc tiu cho trong bng II.10 (vi tng chi ph vn ti l 42000).Bng II.10. Phng n vn ti xut pht

3 1000 7 2500 2 2500 6000

2 4000 5 5 4000

7 2 2000 4 2000

6 3 1500 5 1500

5000 6000 2500

Ta c e13 = 7 2 + 7 3 = +9. Ta tm cch tnh e13 bng cch khc nhanh hn nh trnh by sau y. Trc ht cn xy dng h thng s th v hng v ct {(ui, vj), i = 1, 2, 3; j = 1, 2, 3, 4}. C th gn cho mt th v bt k gi tr 0 (hoc mt gi tr bt k khc), th v ny thng c chn hng hay ct c nhiu s dng nht. Chng hn chn u2 = 0.Cc th v khc c tnh bi cng thc: ui + vij = cij (i, j) s dng. u2 = 0 v1 = 7 (= c21 u2)

v3 = 2 (= c23 u2) v4 = 3 (= c24 u2) u1 = 4 (= c11 v1) u3 = 5 (= c37 v1) v2 = 6 (= c12 u1)Cng thc tng qut tnh cc hiu sut cho cc (i, j) cha s dng l:

eij = cij (ui + vj).Chng hn ta c e13 = c13 (u1 + v3) = 7 (4 + 2) = 9. Cc hiu sut khc c tnh tng t (xem bng II.11).

48

Bng II.11. Tnh ton cc th v v cc hiu sut

v1 = 7 u 1 = 4 u2 = 0 u 3 = 5 3 1000 7 2500 2 2500 6000

v2 = 6 2 5 5 4000 4000 (1)

v3 = 2 7 2 2000 4 2000

v4 = 3 6 3 1500 5 1500 5000 6000 2500

Trong bng II.11 ta thy e22 = 1 < 0. Chn (2, 2) a vo s dng ng vi q = 2500, ta chuyn sang phng n mi v tnh li cc h thng s th v nh trong bng II.12.Bng II.12. Tnh ton cc th v v cc hiu sut cho phng n mi

v1 = 6 u 1 = 3 u2 = 0 u 3 = 4 3 3500 7 2 2500 6000

v2 = 6 2 1500 5 2500 5 4000

v3 = 2 7 2 2000 4 2000

v4 = 3 6 3 1500 5 1500 5000 6000 2500

Chn

u2 = 0 v2 = 5 (= 5 0); v3 = 2 (= 2 0); v4 = 3 (= 3 0); u1= 3 (= 2 5); v1 = 6 (= 3 (3)); u3 = 4 (= 2 6).

Tng chi ph vn ti:

CPVT = (3 3,5 + 2 1,5 + 5 2,5 + 2 2 + 3 1,5 + 2 2,5) 1000= 39500 (tnh cch khc, CPVTmi = 42000 1 2500).

Tip tc tnh ton cc hiu sut: e13 = c13 (u1 + v3) = 7 (3 + 2) = 8; e14 = c14 (u1 +v4) = 6 (3 + 3) = 6; e21 = c21 (u2 + v1) = 7 (0+6) = 1; e32 = c32 (u3 + v2) = 5 (4 + 5) = 4; e33 = c33 (u3 + v4) = 4 (4 + 2) = 6; e34 = c34 (u3 + v4) = 5 (4 + 3) = 6.

49

Ta thy eij 0 (i, j) cha s dng nn iu kin ti u c tho mn. Phng n ti u cho trong bng II.12, vi tng chi ph vn ti nh nht l 39500.

Ch : i vi bi ton vn ti cn cc i ho hm mc tiu th tiu chun dng s l eij 0 (i, j) cha s dng. i vi bi ton vn ti c cm (cung ng khng c s dng) th t cc ph M = + cho cc cm vi bi ton Min hoc M = vi bi ton Max.Gii bi ton vn ti bng phn mm Lingo

gii bi ton vn ti trong Lingo, ta c th s dng cc bi ton mu bng cch nhn vo biu tng Lingo v thc hin cc lnh File > Open > Tran.lng vo bi ton vn ti mu. Sau nhp cc s liu u vo ca bi ton cn gii, chng hn, ca v d xt trong cc mc trn thay cho cc s liu ca bi ton mu (xem hnh II.1).

Hnh II.1. Nhp s liu cho bi ton vn ti

Sau chng ta thc hin LINGO>Solve, kt qu tnh ton s hin ra trn mn hnh (xem hnh II.2).

50

Hnh II.2. Kt qu ca bi ton vn ti

2.

M hnh mng PERT

(Program Evaluation and Review Technique)2.1. Cc khi nim c bn v PERT

Vai tr ca PERT

PERT c th c hiu l phng php hoc k thut theo di v nh gi d n vi mc ch gip cho b my qun l tr li cc cu hi sau y: D n s hon thnh khi no? Mi hot ng ca d n nn c bt u vo thi im no v kt thc vo thi im no? Nhng hot ng no ca d n phi kt thc ng thi hn trnh cho ton b d n b kt thc chm hn so vi k hoch? Liu c th chuyn cc ngun d tr (nhn lc, vt lc) t cc hot ng khng gng sang cc hot ng gng (cc hot ng phi hon thnh ng tin ) m khng nh hng ti thi hn hon thnh d n? Nhng hot ng no cn tp trung theo di? bc u hnh dung v PERT, chng ta xt v d sau y.V d:

Gi s cn thc hin mt d n hoc chng trnh c cc hot ng c lit k trong bng II.13.

51

Bng II.13. Cc hot ng ca mt d n, th t v thi gian thc hin

Hot ng A B C D E F G H I J K L

Hot ng k trc

Thi gian thc hin (tun) 2 2 2 3 4 0 (hot ng gi) 7 6 4 10 3 4

A A E B B D, F C H, J G, I, K

Ta cn lp k hoch thc hin d n trn hon thnh ton b cc hot ng ca d n trong thi gian ngn nht, ng thi phi xc nh c nhng hot ng no cn ch trng (c hiu l cc hot ng gng).V s mng PERTD 3 A 1 C 2 J B E 5 4 H 7 G K 8 L 9 F 6 I

Hnh II.3. S mng PERT

Trn hnh II.3 ta thy mng PERT l mt mng cc nt c nh s c ni vi nhau bi cc cung c mi tn. Mi cung c mi tn biu din mt hot ng ca d n, cn mi nt biu din thi im kt thc mt s hot ng v / hoc thi im bt u ca mt s hot ng khc. Hot ng gi F c k hiu bi cung mi tn vi nt ri c thi gian thc hin bng 0, nhm trnh cho hot ng D v E c cng nt bt u v nt kt thc. Nh vy, trong s mng PERT ta buc phi tun theo quy c: hai hot ng khc nhau th khng c c cng nt bt u cng nh nt kt thc.Xc nh thi gian ti thiu thc hin d n

xc nh thi gian ti thiu thc hin d n, trc ht chng ta nghin cu khi nim thi im bt u sm nht v thi im kt thc sm nht (EST v EFT Earliest start time v Earliest finish time) cho tng hot ng.

52

V d: Hot ng A c ESTA = 0 v EFTA = 2, v

Thi im bt u sm nht l khi bt u khi ng d n, Thi im kt thc sm nht l sau 2 tun. Mi quan h gia EST v FFT l: EFT = EST + thi gian thc hin hot ng. Mt cch tng qut, xc nh EST chng ta c quy tc thi im bt u sm nht: thi im bt u sm nht ca mt hot ng ri mt nt no l thi im mun nht trong cc thi im kt thc sm nht i vi cc hot ng i vo nt . p dng quy tc trn y, c th tnh c ESTK = 12 (do EFTH = 8, EFTJ = 12 v s ln hn l 12) v EFTK = 15. Kt qu tm EST v EFT cho cc hot ng d n c tnh ton tin t nt 1 n nt 9 v c tm tt trong bng II.14 v hnh II.4. Vy thi gian kt thc sm nht d n l sau 19 tun.2

3

A0

2 2E

D6

5

66

F 5

6

I L19

1 00

B

2

4 J

C2

2 2H

G8

9 10 8 15

9

K 15 7 12

2

2

12

Hnh II.4. Tnh EST v EFT cho cc hot ng ca d n Bng II.14. Tnh EST, LST, EFT, LFT v tm ng gng

Hot ng A B C D E F G H I J K L

EST 0 0 0 2 2 6 2 2 6 2 12 15

LST 5 4 0 8 7 11 8 6 11 2 12 15

EFT 2 2 2 5 6 6 9 8 10 12 15 19

LFT 7 6 2 11 11 11 15 12 15 12 15 19

LSTEST (LFTEFT) 5 4 0 6 5 5 6 4 5 0 0 0

Trn cung gng

*

* * *

53

Bc tip theo l xc nh thi im bt u mun nht v thi im kt thc mun nht (LST v LFT Latest start time v Latest finish time) cho tng hot ng.V d: Hot ng L c LSTL = 15 v LFTL = 19, v

Thi im kt thc mun nht l sau 19 tun (nu ta n nh d n phi kt thc sau 19 tun), Thi im bt u mun nht l tun 15 (do hot ng L cn thi gian 4 tun thc hin). Mi quan h gia LST v LFT l: LST = LFT thi gian thc hin hot ng. Mt cch tng qut, xc nh LFT chng ta c quy tc thi im kt thc mun nht: thi im kt thc mun nht ca mt hot ng i vo mt nt no l thi im sm nht trong cc thi im bt u mun nht i vi cc hot ng ri nt . p dng quy tc trn y, c th tnh c LFTA = 7 (do LSTD = 8, LSTE = 7 v s b hn l 7) v LSTA = 5. Kt qu tm LFT v LST cho cc hot ng d n c tnh ton li t nt 9 v nt 1 v c tm tt trong bng II.14 v hnh II.5.7

3

A5

8 7E

D11 F 11

6

11

I

11

11 5 8

14 0

B C2

6

46

G H

L 15 15 8 15 19 K 15 7 12

9

2

2

J12

12

Hnh II.5. Tnh LFT v LST cho cc hot ng ca d n

Ch : Mi cung c mi tn l mt hot ng, nhng c th bao gm nhiu hot ng nh khc. Ni cch khc, bn thn tng hot ng ca d n c th li l mt mng PERT nh.Xc nh hot ng gng, ng gng

Hot ng gng l hot ng m LST EST = LFT EFT = 0, hay [EST, EFT] [LST, LFT] Slack = LST EST = 0 EST = LST ( tr cho php bng 0). EFT = LFT Slack = LFT EFT = 0

Gii thch: Slack ni lng ( tr).

54

Trong v d ang xt, cc hot ng gng l: C J K L (xem bng II.14) v to thnh ng gng (Critical Path). V vy, phng php mng PERT cn c tn l phng php ng gng (CPM Critical Path Method).Xc nh ng gng bng phn mm Lingo

xc nh ng gng bng phn mm Lingo, ta c th s dng cc bi ton mu bng cch nhn vo biu tng Lingo v thc hin cc lnh File > Open > Pert.lng vo bi ton PERT mu. Sau nhp cc s liu u vo ca bi ton cn gii vo thay cc s liu ca bi ton mu, chng hn nh s liu ca v d cho (xem hnh II.6).

Hnh II.6. Nhp s liu cho bi ton PERT

Sau chng ta thc hin LINGO > Solve, kt qu tnh ton s hin trn mn hnh (xem hnh II.7).

Hnh II.7. Kt qu tm cung gng ca bi ton PERT

55

2.2.

S PERT vi s liu ngu nhin

Thi gian thc hin tng hot ng ca d n ni chung l mt lng bin ng kh d on trc, chng ta gi thit chng l cc bin ngu nhin. Gi s ta c cc s liu c tnh v thi gian thc hin cc hot ng ca d n (xem bng II.15) a, m, b. Lc thi gian trung bnh v lch chun thi gian thc hin cc hot ng c c a + 4m + b . tnh theo cng thc t = 6Bng II.15. S liu c tnh v thi gian thc hin cc hot ng

Hot ng A B C D E F G H I J K L

Hot ng k trc

Thi gian c tnh a (sm nht) 1 1 1 1 2 0 3 2 1 4 1 4 m (nhiu kh nng xy ra nht) 2 2 2 2 3 0 6 5 4 9 2 4 b (mun nht) 3 3 3 9 10 0 15 14 7 20 9 4 t (thi gian trung bnh) 2 2 2 3 4 0 7 6 4 10 3 4 ( lch tiu chun, bin thin) 1/3 1/3 1/3 4/3 4/3 0 2 2 1 8/3 4/3 0

A A E B B D, F C H, J G, I, K

Bc tip theo l lp s mng cho d n vi cc thi gian trung bnh t v tm ng gng. ng gng l C J K L bao gm cc hot ng gng C, J, K v L. Cc hot ng ny c tr cho php bng 0, hay ni cch khc, khng cho php s chm tr no. y l cc hot ng cn ht sc ch trng, vic chm thc hin bt c mt hot ng no trong s ny u ko theo s chm tr trong tin ca c d n. T Critical Path (ting Anh) c dch sang ting Vit l ng gng v l do . Thi gian thc hin d n l mt lng ngu nhin tnh theo cng thc: T = TC + TJ + TK + TL. Ta tm k vng ca T (thi gian trung bnh thc hin d n) theo cng thc: m = mT = tC + tJ + tK + tL = 2 + 10 + 3 + 4 = 19 (tun). Tnh lch chun ca thi gian thc hin d n:2 2 2 2 = T = C + J + K + L =

(1/ 3) 2 + (8 / 3) 2 + (4 / 3) 2 + 0 = 3.

Ta coi T (thi gian thc hin d n) l bin ngu nhin tun theo lut chun N(m = 19; = 3).

56

th hm mt xc sut ca T cho trn hnh II.8.

75%

19

21

t

Hnh II.8. ng cong mt chun

tnh P, xc sut thc hin d n trong vng (khng vt qu) 19 tun, ta phi quy T v bin ngu nhin vi phn phi chun tc N(0, 1) nh cho trong ph lc 1. Lc :T m 19 19 P(T 19) = P = P(Z 0) = 0,5 (hay 50%), 3 y Z = (T m)/ l bin ngu nhin tun theo phn phi N(0, 1). Tng t, xc sut thc hin d n trong vng (khng vt qu) 21 tun c tnh nh sau: T m 21 19 P(T 21) = P = P (Z 0,666) = 75%. 3 Ta chuyn sang xem xt vn v tin cy ca thi gian hon thnh d n. Chng hn chng ta mun tr li cu hi sau: Mun thi gian thc hin d n c tin cy 90% th thi gian ti thiu (tnh theo s tun) l bao nhiu? t P (T t) = 90%. Tra bng phn phi chun tc N(0, 1), tm c z = 1,28. V z = (t 19)/3 = 1,28 nn t = 19 + 3. 1,28 23 (tun). Nh vy, d n ang xem xt c kh nng hon thnh vi tin cy ti 90% trong vng (khng vt qu) 23 tun.2.3. iu chnh d n khi k hoch mt s hot ng b ph v

V d: i khi trong qu trnh thc hin d n, k hoch ca mt s hot ng b ph v. Chnh v vy, khi pht hin d n ang b chm so vi k hoch ra ta cn nh li thi gian thc hin (thi gian rt gn) mt s hot ng trong giai on ti. Xt cc d kin cho trong hnh II.9 v bng II.16.

57

2 A

C

4 E

1 B 3 D

5

Hnh II.9. S mng PERT d n cn iu chnh Bng II.16. S liu iu chnh khi k hoach b ph v

Hot ng A B C D E

Thi gian nh mc 6 4 3 8 7

Thi gian rt gn 4 3 2 6 4

Kinh ph b sung/ 1n v thi gian rt gn (triu ng) 2 3 1 1,5 0,5

Sau khi c thi gian nh mc cho cc hot ng nh trong bng II.16, d dng tm c thi gian ti thiu cn thit hon thnh k hoch l 16 (tun). Tuy nhin do yu cu mi, cn rt gn thi gian hon thnh d n trong vng (khng vt qu) 10 (tun). Mun vy ta thc hin cc im sau: Tm thi gian ti thiu d nh thc hin d n (16 tun) v tm ng gng. c tnh thi gian rt gn ti a (ct 3, bng II.16) Khi rt gn thi gian trn ng gng cng phi ch trng ng thi cc cung ng khc. Trn hnh II.9, ta thy cn thc hin A, C v E vi thi gian rt gn ti a (4, 2, 4 tng cc thi gian thc hin cc hot ng gng l 10 tun), ng thi rt gn cc hot ng B v D mc cho php: Phng n 1: rt bt thi gian thc hin hot ng B mt tun v rt bt D mt tun. Phng n 2: khng rt bt B v rt bt D hai tun. Vy khi cn iu chnh thi gian thc hin d n ta cn thay i k hoch ca mt s hot ng theo cc bc nu trn. Tuy c nhiu phng n iu chnh d n, nhng trong vic ph v k hoch cc hot ng ca d n p ng tin mi cn ch v kha cnh chi ph gia tng c mt phng n ti u m bo rt gn c thi gian thc hin vi chi ph nh nht. i vi v d trn ta chn phng n 2.

58

C th p dng phng php tng qut iu chnh d n theo cc mc tiu trn (phng php n hnh cho BTQHTT n v a mc tiu) nh s c trnh by sau y.2.4. Tnh thi gian rt gn ti u bng phng php n hnh

tnh thi gian rt gn bng phng php n hnh (c th s dng cc phn mm my tnh thch hp), ta phi a ra c m hnh ton hc, hay cch khc, cn pht biu c BTQHTT (n hay a mc tiu). Trc ht, cn xc nh cc bin quyt nh. Gi x1, x2, x3, x4, x5 l cc thi im m cc hot ng xy ra (ti cc nt); yA, yB, yC, yD, yE l thi gian cn rt bt cho cc hot ng yu cu mi v y nhanh tin c tho mn. Ta c BTQHTT a mc tiu sau (cn cc tiu ha c thi gian thc hin d n ln tng chi ph gia tng): Mc tiu 1: z1 = x5 Min Mc tiu 2: z2 = 2yA + 3yB + yC + 1,5yD + 0,5yE Min vi cc rng buc: x 2 6 y A + x1 x 4 3 yC + x 2 x 3 4 y B + x1 x5 7 yE + x 4 x 5 8 yD + x 3 x 0,i = 1, 2, 3, 4, 5 i y j 0, j = A, B, C, D, E y A 2, y B 1, yC 1, y D 2, y E 3 x x + 10 1 5 C 2 cch gii m hnh: Chuyn mc tiu 1 thnh rng buc (*). Nu lc BTQHTT khng c phng n kh thi th phi ni lng dn (*): chng hn thay (*) bi x5 x1 + 11. nguyn c hai mc tiu gii theo phng php BTQHTT a mc tiu.2.5. p dng mng PERT trong phn tch chi ph v qun l ti chnh d n

(*)

Trong giai on u ng dng PERT v CPM, cc phng php ny thng c p dng cho bi ton tm thi gian ti thiu thc hin d n, tm cc hot ng gng. Chng t khi c p dng phn tch chi ph, mc d trong cc d n th vic phn tch chi ph (bao gm chi ph trc tip, gin tip v chi ph tin ch) cng rt quan trng. Tuy nhin ngy nay, PERT v CPM c p dng rt rng ri cho cc bi ton dng ny.V d: Chng ta xem xt d n vi cc d kin cho trong bng II.17 v hnh II.10.

59

D 2 A B 1 3 C F

4 E H 8 G 6 I

5

Hnh II.10. Mng PERT cho bi ton phn tch chi ph Bng II.17. D kin cho bi ton PERT chi ph

Hot ng A B C D E F G H I

EST 0 0 3 3 7 4 4 12 5

LST 0 8 9 3 7 10 10 12 11

Thi gian thc hin (thng) 3 2 1 4 5 2 1 3 4

Tng chi ph (triu ng) 30 200 40 20 75 100 75 18 240

Chi ph / mt thng (triu ng) 10 100 40 5 15 50 75 6 60

D thy, thi gian ti thiu hon thnh d n l 15 (thng). Nguyn tc iu hnh ti chnh mt d n l: Lung kinh ph phi c a vo dn dn sao cho p ng c tin d n. Nu kinh ph a vo tha hoc thiu (theo tin ) th phi kp thi iu chnh. Cn nm bt c: nhng hot ng no khng dng ht kinh ph d kin, nhng hot ng no s dng kinh ph nhiu hn d kin c s iu chnh thch hp. Cc bo co nh k cho php kim sot c d n v tin v lung kinh ph. Mun vy, trc ht cn lp bng theo di kinh ph cho d n t thng 1 n thng 15 (xem bng II.18). Phn trn ca tng ng vi cc hot ng gii ngn sm nht, phn di ng vi gii ngn mun nht. Hai hng cui bng dnh cho kinh ph trong tng thng v tng kinh ph cng dn cho ti thng tng ng vi hot ng gii ngn sm nht v gii ngn mun nht.

60

Bng II.18. D kin cho bi ton PERT chi phT. A B C D E F G H I + 110 10 110 10 110 10 220 20 10 10 230 30 45 5 275 35 130 5 405 40 60 115 5 520 45 60 65 5 585 50 60 75 15 660 65 60 60 75 115 735 180 15 155 750 335 15 140 765 475 15 125 780 600 60 6 66 786 666 60 6 66 792 732 60 6 66 798 798 50 75 75 6 6 6 6 6 6 50 50 50 1 10 10 100 2 10 10 100 100 40 40 5 5 5 5 5 5 5 5 15 15 15 15 15 15 15 15 15 15 100 3 10 10 4 5 6 7 8 9 10 11 12 13 14 15

ng gii ngn sm nht min kinh ph kh thi ng gii ngn mun nht

Hnh II.11. th min kinh ph kh thi

Da vo bng II.18, c th v c th min kinh ph kh thi nh trn hnh II.11.

61

Nu tin gii ngn nm ngoi min kinh ph kh thi th cn gp rt a ra cc bin php iu chnh tin gii ngn. Ngoi ra, cng c th iu chnh kinh ph cc hot ng ca d n da vo bng II.18.Lu

Cc vn c bn cn gii quyt khi p dng phng php PERT hay CPM trong theo di v nh gi d n l: Xc nh c s mng PERT ca d n. Tm c ng gng v cc hot ng gng. Tnh c tin cy ng vi cc mc thi hn hon thnh d n khi s liu l ngu nhin. Bit cch iu chnh thi gian rt gn khi tin thc hin d n l chm so vi k hoch. Phn tch chi ph v iu hnh kinh ph d n.3.3.1.

Mt s m hnh mng khcBi ton cy khung ti thiu

Bi ton cy khung ti thiu c nghin cu v ng dng trong nhiu lnh vc (Cng ngh thng tin, in lc, Quy hoch thu li,...). Vn t ra l cn xc nh mt mng ng i ti mi nt ca mng xut pht t mt nt no trong mng, sao cho tng di cc cung ng ny l ngn nht. Phng php tt nht gii bi ton cy khung ti thiu thuc v R. Prim s c trnh by trong mc ny.V d: Mc in cho 7 x trong huyn t ngun pht in n c 7 x vi chi ph ng dy l b nht. S khong cch t ngun cung cp in ti trm in cc x nh trn hnh II.12.

Bi ton t ra l phi pht trin c cy khung hay ng i ti thiu sao cho tng chiu di cc cung ng l b nht. gii ta lp bng II.19 (chiu di cc cung ng c quy gn), trong M l k kiu mt s +, biu th cung ng khng th xy ra trn thc t. Mi hng hay mi ct ca bng u biu th cc nt, chng hn nm trn giao ca hng 2 v ct 7 (cng ging nh nm trn giao ca hng 7 v ct 2) u cha s 9, l khong cch gia hai nt 2 v 7. Mt hng v mt ct c ni l lin thng vi nhau nu nm trn giao ca hng v ct ny cha gi tr khc M.

62

700 4 300 1000 200 3 100 Ngun in (1) 600 500

6

800 5

400 1100 900 2Hnh II.12. S khong cch t ngun in ti cc x Bng II.19. Bng khong cch cc cung ng

7

(Nt hng)

Nt (ct) 1 0 11 1 3 6 10 4 2 11 0 M M M M 9 3 1 M 0 M 5 M M 4 3 M M 0 M 7 M 5 6 M 5 M 0 2 M 6 10 M M 7 2 0 8 7 4 9 M M M 8 0

1 2 3 4 5 6 7

Thut gii Prim

Bc khi to: Lp bng khong cch gia cc nt mng. Trong bng trn, chn ct bt k (v d ct 1, tc l ta chn nt 1 bt u), gch b ct va chn ra khi bng. Cc bc lp:Bc 1: nh du vo hng tng ng (hng cng ch s) vi ct va chn. Trn cc hng c nh du tm c gi tr nh nht. Bc 2: Chn ct tng ng vi va tm c (ct 3 biu din nt chn mi, ghi cung ng va tm c 1 3), ri gch b n i (gch b ct 3). Nu trong bng vn cn cc ct cha gch b ht th quay v bc 1, nu tri li chuyn sang bc kt thc.

63

Bc kt thc: Nu tt c cc ct b gch b ht th dng vi tt c cc cung ng lin thng tm c to nn cy khung ti thiu.Ch : Nhng cu in nghing minh ho cho bc khi to v bc lp u tin. Sau 6 bc lp, qu trnh gii kt thc vi cc cung ng sau: 1 3, 1 4, 1 7, 3 5, 5 6 v 7 2. Tng di cc cung ng ca cy khung ti thiu l = 1 + 3 + 4 + 5 + 2 + 9 = 24. Ngoi ra, c th chn nt khi to l bt c nt no.

Thut ton Prim cn c ng dng trong cc bi ton xc nh chi ph ti thiu nhiu dng khc. Vic chng minh thut gii trn xin dnh li cho ngi c quan tm nghin cu cc vn v thut ton.3.2. Bi ton tm ng i ngn nht v quy hoch ng

Bi ton tm ng i ngn nht

Trong bi ton tm ng i ngn nht, chng ta mun xc nh hnh trnh ngn nht t mt a im xut pht (im gc) i ti im cn n (im ch) trn mt mng lin thng. cho d hiu, chng ta xem xt v d sau y.V d: Bi ton ngi i du lch.

C mt ngi i du lch, xut pht t nt 1 v kt thc hnh trnh nt 10 theo hnh trnh trn hnh II.13.2 100 1 150 4 175 200 3 350 7Hnh II.12. S hnh trnh ng i

300

6 275

200 400

9 100

200 175

5 250 275 125 8 150

10

Ngi du lch xut pht t nt 1. Trong giai on u anh ta ch c quyn (v bt buc) chn mt trong ba nt (thnh ph) 2, 3, 4 vo thm quan. Giai on tip theo, anh ta ch c chn mt trong ba nt 5, 6, 7 du lch. Trong giai on tip ni, anh ta c quyn vo mt trong hai nt 8 hoc 9 trc khi kt thc hnh trnh ti nt 10. Nh vy, trong mi giai on ngi i du lch ch c quyn i vo mt thnh ph (mi thnh ph c coi l mt trng thi ca giai on ). Hy tm cch xc nh ng i ngn nht t nt 1 ti nt 10 tho mn cc iu kin t ra ca bi ton.

64

Nguyn tc ti u Bellman trong quy hoch ng

S dng nguyn tc ti u Bellman trong quy hoch ng gii bi ton ngi du lch, chng ta chia bi ton thnh nhiu giai on, tc l thnh nhiu bi ton nh. Ti mi giai on ta cn tm phng n ti u l cc phng n tt nht ca tnh trng hin c, xt trong mi quan h vi cc phng n ti u tm c ca cc giai on trc. Ta c th gii quyt bi ton dn theo tng giai on theo cch tnh ton tin hoc tnh ton li. gii bi ton ny, ta p dng cch tnh ton li (backward computing) vi cc k kiu v d kin cho trong bng II.20.Bng II.20. Cc d kin ca bi ton quy hoch ng

Giai on Giai on I

u vo 8 9 5

u ra 10 10 8 9

ng i ti u 8 10 9 10 58 69 78

Khong cch ti ch 150 100 400 300 275 600 600 500 700 775 650

Giai on II

6 7 2

5 6 7 2 3 4

26 35 46 12 13 14

Giai on III

3 4 1

Giai on IV

Gii thch: S dng nguyn tc ti u Bellman, tm ng i ngn nht t nt 4 ti nt 10 chng ta tm c phng n ti u l i t nt 4 ti nt 6 cho giai on III (lc ny d(4, 10) = d(4, 6) + Min d(6, 10) = 200 + 300 = 500). iu ny l do hai la chn khc l i t nt 4 ti nt 5 hay 7 th u cho khong cch t nt 4 ti ch l nt 10 ln hn (chng hn nu i qua nt 5 th d(4, 10) = d(4, 5) + Min d(5, 10) = 175 + 400 = 575).

Trong bng II.20, ti giai on IV, ta thy khong cch ngn nht ti ch l 650. i ngc li, t im gc ti im ch ta xc nh c ng i ngn nht l: 1 4 6 9 10 vi tng chiu di l 650.Quy trnh tnh ton tng qut

Trc ht, cn chn c cc bin trng thi (state variables) nh m t trong bng II.21.

65

Bng II.21. Cc bin trng thi ca bi ton quy hoch ng

Bin x4 x3 x2 x1 x0

S trng thi 1 3 3 2 1

Cc trng thi (nt) 1 2, 3, 4 5, 6, 7 8, 9 10

Gi tr c th xy ra ca cc bin trng thi x4 1 x3 = 2 ; x3 = 3; x3 = 4 x2 = 5 ; x2 = 6; x2 = 7 x1 = 8 ; x1 = 9 x0 = 10

Bin trng thi m t trng thi ca h thng trong tng giai on. Xc nh hm mc tiu: t Fi(xi) l khong cch ngn nht ti ch tnh ti giai on i. Theo bng II.20, ta thy: 150 F1(x1) = 100 400 F2(x2) = 300 275 vi x1 = 8 vi x1 = 9 vi x2 = 5 vi x2 = 6 vi x2 = 7

Mc ch ca bi ton l cn tm c gi tr F4(x4) = F4(1). Lp hm truy ton: Fi+1(xi+1) = Min [Fi(xi) + fi(ui)], Min tm theo mi t hp thch hp xi v ui, trong ui l bin iu khin iu khin chuyn trng thi t trng thi xi sang xi+1 v fi(ui) l hiu ng ca bin iu khin tc ng ln hm truy ton (v ln hm mc tiu, nu tnh n bi ton cui cng). Theo biu thc ca hm truy ton ta thy, nu Fi(xi) + fi (ui) l hm phi tuyn th phi dng k thut ti u thch hp tm ra Fi+1(xi+1). Sau y chng ta i tm cc hm truy ton Fi+1(xi+1) vi quy trnh tnh ton li gii bi ton theo tng giai on, nhm cui cng tm ra c F4(x4) = F4(1).Giai on 1: Trong giai on ny, mun chuyn t nt 10 (x0 = 10) v nt 8 (x1 = 8) chng hn, th bin iu khin u0 phi c gi tr 150 (u0 = 150). Hiu ng gy nn bi u0 l f(u0) = 150. iu ny c ngha l nu chuyn t nt 10 ngc v nt 8 th cn i qung ng c chiu di l 150.F0(x0) = 0 x1 = 8 x1 = 9 x0 = 10 + u0 = 150 + u0 = 100 u0 150 100 f0(u0) 150 100 F1(x1) 150 100

Ch : Khng phi bi ton no ui cng trng vi hiu ng fi(ui) ca n. Ni

66

chung, bin iu khin ui c th gy ra hiu ng fi(ui) khc vi ui c v ln cng nh n v o.Giai on 2:x2 5 6 7 x1 = 8 +u1 = 250 x1 = 9 +u1 = 400 +u1 = 200 F1(x1) + f1(u1) x1 = 8 400 x1 = 9 500 300 F2(x2) = Min[F1(x1) + f1(u1)] 400 = 150 + 250 300 = 100 + 200 275 = 150 + 125

+u1 = 125

275

Giai on 3:x3 2 3 4 x2 5 u2 = 275 u2 = 200 u2 = 175 6 u2 = 300 7 x2 = 5 675 600 575 F2(x2) + f2(u2) x2 = 6 600 x2 = 7 F3(x3) = Min [F2(x2) + f2(u2)] 600 600 500

u2 = 200

u2 = 350 u2 = 275

500

625 550

Giai on 4:x4 1 x3 = 2 u3 = 100 x3 = 3 u3 =175 x3 = 4 u3 =150 F3(x3) + f3(u3) x3 = 2 700 x3 = 3 775 x3 = 4 650 F4 (x4) = Min [F3(x3) + f3(u3)] 650

p s: F4(x4) = F4(1) = 650 vi ng i ngn nht trn hnh II.14.x4 = 1 u3 = 150 x3 = 4 b2= x2 = 6 x1 = 9 x0 = 10

u1 = 200

u0 = 100

Hnh II.14. ng i ngn nht 1 4 6 9 10 3.3. p dng quy hoch ng cho mt s bi ton ngnh in

Bi ton 1

Cn phn phi cng sut ti u ca n nh my in vi ph ti tn tht c nh. Bit chi ph ca cc nh my l hm fi(pi) ph thuc vo cng sut pi, vi i = 1, 2, , n. Cn xc nh cc gi tr ca pi sao cho tng chi ph l cc tiu. Vy ta c bi ton ti u sau: Hm mc tiu: z = f1(p1) +....+ fn(pn) Min

67

vi cc rng buc:p1 + p 2 + ... + p n = P 0 pi Pi,max

trong P l tng ph ti, Pi, max l cng sut ti a cho php. Chng hn, vi n = 3 ta c BTQHTT (nguyn) sau y: z = 3p1 + 2p2 + p3 Minp1 + p 2 + p3 = 15 0 pi 6; 0 p 2 6; 0 p3 8

nu bit:f1 (p1 ) = 3p1 f 2 (p 2 ) = 2p 2 f (p ) = p 3 3 3

Chng ta xt phng php gii bi ton ny vi gi thit cc cng sut pi l nguyn. t cc bin trng thi l x1, x2, x3 ; cc bin iu khin l p1, p2, p3 vi quan h nh sau: x1 = p1, x2 = p1 + p2, x3 = p1 + p2 + p3 = 15. Cc hiu ng gy nn bi cc bin iu khin l fi(pi) vi i = 1, 2, 3.x0 = 0 Bin iu khin p1 x1 p2 x2 p3 x3

Thit lp hm truy ton Fi+1 (xi+1) = Min [Fi(xi) + fi+1 (pi+1)]. t F0(x0) = 0, d thy: F1(x1) = Minf1(p1), F2(x2) = Min[f1(p1) + f2(p2)] v F3(x3) = Min[f1(p1) + f2(p2) + f3(p3)] = 3p1 + 2p2 + p3. Mc tiu cui cng l cc tiu ho z = F3(x3). S dng nguyn tc ti u Bellman ta chia bi ton ra cc giai on sau y (vi quy trnh tnh ton tin). Giai on 1: ch xt cng sut p1; Giai on 2: ch xt cng sut p1 v p2; Giai on 3: xt cc cng sut p1, p2 v p3.Giai on 1: (Coi F0(x0) = 0)

68

x1 0 1 2 3 4 5 6

x0 = 0 p1 = 0 p1 = 1 p1 = 2 p1 = 3 p1 = 4 p1 = 5 p1 = 6

f1(p1) = 3p1 0 3 6 9 12 15 18

F1(x1) = Min [F0(x0) + f1(p1)] 0 3 6 9 12 15 18

Giai on 2:x1 x2 0 1 2 3 4 5 6 p2 0 1 2 3 4 5 6 7 8 9 10 11 12 0 1 0 2 1 0 3 2 1 0 0 1 F1(x1) + f2(p2) 2 3 4 5 6 F2 (x2) = Min[F1(x1) + f2(p2)] 0 2 4 6 8 10 12 15 18 21 24 27 30

0 1 0

0 2 4 6 8 10 12

3 5 7 9 11 13 15

6 8 10 12 14 16 18

9 11 13 15 17 19 21

12 14 16 18 20 22 24

15 17 19 21 23 25 27

18 20 22 24 26 28 30

4 3 2 1 0 5 4 3 2 1 6 5 4 3 2 6 5 4 3

2 1 6 5 4 3 2 6 5 4 3

6 5 4 6 5 6

Giai on 3:x2 x3 15 0 6 7 8 9 10 p3 11 4 12 7 5 3 23 F2(x2) + f3(p3) 8 25 9 27 10 29 11 31 12 33 F3(x3) = Min [F2(x2) + f3(p3)] 23

8 7 6

p s: Tng chi ph t gi tr cc tiu l 23, vi p1 = 1, p2 = 6, p3 = 8.

69

x0 = 0 Bin iu khin

x1 = 1 p1 = 1 p2 = 6

x2 = 7

x3 = 15 p3 = 8

Lu

Cc vn c bn cn gii quyt khi p dng phng php quy hoch ng theo nguyn tc Bellman l: Chia bi ton thnh nhiu giai on nh gii bi ton ti u cho tng giai on. Cc yu t ca bi ton quy hoch ng l bin trng thi, bin iu khin, hm truy ton v hm mc tiu. Khi chuyn t mt trng thi no (trong mt giai on) sang trng thi khc (giai on khc) cn c bin iu khin. Mi gi tr ca bin iu khin gy ra mt hiu ng ln hm mc tiu. Tu theo cc bi ton ti u pht sinh trong cc giai on m la chn phng php ti u thch hp. Trong v d ang xt, khi cc hiu ng fi(pi) cho di dng hm tuyn tnh vi cc bin pi nhn cc gi tr ri rc/nguyn th hm truy ton Fi+1 (xi+1) = Min [Fi(xi) + fi+1 (pi+1)] s tnh c bng thut gii da trn bng lit k (nh phng php gii trnh by). Nu fi(pi) phi tuyn vi cc bin pi nhn cc gi tr lin tc th tm Fi+1(xi+1) = Min[Fi(xi) + fi+1(pi+1)] ta c hai cch: Cch 1: ri rc ho theo tng mc. Chng hn vi p1 [0, 6], th coi p1 {0, 1, 2, 3, 4, 5, 6}. Cch 2: p dng phng php ti u thch hp vi bin lin tc (xem chng I) cho hm mc tiu. Chng hn, trong v d trn khi cn tm F2(x2) = Min [F1(x1)+ f2(p2)] = Min[f1(p1) + f2(p2)] = Min [3p1 + 2p2] vi iu kin rng buc: p1 + p2 15 v 0 p1 6, 0 p2 6, c th p dng phng php n hnh.Bi ton 2

Xc nh tuyn ng i ca ng dy truyn ti in t im A n im B, vi cc chng ngi vt khc nhau, sao cho tng chi ph l nh nht. Cc d kin ca bi ton cho trn hnh II.15. Nh vy thit lp s ng truyn ti in th xut pht t A ta c th nh tuyn i ca ng truyn ti in trc ht phi qua mt trong hai im st gn, theo hng bc hay hng ng, vi cc chi ph l 15 v 12. T mt trong hai im ny, chng ta li tip tc xc nh tuyn i cho ng truyn ti in, vi cc chi ph bit... Vy ta c bi ton tm ng i vi chi ph nh nht.

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8 12 10 7 8

9 9 7 2 13 12 12 15 11

13 6 11 4 15 10 16

10 8 9 2

B

10 6 15 A

8 10 11

Hnh II.15. S tuyn i cho dy truyn ti in

Bi ton ny hon ton tng t vi bi ton ngi du lch xt v c th gii bng phng php quy hoch ng (Hng dn: Chia bi ton thnh nhiu giai on nh theo cc ng vi nt t ni trn hnh II.15).

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72

Chng III

GII THIU L THUYT M PHNG V M HNH HNG CH1.1.1.

Mc ch v cc cng c ca m phngKhi nim v m phng ngu nhin

M phng (Simulation) c ng dng rng ri trong kinh t, k thut v nhiu lnh vc khc. Theo T in chnh xc Oxford, bn 1976, "m phng c ngha l gi cch, , lm ra v nh, hnh ng nh, bt chc ging vi, mang hnh thc ca, gi b nh..., lm gi cc iu kin ca tnh hung no thng qua mt m hnh vi mc ch hun luyn hoc tin li". V mt ngha k thut, m phng (hay ni ng hn, phng php m phng) hm cha vic p dng mt m hnh no to ra kt qu, ch khng c ngha l th nghim mt h thng thc t no ang