bài toán vận tải trong toán kinh tế

8/3/2019 Bi ton vn ti trong ton kinh t

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Ths. Nguyen Cong TrThs. Nguyen Cong TrThs. Nguyen Cong Tr

Copyright 2001Copyright 2001Copyright 2001

Ths. Nguyen Cong TrThs. Nguyen Cong TrThs. Nguyen Cong TrCopyright 2001Copyright 2001Copyright 2001

BABAI TOAI TOAN VAN VAN TAN TAII

1.1. BABAI TOAI TOAN VAN VAN TAN TAI DAI DANG TONG TONG QUANG QUATT (Xem)(Xem)

2.2. CACAC TC TNH CHANH CHAT VAT VA TIEU CHUATIEU CHUAN TON TOII UU CUCUAABABAI TOAI TOAN VAN VAN TAN TAII (Xem)(Xem)

3.3. CACAC PHC PHNG PHANG PHAP TM PHP TM PHNG ANG AN CN CCCBIENBIEN AAU TIEN CUU TIEN CUA BAA BAI TOAI TOAN VAN VAN TAN TAII (Xem)(Xem)

4.4. THUATHUAT GIAT GIAI THEI THE V CHO BAV CHO BAI TOAI TOAN VAN VAN TAN TAII (Xem)(Xem)

5.5. CACAC DAC DANG KHANG KHAC CUC CUA BAA BAI TOAI TOAN VAN VAN TAN TAII (Xem)(Xem)

6.6. BABAI TAI TAPP (Xem)(Xem)

CHNG 3

NONOI DUNG BAI DUNG BAI TOAI TOAN VAN VAN TAN TAIIGiaGia ss cacan van van chuyen chuyen mon mot loat loai hai hang hong hoa (xia (ximang, samang, sat thet thep, ...) tp, ...) t mm ieiem cung cam cung cap (trap (tramm

phaphat), kyt), ky hiehieu lau la AA11, A, A22, ..., A, ..., Amm een nn n ieiem tieum tieuthuthu (tra(tram thu), kym thu), ky hiehieu lau la BB11, B, B22, ..., B, ..., Bnn, bie, biet rat rangng(1)(1) SoSo llng hang hang cong co cacac trac tram pham phat At A11, A, A22, ...,, ...,AAmm lalan ln lt lat la aa11, a, a22,..., a,..., amm(2)(2) SoSo llng hang hang cang can n cacac trac tram thu Bm thu B11, B, B22, ...,, ...,BBnn lalan ln lt lat la bb11, b, b22,..., b,..., bnn..(3)(3) Chi phChi ph vavan chuyen chuyen mon mott n v han v hang hong hoa ta ttratram pham phat At A ii een tran tram thu Bm thu Bjj lala ccijij..Hay laHay lap kep ke hoahoach vach van tan tai hai hang hong hoa sao choa sao chototong chi phng chi phvavan tan tai thai thap nhap nhat vat va thothoa man yeua man yeucacau thuu thu phaphat.t.

BABAI TOAI TOAN VAN VAN TAN TAI DAI DANG TONG TONG QUANG QUATT

MO HNH BAMO HNH BAI TOAI TOAN VAN VAN TAN TAIIaat xt x ijij lala soso llng hang hang cang can van van chuyen chuyen tn t tratrammphaphat At A ii een tran tram thu Bm thu Bjj..Ta coTa co totong chi phng chi phvavan tan tai:i:

(1)(1) TraTram pham phat, phat, phat het het hat hang:ng:

(2)(2) TraTram thu, thum thu, thu uu hahang:ng:

(3)(3) Yeu caYeu cau trau tram pham phat, trat, tram thum thu c thoc thoaa

((k can bak can bang thung thu phaphat).t).


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VaVay, mo hy, mo hnh toanh toan cun cua baa bai toai toan van van tan tai (BTVT)i (BTVT)dadang tong tong quang quat nht nhsau:sau:TTm {xm {x ijij} sao cho:} sao cho:


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BABAI TOAI TOAN VAN VAN TAN TAI DI DI DAI DANG BANG BAI TOAI TOAN QHTTN QHTTkhai triekhai trien BTVT van BTVT va xexep hep he rarang buong buoc dc di dai dangnghehe m + n phm + n phng trng trnh cunh cua ma m n bien bien nhn nhsausau

KyKy hiehieu Au Am+n,mm+n,m nn ma trama tran hen he sosocucua hpt tren.a hpt tren.XXTT = (x= (x1111 xx1212 ...... xx1n1n xx2121 xx2222 ... x... x2n2n ... x... xm1m1 xxm2m2 ... x... xmnmn) la) lavect covect cot got gom mm m n than thanh phanh phan; C = (cn; C = (c1111 cc1212...... cc1n1n cc2121 cc2222 ... c... c2n2n ccm1m1 ccm2m2 ...... ccmnmn) la) la vect dovect dongnggogom mm m n than thanh phanh phan; bn; bTT = (a= (a11 aa22 ...... aamm bb11 bb22 ...... bbnn))lala vect covect cot got gom m+ n tham m+ n thanh phanh phan.n.


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BTVT vieBTVT viet dt di dai dang vect vang vect va ma trama tran nhn nhsausau

MoMot vect X thot vect X thoa (*) vaa (*) va (**) go(**) goi lai la phphng ang ann..MoMot P.At P.A aat ct cc tiec tieu thu th gogoi lai la P.A.T.P.A.T.cucua BTVT.a BTVT.MoMot pht phng ang an Xn X c goc goi lai la P.A.C.BP.A.C.Bkhi cakhi caccvect covect cot At Ajj cucua ma traa ma tran hen he soso AA ng vng vi cai caccthathanh phanh phan xn xijij > 0 la> 0 la ooc lac lap tuyep tuyen tn tnh.nh.

MoMot P.A.C.B cut P.A.C.B cua BTVT coa BTVT co nhienhieu nhau nhat lat la m + nm + n 1 tha1 thanh phanh phan dn dng. Neng. Neu mou mot P.A.C.B cut P.A.C.B cua BTVTa BTVTcoco uung m + nng m + n 1 tha1 thanh phanh phan dn dng thng th ccgogoi lai la khong suy biekhong suy bien.n. NgNgc lac lai,i, c goc goi lai laphphng ang an cn cc bien suy biec bien suy bienn..


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BB11 BB22 BBnnTTrraamm tthhuu BBjj

TTrraamm pphhaatt AAiibb11 bb22 bbnn

AA11 cc1111 cc1122 cc11nnaa11 xx1111 xx1122 xx11nn

AA22 cc2211 cc2222 cc22nnaa22 xx2211 xx2222 xx22nn

AAmm ccmm11 ccmm22 ccmmnnaamm xxmm11 xxmm22 xxmmnn

MO TAMO TA BABAI TOAI TOAN DN D I DAI DANG BANG BANG VANG VAN TAN TAII(1)(1) KyKy hiehieu (i, j) lau (i, j) la o treno tren dodong i vang i va cocot jt j..(2)(2) Chi phChi ph vavan chuyen chuyen cn cijij c ghi c ghi gogoc trenc trenben traben trai cui cua o (i, j), la o (i, j), lng hang hang cang can van van chuyen chuyenn

xxijij c ghi c ghi gogoc dc di ben phai ben phai cui cua o (i, j) biea o (i, j) bieuudien tuyedien tuyenn ng vang van chuyen chuyen tn t tratram pham phat At Aiieen tran tram thu Bm thu Bjj..(3)(3) Trong BATrong BANG VANG VAN TAN TAI, moI, mot ot o c goc goi lai la ootreotreoneneu nou no lala o duy nhao duy nhat tren dot tren dong hay tren cong hay tren cot.t.(4)(4) NhNh ng o ng o ng vng vi xi xijij > 0 trong BA> 0 trong BANG VANG VAN TAN TAIIc goc goi lai la o choo chonn, nh, nhng o khang o khac goc goi lai la o loao loaii..(5)(5) MoMot day cat day cac o choc o chon, trongn, trong oo 3 o lien tie3 o lien tieppkhong nakhong nam tren cum tren cung mong mot dot dong hay mong hay mot cot cot tht thc goc goi lai la momot day chuyet day chuyenn..

MO TAMO TA BABAI TOAI TOAN DN DI DAI DANG BANG BANG VANG VAN TAN TAII

(6)(6) MoMot day chuyet day chuyen khen khep kp knn c goc goi lai la momottchu trchu trnhnhhayhay momot vot vongng..

(7)(7) MoMot ma trat ma tran (xn (xijij) la) la momott P.AP.A cucua BTVT nea BTVT neu nou nothoathoa hehe rarang buong buoc. Moc. Mot P.A (xt P.A (xijij) la) lam cm cc tiec tieuuhaham mum muc tieu thc tieu th (x(xijij) la) la P.A.T.P.A.T.. cu. cua baa bai toai toan.n.

(8)(8) MoMot P.A cut P.A cua BTVT khong taa BTVT khong tao thao thanh chu trnh chu trnhnh(vo(vong) thng) th c goc goi lai la PhPhng ang an cn cc bien.c bien.

(9)(9)MoMot P.A.C.B cut P.A.C.B cua BTVT coa BTVT co uu m+nm+n--1 o cho1 o chon thn thc goc goi lai la P.A.C.B khong suy bieP.A.C.B khong suy bienn, ne, neu cou co tthn m+nhn m+n--1 o cho1 o chonn c goc goi lai la P.A.C.B suy bieP.A.C.B suy bienn..


VVDUDU 3.1.3.1.

HHnh 2.1.nh 2.1. HHnh 2.2.nh 2.2. HHnh 2.3.nh 2.3. HHnh 2.4.nh 2.4. HHnh 2.5.nh 2.5.HHnh 2.1. canh 2.1. cac o choc o chon, con, co dadauu , ta, tao thao thanhnh

day chuyeday chuyen, can, cac o (1,1) vac o (1,1) va (4,3) la(4,3) la cacac o treo.c o treo.HHnh 2.2. canh 2.2. cac o choc o chon tan tao thao thanh day chuyenh day chuyen,n,

cacac o (4,1) vac o (4,1) va (3,3) la(3,3) la cacac o treo.c o treo.

HHnh 2.3., Hnh 2.3., Hnh 2.4 vanh 2.4 va HHnh 2.5. canh 2.5. cac o choc o chon tan taoothathanh chu trnh chu trnh, khong conh, khong co o treo.o treo.


TTNH CHANH CHAT 1:T 1: BaBai toai toan van van tan tai luon luon coi luon luon cophphng ang an ton toii u.u.TTNH CHANH CHAT 2:T 2: VVi moi mot pht phng ang an ban bat kyt ky, so, so oochochon cun cua pha phng ang an khong vn khong vt quat qua totong song sotratram pham phat vat va tratram thum thu..

m + nm + n 1 (v1 (vii lala sosoo choo chon cun cua P.A)a P.A)TTNH CHANH CHAT 3:T 3: VVi moi mot pht phng ang an con co uu m+nm+n1 o1 ochochon thn th vvi moi mot o loat o loai bai bat kyt ky cc a vaa vaoophphng ang an se tan se tao thao thanh chu trnh chu trnh vanh va chu trchu trnhnhnanay lay la duy nhaduy nhat.t.TTNH CHANH CHAT 4:T 4: NeNeu lu lng cung ang cung aii vava llng cang cau bu bjjlala sosonguyen thnguyen th babai toai toan con co lli giai giai nguyen.i nguyen.

CACAC TC TNH CHANH CHAT CUT CUA BAA BAI TOAI TOAN VAN VAN TAN TAIIXeXet bat bai toai toan van van tan tai saui sau

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TIEU CHUATIEU CHUAN TON TOII U CUU CUA BAA BAI TOAI TOAN VAN VAN TAN TAII

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BaBai toai toann ooi ngau cui ngau cua BTVTa BTVTTTm {um {uii,v,vjj} sao cho:} sao cho:

VVi cai cac cac capp ooi ngau:i ngau:xx ijij 0 va0 va vv jj uu ii ccijij,, i,ji,jTheoTheo nh lynh ly oo lelech buch bu thth phphng ang an {xn {xijij} cu} cuaaBTVT coBTVT co P.A.T.P.A.T.lala toton tan tai hei he thothong {ung {u ii, v, vjj} sao cho:} sao cho:NeNeu xu xijij > 0 th> 0 th vvjj uuii = c= cijij,,NeNeu vu vjj uuii < c< cijij thth xxijij = 0.= 0.VaVay tieu chuay tieu chuan ton toii u cuu cua BTVT: va BTVT: v jj uuii ccijij,, i,ji,juuii:: c goc goi lai la thethev dov dongng..vvjj:: c goc goi lai la thethev cov cott..

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TIEU CHUATIEU CHUAN TON TOII U CUU CUA BAA BAI TOAI TOAN VAN VAN TAN TAIITren baTren bang vang van tan tai, choi, chon on o aau tien cou tien co ccc phc phvavan chuyen chuyen ben be nhanhat vat va chochon xn xijij nhnhsau:sau:

LaLap lap lai quai qua trtrnh tren cho o tienh tren cho o tiep theo chop theo cho eenneen khi yeu can khi yeu cau trau tram pham phat vat va tratram thum thu ccthoathoa man.man.

BaBang thung thu c vc vi cai cac xc xijij > 0 la> 0 la phphng ang an cn cccbien cubien cua baa bai toai toan.n.

PHPHNG PHANG PHAP CHI PHP CHI PHBEBE NHANHATT

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VV dudu 3.2.3.2. DuDung phng phng phang phap chi php chi ph bebenhanhat, tt, tm phm phng ang an cn cc bien cuc bien cua baa baiitoatoan van van tan tai coi co dadang bang bang saung sau ayay

KieKiem tram tra aaii == bbjj = 175= 175


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P.A.C.B tren khong suy bieP.A.C.B tren khong suy bien, vn, vi giai gia tr Z = 980.tr Z = 980.

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PhPhng phang phap Vogels (1958) cho P.A.C.B khap Vogels (1958) cho P.A.C.B kha tototttheo nghtheo ngha giaa gia tr hatr ham mum muc tieu cuc tieu cua noa no khakha gagannvvi P.A.T.i P.A.T.. Ph. Phng phang phapp c mo tac mo ta nhnhsausau(1) Tren ba(1) Tren bang vang van tan tai, ti, tnh hienh hieu sou sogigia chi pha chi phbebethth hai vhai vi chi phi chi phbebe thth nhanhat.t.

(2) Cho(2) Chon son so lln nhan nhat trong cat trong cac hiec hieu tren vau tren va phanphanphophoi toi toii a cho o coa cho o co chi phchi ph bebe nhanhat mot mot lt lngngxxijij = min(a= min(aii, b, bjj), sau), sau oo ttnh lanh lai hiei hieu sou sododong (cong (cot).t).(3) Qua(3) Qua trtrnh trennh tren c lac lap lap lai choi cho een khi chn khi chcocon lan lai moi mot dot dong hay mong hay mot cot cot duy nhat duy nhat.t.(4) Ba(4) Bang thung thu c vc vi cai cac {xc {x ijij} la} la phphng ang an cn cccbien cubien cua baa bai toai toan.n.

PHPHNG PHANG PHAP VOGELSP VOGELS

VVdudu 3.3:3.3: DuDung phng phng phang phap Vogels, tp Vogels, tmmphphng ang an cn cc bien cuc bien cua baa bai toai toan van van tan taiicoco dadang bang bang saung sau

KieKiem tram tra aaii == bbjj = 175= 175


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(1)(1) TTm P.A.C.B khong suy biem P.A.C.B khong suy bienn aau tien bau tien bangngphphng phang phap chi php chi phbebe nhanhat hoat hoac Vogels.c Vogels.(2)(2) DuDung tieu chuang tieu chuan ton toii u vu vii uujj ccijij,, i,ji,j ee kiekiemmtra P.A.C.B vtra P.A.C.B va ta tmm c.c.(3)(3) NeNeu P.A.C.B thoau P.A.C.B thoa man tieu chuaman tieu chuan ton toii u thu thP.A.C.BP.A.C.B oo lala P.A.T.P.A.T...(4)(4) NeNeu P.A.C.B vu P.A.C.B va ta tm chm cha thoaa thoa man tieuman tieuchuachuan ton toii u thu th ttm cam cach sch saa ooi P.A.C.B cui P.A.C.B cu eecoco P.A.C.B mP.A.C.B mi.i.(5)(5) trtr veve bbc (2), sau moc (2), sau mot sot sobbc lac lap hp hu hau han,n,ta se cota se co P.A.T.P.A.T...PhPhng phang phap tren gop tren goi lai la thuathuat toat toan then thevv

HHNG GIANG GIAI BAI BAI TOAI TOANN

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XAXACC NH P.A.C.BNH P.A.C.B AAU TIENU TIEN(ph(phng phang phap chi php chi phbebe nhanhat hoat hoac Vogels)c Vogels)

ChoChon o van o vao: Maxo: Max ijijXaXacc nh vonh vongng ieieu chu chnhnh

vava aanh danh dau (+); dau (+); dau (u ().).q = min{xq = min{xijij/ (i, j) da/ (i, j) dau (u ()})}

CoCo Them o xThem o xijij=0=0

KeKet thut thuc thuac thuat giat giaii

BaBai toai toan con co P.A.T.P.A.T.

LALAP BAP BANG VANG VAN TAN TAII

TTnh: Vnh: Vjj = U= Uii + C+ C ijijUUii = V= Vjj CCijij

XaXacc nh P.A mnh P.A mii

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SS OO THUATHUAT GIAT GIAI THEI THE VVCUCUA BAA BAI TOAI TOAN VAN VAN TAN TAII

SOSO BBC LAC LAPPLALA HHU HAU HANN

BBc 1. Lac 1. Lap bap bang vang van tan taii(1) Kie(1) Kiem tram tra ieieu kieu kien can ban can bang thung thu phaphat.t.(2) Xa(2) Xacc nh P.A.C.B (banh P.A.C.B (bang phng phng phang phap chi php chi phbebe nhanhat).t).(3) Kie(3) Kiem tra P.A.C.B com tra P.A.C.B co suy biesuy bien hay khongn hay khong

NeNeu P.A.C.B. suy bieu P.A.C.B. suy bien: them van: them vao o (i,j) bao o (i,j) bat kyt kyvvi xi x ijij = 0, khong ta= 0, khong tao thao thanh chu trnh chu trnh.nh.

NeNeu P.A.C.B khong suy bieu P.A.C.B khong suy bien, chuyen, chuyen sang [2]n sang [2]BBc 2. Kiec 2. Kiem tra tm tra tnh tonh toii u cuu cua baa bai toai toann

(1) T(1) Tnh vnh vjj = u= uii + c+ cijijuuii = v= vjj ccijij, trong, trong oo o (i,j) lao (i,j) la o choo chon.n.

THUATHUAT TOAT TOAN THEN THE VV

ChoChon un uii = 0 ta= 0 tai doi dong bang bat kyt ky..(2)(2) aatt ijij = v= vjj uuii ccijij

NeNeuu ijij 0:0: ta cota co P.A.T.P.A.T...NeNeuu ijij > 0: chuye> 0: chuyen sang [3]n sang [3]

BBc 3. Xac 3. Xacc nh vonh vongng ieieu chu chnhnh(1) Cho(1) Chon o van o vao: Maxo: Max ijij (( ijij > 0)> 0)(2) Cho(2) Chon o ran o ra

xaxacc nh vonh vongng ieieu chu chnhnho vao vao seo se cc aanh danh dau (+). Xen keu (+). Xen ke dadauu

((--) va) va dadau (+) tren vou (+) tren vongng ieieu chu chnh.nh.llngng ieieu chu chnh q = min{xnh q = min{xijij/ (i,j) co/ (i,j) co dadau (u (--)})}

THUATHUAT TOAT TOAN THEN THE VVBBc 4. Xac 4. Xacc nh P.A.C.B mnh P.A.C.B mii

Quay veQuay ve bbc [2].c [2].Sau moSau mot sot so bbc lac lap hp h u ha u han, ban, bai toai toan con cophphng ang an ton toii u.u.


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CHUCHU YY..(1) Trong thua(1) Trong thuat giat giai bai bai toai toan van van tan tai, nei, neu Maxu Max ijijaat tat tai nhiei nhieu o, ta chou o, ta chon mon mot o tut o tuy yy y trong sotrong socacaccoo oo lalam om o ieieu chu chnh.nh.(2) Trong P.A.T.(2) Trong P.A.T. ttmm c Xc Xoptopt, ne, neu cou co ijij = 0,= 0, mama(i,j) la(i,j) la o loao loai thi th oo lala dadau hieu hieu bau bai toai toan con co nhienhieuuP.A.T.P.A.T.khakhac.c. ee ttm P.A.C.B.T.m P.A.C.B.T.khakhac, ta choc, ta chon on o(i, j)(i, j) oo lalam om o ieieu chu chnh, ronh, roi ai ap dup dung thuang thuat toat toannthethevv ee xaxacc nh P.A.C.B.T.nh P.A.C.B.T.khakhac Xc X//optopt..(3) Ta(3) Tap php phng ang an ton toii u lau la

X = {X = { XXoptopt + (1+ (1 )X)X//optopt,, 0, 10, 1 }}


VVdudu 3.4.3.4. GiaGiai bai bai toai toan van van tan taii

KieKiem tram tra ieieu kieu kien can ban can bang thu phang thu phattaaii = 40 + 75 + 60 + 70 + 45 = 290= 40 + 75 + 60 + 70 + 45 = 290bbjj = 45 + 55 + 30 + 70 + 50 + 40 = 290= 45 + 55 + 30 + 70 + 50 + 40 = 290

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--22

1010

77

88

+2+2

+1+1

+1+1 +5+5

+1+1

++

--++

-- ++

--++

-- ++

--

q= 20q= 20

BaBangng

11

881010101066991111

4545

2266772288997070

1177665599336060

99131311111010131312127575

66771010998812124040

404050507070303055554545TTPP

1010 3030

707055

20204040

3030 2020 2020

4545

00

7788

11

33

--11

44

--77

55

22

33

+2+2 +1+1 +1+1++

-- ++

-- ++

--++

--

q= 5q= 5

BaBangng

22

881010101066991111

4545

2266772288997070

1177665599336060

99131311111010131312127575

66771010998812124040

404050507070303055554545TTPP

55 3535

55 7070

4545 1515

3030 1515 2525

4545

00

6622 22

11

44

--11

77

--66

55

--22

BaBangng

33

8810101010669911114545

2266772288997070

1177665599336060

99131311111010131312127575

66771010998812124040

404050507070303055554545TTPP Do caDo cacc ijij 00 i,j neni,j nen P.A.T.P.A.T.cucua baa bai toai toan lan la

VaVa ZZminmin = 1.875= 1.875 n v tien v tien ten te..NgoaNgoai ra, bai ra, bai toai toan khong con khong co P.A.T.P.A.T. khakhac vc vkhong cokhong co ijij = 0, v= 0, vi (i, j) lai (i, j) la o loao loaii

0 5 0 0 35 0

0 5 0 70 0 0

45 0 0 0 0 150 0 30 0 15 25

0 45 0 0 0 0

optx


JL H G \ E

- = > 3



6/17

VVdudu 3.5.3.5. GiaGiai bai bai toai toan van van tan taii

KieKiem tram tra ieieu kieu kien can ban can bang thu phang thu phattaaii = 79 + 102 + 70 + 60 = 311= 79 + 102 + 70 + 60 = 311bbjj = 76 + 62 + 88 + 45 + 40 = 311= 76 + 62 + 88 + 45 + 40 = 311

9910101818181812126060335510101717121270704477881111131310210277661515191910107979

40404545888862627676TT

PP


++

--++

--++

-- ++

--

q=30q=30

BaBang 1ng 1

40403030

1515

8888

6464

1414

1212 4848

00

1010 66

--22

1616

55

1313

11

44

+2+2991010181818181212

6060

33551010171712127070

44778811111313102102

77661515191910107979

40404545888862627676TTPP

BaBang 2ng 2

40403030

4545

5858

3434

4444

4242 1818

00

1010 66

--22

1616

55

1313

33

66

9910101818181812126060

33551010171712127070

44778811111313102102

77661515191910107979

40404545888862627676TTPP Do caDo cacc ijij 00 i,j neni,j nen

P.A.T.P.A.T.cucua baa bai toai toan van van tan taii

VaVa ZZminmin = 2.806= 2.806 n v tien v tien ten te..

BaBai toai toan khong con khong co P.A.T.P.A.T. nanao khao khac vc v khongkhongcoco ijij = 0, v= 0, vi (i, j) lai (i, j) la o loao loai.i.

34 0 0 45 0

0 44 58 0 0

0 0 30 0 40

42 18 0 0 0

optx

THUATHUAT GIAT GIAI THEI THE VV

CACAC DAC DANG KHANG KHAC CUC CUA BAA BAI TOAI TOAN VAN VAN TAN TAII

1.1. BABAI TOAI TOAN VAN VAN TAN TAI KHONG CAN BAI KHONG CAN BANGNGTHUTHU PHAPHATT (Xem)(Xem)

2.2. BABAI TOAI TOAN VAN VAN TAN TAI COI CO DADANG HANG HAM MUM MUCCTIEU LATIEU LA MAXMAX (Xem)(Xem)

3.3. BABAI TOAI TOAN VAN VAN TAN TAI COI CO O CAO CAMM (Xem)(Xem)

4.4. BABAI TOAI TOAN VAN VAN TAN TAI XE KHONGI XE KHONG (Xem)(Xem)

1.1. TRTRNG HNG HP 1.P 1. aaii >> bbjjThem traThem tram thu giam thu gia thth BBn+1n+1VVi nhu cai nhu cau thu bu thu bn+1n+1 == aaii bbjjCCc phc phvavan tan tai ci ci,n+1i,n+1 = 0, i = 1, 2, ..., m.= 0, i = 1, 2, ..., m.

2.2. TRTRNG HNG HP 2.P 2. aaii 3



7/17

VVdudu 3.6.3.6. GiaGiai bai bai toai toan van van tan tai saui sau

KieKiem tram tra ieieu kieu kien can ban can bang thung thu phaphattaaii= 165 3



8/17

TTm {xm {x ijij} sao cho:} sao cho:

MO HNH BAMO HNH BAI TOAI TOAN VAN VAN TAN TAIICOCO HAHAM MUM MUC TIEU LAC TIEU LA MAXMAX

1 1

1

1

1 1

max

, 1,

, 1,

0; 0; 0; 0; 1, ; 1, .

m n

ij i j

i j

n

ij i

j

m

ij j

i

m n

ij ij i j i j

i j

Z c x

x a i m

x b j n

x c a b a b i m j n

GioGiong nhng nh babai toai toan QHTT con QHTT co haham mum muc tieu lac tieu lamax, chumax, chung ta cong ta co thethe a baa bai toai toan van van tan tai coi co

haham mum muc tieu Zc tieu Z max vemax ve ZZ// == ZZ min, saumin, sau oodudung thuang thuat toat toan then the vv ee giagiaii. Tuy nhien, chu. Tuy nhien, chungngta cung cota cung co thethe giagiai tri trc tiec tiep bap bai toai toan nan nay bay bangngthuathuat toat toan then thev vv vi moi mot vat vai thayi thay ooi trong thuai trong thuattgiagiai nhi nhsau:sau:1.1. Khi xay dKhi xay dng P.A.C.Bng P.A.C.B aau tien, ta phan phou tien, ta phan phoi toi toiia vaa vao o coo o co ccc phc phlln nhan nhat.t.2.2.Tieu chuaTieu chuan ton toii u lau la vvjj uuii ccijij,, i,ji,j3.3.OO ieieu chu chnh lanh la o coo co {min{min ijij, v, vii ijij < 0}< 0}

THUATHUAT GIAT GIA I BAI BAI TOAI TOAN VAN VAN TAN TAIICOCO HAHAM MUM MUC TIEU LAC TIEU LA MAXMAX

VV dudu 3.7.3.7. MoMot cong ty cot cong ty co 3 x3 x nghienghiep cup cung sang sannxuaxuat mot mot loat loai boi bongng een. Nang suan. Nang suat trong that trong thangngcucua 3 xa 3 xnghienghiep lap lan ln lt lat la AA ii = (650, 1.000, 350)= (650, 1.000, 350)bobong. Hng. Hpp oong cong ty phang cong ty phai giao cho 4 nhai giao cho 4 nhaphan phophan phoi lai la BBjj = (200, 400, 600, 800) bo= (200, 400, 600, 800) bong.ng. nngiagia baban cun cua moi boa moi bongng een tn tngng ng vng vi cai caccnhanha phan phophan phoii c cho bc cho bi ma trai ma tran sau:n sau:vt: 1.000vt: 1.000 oongng

Hay tHay tm kem kehoahoach phan phoch phan phoi hai hang sao cho congng sao cho congtyty aat doanh sot doanh solln nhan nhatt

THUATHUAT GIAT GIAI THEI THE V VV VI HAI HAM MUM MUC TIEU ZC TIEU Z MaxMax

22 25 20 18

30 32 25 28

29 28 25 23

ijc 2323252528282929350350

282825253232303010001000

1818202025252222650650

800800600600400400200200TTPP


250250

200200 400400 400400

400400

350350

00

282832323030

55

3030

1010

--22 --33

--44 --11++

++

++

q = 200q = 200

2323252528282929350350

282825253232303010001000

1818202025252222650650

800800600600400400200200TTPP


450450

200200

400400 600600

200200

150150

00

282832323434

55

3030

1010

--33

--11

++

++

q = 200q = 200

2323252528282929350350

282825253232303010001000

1818202025252222650650

800800600600400400200200TTPP


450450

200200

200200 800800

200200

150150

00

282832323131

22

2727

77

Z = 52.350Z = 52.350

JL H G \ E

- = > 3



9/17

Do caDo cacc ijij 0,0, i, ji, jP.A.T.P.A.T.CUCUA BAA BAI TOAI TOANN

VaVa ZZMaxMax = 52.350= 52.350

0 200 450 0

0 200 0 800

200 0 150 0

optx

THUATHUAT GIAT GIAI BAI BAI TOAI TOAN VAN VAN TAN TAIICOCO HAHAM MUM MUC TIEU LAC TIEU LA MAXMAX BaBai toai toan van van tan tai coi co o cao camm lala babai toai toan van van tan taii

vvi P.A.T.i P.A.T.cucua noa no phaphai thoi thoaa ieieu kieu kien cho trn cho trc.c.

ee giagiai bai bai toai toan nan nay, ta lay, ta lap bap bai toai toan van van tan tai mi mrorong VTng VTMM babang cang cach cho giach cho gia ccc vac van chuyen chuyen n cacac o cac o cam bam bang M, vng M, vi M > 0 li M > 0 ln tun tuy yy y roroi dui dungngthuathuat toat toan then thev. Cov. Co 2 tr2 trng hng hp xap xay ray ra1.1.Trong P.A.T.Trong P.A.T.cucua baa bai toai toan VTn VTMM, ne, neu cau cac o cac o cammcoco xxijij = 0 th= 0 th P.A.T.P.A.T.cucua baa bai toai toan VTn VTMM cung chcung chnhnhlala P.A.T.P.A.T.cucua baa bai toai toan gon goc.c.2.2.Trong P.A.T.Trong P.A.T.cucua baa bai toai toan VTn VTMM, ne, neu cau cac o cac o cammcoco xxijij 0 th0 th babai toai toan gon goc khong coc khong co P.A.T.P.A.T...

BABAI TOAI TOAN VAN VAN TAN TAI COI CO O CAO CAMM

VVdudu 3.8.3.8. GiaGiai bai bai toai toan van van tan tai saui sau ay vay viiNhu caNhu cau trau tram pham phat a = (150, 100, 145, 100)t a = (150, 100, 145, 100)Nhu caNhu cau trau tram thu b = (140, 150, 180)m thu b = (140, 150, 180)Ma traMa tran cn cc vac van chuyen chuyenn

vvii ieieu kieu kienntratram Am A33, A, A44 phaphai phai phat het het hat hang.ng.

KieKiem tram tra ieieu kieu kien can ban can bang thung thu phaphattaaii = 150 + 100 + 145 + 100 = 495= 150 + 100 + 145 + 100 = 495

bbjj = 140 + 150 + 180 = 470= 140 + 150 + 180 = 470LaLap trap tram thu giam thu gia, v, vi bi b44= 25 va= 25 va M > 0 tuM > 0 tuy yy y..


5 4 6

8 5 9

11 6 12

9 7 13

ijc

MM13137799100100

MM1212661111145145

00995588100100

00664455150150

2525180180150150140140TTPP

++

++

00 150150

100100

145145

4040 3535 2525

+3+3 MM--44

+2+2 +3+3 MM--11

+1+1

+1+1

44

11

11

00

99 88 1313 MM q = 25q = 25

BaBang 1ng 1

MM13137799100100

MM1212661111145145

00995588

100100

00664455150150

2525180180150150140140TTPP

++

++

+3+3

+2+2 +3+3

+1+1

+1+1 q = 35q = 35

BaBang 2ng 2

00 150150

7575 2525

145145

6565 3535

44

11

11

00

99 88 1313 11

MM13137799100100

MM1212661111145145

00995588

100100

00664455150150

2525180180150150140140TTPP

++

++

++

+2+2

+4+4

+1+1 q = 40q = 40

BaBang 3ng 3

00 150150

4040 2525

145145

100100

3535

33

00

--33

--11

88 77 99 00

JL H G \ E

- = > 3



10/17

MM13137799100100

MM1212661111145145

00995588100100

00664455150150

2525180180150150140140TTPP

++

++

+4+4 +1+1

+1+1 +1+1 q =105q =105

BaBang 4ng 4

4040 110110

4040

2525

105105

100100

7575

00

11

--22

--44

55 44 1010 11

MM13137799100100

MM1212661111145145

00995588100100

00664455150150

2525180180150150140140TTPP

+2+2

+1+1

++

++

q = 5q = 5

BaBang 5ng 5

4040 55

145145

2525

100100

7575

105105

00

--33

--22

--44

55 44 66 --33

MM13137799100100

MM1212661111145145

00995588100100

00664455150150

2525180180150150140140TTPP

BaBang 6ng 6

4040

55

145145

2525

100100

7070

11011033

00

--11

--11

88 55 99 00 Z =3.285Z =3.285

00++

++

q = 40q = 40P.A.T.P.A.T.khakhacc

Do caDo cacc ijij 00 i,j neni,j nenP.A.C.B.T.P.A.C.B.T.cucua baa bai toai toan van van tan tai tren lai tren la

vava ZZminmin= 3.285= 3.285

40 0 110

0 5 70

0 145 0

100 0 0

optx


MM13137799100100

MM1212661111145145

00995588

100100

00664455150150

2525180180150150140140TTPP

BaBang 7ng 7

4040 55

145145

2525

100100

3030

15015033

00

--11

--11

88 55 99 00Z =3.285Z =3.285

00

P.A.C.B.T.P.A.C.B.T.khakhac cuc cua baa bai toai toan van van tan tai tren lai tren la

vava ZZminmin= 3.285= 3.285TaTap php phng ang an ton toii uuZZoptopt == XXoptopt + (1+ (1 ) X) X//optopt

HayHay

0 0 1 5 0

4 0 5 3 0

0 1 4 5 0

1 0 0 0 0

op tx


4 0 0 1 5 0 4 0

4 0 4 0 5 3 0 4 0

0 1 4 5 0

1 0 0 0 0

op tZ

JL H G \ E

- = > 3



11/17

ieieu kieu kien ran rang buong buoc cuc cua baa bai toai toan van van tan tai xei xekhong lakhong la momot sot sotratram pham phat At Aii phaphai phai phatt uu hahangngcho tracho tram Bm Bjj ((c chc chnh). Xanh). Xacc nh lonh lo trtrnh xenh xe

chachay khong tay khong tai ti t BBjj een An Aii lala t nhat nhat.t.KhiKhi oo tratram pham phat At A ii trtr thathanh tranh tram thu xem thu xekhong, trakhong, tram thu Bm thu Bjj trtr thathanh tranh tram pham phat xe khongt xe khongvava khikhi oo ma trama tran (cn (cijij) la) la ma trama tran khoan khoang cang cachchttngng ng ging gia Aa Aii vava BBjj..

QuiQui c sc s dudung cang cac kyc ky hiehieu nhu nhsau:sau:: l: lng hang hang hong hoa coa co vavan tan tai.i.: l: lng hang hang cung cua xe khong taa xe khong tai.i.: tuye: tuyen xe chan xe chay coy co tatai.i.: tuye: tuyen xe chan xe chay khong tay khong taii

MO HNH BAMO HNH BAI TOAI TOAN VAN VAN TAN TAI XE KHONGI XE KHONG

ijx

xij

THUATHUAT GIAT GIAI BAI BAI TOAI TOAN XE KHONG TAN XE KHONG TAII1.1. LaLap bap bang vang van tan tai ti tngng ng vng vi ma trai ma trannkhoakhoang cang cach. Duch. Dung thuang thuat toat toan then the v tv tm P.A.T.m P.A.T.cucua baa bai toai toan xe khong tan xe khong tai.i.

2.2. TaTao bao bang phong phoi hi hp P.A.T.p P.A.T. cucua baa bai toai toan xen xekhong takhong tai vi vi kei kehoahoach vach van tan taii a cho tra cho trc. Lac. Lapptuyetuyenn ieieuu oong tng tngng ng.ng.3.3. GiaGiam lm lng chenh leng chenh lech gich giaa o troo tronn vava oovuongvuong ee coco babang mng mi thu goi thu gon.n.4.4. LaLap vop vongng ieieuu oong gong gom cam cac o coc o co tatai vai va ookhong takhong tai lien tiei lien tiep nhau, lp nhau, lngng ieieuu oong q=ng q=min{xmin{xijij}, v}, vi xi xijij coco tatai vai va xxijij khong takhong tai. Tri. Tr veve [3].[3].Sau moSau mot sot sobbc lac lap hp hu hau han [3] van [3] va [4], ta se thu[4], ta se thuc kec kehoahoachch ieieuu oong hang hang hong hoa toa toii u.u.

VVdudu 3.9.3.9. MoMot cong ty vat cong ty van tan tai coi co kekehoahoach vach vannchuyechuyen han hang hong hoa theo ha theo hpp oong,ng, c thec the hiehiennqua baqua bang yeu cang yeu cau nhu nhsausau

THUATHUAT GIAT GIAI BAI BAI TOAI TOAN XE KHONG TAN XE KHONG TAII

BB22Cong ty rau quaCong ty rau qua2020BB44CCa haa hang song so445050SaSau riengu riengAA33

BB33CCa haa hang song so331010BB22Cong ty rau quaCong ty rau qua1515BB11CCa haa hang song so112525

DDa haa hauuAA22

BB33CCa haa hang song so333030BB22Cong ty rau quaCong ty rau qua2020CamCamAA11

KyKy hiehieuuNi nhaNi nhannhahang Bng Bjj

LLngng(ta(tan)n)

LoaLoaiihahangng

aa ieiemmcacap hap hang Ang A ii

Cho bieCho biet khoat khoang cang cach gich giaa aa ieiem cungm cungcacap hap hang vang va aa ieiem nham nhan han hang (km)ng (km) c thec thehiehien qua ma tran qua ma tran nhn nhsau:sau:

Hay xaHay xacc nh lonh lo trtrnh vanh van chuyen chuyen han hang hong hoaathothoa yeu caa yeu cau hu hpp oong vang va totong tang tann km xekm xe

chachay khong tay khong tai nhoi nho nhanhat.t.


5243

4625

3412

L

552244337070

446622555050

334411225050

5050404055552525BBjjAAii

BBc 1c 1 (t(tm P.A.T.m P.A.T.cucua baa bai toai toan xe khong tan xe khong tai)i)

ZZminmin= 420 ta= 420 tann kmkm

5050

55

4040

22

11

00

33 33 22 55

BaBang 1ng 1

4545

2525 55

BBc 2c 2 (ta(tao bao bang phong phoi hi hp)p)

AA11 BB22 AA11: 20 T X 1km = 20T: 20 T X 1km = 20T kmkmAA22 BB22 AA22: 5 T X 2km = 10T: 5 T X 2km = 10T kmkmAA33 BB44 AA33: 5 T X 5km = 25T: 5 T X 5km = 25T kmkm

552244337070

446622555050

334411225050

5050404055552525BBjjAAii

5050

55

4040

BaBang 2ng 2

4545

2525 55

20 30

25 15 10

5020

1km1km

2km2km

5km5km

JL H G \ E

- = > 3



12/17

BBc 3c 3 (la(lap tuyep tuyenn ieieuu oong)ng)

AA22 BB11 AA33 BB22 AA11 BB33AA33 BB44 AA22:20Tx10km=200T:20Tx10km=200T--kmkm

552244337070

446622555050

334411225050

5050404055552525BBjjAAii

3030

4040

BaBang 3ng 3

4545

2525

30

25 10 10

4520

q=20q=203km3km 1km1km

2km2km 4km4km


AA22 BB11 AA33 BB44AA22: 5T x 7km = 35T: 5T x 7km = 35Tkm.km.

552244337070

446622555050

334411225050

5050404055552525BBjjAAii

1010

2020

BaBang 4ng 4

2525

55

10

5 10 10

25

q=5q=53km3km

4km4km


AA22 BB22 AA11 BB33 AA33BB44 AA22: 10T x 7km = 70T: 10T x 7km = 70Tkmkm

552244337070

446622555050

334411225050

5050404055552525BBjjAAii

1010

2020

BaBang 5ng 5

2020

10

10 10

20

q=10q=101km1km 2km2km4km4km


AA22 BB33 AA33 BB44AA22 : 10 T x 6km = 60T: 10 T x 6km = 60Tkmkm

552244337070

446622555050

334411225050

5050404055552525BBjjAAii

1010

BaBang 6ng 6

101010

10

q=10q=10

2km2km4km4km

BABANGNG IEIEUU OONG XENG XEAA11 BB22 AA11: 20 T: 20 TAA22 BB22 AA22: 5 T: 5 TAA33 BB44 AA33: 5 T: 5 T

AA22 BB11 AA33 BB22 AA11 BB33AA33 BB44 AA22: 20 T: 20 TAA22 BB11 AA33 BB44 AA22: 5 T: 5 TAA22 BB22 AA11 BB33 AA33

BB44 AA22: 10 T: 10 TAA22 BB33 AA33 BB44 AA22: 10 T: 10 T


1km1km

2km2km

5km5km

3km3km 1km1km2km2km 4km4km

3km3km 4km4km

1km1km 2km2km

4km4km

2km2km 4km4km

Ths. Nguyen Cong TrThs. Nguyen Cong TrThs. Nguyen Cong Tr

Copyright 2001Copyright 2001Copyright 2001

LALAP MO HNH CUP MO HNH CUA BAA BAI TOAI TOAN VAN VAN TAN TAII

[[11]] [[22]]

TM PHTM PHNG ANG AN CN CC BIENC BIEN AAU TIENU TIEN

[[3a3a]] [[3b3b]] [[3c3c**]] [[3d3d]] [[3e3e]]

GIAGIAI BAI BAI TOAI TOAN VAN VAN TAN TAI CAN BAI CAN BANG THUNG THU -- PHAPHATT

[[44]] [[55]] [[66]] [[77]] [[88]] [[99]]

CACAC DAC DANG KHANG KHAC CUC CUA BAA BAI TOAI TOAN VAN VAN TAN TAII

[[10a10a]] [[10b10b]] [[10c10c]]

[[11a11a]] [[11b11b]] [[11c11c]] [[11d11d]]

BABAI TAI TAP CHP CHNG 3NG 3

JL H G \ E

- = > 3



13/17

BAI TAP CHNG 3

LAP MO HNH CUA BAI TOAN VAN TAI DI DANG BAI TOAN QHTT[1] Mot cong ty van tai bien can 110 ngi e botr vao cac nhiem vu: 10 may trng;

25 th may 1; 30 th may 2; 45 th may 3. Phong to chc nhan s cong ty tuyen

c 90 ngi, trong o gom 25 ky s may; 20 ky thuat vien trung cap va 45 congnhan co kinh nghiem. Phong to chc nhan s anh gia trnh o nhan s tng ngvi tng cong viec theo thang iem 5, v du aij = 5 ngha lavi trnh o i co kha nanghoan thanh xuat sac cong viec j (at iem toi a la 5), con aij = 0 la vi trnh o ikhong co kha nang hoan thanh cong viec j (at iem 0), c the hien chi tiet trongbang sau

iem anh gia nang lc (aij)Nhiem vu

Trnh o May trng May 1 May 2 May 3

Ky s 5 4 0 0

Trung cap 3 5 4 0Cong nhan 0 1 5 4

Hay lap kehoach bo tr nhan lc sao cho cong viec at toi u.

[2] Hai oi tuyen bong ban, moi oi co 5 ngi. Qua thong ke nhieu tran au trong quakh, ngi ta d oan xac suat thang cuoc moi au thu cua moi oi c the hienqua bang sau

oi II

oi Iau thu 1 au thu 2 au thu 3 au thu 4 au thu 5

au thu 1 0,4 0,5 0,6 0,7 0,8

au thu 2 0 0,3 0,4 0,4 0,7au thu 3 0,2 0,6 0,4 0,3 0,5

au thu 4 0,6 0,3 0,4 0,7 0,6

au thu 5 0 0,2 0,3 0,4 0,6

Gia s cac au thu cua oi I c quyen chon thi au vi cac tuyen thu oi II. Hay sapxep cac au thu cua oi I sao cho xac suat thang toan oan cua oi I cao nhat.

TM PHNG AN CC BIEN AU TIEN[3] Tm phng an cc bien au tien bang hai phng phap chi ph be nhat va phng

phap Vogels cua cac bai toan van tai sau ay

a) BjAi

20 25 30 15 b) BjAi

3 5 10 14 c)* Bj

Ai10 30 50

40 4 5 1 2 10 1 3 7 1 25 7 6 5

20 3 4 7 8 15 2 4 2 3 10 2 1 4

30 2 6 9 3 7 6 5 4 1 45 3 5 2

JL H G \ E

- = > 3



14/17

d) BjAi

40 30 20 50e) Bj

Ai30 20 25 35 40

30 3 7 4 6 30 13 7 6 2 12

40 4 6 2 5 20 5 1 10 5 11

70 1 5 7 8 40 10 5 3 7 14

60 6 3 2 11 10

GIAI BAI TOAN VAN TAI BANG THUAT TOAN THE V[4] Giai bai tap [3], vi phng an cc bien au tien thu c bang phng phap chi

ph be nhat.

s: a) va Z0 0 30 10

0 20 0 0

20 5 0 5optx min = 215; b) va Z

3 0 0 7

0 5 10 0

0 0 0 7optx min = 57;

c)*

va Z = 245; d) va Z = 510;

e) va Z = 800.

0 20 5

0 10 00 0 45

optx min

0 0 0 30

0 0 20 2040 30 0 0

optx min

0 0 0 30 0

10 10 0 0 0

0 10 25 5 0

20 0 0 0 40

optx min

[5] a) Giai bai toan van tai

BjAi

50 160 120 80

220 5 4 3 10

100 5 9 7 12

90 10 6 8 15

b) Bai toan co phng an toi u khac hay khong?

s: a) va z0 70 120 30

50 0 0 50

0 90 0 0optx min = 2330; b) khong co PATU khac.

[6] a) Giai bai toan van tai

BjAi

20 100 45 15

90 10 6 4 1

40 3 4 2 5

50 9 4 3 7

b) Bai toan co P.A.T. khac hay khong? Neu co, hay ch ra tap phng an toi u.

JL H G \ E

- = > 3



15/17

s: a) ; b) Z0 50 25 15

20 0 20 0

0 50 0 0optx

0 30 45 15

20 20 0 0

0 50 0 0optx min

= Z/min =715;

.0 30 20 45 20 15

20 20 20 20 0 , 0,1

0 50 0 0optT

[7] Cho bai toan van tai co dang sau ay:12011020

ia 5060304070jb

23412

54385

67524

ijc

a) Tm phng an toi u cua bai toan tren.

b) Theo anh (ch) dau hieu nao cho ta biet bai toan van tai co nhieu phng antoi u? Phng an toi u tm c cau a) co duy nhat khong? Neu co hay ch

ra phng an cc bien toi u khac.c) Tm tap cac phng an toi u va ch ra 3 phng an toi u khac nhau.

a) s: va z0 20 0 0 0

0 0 30 60 20

70 20 0 0 30opt

x min=690; b) s:0 20 0 0 0

20 0 30 60 0

50 20 0 0 50opt

x

[8] Cho bai toan van tai co dang sau ay:38, 45, 66, 45ia 52, 45, 38, 59jb

9 5 6 14

10 7 9 1510 10 6 7

4 8 13 14

ijc

a) Tm phng an toi u cua bai toan tren.

b) Phng an toi u va tm c co duy nhat khong? (co giai thch). Ch ramot phng an toi u khac? (neu co).

a) s: va Z

0 7 31 0

7 38 0 0

0 0 7 59

45 0 0 0

optx min = 1.192; b) P.A.T.. tren duy nhat.

[9] Giai bai tap [1], [2].

JL H G \ E

- = > 3



16/17

CAC DANG KHAC CUA BAI TOAN VAN TAI[10] Giai cac bai toan van tai sau ay va tm phng an toi u khac (neu co)

a)

BjAi

60 60 80 100

80 4 5 6 12

70 10 3 9 5

100 6 4 7 9

s: va Z60 0 20 0

0 0 0 70

0 60 40 0opt

x min = 1.230; P.A.T.. tren duy nhat.

b) Tram phat: ; Tram thu100 20 30 50ia 70 60 25 50jb

Ma tran cc ph van tai

10 14 24 8

30 20 18 14

2 12 6 7

8 16 14 36

ijc

s: Z

0 50 0 50

0 5 15 0

20 0 10 0

50 0 0 0

optx min = 1.970; va Z

0 55 0 45

0 0 15 5

20 0 10 0

50 0 0 0

optx/

min = 1.970.

c) Tram phat: ; tram thu :79, 50, 60, 50ia 46, 45, 76, 20, 52jb

Ma tran cc ph van tai10 1 5 13 85 6 10 8 13

3 2 8 9 6

13 5 7 10 13

ijc

s: Z

0 45 34 0 0

38 0 0 12 0

8 0 0 0 52

0 0 42 8 0

optx min = 1.211; P.A.T.. tren duy nhat.

[11] Giai cac bai toan van tai co o cam sau ay va tm phng an toi u khac (neu co)

a) Tram phat: ; tram thu :504090ia 4510020jb


349

243

4610

ijc

ieu kien tram A3 phai phat het hang.

b) Tram phat: ; tram thu :100 80 50ia 65 90 50 30jb

JL H G \ E

- = > 3



17/17


10 9 12 7

9 11 10 15

8 7 14 12

ijc

ieu kien tram B2 phai thu u hang.

c) Tram phat: ; tram thu :90,100,220i

a 80,120,160,50j

b


158610

12795

10345

ijc

ieu kien tram phat A3 khong c phat cho tram thu B2.

d) Tram phat: ; tram thu :504090ia 4510020jb


349

243

4610

ijc

ieu kien tram thu B2 khong c thu cua tram phat A1.

JL H G \ E

- = > 3

bài toán vận tải trong toán kinh tế

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