Bài giảng Quy Hoạch Tuyến Tình

7/31/2019 Bi ging Quy Hoch Tuyn Tnh

1/94

B GIO DC V O TOTRNG I HC NG THP

NGUYN VN HNG

GIO TRNH QUY HOCH TUYN TNH CHO TON

(LU HNH NI B)

NG THP-2011


2/94

M U

Quy hoch tuyn tnh l mt mn hc ng dng, ra i t cui nhng nm 40 cath k XX. Ngay t khi mi ra i n tm c nng ng dng a dng trong nhiulnh vc hot ng khc nhau ca i sng kinh t x hi."Gio trnh quy hoch tuyn tnh" trnh by nhng ni dung c bn ca l thuyt quyhoch tuyn tnh v cc phng php gii. Ti liu ny c bin son cho cc sinhvin ngnh Ton.

Gio trnh bao gm 5 chng:

Chng 1. a ra bi ton quy hoch tuyn tnh dng tng qut da trn c sxy dng t cc bi ton thc t. a ra phng php gii bi ton quy hoch tuyn

tnh bng th.Chng 2. Kho st tnh cht c bn v gii tch li v ng dng ca n. Tip

l nghin cu tnh cht tp phng n ca bi ton quy hoch tuyn tnh.

Chng 3. Chng ny trnh by c s l thuyt ca phng php n hnh, t xy dng phng php v gii cc bi ton quy hoch tuyn tnh.

Chng 4. Kho st bi ton quy hoch tuyn tnh i ngu, nu cch xy dngbi ton v trnh by cc tnh cht v a ra phng php gii bi ton.

Chng 5. Trnh by bi ton vn ti tng qut, t cc tnh cht c bn v sau xy dng thut ton gii cho bi ton.

Trong tt c cc chng a ra u c nhng v d minh ha c th cho tng dngbi ton, sau cui ca mi chng u c h thng bi tp kh a dng v phong ph.V nhiu l do, chc chn bi ging khng trnh khi nhng sai xt. Chng ti mongc s ng gp ca ng nghip v cc bn sinh vin.

Tc gi

2


3/94

Mc lc

1 Bi ton quy hoch tuyn tnh 51.1 Mt vi bi ton thc t dn n bi ton quy hoch tuyn tnh . . . . 5

1.1.1 Bi ton lp k hoch sn xut . . . . . . . . . . . . . . . . . . 5

1.1.2 Bi ton xc nh khu phn thc n . . . . . . . . . . . . . . . 7

1.1.3 Bi ton vn ti . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2 Bi ton quy hoch tuyn tnh . . . . . . . . . . . . . . . . . . . . . . . 9

1.2.1 Bi ton QHTT dng tng qut . . . . . . . . . . . . . . . . . . 9

1.2.2 Bi ton QHTT dng chnh tc . . . . . . . . . . . . . . . . . . 10

1.2.3 Bi ton QHTT dng chun tc . . . . . . . . . . . . . . . . . . 12

1.3 Phng php hnh hc gii bi ton QHTT . . . . . . . . . . . . . . . 12

1.3.1 Nhn xt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.3.2 Thut ton th . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Tnh cht ca bi ton quy hoch tuyn tnh 192.1 Mt s kin thc c bn v gii tch li . . . . . . . . . . . . . . . . . . 19

2.1.1 Tp hp li . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.1.2 Hm li . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.1.3 Mt s ng dng ca hm li . . . . . . . . . . . . . . . . . . . 22

2.2 Tnh cht tp phng n ca bi ton QHTT . . . . . . . . . . . . . . 26

3 Phng php n hnh 373.1 C s ca phng php n hnh . . . . . . . . . . . . . . . . . . . . . 37

3.2 Cng thc bin i bng n hnh . . . . . . . . . . . . . . . . . . . . . 40

3.3 Thut ton n hnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.4 Thut ton n hnh m rng . . . . . . . . . . . . . . . . . . . . . . . 46

3.5 Bi ton quy hoch tuyn tnh cha tham s . . . . . . . . . . . . . . . 51

3


4/94

4 Bi ton quy hoch tuyn tnh i ngu 584.1 Cc khi nim v bi ton i ngu . . . . . . . . . . . . . . . . . . . . 58

4.1.1 Cp bi ton i ngu i xng . . . . . . . . . . . . . . . . . . 58

4.1.2 Cp bi ton i ngu khng i xng . . . . . . . . . . . . . . 594.1.3 S i ngu tng qut . . . . . . . . . . . . . . . . . . . . . 61

4.2 Tnh cht ca cp bi ton i ngu . . . . . . . . . . . . . . . . . . . 62

4.2.1 Cc nh l i ngu . . . . . . . . . . . . . . . . . . . . . . . . 62

4.2.2 Cch gii bi ton i ngu . . . . . . . . . . . . . . . . . . . . 63

4.3 Phng php n hnh i ngu . . . . . . . . . . . . . . . . . . . . . . 66

4.3.1 C s l lun ca phng php . . . . . . . . . . . . . . . . . . 66

4.3.2 Thut ton n hnh i ngu . . . . . . . . . . . . . . . . . . . 67

5 Bi ton vn ti 755.1 Bi ton vn ti tng qut . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.1.1 Xy dng bi ton . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.1.2 Tnh cht ca bi ton vn ti . . . . . . . . . . . . . . . . . . . 76

5.2 Phng php tm phng n cc bin xut pht . . . . . . . . . . . . . 77

5.2.1 Phng php cc tiu cc ph . . . . . . . . . . . . . . . . . . 77

5.2.2 Phng php gc Ty -Bc . . . . . . . . . . . . . . . . . . . . 775.2.3 Phng php F- Ghen . . . . . . . . . . . . . . . . . . . . . . . 78

5.3 Thut ton th v . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

5.3.1 C s l lun ca thut ton . . . . . . . . . . . . . . . . . . . . 79

5.3.2 Thut ton th v . . . . . . . . . . . . . . . . . . . . . . . . . . 80

5.4 Cc dng khc ca bi ton vn ti . . . . . . . . . . . . . . . . . . . . 84

5.4.1 Bi ton vn ti khng cn bng thu pht . . . . . . . . . . . . 84

5.4.2 Bi ton vn ti c hm mc tiu cc i . . . . . . . . . . . . . 87

5.4.3 Bi ton vn ti c cm . . . . . . . . . . . . . . . . . . . . . 88

Ti liu tham kho . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 0 1

4


5/94

Chng 1

Bi ton quy hoch tuyn tnh

1.1 Mt vi bi ton thc t dn n bi ton quyhoch tuyn tnh

1.1.1 Bi ton lp k hoch sn xut

a) Bi tonMt c s sn xut hai loi sn phm A v B t cc nguyn liu I, II, III, chi ph tngloi nguyn liu v li ca n v sn phm cng nh d tr nguyn liu cho trongbng sau y.

N.liuS.phm A B D trI 2 1 8

II 0 1 0

III 1 0 3

Li 3 5

Hy lp bi ton th hin k hoch sn xut sao cho tng s li ln nht trn c s dtr nguyn liu c.

M hnh bi tonGi x1, x2 ln lt l s sn phm A v B c sn xut (x1, x2 0).

Max(3x1 + 5x2)

vi iu kin

2x1 + x2 8x2 4

x1

3

x1, x2 0

5


6/94

b) Tng qutMt cng ty d nh sn xut n sn phm Sj, (j = 1,..n) t m nguyn liu Ni, (i =1, ...m), li nhun mi sn phm v chi ph nguyn liu sn xut cho mi sn phmcho bng sau:

N.LiuS.Phm S1 S2 ... S n D trN1 a11 a12 ... a1n b1N2 a21 a22 ... a2n b2... ... ... ... ... ...

Nm am1 am2 ... amn bmLi nhun C1 C2 ... C n

Yu cu hy lp k hoch sn xut trong iu kin nguyn liu d tr hin c v cc

iu kin khc khng cn quan tm, sao cho x nghip thu dc nhiu li nhun nht.M hnh bi tonGi xj s lng sn phm th j cn sn xut, j=1,...n. Khi ta c:

a11x1 + a12x2 + ... + a1nxn b1a21x1 + a22x2 + ... + a2nxn b2...............................................

am1x1 + am2x2 + ... + amnxn bm

Gi f(x) l tng li nhun

f(x) = c1x1 + c2x2 + ... + cnxn

Nh vy ta c bi ton

M ax(f(x) =n

j=1

cjxj)

vi iu kin

a11x1 + a12x2 + ... + a1nxn b1a21x1 + a22x2 + ... + a2nxn b2...............................................

am1x1 + am2x2 + ... + amnxn bmxj 0, j = 1, 2, 3..

Hay

M ax(f(x) =

nj=1 c

jxj)

6


7/94

vi iu kin

n

j=1aijxj b1, i = 1, 2...m

xj 0, j = 1, 2, 3..

1.1.2 Bi ton xc nh khu phn thc n

a) Bi tonGi s sinh sng trong mt ngy m, mi ngi cn t nht 70g Protit, 30g Lipitv 420g Gluxit. Hm lng cc cht trn c trong 1g thc n A v B nh sau:

Cht D.DngThc n A BProtit(g) 0, 1 0, 2

Lipit(g) 0, 1 0, 1

Gluxit(g) 0, 7 0, 6

Ngoi ra, bit gi ca mi gam thc n A v B tng ng l 4 v 6. Hy xc nhkhi lng thc n l bao nhiu sao cho va m bo cht dinh dng, va s tinmua l nh nht.M hnhGi x1, x2 ln lt l s gam thc n A v B c sn xut (x1, x2 0).

M in(4x1 + 6x2)

vi iu kin

0, 1x1 + 0, 2x2 700, 1x1 + 0, 1x2 300, 7x1 + 0, 6 420x1, x2 0

b) Tng qutTa th hin bi ton qua bngMt x nghip cn mua n loi thc n Tj, (j = 1,..n) vi m cc cht dinh dngDi, (i = 1, ...m), gi mua mi loi thc n v cc cht dinh dng cho di bng sau:

C.D.DngT.Phm T1 T2 ... T n S.L.Ti thiuD1 a11 a12 ... a1n b1D2 a21 a22 ... a2n b2... ... ... ... ... ...

Dm am1 am2 ... amn bmGi mua c1 c2 ... cn

7


8/94

Yu cu hy lp k hoch xc nh khu phn thc n sao cho va m bo cht dinhdng, va chi ph s tin mua l nh nht.M hnhTm (x1, x2,...,xn) sao cho

Min(f(x) =n

j=1

cjxj)

vi iu kin

nj=1

aijxj bi, i = 1, 2...m

xj 0, j = 1, 2, 3..trong xj l s n v thc n cn mua, bt ng thc trn l rng buc lng cht dinhdng cc loi c trong khi lng mua cn tm khng c t hn lng ti thiu cnthit

1.1.3 Bi ton vn ti

a) Bi tonC mt loi hng cn c vn chuyn t hai kho (trm pht) A1 v A2 ti 3 ni tiuth (trm thu) B1, B2, B3. Bng sau cho bit s lng hng cn vn chuyn i mikho v s lng hng cn nhn mi ni tiu th v cc ph vn chuyn mt n vhng t mi kho ti ni tiu th tng ng

Trm phtTrm thu B1 B2 B3 Lng phtA1 5 2 3 30

A2 2 1 1 75

Lng thu 35 25 45

Hy lp k hoch vn chuyn tha mn mi yu cu thu pht sao cho chi ph vnchuyn l nh nhtM hnh bi tonGi xij, i = 1, 2;j = 1, 2, 3 l lng hng cn vn chuyn t kho Ai ti ni tiu th Bjth m hnh ton hc ca bi ton s l. Tm xij, i = 1, 2;j = 1, 2, 3 sao cho

Min(f(x) = 5x11 + 2x12 + 3x13 + 2x21 + x22 + x23)

8


9/94

vi iu kin

x11 + x12 + x13 = 30

x21 + x22 + x23 = 75

x11 + x21 = 35

x12 + x22 = 25

x13 + x23 = 45

xij 0, i = 1, 2;j = 1, 2, 3

b) Tng qutCn vn chuyn 1 loi hng t m ni pht n n ni thu, ni pht th i c s lnghng ai(i = 1, , , m). Ni nhn th j cn s lng hng l bj(j = 1, ...n). Cc ph vn

chuyn mt n v hng t i n j l cij. Nn vn chuyn nh th no m bo phtht v nhn s hng m c tng cc ph vn chuyn b nht.M hnh bi tonGi xij l s lng hng cn vn chuyn t i n j, suy ra bi ton quy hoch tuyntnh

Min(mi=1

nj=1

xij)

vi iu kin

nj=1

xij = bj i = 1, 2...m

mi=1

xij = ai j = 1, 2...n

xij 0, cij 0, bj, ai 0, i = 1, 2...m, j = 1, 2...n

1.2 Bi ton quy hoch tuyn tnh

1.2.1 Bi ton QHTT dng tng qut

Tm x = (x1, x2,...,xn) sao cho:

M in(M ax) {f(x) =n

j=1

cjxj}, (1.1)

9


10/94

vi iu kin

n

j=1aijxj bi, i = 1,...,k (1.2)

nj=1

aijxj bi, i = k + 1,...,k + m1 (1.3)n

j=1

aijxj = bi, i = k + m1 + 1,...,m (1.4)

xj 0, j = 1, 2,...,r,r n (1.5)

Bi ton (1.1) (1.5) c gi l bi ton quy hoch tuyn tnh dng tng qut,f(x) gi l hm mc tiu, cc iu kin (1.2) (1.5) c gi l iu kin buc (hociu kin rng buc). Mi vc t X Rn tha mn iu kin buc c gi l 1 phng n ca bi tonquy hoch. Tp hp tt c cc phng n gi l tp phng n v c k hiu l M. Mt phng n lm cc tiu (cc i) th c gi l phng n ti u (p.a.t.u) Bi ton quy hoch tuyn tnh gi l gii c nu c t nht mt phng n ti u.

1.2.2 Bi ton QHTT dng chnh tc

Bi ton dng chnh tc l bi ton c dng

M in(Max) {f(x) =n

j=1

cjxj}, (1.6)

vi iu kin

nj=1

aijxj = bi, i = 1, 2,...,m (1.7)

xj 0, j = 1, 2,...,n (1.8)

Ta c th d dng a bt k bi ton quy hoch tuyn tnh no v dng chnh tcnh quy tc sau:

+ Nu c maxf(X) th ta i thnh min{f(X)}. Sau khi tm c phng n tiu, ta c maxf(X) = min{f(X)} v ngc li.+ Nu c bt ng thc

nj=1

aijxj bi hocn

j=1

aijxj bi th ta a thm n ph(bin

ph) xn+i 0, vi h s hm mc tiu cn+i = 0 ta cn

j=1

aijxj xn+i = bi hocn

j=1

aijxj + xn+i = bi

+ Nu xk cha rng buc v du, th c th thay n bi hai n

xk, x

k 0, xk = xk xk

10


11/94

Ch :1. thun tin cho vic gii bi ton ta xem bi ton dng chnh tc l bi ton min2. Mt s cch vit khc ca bi ton quy hoch tuyn tnh

Dng ma trn

M in(CX)

vi iu kin AX = B

X 0

trong

C = (c1, c2,...,cn)1n l ma trn hng cp 1 nX = (x1, x2,...,xn)Tn1

B = (b1, b2,...,bm)Tm1 l cc ma trn ct

V ma trn A = (aij) l ma trn cp m n Dng vc t

MinC, Xvi iu kin

x1A1 + ... + xnAn = A 0x1, x2,...,xn 0

trong C = (c1, c2,...,cn) Rn, X = (x1, x2,...,xn) Rn

A0 = (b1, b2,...,bm) Rm, Aj = (a1j,...,amj) Rm, j = 1, 2, ...nV d 1. a bi ton sau y v dng chnh tc

Max(f(x) = x1 + x2 + x3)

vi iu kin

x1 + x2 + x3 = 5

x1 2x2 + x3 3x1 + x2 x3 4x1, x3 0

Ta thay maxf(X) bi min{f(X)}, a thm n ph x4, x5 0 v thay x2 bix2 = x

2 x2, x2, x2 0 ta c

M in(x1 x2 + x2 x3)

11


12/94

vi iu kin

x1 + x2 x2 + x3 = 5

x1

2(x2

x2) + x3 + x4 = 3

x1 + x2 x2 x3 x5 = 4

x1, x3, x2, x

2, x4, x5 0

1.2.3 Bi ton QHTT dng chun tc

Bi ton dng chun tc l bi ton dng

M in(Max) {f(x) =n

j=1

cjxj}, (1.9)

vi iu kin

nj=1

aijxj ()bi, i = 1, 2,...,m (1.10)

xj 0, j = 1, 2,...,n (1.11)

1.3 Phng php hnh hc gii bi ton QHTT

1.3.1 Nhn xt

Xt trong khng gian R2 vi h trc ta xoy.+ Phng trnh ax + by = c c biu din bi 1 ng thng vi vc t php tuyn(a, b).+ Cc im (x, y) tha mn ax + by c (ax + by c) nm trn na mt phng cb l l ng thng ax + by = c.+ Phng trnh ax + by = c khi m a, b khng i c thay i th ta c 1 h ngthng song song c vc t ch phng l (-b, a).V vy m t hnh nh hnh hc cu bi ton quy hoch tuyn tnh trong R2 qua thutton sau:

1.3.2 Thut ton th

Xt bi ton quy hoch tuyn tnh

Min(Max) {f(X) = c1x + c2y}

vi iu kin

aix + biy di, i = 1, 2,...,mx, y 0

Trn h trc ta vung gc xoy, ta thc hin thut ton.

12


13/94

Bc 1. Biu din cc iu kin buc ln h trc ta xoy, xc nh tp phngn M.

Bc 2. biu din phng ca hm mc tiu c1x + c2y = f. Ta cho f mt gi tr f0

c th no c c1x + c2y = f0.+ Gi c1x + c2y = f0 l ng mc.

+ Ta xc nh hng tng (gim) ph thuc vo bi ton max hay min (bi tonmax tnh tin theo hng tng v ngc li) nh sau: Ta ly I(x0, y0) ph thuc votp phng n M, I(x0, y0) khng thuc ng mc(thng thng ly I 0 gc ta).

Tnh f(I) = f0Nu f0 < f0 th hng ng mc ti im I l hng tng.Nu f0 > f0 th hng ng mc ti im I l hng gim.

Bc 3. Ta tnh tin ng mc theo hng tng hoc gim ph thuc vo bi ton.im ti hn cui cng ca tp phng n s l phng n ti u ca bi ton cho.

V d 2. Lp k hoch sn sut

M ax(3x + 5y)

vi iu kin

2x + y 8

y 4x 3x, y 0

Gii

+ Min phng n l a gic li OABCD+ V ng mc 3x + 5y = 15Chn I

O(0, 0)

f(I) = 0 nh vy hng

t ng mc ti O l hng gim v

f(max) = f(B) = f(2, 4) = 3.2 + 5.4 = 26

Vy phng n ti u ca bi ton l X(2, 4).

V d 3. Gii bi ton

Max(f = x1 x2)

13


14/94

vi iu kin

2x1 + x2 4x1

2x2

2

x1, x2 0

Hng dn.

+ V ng mc x1 x2 = 1+ Chn I O(0, 0) f(I) = 0+ f(I) < 1 suy ra hng ng mc ti imI l hng gim. Do vy bi ton khng cphng n ti u.

V d 4. Gii bi ton

M ax(f = 4x1 + 5x2)

vi iu kin

2x1 + x2 8x1 + 2x2 7x2 3x1, x2 0

Hng dn. Xt ng mc 4x1 + 5x2 = 10 fmax = 22 ti X(3, 2).

14


15/94

BI TP CHNG 1

Bi tp 1. Gi s c hai kho hng, cha 30 tn v 40 tn hng. C 3 ca hng ckh nng tip nhn 20 tn, 25 tn v 35 tn. Cc ph tr 1 tn hng t kho cho bi

KhoC.Hng A(20) B(25) C(35)I : 30 3 5 4

II : 40 6 7 1

Tm cch tr hng tha mn 2 yu cu:- t tin nht.- Gii ta kho.- Yu cu lp m hnh bi ton.

Bi tp 2. Trong mt chu k k hoch, nh my s dng hai loi vt liu V1, V2 sn xut 3 loi sn phm S1, S2, S3. Lng vt liu Vi dng sn xut 1 n v snphm Sj, gi bn 1 n v sn phm Sj, s lng vt liu mi loi nh my c, cho bibng sau:

VLSP S1 S2 S3 S lng cV1 4 2 5 10.000

V2 2 6 3 14.000

Gi bn 12nd 8nd 14nd

Vy lp k hoch sn xut, xc nh s lng sn phm mi loi cn sn xut, sao chotng thu nhp ln nht. Yu cu: Lp m hnh bi ton.

Bi tp 3. Mt x nghip c th sn xut c 3 loi sn phm, k hiu l: A; B; C.nh mc hao ph nguyn liu, vn, lao ng (gi cng) v li nhun thu c tnhcho 1 n v sn phm mi loi trong bng sau y:

Sn phm A B C Mc huy ng ti aN.Liu (kg) 2 3 3 150

Vn (1.000d) 1 3 5 120

L.ng(gi cng) 4 8 1 100Li nhun (1.000d) 2 3 5

Hy lp m hnh bi ton tm phng n sn xut sao cho trong phm vi s nguynliu, vn, gi cng huy ng c, xi nghip t li nhun cao nht.

Bi tp 4. Mt x nghip sn xut 3 loi sn phm A, B, C vi cc s liu sau:

Loi sn phm A B CGi bn (1000/v) 32 50 58

Chi ph sn xut (1000/v) 20 30 40Thi gian hon tt sn phm (gi/v) 1 2 3

15


16/94

Bit rng x nghip hin c s vn dng cho sn xut l 3 triu ng. Qu thi giansn xut l 180 gi. Theo cc hp ng k vi khch hng, yu cu sp A phi clng sn xut t nht l 100 v. Gi s mi sp sn xut ra u tiu th c ht. Lpk hoch sn xut cho tng li nhun ln nht.

Bi tp 5. Mt cng ty mun thc hin chin lc qung co mt loi sn phm camnh vi chi ph c tnh l 150 ngn ng/thng. C 3 loi hnh qung co c chnl: Tivi, Bo, i vi cc d liu nh sau:

Loi hnh qung co Tivi Bo iChi ph mt ln qung co (triu ng) 2 1, 5 0, 8

S ln qung co ti a trong thng 65 50 60D on s tip nhn qung co mi ln(ngn ngi) 10 15 8

Chin lc qung co phi c t nht 30 ln qung co trn Tivi trong mt thng. Hylp m hnh bi ton xc nh k hoch qung co ti u.

Bi tp 6. Mt doanh nghip sn xut bnh ko sn xut 4 loi ko A, B, C, D. Bitrng nguyn liu hin c, nh mc tiu hao nguyn liu cho 1 kg ko v li thu ckhi bn 1kg ko cho bng sau:

Nguyn liu S lng(kg) inh mc tiu hao(kg)A B C D

ng 1000 0, 6 0, 2 0,2 0,7

Sa 300 0, 3 0, 7 0,8 0,1C ph 80 0, 1 0, 1 0 0,2Li() 500 4500 3000 4000

Gi s cc nguyn vt liu khc khng tnh n v sn xut bao nhiu th tiu th htby nhiu. Hy lp m hnh bi ton sao cho li sut thu c l ln nht.

Bi tp 7. Mt hp cht c ch to t cc n cht A, B, C, D, E. Cc n chtny c ly t qung I, II, II, IV. Hm lng n cht(n v n cht) c chatrong 1 n v(v) qung mi loi v gi 1 n v qung mi loi c cho di bngsau:

n chtQung I II III IV A 3 4 1, 5 0

B 0 3 2 1

C 1 2 1, 5 2

D 2 0 3 1, 5

E 0, 01 0, 03 0, 07 0

Gi(ngn /v) 7 6 8 5

Bit rng 1 n v hp cht cn t nht 12 v n cht A; 8 v cht B; nhiu nht 6v cht C; t nht 7 v cht D; 0,5 v cht E. Hy lp m hnh bi ton xc nh s

16


17/94

lng mua mi loi qung bao nhiu n v tng chi ph v nguyn liu cho 1 nv hp cht l thp nht.

Bi tp 8. Mt c s t nhn sn xut giy vi d liu v 1 i dy c cho di

bng sau:Giy n ng Giy ph n Giy tr em

Gi bn (ngn ng) 70 50 25Chi ph vt liu(ngn ) 25 15 10Thi gian hon tt(gi) 3 2, 5 1, 5

C s c 10 ngi th lm vic 8h/ngy. Lng cho 1 th l 1,5 triu ng/thng.Ngi ch c s ny tnh rng vo thng ti, tng gi tr vt liu m ng huy ngc l 7 triu ng. Gi s rng giy sn xut ra u tiu th ht. Hyax lp m hnh

bi ton sao cho li nhun thu c khi bn l giy sn xut trong thng ti l caonht.

Bi tp 9. a bi ton QHTT sau y v dng chnh tc:

a.

Min(4x1 + 3x2 + 4x3)

vi iu kin

2x1 + 3x2 + 4x3 6x1 + 3x2 + 4x3 = 7

3x1 + 2x2 + x3 9xj 0, j = 1, 2, 3

b.

Max(2x1 + 3x2 + 4x3 + x4)

vi iu kin

3x1 + x2 + 2x3 + x4 = 2

2x1 x3 3x4 1x1 x3 2x4 4xj 0, j = 1, 2, 3

c.

Min(x1 + 3x2 2x3)

17


18/94

vi iu kin

3x1 + x2 2x3 72x1

4x2 + x3 = 12

4x1 + 3x2 8x3 10x1 0, x3 0

Bi tp 10. Gii bi ton sau bng phng php th.a.

Min(x1 + x2)

vi iu kin

2x1 + x2 2x1 2x2 2x1 + x2 5x1, x2 0

b.

Max(x1 + x2)

vi iu kin

2x1 + x2 2x1 2x2 2x1, x2 0

18


19/94

Chng 2

Tnh cht ca bi ton quy hochtuyn tnh

2.1 Mt s kin thc c bn v gii tch li

2.1.1 Tp hp li

a) nh ngha. T hp li . Cho k im x1, x2,...,xk Rn. Khi

x =ki=1

ixi, i 0ki=1

i = 1

ta ni x l t hp li ca h im cho. on thng. Cho x1, x2 Rn, tp hp

x1x2 = {x Rn : x = 1x1 + 2x2, 1, 2 0, 1 + 2 = 1}

c gi l on thng ni hai im cho. Tp hp li. Tp hp M Rn c gi l tp hp li nu on thng ni hai imbt k thuc M, nm trn trong M, ngha l vi

x1, x2 M, x = x1 + (1 )x2, 0 .1

th x M. im cc bin. Cho tp li M Rn. im x M c gi l im cc bin caM, nu khng tn ti x1, x2 M m

x = x1 + (1 )x2, 0 < < 1.

Ngha l x khng th l im trong ca bt k on thng no thuc M. im ccbin ca tp li cc phng n gi l phng n cc bin.

19


20/94

Tp M = {x Rn :n

j=1

aijxj bi, i = 1,...,m} c gi l tp li a din

Tp li a din M khc rng v b chn c gi l a din li

Hay trong Rn

, tp cc im l t hp li ca mt s hu hn im cho trc c gil a din li sinh bi cc im y.b) Tnh cht.Tnh cht 1. Giao ca cc tp li l mt tp li.Tnh cht 2. Tng (Hiu) ca cc tp li l mt tp li.Tnh cht 3. NuM l tp hp li th n cha mi t hp li ca hu hn im catp hp .

V d 1. Chng minh rng tp hp sau l tp li:

D = {(x1, x2) R2

, x1 + 2x2 50, x1 40, x2 20, x1, x2 0}Chng minh. Ly bt k x = (x1, x2) D, y = (y1, y2) D.Ta chng minh z = x + (1 )y D, [0, 1]Tht vy: Gi z = (z1, z2), z1 = x1 + (1 )y1, z2 = x2 + (1 )y2, [0, 1].Ta c:

z1 + 2z2 = x1 + (1 )y1 + 2[x2 + (1 )y2]= (x1 + 2x2) + (1 )(y1 + 2y2) 50 + (1 )50 = 50

z1 = x1 + (1 )y1 40 + (1 )40 = 40z2 = x2 + (1 )y2 20 + (1 )20 = 20z1 = x1 + (1 )y1 0 + (1 )0 = 0z2 = x2 + (1 )y2 0 + (1 )0 = 0

Vy suy ra z D nn D l tp li.

2.1.2 Hm li

a) nh ngha. Cho hm f xc nh trn tp li M, hm f c gi l hm linu x, y M, [0, 1] ta c:

f(x + (1 )y) f(x) + (1 )f(y)du bng = xy ra khi = 0 hoc = 1 v f c gi l hm li cht. Ngc li,

f(x + (1 )y) > f(x) + (1 )f(y)th f c gi l hm lm (lm cht).Nu f(x) l hm li th

f(x) l hm lm.

b) Tnh cht ca hm li

20


21/94

Tnh cht 1. Hm tuyn tnh l hm va li va lm.

Tnh cht 2. Hm mt bin s f(x) xc nh v lin tc, c o hm cp 2 trnkhong (a, b). Khi f(x) l hm li trn khong (a, b) nu f(x) 0, x (a, b)

Tnh cht 3. (Bt ng thc Jensen.)Hm f(x) l hm li trn tp li M khi v ch khi

f(ki=1

ix(i))

ki=1

if(x(i)), x(i) M (2.1)

trong i 0,ki=1

i = 1

Chng minh:Gi s c bt ng thc (2.1), khi vi k=2 th f l hm li.

Ngc li, gi sf(x) l hm li trn M, ta xt

x =ki=1

ix(i), i 0,

ki=1

i = 1

Do M li nn x M. Ta chng minh (2.1) theo quy npVi k = 2 bt ng thc lun ng.Gi s bt ng thc ng vi k tc l

f(ki=1

ix(i))

ki=1

if(x(i))

ta chng minh ng vi k + 1, tc l

f(k+1i=1

ix(i))

k+1i=1

if(x(i))

Tht vy, khng mt tnh tng qut, gi s0 < k+1 < 1. Khi

f(

k+1i=1

ix(i)

) = f(

ki=1

ix(i)

+ k+1x(k+1)

)

= f((1 k+1)ki=1

i1 k+1x

(i) + k+1x(k+1))

Ta t i = i1k+1 , vi mi i 0,ki=1

i = 1. Do theo gi thit bt ng thc

ng vi k ta c

f(

ki=1

ix(i)

) k

i=1if(x

(i)

) =

1

1 k+1k

i=1if(x

(i)

)

21


22/94

T ta c

f(k+1

i=1 ix(i)) = f((1 k+1)

k

i=1i

1

k+1

x(i) + k+1x(k+1))

k+1i=1

if(x(i))

T ta suy ra iu phi chng minh.

H qu 1. Nu hm sy = f(x) li trn tp li M, vix1, x2,...xn M th ta c btng thc

f(x1 + x2 + ... + xn

n) f(x1) + f(x2) + ... + f(xn)

n, n 2 (2.2)

H qu 2. Nu hm sy = f(x) li trn tp li M, vix1, x2,...xn M th ta c btng thc

f(m1x1 + m2x2 + ... + mnxn

m1 + m2 + ... + mn) m1f(x1) + m2f(x2) + ... + mnf(xn)

m1 + m2 + ... + mn(2.3)

Tnh cht 4. Hm lin tc f l hm li khi v ch khi tha mn

f(X+ Y

2) 1

2f(X) +

1

2f(Y)

2.1.3 Mt s ng dng ca hm li

V d 2. Chng minh bt ng thc Csi

x1 + x2 + ... + xnn

nx1.x2...xn, (xi 0, i = 1, 2,...,n)

Chng minh.+ Nu tn ti i sao cho xi = 0 th bt ng thc ng+ Gi sxi > 0,

i, ta xt hm li f(x) =

lnx. Theo bt ng thc (2.2) ta c

f(x1 + x2 + ... + xn

n) f(x1) + f(x2) + ... + f(xn)

n

ln( x1 + x2 + ... + xnn

) ln(x1) ln(x2) ... ln(xn)n

ln( x1 + x2 + ... + xnn

) ln(x1.x2..xn) 1n

x1 + x2 + ... + xnn

nx1.x2..xn

Du bt ng thc xy ra khi v ch khi x1 = x2 = ... = xn

22


23/94

V d 3. Chng minh bt ng thc Bunhiacpxki

(a1b1 + a2b2 + ... + anbn)2 (a21 + a22 + ...a2n)(b21 + b22 + ... + b2n),

trong a1, a2,...,an; b1, b2,....,bn l 2n s ty Chng minh.Ta xt hm li f(x) = x2, vi mi x R. Theo bt ng thc (2.3) ta c

(m1x1 + m2x2 + ... + mnxn

m1 + m2 + ... + mn)2 m1x

21 + m2x

22 + ... + mnx

2n

m1 + m2 + ... + mn

(m1x1 + ... + mnxn)2 (m1 + ... + mn)(m1x21 + ... + mnx2n)

t mi = b2i v xi =aibi

ta c

(b21a1b1+ b22a2b2

+ ... + b2nanbn)2 (b21 + b22 + ... + b2n)(b21a

2

1b21

+ b22a2

2b21

+ ... + b21a2

nb2n

)

(a1b1 + a2b2 + ... + anbn)2 (a21 + ... + a2n)(b21 + ... + b2n)

Du bng xy ra khi v ch khi a1b1

= a2b2

= ... = a2b2

. Suy ra iu phi chng minh.

V d 4. Chng minh bt ng thc Mincpxki

n

a1a2...an +n

b1b2...bn n

(a1 + b1)(a2 + b2)...(an + bn),

trong a1, a2,...,an; b1, b2,....,bn l 2n s dng

Chng minh.Xt hm li f(x) = ln(1 + ex). Khi p dng bt ng thc Jensen (2.2) ta c

f(x1 + x2 + ... + xn

n) f(x1) + f(x2) + ... + f(xn)

n,

Chn xi = ln biai , i = 1,..n ta c

ln(1 + eln

b1a1

+...+lnbnan

n ) ln(1 + eb1a1 ) + ln(1 + e

b2a2 ) + ... + ln(1 + e

bnan )

n

ln(1 + eln nb1.b2...bna1a2...an ) ln(1 +b1a1 ) + ... + ln(1 +

bnan )

n

n

a1.a2...an +n

b1.b2...bnn

a1.a2...an n

(a1 + b1)...(an + bn)

a1...an

na1a2...an + n

b1b2...bn n

(a1 + b1)(a2 + b2)...(an + bn),

Suy ra iu phi chng minh.

V d 5. Chng minh bt ng thc Holder

a1b1 + a2b2 + ... + anbn ap1 + ap2 + ... + apn 1pbq1 + bq2 + ... + bqn 1q23


24/94

vi ai, bi > 0;p, q > 0; 1p +1q

= 1

Chng minh.Xt hm li f(xp), (x > 0, p > 1). Theo bt ng thc Jensen (2.3) ta c

m1x1 + m2x2 + ... + mnxn

m1 + m2 + ... + mn

p m1xp1 + m2xp2 + ... + mnxpn

m1 + m2 + ... + mn

hay l

m1x1 + m2x2 + ... + mnxnm1 + m2 + ... + mn

m1xp1 + m2xp2 + ... + mnxpn

m1 + m2 + ... + mn

1p

T suy ra

m1x1 + m2x2 + ... + mnxn

m1xp1 + ... + mnxn

1p

m1 + ... + mnp1

p ()

Mt khc do 1p

+ 1q

= 1 nn p1p

= 1q. thay vo (*) ta c

m1x1 + m2x2 + ... + mnxn

m1xp1 + ... + mnxn

1p

m1 + ... + mn 1

q

Hay

ni=1

mixi ni=1

mixpi 1

p n

i=1

mi 1

q

t mi = bqi , xi = aib1qi ta c

ni=1

bqiaib1qi

ni=1

bqi

aib

1qi

p 1p ni=1

bqi

1q

Hay l

ni=1

aibi

ni=1

api bp+qpqi

1p

ni=1

bqi

1q

Mt khc li do 1p

+ 1q

= 1 nn p + qpq = 0, suy ra bp+qpq = b0 = 1Vy

a1b1 + a2b2 + ... + anbn

ap1 + ap2 + ... + a

pn

1p

bq1 + bq2 + ... + b

qn

1q

iu phi chng minh

24


25/94

V d 6. Chng minh bt ng thc Svacx Cho 2n s thc a1,...,an; b1,...,bn trong bi > 0, i = 1, 2...n ta c

a21

b1+

a22

b2+ ... +

a2n

bn (a1 + a2 + ... + an)

2

b1 + b2 + ... + bn

Chng minh.Xt hm li f(x) = x2, x R. Theo bt ng thc Jensen (2.3) ta c

(m1x1 + m2x2 + ... + mnxn

m1 + m2 + ... + mn)2 m1x

21 + m2x

22 + ... + mnx

2n

m1 + m2 + ... + mn

Hay

(m1x1 + m2x2 + ... + mnxn)2

m1 + m2 + ... + mn m1x21 + m2x22 + ... + mnx2n

t mi = bi v xi = aibi . Khi ta c

(a1 + a2 + ... + an)2

b1 + b2 + ... + bn a

21

b1+

a22b2

+ ... +a2nbn

Suy ra iu phi chng minh.

V d 7. Cho a,b,c > 0. Chng minh rng

(b + c)a(c + a)b(a + b)c

2

3(a + b + c)

a+b+c

Chng minh.Ly logarit hai v, bt ng thc tng ng vi

aln(b + c) + bln(c + a) + cln(a + b)

a + b + c ln

2

3(a + b + c)

Xt hm li f(x) = ln(a + b + c x), vi 0 < x < a + b + c. Khi p dng bt ngthc Jensen (3) ta c

fa.a + b.b + c.c

a + b + c

af(a) + bf(b) + cf(c)

a + b + c

ln

a + b + c a2 + b2 + c2

a + b + c

aln(b + c) bln(a + c) cln(a + b)

a + b + c

ln2ab + 2bc + 2ac

a + b + c

aln(b + c) + bln(a + c) + cln(a + b)

a + b + c

25


26/94

Do hm y = lnx l ng bin trn (0, ), nn ta ch cn chng minh2

3(a + b + c) 2ab + 2bc + 2ac

a + b + c(*)

Tht vy (*) tng ng vi

(a + b + c)2 3(ab + bc + ac)a2 + b2 + c2 ab + bc + ca(a b)2 + (b c)2 + (c a)2 0

lun lun ng nn (*) ng, t suy ra

ln2

3

(a + b + c) ln

2ab + 2bc + 2ac

a + b + c

aln(b + c) + bln(a + c) + cln(a + b)a + b + c

Vy ta suy ra

(b + c)a(c + a)b(a + b)c

2

3(a + b + c)

a+b+c

iu phi chng minh. Du = xy ra khi v ch khi a = b = c

2.2 Tnh cht tp phng n ca bi ton QHTT

Khng mt tnh tng qut ta xt bi ton quy hoch tuyn tnh dng chnh tc

Min(Max) {f(x) =n

j=1

cjxj}, (1.12)

vi iu kin

nj=1

aijxj = bi, i = 1, 2,...,m (1.13)

xj 0, j = 1, 2,...,n (1.14)

+ H (1.13) c ng m phng trnh c lp tuyn tnh+ Mi bi 0, i =1, 2, ..., m+ m < n (nu m n th tp phng n c nhiu nht mt im, do vy vic ti uha l tm thng). K hiu M l tp phng n ca bi ton (1.12), (1.13), (1.14).

Tnh cht 1. Phng n X0 gi l phng n cc bin nu khi th X0 vo h rng

buc tha mn "cht" n rng buc c lp tuyn tnh (n l s n s).+ Rng buc th i gi l "cht" vi phng n X, nu th X vo cho ta mt ng thc

26


27/94

ng.+ Rng buc th i gi l "lng" vi phng n X nu khi th X vo cho ta 1 bt ngthc ngt (>,


28/94

V d 8. Cho bi ton quy hoch tuyn tnh

Min(f(x) = 8x1 + 2x2 + 9x3 x4)

vi iu kin

3x1 + 2x3 x4 14x1 4x2 2x4 = 8x1 + 7x2 + x3 + 3x4 7x1 0, x2 0, x3 0

Hy xt xem vc t X0 = (0, 1, 6, 2) c phi l phng n cc bin ca bi ton cho hay khng?Gii. Thay X0 vo h rng buc ca bi ton ta c

3x1 + 2x3 x4 = 14x1 4x2 2x4 = 8 x1 + 7x2 + x3 + 3x4 = 7

x1 = 0

x2 = 1 < 0x3 = 6 > 0

Nh vy X0 l mt phng n ca bi ton Phng n X0 tha mn 4 rng buc cht, s rng buc cht ng bng s bin cabi ton v nh thc ca ma trn cc h s ng vi 4 rng buc cht l:

3 0 2 11 4 0 2

1 7 1 31 0 0 0

=

0 2 1

4 0 27 1 3

= 0

Suy ra h 4 rng buc cht ph thuc tuyn tnh, nn phng n X0 khng phi l

phng n cc bin ca bi tonVn cu hi nh trn nhng vi vc t X1 = (4, 0, 0, 2) Thay X1 vo h rng buc ca bi ton ta c

3x1 + 2x3 x4 = 14x1 4x2 2x4 = 8 x1 + 7x2 + x3 + 3x4 = 10 < 7

x1 = 4 > 0

x2 = 0

x3 = 0

28


29/94

Nh vy X1 l mt phng n ca bi ton Phng n X1 tha mn 4 rng buc cht, s rng buc cht ng bng s bin cabi ton v nh thc ca ma trn cc h s ng vi 4 rng buc cht l:

3 0 2 11 4 0 20 1 0 0

0 0 1 0

=

3 0 11 4 20 1 0

=3 11 2

= 5 = 0

Nh vy h rng buc cht c lp tuyn tnh. Do X1 l phng n cc bin(khngsuy bin) ca bi ton

Vn cu hi nh trn nhng vi vc t X2 = (0, 0, 5, 4) Thay X2 vo h rng buc ca bi ton ta c

3x1 + 2x3 x4 = 14x1 4x2 2x4 = 8 x1 + 7x2 + x3 + 3x4 = 7

x1 = 0

x2 = 0

x3 = 5 > 0

Nh vy X2 l mt phng n ca bi ton

Phng n X2 tha mn 5 rng buc cht, s rng buc cht nhiu hn s bin cabi ton. Xt 4 rng buc cht trong s 5 rng buc cht trn v nh thc ca matrn cc h s ng vi 4 rng buc cht l:

1 4 0 21 7 1 31 0 0 0

0 1 0 0

=

1 0 2

1 1 31 0 0

= 2 = 0

Nh vy h 4 rng buc cht c lp tuyn tnh. Do X2 l phng n cc bin suy

bin ca bi tonCng cu hi nh trn nhng vi vc t X3 = (6, 0, 0, 1) Thay X3 vo h rng buc ca bi ton ta c

3x1 + 2x3 x4 = 19 > 14x1 4x2 2x4 = 8 x1 + 7x2 + x3 + 3x4 = 9 < 7

x1 = 6 > 0

x2 = 0

x3 = 0

29


30/94

Nh vy X3 l mt phng n ca bi ton Phng n X3 tha mn 3 rng buc cht, s rng buc cht t hn s bin ca biton.Do X3 khng phi l phng n cc bin ca bi tonV d

9. Cho bi ton QHTTMin(6x1 + 3x2 + 5x3 + 2x4)

vi iu kin

3x1 + 2x2 + x3 + 2x4 15x1 + 2x2 + 2x3 + 3x4 = 10

2x1 + x2 + 2x3 + x4 12xj 0, j = 1, 2, 3, 4

Cho vc t X = (2, 0, 4, 0)a) X c phi l phng n cc bin ca bi ton khng?b) Chng t rng bi ton c phng n D = nhng hm mc tiu khng b chn.Hng dna) Tha mn 4 rng buc cht(rng buc 2, 3 v x2 = 0, x4 = 0) v h c s gm 4 vct {x1, x2, x3, x4} c lp tuyn tnh. Vy X l PACB khng suy bin.b) T h rng buc ta rt x1, x2, x3 theo x4, x5, x6 ta c

x1 = 24/7 + 5/7x4 2/7x5 + 2/7x6x2 = 10/7 9/7x4 2/7x5 5/7x6x3 = 13/7 4/7x4 + 3/7x5 + 4/7x6

Ta cho x4 = x5 = x6 = 0 th c phng n X = (24/7, 10/7;13/7;0) suy ra tpphng n D = . Thay x1, x2, x3 vo f ta c

f(x) = 239/7 3/7x4 + 21/7x5 17/7x6

c nh x5; x6 v cho x4 + f nn hm mc tiu khng bi chn.pcm.V d 10. Cho bi ton QHTT

Min(4x1 6x2 + 3x3)vi iu kin

2x1 + 4x2 x3 03x1 5x2 + 2x3 1x1 2x3 23x2 + x3 2x1

x2

2

xj 0;j = 1, 2, 3

30


31/94

Hy chng t X0 = (2, 1, 0) l phng n cc bin ti u ca bi ton cho.Hng dn Ta chng minh X0 l phng n cc binTa thay X0 vo h tha mn 3 rng buc cht v nh thc ca ma trn cc h s ngvi 3 rng buc cht c lp tuyn tnh. Do X0 l phng n cc bin khng suybin. Ta chng minh X0 l phng n ti u(chng minh hm mc tiu b chn di).Tht vy ta nhn bpt (2) vi 2 sau cng vi bpt (1) theo v ta c

f(x) = 4x1 6x2 + 3x3 2Vy hm mc tiu b chn di bi 2 nn suy ra iu phi chng minh.V d 11. Cho bi ton QHTT sau

M ax(x2 + 2x3 2x4 2x5)vi iu kin

2x1 + 3x2 + x3 + x5 44x1 5x2 + 3x4 x5 6x1 + 2x2 + 2x3 x4 = 3xj 0, j = 1, 2, 3, 4, 5

Chng minh rng X0 = (1, 2, 0, 0, 0) l phng n cc bin ti u suy bin.

Hng dn Ta chng minh X0 l phng n cc bin suy binTa thay X0 vo h thy c 6 rng buc cht vy X0 l phng n cc bin suy bin. Ta chng minh hm mc tiu b chn trn. Tht vy, nhn bpt (1) vi 2, cng theov vi (2) ta c

x2 + 2x3 + 3x4 + x5 8f(x) = x2 + 2x3 2x4 2x5 2 5x4 3x5 2

V x4, x5

0 vy f b chn trn bi 2. Vy X0 l phng n cc bin suy bin.

V d 12. Cho bi ton QHTT

Min(4x1 5x2 + 2x3)vi iu kin

x1 + 3x2 3x3 = 17x1 x2 + 2x3 24x1 + 5x2 2x3 432x1

x2 + 8x3 =

1

x3 0

31


32/94

1) Xc nh tp cc phng n, PACB2) Chng t bi ton gii c.Hng dn1) Thm cc n ph ri gii h2) T phng trnh 1,4 rt x1, x2 th vo bt phng trnh cn li xc nh x3 vtin hnh xt f

V d 13. Chng t bi ton sau gii c

M ax(x1 + x2 + 2x3 + x4)

vi iu kin

2x1 x2 + 4x3 x4 10x1 2x2 + x4 134x1 x2 + 3x3 82x1 + x3 5x4 53x1 + 2x2 2x3 + x4 = 0xj 3;j = 1, 3, 4

Hng dnRt x2 t phng trnh 5 vo f, va s dng xj 3;j = 1, 3, 4 ta suy ra f(x) b chntrn


Min(2x1 + x2 + x3 + 5x4)vi iu kin

x1 = 5

3x1 2x2 9x3 8

x3 + 5x4 =

13

1) Hy ch ra mt PACB v chng t bi ton khng gii c2) Nu thay c2 = 4, chng t bi ton gii c v tm mt phng n ti u.Hng dn1) Thm cc n ph ri gii h phng trnh v suy ra PACBT h phng trnh rt x1, x2 th vo f(x) ta c

f(x) = 23 + x2 + 2x3; x2 3; x4 1Ta c nh x4 cho x2 do bi ton khng gii c2) Khi c2 =

4

f(x) =

23

4x2 + 2x3; x2

3; x4

1

f(x)

23 Vy f(x)

b chn di bi -23 khi phng n ti u chnh l X0 = (5, 3, 8, 1).

32


33/94

BI TP CHNG 2

Bi tp 1. Hy chng minh rng:a) Giao ca tp li l mt tp li

b) Tng(Hiu) ca cc tp li l mt tp li.Bi tp 2. Gi sf : D(f) R, D(f) Rn l tp li. Chng minh rng f l litrn D(f) khi v ch khi vi x, y D(f) th hm

() = f[x + (1 )y]

l hm li theo trn [0, 1].

Bi tp 3. Gi sf : D(f) R l hm li, D(f) Rn l tp li. Gi sg : D(g) Rl n iu tng v li trn D(g) R v f(D(f)) D(g).

Chng minh rng hm hp F : D(f) R, trong F(x) = g(f(x)) l hm li trnD(f).Bi tp 4. Trong mt phng vi h trc ta vung gc xoy, on thng AB cbiu din:

AB = {X(x, y)|X = A + (1 )B, [0, 1]}cho hm tuyn tnh f xc nh trn tam giaics ABC. Chng minh rng:a) X(x, y) ABC c

X = 1A + 2B + 3C, 1, 2, 3 0, 1 + 2 + 3 = 1

b) Nu f(A) = min{f(A), f(B), f(C)} th f(A) = min f(X), X ABCBi tp 5. Hy kim tra xem cc hm s su y c phi l hm li hay khng?a) f(x) = |x| vi mi x R.b) f(x) = ex vi mi x R.c) f(x) = (x 1)3 vi 1 x < +.d) f(x) = (x 1)3 vi 0 x < +.e) f(x) = logx vi 0 < x < +.f) f(x) = ex2 vi 0 x < +.

Bi tp 6. S dng cc tnh ca hm li chng minh cc bi ton sau:1. Cho cc s dng a1,...,an, a1 + a2 + ... + an = 1. Chng minh rng

1

ak1+

1

ak2+ ... +

1

akn nk+1, (k > 0)

2. Cho n > 1 v x1, x2, x3,...,xn > 0. Chng minh rng

xn1 + xn2 + ... + x

nk

k

x1 + x2 + ... + xkk

n

33


34/94

3. Cho k l s nguyn dng v x1, x2, x3,...,xn > 0, n > 1. Chng minh rng

(x1 + x2 + ... + xn)2k n2k1(x2k1 + x2k2 + ... + x2kn )

4. Cho a,b,c > 0. Chng minh rng

(b + c)a(c + a)b(a + b)c 23

(a + b + c)a+b+c

5. Cho a1, a2,...,an > 0. Chng minh rng

aa11 .aa22 ...a

ann

a1 + a2 + ... + an

n

a1+a2+...+an

Bi tp 7. Cho bi ton QHTTMin(7x1 + 6x2 + 4x3 + 3x4)

vi iu kin

x1 1x1 + x2 2x1 + x2 + x3 3x1 + x2 + x3 + x4

4

xj 0, j = 1, 2,...4

a)Phng n X = (1, 1, 1, 1) c phi phng n cc bin ca bi ton, c suy bin haykhng?b) Phng n X = (4, 0, 0, 0) c phi phng n cc bin ca bi ton, c suy binhay khng?c) Chng t hm mc tiu b chn di bi 20.

Bi tp 8. Cho bi ton QHTT

Min(x1 + 2x2 + x3 + 3x4)

vi iu kin

x1 + x2 + x3 + 2x4 = 10

x1 + 3x2 + 3x4 = 9xj 0, j = 1, 2,...4

a)Phng n X = (1, 1, 1, 1) c phi phng n cc bin khnmg suy bin hay khng?b) Phng n X = (4, 0, 0, 0) c phi phng n cc bin ca bi ton khng?c) Chng t hm mc tiu b chn di.

34


35/94


M in(3x1 4x2 + 3x3 + x4 + 8x5 4x6)

vi iu kin

2x1 + x3 + 2x5 3x6 = 2x2 x3 + x4 2x5 + x6 = 2x1 2x2 + 3x4 + 5x5 x6 = 5

a)Xc nh tp cc phng n ca bi ton, v chng t bi ton khng gii c.b) Khi hm mc tiu f(x) = 3x1 4x2 + 3x3 + x4 + 11x5 6x6 min th c kt lung v bi ton?Nu c im ca tp cc phng n.

Bi tp 10. Cho bi ton QHTTM in(3x2 + x3 + x4 + 2x5 + x6)

vi iu kin

x1 2x2 + x3 2x5 203x2 x3 + x4 + x5 11x2 + 4x3 + 3x5 + x6 3xj

0, j = 2, 3, 5

Hy chng t bi ton gii c, v tm mt phng n cc bin ti u.


max(4x1 5x2 + 2x3)

vi iu kin

x1 + 3x2 3x3 = 17

x1 x2 + 2x3 24x1 + 5x2 2x3 432x1 x2 + 8x3 = 1x3 > 0

a)Xc nh tp cc phng n ca bi ton, phng n cc bin.b) Chng t bi ton gii c.


M ax(x1 + x2 + x3 + 2x4)

35


36/94

vi iu kin

2x1 x2 x3 + x4 10

x1 + 5x3

2x4

15

4x1 + 3x2 + x3 x4 8x1 + 2x2 + 3x3 + x4 3xj 0, j = 1, 2, 3, 4

Chng t bi ton gii c.


Min(2x1 3x2 + x3)

vi iu kin

x1 2x2 + x3 2x1 + 5x3 172x2 9x3 43x1 x2 + 3x3 6

Chng t bi ton c phng n ti u.


Min(5x1 9x2 + 15x3 + 7x4 + 6x5)

vi iu kin

x1 + 3x2 x3 x4 + x5 14x1 + x3 + 2x4 x5 = 4x1 x2 + x3 2x5 1xj 0, j = 2,.., 5

Hy xt xem phng n X = (0, 1, 0, 2, 0) c phi l phng n cc bin ti u ca biton cho hay khng?

36


37/94

Chng 3

Phng php n hnh

tng chung: Xut pht t phng n cc bin X0, kim tra X0 c ti u haykhng?+ Nu X0 l phng n ti u th dng.+ Nu X0 cha l phng n ti u th ta i xy dng phng n cc bin mi X1 tthn X0.

Qu trnh tip tc nh vy, ta c dy cc phng n cc bin tt dn (theo nghagi tr hm mc tiu gim dn t phng n cc bin trc n phng n cc binsau) X0, X1,..,Xk. Sau hu hn bc lp ta tm c phng n ti u hoc pht hinra bi ton v nghim. chnh l ni dung ca qu trnh xy dng dy cc phng

n cc bin tt dn, cn gi l phng php n hnh

3.1 C s ca phng php n hnh

Khng mt tnh tng qut, gi thit bi ton QHTT cho di dng

Min(CX), (1.15)

vi iu kinAX = B (1.16)

X 0 (1.17)+ H c m phng trnh c lp tuyn tnh

+ bi 0, i = 1, 2, ...m+ m < n+ (1.16), (1.17), (1.18) khng suy bin. Khi d dng tm c phng n cc binxut pht

X0 = (x0i , x

02,...,x

0m, 0,..., 0); x

0i = bi, i = 1, 2,...,m

vi c s lin kit A1A2...Am l ma trn n v B= (A1A2...Am) hng m.

K hiu XB = (x01, x

02,...,x

0m), C

= (c1, c2,...,cm)

37


38/94


39/94

nh l 3.1.2. Nu ti phng n cc bin X0, k > 0 vaik 0 (i = 1, 2, ...m)th bi ton khng c phng n ti u.

Chng minh. Theo gi thit X0 l phng n nn

x01A1 + x02A2+,...,x

0mAm = A0 ()

Mt khc theo s biu din ca vc t Ak qua c s ta c

Ak = a1kA1 + a2k + ... + amkAm. ()

Nhn hai v ca phng trnh () vi > 0, sau ly () () ta c

(x01 a1k)A1 + (x02 a2k)A2 + ... + (x0m amk)Am + Ak = A0

Theo gi thit mi aik 0 nn vi mi > 0 ta lun c x0i aik 0, ngha l

X1 = (x01 a1k, x02 a2k,...,x0m amk, 0,...,,..., 0)

l mt phng n khc vi X0, vi mi > 0 ln ty . Khi ta c

f(X1) = f(X0) kV k > 0, > 0 ln ty , nn f(X1) b ty , iu chng t hm mc tiu khngb chn trn tp phng n. Vy suy ra iu phi chng minh.

nh l 3.1.3. Nu ti phng n cc bin X0, tn tik> 0 v aik > 0 th xydng c phng n cc bin mi X1 tt hn X0.

Chng minh. T chng minh ca nh l 1.5.1 cho ta thy rng X1 l phngn th cn c iu kin

x0i aik 0 0 < x0iaik

, aik > 0

Chn 0 = minaik>0

x0iaik

, gi s ti ch s i = 1, tc l 0 =x01

a1k> 0. Khi ta c phng

n mi X1 c dngX1 = 0, x

02 0a2k,...,x0m 0amk, 0,...,0,..., 0)

X1 c m ta dng v v h {A2,...,Am, Ak} c lp tuyn tnh, nn X1 l phngn c bin. ng thi lc ny

f(X1) = f(X0) 0kKhi f(X1) < f(X0), tc l X1 tt hn X0. Suy ra iu phi chng minh

39


40/94

3.2 Cng thc bin i bng n hnh

a) Cng thc bin i.

Vn t ra l:+ Ti phng n cc bin X0 ta suy ra c x0i , aij, j, f(X0).+ Ti phng n cc bin X1 ta cn tm cc i lng tng ng x1i , a1ij, 1j , f(X1).+ Chn

0 = minaik

x0iaik

=x0lalk

Khi x1 c xc nh bi cng thc

x1i =

x0l

alknu i = k

x0i

x0l

alkaik nu i

= k

(1.24)

f(X1) = f(X0) 0k (1.25)

Theo i s tuyn tnh khi i c s

a1ij =

alkalk

nu i = kaij aljalk aik nu i = k

(1.26)

1j = j aljalk

k = (mi=1

cia1ij cj) (1.27)

Ch : Ta thng chn k = max1

j>0

j th tm c phng n ti u nhanh hn.

b) Bng n hnh

Bng ACB H s b x1 x2 ... xj ... xnxi ci (P A) c1 c2 ... cj ... cn... ... ... ... ... ... ... ... ...

I xi ci bi ai1 ai2 ... aij ... ain... ... ... ... ... ... ... ... ...

f(X) 1 2 ... j ... n

trong + aij l phn t trc. Ti bng I tng ng vi phng n cc bin xut pht X0 ctxj c gi l ct xoay, hng xi c gi l hng xoay, vc t xj i vo thay th cho

vc t xj i ra.+ j =Cc phn t ct ci nhn tng ng vi cc phn t ct xj cng li c bao

40


41/94

nhiu tr i phn t u ct

V D 1 = (xi cc ptx1) c1

+ Cc phn t hng j bng mi = cc phn t hng i bng c chia cho phn t trc.+ f(X)= cc phn t ct ci nhn tng ng vi cc phn t ct bi ri sau cng li.+ Cc phn t hng i bng mi= cc phn t hng k bangr mi nhn vi (aik)(c),sau cng vi cc phn t hng i bng c.

3.3 Thut ton n hnh

Thut ton n hnh.Bc 1. Tm c s v phng n cc bin xut pht X0 ( chn c s n v).

Bc 2. Kim tra j 0, j = 1, 2,..n?.+ Nu c ta i n kt lun X0 l phng n ti u.+ Nu khng ta chuyn sang bc 3.Bc 3. Kim tra k > 0 m aik 0?+ Nu c th tr li bi ton khng c phng n ti u.+ Nu khng th chuyn sang bc 4.Bc 4. Chn k = max

j>0j, xk vo c s.

Bc 5. Chn 0 = minaik>0

b0iaik

=b0lalk

,xl ra khi c s.

Bc 6. Xy dng X1 theo cng thc (1.24) (1.27).b1 c xc nh bi cng thc

b1i =

b0l

alknu i = k

b0i b0l

alkaik nu i = k

(1.24)

f(X1) = f(X0) 0k (1.25)

Theo i s tuyn tnh khi i c s

a1ij =

alkalk

nu i = kaij aljalk aik nu i = k

(1.26)

1j = j aljalk

k = (mi=1

cia1ij cj) (1.27)

Gn X0 := X1 tr v bc 2.Ch :1. Nu gii bi ton Max bng phng php n hnh m khng a v bi

41


42/94

ton Min th ta cng lm tng t nh bi ton Min nh sau:Bc 1. Tm c s v phng n cc bin xut pht X0 ( chn c s n v).Bc 2. Kim tra j 0, j = 1, 2,..n?.+ Nu c ta i n kt lun X0 l phng n ti u.+ Nu khng ta chuyn sang bc 3.Bc 3. Kim tra k < 0 m aik 0?+ Nu c th tr li bi ton khng c phng n ti u.+ Nu khng th chuyn sang bc 4.Bc 4. Chn k = min

j0

x0iaik

=x0lalk

,xl ra khi c s.

Bc 6. Xy dng X1 theo cng thc (2.11) (2.14).Gn X0 := X1 tr v bc 2.

2. a) i vi bi ton Min+) Bi ton n hnh c phng n ti u duy nht nu j < 0 ti cc xj khngnm trong c s ca phng n ti u.+) Ngc li, nu c nhiu phng n ti u th j = 0 m xj ngoi c s.b) i vi bi ton Max+) Bi ton n hnh c c phng n ti u duy nht nu j > 0 ti cc xj khngnm trong c s ca phng n ti u.+) Ngc li, nu c nhiu phng n ti u th j = 0 m xj ngoi c s.c) Nu j = 0 m xj ngoi c s, th ta tm phng n ti u khc bng cch aj = 0 m xj ngoi c s vo c s,v a j = 0 m xl trong c s ra ngoi c sbng cch

Chn 0 = minaik>0

x0iaik

=x0lalk

3. Nu bi ton c 2 phng n ti u khc nhau th bi ton c v s phng n tiu.

V d 1. Gii bi ton QHTT sau:

M in(f(x) = x1

x2

+ x3

+ x4

+ x5

x6)

vi iu kin

x1 + x4 + 6x6 = 9

3x1 + x2 4x3 + 2x6 = 2x1 + 2x2 + x5 + 2x6 = 6

xj 0, j = 1, 2, ..., 6

Gii. Chn cc n c s(c bn){

x4, x2, x5

}l cc vc t n v nn ta c phng n

42


43/94

cc bin xut pht X0 = (0, 2, 0, 9, 6, 0) v c bng n hnh sau:

Bng ACB H s PA x1 x2 x3 x4 x5 x6xi ci bi 1

1 1 1 1

1

x4 1 9 1 0 0 1 0 6I x2 1 2 3 1 -4 0 0 2

x2 1 6 1 0 2 0 1 213 2 0 5 0 0 7

x4 1 3 -8 -3 12 1 0 0II x6 -1 1 3/2 1/2 -2 0 0 1

x5 1 4 -2 -1 6 0 1 06 25/2 7/2 19 0 0 0

x3 1 1/4 -2/3 -1/4 1 1/12 0 0

III x6 -1 3/2 1/6 0 0 1/6 0 1 x5 1 5/2 2 1/2 0 -1/2 1 05/4 1/6 5/4 0 19/12 0 0

x3 1 3/2 1/3 0 1 -1/6 1/2 0IV x6 -1 3/2 1/6 0 0 1/6 0 1

x5 1 5 4 1 0 -1 2 05 29/6 0 0 1/3 -5/2 0

Ti bng IV c j 0, j = 1,..,n nn ta c phng n ti u l X = (0, 5, 3/2, 0, 0, 3/2)v fmin =

5

V d 2.

M in(f(x) = x1 2x2 x3)

vi iu kin

x1 + x2 2x3 + x4 = 2x1 x2 + x3 + x5 = 1x2 + x3 + x6 = 5

2x1 x2 + x7 = 2xj 0, j = 1, 2, ..., 6, 7

Gii. Chn cc n c s {x4, x5, x6, x7} v phng n cc bin xut pht X = (0, 0, 0, 2, 1, 5, 2)

43


44/94

v bng n hnh.

Bng ACB H s PA x1 x2 x3 x4 x5 x6 x7xi ci bi 1 2 -1 0 0 0 0

x4 0 2 1 1 -2 1 0 0 0I x5 0 1 1 1 1 0 1 0 0x6 0 5 0 1 1 0 0 1 0x7 0 2 1 -1 0 0 0 0 0

0 1 2 1 0 0 0 0 x2 -2 2 1 1 -2 1 0 0 0

II x5 0 3 2 0 1 1 1 0 0 x6 0 3 -1 0 3 -1 0 1 0

x7 0 4 2 0 -2 1 0 0 1

4

3 0 5

2 0 0 0

x2 -2 4 1/3 1 0 1/3 0 2/3 0III x5 0 3 2 0 1 1 1 0 0

x3 0 4 5/3 0 0 2/3 1 1/3 0x7 0 6 4/3 0 0 1/3 0 2/3 1

9 4/3 0 0 1/3 0 5/3 0Ti bng III , j 0 v phng n ti u l

X = (0, 4, 1, 0, 4, 0, 6) , fmin = 9

V d 3. Gii bi ton QHTT sauMin(f(x) = x1 + x2 + x3)

vi iu kin

x1 x4 2x6 = 2x2 + x4 3x5 + x6 = 4x3 + 2x4 5x5 + 6x6 = 6xj 0, j = 1, 2, ..., 6

p s X = (5, 1, 0, 3, 0, 0), fmin = 6V d 4. Cho bi ton QHTT sau

Min(f(x) = 5x1 + 4x2 + 5x3 + 2x4 + x5 + 3x6)

vi iu kin

2x1 + 4x2 + 3x3 + x4 = 152

4x1 + 2x2 + 3x3 + x5 = 60

3x1 + x3 + x6 = 36

xj 0, j = 1, 2, ..., 6

44


45/94

a) Gii bi ton trn bng phng php n hnhb) Phng n ti u cu a) c phi l duy nht khng?Tm tp hp tt c cc phngn ti u ca bi tonc) Tm phng n ti u ca bi ton khi thay x1 = 6Gii.a) Chn cc n c s {x4, x5, x6} v phng n cc bin xut pht X = (0, 0, 0, 152, 60, 36)v bng n hnh.

Bng ACB H s PA x1 x2 x3 x4 x5 x6xi ci bi 5 4 5 2 1 3x4 2 152 2 4 3 1 0 0

I x5 1 60 4 2 3 0 1 0 x6 3 36 3 0 1 0 0 1

472 12 6 7 0 0 0x4 2 128 0 4 7/3 1 0 -2/3II x5 1 12 0 2 5/3 0 1 -4/3

x1 5 12 1 0 1/3 0 0 1/3328 0 6 3 0 0 4

x4 2 104 0 0 -1 1 -2 2III x2 4 6 0 1 5/6 0 1/2 -2/3

x1 5 12 1 0 1/3 0 0 1/3292 0 0 2 0 3 0

Ta thy vi j 0 nn phng n ti u ca bi ton l X2 = (12, 6, 0, 104, 0, 0) vfmin = 292

b) Nhn vo bng 3 ta thy 6 = 0 m x6 khng nm trong h c s (khng l n cbn) nn bi ton c nhiu phng n ti u. tm tp hp tt c cc phng n ti u khc ta lm nh sau: Tm 0 = min{104/2, 36} =36 > 0 cho x6 = [0, 36], x3 = x5 = 0. Khi tp phng n ti u ca bi ton l

X = X2 ai6 = (12 13

; 6 +2

3;0;104 2; 0; )

c) Khi x1 = 6 = 12 13 = 18, phng n ti u l X = (6, 18, 0, 68, 0, 18) vfmin = 292

Ch : Ta c th dng bng n hnh tm phng n ti u khc ca bi ton, l tm cc phng n ti u khc ta a x6 vo c s v x1 ra ngoi c s v 0 =min{104/2, 36} = 36 v phng n ny cng khng duy nht. Ta c thm bng n

45


46/94

hnh l

Bng ACB H s PA x1 x2 x3 x4 x5 x6xi ci bi 5 4 5 2 1 3

x4 2 32 -6 0 -3 1 -2 0IV x2 3 30 2 1 3/2 0 1/2 0x6 4 36 3 0 1 0 0 1

292 0 0 -2 0 -3 0

Phng n ti u l X = (0, 30, 0, 32, 0, 36) v fmin = 292

3.4 Thut ton n hnh m rng

C s l lun.Xt bi ton gc

M in {f(x) =n

j=1

cjxj}, (1.28)

vi iu kin

nj=1

aijxj = bi, i = 1, 2,...,m (1.29)

xj

0, j = 1, 2,...,n (1.30)

+ Nu bi ton trn ma c c s (x1, x2,..,xn) l c s n v cp m Tm phngn cc bin ban u x0i = bi+ Nu bi ton khng c c s n v cp m ta xt bi ton sau:

Min {f(x) =n

j=1

cjxj + Mmi=1

xi}, (1.31)

vi iu kin

nj=1

aijxj + xn+i = bi, i = 1, 2,...,m (1.32)

xj 0, j = 1, 2,...,n + m (1.33)

vi M > 0 ty .

Bi ton (1.31) (1.33) c gi l bi ton m rng(hay cn gi l bi ton M)n xn+i gi l n giBi ton (1.31) (1.33) c c s n v (xn+1, xn+2,..,xn+m). Khi pacb ban ul:

X0 = (0, 0, ..., 0, b1, b2,...,bm) = (0, X0)

v bi ton (1.31) (1.33) gii c bng phng php n hnh.

46


47/94

Mi quan h gia bi ton gc v bi ton m rng1) Nu bi ton M khng c phng n ti u(theo ngha hm mc tiu gim v hn),th bi ton gc cng khng c phng n ti u.2) Bi ton gc c phng n ti u X khi v ch khi bi ton M c phng n tiu. V n xy ra 3 trng hp sau:a)Nu trong h n c bn khng c n gi th thu c PATU ca bi ton gc bngcch b i m thnh phn ng vi n gi ca PATU ca bi ton Mb) Nu tn ti n gi trong h nhng chng bng 0 th bi ton gc c phng n tiu chnh l phng n ti u ca bi ton M m b i cc thnh phn ng vi n gi.c) Nu trong phng n ti u ca bi ton M c thnh phn ng vi n gi mxn+i > 0 th bi ton gc khng c phng n ti u.Ch .1) Trong bi ton M gi tr f(x) v j ca bng n hnh c chia lm 2 phn

j = j + Mj

chia hng j bi hai hngj

j jKhi xt ti j ta xt j trc, sau cn thit mi xt n j . V khi nh gi j tanh gi nh sau: V M ln ty nn

j = j + Mj > 0 j > 0 hoc j = 0 v j > 0j = j + Mj < 0 j < 0 hoc j = 0 v j < 0

2) TRong bng n hnh khng cn thit phi ghi cc ct ng vi cc n gi

3) Nu xt bi ton Max th h s ca cc n gi trong hm mc tiu ca bi ton mrng l (-M)V d 5. Gii bi ton

Min(f(x) = x1 2x2 + x3)vi iu kin

x1 + 4x2 2x3 + x4 = 6x1 + x2 + 2x5 x5 = 62x1

x2 + 2x3 = 4

xj 0, j = 1, 2, 3, 4, 5Ta thy bi ton cha c c s n v nn ta a thm n gi

Min(f(x) = x1 2x2 + x3 + Mx6 + Mx7)vi iu kin

x1 + 4x2 2x3 + x4 = 6x1 + x2 + 2x5 x5 + x6 = 62x1

x2 + 2x3 + x7 = 4

xj 0, j = 1, 2, 3, 4, 5, 6, 7

47


48/94

a bi ton (M) chn c s lin kt l (x4, x6, x7) suy ra pacb ban u X0 =(0, 0, 0, 6, 0, 6, 4). Ta c bng n hnh.

Bng ACB H s PA x1 x2 x3 x4xi ci bi 1 2 -1 0x4 0 6 -1 4 -2 1

I x6 M 6 1 1 2 0 x7 M 4 2 -1 2 0

j 0 1 2 -1 0j 10 3 0 4 0

x4 0 10 1 3 0 1II x6 M 2 1 2 0 0

x3 1 2 1 -1/2 1 0

j 2 2 3/2 0 0j 2 1 2 0 0 x4 1 7 5/2 0 0 1

III x2 2 1 1/2 1 0 0x3 1 5/2 3/4 0 1 0

j 1/2 11/4 0 0 0x1 -1 14/5 1 0 0 2/5

IV x2 2 12/5 0 1 0 1/5x3 1 2/5 0 0 1 -3/10

j

-36/5 0 0 0

11/10

Ti bng IV c j 0, (j = 1, ..5) nn phng n ti u l

X = (14/5, 12/5, 2/5, 0), fmin = 36/5

V d 6. Gii bi ton

Min(f(x) = 6x1 + 3x2 + x3)

vi iu kin

2x1 + 5x2 + x3 104x1 3x2 + 2x3 = 162x1 + 4x2 + x3 = 8

xj 0, j = 1, 2, 3

Ta a bi ton v dng chnh tc

Min(f(x) = 6x1 + 3x2 + x3)

48


49/94

vi iu kin

2x1 + 5x2 + x3 + x4 = 10

4x1

3x2 + 2x3 = 16

2x1 + 4x2 + x3 = 8

xj 0, j = 1, 2, 3, 4

Bi ton cha c h c s n v nn ta thm 2 n gi x5, x6. Lc ta c bi ton sau

Min(f(x) = 6x1 + 3x2 + x3 + M(x5 + x6))

vi iu kin

2x1 + 5x2 + x3 + x4 = 10

4x1 3x2 + 2x3 + x5 = 162x1 + 4x2 + x3 + x6 = 8

xj 0, j = 1, 2, 3, 4, 5, 6

Khi ta c c s n v l {x4, x5, x6} v phng n cc bin ban u l X0 =(0, 0, 0, 10, 16, 8). Gii bi ton bng bng n hnh sau

Bng ACB H s PA x1 x2 x3 x4xi ci bi 6 3 1 0x4 0 10 2 5 1 1

I x5 M 16 4 3 2 0 x6 M 8 2 4 1 0

j 0 6 3 -1 0j 24 6 1 3 0

x4 0 2 0 1 2 1II x5 M 0 0 11 0 0

x1 6 4 1 2 1/2 0j 24 0 9 2 0

j 0 0 11 0 0x4 0 2 0 1 0 1

III x5 M 0 0 11 0 0x3 1 8 2 4 1 0

j 8 4 1 0 0j 0 0 11 0 0

Ta thy ti bng 3 c j 0 nn phng n ti u l X = (0, 0, 8) v fmin = 8V d 7. Gii bi ton

Max(f(x) = 2x1 x2 + x3)

49


50/94

vi iu kin

4x1 x2 + 2x3 122x1 + 2x2 x3 10x1 2x2 1/2x3 = 20xj 0, j = 1, 2, 3

Ta a bi ton v dng chnh tc

Min(f(x) = 2x1 + x2 x3)vi iu kin

4x1 x2 + 2x3 x4 = 122x1 + 2x2 x3 + x5 = 10x1 2x2 1/2x3 = 20xj 0, j = 1, 2, 3, 4, 5

Bi ton cha c c s n v nn ta thm n gi x6, x7

Min(f(x) = 2x1 + x2 x3 + M(x6 + x7))vi iu kin

4x1 x2 + 2x3 x4 + x6 = 122x1 + 2x2 x3 + x5 = 10x1 2x2 1/2x3 + x7 = 20xj 0, j = 1, 2, 3, 4, 5, 6, 7

Khi c s n v l {x6, x5, x7} v phng n cc bin ban u l X0 = (0, 0, 0, 10, 12, 20).Ta c bng n hnh

Bng ACB H s PA x1 x2 x3 x4 x5xi ci bi 2 1 -1 0 0x6 M 12 -4 -1 2 -1 0

I x5 0 10 2 2 1 0 1

x7 M 20 1 -2 1/2 0 0j 0 2 1 1 0 0j 32 3 3 3/2 1 0

x3 1 6 -2 -1/2 1 -1/2 0I x5 0 16 4 3/2 0 -1/2 1

x7 M 23 0 -9/4 0 1/4 0j -6 0 1/2 0 1/2 0j 23 0 9/4 0 1/4 0

Phng n ti u ca bi ton M l X = (0, 0, 6, 0, 16, 0, 23) nhng trong phng n

ti u ca bi ton c n gi x7 = 23 > 0 nn bi ton gc cho khng c phngn ti u.

50


51/94

3.5 Bi ton quy hoch tuyn tnh cha tham s

V d 8. Gii bi ton quy hoch tuyn tnh sau theo tham s

Min(f(x) = 7x1 + 2x2 + ( 15)x3 + (10 )x4 2x5 4x6)vi iu kin

x1 4x3 + x4 + x6 = 22x1 + x2 4x3 2x6 = 2x1 3x3 + x5 + 2x6 = 3xj 0, j = 1, 2, ..., 6

Bng ACB H s PA x1 x2 x3 x4 x5 x6xi ci xi 7 2 15 10 -2 4x4 10 2 1 0 -4 1 0 1

I x2 2 2 2 1 -4 0 0 2x5 -2 3 -1 0 -3 0 1 2

18 2 9 0 3 27 0 0 6 x4 10 1 0 -1/2 -2 1 0 2

II x1 7 1 1 1/2 -2 0 0 1x5 -2 4 0 1/2 -5 0 1 1

9 09

2 +

2 9 + 0 0 15 2 x6 4 1/2 0 1/4 -1 1/2 0 1

III x1 7 3/2 1 1/4 -3 1/2 0 0x5 -2 7/2 0 3/4 -4 1/2 1 0

3/2 0 3/4 6 152

0 0

Ta xt bng I, ti 3 = 3 27, nu 3 27 > 0 > 9 th bi ton khng cphng n ti uNu = 9, ta c phng n ti u X = (0, 2, 0, 2, 3, 0) v fmin = 0Nu < 9, th ta chuyn sang bng II, y ta xt h iu kin: 2 =

9/2 + /2

0, 3 = 9 + 0, 6 = 15 2 0, suy ra 15/2 9. Khi ta c phng nti u l X = (1, 0, 0, 1, 4, 0) v fmin = 9 Nu < 15/2, th ta chuyn sang bc 3, y ta cng xt h iu kin 3 = 6 0, 4 = 15/2 + 0, suy ra 6 15/2. Khi ta c phng n ti u lX = (3/2, 0, 0, 0, 7/2, 1/2) v fmin = 3/2Nu < 6, bi ton khng c phng n ti u.Tm li:Nu > 9 hoc < 6 bi ton khng c phng n ti uNu 15/2 9 bi ton c phng n ti u l X = (1, 0, 0, 1, 4, 0); fmin = 9

Nu 6 15/2, bi ton c phng n ti u l X

= (3/2, 0, 0, 0, 7/2, 1/2) vfmin = 3/2

51


52/94


Min(f(x) = x1 + x2 + x3)

vi iu kin

x1 x4 2x6 = 2x2 + x4 3x5 + x6 = 4x3 + 2x4 5x5 + 6x6 = xj 0, j = 1, 2,..., 6

a) Vi = 1, = 6 gii bi ton bng phng php n hnh.b) K hiu patu cu a) l X0 ti ( = 6), coi l tham s. Tm tt c cc gi tr ca

X ti u.c) Ti va tm c cu b), coi l tham s, hy tm tt c cc gi tr ca c phng n ti u ca bi ton cho.Hng dn.a) Chn c s {A1, A2, A3} suy ra phng n cc bin xut pht l X0 = (2, 4, 6, 0, 0, 0).Ta c bng sau:

Bng ACB H s PA x1 x2 x3 x4 x5 x6xi ci xi 1() 1 1 0 0 0x1 1() 2 1 0 0 -1 0 -2

I x2 1 4 0 1 0 1 -3 1 x3 1 6() 0 0 1 2 -5 612 0 0 0 2 -8 5

x1 1() 4 1 0 1/3 -1/3 10/6 0II x2 1 3 0 1 -1/6 2/3 -13/6 0

x6 0 1(/6) 0 0 1/6 1/3 -5/6 17 0 0 5

313

236

0

x1 1() 5 1 0 3/6 0 -51/6 1III x2 1 1 1 0 3/6 0 -15/6 1

x4 0 3(/2) 0 0 3/6 1 -15/6 36 0 1 1 0 3 1

Phng n ti u l: X = (5, 1, 0, 3, 0, 0), fmin = 6.b) Ti bng I cf(X) = 10 + 2 , 4 = 3 , 6 = 7 2.Ti bng II cf(X) = 4 + 3, 3 = 7/6 /3, 4 = 23 , 5 = 13106 .Ti bng III cf(X) = 5 + 1, 3 =

9+36

, 5 =315

6, 6 =

2 +

52


53/94

pacb bng III l ti u th

3 05

0

6 0 1

5

2

c) ta xi l patu ca bi ton th bi 0 2 0 0V d 10. Tng t ta c bi ton l:

M ax(f(x) = 10x1 + 3x2 3x3 + x4 + 2x5 + 4x6)

vi iu kin

x1 + x2 x3 + x5 + 2x6 = 22x1 + x3 x4 + 2x5 + x6 = 38x1 + 6x3 6x4 + 3x5 + 4x6 = 18xj 0, j = 1, 2,..., 6

53


54/94

BI TP CHNG 3

Bi tp 1. Gii bi ton sau

Max(2x2 + x3)

vi iu kin

x1 + x2 45x1 + 2x2 + x3 4x2 5xj 0, j = 1, 2, 3

Bi tp 2. Gii bi ton sau:

Max(10x1 + 3x2 2x3)

vi iu kin

x1 + x2 x3 = 22x1 + x3 x4 = 38x1 6x3 + 6x4 = 19xj 0, j = 1, 2, 3, 4

Bi tp 3. Gii bi ton sau:Max(x1 + x2 + 3x3 + 2x4)

vi iu kin

x1 + 2x2 + x3 + 2x4 102x1 + x2 + 3x3 + 4x4 = 9

x1 + 2x2 + 2x3 + x4 8xj

0, j = 1, 2, 3, 4


Min(x1 x2 3x3 2x5)

vi iu kin

2x1 x2 x4 = 6x3 2x4 + 2x5 = 43x1

x4 + x5

10

xj 0, j = 1, 2, 3, 4, 5

54


55/94


Max(2x1 + 3x2 + 4x3 + x4)

vi iu kin

x1 + x2 + x3 + x4 52x1 + 2x2 + 3x3 = 18

2x1 + x2 + 2x4 8xj 0, j = 1, 2, 3, 4


M ax(x1 2x2 + x3)

vi iu kin

x1 + 3x2 x3 42x2 + 3x3 = 9

x1 x2 + x3 3xj 0, j = 1, 2, 3

Bi tp 7. Cho bi ton sau:

Max(2x1 x2 + 6x3 + 4x4)

vi iu kin

x1 + 2x2 4x3 x4 2x1 + x2 + 2x3 + 6x4 13x1 + x2 + x3 + 5x4 12xj 0, j = 1, 2, 3, 4

a) Gii bi ton trn.b) Chng minh bi ton c v s phng n ti u, tm tp hp tt c cc phng n.c) Tm phng n ti u ca bi ton sao cho v tri ca rng buc (3) ng bng 4.Bi tp 8. Cho bi ton sau:

Min(4x1 + x2 + x3 + x4)

vi iu kin

2x2 + x3 + x4 = 16

4x2 + 2x5 + x6 = 8

x1 + x2 + 2x4 + x5 = 2

xj 0, j = 1, 2, 3, 4, 5

55


56/94

a) Gii bi ton trn.b) Chng minh bi ton c v s phng n ti u, tm tp hp tt c cc phng n .

Bi tp 9. Cho bi ton sau:Max(x1 3x2 4x3 + x4 + 5x5)

vi iu kin

2x1 + x2 3x3 + 2x4 = 30x2 x3 + x4 x5 = 233x1 2x2 + x3 + x4 + 4x5 x6 = 10xj 0, j = 1, ..., 6

a) Gii bi ton trn.b) Chng minh bi ton c v s phng n ti u, tm tp hp tt c cc phng n .


Min(3x1 + x2 + 7x3 + 5x4)vi iu kin

x1 + 2x2 3x3 + x4 37

x1

x2

+ x3

+ 3x4

=

5

2x1 + 3x2 x3 x4 25xj 0, j = 1, ..., 4

a)Chng t X0 = (14, 0, 0, 3) l phng n cc bin.b) Gii bi ton trn.c) Cho kt lun v bi toan Max.


M in(x1 2x2 3x3 + x4)vi iu kin

x1 + 2x2 + 3x3 = 12

2x1 + x2 + 5x3 = 10

x1 + 2x2 + x3 + x4 = 20

xj 0, j = 1, ..., 4a)Chng t X0 = (0, 5, 4, 6) l phng n cc bin.

b) Gii bi ton trn xut pht t pacb X0.

56


57/94

Bi tp 12. Gii v bin lun bi ton theo tham s .a.

M in(M ax)(x1 + 12x2 + (4

)x3 + 10x4)

vi iu kin

2x1 + 3x2 + x3 + x4 402x1 + x2 + 3x3 + 2x4 = 20

x1 + 2x2 + x3 + 2x4 24xj 0, j = 1, ..., 4

b.

Min(M ax)(2x1 + x2 + (3 )x3)

vi iu kin

x1 + 2x2 + 3x3 122x1 + 3x2 + x3 92x1 + x2 + 2x3 15xj 0, j = 1, 2, 3

c.Min(Max)((5 + 2)x1 + (3 )x2 + (3)x3)

vi iu kin

2x1 + 3x2 + x3 6x1 + 2x2 + 2x3 82x1 + x2 + 2x3 4xj

0, j = 1, 2, 3

57


58/94

Chng 4

Bi ton quy hoch tuyn tnh ingu

4.1 Cc khi nim v bi ton i ngu

4.1.1 Cp bi ton i ngu i xng

Cho bi ton QHTT (P)

Min

{f(x) =

n

j=1

cjxj

}

vi iu kin

nj=1

aijxj bi, i = 1, 2,...,m

xj 0, j = 1, 2,...,n

Ta ni bi ton (P) gi l bi ton gc.Bi ton (D) sau s c gi l bi ton i ngu ca bi ton gc.

M ax {g(y) =mi=1

biyi}

vi iu kin

mi=1

aijyi cj, j = 1, 2,...,n

yi 0, i = 1, 2,...,mBi ton (P) cng bi ton (D) to thnh cp bi ton i ngu.Bi ton (P) hoc (D) c th vit c di dng ma trn

M in

{CX : AX

b, X

0

}Max{Y b : Y A c, Y 0}

58


59/94

V d 1. Cho bi ton (P)

Min(f(x) = 2x1 x2 + x3)

vi iu kin

x1 + 2x2 x3 22x1 + x2 + 2x3 5xj 0, j = 1, 2, 3

Tm bi ton i ngu (D) v ch ra cc cp iu kin i nguBi ton i ngu (D) l

M ax(g(y) = 2y1 + 5y2)

vi iu kin

y1 + 2y2 22y1 y2 1y1 + 2y2 1y1, y2 0,

Cc cp iu kin i ngu l

x1 0 y1 + 2y2 2x2 0 2y1 y2 1x3 0 y1 + 2y2 1y1 0 x1 + 2x2 x3 2y2 0 2x1 + x2 + 2x3 5

4.1.2 Cp bi ton i ngu khng i xngCho bi ton gc (P) l dng chnh tc

M in {f(x) =n

j=1

cjxj},

vi iu kin

nj=1

aijxj = bi, i = 1, 2,...,m

xj 0, j = 1, 2,...,n

59


60/94

th bi ton i ngu (D) l:

M ax {g(y) =m

i=1biyi}

vi iu kin

mi=1

aijyi cj, j = 1, 2,...,n

yi R

Cc cp bt phng trnh tng ng xj 0 vmi=1

aijyi cj, j = 1, 2,...,n c gi l

cc cp iu kin i nguCh :1) i ngu ca bi ton i ngu l bi ton gc2) Nu bi ton gc (P) cho di dng

M ax {f(x) =n

j=1

cjxj},

vi iu kin

nj=1

aijxj = bi, i = 1, 2,...,m

xj 0, j = 1, 2,...,n

th bi ton i ngu (D) l:

M in {g(y) =mi=1

biyi}

vi iu kin

mi=1

aijyi cj, j = 1, 2,...,n

yi R

Cc cp iu in i ngu tng ng s l xj

0 vm

i=1

aijyi

cj, j = 1, 2,...,n

V d 2. Cho bi ton (P)

Min(f(x) = 13x1 3x2 4x3 + 19x4)

vi iu kin

2x1 + x2 x3 + 3x4 = 44x1 + 2x2 + 3x3 + x4 = 233x1

x2 + x3 + 6x4 = 96

xj 0, j = 1, 2, 3, 4

60


61/94

Hy vit bi ton i ngu (D) ca bi ton (P) v ch ra cp iu kin i nguBi ton i ngu (D) s l

Max(g(y) = 44y1 + 23y2 + 96y3)

vi iu kin

2y1 y2 + 3y3 13y1 + 2y2 y3 3y1 + 3y2 + y3 43y1 + y2 + 6y3 19


x1

0

2y1

y2 + 3y3

13

x2 0 y1 + 2y2 y3 3x3 0 y1 + 3y2 + y3 4x4 0 3y1 + y2 + 6y3 19

4.1.3 S i ngu tng qut

Bi ton gc (i ngu) Bi ton i ngu (Gc)X = (x1, x2,...,xn) Rn Y = (y1, y2,...,ym) Rmf(x) =

nj=1

cjxj min f(x) =mi=1

biyi maxn

j=1

aijxj = bi, i I1 yj khng ph thuc vo du (i I1)n

j=1

aijxj bi, i I2 yi 0(i I2)n

j=1

aijxj

bi, i

I3 yi

0(i

I3)

xj khng rng buc du (j J1)mi=1

aijyi = cj , j J1

xj 0 (j J2)mi=1

aijyi cj, j J2

xj 0 (j J3)mi=1

aijyi cj, j J3

trong

I1 = {1, ...m1}, I2 = {m1 + 1,..m2}, I3 = {m2 + 1,..m}, I1 I2 I3 = {1, ...m}J1 = {1, 2,..n1}, J2 = {n1 + 1,..n2}, J3 = {n2 + 1,..n}, J1 J2 J3 = {1, ...n}

61


62/94

4.2 Tnh cht ca cp bi ton i ngu

4.2.1 Cc nh l i ngu

nh l 4.2.1. Cho X l phng n ca bi ton (P) v Y l phng n ca bi ton(D) th khi ta c f(X) g(Y)Chng minh.

nh l 4.2.2. Cho X, Y ln lt l phng n ca bi ton (P), (D) v tha mnf(X) = g(Y) th X l phng n ti u ca (P) vY l phng n ti u ca (D)Chng minh.

nh l 4.2.3. (nh l i ngu).1) Nu mt trong hai bi ton ca cp bi ton i ngu c patu th bi ton kia cngc phng n ti u, ng thi min f(X) = max g(Y).2) Nu hm mc tiu ca bi ton ny khng b chn trn tp phng n th bi tonkia c tp phng n bng rng.Chng minh.

H qu:Nu tp phng n ca bi ton gc v bi ton i ngu khc rng th chng u cphng n ti u.

nh l 4.2.4. ( nh l lch b yu)Phng n X, Y ca bi ton i ngu ti u khi v ch khi.

a)mi=1

aijy

i = cj, nuxj > 0, (x

j < 0)

(hoc xj = 0, numi=1

aijyi < cj, (

mi=1

aijyi > cj)).

b)n

j=1

aijx

j = bi, nu yi > 0, (y

i < 0)

(hoc y

i = 0 nu

n

j=1 aijx

j > bi (

n

j=1 aijx

j > bi))Chng minh.

nh l 4.2.5. (nh l lch b).

Phng n X, Y l phng n ti u khi v ch khimi=1

aijy

i = cj, nuxj > 0 (hoc

xj = 0, numi=1

aijy

i < cj).

Chng minh.

62


63/94

4.2.2 Cch gii bi ton i ngu

T cc nh l i ngu v t li gii ca bi ton gc ta c th tm c li giica bi ton i ngu.

1. Nu bi ton gc(P) khng c phng n ti u th bi ton i ngu (D) cngkhng c phng n ti u.2. Nu bi ton gc c phng n ti u th bi ton i ngu cng c phng n tiu v gi tr ti u ca hai bi ton bng nhau.3. Phng n ti u ca bi ton i ngu tm c t phng n ti u X0 ca biton gc nh nh l lch b yu nh sau:Bc 1. Lp h phng trnh ti u ca bi ton i ngu.

Th phng n ti u X0 ln lt vo cc bt ng thc ca bi ton gc: Nurng buc no tha mn X0 vi du bt ng thc tht s th rng buc i ngu ca

n trong bi ton i ngu phi tha mn vi du bng, ngha l rng buc trthnh 1 phng trnh. T ta suy ra c 1 h phng trnh theo cc bin ca biton i ngu, ta gi h phng trnh ny l h phng trnh ti u ca bi ton ingu.Bc 2. H phng trnh ti u l mt h phng trnh tuyn tnh nn ta s dng ccphng php gii h phng trnh tuyn tnh tm nghim ca h phng trnh tiu.Bc 3. Tm phng n ti u ca bi ton i ngu.Th nghim ca h phng trnh ti u tha mn cc rng buc cn li ca bi ton

i ngu l tp phng n tri u ca bi ton i ngu.V d 3. Cho bi ton quy hoch tuyn tnh sau

Min(f(x) = 3x1 + 4x2 + x3)

vi iu kin

3x1 + 2x2 4x3 152x1

x2

5x3

8

4x1 + 2x2 + 2x3 10x1, x1 0, x3 0

Cho bit bi ton trn c phng n ti u l X0 = (7, 0, 9). Hy lp v gii biton i ngu ca bi ton trn.

Gii

Bi ton i ngu s l

Max(g(y) = 15y1 + 8y2 + 10y3)

63


64/94

vi iu kin

3y1 + 2y2 + 4y3 32y1

y2 + 2y3

4

4y1 5y2 + 2y3 1y1, y2, y3 0,

Do bi ton cho c phng n ti u l X0 = (7, 0, 9) nn theo lch b yu tac

y2 = 0( do rng buc 2 ca bt gc ln thc s)

3y1 + 2y2 + 4y3 = 3

4y1 5y2 + 2y3 = 1

Gii h phng trnh trn ta c (y1 = 1/5, y2 = 0, y3 = 9/10) nghim ny tha mncc rng buc cn li ca bi ton i ngu nn bi ton i ngu c li gii ti u lY = (1/5, 0, 9/10), g(Y) = 12.

V d 4. Tm phng n ti u ca bi ton i ngu. Bit bi ton tom gc cphng n ti u X = (0, 7/2, 1) v c dng

Min(f(x) = 12x1 + 8x2 + x3)

vi iu kin

x1 + 2x2 + x3 = 8

2x1 + 2x2 + 3x3 102x1 + 3x2 + 3/2x3 = 12

xj 0, j = 1, 2, 3

GiiBi ton i ngu:

max(g(y) = 8y1 + 10y2 + 12y3)

vi iu kin

y1 + 2y2 + 2y3 122y1 + 2y2 + 3y3 8y1 + 3y2 + 3/2y3 1y2 0

V: x2 = 7/2 > 0 2y1 + 2y2 + 3y3 = 8

64


65/94

x3 = 1 > 0 y1 + 3y2 + 3/2y3 = 1Khi thay vo h cho th u tha mn du bng bi ton gc nn khng suy ray2 = 0.

Vy gii h phng trnh tao bi 2 phng trnh trn theo y3 ta cy1 =

11 3y32

, y2 = 3/2, y3 R

Thay y1, y2 vo bt phng trnh th 3 ta suy ra c y3 19. Khi ta c phngn

Y = (11 3y3

2, 3/2, y3)

Thay vo hm mc tiu ca bi ton i ngu ta c f(X) = g(Y) = 29.Vy phng n ti u ca bi ton i ngu l

Y = ( 11 3y32 , 3/2, y3); y3 19V d 5. Cho bi ton quy hoch tuyn tnh sau y:

Max(f(x) = 15x1 + 12x2 + 16x3 + 10x4)

vi iu kin

x1 + 3x2 + x3 + x4 202x1 + x2 + 3x3 + 2x4 = 20

x1

+ 2x2

+ x3

+ 2x4

24

xj 0, j = 1, 2, 3, 4a) Gii bi ton trn bng phng php n hnhb) Vit bi ton i ngu v xc nh cc cp rng buc i nguc) Tm phng n ti u ca bi ton i ngu bng nh l lch b yu

Giia) Dng bng n hnh gii bi ton trn, ta c phng n ti u

X = (0, 4, 0, 8), fmax = 128

b) Bi ton i ngu l

Min(g(y) = 20y1 + 20y2 + 24y3)

vi iu kin

y1 + 2y2 + y3 153y1 + y2 + 2y3 12y1 + 3y2 + y3 16y1 + 2y2 + 2y3

10

y1 0, y3 0

65


66/94


x1 0 y1 + 2y2 + y3 15x2

0

3y1 + y2 + 2y3

12

x3 0 y1 + 3y2 + y3 16x4 0 y1 + 2y2 + 2y3 10y1 0 x1 + 3x2 + x3 + x4 20y3 0 x1 + 2x2 + x3 + 2x4 24

c) Theo nh l d lch b yu ta cDo x2 = 4 > 0 nn suy ra 3y1 + y2 + 2y3 = 12Do x4 = 8 > 0 nn suy ra y1 + 2y2 + 2y3 = 10

Thay X vo h rng buc gc u xy ra du bng.Nh vy ta gii h 2 phng trnhtrn theo y1 ta c

3y1 + y2 + 2y3 = 12

y1 + 2y2 + 2y3 10y1 R

y2 = 2 + 2y1y3 = 7 (5/2)y1y1 R

Thay y2, y3 vo cc bt phng trnh cn li ta suy ra c y1 24/5. Sau ta lithay Y = (y1, 2 + 2y1, 7 (5/2)y1) vo g(y), ta c g(Y) = f(X) = 128. Vy tpphng n ti u ca bi ton i ngu l

Y = (y1, 2 + 2y1, 7 (5/2)y1), g(Y) = 128

4.3 Phng php n hnh i ngu

tng chung: Ta s dng bi ton i ngu gii bi ton gc, gii bi toni ngu nhng kt qu th hin trn bi ton gc, vic gii bi ton i ngu cm t trn bng n hnh v gi l phng php n hnh i ngu.

4.3.1 C s l lun ca phng php

Xt bi ton gc

M in(CX)

vi iu kin

AX = bX 0

66


67/94

Ta c bi ton i ngu

M ax(Y b)

vi iu kin Y A CY 0

Xut pht t c s B = (xj)jI no ca ma trn A.K hiu C = (cjI). Xt Y = CB1 l phng n ti u ca bi ton i ngu vij 0, j = 1, ...n, trong j = CXj Cj.

nh l 4.3.1. Nu ti phng n Y = C

B1 m X = B

1b 0, th X l phngn ti u ca bi ton gc (v Y l phng n ti u cu bi ton i ngu)

Ngc li, iu kin ca nh l trn cha c tha mn, trong trng hp nyta gi X = B1b l 1 gi phng n ti u ( c j 0).

K hiu i l hng i ca ma trn B1. (Ch rng khi B1 th hng i cng cdng vc t n v)

nh l 4.3.2. Y = Y i s l phng n ca bi ton i ngu nu tha mn

0 v aij

j ,

j

nh l 4.3.3. Nu ti phng n Y = CB1 (tc l c j 0) c xl < 0 (c trongX = B1) v aij 0, j = 1, 2,..n th hm mc tiu ca bi ton i ngu khng bchn trn tp phng n (bi ton gc khng c phng n ti u)

nh l 4.3.4. Nu ti phng n Y = CB1, tn ti xl < 0 v tn ti alj < 0 thxy dng c phng n mi Y tt hn Y.

4.3.2 Thut ton n hnh i ngu

Bc 0. Xy dng gi phng n ban u.

Xut pht t h m vc t c lp tuyn tnh c B = (xj), vi j J, |B| = m, saocho j 0, j = 1, 2,..,n.Tm X = (X, 0) trong X = CB1, xj = 0, j / J.

Bc 1. Kim tra tnh ti u: xi 0, i J?+ Nu c, bi ton c phng n ti u l X.

+ Ngc li, chuyn sang bc 2.

Bc 2. Kim tra tnh v nghim. Ti xi < 0, kim tra aij

0,j = 1, 2,..n?.

+ Nu c bi ton khng c phng n ti u.

67


68/94

+ Ngc li, chuyn sang bc 3.

Bc 3. Xc nh vc t ra khi c s

Tn ti xi < 0, aij < 0Chn xl = minxi


69/94

Bc 1. Tm c s {x3, x6, x2}.Bc 2. Ta thy khng c xj 0, j I.Bc 3. xl < 0 v xij < 0 v x6 ra khi c s.Bc 4. Chn phn t trc alk

= minalj


70/94

max(g(y) = 2y1 + 4y2 + y3 + y4)

vi iu kin

y1 + 3y2 + y3 + y4 12y1 + y2 3y2 + 4y3 + y4 3yi 0, i = 1...4

Bi ton dng chnh tc

max(g(y) = 2y1 4y2 y3 y4)vi iu kin

y1 + 3y2 + y3 + y4 + y5 = 1

2y1 + y2 + y6 = 3

y2 + 4y3 + y4 + y7 = 3

yi 0, i = 1...7

S dng phng php n hnh gii bi ton trn vi c s n v l {x5, x6, x7}phng n cc bin xut pht Y0 = (1, 1, 3) Khi gii ra ta c phng n ti ul Y = (1, 0, 3/4, 0) gmax = 11/4 Theo nh l lch b ta cV y1 = 1 > 0 nn x1 + 2x2 = 2 (1)y3 > 0 nn 4x3 = 1 (2)v 2y1 + y2 = 2 < 3 nn x2 = 0 (3)T (1), (2) (3) ta c X = 2, 0, 1/4 v fmax = 11/4

V d 8. Gii bi ton QHTT sau y bng cch gii bi ton i ngu ca n

Min(f(x) = 28x1 + 24x2 + 12x3)

vi iu kin

2x1 + x2 + x3 12x1 + 3x2 + x3 15x1 + x3 63x1 + 2x2 9xj 0, j = 1, 2, 3

70


71/94

BI TP CHNG 4

Bi tp 1. Cho bi ton quy hoch tuyn tnh sau

Min(f(x) = 2x1 + 5x2 + 4x3 + x4

5x5)

vi iu kin

x1 6x2 2x4 9x5 = 322x2 + x3 + 1/2x4 + 3/2x5 = 30

3x2 + x5 363x1 + 2x2 9xj 0, j = 1, 2, 3, 4, 5

a) Gii bi ton trn bng phng php n hnhb) Tm phng n ti u ca bi ton i ngu


Max(f(x) = 2x1 + 6x2 + 4x3 2x4 + 3x5)

vi iu kin

x1 + 2x2 + 4x3 = 52

4x2 + 2x3 + x4 = 60

3x2 + x5 = 36

xj 0, j = 1, 2, 3, 4, 5

a) Gii bi ton trn bng phng php n hnhb) Tm phng n ti u ca bi ton i ngu


Min(f(x) = 5x1 9x2 + 15x3 + 7x4 + 6x5)

vi iu kin

x1 + 3x2 x3 x4 + x5 14x1 + x3 + 2x4 x5 = 4x1 x2 + x3 2x5 1xj 0, j = 1, 2, 3, 4, 5

a) Hi X0 = (0, 1, 0, 2, 0) c phi l phng n cc bin khng?b) Vit bi ton i ngu ca bi ton trn, tm tt c cc phng n ca bi ton inguc) Nu thay c = 10/3 th c kt lun g cho X0?

71


72/94


M in(f(x) = 3x1 2x2 2x3 + 2x4 + x5)

vi iu kin

2x1 x3 + x4 4x5 155x1 + 3x2 + 2x3 + 3x5 = 0x2 + x3 3x4 8xj 0, j = 1, 2, 3, 4, 5

a) Chng t bi ton trn lun c phng n ti u vi mi vc t bb) Vit bi ton i ngu ca bi ton trn, tm tt c cc phng n ca bi ton ingu

c) Khi f(x) M ax th bi ton c phng n ti u hay khng?Bi tp 5. Cho bi ton quy hoch tuyn tnh sau

M ax(f(x) = 3x1 2x2 + x3 3x4)vi iu kin

x1 + x2 + 2x3 3x4 10x1 4x2 + x3 3x4 = 113x1

x3 + 2x4

5

xj 0, j = 1, 2, 3, 4

a) Vit bi ton i ngu v ch ra cc cp iu kin i ngub) Gii bi ton gc v bi ton i nguc) Nu thay rng buc (1) bi du "=" th cu tra li ca cp bi ton trn snh th no?


M ax(f(x) = 3x1 2x2 + x3 3x4)

vi iu kin

x1 + x2 + 2x3 3x4 = 10x1 4x2 + x3 3x4 = 113x1 x3 + 2x4 5xj 0, j = 1, 2, 3, 4

a) Chng minh bi ton i ngu c v s phng n ti u, bit X = (4, 0, 7, 0) l 1phng n ti u ca bi ton gc

b) Tm mt phng n ti u ca bi ton i ngu lm cho v tri ca rng buc tht ng bng 1. Phng n c phi l phng n cc bin hay khng?

72


73/94


Max(f(x) = 7x1 3x2 4x3 5x4)

vi iu kin

x1 + 2x2 + 2x3 2x1 + x2 + 3x3 + x4 62x1 + x2 + x3 = 23

xj 0, j = 1, 2, 3, 4

a) Vit bi ton i ngu v ch ra cc cp iu kin rng bucb) Tm mt phng n ti u ca cp bi ton i ngu

Bi tp 8. Cho bi ton quy hoch tuyn tnh sauMin(f(x) = x1 + x2 + x3)

vi iu kin

4x1 2x2 + 5x3 + x4 8x1 + 3x2 + x3 + 2x4 213x1 x2 + 3x3 + x4 = 8xj

0, j = 1, 2, 3, 4

a) Vit bi ton i ngu v ch ra cc cp iu kin rng bucb) Tm mt phng n ti u ca bi ton gc sau suy ra patu ca bi ton ingu


Min(f(x) = 4x1 + 6x2 5x3 + 2x4 3x5)

vi iu kin

3x1 + 2x2 x3 + x5 = 20x1 x2 + x4 + x5 142x2 3x3 + x4 2x5 = 8xj 0, j = 1, 2, 3, 4, 5

a) Vit bi ton i ngu v ch ra cc cp iu kin rng bucb) Tm mt phng n ti u ca bi ton gc sau suy ra patu ca bi ton inguc) Phng n ti u va tm c c duy nht hay khng? Nu khng duy nht hytm phng n ti u cc bin th hai.

73


74/94


Min(f(x) = 5x1 2x2 x3 x4 + x5)vi iu kin

2x1 + 3x2 x3 + x4 232x1 + x2 + 2x3 + x5 = 11x1 + 4x3 + 2x5 18xj 0, j = 1, 2, 3, 4, 5

a) Vit bi ton i ngu v ch ra cc cp iu kin rng bucb) Tm mt phng n ti u ca bi ton gc sau suy ra patu ca bi ton ingu

c) Phng n ti u va tm c c duy nht hay khng? Nu khng duy nht hytm phng n ti u khng cc bin.


Min(f(x) = 3x1 + 5x2 + 2x3 + 3x4)

vi iu kin

3x1 + 2x2 + 4x3 + 9x4 82x1 + x2 + 3x3 + 7x5 = 6

x1 + 2x2 + 3x3 + 5x4 6xj 0, j = 1, 2, 3, 4

a) Vit bi ton i ngu v ch ra cc cp iu kin rng bucb) Tm mt phng n ti u ca bi ton gcc) Chng t cc vc t sau X0 = (13 , 0, 1,

13

) v Y0 = (5, 6, 0) ln lt l phng nti u ca bi ton gc v bi ton i ngu


M in(f(x) = 3x1

x2 + 4x3

5x4)

vi iu kin

2x1 + x2 + x3 4x4 33x1 2x2 4x3 + 2x4 2x1 2x2 + 3x3 4x4 = 31xj 0, j = 1, 2, 3, 4

a) Vit bi ton i ngu v ch ra cc cp iu kin rng bucb) Tm mt phng n ti u ca cp bi ton ni trn

c) Chng minh bi ton i ngu ch c 1 phng n

74


75/94

Chng 5

Bi ton vn ti

5.1 Bi ton vn ti tng qut5.1.1 Xy dng bi ton

Bi ton t raGi s cn vn chuyn 1 loi hng t m ni pht A1, A2,...,Am n n ni thu B1, B2,...,Bn,ni pht thAi c s lng hng ai(i = 1, , , m). Ni nhn thBj cn s lng hngl bj(j = 1, ...n). Cc ph vn chuyn mt n v hng tAi n Bj l cij. Nn vnchuyn nh th no m bo pht ht v nhn s hng m c tng cc ph vn

chuyn b nht.M hnh bi ton.

Gi xij l s lng hng cn vn chuyn t Ai n Bj v xij 0, ta cng c cc iukin rng buc sau:

- Tng lng hng pht l:n

j=1

xij = ai, i = 1,...,m

- Tng lng hng t cc trm pht n trm thu l:mi=1

xij = bi, j = 1,...,n

- Tng chi ph vn chuyn l: Z = Min(

mi=1

nj=1

xij)

Nh vy ta c bi ton sau:Tm xij , i = 1, m , j = 1, n

Min(mi=1

nj=1

xij) (3.1)

75


76/94

vi iu kin

n

j=1xij = ai, i = 1,...,m (3.2)

mi=1

xij = bj, j = 1,...,n (3.3)

xij 0, cij 0, bj, ai 0, i = 1, 2...m, j = 1, 2...n (3.4)

ng thi tha mn iu kin cn bng thu pht, ngha l:mi=1

ai =n

j=1

bj

Bi ton vn ti l mt bi ton quy hoch tuyn tnh v vy c th dng thut ton

n hnh gii nhng do cu to c bit ca n nn ta c th c mt phng phpgii n gin v thun tin hn, n th hin bng sau:

Thu b1 b2 ... bj ... bnPht

a1 c11 c12 ... c1j ... c1nx11 x12 ... x1j ... x11

... ... ... ... ... ... ...ai ci1 ci2 ... cij ... cin

xi1 xi2 ... xij ... xin... ... ... ... ... ... ...am cm1 cm2 ... cmj ... cmn

xm1 xm2 ... xmj ... xmn

- (i, j) c xij > 0 gi l chn, cc khng chn gi l loi.- Mt phng n cc bin(c bn) gi l khng suy bin, nu c m+n-1 chn.- Mt dy cc lin tip sao cho 2 vit lin nhau th cng hng hoc cng ct.- Khng c 3 no vit cng 1 hng hoc cng 1 ct th c gi l 1 dy chuyn. Mtdy chuyn khp kn c gi l 1 vng v k hiu l V.

Cc vng thng gp

5.1.2 Tnh cht ca bi ton vn ti

Tnh cht 1. Bi ton vn ti lun c phng n ti u.

H qu 3. S chn trong mt phng n cc bin ca bi ton vn ti l m+n-1.Tnh cht 2. iu kin cn v h cc vct ct ca ma trn cc h s ca biton vn ti c lp tuyn tnh l cc tng ng khng to thnh vng.

H qu 4. Mt phng n cc bin ca bi ton vn ti l phng n cc bin khi vch khi cc tng ng vi cc thnh phn dng khng to thnh vng.

76


77/94

H qu 5. Cho phng n cc bin X vi tp hp H gm m+n-1 chn khng lpthnh vng (l, k) / H. Khi tp hp

H1 = H

{(l, k)

}s lp thnh vng V duy nht, v ng thi, nu (p, q) V thH2 = V \ {(p, q)} to thnh m+n-1 khng lp thnh vng.

5.2 Phng php tm phng n cc bin xut pht

5.2.1 Phng php cc tiu cc ph

Bc 1. Chn c cc ph nh nht v phn phi mt lng hng ti a c thc c vo .

Bc 2. Xa cc trn hng hoc ct ht kh nng phn phi v tr v bc 1vi nhng cha b xa.

V d 1.

Thu 25 80 120 45 30Pht

70 7 2 9 12 6

70

85 8 6 4 3 9 40 45

35 5 3 6 7 11

10 25 110 11 5 10 8 1

25 55 30

+ Ta thy c45 = 1 l nh nht nn ta phn phi mt lng ti a l 30 ct 5. Dovy hang 4 d 80, xa tt c cc hng ct 5 bi du ().

+ Tng t ta ta c bng trn.

5.2.2 Phng php gc Ty -Bc

Bc 1. Phn phi 1 lng hng ti a c th c vo gc ty bc ca bng.

Bc 2. Xa cc trn hng hoc ct ht kh nng phn phi v tr li bc1 vi nhng cha b xa.

77


78/94

V d 2.

Thu 25 80 120 45 30Pht

70 7 2 9 12 625 45

85 8 6 4 3 9

35 50 35 5 3 6 7 11

35 110 11 5 10 8 1

35 45 30

5.2.3 Phng php F- GhenL phng php xy dng phng n cc bin xut pht tt nht v n rt gn vi

phng n ti u ca bi ton v gim i c mt s hu hn bc lp.

Bc 1. Chn c cc ph nh nht trong 1 hng hoc ct c lch cc ph lnnht theo ngha lch cc ph ca mt hng hoc ct l hiu s gia cc ph thpnht v cc ph thp nh trong hng hoc ct .

Bc 2. Phn phi mt lng hng ti a c th vo c cc ph b nht nu bc 1.

Bc 3. Xa hng hoc ct ht kh nng phn phi v tr li bc 1 vi nhng cha b xa.

Ch : Nu bc 1 c nhiu hng hoc ct c lch cc ph ln nht nh nhauth u tin phn phi ti a c th c vo c cc ph b nht trong cc hng hocct y.

V d 3.

Thu 25 80 120 45 30Pht

70 7 2 9 12 6 4,5 70

85 8 6 4 3 9 1,1,1,2,2 40 45

35 5 3 6 7 11 2,2,2,2,125 10

110 11 5 10 8 1 4,3,3,5,1 10 70 30

2,2,3,3,3 1,1,2,2 2,2,2,2,2 4,4,4 5

78


79/94

5.3 Thut ton th v

5.3.1 C s l lun ca thut ton

Xt bi ton

min(mi=1

nj=1

xij) (3.1)

vi iu kin

nj=1

xij = ai, i = 1,...,m (3.2)

mi=1

xij = bj, j = 1,...,n (3.3)

xij 0, cij 0, bj, ai 0, i = 1, 2...m, j = 1, 2...n (3.4)

Ta c bi ton i ngu ca bi ton trn l:

max(mi=1

aiui +n

j=1

bjvj)

vi iu kin ui + vj ciji = 1, 2...m, j = 1, 2...n

t ij = ui + vj cij .nh l 5.3.1. (Du hiu ti u).Nu ti phng n cc bin X. ui, vj sao cho ij = 0 ti chn v ij < 0, (i, j) thX l phng n ti u.nh l 5.3.2. Nu ti phng n cc bin X0, tn ti lk > 0 th xy dng cphng n cc bin mi X1 tt hn X0.

Ch :1) Chn lk = max

ij>0ij a (l, k) vo lm chn.

2) Theo nh l 3.3.1. Mun tm ui, vj ta i gii phng trnhui + vj cij = 0, (i, j) HTa c m+n n v m+n-1 phng trnh. Do cho 1 n no bt k bng 0 (n noxut hin nhiu =0) suy ra tm cc n cn li.

79


80/94

5.3.2 Thut ton th v

gii bi ton vn ti ta lm theo cc bc sau:Bc 1.

+ Tm phng n cc bin xut pht X0 c m+n-1 chn+ Tm ui, vj ti cc chn+ Tm ij ti cc loi.Bc 2. Kim tra ij 0+ Nu c ta kt lun X0 l phng n ti u.+ Nu khng chuyn sang bc 3.Bc 3. Chn lk = max

ij>0ij a (l, k) vo lm chn.

Lp vng V, Vc, Vl nh s trn vng tnh t (l, k).Bc 4: Chn xpq = min

xijVc

xij, a (p, q) ra khi vng.

Bc 5. Xy dng X1 theo cng thc

x1ij =

xij vi (i, j) / Vxij xpq vi (i, j) Vcxij + xpq vi (i, j) Vl

V d 4. Gii bi ton bng phng php th v

Thu 8 10 5 5Pht12 5 3 7 4

2 106 6 3 5 2

1 510 5 3 6 7

5 5

Bc 1. Bi ton cn bng thu pht, s dng phng php cc tiu tm

phng n cc bin xut pht, v c 4+3-1=6 chn khng lp thnh vng.+ Tm ui, vj ti cc chn ui + vj = cij.

u1 + v1 = 5

u1 + v2 = 3

u2 + v1 = 6

u2 + v3 = 5

u3 + v1 = 2

u3 + v4 = 1

u1 = 0

v1 = 5

v2 = 3

u2 = 1

u3 = 3v3 = 4

v4 = 4

80


81/94

+ Tnh ij ti cc loi l:

13 = 3, 14 = 0, 22 = 1, 33 = 3, 32 = 6, 24 = 3

Bc 2. Ta thy khng c ij 0.Bc 3. Chn max{22, 24} = 24 = 3a (2,4) vo lm chn .+ Lp vng

V = {(2, 4), (3, 4), (3, 1), (2, 1)}Bc 4. Chn

xpq = minxijVc

{(2, 4), (2, 1)} = x21 = 1

a (2, 1) ra khi vng.

Bc 5. Xy dng X1 theo cng thc bc 5 ta c bngThu 8 10 5 5

Pht12 5 3 7 4 u1 = 0

2 106 6 3 5 2 u2 = 2

5 110 5 3 6 7 u3 = 3

6 4v1 = 5 v2 = 3 v3 = 7 v4 = 4

Tr li bc 1:+ Tnh ui, vj ti cc chn c cc gi tr trn bng.+ Tnh ij 0Vy

X1 =

2 10 0 00 0 5 1

6 0 0 4

l phng n ti u v fmin = 83

Nhn xt:- Nu bi ton vn ti khng suy bin th thut ton th v s kt thc sau hu hnbc.- Nu bi ton suy bin th ta phi b sung sao cho m + n 1 chn trong chn b sung c lng phn phi bng 0 v khng lp thnh vng.

81


82/94

V d 5. Gii bi ton vn ti vi s liu cho nh sau:

Thu 50 75 50 25Pht

50 8 12 16 11

40 7 10 9 9

50 10 12 9 10

60 9 14 10 15

Gii. Ta tm phng n cc bin xut pht bng phng php Fogel

Thu 50 75 50 25Pht

50 8 12 16 11 u1 = 0

5040 7 10 9 9 u2 = 1

0- 40+50 10 12 9 10 u3 = 1

25 2560 9 14 10 15 u4 = 3

2+ 10- 50v1 = 8 v2 = 11 v3 = 7 v4 = 9

Bc 1. Bi ton cn bng thu pht, s dng phng php Fogel tm phng ncc bin xut pht, v ta thy 6 chn cha m+n-1 chn nn ta b sung thm1 chn l: (2, 1)( chn khng lp thnh vng), lc ny ta c m+n-1=7 chnkhng lp thnh vng.+ Tm ui, vj ti cc chn ui + vj = cij.

u1 + v1 = 8u2 + v1 = 7

u2 + v2 = 10

u3 + v2 = 12

u3 + v4 = 10

u4 + v2 = 14

u4 + v3 = 10

u1 = 0

v1 = 8

v2 = 11

u2 = 1u3 = 1

v3 = 7

v4 = 9

u4 = 3

+ Tnh ij ti cc loi v c 41 = 2 > 0Bc 2. Ta thy khng c ij 0.

82


83/94

Bc 3. Chn max{22, 24} = 24 = 3a (2,4) vo lm chn .+ Lp vng

V ={

(4, 1), (4, 2), (2, 2), (2, 1)}Bc 4. Chn

xpq = minxijVc

{(4, 2), (2, 1)} = x21 = 0

a (2, 1) ra khi vng v khi ta c phng n X1,

Documents

Bài giảng Quy Hoạch Tuyến Tình