quy hoach tuyen tinh bai giang

Bi ging quy hoch ton

Chng 1. BI TON QUY HOCH TUYN TNH PHNG PHP HNH HC

1.1. Cc bi ton thc t

1.1.1. Bi ton lp k hoch sn xut

a) V d sn xut ko v bnh cn 2 th nguyn liu chnh l ng v bt m, vi tr lng hin c l 0,9kg ng v 1,1 kg bt m. 1kg ko cn 0,5 kg ng v 0,3 kg bt m; 1kg bnh cn 0,2kg ng v 0,4 kg bt m. Gi 1kg ko l 10000; 1kg bnh l 20000. Hy lp k hoch sn xut sao cho tng gi tr sn phm ln nht. Gi x1 l s kg ko c sn xut; x2 l s kg bnh c sn xut. C m hnh ton hc: f(x) = 10000x1 +20000x2 max

++

0,1.14.03.09.02.05.0

21

21

21

xxxxxx

b)Tng qut sn xut n loi sn phm khc nhau cn m loi yu t sn xut

vi tr lng hin c l b1, b2, ..., bm. H s hao ph yu t i ( i=1..m ) cho 1 n v sn phm j (j=1..n) l aij. Gi 1 n v sn phm j l cj (j=1..n). Hy lp k hoch sn xut trn c s cc yu t sn xut hin c sao cho tng gi tr sn phm ln nht. Gi xj l s sn phm j c sn xut, f(x) l tng doanh thu ng vi k hoch sn xut x = (x1,x2, ..., xn). C m hnh ton hc:

f(x) = c=

n

j 1jxj max

==

=)..1(0

)..1(1

njx

mibxa

j

ij

n

jij


1.1.2. Bi ton vn ti

C m kho hng cha cng 1 loi hng ha vi s lng kho i l ai (i=1..m). ng thi c n ca hng vi nhu cu ca hng j l bj (j=1..n). Chi ph vn chuyn 1 n v hng t kho i n ca hng j l cij. Hy lp k hoch vn chuyn sao cho tha mn nhu cu cc ca hng v chi ph vn chuyn thp nht. Gi xij l s lng hng chuyn t kho i n ca hng j f(x) l tng chi ph theo k hoch vn chuyn x. M hnh ton hc:

f(x) = c=

m

i 1 =

n

j 1ijxij min

====

=

=

=

)..1,..1(0

)..1(

)..1(

1

1

njmix

njbx

miax

ij

j

m

iij

i

n

jij

1.1.3. Bi ton xc nh khu phn

C n loi thc n gia sc, gi 1 n v thc n j l c (j=1..n). Gia sc cn m cht dinh dng vi nhu cu ti thiu cht i l bi (i=1..m). Bit hm lng cht i c trong 1 n v thc n j l aij. Hy xc nh khu phn thc n cho gia sc sao cho chi ph thp nht ng thi m bo cc cht dinh dng cho gia sc. Gi xj l lng thc n j c trong khu phn, f(x) l gi khu phn x = (x1,x2, ..., xn). C m hnh ton hc sau:

f(x) = c=

n

j 1jxj min

==

=)..1(0

)..1(1

njx

mibxa

j

ij

n

jij

1.2. Bi ton qui hoch tuyn tnh

Xt bi ton


(1) f(x) = c=

n

j 1jxj min

(2)

+=

+=

==

=

=

=

)..1(

)..1(

)..1(

1

1

1

mkibxa

kpibxa

pibxa

ij

n

jij

ij

n

jij

ij

n

jij

Bi ton (1,2) gi l bi ton quy hoch tuyn tnh dng tng qut, k hin l (d,f). * f(x) gi l hm mc tiu. * H (2) gi l h rng buc. * Ma trn A = (aij)mxn gi l ma trn s liu. * Vect C = (cj)n gi l h s hm mc tiu. Mi b s x=(x1, x2, ..., xn) tha mn h rng buc (2) gi l phng n, k hiu x d. Phng n lm cho hm mc tiu f(x) t cc tr cn tm gi l phng n ti u, hay l nghim ca bi ton (d,f) .

1.3. Phng php hnh hc

Phng php hnh hc dng gii bi ton (d,f) 2 n, hoc nhiu hn 2 n nhng c th a v bi ton 2 n tng ng. Xt bi ton

f(x) = ax +by min (max) (d) { )..1( miciybxa ii =+

Min d d l giao cc na mt phng, hay l mt a gic. Bi ton c th pht biu bng hnh hc nh sau: Tm trong h ng thng song song ax+ by = f gi l h ng mc ,mt ng mc ng vi f nh nht (ln nht) c t nht 1 im chung vi min d. V d 1.1

f(x,y) = x + 2y max

++

0,1143925

yxyxyx


y A(0,11/4) B(1,2) d O C(9/5,0) x

Qua hnh v thy ng thng qua A(0, 4

11) ng vi f ln nht. Vy nghim l x1=0,

x2= 411 v fmax = 2

11 .

Nhn xt - Nghim l nh ca a gic. - Nu hm mc tiu l f(x,y) = 3x + 4y th nghim l c on thng AB. - Gi tr f ca h ng mc tng theo chiu ca php vect.

V d 1.2

f(x,y) = x + y max

0,221

yxyx

yx

d A(0,1) O B(2,0) Theo hnh v, hm mc tiu khng b chn trn trong min d nn bi ton v nghim.

---oOo---

Bi ging Quy hoch ton hc Trang 5 ________________________________________________________________________

1.4. Bi tp Gii cc bi ton sau bng phng php hnh hc 1. f(x) = x + 2y max 2. f(x) = 5x - 3y min

3 63 4 1

0 0

x yx y

x y

+ +

,

2x yx yx y

+ +

2 43 60 0,

3. f(x) = 3x + y max 4. f(x) = 2x + 3y +10 max

+

+

3 63 5 1

0 0

x yx y

x y,5

3 64

2 40 0

x yx yx yx y

+ + +

,

5. f(x) = 2x + 5y max 6. f(x) = x + 3y max

2 283

2 10 0

x yx yx y

x yx y

+ + +

+

,

2

x yx yx y

+ +

3 64

0 0,

7. f(x) = x + 2y max 8. f(x) = 2x + 3y min

x y

x yx y

+ +

82 1

0 0,4

x yx yx y

x y

+ + +

2 83 63 4 1

0 0,2

00

9. f(x) = 5x1 + 2x2 + 3x3 max 10. f(x) = 2x1 + x3 min

x x xx x xx x x

x x x

1 2 3

1 2 3

1 2 3

1 2 3

12 5 3 44 3 2

0 0

+ + =+ + + +

, ,

x x xx x x

x x x

1 2 3

1 2 3

1 2 3

12 2 3

0 0

+ + =+ +

, ,

***********************

________________________________________________________________________ GV: Phan Thanh Tao


Chng 2. PHNG PHP N HNH 2.1. Dng chnh tc v dng chun tc

2.1.1. nh ngha

Trong thc t, a s cc bi ton c iu kin khng m ca cc n. T c nh ngha dng chnh tc l bi ton (d,f) nh sau:

f(x) = c=

n

j 1jxj min (1)

===

=)3()..1(0

)2()..1(1

njx

mibxa

j

ij

n

jij

(2) gi l rng buc cng bc, (3) gi l rng buc t nhin.

Vi bi ton (d,f) chnh tc, c th gi s m n. Mt trng hp c bit ca dng chnh tc l ma trn s liu A = (aij)mxn c cha m vect ct l m vect n v ca khng gian Rm v bi 0 (i=1..m) gi l dng chun tc. Khng mt tnh tng qut, c th nh ngha bi ton (d,f) chun tc nh sau:

f(x) = c=

n

j 1jxj min

===+

+=)..1(0

)..1(1

njx

mibxax

j

ij

n

mjiji

trong bi 0 (i=1..m).

2.1.2. Cc php bin i

Cc php bin i sau a bi ton (d,f) bt k v dng chnh tc tng ng gii, v t suy ra nghim ca bi ton ban u. a/ f(x) max g(x) = -f(x) min b/ vi x

=

n

jijij bxa

1

=+ =+

n

jiinjij bxxa

1n+i0

vi x=

n

jijij bxa

1

=+ =

n

jiinjij bxxa

1n+i0

xn+i gi l n ph. C kt lun sau: Nu x = (x1, x2, ..., xn, xn+1, ..., xn+m) l nghim ca bi ton chnh tc bin i th x=(x1, x2, ..., xn) l nghim bi ton gc.

________________________________________________________________________ GV: Phan Thanh Tao


c/ Nu n xj khng rng buc v du th c thay bng hiu hai n khng m. Ngha l t xj =xj xj vi xj0, xj0. d/ Trng hp bi < 0 th nhn hai v phng trnh cho -1 c c bi>0. Vy: Mi bi ton quy hoch tuyn tnh u c th a v bi ton dng chnh tc tng ng. Hn na c th cc h s t do bi trong h rng buc l khng m.

2.1.3. Phng n c bn

Xt bi ton (d,f) dng chnh tc

f(x) = c=

n

j 1jxj min

===

=)..1(0

)..1(1

njx

mibxa

j

ij

n

jij

t Aj = (a1j , a2j , ... , amj ) l vect ct th j trong ma trn Amxn

b = (b1, b2, ... , bm) l ct h s t do. Gi s x = ( x 1, x 2,..., x n) l phng n ca bi ton th h vect { Aj / x j > 0 } gi l h vect lin kt vi phng n x. nh ngha

x d l phng n c bn nu h vct lin kt vi x c lp tuyn tnh. n xj gi l n c bn nu x j > 0. Nhn xt:

- Phng n c bn c ti a m thnh phn dng. Phng n c bn c ng m thnh phn dng gi l khng suy bin. Ngc li gi l suy bin. Bi ton c phng n c bn suy bin gi l bi ton suy bin.

- S phng n c bn ca mt bi ton (d,f) l hu hn. - Vi bi ton dng chun tc th c phng n c bn l xo = (b1, b2, ... ,bm,0,...,0).

2.1.4. Cc tnh cht

Tnh cht 1 Bi ton (d,f) ch xy ra 1 trong 3 trng hp sau: a) V nghim b) C 1 nghim duy nht c) V s nghim. Tnh cht 2 Nu hm mc tiu f(x) l chn di (trn ) i vi bi ton dng min (max) trn tp phng n d th bi ton (d,f) c nghim.

________________________________________________________________________ GV: Phan Thanh Tao


Tnh cht 3 Nu bi ton (d,f) c nghim th c nghm l phng n c bn.

2.2. Phng php n hnh

2.2.1. Ni dung

Xut pht t phng n c bn no , tm cch nh gi n. Nu cha ti u th chuyn sang phng n c bn mi tt hn. Nu bi ton c nghim th sau hu hn bc s tm c phng n c bn ti u. Hn na du hiu v nghim cng c th hin trn thut ton . V d 2.1 Xt bi ton (d,f) dng chun tc: f(x) = x1 -2x2 +3x3 -2x4 min

==++=+

)4..1(05432

421

321

jxxxx

xxx

j

C phng n c bn xo = (0, 0, 4, 5) v f(xo)=2 vi x3, x4 l n c bn. nh gi: x=(x1,x2,x3,x4) d :

==++=+

)4..1(05432

421

321

jxxxx

xxx

j

==+=

)4..1(05

324

214

213

jxxxx

xxx

j

f(x) = x1 -2x2 +3x3 -2x4 = x1 -2x2 +3(4-2x1 +3x2) -2(5-x1 -x2) = 2 -3x1 +9x2

= 2-1x1 - 2x2V x1, x2 0 nn nu 1, 2 0 th f(x)2 v xo l phng n ti u. Tuy nhin, y 1=3>0 nn xo cha phi l nghim. Th chn x1, x4 lm n c bn , cho x2=0 v x3=0. C

=+=

542

41

1

xxx

x 1=2 v x4=3. R rng A1, A4 c lp tuyn tnh nn c phng n c bn l x = (2, 0, 0, 3) v f( x ) = - 4. nh gi: x=(x1,x2,x3,x4) d :

________________________________________________________________________ GV: Phan Thanh Tao


=++=+

5432

421

321

xxxxxx

+=

+=

324

321

21

253

21

232

xxx

xxx

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

23 x2 - 2

1 x3) -2x2 +3x3 -2(3- 25 x2 + 2

1 x3)

= - 4 +29 x2 + 2

3 x3 (= -4-2x2 - 3x3)

-4 V x2, x3 0 nn x l phng n ti u (2, 30).

2.2.2. Bng n hnh

Cho bi ton (d,f) chun tc:

f(x) = c=

n

j 1jxj min

===+

+=)..1(0

)..1(1

njx

mibxax

j

ij

n

mjiji


j=1..n t j = c=

m

i 1iaij - cj v gi l c lng ca n xj i vi phng n c bn

xo=(b1, b2, , bm, 0, , 0) vi f(xo)= c=

m

i 1ibi

Lu : i = 0 , i=1..m C bng n hnh sau: H s

n CB

P/n x1c1

x2c2

xmcm

xm+1cm+1

xscs

xncn

c1 x1 b1 1 0 0 a1,m+1 a1s a1nc2 x2 b2 0 1 0 a2,m+1 a2s a2n cr xr br 0 0 0 ar,m+1 ars arn cm xm bm 0 0 1 am,m+1 ams amn f(x) 1 2 m m+1 s n

________________________________________________________________________ GV: Phan Thanh Tao


2.2.3. C s l lun

Cho bi ton (d,f) chun tc:

f(x) = c=

n

j 1jxj min

===+

+=)..1(0

)..1(1

njx

mibxax

j

ij

n

mjiji


j=1..n t j = c=

m

i 1iaij - cj

C phng n c bn xo=(b1, b2, , bm, 0, , 0) vi f(xo)= c=

m

i 1ibi

nh l 1 ( Du hiu ti u) Nu j 0 vi mi j = 1..n th xo l phng n ti u. Chng minh

C f(xo)= c=

m

i 1ibi

x=(xj)n d : xi + a+=

n

mj 1ijxj =bi (i=1..m) xi = bi - a

+=

n

mj 1ijxj (i=1..m)

f(x) = c=

n

j 1jxj

= c=

m

i 1ixi + c

+=

n

mj 1jxj

= c=

m

i 1i(bi - a

+=

n

mj 1ijxj) + c

+=

n

mj 1jxj

= c=

m

i 1ibi - ( c

+=

n

mj 1

=

m

i 1iaij-cj) xj

= f(xo) - +=

n

mj 1j xj

f(xo) : v j 0 v xj 0 (j=m+1..n) nh l 2 ( Du hiu v nghim) Nu k >0 v aik0 i = 1..m th bi ton v nghim. Chng minh V i = 0 , i=1..m v k >0 nn c k>m.

________________________________________________________________________ GV: Phan Thanh Tao


>0, xt b s x=(xj)n vi

+===

==

),..1(0

)..1(

kjnmjxx

miabx

j

k

ikii

i=1..m: xi + a+=

n

mj 1ijxj = (bi-aik) + aik= bi (1)

xk= >0 nn xj0 j=m+1..n i=1..m: xi = bi-aik bi 0. V >0 v aik0. Vy xj0 j=m+1..n (2) (1) v (2) c x d

f(x) = c=

n

j 1jxj

= c=

m

i 1ixi + c

+=

n

mj 1jxj

= c=

m

i 1i(bi-aik) + c

+=

n

mj 1jxj

= c=

m

i 1ibi - c

=

m

i 1iaik+ck

= f(xo) ( c=

m

i 1iaik-ck )

= f(xo) k Cho x + th f(x) - trn d. Hay f(x) khng b chn di trn d. Vy bi ton v nghim. nh l 3 ( iu chnh phng n) Nu k >0, aik>0 th c th tm c phng n c bn mi tt hn xo, trong trng hp bi ton khng suy bin. Chng minh: Gi s s = max {j} vi j>0 (j=1..n). Theo gi thit a is>0 t =min {

is

i

ab } vi ais> 0 . C >0 do bi ton khng suy bin.

Gi s =rs

r

ab , c

rs

r

ab

is

i

ab

Xt b s x =(xj)n vi

________________________________________________________________________ GV: Phan Thanh Tao


+===

==

),..1(0

)..1(

sjnmjxx

miabx

j

s

isii

i=1..m: xi + a+=

n

mj 1ijxj = (bi-ais) + ais= bi (1)

xs= >0 nn xj0 j=m+1..n i=1..m: xi = bi-ais = bi-

rs

r

ab ais 0. V

is

i

ab

rs

r

ab (i=1..m) v ais >0.

Vy xj0 j=m+1..n (2) (1) v (2) c x d

C xr = br-ars = br-rs

r

ab ars=0. Vy xr l n khng c bn.

H vect lin kt xo l m vect n v {A1, A2, , Am}. Vy h vect lin kt x l h con ca {A1, A2, , Am}U {As}\{Ar}. Gi s h vect lin kt x ph thuc tuyn tnh th h {A1, A2, , Am} {AU s}\{Ar} ph thuc tuyn tnh.

Nn ki0 sao cho: + k=

m

rii

ii Ak1

sAs = ( vect khng)

Nu ks=0 th k i0 (i=1..m) sao cho: = . Mu thun v {A=

m

rii

ii Ak1

1, A2, , Am} l h

vect n v. Vy ks0 v + k=

m

rii

ii Ak1

sAs =

hay As = -=

m

rii

is

i Akk

1(3)

Ngoi ra, As = (a1s , a2s , ... , ams ) = a=

m

i 1isAi (4)

Tr (4) cho (3) c

=

+m

rii

is

iis Ak

ka

1)( + arsAr = .

Do {A1, A2, , Am} l h c lp tuyn tnh nn c ars=0 (mu thun). Vy h vect lin kt x l h c lp tuyn tnh. Hay x l phng n c bn.

f( x ) = c=

n

j 1jxj

________________________________________________________________________ GV: Phan Thanh Tao


= c=

m

i 1ixi + c

+=

n

mj 1jxj

= c=

m

i 1i(bi-ais) + c

+=

n

mj 1jxj

= c=

m

i 1ibi - c

=

m

i 1iais+cs

= f(xo) ( c=

m

i 1iais-cs )

= f(xo) s < f(xo) , v >0 v s>0. Hay phng n c bn tt hn phng n c bn xo mt lng s.

2.2.4. Cc bc ca thut ton n hnh

Bc 1 Kim tra tnh ti u ca phng n xo=(b1, b2, , bm, 0, , 0)

* Nu j 0 j = 1..n th xo l phng n ti u v fmin=f(xo)= c=

m

i 1ibi.

* Nu k>0 th chuyn sang bc 2. Bc 2 Kim tra iu kin v nghim * Nu k >0 v aik0 vi mi i = 1..m th bi ton v nghim. * Nu k >0, aik>0 th chuyn sang bc 3. Bc 3 Tm n thay th v n loi ra * Nu s = max {j} vi j>0 (j=1..n) th a xs a vo tp n c bn .

* Nu rs

r

ab =min {

is

i

ab } vi ais> 0 th loi xr ra khi tp n c bn .

* Chuyn sang bc 4. Bc 4 Bin i bng n hnh

* Bin i bng n hnh theo cng thc sau:

=

=

)('

'

riaaa

aa

aa

a

isrs

rjijij

rs

rjrj

==

)('

'

riaabbb

b

isrs

rii

r

* Tnh li cc gi tr j, quay li bc 1.

________________________________________________________________________ GV: Phan Thanh Tao


Qu trnh ny c th m t nh vic bin i s cp v hng trn ma trn b sung ca h rng buc sao cho vect As tr thnh vect n v th r, v cc vect n v khc vn gi nguyn. Nhn xt Cc cng thc bin i cho aij cng ng cho c bi v j nu xem b l ct th 0 v l hng th m+1 ca ma trn s liu Amxn. V d 2.2 f(x) = 5x1 +4x2 + 5x3 +2x4 +x5 + 3x6 min

==++=+++=+++

)6..1(03636032452342

631

5321

4321

jxxxx

xxxxxxxx

j

H s

n CB

P/n x15

x24

x35

x42

x51

x63

2 x4 52 2 4 3 1 0 0 1 x5 60 4 2 3 0 1 0 3 x6 36 3 0 1 0 0 1 272 12 6 7 0 0 0 2 x4 28 0 4 7/3 1 0 -2/3 1 x5 12 0 2 5/3 0 1 -4/3 5 x1 12 1 0 1/3 0 0 1/3 128 0 6 3 0 0 -4 2 x4 4 0 0 -1 1 -2 2 4 x2 6 0 1 5/6 0 1/2 -2/3 5 x1 12 1 0 1/3 0 0 1/3 92 0 0 -2 0 -3 0 j 0 j =1..6, x opt= (12, 6, 0, 4, 0, 0) v fmin=92 V d 2.3 f (x) = 3x1 -2x2 +2x3 - x4 min

==++=+

)4..1(013212

432

421

jxxxxxxx

j

________________________________________________________________________ GV: Phan Thanh Tao


H s

n CB

P/n x13

x2-1

x32

x4-1

3 x1 1 1 1 0 -2 2 x3 1 0 -2 1 3 5 0 0 0 1 3 x1 5/3 1 -1/3 2/3 0 -1 x4 1/3 0 -2/3 1/3 1 14/3 0 2/3 -1/3 0 C 2=2/3>0 v trn ct ny khng c s dng nn bi ton v nghim.

2.2.5. Bi ton n ph

Cc php bin i a bi ton (d,f) v dng chnh tc

vi x=

n

jijij bxa

1

=+ =+

n

jiinjij bxxa

1n+i0

vi x=

n

jijij bxa

1

=+ =

n

jiinjij bxxa

1n+i0

xn+i gi l n ph. C kt lun sau: Nu x = (x1, x2, ..., xn, xn+1, ..., xn+m) l nghim ca bi ton chnh tc bin i th x=(x1, x2, ..., xn) l nghim bi ton gc.

V d 2.4 f (x) = -x1 +3x2 -2x3 max

=++

=++

)4..1(0108341242

723

321

321

4321

jxxxx

xxxxxxx

j

Bi ton chnh tc tng ng g (x) = x1 -3x2 +2x3 min

==+++

=+++=++

)6..1(010834

1242723

6321

5321

4321

jxxxxx

xxxxxxxx

j

Trong x5, x6 l n ph. y l bi ton (d,f) chun tc nn c a vo bng n hnh gii.

________________________________________________________________________ GV: Phan Thanh Tao


H s

n CB

P/n x11

x2-3

x32

x40

x50

x60

0 x4 7 3 -1 2 1 0 0 0 x5 12 -2 4 1 0 1 0 0 x6 10 -4 3 8 0 0 1 0 -1 3 -2 0 0 0 0 x4 10 5/2 0 9/4 1 1/4 0 -3 x2 3 -1/2 1 0 1/4 0 0 x6 1 -5/2 0 29/4 0 -3/4 1 -9 1/2 0 -11/4 0 -3/4 0 1 x1 4 1 0 9/10 2/5 1/10 0 -3 x2 5 0 1 7/10 1/5 3/10 0 0 x6 11 0 0 19/2 1 -1/2 1 -11 0 0 -16/5 -1/5 -4/5 0 j 0 j =1..6, x opt= (4, 5, 0, 0, 0, 11) v fmin=-11. Vy nghim bi ton gc l xopt= (4, 5, 0, 0) v fmax=11. Nu cc gi tr min/max t ti nhiu v tr th chn ty mt v tr bt k trong s . Thng thng chn ch s nh nht.

2.2.6. Bi ton n gi

Cho bi ton (d,f) dng chnh tc:

f(x) = c=

n

j 1jxj min

===

=)..1(0

)..1(1

njx

mibxa

j

ij

n

jij

trong bi0 (i=1..m). Xt bi ton:

f ( x ) = c=

n

j 1jxj + M x

=

m

i 1n+i min

+===+ +

=

)..1(0

)..1(1

mnjx

mibxxa

j

iinj

n

jij

vi M l s dng kh ln ( M).. Bi ton ny gi l bi ton m rng ca bi ton trn, hay bi ton M.

________________________________________________________________________ GV: Phan Thanh Tao


Vi bi ton M c ngay phng n c bn ban u vi xn+i(i=1..m) l cc n c bn . Dng thut ton n hnh gii. xn+i gi l cc n gi. Sau khi gii bi ton M, c c quan h gia bi ton M v bi ton (d,f) nh sau:

Nu bi ton M v nghim th bi ton (d,f) v nghim. Nu bi ton M c nghim x = (x1, x2, ..., xn, 0,...,0) th x = (x1, x2, ..., xn) l

nghim ca bi ton (d,f). Nu bi ton M c nghim x = (x1, x2, ..., xn+m) v xn+m)>0 th bi ton (d,f) v

nghim. Tin trnh gii bi ton M l loi dn cc n gi ra khi tp n c bn cho n khi loi tt c l bt u gii bi ton gc. Nn t c th khng cn tnh cho cc ct n gi. Nu cui cng khng loi c cc n gi m nhn gi tr 0 th bi ton gc cng c nghim. y gi s bi ton (d,f) trong ma trn s liu A khng c vect n v no. Tuy nhin, ch cn thm mt s (


==++=+++=++

)7..1(042262624

7321

65321

4321

jxxxxx

xxxxxxxxx

j

y l bi ton dng chun tc nn c a vo bng n hnh gii. H s n

CB P/n x1

-1 x2-2

x31

x40

x50

x6M

x6M

0 x4 6 -1 4 -2 1 0 0 0 M x6 6 1 1 2 0 -1 1 0 M x7 4 2 -1 2 0 0 0 1 3M+1 2 4M-1 0 -M 0 0 0 x4 10 1 3 0 1 0 0 1 M x6 2 -1 2 0 0 -1 1 -1 1 x3 2 1 -1/2 1 0 0 0 1/2 -M+2 2M+3/2 0 0 -M 0 -2M+1/2

0 x4 7 5/2 0 0 1 3/2 -3/2 5/2 -2 x2 1 -1/2 1 0 0 -1/2 1/2 -1/2 1 x3 5/2 3/4 0 1 0 -1/4 1/4 1/4 -9/2 11/4 0 0 0 3/4 -M-3/4 -M+5/4

-1 x1 14/5 1 0 0 2/5 3/5 -3/5 1 -2 x2 12/5 0 1 0 1/5 -1/5 1/5 0 1 x3 2/5 0 0 1 -3/10 -7/10 7/10 -1/2 -36/5 0 0 0 -11/10 -9/10 -M+9/10 -M-3/2 Nghim bi ton M l x = (14/5, 12/5, 2/5, 0, 0,0,0), n gi b loi t bng th 3. Nghim bi ton gc chnh tc l x = (14/5, 12/5, 2/5,0,0), vi x4, x5 l n ph, nn c nghin bi ton gc l xopt= (14/5, 12/5, 2/5,0,0) v fmax = 36/5 V d 2.6 f (x) = 8x1 -6x2 -2x3 max

==+=++

)3..1(04344434

321

321

jxxxxxxx

j

Bi ton M tng ng: f ( x ) = -8x1 +6x2 +2x3 +Mx4 +Mx5 min

==++=+++

)5..1(04344434

5321

4321

jxxxxx

xxxx

j

________________________________________________________________________ GV: Phan Thanh Tao


H s

n CB

P/n x1-8

x26

x32

x4M

x5M

M x4 4 4 3 4 1 0 M x5 4 4 1 -3 0 1 8M+8 4M-6 M-2 0 0 -8 x1 1 1 3/4 1 1/4 0 M x5 0 0 -2 -7 -1 1 0 -2M-12 -7M-10 -2M-2 0 Nghim bi ton M l X= (1,0,0,0,0) n gi x5 cn l n c bn nhng nhn gi tr 0 nn nghim bi ton gc l x = (1,0,0) v fmax = 8 V d 2.7

f (x) = -8x1 +6x2 +2x3 min

==+=++

)3..1(05344434

321

321

jxxxxxxx

j

Bi ton M tng ng: f ( x ) = -8x1 +6x2 +2x3 +Mx4 +Mx5 min

==++=+++

)5..1(05344434

5321

4321

jxxxxx

xxxx

j

H s n

CB P/n x1

-8 x26

x32

x4M

x5M

M x4 4 4 3 4 1 0 M x5 5 4 1 -3 0 1 8M+8 4M-6 M-2 0 0 -8 x1 1 1 3/4 1 1/4 0 M x5 1 0 -2 -7 -1 1 0 -2M-12 -7M-10 -2M-2 0

n gi x5 cn l n c bn nhng nhn gi tr x5 =1>0 nn nn bi ton gc v nghim.

________________________________________________________________________ GV: Phan Thanh Tao


2.3. Ci t thut ton n hnh

2.3.1. Khai bo d liu

a) Xem b l ct 0, c l hng 0 v l hng m+1 ca ma trn s liu a v f(xo)=a[m+1][0] vi a[0][0]=0. Cc gi tr bi, cj, j v f(x) bin i thao cng cng thc vi aij. Ngha l: bi=a[i][0], cj=a[0][j], j=a[m+1][j]. Ch cn khai bo mt ma trn A nh sau: float a[m+2][n+1]; b) Cc mng nh du tp n c bn: int cb[n+1], acb[m]; vi ngha:

xj l n c bn cb[j]=1 v

xj n c bn th i acb[i]=j Tp n c bn ban u gm m n c nhp t bn phm

2.3.2. Tnh cc c lng j

for (j=1; j


2.3.5. Tm n loi ra

r=1; while (a[i][j]


2.4. Bi tp Gii cc bi ton sau bng phng php n hnh 1. f(x) = 7x1 - 5x2 - 3x3 max

4x + x - 3x 154x + 3x + 5x 12x + 2x + 3x 10 x 0 (j = 1..3)

1 2 3

1 2 3

1 2 3

j

===

2. f(x) = -5x1 - 4x2 + 5x3 - 3x4 max

2x + 4x + 3x x 424x - 2x + 3x 243x + x 15 x 0 (j = 1..4)

1 2 3 4

1 2 3

1 3

j

+ =

3. f(x) = 7x1 +15x2 + 5x3 min

3x - 2x - 4x-x + 4x + 3x -32x + x + 8x 2 x 0 (j = 1..3)

1 2 3

1 2 3

1 2 3

j

1

4. f(x) = 2x1 +17x2 +18x3 max

6x + 4x + 7x

8x + 4x 30

x 0 (j = 1..3)

1 2 3

1 3

j

50

5. f(x) = 3x1 -x2 +2x3 x4 +5x5 max

2x - x + x + 2x + x

4x - 2x + x

x - x + 2x + x

x + x + 2x + x

x 0 (j = 1..5)

1 2 3 4 5

1 2 3

1 2 3 5

1 2 3 4

j

=

1720

18100

________________________________________________________________________ GV: Phan Thanh Tao


6. f(x) = -5x1 - 4x2 + 5x3 - 3x4 max

2x + 4x + 3x x 424x - 2x + 3x 243x + x 15 x 0 (j = 1..4)

1 2 3 4

1 2 3

1 3

j

+ =

7. f(x) = 8x1 + 7x2 + 9x3 ----> min

4 5 33 6 42 4 8

0 1 3

1 2 3

1 2 3

1 2 3

x x xx x xx x x

x jj

+ =+ + + =

=

, ..

69

763

2

8. f(x) = 2x1 - x2 + 3x3 ----> min

7x + 3x + 9x 52x - x - 8x6x + 4x + 2x

1 2 3

1 2 3

1 2 3

= =

=

1820

0 1 3x jj , .. 9. f(x) = 3x1 + 2x2 - 4x3 ----> min

5 6 84 3 42 7 2

0 1 3

1 2 3

1 2 3

1 2 3

x x xx x xx x x

x jj

+ =+ + =

=

, ..

10. f(x) = 3x1 - x2 + 5x3 ----> min

2x + 3x + 7x 5

5x - 2 x - x

6x + 2x + x

1 2 3

1 2 3

1 2 3

=

= =

4 1

10

0 1 3x jj , ..

11. f(x) = x1 + 2x2 + x3 max

x - 4 x + 2x -6x + x + 2x 52x - x + 2x 3 x 0 (j = 1..3)

1 2 3

1 2 3

1 2 3

j

==

***********************

________________________________________________________________________ GV: Phan Thanh Tao


Chng 3. BI TON I NGU 3.1. Cc bi ton thc t

3.1.1. Bi ton lp k hoch sn xut

Mt nh my sn xut hai loi sn phm A, B gm hai phn xng vi nng sut nh sau: Phn xng I : 1 nghn sn phm A + 4 nghn sn phm B trong 1 nm. v Chi ph 16 triu ng. Phn xng II : 3 nghn sn phm A + 1 nghn sn phm B trong 1 nm. v Chi ph 15 triu ng. K hoch Nh nc giao cho nh my l: 1 nghn sn phm A + 2 nghn sn phm B. Hy lp k hoch sn xut sao cho tng chi ph thp nht ng thi m bo k hoch nh nc giao cho nh my. Gi x1 l thi gian phn xng I sn xut ( n v nm), x2 l thi gian phn xng II sn xut ( n v nm) Tng chi ph ca k hoach sn xut x=(x1, x2) l f(x) = 16x1 + 15x2 (triu ng) M hnh ton hc: f(x) = 16x1 + 15x2 min

(d)

++

02413

2,1

21

21

xxxxxx

3.1.2. Bi ton nh gi sn phm

Vi nng sut hai phn xng ca nh my nh bi ton trn . Nh my sn xut c 1 nghn sn phm A v 2 nghn sn phm B. Hy nh gi tr cho 1 sn phm A v 1 sn phm B sao cho tng gi tr ca sn phm: phn xng I khng vt qu chi ph l 16 triu ng/nm v phn xng II khng vt qu chi ph l 15 triu ng/nm, v tng gi tr sn phm ca nh my ln nht. Gi y1(nghn ng) l gi tr n v sn phm A,

y2(nghn ng) l gi tr n v sn phm B Tng gi tr sn phm theo k hoch nh gi y=( y1, y2) l

g(y) = y1+2y2(nghn ng) M hnh ton hc: g(y) = y1+2y2 max

________________________________________________________________________ GV: Phan Thanh Tao


( d~ )

++

0153164

2,1

21

21

yyyyyy

x2

(4,3) d

(5/11, 2/11) O O x1 fmin=f(5/11, 2/11)= 10 (triu ng) gmax=g(4, 3)= 10 (triu ng) Nhn xt: fmin= gmax

3.2. Bi ton i ngu

3.2.1. i ngu khng i xng

Cho bi ton (d,f) dng chnh tc

(1) f(x) = c=

n

j 1jxj min

(d)

===

=)..1(0

)..1(1

njx

mibxa

j

ij

n

jij

Cng vi bi ton (I), xt bi ton ( d~ , g) nh sau:

( 1~ ) g(y) = b=

m

i 1iyi max

( d~ )

==

=)..1(dotu

)..1(1

miy

njcya

i

ji

n

jji

( 1~ ) gi l bi ton i ngu ca bi ton (1).

________________________________________________________________________ GV: Phan Thanh Tao


Bi ton i ngu ca bi ton ( D, f ) bt k l bi ton i ngu ca bi ton dng chnh tc tng ng vi n. Nu xem ( 1~ ) l bi ton gc th ( 1 ) l bi ton i ngu ca n. V mt hnh thc, cp ( 1, 1~ ) gi l cp bi ton i ngu khng i xng. Cch thnh lp

- Bi ton gc dng chnh tc. - H s hm mc tiu ca bi ton ny l h s t do trong h rng buc ca bi

ton kia. - Ma trn s liu chuyn v cho nhau. - Bi ton i ngu l bi ton max v rng buc l .

V d

f(x) = x1 + 2x2+3x3 min

(d)

==+

=++

)3..1(05432

1

321

321

jxxxxxxx

j

Bi ton i ngu ( d~ , g) g(y) = y1-5y2 max

( d~ )

++

dotuyyyyyyyy

2,1

21

21

21

342312

3.2.2. i ngu i xng

Cho bi ton (d,f) dng sau

(2) f(x) = c=

n

j 1jxj min

(d)

==

=)..1(0

)..1(1

njx

mibxa

j

ij

n

jij

Bi ton dng chnh tc tng ng

f(x) = c=

n

j 1jxj min

+=== +

=

)..1(0

)..1(1

nmjx

mibxxa

j

iinj

n

jij

xn+i l n ph.

________________________________________________________________________ GV: Phan Thanh Tao


Bi ton i ngu

( 2~ ) g(y) = b=

m

i 1iyi max

( d~ )

==

=)..1(0

)..1(1

miy

njcya

i

ji

n

jji

hay

( 2~ ) g(y) = b=

m

i 1iyi max

( d~ )

==

=)..1(0

)..1(1

miy

njcya

i

ji

n

jji

Ngc li nu xem Nu xem ( 2~ ) l bi ton gc th ( 2 ) l bi ton i ngu ca n. V mt hnh thc, cp ( 2, 2~ ) gi l cp bi ton i ngu i xng. Cch thnh lp

- H s hm mc tiu ca bi ton ny l h s t do trong h rng buc ca bi ton kia.

- Ma trn s liu chuyn v cho nhau. - Bi ton min rng buc l v bi ton max rng buc l . - C hai bi ton u c rng buc cc n khng m.

V d 3.1

f(x) = 3x1 + 2x2+x3 min

(d)

=++

+

)3..1(01427654

432

321

321

321

jxxxx

xxxxxx

j

Bi ton i ngu

g(y) = 4y1-6y2 +y3 max

( d~ )

=++

++

)3..1(01453224

372

321

321

321

iyyyyyyy

yyy

i

Nhn xt Vi bi ton ( d~ ,g) ch cn a v dng chnh tc th tr thnh dng chun tc.

________________________________________________________________________ GV: Phan Thanh Tao


3.2.3. S tucker

T hai cp bi ton i ngu ( 1, 1~ ) v ( 2, 2~ ) c s Tucker vit bi ton i ngu ca bi ton bt k nh sau

Bi ton gc

f(x) = c=

n

j 1jxj min

Bi ton i ngu

g(y) = b=

m

i 1iyi max

=

n

j 1aijxj =bi (i=1..p)

yi t do (i=1..p)

=

n

j 1aijxj bi (i=p+1..m)

yi 0 (i=p+1..m)

xj0 (j=1..q) =

m

i 1aijyi =cj(j=1..q)

xj t do (j=q+1..n) =

m

i 1aijyi cj (j=q+1..n)

Lu Bi ton min khng c rng buc v Bi ton max khng c rng buc . V d

f(x) = 2x1 + x2+4x3 min

(d)

=++

+

0,2223553

432

31

321

321

321

xxxxx

xxxxxx

Bi ton i ngu

g(y) = 4y1-5y2 +2y3 max

( d~ )

+=+

++

0,4253123

232

21

321

321

321

yyyyyyyy

yyy

3.3. Cc nguyn l i ngu

Xt cp bi ton i ngu (d,f) v ( d~ ,g) vi f(x)min v g(y)max. C cc nguyn l sau

________________________________________________________________________ GV: Phan Thanh Tao


3.3.1. Nguyn l 1

a) xd, y d~ : f(x) g(y). b) xo d, y o d~ : f(xo) =g(yo) => f(xo) = f min v g(yo)= gmax . Chng minh a) xd, y d~ : f(x) = c

=

n

j 1jxj

( a=

n

j 1

=

m

i 1ijyi)xj

( a=

m

i 1

=

n

j 1ijxj) yi

b=

m

i 1iyi =g(y)

b) xo d, y o d~ : f(xo) =g(yo) xd: f(x) g(yo) =f(xo) =>f(xo) = fmin y d~ : g(y) f(xo)=g(yo) =>g(yo) = gmax3.3.2. Nguyn l 2

Nu bi ton ny c nghim th bi ton kia cng c nghim v cp nghim tho mn iu kin cn bng fmin = gmax.

3.3.3. Nguyn l 3 ( lch b)

Cho xd, y d~ . iu kin cn v x, y l nghim tng ng ca cp bi ton i ngu l:

(1)

==

=>

=

=n

jijiji

i

n

jijij

bxay

ybxa

1

1

0

0

(2)

=>

=0 th a=

m

i 1ijyi =cj

2) Nu a=

m

i 1ijyi 0 th a=

n

j 1ijxj =bi

4) Nu a=

n

j 1ijxj >bi th yi=0

C th pht biu cch khc v nguyn l lch b nh sau: iu kin cn v x , y l nghim tng ng ca cp bi ton i ngu l : trong cc cp iu kin i ngu, nu iu kin ny xy ra vi bt ng thc thc s th iu kin kia xy ra vi du bng.

3.4. ngha kinh t

Xt cp i ngu i xng ( 2, 2~ )

3.4.1. ngha bi ton (2)

C n cch khc nhau sn xut m loi sn phm. Cch th j s dng cng 1 cho aij n v sn phm loi i (i=1..m) v chi ph cj (j=1..n). Hy tm cng xj cn s dng cho tng cch sn xut, tng s n v ca sn phm loi ic sn xut ra t ra bng bi (i=1..m) v tng chi ph sn xut l t nht. x = (xj )n : phng n sn xut

3.4.2. ngha bi ton ( 2~ )

Cng iu kin vi bi ton (2 ) . Gi s sn xut c bi sn phm i (i=1..m) . Hy nh gi tr yi cho mi n v sn phm loi i (i=1..m), m bo tng gi tr sn phm sn xut theo cch j khng vt qu chi ph sn xut l cj (j=1..n) ng thi tng gi tr sn phm l ln nht.

________________________________________________________________________ GV: Phan Thanh Tao


y = ( yi ) : phng n nh gi.

3.4.3. ngha nguyn l lch b

iu kin cn v phng n sn xut x=(xj)n v phng n nh gi y=(yi)m ng thi ti u l: 1/ Nu mt cch sn xut c s dng (xj >0) th tng gi tr sn phm c sn

xut theo cch y phi ng bng chi ph ( a=

m

i 1ijyi =cj).

2/ Nu mt loi sn phm c gi tr ( yi> 0 ) th tng s sn phm c sn xut phi

ng bng nhu cu ( a=

n

j 1ijxj =bi)

---oOo---

3.5. Bi tp Gii cc bi ton sau bng phng php n hnh. Vit bi ton i ngu ca chng. Da vo nguyn l lch b tm nghim bi ton i ngu. 1. f(x) = -5x1 - 4x2 + 5x3 - 3x4 max

2x + 4x + 3x x 424x - 2x + 3x 243x + x 15 x 0 (j = 1..4)

1 2 3 4

1 2 3

1 3

j

+ =

2. f(x) = 2x1 +17x2 +18x3 max

6x + 4x + 7x

8x + 4x 30

x 0 (j = 1..3)

1 2 3

1 3

j

50

3. f(x) = -5x1 - 4x2 + 5x3 - 3x4 max

2x + 4x + 3x x 424x - 2x + 3x 243x + x 15 x 0 (j = 1..4)

1 2 3 4

1 2 3

1 3

j

+ =

4. f(x) = 8x1 + 7x2 + 9x3 ----> min

________________________________________________________________________ GV: Phan Thanh Tao


4 5 33 6 42 4 8

0 1 3

1 2 3

1 2 3

1 2 3

x x xx x xx x x

x jj

+ =+ + + =

=

, ..

69

Vit bi ton i ngu cc bi ton sau. Gii cc bi ton i ngu bng phng php n hnh. Da vo nguyn l lch b tm nghim ca bi ton gc. 1. f(x) = 7x1 +15x2 + 5x3 min

3x - 2x - 4x-x + 4x + 3x -32x + x + 8x 2 x 0 (j = 1..3)

1 2 3

1 2 3

1 2 3

j

1

2. f(x) = x1 + 2x2 + x3 max

x - 4 x + 2x -6x + x + 2x 52x - x + 2x 3 x 0 (j = 1..3)

1 2 3

1 2 3

1 2 3

j

==

***********************

________________________________________________________________________ GV: Phan Thanh Tao


Chng 4. BI TON VN TI 4.1. Bi ton vn ti dng chnh tc

4.1.1. nh ngha

Cn vn chuyn mt loi hng ho t m trm pht A1, A2, ..., Am n n trm thu B1, B2, ..., Bn. Lng hng cn chuyn i tng ng l a1 , a2 , ... , am ; yu cu ca cc trm thu tng ng l b1 , b2 , ... , bn . Chi ph vn chuyn 1 n v hng ho t trm pht Ai n trm thu Bj l cij . Hy lp k hoch vn chuyn sao cho tng chi ph thp nht v ng thi m bo cc yu cu ca trm thu v pht. Gi xij l lng hng chuyn t Ai n Bj Tng chi ph theo k hoch vn chuyn x={ xij }mxn l

f(x) = c=

m

i 1

=

n

j 1ijxij

M hnh ton hc:

f(x) = c=

m

i 1 =

n

j 1ijxij min

====

==

=

=

)..1,..1(0

)..1(

)..1(

1

1

njmix

njbx

miax

ij

j

m

iij

i

n

jij

y l bi ton (d, f ) dng chnh tc: Bi ton i ngu l:

g(u,v) = v=

n

j 1j - u

=

m

i 1i max

{ )..1,..1( njmicuv ijij == Cp iu kin i ngu l: xij 0 vj - ui cijT , theo nguyn l lch b c tiu chun ti u cho phng n x={ xij }mxn l l tn ti h thng s {ui, vj } tho mn:

(*) >=

==0

)..1,..1(

ijijij

ijij

xneucuvnjmicuv

ui gi l th v dng; vj gi l th v ct. H thng {ui , vj }m+n gi l h thng th v.

________________________________________________________________________ GV: Phan Thanh Tao


Vy, gii bi ton vn ti cn tm phng n c bn v kim tra tnh ti u qua h thng th v. Da vo ngha chung ca bi ton i ngu c th xem th v dng ui l gi tr 1 n v hng ti trm pht Ai, th v ct vj l gi tr ti trm thu Bj. Cng thc (*) mang ngha kinh t l: Trong mi phng n vn chuyn tt nht th chnh lch gi tr hng ti trm pht v trm thu u khng vt qu chi ph vn chuyn trc tip gia hai ni. V nu hng vn chuyn t Ai n Bj th gi tr hng ti Bj ng bng gi tr ti Ai cng chi ph vn chuyn.

4.1.2. iu kin cn bng thu pht

t: a = a=

m

i 1i gi l tng pht

b = b=

n

j 1j gi l tng thu

Bi ton vn ti dng chnh tc cn bng thu pht (a=b) lun lun c nghim.

Xt x={ xij }mxn vi xij = aba ji =

bba ji

c xij > 0 (i=1..m, j=1..n)

=

n

j 1xij=

=

m

i 1 bba ji = a

=

m

i 1i (i=1..m)

=

m

i 1xij=

=

m

i 1 aba ji = b

=

m

i 1j (j=1..n)

Vy xd Nu a b th bi ton khng c phng n . V vy bi ton lun c kho st vi gi thit a = b, gi l iu kin cn bng thu pht. Trong iu kin ny bi ton lun lun c nghim (d v f b chn di trn d).

4.2. Bng phn phi v tnh cht

4.2.1. Bng phn phi

Cho bi ton vn ti dng chnh tc

f(x) = c=

m

i 1

=

n

j 1ijxij min

________________________________________________________________________ GV: Phan Thanh Tao


====

==

=

=

)..1,..1(0

)..1(

)..1(

1

1

njmix

njbx

miax

ij

j

m

iij

i

n

jij

y l bi ton (d,f) dng c bit nn c a vo bng phn phi gii theo thut ton ring. Bng phn phi: Pht\Thu b1 b2 bj bn

a1 c11 c12 c1j c1na2 c21 c22 c2j c2n .. .. .. .. ai ci1 ci2 cij cin .. .. .. .. am cm1 cm2 cmj cmn

4.2.2. Tnh cht

Xt bng phn phi mxn (m hng n ct) hng i , ct j gi l (i,j). Dy chuyn l dy c dng: 2 lin tip nm trn cng 1 hng hay 1 ct, 3 lin

tip khng cng nm trn cng 1 hng hay 1 ct. Chu trnh (vng) l dy chuyn khp kn. Cc dng vng thng gp:

Nhn xt: S trn mt vng l s chn.

________________________________________________________________________ GV: Phan Thanh Tao


Tnh cht 1 Vi bng phn phi mxn th s ti a khng to thnh vng l m+n-1. Tnh cht 2 C m+n-1 khng to thnh vng th thm vo 1 bt k s cha 1 vng duy nht v ngc li b i 1 bt k trong vng s cn li m+n-1 khng cha vng. Gi xij l lng hng phn phi cho ( i,j ). Cho phng n x={ xij }mxn. (i,j ) gi l chn nu xij > 0 ; ngc li gi l loi. Tnh cht 3 Phng n c bn l phng n c tp hp chn khng cha vng.

4.3. Thut ton th v

4.3.1. Ni dung

Xut pht t phng n c bn ban u, dng h thng th v kim tra tiu chun ti u, nu cha ti u th chuyn sang phng n c bn mi ti u. Sau hu hn bc c c phng n ti u.

4.3.2. Xy dng phng n c bn ban u

* Nguyn tc phn phi ti a Khi chn (i,j ) phn phi, phn phi ti a vo (i,j ) l t xij = min (ai , bj). Sau khi phn phi vo (i,j), loi hng i hoc ct j tha mn nhu cu ra khi bng n hnh. Tip tc phn phi cho bng mi, cho n khi tha mn nhu cu ca tt c cc trm th c c phng n c bn ban u. V.bng nguyn tc ti a, tp cc chn khng to thnh vng. * Cc phng php phn phi a/ Phng php phn phi theo chi ph nh nht u tin phn phi cho c chi ph cij nh nht. b/ Phng php gc ty bc u tin phn phi cho (1,1), gc ty bc.

c/ Phng php xp x Fghen u tin phn phi c cc ph nh nht thuc hng (ct) c chnh lch ln nht gia cc ph nh nht v nh.

Nhn xt: Phng php gc ty bc, tuy c phng n c bn ban u xa phng n ti u, nhng thun tin trong ci t. Phng php Fghen c phng n c bn ban u gn phng n ti u, v c tnh n bc sau. n gin v t bc lp, cc v d c trnh by theo phng php chi ph nh nht. Tuy nhin, trong hng dn ci t th dng phng php gc ty bc.

________________________________________________________________________ GV: Phan Thanh Tao


4.3.3. Cc bc ca thut ton th v

Bc 1 Xy dng phng n c bn ban u Xy dng phng n c bn ban u vi tp S cc chn gm m+n-1 khng cha

vng bng bt k phng php no. Vi bi ton khng suy bin, mi chn ch c mt trm tha mn v c loi ra

khoi bng phn phi, tnh li nhu cu v phn phi tip. Ring cui cng, do cn bng thu pht nn c hai trm u tha mn.. Vy c ng m+n-1 chn khng cha vng.

Bc 2 Xy dng h thng th v {ui , vj}. Gii h vj - ui = cij nu (i ,j) S. y l h m+n-1 phng trnh, m+n n nn c th chn 1 n tu , thng thng chn cho u1=0. Cc th v cn li c tnh theo cng thc sau vj = ui + cij vi ui bit v (i,j) S ui = vj - cij vi vj bit v (i,j) S Bc 3 Kim tra tiu chun ti u. t ij = vj - ui - cij. C ij =0 nu (i,j) S Nu ij 0 (i,j) th x={xij}mxn l phng n ti u. Nu (i,j)>0 th chuyn sang bc 4. Bc 4 iu chnh phng n Nu rk = max ij th (r, k) gi l iu chnh. S {(r, k)} cha mt vng duy nht, gi l vng iu chnh V. Tm vng V bng cch nh s chn l bt u t (r, k). Gi Vl l tp c ch s l v Vchn l tp c ch s chn. t q= min {xc}, vi xc l xij vi (i,j) Vchn. q gi l lng iu chnh. Nu xiojo =q th (io, jo) tr thnh loi trong phng n c bn mi. Vi bi ton suy bin, lng iu chnh q c th t ti nhiu , khi ch loi 1 , cc cn li tr thnh chn 0. Bin i phng n x thnh x' theo cng thc sau: xij=xij + q nu (i,j) Vl xij=xij - q nu (i,j) VchnQuay li bc 2. Sau hu hn bc c phng n ti u.

________________________________________________________________________ GV: Phan Thanh Tao


V d 4.1 a=b=130 ai\bj 40 70 20 vj

80 2010

409

202

0

30 4

3

201

6

20 202

6

2

8

ui 10 9 7

ai\bj 40 70 20 vj

80 20 10

20 9

+ 2

0

30 4

20 3

20 1

6

20 20 2

6

2

8

ui 10 9 7 f(x)=830 ij 0 (i,j), fmin=730 V d 4.2 a=b=311

ai\bj 76 62 88 45 40 vj

79 3410

19

15

45 6

7

0

102 13

4411

58 8

7

4

5

70 12

17

30 10

5

403

3

60 4212

1818

18

10

9

-2

ui 10 16 13 6 6

ai\bj 76 62 88 45 40 vj

79 64 10

19

15

15 6

7

0

102 13

14 11

88 8

7

4

5

70 12

17

+ 10

30 5

403

1

60 12 12

48 18

18

10

9

-2

ui 10 16 13 6 4 F(x)= 2866 Fmin= 2806

4.4. Cc dng khc

4.4.1. Khng cn bng thu pht

Nu a b th bi ton khng c phng n. Tuy nhin thc t khng i hi phi ch ht hng t cc trm pht mi p ng yu cu cc trm thu; hay tri li c th lng hng cc trm pht khng p ng c yu cu cc trm thu. Khi ta thm trm thu gi Am+1 hoc trm pht gi Bn+1bng chnh lch gia tng thu v tng pht: am + 1 = b - a nu ab Lng hng vn chuyn t trm pht Ai n trm thu gi Bn+1, ngha l lng hng c gi li Ai , v lng hng chuyn t trm pht gi Am+1 n trm thu Bj ngha l ti Bj lng hng y khng c tho mn. Tt nhin cc ph cc trn trm gi bng khng.

________________________________________________________________________ GV: Phan Thanh Tao


Cc ph cc trm gi bng khng, thng nh nht, nhng vn u tin phn phi cho cc c cc ph dng nh nht cc khng phi ca trm gi, cui cng hng cn li phn phi vo cc ca trm gi. V d 4.3 a=120, b=100. Thm tram thu gi vi a4=20. ai\bj 30 40 30

20 6 4 3 40 8 3 7 40 7 5 6 20 5 1 2

ai\bj 30 40 30 20 vj

20 6

4

203

0

0

40 8

303

7

10 0

-1

40 307

5

6

10 0

-3

20 5

101

102

0

1

ui 4 2 3 -1

ai\bj 30 40 30 20 vj

20 6

4

20 3

0

0

40 8

20 3

7

20 0

-1

40 30 7

5

10 6

+ 0

-3

20 5

20 1

0 2

0

1

ui 4 2 3 -1 F(x)= 410 Fmin= 390

4.4.2. Suy bin

Vi bi ton suy bin, khi xy dng phng n c bn u tin c th ng thi tha mn c hai trm thu v pht. c c ng m+n-1 chn, ch lai mt trm, trm cn li xem nh vn cn nhu cu 0. Khi phn phi cho trm ny to nn chn 0, ngha l chn vi lng hng phn phi l 0. Hay lng iu chnh q c th t ti nhiu , khi ch loi 1 , cc cn li tr thnh chn 0.

________________________________________________________________________ GV: Phan Thanh Tao


V d 4.4

F(x)= 885

F(x)= 810

ai\bj 30 20 25 35 40 vj

30 18

7

6

302

12

0

20 0 5

20 1

10

5

11

0

40 10

5

+ 3

57

3514

-5

60 30 6

3

25 2

11

510

-1

ui 5 1 1 2 9

ai\bj 30 20 25 35 40 vj

30 18

7

6

302

12

0

20 0 5

20 1

10

5

11

0

40 10

+ 5

25 3

57

1014

-5

60 30 6

3

2

11

3010

-1

ui 5 1 -2 2 9

ai\bj 30 20 25 35 40 vj

30 18

7

6

302

12

0

20 10 5

10 1

10

5

11

-1

40 10

10 5

25 3

57

14

-5

60 20 6

3

2

11

4010

-2

ui 4 0 -2 2 8

Fmin=800

________________________________________________________________________ GV: Phan Thanh Tao


4.4.3. Dng cc i

Xt bi ton vn ti dng max:

f(x) = q=

m

i 1

=

n

j 1ijxij max

====

==

=

=

)..1,..1(0

)..1(

)..1(

1

1

njmix

njbx

miax

ij

j

m

iij

i

n

jij

a v dng chnh tc tng ng bng cch t cij = - qij (i=1..m, j=1..n)

g(x)= q=

m

i 1 =

n

j 1ijxij min

====

==

=

=

)..1,..1(0

)..1(

)..1(

1

1

njmix

njbx

miax

ij

j

m

iij

i

n

jij

C fmax= -gmin. V d 4.5 Phn phi lao ng. Mt cng ty vn ti bin cn tuyn 110 ngi b tr 10 ngi lm my trng (MT), 25 th 1, 30 th 2 v 45 th 3. Phng t chc tm c 90 ngi gm 25 k s (KS), 20 trung cp (TC) v 45 cng nhn (CN). Kh nng cn b c nh gi theo cng vic qua bng sau Cng vic MT Th

1 Th 2

Th 3

KS 5 4 0 0 TC 3 5 4 0 CN 0 1 5 4

Trnh Cn b tr sao cho s dng ti a nng lc ca mi ngi. y l bi ton vn ti dng max. Khng cn bng thu pht. a vo trm pht gi: a4= 110 - 90 = 20

________________________________________________________________________ GV: Phan Thanh Tao


ai\bj 10 25 30 45 vj

25 10 -5

5 -4

0

100

0

20 -3

20 -5

+ -4

0

1

45 0

-1

30 -5

15-4

4

20 0

0

0

200

0

ui -5 -4 -1 0

ai\bj 10 25 30 45 vj

25 10 -5

5 -4

0

100

0

20 -3

10 -5

10 -4

0

1

45 0

-1

20 -5

25-4

2

20 0

0

0

200

-2

ui -5 -4 -3 -2

y l phng n ti u Vy c phng n phn phi lao ng ti u nh sau: 10 k s lm My trng 15 k s lm Th 1 10 Trung cp lm Th 1 10 Trung cp lm Th 2 20 Cng nhn lm Th 2 25 Cng nhn lm Th 3 V d 4.6 Bi ton phn phi t trng C 3 loi rung A, B, C vi din tch tng ng l 20, 25, 30 ha trng 3 loi la I, II, III vi din tch theo k hoch l 15, 30, 30 ha tng ng. Hy tm phng n phn phi t trng sao cho tng sn lng cao nht ng thi m bo k hoch. Bit sn lng la trn tng loi t cho trong bng sau (tn/ha)

________________________________________________________________________ GV: Phan Thanh Tao


y l bi ton vn ti dng max

fmax=770 V d 4.7 Bi ton b nhim

Cn phn n vic cho n ngi. Ngi i lm vic j th nng sut l cij (i,j=1..n). Hy phn cng vic cho n ngi tng nng sut cao nht. t xij=1 nu ngi i lm vic j; ngc li t xij=0. Bi ton ny cn gi l bi ton quy hoch nguyn 0-1. V suy bin nn c thut ton khc tin hn. Bng nng sut c cho nh sau

la t

I 15

II 30

III 30

A(25) 12 8 8 B(25) 8 10 9 C(30) 8 10 10

ai\bj 15 30 30 vj

25 15 -12

-8

5 -8

0

20 -8

25 -10

-9

2

45 5 -8

-10

25 -10

2

ui -12 -8 -8

Vic Ng

1

2

3 4

A 5 2 6 4 B 3 7 5 6 C 4 1 5 2 D 8 6 7 3

________________________________________________________________________ GV: Phan Thanh Tao


f(x)=23

f(x)=24

ai\bj 1 1 1 1 vj

1 -5

-2

1 -6

0-4

0

1 -3

1 -7

-5

0-6

2

1

-4

-1

+ -5

1-2

-2

1 1 -8

-6

0 -7

-3

1

ui -7 -5 -6 -4

ai\bj 1 1 1 1 vj

1 -5

-2

-6

1-4

0

1 -3

1 -7

-5

0-6

2

1

-4

-1

1 -5

0-2

-2

1 1 -8

+ -6

0 -7

-3

0

ui -8 -5 -7 -4

ai\bj 1 1 1 1 vj

1 -5

-2

-6

1-4

0

1 -3

1 -7

-5

0-6

2

1

-4

-1

1 -5

0-2

-2

1 1 -8

0 -6

-7

-3

1

ui -7 -5 -7 -4

fmax=24

________________________________________________________________________ GV: Phan Thanh Tao


4.4.4. Bi ton xe rng

Bi ton xe rng ng dng thng xuyn trong thc t, nn c xem l mt dng c bit ca bi ton vn ti V d 4.8 Cng ty vn ti cn hon thnh hp ng ch hng sau: 1) Than: Kim Lin Ngc Hi: 50 tn 2) Xi mng: Ga H Ni Chung: 24 tn 3) Xi mng: Ga H Ni Ba th: 10 tn 4) Sn: Mai Lnh H ng: 8 tn 5) Mui: Thng Tn H ng: 42 tn 6) Mui: Thng Tn Trc Sn : 8 tn 7) Ng: Kim bi H ng: 34 tn Hy lp k hoch vn chuyn sao cho tng s tn xe rng t nht. Vi c ly cc a im nh sau:

Ngc hi Chung Ba th H ng Trc Sn

Kim Lin 11 27 40 10 21 Ga H Ni 12 28 41 11 22 Mai Lnh 18 18 31 7 4 Thng Tn 6 34 35 17 28 Kim Bi 26 2 15 15 20

Cc ph l c ly Ni c hng l ni thu xe rng Ni cn hng l ni phat xe rng Trm thu xe rng Trm pht xe rngKim lin: 50 Ngc Hi: 50 Ga H Ni: 34 Chung: 24 Mai Lnh: 8 Ba Th: 10 Thng Tn: 50 H ng: 84 Kim Bi: 34 Trc Sn: 8 y l bi ton vn ti dng cc tiu cn bng thu pht.

________________________________________________________________________ GV: Phan Thanh Tao


F(x)= 1404

ai\bj 50 34 8 50 34 vj

50 11

12

18

506

25

0

24 27

28

18

34

242

11

10

40

41

31

35

1015

-2

84

50 10

34 11

0 7

17

+ 15

-4

8 21

22

8 4

028

020

-1

ui 6 7 3 6 13

ai\bj 50 34 8 50 34 vj

50 11

12

18

506

25

0

24 27

28

18

34

242

11

10

40

41

31

35

1015

-2

84

50 10

34 11

7

17

015

-2

8 21

22

8 4

028

020

-1

ui 8 9 3 6 13

Fmin= 1404 Bng phn phi xe rng vi tng tnxkm xe rng t nht l: Tuyn ng S tn xe rng Ngc hi Thng tn 50 Chung Kim bi 24 Ba th Kim bi 10 H ng Kim lin 50 H ng Ga H ni 34 Trc sn Mai lnh 8

________________________________________________________________________ GV: Phan Thanh Tao


Kt hp cc trm c ngun xe (Ga H ni, Bn xe Kim lin), c th phn phi l trnh ti u nh sau: 1. Ga H ni (24 xi mng) Chung Kim bi (24 ng) H ng Ga H ni. 2. Ga H ni (10 xi mng) Ba th Kim bi (10 ng) H ng Ga H ni. 3. Kim lin (42 than) Ngc hi Thng tn (42 mui) H ng Kim lin. 4. Kim lin (8 than) Ngc hi Thng tn (8 mui) Trc sn Mai lnh (8 sn) H ng Kim lin.

4.4.5. Bi ton cm

Do yu cu k thut, phi hn ch khng c vn chuyn trn mt s tuyn ng no . Khi ta xem cc ph ca (i,j) b cm l cij= M kh ln( M). Tip tc thut ton th v bnh thng. V d 4.9

ai\bj 72 45 9 vj

22 5

22 3

7

0

60 60 1

0 2

M

1

5 M

3

5 4

1

23 M

23 2

5

1

16 12 3

+ 3

4 6

-1

ui 2 3 5

ai\bj 72 45 9 vj

22 5

22 3

7

0

60 60 1

2

M

1

5 M

3

5 4

1

23 M

23 2

5

1

16 12 3

0 3

4 6

-1

ui 2 3 5

________________________________________________________________________ GV: Phan Thanh Tao


4.5. Ci t thut ton th v

4.5.1. Khai bo d liu

a) Ma trn cc ph C=(cij)mxn , mng hng pht A=(ai)m, mng hng thu B=(bj)n , phng n X=( xij)mxnint a[m], b[n], c[m][n], x[m][n]; b) H thng th v {ui, vj}. int u[m], v[n]; c) Mng S cc chn v vng iu chnh V c khai bo l cc ma trn 0/1 nh du nh sau: int S[m][n], V[m][n]; vi ngha:

S[i][j]=1 (i,j) S v

V[i][j]=1 (i,j) V 4.5.2. Xy dng phng n c bn ban u

Tm phng n c bn ban u bng nguyn tc phn phi ti a v phng php gc ty bc. Cc mng nh du cc trm tha mn cha ( loi khi bng phn phi). int aa[m], bb[n];. vi ngha:

Trm Ai tha mn aa[i]=0 v

Trm Bj tha mn bb[j]=0 void phanphoi() { int i, j, dem=0; for (em=0; dem

Bi ging Quy hoch ton hc Trang 49 ________________________________________________________________________ bb[j]=1; a[i]-=b[j]; x[i][j]=b[j]; } } }

4.5.3. Xy dng h thng th v

n gin, lp nhiu nht m+n-1 ln cho vic kim tra c bng phn phi. Cc mng nh du cc th v ui, vj c tnh cha int uu[m], vv[n];. vi ngha:

ui c uu[i]=1 v

vj c vv[j]=1 void Thevi() { int i, j, uu[m]={0}, vv[n]={0}, dem; u[1]=0; uu[1]=1; for (dem=0; dem


4.5.5. Tm vng iu chnh

treo trong tp V l mt mnh trn dng hoc ct. Thut ton tm vng iu chnh V duy nht trn tp S bng cch xa tt c cc treo cho n khi khng cn th tp cn li l vng V cn tm. int TimVongDC( ) { int i,j,done=0,dem; for (i=0; i

Bi ging Quy hoch ton hc Trang 51 ________________________________________________________________________ j=0; while (j


10 45 65 120 25 10 7 9 8 120 4 5 2 3 60 1 2 6 2

***********************

________________________________________________________________________ GV: Phan Thanh Tao


Chng 5. PHNG PHP HUNGARY Phng Php ny c nh ton hc Hungary Egervary cng b trong mt bi bo

nm 1931, 16 nm trc khi phng php n hnh ca Dantzig ra i, nhng khng ai bit n, cho n nm 1953 nh ton hc M Kuhn dch bi bo v t tn l Phng php Hungary.

5.1. Bi ton b nhim

Cn phn n vic cho n ngi. Ngi i lm vic j th chi ph l cij (i,j=1..n). Hy phn cng vic cho n ngi tng chi ph thp nht.

t xij=1 nu ngi i lm vic j; ngc li t xij=0. M hnh ton hc:

g(x) = c=

n

i 1

=

n

j 1ijxij min

==

==

=

==

1..n)=j(i, 1hay 0ijx 1

)..1( 1

1)..1( 1

n

injijx

n

jniijx

Ma trn C=(cij)nxn gi l ma trn chi ph. Thc s c th b hn ch xij=0 hoc 1 thay xij l s t nhin th mi rng buc m bo xij=0 hoc 1. Do , rng buc xij=0 hoc 1 c vit li l xij0, nguyn. y l m hnh thc s ca bi ton vn ti. C th dng thut ton th v gii. Vi thut ton ny c 2n-1 chn. Tuy nhin ch c n chn khc 0, v bi ton suy bin. V vy c th c nhiu bc lp m phng n mi khng tt hn. R rng mi phng n l mt hon v ca cc s 1..n. V d hon v (4,2,1,3) ngha l ngi 1 lm vic 4, ngi 2 lm vic 2, ngi 3 lm vic 1 v ngi 4 lm vic 3. Mt cch vit hon v dng ma trn M=(mij)nxn, vi mij=1 khi v ch khi ngi i lm vic j. nh l 5.1. Nu ma trn chi ph ca bi ton b nhim c cc phn t khng m v c t nht n s 0, th mt phng n ti u tn ti nu n s 0 nm trong cc v tr cc s 1 ca ma trn hon v Pnxn. Ma trn P biu din phng n ti u. R rng, mi phng n u c tng chi ph khng nh hn 0, nn 0 l nh nht.

________________________________________________________________________ GV: Phan Thanh Tao


nh l ny cung cp mt mc tiu ca thut ton. Chng ta s chng t rng c th thay i chi ph m khng thay i li gii. Thut ton s trnh by cch sa i ma trn chi ph c cha cc s 0 trn mi dng v mi ct. nh l 5.2. Gi s ma trn chi ph l C=(cij)nxn. Gi s X=(xij)nxn l phng n ti u. Gi C l ma trn c c t C bng cch cng hng s vo dng th r. Th X cng l phng n ti u ca bi ton mi xc nh bi C. Chng minh Hm mc tiu ca bi ton mi l

g(x) = c=

n

i 1 =

n

j 1ijxij = c

=

n

rii 1

=

n

j 1ijxij + (c

=

n

j 1rj+)xrj

= c=

n

i 1

=

n

j 1ijxij + x

=

n

j 1rj

= c=

n

i 1

=

n

j 1ijxij +

v mi dng c tng bng 1. Do gi tr nh nht cho g(x) nhn c khi f(x) nh nht. Hay hai bi ton cng phng n ti u. Pht biu tng t cho vic cng thm hng s vo ct. Do , chin thut l sa i C bng cch cng thm vo mi dng/ct cc hng s.

V d 5.1. Gi s bi ton b nhim c ma trn chi ph

C=

4 5 2 5 3 1 1 4 12 3 6 3 12 6 5 9

Bt u rt gn mi dng c s 0 bng cch tr mi dng cho s nh nht trn dng .

2 3 0 3 2 0 0 3 9 0 3 0 7 1 0 4

Ct 1 khng c s 0. Tr ct 1 cho 2 c By gi c t nht 1 s 0 trn mi dng/ct.

0 3 0 3 0 0 0 3 7 0 3 0 5 1 0 4

________________________________________________________________________ GV: Phan Thanh Tao


Th gn cc s 1 cho ma trn hon v. Thc ra, khng cn ma trn hon v, m ch cn nh du * ti cc s 0 trn ma trn chi ph biu hin mt b nhim. Phi nh du * ti v tr (4,3) v dng 4 ch c mt s 0. Cn li bt u t dng 1 l (1,1), dng 2 l (2,2), dng 3 l (3,4).

0* 3 0 3 0 0* 0 3 7 0 3 0* 5 1 0* 4

May thay y l phng n ti u, v tng chi ph l: 4 + 1 + 3 + 5 = 13. Tuy nhin, khng phi lun gp may nh v d 1. V d 5.2

4 1 3 4 5 6 2 9 6 5 8 5 7 6 2 3

Tr mi dng cho phn t nh nht, c c

3 0 2 3 3 4 0 7 1 0 3 0 5 4 0 1

Tr mi ct cho phn t nh nht, c c

2 0 2 3 2 4 0 7 0 0 3 0 4 4 0 1

To mt cch nh du * tng dng, c c

2 0* 2 3 2 4 0* 7 0* 0 3 0 4 4 0 1

________________________________________________________________________ GV: Phan Thanh Tao


Ma trn ny khng biu din cch b nhim y ; ngi 4 cha c vic. C hai trng hp: hoc l khng th hon thnh vic nh du * cho cc s 0, hoc l c nhng thut ton khng tm ra. Lu u tin l cc s 0 trong ma trn nxn c tnh cht l tt c cc s 0 c th c ph bi n dng/ct. V d, chn n ct ph c ma trn. Gi s cc s 0 c th ph vi k dng/ct, k


Trong v d 5.2, v c th ph tt c cc s 0 vi 3 dng/ct, nn theo nh l trn th nhiu nht l 3 s 0 c nh du *. Ngha l khng th nh du * cho 4 s 0, v nh du * nhiu s 0 hn phi dng th tc m t trn. Mt kh nng khc khng hon thnh b nhim l thut ton tm theo dng tht bi. V d 5.3.

4 2 9 7 7 8 5 6 3 3 4 1 7 5 2 6

Tr mi dng, ri tr mi ct nh trn, c c

0 0 7 5 0 3 0 1 0 2 3 0 3 3 0 4

Trn mi dng, nh du * cho phn t 0 u tin khng nm trn ct nh du trc , c c

0* 0 7 5 0 3 0* 1 0 2 3 0* 3 3 0 4

vic b nhim cha hon thnh. Tuy nhin c mt cch nh du hon thnh l

0 0* 7 5 0* 3 0 1 0 2 3 0* 3 3 0* 4

Do phi khai trin mt thut ton tm ra cch nh du *. Cc bc ca thut ton nh sau:

________________________________________________________________________ GV: Phan Thanh Tao


Bc 1. Tr mi dng, mi ct mi dng v mi ct c t nht mt s 0. Bc 2. Trn mi dng, nh du * cho phn t 0 u tin khng nm trn ct nh du trc . Nu n s 0 c nh du th hon thnh b nhim, dng; c c phng n ti u. Bc 3. Gi s nh du t hn n s 0, xc nh c cch khc nh du hon thnh khng. Nu c th dng sau khi b nhim li. Bc 4. Nu vic nh du li khng hon thnh th tm k dng/ct (k


C hai cch xy dng dy ny c th kt thc. Mt cch khi thay 0 thnh 0* v ngc li. Cch khc l khi khi tt c cc c xo v chng nm trn ct cn thit. Trong trng hp ny, khng th b nhim, v phi sang bc 4. V d 5.4.

8 7 9 9 5 2 7 8 6 1 4 9 2 3 2 6

C=

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

C= nh du 0 bt u t dng 1 bc 2. Thy dng 2 l dng u tin khng c 0 trn ct cha nh du. im ny,

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

C= Ri bt u xy dng li dy 0 v 0* theo bc 3. u tin l (2,2), cha 0. Tm 0* trn ct 2 (1,2). Tm 0 trn dng 1 (1,4). Trn ct 4 khng c . Ri vo trng hp A. Dy l: (2,2), (1,2), (1,4). i 0 thnh 0* v ngc li, c

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

C= tng c 0* mt phn t. By gi lp li bc 3 cho dng 3. Khng c 0* trn dng 3; do xy dng dy bt u t (3,2). 0* trn ct 2 (2,2). Nhng khng c 0 trn dng 2. Ri vo trng hp B v ct 2 l ct cn thit. Dy l: 0 ti (3,2), 0* ti (2,2).

V tt c cc trong dy u nm trn ct cn thit nn phi sang bc 4 xc nh cc dng cn thit.

________________________________________________________________________ GV: Phan Thanh Tao


Mt dng c gi l cn thit nu c 0* trn ct khng cn thit. Bt u vi dng 1, tm 0* trn ct 4, v vy dng 1 l dng cn thit. Dng 2 c 0* trn ct cn thit, ct 2. Do dng 2 khng cn thit. Dng 3 khng c 0* v vy khng cn thit. Dng 4 c 0* trn ct 1 nn l dng cn thit. Chi tit bc 4

Ph mi dng v ct cn thit vi mt ng cho ra k ng ph nh m t trc y. Th tc ny t ng ph tt c cc phn t 0 ca C. C c ph nh sau

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

C= Sang bc 5. Tr 3 vo mi phn t khng ph, cng 3 vo mi phn t ph kp. C quay li bc 2 l

1 3 2 0 0 0 2 1 2 0 0 3 0 4 0 2

C= Hon thnh b nhim bc 2 nh sau

1 3 2 0* 0* 0 2 1 2 0* 0 3 0 4 0* 2

By gi c th vit li bc 3 trong thut ton nh sau: Gi s khng c 0 trn dng i0 c nh du v c mt phn t 0 ti (i0,j0). Xy dng mt dy theo ct ri theo dng xen k t 0 n 0* dn 0, n 0*, ... nh sau: (A) Nu ang 0 ti (ik,jk) tm 0* trn ct jk. Nu c th ni vo dy. Nu khng c th thay i trong dy vi 0 thnh 0* v 0* thnh 0 v tm dng tip theo khng c 0*. (B) Nu ang 0* ti (ik+1,jk) tm 0 trn dng ik+1. Nu c th ni vo dy. Nu khng c th nh du jk l ct cn thit v xo cc (ik,jk) v (ik+1,jk) ra khi dy. Nu c nhiu th xem l 0* (ik, jk-1). Lp li trng hp B. Nu khng c nhiu trong

________________________________________________________________________ GV: Phan Thanh Tao


dy, tm phn t 0 trn dng i0 khng nm trn ct cn thit. Nu tm thy trn ct j0 th lp li bc 3 bt u t (i0,j0). Nu khng c th sang bc 4. Thut ton ny gi l phng php Hungary vi cng trnh ca hai nh ton hc Konig v Egervary. Phng php Hungary gi s bi ton dng cc tiu. Tuy nhin c th sa i mt t gii cho bi ton cc i nh sau: Nu nhn -1 cho mi chi ph th chi ph dng thnh m v khng p dng c tiu chun ti u ca nh l 5.3. s dng phng php Hungary, sau khi nhn -1, cng thm chi ph ln nht trn ma trn gc, th tt c chi ph u khng m. V d 5.5. Gi s bi ton b nhim chi ph dng cc i cho bi

3 7 4 6 5 2 8 5 1 3 4 7 6 5 2 6

C = Ly chi ph ln nht l 8 tr cho tt c chi ph, c bi ton dng cc tiu tng ng l 5 1 4 2

3 6 0 3 7 5 4 1 2 3 6 2

---oOo---

________________________________________________________________________ GV: Phan Thanh Tao


5.2. Bi tp Gii cc bi ton b nhim dng min sau y: 1. 2.

4 5 3 8 9 2 4 3 1 2 6 5

10 20 15 6 5 11 8 7 12 4 10 5 10 18 2 10 3. 4.

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

9 3 5 7 5 2 8 6 6 3 4 2 5 4 8 3

Gii cc bi ton b nhim dng max sau y: 5. 6. 3 1 2 5 5

2 7 8 6 2 9 3 7 3 1 5 6 2 1 3 9 2 5 4 6

2 1 5 3 8 7 6 7 5 2 4 6 6 3 3 2

*********************

________________________________________________________________________ GV: Phan Thanh Tao


MC LC

Chng 1. BI TON QUY HOCH TUYN TNH PHNG PHP HNH HC................................................................................ 1

1.1. Cc bi ton thc t......................................................................................................... 1 1.1.1. Bi ton lp k hoch sn xut...................................................................................... 1 1.1.2. Bi ton vn ti ............................................................................................................. 2 1.1.3. Bi ton xc nh khu phn......................................................................................... 2

1.2. Bi ton qui hoch tuyn tnh ......................................................................................... 2 1.3. Phng php hnh hc .................................................................................................... 3 1.4. Bi tp ............................................................................................................................. 5

Chng 2. PHNG PHP N HNH................................................................................ 6 2.1. Dng chnh tc v dng chun tc................................................................................... 6

2.1.1. nh ngha..................................................................................................................... 6 2.1.2. Cc php bin i.......................................................................................................... 6 2.1.3. Phng n c bn.......................................................................................................... 7 2.1.4. Cc tnh cht ................................................................................................................. 7

2.2. Phng php n hnh .................................................................................................... 8 2.2.1. Ni dung........................................................................................................................ 8 2.2.2. Bng n hnh............................................................................................................... 9 2.2.3. C s l lun ............................................................................................................... 10 2.2.4. Cc bc ca thut ton n hnh............................................................................... 13 2.2.5. Bi ton n ph ........................................................................................................... 15 2.2.6. Bi ton n gi ............................................................................................................ 16

2.3. Ci t thut ton n hnh ........................................................................................... 20 2.3.1. Khai bo d liu.......................................................................................................... 20 2.3.2. Tnh cc c lng j ................................................................................................. 20 2.3.3. Kim tra ti u v tm n thay th .............................................................................. 20 2.3.4. Kim tra v nghim .................................................................................................... 20 2.3.5. Tm n loi ra .............................................................................................................. 21 2.3.6. Bin i bng .............................................................................................................. 21

2.4. Bi tp ........................................................................................................................... 22 Chng 3. BI TON I NGU....................................................................................... 24

3.1. Cc bi ton thc t....................................................................................................... 24 3.1.1. Bi ton lp k hoch sn xut.................................................................................... 24 3.1.2. Bi ton nh gi sn phm ........................................................................................ 24

3.2. Bi ton i ngu .......................................................................................................... 25 3.2.1. i ngu khng i xng ........................................................................................... 25 3.2.2. i ngu i xng ...................................................................................................... 26 3.2.3. S tucker ................................................................................................................ 28

3.3. Cc nguyn l i ngu................................................................................................. 28 3.3.1. Nguyn l 1................................................................................................................. 29 3.3.2. Nguyn l 2................................................................................................................. 29 3.3.3. Nguyn l 3 ( lch b)............................................................................................ 29

3.4. ngha kinh t.............................................................................................................. 30 3.4.1. ngha bi ton (2) .................................................................................................... 30

________________________________________________________________________ GV: Phan Thanh Tao


3.4.2. ngha bi ton ( 2~ )................................................................................................... 30 3.4.3. ngha nguyn l lch b ..................................................................................... 31

3.5. Bi tp ........................................................................................................................... 31 Chng 4. BI TON VN TI .......................................................................................... 33

4.1. Bi ton vn ti dng chnh tc ..................................................................................... 33 4.1.1. nh ngha................................................................................................................... 33 4.1.2. iu kin cn bng thu pht........................................................................................ 34

4.2. Bng phn phi v tnh cht.......................................................................................... 34 4.2.1. Bng phn phi ........................................................................................................... 34 4.2.2. Tnh cht ..................................................................................................................... 35

4.3. Thut ton th v ........................................................................................................... 36 4.3.1. Ni dung...................................................................................................................... 36 4.3.2. Xy dng phng n c bn ban u ......................................................................... 36 4.3.3. Cc bc ca thut ton th v.................................................................................... 37

4.4. Cc dng khc ............................................................................................................... 38 4.4.1. Khng cn bng thu pht ............................................................................................ 38 4.4.2. Suy bin ...................................................................................................................... 39 4.4.3. Dng cc i ............................................................................................................... 41 4.4.4. Bi ton xe rng.......................................................................................................... 45 4.4.5. Bi ton cm ............................................................................................................ 47

4.5. Ci t thut ton th v ................................................................................................ 48 4.5.1. Khai bo d liu.......................................................................................................... 48 4.5.2. Xy dng phng n c bn ban u ......................................................................... 48 4.5.3. Xy dng h thng th v............................................................................................ 49 4.5.4. Kim tra ti u ............................................................................................................ 49 4.5.5. Tm vng iu chnh ................................................................................................... 50 4.5.6. Bin i bng .............................................................................................................. 50

4.6. Bi tp ........................................................................................................................... 51 Chng 5. PHNG PHP HUNGARY ............................................................................. 53

5.1. Bi ton b nhim ......................................................................................................... 53 5.2. Bi tp ........................................................................................................................... 62

________________________________________________________________________ GV: Phan Thanh Tao

quy hoach tuyen tinh bai giang

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