QUYHOACHTUYENTINH ch6-CacBaiToanDacBiet

8/4/2019 QUYHOACHTUYENTINH ch6-CacBaiToanDacBiet

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Chng 6 Bi ton QHTT c bit 1 GV. Nguyen Vu Quang

CHNG 6: CC BI TON QUI HOCH TUYN TNHC DNG C BIT

I. Bi ton Vn ti (transportation problem)1.1 Thit lp bi ton

Dng quy hoch tuyn tnh tng qut:

iu kin c vng nghim kh d:

21 i m

s1 s2 si sm

n2 j1

d1 d2 djdn

cij,, xij

cij : chi ph vn chuyn 1 n v hng ha tim cung i n im cu jxij : l lng hng vn chuyn tim i n im j

jix

njdx

misxts

xcZMinimize

ij

m

i

jij

n

j

iij

m

j

ijij

n

i

,0

,...,2,1

,...,2,1.

1

1

11

=

=

=

=

=

==

== n

j

j

m

i

i ds11


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Dng cn bng ca bi ton:

V d:Mt doanh nghip c 3 kho cha sn phm XYZ cn phivn chuyn n 4 im bn l. Chi ph vn chuyn / snphm nh sau:

M hnh qui hoch tuyn tnh ca bi ton?

jix

njdx

misxts

xcZMinimize

ij

m

i

jij

n

j

iij

m

j

ijij

n

i

,0

,...,2,1

,...,2,1.

1

1

11

==

==

=

=

=

==

Khi tng cung = tng cu: ==

=n

j

j

m

i

i ds11

1208060100Nhu cu

10018673

140510762

12069751

4321

Cung cpim bn l

Kho

1208060100Nhu cu

10018673

140510762

12069751

4321

Cung cpim bn l

Kho


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Dng khng cn bng ca bi ton:

1. Nu , thm vo im n gi cn bng

2. Nu , thm vo im ngun gi cn bng

==

>n

j

j

m

i

i ds11

smm

dndjd2d1Nhu cu

siciji

s22

s11

nj21

Cung

cp

im nimngun

dn+1 = si - dj

n+1

dn+1

0

0

0

0

==

phi

gn gi tr 0 vo 1 rng no (tham kho ti liu)

A. Phng php duyt tun t(Stepping-Stone Method)0. Tm li gii ban u ( trnh by)1. Xt mi rng ij, tnh chs ci tin Iij2. Nu mi Iij 0, li gii ang c l ti u

Nu khng, chn c Iij m b nht v iu chnh xijxung quang ny

3. Quay tr li bc 1, cho n khi mi Iij khng m

V d: sau khi c li gii ban u theo phng php gcty bc, xt cc rng (1,3), (1,4), (2,1), (3,1), (3,2), (3,3)

1. V 1 ng i khp kn qua cc rng, ni lin cc khc c gn gi tr. (v d (3,1) xem hnh di)

nh du + hoc xen k ti mi gc ca ng i(bt u du + ti rng) (xem hnh di)

Chs ci tin Iij = tng i s chi ph vn chuyn vidu xc nh theo bc trn. (I31 = 7+7+5-5-7-1 = 6)

Kho im bn l Tng1 2 3 4 cung

5 7 9 6

100 20 1201

6 7 10 5

40 80 20 1402

7 6 8 1

100 1003

Tngcu 100 60 80 120 360

+

+

+


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Tng t: I13 = 1; I14 = 1; I21 = 1; I32 = 3; I33 = 2

Vy (1,3) c chs ci tin m. iu chnh nh sau:

2. Xc nh xij nh nht trong cc du () gi l xmin

Ly xij cc d

u () tr

xminLy xij cc du (+) cng xmin

7 9

20

7 10

40 80

3. Tnh cc Iij cho cc rng mi, u khng m, biton t ti u.

+

+


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B. Phng php phn phi ci tin (Modified Distribution method)

Thay v tnh Iij cho mi rng => tm gin tip qua h ccbin i ngu ui v vj

Gi ui v vj l cc bin i ngu ng vi cc im ngun iv im n j, ta s c

Iij = cij (ui + vj) = 0 ti tt c (i,j) c gn gi tr

Iij = cij (ui + vj) 0 ti cc rng

Cho u1 = 0, da vo h phng trnh xc nh c t cc gn gi tr, gii tt c ui v vj v xc nh Iij cho cc rng.

V d:

I13 = c13 (u1 + v3) = 20 + (0+2) = 18 I14 = -2I21 = c21 (u2 + v1) = -5 I31 = -15I32 = 9 I33 = 9

c ch s m b nht l (3,1), iu chnh li ny nh trnhby trong phng php A, sau tnh li cc bin i ngu ui, vj


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I11 = c11 (u1 + v1) = 15 I21 = 10 I13 = 18I32 = 9 I14 = -2 I33 = 9

(1,4) c ch s ci tin b m, iu chnh li ny nh trnhby trong phng php A, sau tnh li cc bin i ngu ui, vj


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1.3 Mt s trng hp c bit (tham kho ti liu)a. Bi ton vn ti a nghimb. Bi ton vn ti c hm mc tiu cc ic. Bi ton vn ti vi rng buc vng i

II. Bi ton phn cng (Assignment problem)2.1 Thit lp bi ton

Phn cng n cng vic cho n cng nhn, cij l chi ph cng nhn i thc hin cng vic j

Cng vicCng

nhn 1 j n

1i cij

n

Gi xij l bin s cho bit cng vic j c c giao cho cng

nhn i hay khng.

Dng quy hoch tuyn tnh ca bi ton:

2.2 Phng php HungarianV bng chi ph, cng vic c biu din theo ctBc 0: Xc nh ma trn rt gim

1. Tr mi hng cho s nh nht trn hng

2. Tr mi ct cho s nh nht trong ct Bc 1: Xc nh s ti u

1. V cc ng thng i qua cc gi tr 0 sao chos lng ng thng l t nht

2. Nu sng thng = n, tm c nghim ti u.Nu khng th thc hin bc 2.

Lu : nghim c xcnh qua cc gi tr0c lp (xem v dtrang sau)


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Bc 2: Bin i1. Chn s nh nht k khng nm trn cc ng

thng. Tr tt c cc s khng nm trn ngthng cho k, cng tt c cc s nm trn giao

im cc ng thng cho k.2. Quay li bc 1.

V d: chi ph trm ngn ngCng vicCng

nhn 1 2 3 4 51 2 3 5 1 42 -1 1 3 6 23 -2 4 3 5 04 1 3 4 1 45 7 1 2 1 2

Cng vicCngnhn 1 2 3 4 5

1 1 2 4 0 32 0 2 4 7 33 0 6 5 7 2

4 0 2 3 0 35 6 0 1 0 1


1 1 2 3 0 22 0 2 3 7 23 0 6 4 7 14 0 2 2 0 2

5 6 0 0 0 0


1 1 1 2 0 12 0 1 2 7 13 0 5 3 7 04 0 1 1 0 15 7 0 0 1 0


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1 1 1 2 0 12 0 1 2 7 13 0 5 3 7 0

4 0 1 1 0 15 7 0 0 1 0


1 1 0 1 0 02 0 0 1 7 03 1 5 3 8 04 0 0 0 0 05 8 0 0 2 0

x12 = x21 = x35 = x44 = x53 = 1, xij khc = 0; Z = 5 trm ngnx14 = x22 = x35 = x41 = x53 = 1x14 = x21 = x35 = x42 = x53 = 1x14 = x21 = x35 = x43 = x52 = 1

Nu vng thng khc th kt qu th no???Cng vicCng

nhn 1 2 3 4 51 1 1 2 0 12 0 1 2 7 13 0 5 3 7 04 0 1 1 0 15 7 0 0 1 0


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Trng hp hm mc tiu max

Nu bi ton c pij l li nhun/ hiu sut khi phn cng, tachuyn bi ton max thnh min bng cch:

i du pij hoc Ly pij ln nht tr cho tt c cc pij cn li

V d:Khu vc

i tuA B C D

1 20 60 50 552 60 30 80 75

3 80 100 90 804 65 80 75 70

Chuyn thnh bng min chi ph (c hi)Khu vc

i tuA B C D

1 80 40 50 452 40 70 20 25

3 20 0 10 204 35 20 25 30

Khu vci tu

A B C D1 40 0 10 52 20 50 0 53 20 0 10 20

4 15 0 5 10

Khu vci tu

A B C D1 25 0 10 02 5 50 0 03 5 0 10 154 0 0 5 5

Nghim ti u: 1-D, 2-C, 3-B, 4-A, Z = 300 triu


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III. Bi ton dng chy ti a (Maximum Flow problem)

3.1 Bi ton c cung nh hng:

M hnh bi ton:cij: dung lng ti a t nt i n nt jxij: lng vn chuyn t nt i n nt jf: l tng lng vn chuyn t nt 1 n nt cui cng mHm mc tiu: Max f

Rng buc: lng ra khi nt 1 = f lng vo nt m = f lng vo nt i (khc 1 v m) = lng ra khi nt icij xij 0

Cch gii: gii thut Ford-Fulkerson

[i

, vj]: nt j nhn thm/ gi

m b

t l

c vj t

nt iRing nt 1 gn nhn [_, ] : nt 1 c kh nng cung v hn

V d: tip theo v d trn

Khi nim tp ct: cha nt ngun, cha nt n

2

3

1

4

5

6

9

82

4

42

5

6

5


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V d:

Cc trng hp c nhiu nt ngun v nt n:

Thm nt ngun gi 0:c0A = cA1+cA2; c0B = cB1+cB2+cB3; c0C = cC2+cC3

Thm nt n gi n:c14+c24 = c4n; c15+c25+c35 = c5n ; c26+c36 = c6n

1

2

A 4

5

63

B

C

n0

2

3

1

4

5

67

109

5

16

3

5

9


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3.2 Bi ton c cung khng nh hng:

Cch gii: tng t vi bi ton c cung nh hng

4

2

1

3

5

10

20

5 10

30

40

20

20

30

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