哈工大運(yùn)籌學(xué)

minz2x14x16x22x12x2x1.x2(3)maxzx16x110x25x13x2

maxz3x12x1x23x14x2x1.x2(4)maxz5x12x1x22x13x2x1.x2 maxz10x13x14x25x12x2

maxz10x15x20x30x43x14x2x35x12x2x4x1 x3,求解過(guò) 5 500c 3410 201500 x32/50110 10010 015/1410 1000

00 0 98 8 8/ x

/50 0

3/2x 21000210000510020105110012100005100410010010000015/7/1003/20101/3/20001/5x26x12x21x2a 6136131001200

11/2 1/2 7j221 x1 x3000603 0010000010021011 2011121 11 01/2 5 5 x2 2/3101/34/3052/35/3041/30400100000101/418/416/411/4100543 學(xué)型 28004500007300 甲乙丙5796847售價(jià)（元/件）2.2寫(xiě)出下列線性規(guī)劃的對(duì)偶 z2x12x24xx13x24x32x1x23x3x14x23x3x1x20x3無(wú)約 (3)minz i1j

max5x16x2x12x22x3x15x2x34x17x23x3無(wú)約束x30x3n(4)maxznj xij

i1,

j

aijxj

i ,m1m

aijxj

ij

j1,

j

im

1 ,xij

i1, m

j1, n

xj

j ,n1njn11 ,maxzx1x1x2x32x1x2x3xjmaxz2x14x2x3x13x2x42x1x2x2x3x4x1x2x3 2已知原問(wèn)題的最優(yōu)解為2,2,4,02.6分別把A、Bmaxzcx1cx2 a1nxn a2nxn a3nxn

maxzcx1cx2 5a1nxn/5a21x11/5a22x2 a2nxn1 (a3n3a1n)xnj0,xj1, j0,xj1, nminwb1y1b2y2b3a11y1a21y2a31y3a12y1a22y2a32y3

對(duì)偶問(wèn)?1?15a12?11/5a22y?2(a323a12)?3 a1ny1a2ny2a3ny3y1y3無(wú)約束，y2

n?2?3等價(jià)?1?1?3)?1?3)?3 ?1?3)2n??3

?1?3y21/5?y3?2.7利用對(duì)偶單純形法3(1)minz4x112x23x13x32x22x3

xj0,j

2 maxz4x112x2x13x3x42x22x3x5xj0,j

cj 5/cj 3/cj

1/ 1/ 1/3 1/ 已知線性規(guī)劃問(wèn)maxz2x1x2x1x2x3(1)目標(biāo)函數(shù)變?yōu)?2)第一個(gè)約束條件右 6變?yōu)?3)增加一個(gè)約束條件-

10

b

,b1b

4 cj

4

0

110

0 cj

31

bb633x(3,0,0,0,7),z

cj

cj

8/

2/ 2/

10/cj

1/ 1/ 1/ 4/37/31/

cj 10/ 22/ 8/cj

2/ 2/ 1/ 8/ 2/ 1/ 1/ 1/ 8/ 8/ 1/maxz4x13x2x12x2x32x13x23x3x1,x2,x3minw10y110y2y12y22y13y2y13y2 考慮如下線性規(guī)劃問(wèn)題

minz60x140x2803x12x2x34x1x23x32x12x22x3

maxw2y14y23y14y22y3

y13y22y3 0004001110001000010 80/3020/32430000321000212010013200124300000004110 00101000101000010002.9己知線性規(guī)劃問(wèn)題用單純32000032000004/012/000102/00030011002/002/010cj004/00x12x2x1x2x2(1)第①,②個(gè)約束條件的右端項(xiàng)分別由6變?yōu)?,由8變?yōu)?2)目標(biāo)函數(shù)變?yōu)?3)增加一個(gè)變量x3，其在目標(biāo)函數(shù)中的系數(shù)c3=4,系數(shù)(4)問(wèn)題中x2的系數(shù)列向量變?yōu)?5)增加一個(gè)新的約束條件x4≤4(1)minzx1x2x32x1x23x4xj

將λ 將λ

1/

1/

3/

cj 1/

23/

0cj0cjzjcj25012 1/ 1/ 1/ 3/ 1/ 3/

cjcj002 1012 2103cj00423 23

1+λ/3-4≤0,-λ/3+4/5≤0,-λ/3-解得所需勞動(dòng)力，材料等有關(guān)數(shù)劇見(jiàn)c635345產(chǎn)品利潤(rùn)（元件314(1)確定獲利最大的產(chǎn)品生產(chǎn)設(shè)生產(chǎn)Α,Β,c產(chǎn)品的數(shù)量(1)確定獲利最大的產(chǎn)品生產(chǎn)設(shè)生產(chǎn)Α,Β,c產(chǎn)品的數(shù)量x1,x2,x3件,得數(shù)學(xué)模型并求解6x13x25x33x14x25x3x1,x2,x331431400xx0x6351003401cj314000x014x63/4/101/cj3/11/004/3511/01/1/4x30111/2/cj001/3/

(3)如果設(shè)計(jì)一種新產(chǎn)D,其問(wèn)是否值得生產(chǎn)解σ=3-(1/5,3/5)(8,2)=1/5>0,值得生產(chǎn) (3)如果設(shè)計(jì)一種新產(chǎn)D,其問(wèn)是否值得生產(chǎn)解σ=3-(1/5,3/5)(8,2)=1/5>0,值得生產(chǎn)要想使最優(yōu)解不變，基變量值≥05-λ/3≥0,3+2λ/5≥0解得552792最優(yōu)解不變(1)單位運(yùn)價(jià)表中第r行最優(yōu)解不變(1)單位運(yùn)價(jià)表中第r行的每個(gè)cij都加上一個(gè)常數(shù) ij(2)單(2)單位運(yùn)價(jià)表中第列cij icijui iuiii都加上一個(gè)常數(shù)vj

其

jK>0最K>0最優(yōu)解不產(chǎn)5055銷(xiāo)51792第1第2第30020203M04M02MM01MM0333317(3)若x1≤2,則x2≥1;否則(3)若x1≤2,則x2≥1;否則(2)變量x只能取0,3,5,7中的一(4)以下四個(gè)約束條件中至少滿足兩x1+x2≤5x1≤2x3≥2(1)x(1)x1+x2≤2或((22444100128530204030203502

3

4

4 ABCDE甲乙丙丁 P1:利潤(rùn)指標(biāo)為每月16000

P2充分利用生產(chǎn)能力

6x18x210x3dd

ddd P4:產(chǎn)量以銷(xiāo)量為

x1dd x2dd x3dd minzp1dp2d

p3dp4(d

ddd

dd 量為6，8，6，10t11

x11x12x13x14d 2x21x22x23x24d 2

x

x32x33x34d

xxxx

x21 x31d45 x x d456 x x d67 x x d7

386

x34d8 365224418 365224414363 i1jminzp1dp2d p3(d p4 (1)minzp1dp2dp3d x12x2ddx12x2d

d x12x2d

dx1,x20;d

i minzp1dp2dp3(d 6x12x2dd x1x2dd 5x1d3d3x1,x20;d (1)minzp1dp2dp3(d 3x1x2x3dd x1x22x3ddx1x2

x3d

10102233 1 2 3 100001100100 100001010000 00010011000002 xi0;d,d

x1

d1304d130411300102001001020011 d3 d3

0501300000000100030010231001131001112d1 d1

151 1/21/21/2

5013/2 01/21/21/2000112000000100000000011 0 x1

d d d3 d3

50

1/21/21/21 151

1/41/41/4

015/4

1/43/4 000000000000000100p30

1/21/2

1/2 00 30100111200102001001000100000000000100000000011F、B、J、A最后到達(dá)M,請(qǐng)標(biāo)ABABCDEF1***2**3**4***5***6**7***8**9******BEAFDC278578573212321342223224476545858239843 7 5(5)3(3)s

b4(4)2(0) e A B302ED2B3 C34-31001-010d001c-000 2minz2x1x22x3x1x2x3x1x2x3x10x20x33(1)求a-g（2）2c01ade01b-fg4maxz5x13x2536-00801001536-00801001512000-401-100-0002x1x23x3x1x2x3(1)寫(xiě)出對(duì)偶問(wèn)題5某一實(shí)際的問(wèn)題可以敘述如下：6已知 ABABCDE137已知線性規(guī)劃問(wèn)minz2x1x2x1x2x3x1x2kx3x10x20x3無(wú)約最優(yōu)解為x15x20x3(1)求k的值（2）8--0--000 00-13-15x1100-33 01000006x1+x2-x2- x11 x50 Cj-0 Maxz=2x1- 6變?yōu)槿绫硭镜膯?wèn)題，若產(chǎn)地I有一個(gè)單位物資未運(yùn)出，則將發(fā)生費(fèi)用，假定1,2,3產(chǎn)地單位物資位，產(chǎn)地3的物資至少運(yùn)出27個(gè)單位，試求此問(wèn)題AB12 3020某公司下屬三個(gè)小型煤礦A1,A2,A3，每天煤炭的生產(chǎn)量P1：A1產(chǎn)地因庫(kù)存限制，應(yīng)盡量全部調(diào)365224414363用匈牙利法求解指派 5 5 2寫(xiě)出下題的動(dòng)態(tài)規(guī)劃的基本2x1(1)maxz5x1x1x1x2x1,x2

29x22x 一、填空(10分個(gè)基，該基對(duì)應(yīng)的決策變量