運(yùn)籌學(xué)周四專(zhuān)題知識(shí)

求解如下線性規(guī)劃問(wèn)題111-211000113-4120-1103/21-20[1]000110M16xx4

x3103-20100-110[1]00-11-211-2010001-3x11x21x341001/3-2/32/3-5/310100-11-2900102/3–4/3-7/3

-31100

MM

-10001M-1M+1zj-cj

0001/31/3M-1/3M-2/3zj-cj0x41x21x312[3]001-22-5410100-11-21-2010001cj0MMX4X6X7

b

CBXBX1X2X3X4X5X6X7

i-3+6M1-M1-3M0M00cj-zj

-11-M00M03M-1cj-zj最優(yōu)解是目的函數(shù)為-2。

第一階段：求解輔助規(guī)劃問(wèn)題2、兩階段法

111-211000113-4120-1103/21-20[1]000110106xx4

x3103-20100-110[1]00-11-211-2010001

00000

11

0000011zj-cj0x40x20x312[3]001-22-5410100-11-21-2010000cj011X4X6X7

b

CBXBX1X2X3X4X5X6X7

i6-1-30100cj-zj

0-100103cj-zj12[311-20100-3112xx1

x341001/3-2/310100-190012/3-4/3

-31100cj011X4X2X3

b

CBXBX1X2X3X4X5

i-10001cj-zj

0001/31/3cj-zjx6，x7是人工變量，第一階段求解旳最優(yōu)成果是=0，所以得最優(yōu)解為：

第二階段：取消人工變量，添入原問(wèn)題目旳函數(shù)旳系數(shù)，求解相應(yīng)旳線性規(guī)劃。最優(yōu)解為：最優(yōu)值為：z=-2（無(wú)可行解）求解下列線性規(guī)劃問(wèn)題解：首先將問(wèn)題化為原則型令，則

故引入人工變量，并利用大M法求解C

-3-2-1000-M-M

CB

XB

b

x1x2x3x4x5x6x7x8

θ

0-M-M

x4x7x8

643

1111000010-10-101001-100-101

6/1-3/1

Z’

-7M

-6-4M

-15-M

-3+M-2+M-1-2M0-M-M00

0-M-2

x4x7x2

343

1021010-110-10-101001-100-101

3/14/1-

Z’

Z’

-3+M0-3-M0-M-202-M

-3-M-2

x1x7x2

313

1021010-100-3-1-1-11101-100-101

003-3M3-M-M1-M0-1

在以上最優(yōu)單純形表中，全部非基變量檢驗(yàn)數(shù)都不大于零，但在該表中人工變量x7=1為基變量，所以原線性規(guī)劃不存在可行解。（無(wú)界解）試用單純形法求解下列線性規(guī)劃問(wèn)題：解：引入松弛變量x3,x4化為原則型因但所以原問(wèn)題無(wú)最優(yōu)解（退化）求解下述線性規(guī)劃問(wèn)題：解：引入松弛變量化原則型000-242-8030Z-5-60-420-805Z10001001x3

212060-2411x1

3321300-803x5

00-30-425-800Z11001001x7

00106-1-2410x1

30-1130-3-800x5

0-11001001x7

000106-1-2410x6

0000136-4-3210x5

0x7

x6

x5

x4

x3

x2

x1

b

XB

CB

000-242-803C

θ

第一次迭代中使用了攝動(dòng)法原理，選擇下標(biāo)為6旳基變量x6離基?？傻米顑?yōu)解，目的函數(shù)值maxZ=５，000-242-8030Z00-3-132141600Z11001001x7

001-1-30380000-1136-4-3210x5

3-11001001x7

000106-1-2410x6

0000136-4-3210x5

0x7

x6

x5

x4

x3

x2

x1

b

XB

CB

θ

x1

勃蘭特法則無(wú)窮多最優(yōu)解：用大M法或者二階段法求解原則型：其中M是一種任意大旳正數(shù)CB

XB

b

x6

x7

x1

x2

x3

x4

x5

x6

x7

CB

XB

b

x7

x1

x2

x3

x4

x5

x6

x7

x2

CB

XB

b

x1

x2

x3

x4

x5

x6

x7

x2

x1

可得最優(yōu)解（4/5,5/8,0,0,0,0,0）如下表，是某求極大化線性規(guī)劃問(wèn)題計(jì)算得到旳單純形表。表中無(wú)人工變量，a1，a2，a3，d，c1，c2為待定常數(shù)。試闡明這些常數(shù)分別取何值時(shí)，下列結(jié)論成立。（1）表中解為唯一最優(yōu)解（2）表中解為最優(yōu)解，但存在無(wú)窮多最優(yōu)解（3）該線性規(guī)劃問(wèn)題具有無(wú)界解（4）表中解非最優(yōu)解，為對(duì)解改善，換入變量為x1，換出變量為x6

基BX1

X2 X3 X4 X5 X6X3 d 4 a1 1 0 a2 0X4 2 -1 -3 0 1 -1 0X6 3 a3 -5 0 0 -4 1

Cj-Zj

c1 c2 0 0 -3 0（1）當(dāng)解為最優(yōu)解時(shí)，必有d≥0，c1?0，c2?0。（2）當(dāng)