單純形法練習(xí)題

用單純形法求解下列線性規(guī)劃1、解：將線性規(guī)劃標(biāo)準(zhǔn)化令得到一組基本可行解，即：利用單純形法進(jìn)行求解Cj27404500bi基變量CBx1（2）x2（3）x3（1）x4x5x4012[3]10100x5022301120zj00000z=0λj27404500x3451/32/311/30100/3x50[1]00-1120zj153045150z=1500λj1201000-1500x3450[2/3]12/3-1/380/3x127100-1120zj273045312z=1740λj01000-30-120x240013/21-1/240x127100-1120zj274060137λj00-15-13-7z=2140最優(yōu)解為：，最優(yōu)目標(biāo)函數(shù)值為：。2、解：將線性規(guī)劃標(biāo)準(zhǔn)化令得到一組基本可行解，即：利用單純形法進(jìn)行求解Cj1020000bi基變量CBx1（2）x2（1）x3x4x5x301110050x401001020x502[6]001120zj000000λj1020000x302/3010-1/630x40[1]001020x2201/31001/620zj20/3200020/6z=400λj10/3000-10/3x30000-2/3-1/650/3x1101001020x220010-1/31/640/3zj1020010/310/3z=1400/3λj000-10/3-10/3最優(yōu)解為：，最優(yōu)目標(biāo)函數(shù)值為：。3、解：將線性規(guī)劃標(biāo)準(zhǔn)化令得到一組基本可行解，即：利用單純形法進(jìn)行求解Cj35000bi基變量CBx1（2）x2（1）x3x4x5x30101004x400[2]01012x503200118zj000000λj35000x30101004x250101/206x50[3]00-116zj0505/20z=30λj300-5/20x300011/3-1/32x250101/206x13100-1/31/32zj3503/21z=36λj000-3/2-1最優(yōu)解為：，最優(yōu)目標(biāo)函數(shù)值為：。4、

解：將線性規(guī)劃標(biāo)準(zhǔn)化令得到一組基本可行解，即：利用單純形法進(jìn)行求解Cj2-11000bi基變量CBx1（2）x2（1）x3x4x5x6x4031110060x50[1]-1201010x6011-100120zj000000z=0λj2-11000x4004-51-3030x121-1201010x600[2]-30-1110zj2-24020z=20λj01-30-20x400011-1-210x12101/201/21/215x2-101-3/20-1/21/25zj2-15/203/21/2z=25λj00-3/20-3/2-1/2最優(yōu)解為：，最優(yōu)目標(biāo)函數(shù)值為：。5、解：將線性規(guī)劃標(biāo)準(zhǔn)化令得到一組基本可行解，即：利用單純形法進(jìn)行求解Cj3400bi基變量CBx1（2）x2（1）x3x4x3011105x401[2]016zj0000z=0λj3500x30[1/2]01-1/22x241/2101/23zj2402z=12λj100-2x13102-14x2401-111zj3421z=16λj00-2-1最優(yōu)解為：，最優(yōu)目標(biāo)函數(shù)值為：。6、解：將線性規(guī)劃標(biāo)準(zhǔn)化令得到一組基本可行解，即：利用單純形法進(jìn)行求解Cj132000bi基變量CBx1x2（1）x3(2)x4x5x6x401[2]-21002x502110103x60-1010014zj000000-z=0λj132000x231/21-11/2001x503/20[2]-1/2102x60-1010014zj3/23-33/200-z=3λj-1/205-3/200x235/4101/41/202x323/401-1/41/201x60-7/4001/4-1/213zj21/4321/45/20-z=8λj-17/4-30-1/4-5/20最優(yōu)解為：，最優(yōu)目標(biāo)函數(shù)值為：。7、解：將線性規(guī)劃標(biāo)準(zhǔn)化令得到一組基本可行解，即：利用單純形法進(jìn)行求解Cj23100bi基變量CBx1（2）x2（1）x3x4x5x301-21105x402-12016zj000000λj231008、解：將線性規(guī)劃標(biāo)準(zhǔn)化令得到一組基本可行解，即：利用單純形法進(jìn)行求解Cj1520000bi基變量CBx1（2）x2（1）x3x4x5x303210080x402[3]01080x501100135zj00000z=0λj1520000x30[5/3]01-2/3080/3x2202/3101/3080/3x501/300-1/3125/3zj40/320020/30z=1600/3λj5/3