Question

两阶段法求解,mins=2x1-x2+x3( x1+2x2- x3=12x1+ x2+ x3=5x1- x2+2x3=4

Answer 1

用两阶段法求解

min f=2x1-x2+x3

s.t x1+2x2- x3=1

2x1+ x2+ x3=5

x1- x2+2x3=4

xi>=0,i=1,2,3

引入人工变量x4,x5,x6,使

min g=x4+x5+x6

s.t x1+2x2-x3+x4=1

2x1+x2+x3+x5=5

x1-x2+2x3+x6=4

xi>=0,i=1,...,6

其单纯形表是

基变量 x1 x2 x3 x4 x5 x6 右项

-g 0 0 0 1 1 1 0

x4 1 2 -1 1 0 0 1

x5 2 1 1 0 1 0 5

x6 1 -1 2 0 0 1 4

-f 2 -1 1 0 0 0 0

将x4,x5,x6这三行乘以-1加到-g这行,得表

基变量 x1 x2 x3 x4 x5 x6 右项

-g -4 -2 -2 0 0 0 -10

x4 1 2 -1 1 0 0 1

x5 2 1 1 0 1 0 5

x6 1 -1 2 0 0 1 4

-f 2 -1 1 0 0 0 0

因为-4