Question

运筹学单纯形或改进单纯形法 maxz=3x1+x2 x1+x2≤4 -x1+2x2≤3 5x1+x2≤15 x1 x2≥0 ...

运筹学单纯形或改进单纯形法
maxz=3x1+x2
x1+x2≤4
-x1+2x2≤3
5x1+x2≤15
x1 x2≥0
x为字母

Answer 1

cj 3 1 0 0 0
CB XB B-1b x1 x2 x3 x4 x5 θi
0 x3 4 1 1 1 0 0 4
0 x4 3 -1 2 0 1 0 /
0 x5 15 (5) 1 0 0 1 3
cj-zj 3 1 0 0 0
0 x3 1 0 (4/5) 1 0 -1/5 5/4
0 x4 6 0 11/5 0 1 1/5 30/11
3 x1 3 1 1/5 0 0 1/5 15
cj-zj 0 2/5 0 0 -3/5
1 x2 5/4 0 1 5/4 0 -1/4
0 x4 13/4 0 0 -11/4 1 3/4
3 x1 11/4 1 0 -1/4 0 1/4
cj-zj 0 0 -1/2 0 -1/2 所以最优解为(11/4,5/4,0,13/4,0)T
z*=19/2

Answer 2

用清华版的运筹学基础里的单纯形表计算就可以了啊