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用两阶段法求解下列问题: (1) min f=2x1+x2-x3-x4 s.t.x1-

2022-08-12 13:56:54 问答库 阅读 196 次

问题详情

用两阶段法求解下列问题:
(1) min f=2x1+x2-x3-x4
s.t.x1-x2+2x3-x4=2,
2x1+x2-3x3+x4=6,
x1+x2+x3+x4=7,
xj≥0(j=1,2,3,4);
(2)max z=10x1+15x2+12x3,
s.t.5x1+3x2+x3≤9,
-5x1+6x2+15x3≤15,
2x1+x2+x3≥5,
x1,x2,x3≥0;
(3)max z=2x1-x2+2x3,
s.t.x1+x2+x3≥6,
-2x1+x3≥2,
2x2-x3≥0,
x1,x2,x3≥0;
(4)max z=5x1+3x2+6x3,
s.t.x1+2x2+x3≤18,
2x1+x2+3x3≤16,
x1+x2+x3=10,
x1,x2≥0,x3无符号限制.

参考答案

(1)x*=(3,0,1,3)T,f*=2.
(2)无可行解.
(3)无有限最优解.
(4)x*=(14,0,-4)T,z*=46.

考点:阶段,问题