运筹学清华大学出版社(胡运权着)课后答案(...

运筹学清华大学出版社(胡运权着

答解题习答解题习 版二第 程教学筹运 版二第 程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 2 e ga p 2 e ga p

)4 (

)3 (

0 2 x, 1x 2 2 x 3 1x 2 . ts 2 2 x 1x 2 x 6 1x 5 Z xam2

8 2x 5 01 1x 5 . ts 021 2 x 01 1x 6 x 1x Z xam2

)2 (

)1(

2

0 2 x, 1x 21 2 x 4 1x 3 . ts 2 2 x 1x 2 x 2 1x 3 Z xam

2

0 2 x, 1x 4 2 x 2 1 x 2 . ts 6 2 x 6 1x 4 x 3 1 x 2 Z nim

。解行可无是还解界无、解优最多穷无、解优最一惟有具题问出指并。题问划规性线列下解求法解图用 1.1

程教学筹运程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 3 e ga p 3 e ga p

解无题问该

)2(

2

解优最个一是3 Z,

0 2 x, 1x 2 1 2 x 4 1 x 3 . ts 2 2 x 1x 2 x 2 1 x 3 Z xa m

3 2 x,1 1x 1

2

解优最多穷无 0 2 x, 1x 4 2 x 2 1 x 2 . ts )1( 6 2 x 6 1x 4 x 3 1 x 2 Z n im

程教学筹运程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 4 e ga p 4 e ga p

解界无有题问该

)4 (

61 Z,6 2 x, 01 1x 解优最一唯 )3 (

0 2 x, 1x 2 2 x 3 1x 2 . ts 2 2 x 1x 2 x 6 1x 5 Z xam2

8 2x 5 01 1x 5 . ts 021 2 x 01 1x 6 x 1x Z xam2

程教学筹运程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 5 e ga p 5 e ga p

)2(

束约无 3 x, 0 2 x, 0 1 x 6 3 x 2 x 1 x 2 ts 4 3 x 2 x 1x x 3 2 x 2 1 x 2 Z nim3

)1(

4

束约无 4 x,0 3x, 2 x, 1x 2 4 x 3x 2 x 3 1x 2 ts . 41 4 x 2 3 x 2 x 1x 2 4 x 3 x 2 2 x 1x 4 x 5 3x 2 2 x 4 1x 3 Z nim

。式形准标成化题问划规性线述下将

2 .1

程教学筹运程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 6 e ga p 6 e ga p

0 6 x, 24 x, 14 x, 3x, 2 x, 1x 2 6 x 24 x 14 x 3x 2 x 3 1x 2 ts 41 5x 24 x 2 14 x 2 3x 2 x 1x 2 24 x 14 x 3x 2 2 x 1x 4 x 5 14 x 5 3x 2 2 x 4 1x 3 Z xam24

4

束约无4x,0 3x, 2x,1x 2 4x 3x 2x3 1x2 ts . )1( 41 4x2 3x 2x 1x 2 4x 3x2 2x 1x4 x5 3x2 2x4 1x3 Znim

程教学筹运程

教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 7 e ga p 7 e ga p

23

0 4 x, 23 x, 13 x, 2 x, 1 x 6 4 x 23 x 13 x 2 x 1x 2 ts 4 23 x 13 x 2 x 1x x 3 13 x 3 2 x 2 1x 2 Z xam

)2(

束约无 3 x, 0 2 x, 0 1x 6 3 x 2 x 1x 2 ts 4 3 x 2 x 1x x 3 2 x 2 1x 2 Z nim3

程教学筹运程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 8 e ga p 8 e ga p

)2(

4

) 4 ,1 j (, 0 j x 3 4 x 2 3 x 2 x 2 1x 2 ts 7 4 x 4 3 x 3 2 x 2 1x x 2 3 x 3 2 x 2 1x 5 Z nim

)1(

6, ,1 j 0 j x , 0 6 x 1x 3 ts 01 5 x 2 3 x 4 2 x 1x 8 9 4 x 3 3 x 6 2 x 3 1x 21 x 2 2 x 1x 3 Z xam3

。解优最定确并 解行可基是些哪出指 解基有所出找题问划规性线述下对 3 .1

程教学筹运程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 9 e ga p 9 e ga p

52.2 0 3 3 Z

52.2 0 0 06x

0 0 0 32x

57.0 0 0 01x

2 0 0 5 3 0 8 0 5.1 5.3 0 0 5x 4x 3x解行可基

)1(

, 6, ,1 j 0 j x 0 6 x 1x 3 ts 01 5 x 2 3 x 4 2 x 1 x 8 9 4 x 3 3 x 6 2 x 3 1x 21 x 2 2 x 1x 3 Z xam3

程教学筹运程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 0 1 e ga p 0 1 e ga p

5/34 5 5 Z

0 1 04x

0 0 5.02x

5/2 0 01x

5/11 1 2 3x解行可基

)2(

4

) 4 ,1 j (, 0 j x 3 4 x 2 3 x 2 x 2 1x 2 ts 7 4 x 4 3 x 3 2 x 2 1x x 2 3 x 3 2 x 2 1x 5 Z nim

程教学筹运程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 1 1 e ga p 1 1 e ga p

)1(

2

0 2 x, 1x 8 2 x 2 1x 5 . ts 9 2 x 4 1x 3 x 5 1x 01 Z xam

。点顶一哪的域行可中法解图应对解行可基各的中表形纯单出指照对并 题问划规性线述下解求法形纯单和法解图用别分 4.1

程教学筹运程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 2 1 e ga p 2 1 e ga p

)2(

0 2 x, 1x 42 2 x 2 1x 6 . ts 51 2 x 5 1x 3 x 1x 2 Z xam2

程教学筹运程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS j

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 3 1 e ga p 3 1 e ga p

c41/01-d41/3 c41/2+d41/553/01 41/24/34x 0

0 0 12x d

0 1 01x c

1 2/3 b

1x 2x

c dBC

41/53x 0

基→jC

下如表形纯单终最到得 解。优

最到达数函标目使次依点顶个每的域行可题问该使 化变何如值的d,c论讨 2xd+ 1xc= Z xam为变数函标目若 中)1(题上 5.l

程教学筹运程教学筹运

运筹学清华大学出版社(胡运权着

tnemeganaM fo loohcS tnemeganaM fo loohcS

1 1 0 2 y r a u n aJ 6 1 1 0 2 y r a u n aJ 6 4 1 e ga p 4 1 e ga p

。点原的中图为解优最时0于小d且01/3于小d/c当或时0于等于小c且2/5于大 d/c当 点 C的中图为解优最时 0于大 d且 01/3于小 d/c当 点B的中图为解优最时0于等于大c且2/5于大d/c当 点 A的中图 …… 此处隐藏：3752字，全部文档内容请下载后查看。喜欢就下载吧 ……