运筹学习题集02.docx
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运筹学习题集02.docx
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运筹学习题集02
三、用图解法求解以下线性规划问题:
(1)
max
z=
x1
+3x2
s.t.
x1
+x2
≤10
-2x1
+2x2
≤12
x1
≤7
x1,
x2
≥0
x2
10
(2,8)
6
x1
-60710
最优解为(x1,x2)=(2,8),maxz=26
(2)
min
z=
x1
-3x2
s.t.
2x1
-x2
4
x1
+x2
3
x2
5
x1
4
x1,
x2
0
x2
5
3
x1
0234
最优解为(x1,x2)=(0,5),minz=-15
(3)
max
z=
x1
+2x2
s.t.
x1
-x2
1
x1
+2x2
4
x1
3
x1,
x2
0
x2
2
x1
01234
多个最优解,两个最优极点为(x1,x2)=(2,1),和(x1,x2)=(0,2),maxz=5
(4)
min
z=
x1
+3x2
s.t.
x1
+2x2
4
2x1
+x2
4
x1,
x2
0
x2
x1=0
4
x4=0
2
x3=0
x2=0x1
024
最优解为(x1,x2)=(4,0),minz=4
八、用单纯形表求解以下线性规划问题
(1)
max
z=
x1
-2x2
+x3
s.t.
x1
+x2
+x3
≤12
2x1
+x2
-x3
≤6
-x1
+3x2
≤9
x1,
x2,
x3
≥0
解:
标准化,将目标函数转变成极小化,引进松弛变量x4,x5,x60,得到:
min
z’=
-x1
+2x2
-x3
s.t.
x1
+x2
+x3
+x4
=12
2x1
+x2
-x3
+x5
=6
-x1
+3x2
+x6
=9
x1,
x2,
x3,
x4,
x5,
x6
≥0
列出初始单纯形表
z’
x1
x2
x3
x4
x5
x6
RHS
z’
1
1
-2
1
0
0
0
0
x4
0
1
1
[1]
1
0
0
12
12/1
x5
0
2
1
-1
0
1
0
6
--
x6
0
-1
3
0
0
0
1
9
--
选取x3为进基变量,确定x4为离基变量
z’
x1
x2
x3
x4
x5
x6
RHS
z’
1
0
-3
0
-1
0
0
-12
x3
0
1
1
1
1
0
0
12
12/1
x5
0
[3]
2
0
1
1
0
18
18/3
x6
0
-1
3
0
0
0
1
9
--
得到最优解(x1,x2,x3,x4,x5,x6)=(0,0,12,0,18,9),minz’=-12,maxz=12
由于其中非基变量x1在目标函数中的系数为0,x1进基,x5离基,可以得到另一最优解:
z’
x1
x2
x3
x4
x5
x6
RHS
z’
1
0
-3
0
-1
0
0
-12
x3
0
0
1/3
1
2/3
-1/3
0
6
x1
0
1
2/3
0
1/3
1/3
0
6
x6
0
0
11/3
0
1/3
1/3
1
15
新的最优解为(x1,x2,x3,x4,x5,x6)=(6,0,6,0,0,15),minz’=-12,maxz=12
原问题最优解的全体为:
,(0≤≤1),都有maxz=12
(2)
min
z=
-2x1
-x2
+3x3
-5x4
s.t.
x1
+2x2
+4x3
-x4
≤6
2x1
+3x2
-x3
+x4
≤12
x1
+x3
+x4
≤4
x1,
x2,
x3,
x4
≥0
引进松弛变量x5,x6,x70,得到
min
z=
-2x1
-x2
+3x3
-5x4
s.t.
x1
+2x2
+4x3
-x4
+x5
=6
2x1
+3x2
-x3
+x4
+x6
=12
x1
+x3
+x4
+x7
=4
x1,
x2,
x3,
x4,
x5,
x6,
x7
≥0
初始单纯形表为
z
x1
x2
x3
x4
x5
x6
x7
RHS
z
1
2
1
-3
5
0
0
0
0
x5
0
1
2
4
-1
1
0
0
6
--
x6
0
2
3
-1
1
0
1
0
12
12/1
x7
0
1
0
1
[1]
0
0
1
4
4/1
x4进基,x7离基
z
x1
x2
x3
x4
x5
x6
x7
RHS
z
1
-3
1
-8
0
0
0
-5
-20
x5
0
2
2
5
0
1
0
1
10
10/2
x6
0
1
[3]
-2
0
0
1
-1
8
8/3
x4
0
1
0
1
1
0
0
1
4
--
x2进基,x6离基
z
x1
x2
x3
x4
x5
x6
x7
RHS
z
1
-10/3
1
-22/3
0
0
-1/3
-14/3
-68/3
x5
0
4/3
0
19/3
0
1
-2/3
5/3
14/3
x2
0
1/3
1
-2/3
0
0
1/3
-1/3
8/3
x4
0
1
0
1
1
0
0
1
4
最优解为:
(x1,x2,x3,x4,x5,x6,x7)=(0,8/3,0,4,14/3,0,0),minz=-68/3
(3)
min
z=
3x1
-x2
s.t.
-x1
-3x2
≥-3
-2x1
+3x2
≥-6
2x1
+x2
≤8
4x1
-x2
≤16
x1,
x2
≥0
将第一、第二个约束条件两边同乘以-1;引进松弛变量x3,x4,x5,x6≥0,得到
min
z=
3x1
-x2
s.t.
x1
+3x2
+x3
=3
2x1
-3x2
+x4
=6
2x1
+x2
+x5
=8
4x1
-x2
+x6
=16
x1,
x2,
x3,
x4,
x5,
x6
≥0
初始单纯形表为:
z
x1
x2
x3
x4
x5
x6
RHS
z
1
-3
1
0
0
0
0
0
x3
0
1
[3]
1
0
0
0
3
3/3
x4
0
2
-3
0
1
0
0
6
--
x5
0
2
1
0
0
1
0
8
8/1
x6
0
4
-1
0
0
0
1
16
--
x2进基,x3离基,
z
x1
x2
x3
x4
x5
x6
RHS
z
1
-10/3
0
-1/3
0
0
0
-1
x3
0
1/3
1
1/3
0
0
0
1
x4
0
3
0
1
1
0
0
9
x5
0
5/3
0
-1/3
0
1
0
7
x6
0
13/3
0
1/3
0
0
1
17
最优解为:
(x1,x2,x3,x4,x5,x6)=(0,1,0,9,7,17),minz=-1
用两阶段法求解以下线性规划问题
(1)
max
z=
x1
+3x2
+4x3
s.t.
3x1
+2x2
≤13
x2
+3x3
≤17
2x1
+x2
+x3
=13
x1,
x2,
x3
≥0
解:
将目标函数转化成极小化,引进松弛变量x4,x5,x6≥0,得到
min
z’=
-x1
-3x2
-4x3
s.t.
3x1
+2x2
+x4
=13
x2
+3x3
+x5
=17
2x1
+x2
+x3
=13
x1,
x2,
x3,
x4,
x5,
≥0
引进人工变量x60,构造辅助问题:
min
z’’=
x6
s.t.
3x1
+2x2
+x4
=13
x2
+3x3
+x5
=17
2x1
+x2
+x3
+x6
=13
x1,
x2,
x3,
x4,
x5,
x6
≥0
列出辅助问题的系数矩阵表:
z’’
x1
x2
x3
x4
x5
x6
RHS
z’’
1
0
0
0
0
0
-1
0
x4
0
3
2
0
1
0
0
13
x5
0
0
1
3
0
1
0
17
x6
0
2
1
1
0
0
1
13
消去基变量x6在目标函数中的系数,并开始单纯形叠代:
z’’
x1
x2
x3
x4
x5
x6
RHS
z’’
1
2
1
1
0
0
0
13
x4
0
[3]
2
0
1
0
0
13
13/3
x5
0
0
1
3
0
1
0
17
--
x6
0
2
1
1
0
0
1
13
13/2
x1进基,x4离基,
z’’
x1
x2
x3
x4
x5
x6
RHS
z’’
1
0
-1/3
1
-2/3
0
0
13/3
x1
0
1
2/3
0
1/3
0
0
13/3
--
x5
0
0
1
3
0
1
0
17
17/3
x6
0
0
-1/3
[1]
-2/3
0
1
13/3
13/3
x3进基,x6离基,
z’’
x1
x2
x3
x4
x5
x6
RHS
z’’
1
0
0
0
0
0
-1
0
x1
0
1
2/3
0
1/3
0
0
13/3
x5
0
0
2
0
2
1
0
4
x3
0
0
-1/3
[1]
-2/3
0
1
13/3
辅助问题已经获得最优解,且minz’’=0,因而可以转入第二阶段,其系数矩阵表为:
z’
x1
x2
x3
x4
x5
RHS
z’
1
1
3
4
0
0
0
x1
0
1
2/3
0
1/3
0
13/3
x5
0
0
2
0
2
1
4
x3
0
0
-1/3
1
-2/3
0
13/3
消去基变量x1,x3在目标函数中的系数:
z’
x1
x2
x3
x4
x5
RHS
z’
1
0
11/3
0
7/3
0
-65/3
x1
0
1
2/3
0
1/3
0
13/3
13/2
x5
0
0
[2]
0
2
1
4
4/2
x3
0
0
-1/3
1
-2/3
0
13/3
--
x2进基,x5离基
z’
x1
x2
x3
x4
x5
RHS
z’
1
0
0
0
-4/3
-11/6
-29
x1
0
1
0
0
-1/3
-1/3
3
x2
0
0
1
0
2
1/2
2
x3
0
0
0
1
-1/3
1/6
5
得到原问题的最优解:
(x1,x2,x3)=(3,2,5),minz’=-29,maxz=29
(2)
max
z=
2x1
-x2
+x3
s.t.
x1
+x2
-2x3
≤8
4x1
-x2
+x3
≤2
2x1
+3x2
-x3
≥4
x1,
x2,
x3
≥0
解:
将问题化为标准化形式:
min
z’=
-2x1
+x2
-x3
s.t.
x1
+x2
-2x3
+x4
=8
4x1
-x2
+x3
+x5
=2
2x1
+3x2
-x3
-x6
=4
x1,
x2,
x3,
x4,
x5,
x6
≥0
引进人工变量x7≥0,构造辅助问题:
min
z’’=
x7
s.t.
x1
+x2
-2x3
+x4
=8
4x1
-x2
+x3
+x5
=2
2x1
+3x2
-x3
-x6
+x7
=4
x1,
x2,
x3,
x4,
x5,
x6,
x7
≥0
辅助问题的系数矩阵为:
z’’
x1
x2
x3
x4
x5
x6
x7
RHS
z’’
1
0
0
0
0
0
0
-1
0
x4
0
1
1
-2
1
0
0
0
8
x5
0
4
-1
1
0
1
0
0
2
x7
0
2
3
-1
0
0
-1
1
4
消去基变量x7在目标函数中的系数
z’’
x1
x2
x3
x4
x5
x6
x7
RHS
z’’
1
2
3
-1
0
0
-1
0
4
x4
0
1
1
-2
1
0
0
0
8
8/1
x5
0
4
-1
1
0
1
0
0
2
--
x7
0
2
[3]
-1
0
0
-1
1
4
4/3
x2进基,x7离基,
z’’
x1
x2
x3
x4
x5
x6
x7
RHS
z’’
1
0
0
0
0
0
0
-1
0
x4
0
1/3
0
-5/3
1
0
1/3
-1/3
20/3
x5
0
14/3
0
2/3
0
1
-1/3
-1/3
10/3
x2
0
2/3
1
-1/3
0
0
-1/3
1/3
4/3
得到辅助问题的最优解,且minz’’=0,转入第二阶段。
建立系数矩阵表:
z’
x1
x2
x3
x4
x5
x6
RHS
z’
1
2
-1
1
0
0
0
0
x4
0
1/3
0
-5/3
1
0
1/3
20/3
x5
0
14/3
0
2/3
0
1
-1/3
10/3
x2
0
2/3
1
-1/3
0
0
-1/3
4/3
消去基变量x2在目标函数中的系数,得到
z’
x1
x2
x3
x4
x5
x6
RHS
z’
1
8/3
0
2/3
0
0
-1/3
4/3
x4
0
1/3
0
-5/3
1
0
1/3
20/3
--
x5
0
14/3
0
[2/3]
0
1
-1/3
10/3
10/2
x2
0
2/3
1
-1/3
0
0
-1/3
4/3
--
x3进基,x5离基
z’
x1
x2
x3
x4
x5
x6
RHS
z’
1
-2
0
0
0
-1
0
-2
x4
0
12
0
0
1
5/2
-1/2
15
x3
0
7
0
1
0
3/2
-1/2
5
x2
0
3
1
0
0
1/2
-1/2
3
原问题的最优解为:
(x1,x2,x3,x4,x5,x6)=(0,3,5,15,0,0),minz’=-2,maxz=2。
(3)
min
z=
x1
+3x2
-x3
s.t.
x1
+x2
+x3
≥3
-x1
+2x2
≥2
-x1
+5x2
+x3
≤4
x1,
x2,
x3
≥0
解:
引进松弛变量x4,x5,x6≥0,得到
min
z=
x1
+3x2
-x3
s.t.
x1
+x2
+x3
-x4
=3
-x1
+2x2
-x5
=2
-x1
+5x2
+x3
+x6
=4
x1,
x2,
x3,
x4,
x5,
x6
≥0
引进人工变量x7,x8≥0,构造辅助问题
min
z’=
x7
+x8
s.t.
x1
+x2
+x3
-x4
+x7
=3
-x1
+2x2
-x5
+x8
=2
-x1
+5x2
+x3
+x6
=4
x1,
x2,
x3,
x4,
x5,
x6,
x7,
x8
0
列出辅助问题系数矩阵表
z’
x1
x2
x3
x4
x5
x6
x7
x8
RHS
z’
1
0
0
0
0
0
0
-1
-1
0
x7
0
1
1
1
-1
0
0
1
0
3
x8
0
-1
2
0
0
-1
0
0
1
2
x6
0
-1
5
1
0
0
1
0
0
4
消去基变量x7,x8在目标函数中的系数,得到
z’
x1
x2
x3
x4
x5
x6
x7
x8
RHS
z’
1
0
3
1
-1
-1
0
0
0
5
x7
0
1
1
[1]
-1
0
0
1
0
3
3/1
x8
0
-1
2
0
0
-1
0
0
1
2
--
x6
0
-1
5
1
0
0
1
0
0
4
4/1
x3进基,x7离基,
z’
x1
x2
x3
x4
x5
x6
x7
x8
RHS
z’
1
-1
2
0
0
-1
0
-1
0
2
x3
0
1
1
1
-1
0
0
1
0
4
4/1
x8
0
-1
2
0
0
-1
0
0
1
2
2/2
x6
0
-2
[4]
0
1
0
1
-1
0
1
1/4
x2进基,x6离基
z’
x1
x2
x3
x4
x5
x6
x7
x8
RHS
z’
1
0
0
0
-1/2
-1
-1/2
-1/2
0
3/2
x3
0
3/2
0
1
-5/4
0
-1/4
5/4
0
15/4
x8
0
0
0
0
-1/2
-1
-1/2
1/2
1
3/2
x2
0
-1/2
1
0
1/4
0
1/4
-1/4
0
1/4
得到辅助问题的最优解,但minz=3/2>0,因此原问题无可行解。
第二章对偶和灵敏度分析
一、写出以下问题的对偶问题
(1)
max
z=
-x1
+2x2
s.t.
3x1
+4x2
≤12
2x1
-x2
≥2
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- 运筹学 习题集 02