�12 x1 � 3 x2 � 6 x3 � 3 x4 � 9
(1)
st
��8 ��3
x1 x1
� �
x2 x6
� 4 x3 �0
�
2 x5
� 10
�� x j � 0�, j � 1,� ,6�
min Z � 5 x1 � 2 x2 � 3 x3 � 2 x4
� x1 � 2 x2 � 3 x3 � 4 x4 � 7
运筹学教程�第二版� 习题解答
运筹学教程
1.1 用图解法求解下列线性规划问题。并指出问 题具有惟一最优解、无穷多最优解、无界解还是无可 行解。
min Z � 2 x1 � 3 x2 � 4 x1 � 6 x2 � 6
(1) st .�� 2 x1 � 2 x2 � 4 �� x1 , x2 � 0
Z
0
0.5
2
0
5
0
0
1
1
5
2/5
0
11/5
0
43/5
page 10 6 January 2011
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运筹学教程
1.4 分别用图解法和单纯形法求解下述线性规划 问题�并对照指出单纯形表中的各基可行解对应图解 法中可行域的哪一顶点。
max Z � 10 x1 � 5 x2 �3 x1 � 4 x2 � 9
max Z � x1 � x2 �6 x1 � 10 x2 � 120 (3) st.�� 5 � x1 � 10 �� 5 � x2 � 8
max Z � 3x1 � 2 x2 �2 x1 � x2 � 2
(2) st.��3x1 � 4 x2 � 12 �� x1, x2 � 0
max Z � 5x1 � 6 x2 � 2 x1 � x2 � 2
(4) st.��� 2 x1 � 3x2 � 2 �� x1, x2 � 0
page 2 6 January 2011
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page 3 6 January 2011
min Z � 2 x1 � 3 x2
� 4 x1 � 6 x2 � 6 (1) st .��2 x1 � 2 x2 � 4
page 14 6 January 2011
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运筹学教程
l.6 考虑下述线性规划问题�
max Z � c1 x1 � c2 x2 � a11 x1 � a12 x2 � b1
st .��a21 x1 � a22 x2 � b2 �� x1, x2 � 0
式中�1≤c1≤3, 4≤c2≤6, -1≤a11≤3, 2≤a12≤5, 8≤b1≤12, 2≤a21≤5, 4≤a22≤6, 10≤b2≤14,试确定 目标函数最优值的下界和上界。
� �
2x4 �14 x3 � x4 �
. 2
��x1, x2, x3 � 0, x4无约束
max Z � 3x1 � 4 x2 � 2 x3 � 5 x41 � 5 x42
� � 4 x1 � x2 � 2 x3 � x41 � x42 � 2
st
�� ���
x1 � x2 � x3 2 x1 � 3x2 �
(1)
st
�� x1 � x2 � x3 ��� 2 x1 � 3x2
� �
2 x4 x3 �
� 14 x4 �
. 2
�� x1,
x2 ,
x3
�
0,
x
无约束
4
min Z � 2 x1 � 2 x 2 � 3 x3
� � x1 � x2 � x3 � 4
(2)
st
� �
� 2 x1 � x2 � x3 � 6
该题是无穷多最优解。
最优解之一�x1
�
9 5
,
x2
�
4 5
,
x3
�
0,
Z
�
6
page 19 6 January 2011
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运筹学教程
maxZ � 4x1 � x2
�3x1 � x2 � 3 (3) st����4x1x1��23x2x2��xx4 3��46
��xj � 0�, j �1,�,4�
(2)
st
� �
2
x1
�
2 x2
�
x3
�
2 x4
�
3
� �
x j � 0, ( j � 1,� 4)
page 8 6 January 2011
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运筹学教程
x1 0 0 0 0.75
max Z � 3 x1 � x2 � 2 x3
�12 x1 � 3 x2 � 6 x3 � 3 x4 � 9
c x1 1 1
�j
0
d
0
0
x2
x3
x4
1
5/14
-3/40Fra bibliotek-2/14
10/35
0 -5/14d+2/14c 3/14d-10/14c
page 13 6 January 2011
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当c/d在3/10到5/2之间时最优解为图中的A点�当 c/d大于5/2且c大于等于0时最优解为图中的B点�当c/d 小于3/10且d大于0时最优解为图中的C点�当c/d大于 5/2且c小于等于0时或当c/d小于3/10且d小于0时最优解 为图中的原点。
�� x1 , x2 � 0
无穷多最优解�
x1
� 1, x2
�
1 ,Z
3
�
3是一个最优解
max Z � 3 x1 � 2 x 2
� 2 x1 � x2 � 2 ( 2 ) st .��3 x1 � 4 x 2 � 12
�� x1 , x 2 � 0
该问题无解
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page 22 6 January 2011
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运筹学教程
1.9 若X(1)、X(2)均为某线性规划问题的最优解� 证明在这两点连线上的所有点也是该问题的最优解。
max Z � C T X
设 X (1)和 X ( 2 )满足� � AX � b
� �
X
�0
对于任何 0 � a � 1, 两点连线上的点 X 满足�
X � aX (1) � (1 � a ) X ( 2 )也是可行解�且
C T X � C T aX (1) � C T (1 � a ) X ( 2 )
� C T aX (1) � aC T X ( 2 ) � C T X ( 2 )
� 2 x41 � 2 x42 x3 � x41 � x42
� �
x5 x6
� �
14 2
��
x1, x2 , x3 , x41 , x42 , x6 � 0
page 6 6 January 2011
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运筹学教程
min Z � 2 x1 � 2 x2 � 3 x3
page 17 6 January 2011
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运筹学教程
l.7 分别用单纯形法中的大M法和两阶段法求解 下列线性规划问题�并指出属哪—类解。
maxZ � 3x1 � x2 � 2x3
�x1 � x2 � x3 � 6
(1)
st
��� 2x1 � x3 � ��2x2 � x3 � 0
(4) st.��� 2 x1 � 3x2 � 2 �� x1, x2 � 0
该问题有无界解
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运筹学教程
1.2 将下述线性规划问题化成标准形式。
min Z � �3x1 � 4 x2 � 2 x3 � 5x4
�4 x1 � x2 � 2 x3 � x4 � �2
项目
X1
X2
X3
X4
X5
X4
6
(b) (c) (d)
1
0
X5
1
-1
3
(e)
0
1
Cj�Zj
a
-1
2
0
0
X1
(f)
(g)
2
-1 1/2 0
X5
4 (h) (i)
1
1/2 1
Cj�Zj
0
-7
j
k
(l)
b=2, c=4, d=-2, g=1, h=0, f=3, i=5, e=2, l=0,
a=3, j=5, k= -1.5
page 4 6 January 2011
max Z � x1 � x2 �6 x1 � 10 x2 � 120 (3) st.�� 5 � x1 � 10 �� 5 � x2 � 8 唯一最优解� x1 � 10, x2 � 6, Z � 16
max Z � 5x1 � 6 x2 � 2 x1 � x2 � 2
(1)
st
��8 ��3
x1 x1
� �
x2 x6
� �
4 x3 0
�
2 x5
�
10
�� x j � 0�, j � 1,� ,6�
基可行解
x2 x3 x4 x5 x6 3 0 0 3.5 0
0 1.5 0 8 0
00350
0 0 0 2 2.25