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变尺度法
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‎Chapter1/chapter1.tex‎

Lines changed: 4 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -13,10 +13,10 @@ \section{绪论}
1313
\addConstraint{x}{\geq 0}{.}
1414
\end{maxi*}
1515
\begin{align*}
16-
f'&=2(a-2x)(-2)x+(a-2x)^2
17-
&=-4(ax-2x^2)+a^2+4x^2-4ax
18-
&=-8ax+12x^2+a^2
19-
&=(2x-a)(6x-a)=0
16+
f'&=2(a-2x)(-2)x+(a-2x)^2\\
17+
&=-4(ax-2x^2)+a^2+4x^2-4ax\\
18+
&=-8ax+12x^2+a^2\\
19+
&=(2x-a)(6x-a)=0\\
2020
x=a/6
2121
\end{align*}
2222
\end{solution}

‎Chapter2/chapter2-2.tex‎

Lines changed: 83 additions & 54 deletions
Original file line numberDiff line numberDiff line change
@@ -1,3 +1,5 @@
1+
\section{线性规划}
2+
13
\subsection{单纯形法}
24

35
\subsubsection{单纯形法求解}
@@ -79,7 +81,7 @@ \subsubsection{单纯形法求解}
7981
\end{maxi*}
8082
\end{problem}
8183
\begin{solution}
82-
引入松弛变量，化为标准形式：
84+
引入松弛变量$x_4,x_5,x_6$，化为标准形式:
8385
\begin{maxi*}|s|
8486
{}
8587
{-x_1 + 3x_2 + x_3 + 0x_4 + 0x_5 + 0x_6}
@@ -116,17 +118,7 @@ \subsubsection{单纯形法求解}
116118
\end{simplex}
117119
\end{center}
118120
故最优解$(\frac{78}{25},\frac{114}{25},\frac{11}{10})$，最大值为
119-
\begin{align*}
120-
f_{max}&=-1\times\frac{78}{25}+3\times\frac{114}{25}+1\times\frac{11}{10}\\
121-
&=\dfrac{-78+114\times3}{25}+\dfrac{11}{10}\\
122-
&=\dfrac{-78+342}{25}+\frac{11}{10}\\
123-
&=\dfrac{264}{25}+\dfrac{11}{10}\\
124-
&=\dfrac{264*4+110}{100}\\
125-
&=\dfrac{1056+110}{100}\\
126-
&=\dfrac{1166}{100}\\
127-
&=\dfrac{583}{50}
128-
\end{align*}
129-
121+
$$f_{max}=-1\times\frac{78}{25}+3\times\frac{114}{25}+1\times\frac{11}{10}=\dfrac{583}{50}$$
130122
\end{solution}
131123
\begin{problem}{1.4}
132124
\begin{mini*}|s|
@@ -142,7 +134,7 @@ \subsubsection{单纯形法求解}
142134
\end{problem}
143135
\begin{solution}
144136
求${3x_1 - 5x_2 - 2x_3 - x_4}$的最小解，即求${-3x_1 + 5x_2 + 2x_3 + x_4}$的最大解.\\
145-
引入松弛变量，化为标准形式：
137+
引入松弛变量$x_5,x_6,x_7$，化为标准形式：
146138
\begin{maxi*}|s|
147139
{}
148140
{-3x_1 + 5x_2 + 2x_3 + x_4 + 0x_5 + 0x_6 + 0x_7}
@@ -190,7 +182,7 @@ \subsubsection{单纯形法求解}
190182
\end{problem}
191183
\begin{solution}
192184
求${-3x_1 - x_2}$的最小解，即求${3x_1 + x_2}$的最大解.\\
193-
引入松弛变量，化为标准形式：
185+
引入松弛变量$x_5$，化为标准形式：
194186
\begin{maxi*}|s|
195187
{}
196188
{3x_1 + x_2 + 0x_3 + 0x_4 + 0x_5}
@@ -239,22 +231,24 @@ \subsubsection{转化为标准形式并列出单纯形表}
239231
\end{maxi*}
240232
\end{problem}
241233
\begin{solution}
234+
标准形式要求$b\geq0$，故约束条件二两边同乘-1,\\
235+
引入松弛变量$x_4,x_5$，化为标准形式：
242236
\begin{maxi*}|s|
243237
{}
244238
{2x_1 - 2x_2 - 3x_3 + 0x_4 + 0x_5}
245239
{}
246240
{}
247-
\addConstraint{2x_1 - x_2 + 2x_3 + x_4}{= 2}
248-
\addConstraint{x_1 - 2x_2 + 3x_3 - x_5}{= 2}
241+
\addConstraint{2x_1 - x_2 + 2x_3 + x_4\ph}{= 2}
242+
\addConstraint{x_1 - 2x_2 + 3x_3 \ph - x_5}{= 2}
249243
\addConstraint{x_1, x_2, x_3,x_4,x_5}{\geq 0}{.}
250244
\end{maxi*}
251245
\begin{center}
252246
\begin{simplex}{}
253-
c_j \rightarrow &&& 2 & -2 & -3 & 0 & 0 \\
254-
C_B & X_B &b & x_1 & x_2 & x_3 & x_4 & x_5\\
255-
0 & x_4 &2 & 2 & -1 & 2 & 1 & 0 \\
256-
0 & x_5 &2 & 1 & -2 & 3 & 0 & -1 \\
257-
c_j - z_j &&& 2 & -2 & -3 & 0 & 0 \\
247+
c_j \rightarrow &&& 2 & -2 & -3 & 0 & 0 \\
248+
C_B & X_B &b & x_1 & x_2 & x_3 & x_4 & x_5 \\
249+
0 & x_4 &2 & 2 & -1 & 2 & 1 & 0 \\
250+
0 & x_5 &2 & 1 & -2 & 3 & 0 & -1 \\
251+
c_j - z_j &&& 2 & -2 & -3 & 0 & 0 \\
258252
\end{simplex}
259253
\end{center}
260254
\end{solution}
@@ -272,24 +266,30 @@ \subsubsection{转化为标准形式并列出单纯形表}
272266
\end{mini*}
273267
\end{problem}
274268
\begin{solution}
275-
\begin{mini*}|s|
269+
求原问题的最小解，即求$3x_1 - 4x_2 + 2x_3 - 5x_4$最大解，\\
270+
标准形式要求$b\geq0$，故约束条件一两边同乘-1,\\
271+
用非负变量之差$x_5-x_6$代替$x_4$，引入松弛变量$x_7,x_8$，化为标准形式：
272+
\begin{maxi*}|s|
276273
{}
277-
{-3x_1 + 4x_2 - 2x_3 + 5x_5 - 5x_6 + 0x_7 + 0x_8}
274+
{3x_1 - 4x_2 + 2x_3 - 5x_5 + 5x_6 + 0x_7 + 0x_8}
278275
{}
279276
{}
280277
\addConstraint{-4x_1 + x_2 - 2x_3 + x_5 - x_6 \ph \ph}{=2}
281278
\addConstraint{x_1 + x_2 - x_3 + 2x_5 - 2x_6 + x_7 \ph}{= 14}
282279
\addConstraint{-2x_1 + 3x_2 + x_3 - x_5 + x_6 \ph - x_8}{= 2}
283280
\addConstraint{x_1, x_2, x_3,x_5,x_6,x_7,x_8}{\geq 0}{.}
284-
\end{mini*}
281+
\end{maxi*}
282+
为了列出单纯形表，需要化出一个基，引入人工变量$x_9$，以大M法求解
285283
\begin{center}
286-
\begin{simplex}{}
287-
c_j \rightarrow &&& -3 & 4 & -2 & 5 & -5 & 0 & 0 \\
288-
C_B & X_B &b & x_1 & x_2 & x_3 & x_5 & x_6 & x_7 & x_8\\
289-
0 & x_3 &2 & -4 & 1 & -2 & 1 & -1 & 0 & 0 \\
290-
2 & x_4 &14 & 1 & 1 & -1 & 2 & -2 & 1 & 0 \\
291-
1 & x_5 &2 & -2 & 3 & 1 & -1 & 1 & 0 & -1 \\
292-
c_j - z_j &&& 8 & 9 & 0 & 0 & 0 & 0 & 0 \\
284+
\begin{simplex}{
285+
columns={10mm}
286+
}
287+
c_j \rightarrow &&& 3 & -4 & 2 & -5 & 5 & 0 & 0 & -M \\
288+
C_B & X_B &b & x_1 & x_2 & x_3 & x_5 & x_6 & x_7 & x_8& x_9\\
289+
-M & x_9 &2 & -4 & 1 & -2 & 1 & -1 & 0 & 0 & 1 \\
290+
0 & x_7 &14 & 1 & 1 & -1 & 2 & -2 & 1 & 0 & 0 \\
291+
0 & x_8 &2 & -2 & 3 & 1 & -1 & 1 & 0 & -1 & 0 \\
292+
c_j - z_j &&& 3-4M& M-4 & 2-2M& M-5 & 5-M & 0 & 0 & 0 \\
293293
\end{simplex}
294294
\end{center}
295295
\end{solution}
@@ -476,6 +476,7 @@ \subsubsection{分别用大M法与两阶段法求解并对比}
476476
(1)利用大M法求解：
477477
\begin{center}
478478
\begin{simplex}{
479+
columns={10mm},
479480
hline{6,7,10,11,14,15} = {0.08em},
480481
cell{6,10,14}{1} = {c=3,r=1}{c},
481482
}
@@ -588,11 +589,11 @@ \subsubsection{分别用大M法与两阶段法求解并对比}
588589
0 & x_2 &6/5 & 0 & 1 & -3/5& 0 & -4/5& 3/5 \\
589590
0 & x_4 &1 & 0 & 0 & 1 & 1 & 1 & -1 \\
590591
c_j - z_j &&& 0 & 0 & 0 & 0 & -1 & -1 \\
591-
\end{simplex}
592-
\begin{simplex}{
593-
hline{6,7} = {0.08em},
594-
cell{6}{1} = {c=3,r=1}{c},
595-
}
592+
\end{simplex}
593+
\begin{simplex}{
594+
hline{6,7} = {0.08em},
595+
cell{6}{1} = {c=3,r=1}{c},
596+
}
596597
c_j \rightarrow &&& -4 & -1 & 0 & 0 \\
597598
C_B & X_B &b & x_1 & x_2 & x_3 & x_4 \\
598599
-4 & x_1 &3/5 & 1 & 0 & 1/5 & 0 \\
@@ -620,6 +621,34 @@ \subsubsection{不限方法求解}
620621
\addConstraint{x_1, x_2, x_3}{\geq 0}{.}
621622
\end{mini*}
622623
\end{problem}
624+
\begin{solution}
625+
引入松弛变量$x_4,x_5$，化为标准形式：
626+
\begin{maxi*}|s|
627+
{}
628+
{-4x_1 - 6x_2 - 18x_3 + 0x_4 + 0x_5}
629+
{}
630+
{}
631+
\addConstraint{x_1 \ph + 3x_3 - x_4\ph}{= 3}
632+
\addConstraint{\ph x_2 + 2x_3\ph - x_5}{= 5}
633+
\addConstraint{x_1, x_2, x_3, x_4, x_5}{\geq 0}{.}
634+
\end{maxi*}
635+
\begin{center}
636+
\begin{simplex}{
637+
hline{5,6} = {0.08em},
638+
cell{5}{1} = {c=3,r=1}{c},
639+
}
640+
c_j \rightarrow &&& -4 & -6 & -18 & 0 & 0 \\
641+
C_B & X_B &b & x_1 & x_2 & x_3 & x_4 & x_5 \\
642+
-4 & x_1 &3 & 1 & 0 & [3] & -1 & 0 \\
643+
-6 & x_2 &5 & 0 & 1 & 2 & 0 & -1 \\
644+
c_j - z_j &&& 0 & 0 & 6 & -4 & -6 \\
645+
-18 & x_3 &1 & 1/3 & 0 & 1 & -1/3& 0 \\
646+
-6 & x_2 &3 & -2/3& 1 & 0 & 2/3 & -1 \\
647+
c_j - z_j &&& -2 & 0 & 0 & -2 & -6 \\
648+
\end{simplex}
649+
\end{center}
650+
故$(0,3,1)$为最优解，最小值为$f_{min}=4\times0+6\times3+18\times1=36$.
651+
\end{solution}
623652
\begin{problem}{2.2}
624653
\begin{maxi*}|s|
625654
{}
@@ -766,7 +795,7 @@ \subsubsection{对偶理论基础}
766795
由互补松弛性$(X_s^T\hat{Y} = 0)$得
767796
$$y_4^*=0,y_1^*,y_2^*,y_3^*>0$$
768797
又因$x_1^*,x_2^*,x_3^*>0$,由互补松弛性$(\hat{X^T}Y_s=0)$得对偶问题的1、2、3个约束条件应满足严格等式，有
769-
$$\left\{
798+
$$\left\{
770799
\begin{aligned}
771800
y_1 + 3y_2\ph &=8\\
772801
2y_1 + y_2\ph &=6\\
@@ -796,22 +825,22 @@ \subsubsection{对偶单纯形法求解}
796825
\end{problem}
797826
\begin{solution}
798827
\begin{center}
799-
\begin{simplex}{
800-
hline{5,6,8,9} = {0.08em},
801-
cell{5,8}{1} = {c=3,r=1}{c},
802-
}
803-
c_j \rightarrow &&& -4 & -12 & -18 & 0 & 0 \\
804-
C_B & X_B &b & x_1 & x_2 & x_3 & x_4 & x_5 \\
805-
0 & x_4 &-3 & -1 & 0 & -3 & 1 & 0 \\
806-
0 & x_5 &-5 & 0 & [-2]& -2 & 0 & 1 \\
807-
c_j - z_j &&& -4 & -12 & -18 & 0 & 0 \\
808-
0 & x_4 &-3 & -1 & 0 & [-3]& 1 & 0 \\
809-
-12 & x_2 &2/5 & 0 & 1 & 1 & 0 & -1/2\\
810-
c_j - z_j &&& -4 & 0 & -6 & 0 & -6 \\
811-
-18 & x_3 &1 & 1/3 & 0 & 1 & -1/3& 0 \\
812-
-12 & x_2 &3/2 & -1/3& 1 & 0 & 1/3 & -1/2\\
813-
c_j - z_j &&& -2 & 0 & 0 & -2 & -6 \\
814-
\end{simplex}
828+
\begin{simplex}{
829+
hline{5,6,8,9} = {0.08em},
830+
cell{5,8}{1} = {c=3,r=1}{c},
831+
}
832+
c_j \rightarrow &&& -4 & -12 & -18 & 0 & 0 \\
833+
C_B & X_B &b & x_1 & x_2 & x_3 & x_4 & x_5 \\
834+
0 & x_4 &-3 & -1 & 0 & -3 & 1 & 0 \\
835+
0 & x_5 &-5 & 0 & [-2]& -2 & 0 & 1 \\
836+
c_j - z_j &&& -4 & -12 & -18 & 0 & 0 \\
837+
0 & x_4 &-3 & -1 & 0 & [-3]& 1 & 0 \\
838+
-12 & x_2 &2/5 & 0 & 1 & 1 & 0 & -1/2\\
839+
c_j - z_j &&& -4 & 0 & -6 & 0 & -6 \\
840+
-18 & x_3 &1 & 1/3 & 0 & 1 & -1/3& 0 \\
841+
-12 & x_2 &3/2 & -1/3& 1 & 0 & 1/3 & -1/2\\
842+
c_j - z_j &&& -2 & 0 & 0 & -2 & -6 \\
843+
\end{simplex}
815844
\end{center}
816845

817846
故$(0,\frac{3}{2},1)$为最优解，最小值为$f_{min}=4\times0+12\times\frac{3}{2}+18\times1=36$.
@@ -1016,7 +1045,7 @@ \subsubsection{对偶单纯形法求解}
10161045
\end{problem}
10171046
\begin{solution}
10181047
利用等式约束条件，构造可行基，改造原问题为：
1019-
\begin{maxi*}|s|
1048+
\begin{maxi*}|s|
10201049
{}
10211050
{-4x_1 - 3x_2 - 5x_3 - x_4 - 2x_5}
10221051
{}

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