目标函数最优值与向量(b)的关系
在这一部分,我们分析目标函数最优值与向量(b)的关系。
首先,令
[P(b)={x\mid Ax=b,,x\geq 0} ]
为一个可行的集合。令
[S={b\mid P(b)是非空的}. ]
并且观测
[S={Ax\mid x\geq 0}, ]
注意,(S)是一个凸集。对于任意(b\in S),定义
[F(b)=\min_{x\in P(b)}, c^Tx, ]
它是关于(b)的最优值函数。
在这一部分,我们假设对偶可行域({p\mid p^TA\leq c^T})是非空的。那么,由对偶理论可知对所有(b\in S),最优值(F(b))是有限的。我们的目标是研究(F(b))的结构。
首先固定(b^\in S)。假设存在一个非退化的原始最优基本可行解,记(B)为相对应的最优基矩阵,那么基本变量(x_B)的值此时为(x_B=B{-1}b),并且由非退化知(x_B > 0)。另外,判断数向量也是非负的。如果此时改变(b*)为(b),并且假设(b-b*)足够小,(B^{-1}b)维持为正值,那么原来的解对于新的问题仍然是一个基本可行解。而判断数由于并不受(b)影响,因此仍然保持非负。综上,(B)也是新问题的一个最优解,而新问题的最优值为
[F(b)=cT_BB{-1}b=p^Tb,\quad b足够接近b^*, ]
其中(pT=cT_BB^{-1})是对偶问题的最优解。这说明(F(b))是关于(b)的线性函数,并且梯度为(p)。下面叙述(F(b))的一个全局性质。
定理: 最优值函数(F(b))是关于(b\in S)的凸函数。
证:
令(b1)和(b2)为(S)中的两个元素。对于(i=1,2),令(x^i)为问题
[\min_{x}{c^Tx\mid Ax=b^i,,x\geq 0} ]
的最优解。因此,(F(b1)=cTx^1), (F(b2)=cTx^2)。对于(\lambda \in [0,1]),向量
[y=\lambda x^1+(1-\lambda )x^2 ]
非负,且满足(Ay=\lambda b^1+(1-\lambda )b^2)。这说明当线性规划问题右侧的向量为(\lambda b^1+(1-\lambda )b^2)时,(y)是一个可行解。因此,
[\begin{align*} &F(\lambda b^1+(1-\lambda )b^2) \ \leq\ & c^Ty \ =\ &\lambda cTx1+(1-\lambda )cTx2 \ =\ & \lambda F(b^1)+(1-\lambda )F(b^2). \end{align*} ]
证毕。
下面我们从对偶的角度重新观测上面的定理。考虑对偶问题:
[\begin{align*} \max \quad &p^Tb \ \text{s.t.} \quad &p^TA\leq c^T, \end{align*} ]
注意我们之前假设它是可行的。对于任意(b\in S),(F(b))是有限的,并且根据强对偶定理可知它的值等于对偶问题的最优目标函数值。令(p1,p2,\cdots,pN)为对偶可行域的极点。(由于(A)是行线性无关的,因此(A)的列的扩张为(\mathbb{R}m),由线性规划中的几何(四 )中的定理可知对偶问题至少存在一个极点)。由于对偶问题的最优解必然是落在一个极点上,因此
[F(b)=\max_{i=1,\cdots,N} (pi)Tb,\quad b\in S. ]
这说明,(F)等价于有限个线性函数的最大值。因此,它是一个分段线性凸函数。当(F)在线性部分时,存在唯一解,而当(F)在不可导的部分时(两个线性相接的部分),说明解不是唯一的,更进一步说明原始问题的解是退化的(这是因为当原问题非退化时,由前面叙述的局部性质可知,此时它是线性的)。
下面考虑一种特殊情形,将(b)变为(b=b^+\theta d),其中(b*)和(d)固定。令(f(\theta)=F(b+\theta d))为系数为(\theta)时的最优值。因此,
[f(\theta)=\max_{i=1,\cdots,N} (pi)T(b^+\theta d),\quad b^+\theta d\in S. ]
对偶最优解集
这一部分说明任意对偶最优解都可以看作是(F)的“广义梯度”。
定义: 令(F)为定义在凸集(S)上的凸函数,存在(b^\in S)。我们称向量(p)是(F)在点(b^)的次梯度 (subgradient) 是指满足
[F(b*)+pT(b-b^)\leq F(b),\quad \forall b\in S. ]
注意,如果(b^)在交接点,那么将存在多个次梯度。下图分别给出两种不同情况的次梯度。
定理: 假设问题
[\min_{x}{c^Tx\mid Ax=b^,,x\geq 0} ]
可行并且最优值有界。那么,向量(p)是对偶问题最优解当且仅当(p)是(F)在点(b^)的次梯度。
证:
回忆定义在集合(S)上的函数(F)。假设(p)是对偶问题的最优解,那么由强对偶定理可知(pTb=F(b^))。考虑任意(b\in S),对于任意可行解(x\in P(b)),由弱对偶定理可得(p^Tb\leq c^Tx)。给这个不等式两边关于(x\in P(b))取最小值,可得(p^Tb\leq F(b))。因此
[pTb-pTb^\leq F(b)-F(b^). ]
这说明(p)是(F)在(b^*)的次梯度。
下面证另一面,记(p)为(F)在(b^*)的次梯度,那么
[F(b*)+pT(b-b^*)\leq F(b),\quad \forall b\in S. ]
选择使得(Ax=b)及(x\geq 0)成立的(x\in P(b))。已知(F(b)\leq c^Tx),因此
[pTAx=pTb\leq F(b)-F(b*)+pTb^\leq cTx-F(b)+pTb. ]
由于上式对任意的(x\geq 0)都成立,因此必然有(p^TA\leq cT)(因为如果不满足的话,我们总可以通过扩大x的值使得上述不等式不成立),这说明(p)是对偶问题的一个可行解。另外,令(x=0),可得(F(b)\leq pTb)。由弱对偶可知,对于任意可行解(q)都有(qTb\leq F(b^)\leq pTb),这说明(p)是对偶问题最优解。证毕。
目标函数最优值与向量(c)的关系
假设原问题的可行域非空。定义对偶可行域
[Q©={p\mid p^TA\leq c^T}, ]
并且令
[T={c\mid Q©非空}. ]
若(c^1\in T)并且(c^2\in T),那么存在(p1)和(p2)使得((p1)TA\leq (c1)T)及((p2)TA\leq (c2)T)。对于(\lambda \in [0,1]),我们有
[\left(\lambda (p1)T+(1-\lambda) (p2)T\right)A\leq \lambda (c1)T+(1-\lambda) (c2)T, ]
这说明(\lambda (c1)T+(1-\lambda) (c2)T\in T),与此同时也说明(T)是一个凸集。
若(c\notin T),对偶问题可行域为空说明原问题的最优值为(-\infty)(因为我们假设原问题可行域非空)。若(c\in T),那么最优值一定有界。因此,原问题的最优解(G©)有界当且仅当(c\in T)。
记(x1,x2,\cdots,x^N)是原问题可行域的基本可行解,显然它与(c)无关。由于标准型问题的最优解总是出现在极点,因此
[G©=\min_{i=1,\cdots,N}\ cTxi. ]
这说明(G©)是有限个线性方程的最小值,因此它是一个分段线性凹函数。如果对于(c^\in T),原问题有唯一的最优解(xi),那么((c)Txi < (c*)Tx^j)对任意(j\not= i)都成立。假设©非常接近(c^*), 不等式(cTxi < cTxj)对于(j\not= i)仍然成立。这说明(xi)仍然是原问题的唯一最优解且最优值为(cTxi)。因此,局部的有(G©=cTx^i)。另外,(G)在交叉点时对应多个原问题。
下面的定理总结上面提到的结论。
定理: 考虑可行域非空的标准型的线性规划问题。
集合(T)是凸集,且(T)中©对应的最优值有界。
最优值函数(G©)是关于(c\in T)的凹函数。
如果与©有关的原问题存在唯一最优解(x*),那么(G)在©的邻域内是线性的且梯度等于(x*)。
参数规划
固定(A,B,C)以及一个与©维度相同的向量(d)。对于任意实数(\theta),考虑问题
[\begin{align*} \min \quad &(c+\theta d)^Tx \ \text{s.t.}\quad &Ax=b \ &\ \ \ x\geq 0, \end{align*} ]
记(g(\theta))为这个问题的最优值,并且假设可行域非空。对于最优值有界的(\theta),有
[g(\theta)=\min_{i=1,\cdots,N}\ (c+\theta d)Txi, ]
其中(x1,\cdots,xN)是可行域的极点,如下图所示,(g(\theta))是关于(\theta)的一个分段线性凹函数。为了更进一步观察(g(\theta)),我们先观察一个例子。
例: 考虑问题
然后引入松弛变量,是的上述问题转化为标准型问题,接着令松弛变量为基本变量。这样的方式确定了一个基本可行解,然后有如下的单纯形表
若(-3+2\theta\geq 0)且(2-\theta\geq 0),那么所有的判断数都非负,说明此时得到了最优基本可行解。也即,
[g(\theta)=0, \quad 若3/2\leq \theta\leq 3. ]
若(\theta)增加到高于3,那么(x_2)的判断数此时为负值。此时令(x_2)进入基矩阵,(x_4)出基矩阵,得到如下的单纯形表
如果(3\leq \theta\leq 5.5/1.5),那么表中的判断数都是非负,此时有
[g(\theta)=7.5-2.5\theta,\quad 若3\leq \theta\leq 5.5/1.5. ]
如果(\theta)增加的超过(5.5/1.5),那么(x_3)的判断数变为负值。如果我们想要(x_3)进基,但是不存在正的pivot元素,因此此时原问题无界且(g(\theta)=-\infty)。
现在我们回到第一个表再考虑另一面,如果(\theta)增加的小于(3/2),那么(x_1)的判断数就变为负值,此时使得(x_1)进基,(x_5)出基,得到
如果(5/4\leq \theta\leq 3/2),那么得到最优解,且
[g(\theta)=-10.5+7\theta,\quad 若5/4\leq\theta\leq 3/2. ]
而对于(\theta < 5/4),(x_3)的判断数变为负值,但是所对应的列不存在正的pivot元素,因此这时的最优值为(-\infty)。
我们将上述的结果汇总绘制得到如下的图:
接下来,我们总结上面例子的一般步骤。假设现在我们有一个基本可行解以及它所对应的基矩阵(B),并且假设这个基矩阵对于满足(\theta_1\leq \theta\leq\theta_2)的(\theta)是最优的。令(x_j)为当(\theta >\theta_2)时,它的判断数为负值。又因为当(\theta_1\leq \theta\leq\theta_2)时,(x_j)的判断数为正值,因此当(\theta =\theta_2)时,(x_j)的判断数为0。然后此时考虑令(x_j)进基。
若列(B^{-1}A_j)不存在正的pivot元素,那么对于(\theta > \theta_2),(x_j)的判断数为负值,因此此时的最优值为(-\infty)。
若列(B^{-1}A_j)至少存在一个正元素,那么我们可以进行基变换得到新的基矩阵(\overline B)。当(\theta = \theta_2)时,进基变量的判断数为0,则新的基矩阵对应的目标函数值不变。并且,对于旧基矩阵,目标函数值是最优值,那么对于新的基矩阵,这个值仍然为最优值。对于(\theta < \theta_2),由于进基变量的判断数为正值,因此新的基矩阵对应的目标函数值不会小于原基矩阵对应的最优值(注:判断数为负值时是使得目标函数值下降的方向)。这说明新的基矩阵在(\theta < \theta_2)时不可能是最优基矩阵。那么,上述的叙述表明新的基矩阵为最优基矩阵是在(\theta)满足(\theta_2 \leq \theta \leq \theta_3)时成立的。重复上述的操作,我们可以得到一组基矩阵以及相对应的满足最优性的(\theta)的范围。
注意,当一个基矩阵在满足(\theta \in [\theta_i,\theta_{i+1}])时是最优的,那么对于(\theta \geq \theta_{i+1}),这个基矩阵将不可能是最优的(存在判断数为负值,说明存在了下降方向)。因此,若(\theta_{i+1} > \theta_{i})对于所有(i)都成立,那么同一个基矩阵不会出现两次,并且(\theta)的所有范围也将会在有限次迭代后得到。注意,每一次迭代都会得到一个新的交界点。
如果存在(\theta_i = \theta_{i+1}),说明上面提到的操作可能会出现循环(注:这种情形仅会在原问题退化时出现),但是可以使用一些反循环的技巧避免循环的出现。
敏感性分析总结
若加入新变量,检查判断数是否为负值,若是,原单纯形表中加入新的列继续使用单纯形法。
若加入新约束,检查约束是否还满足,若不满足,构造辅助问题继续求解。
若(b)或(c)改变了(\delta),得到原基矩阵维持最优性的(\delta)的范围。
若(A)中元素变化(\delta),分析与上述类似。不过,如果变化的是基本列中的元素,分析过程稍有复杂。
假设对偶问题可行域非空,最优值是关于(b)的分段线性凸函数(前提(b)使得原问题存在可行解)。另外,最优值的次梯度与对偶问题的最优解有关。
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假设原问题可行域非空,最优值是关于向量(c)的分段线性凹函数(前提(c)使得原问题有界)。
如果目标函数中系数是参数(\theta)的仿射函数,那么可通过一系列操作得到所有(\theta)的最优值。