如果你也在 怎样代写运筹学Operations Research这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。假设检验Hypothesis是假设检验是统计学中的一种行为，分析者据此检验有关人口参数的假设。分析师采用的方法取决于所用数据的性质和分析的原因。假设检验是通过使用样本数据来评估假设的合理性。
运筹学（Operation）是近代应用数学的一个分支。它把具体的问题进行数学抽象，然后用像是统计学、数学模型和算法等方法加以解决，以此来寻找复杂问题中的最佳或近似最佳的解答。
二战中运筹学的应用
在二战时期，作战研究被定义为 “一种科学方法，为执行部门提供有关其控制的行动的决策的量化依据”。它的其他名称包括作战分析（英国国防部从1962年开始）和定量管理。
在第二次世界大战期间，英国有近1000名男女从事作战研究。大约有200名作战研究科学家为英国军队工作。
帕特里克-布莱克特在战争期间为几个不同的组织工作。战争初期，在为皇家飞机研究所（RAE）工作时，他建立了一个被称为 “马戏团 “的团队，帮助减少了击落一架敌机所需的防空炮弹数量，从不列颠战役开始时的平均超过20,000发减少到1941年的4,000发。
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我们提供的假设检验Hypothesis及其相关学科的代写，服务范围广, 其中包括但不限于:
- 商业分析 Business Analysis
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- 数据挖掘/数据科学/大数据 Data Mining / Data Science / Big Data
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- 地理/地理信息科学 Geography/Geographic Information Science
- 图论 Graph Theory
- 工业工程 Industrial Engineering
- 库存控制 Inventory control
- 数学建模 Mathematical Modeling
- 数学优化 Mathematical Optimization
- 概率和统计 Probability and statistics
- 排队论 Queueing theory
- 社交网络/交通预测模型 Social network/traffic prediction modeling
- 随机过程 Stochastic processes
- 供应链管理 Supply chain management
运筹学代写
数学代写|运筹学作业代写OPERATIONS RESEARCH代考|dynamic path
As we have mentioned above, we will solve the problem by searching for the shortest dynamic f-augmenting path in the residual network successively. To ensure that the algorithm can be finished within finite steps, we need a condition that the network should not contain any “negative cycle”. The following definition introduce the concept of negative cycle in a time-varying network.
Definition 4.2 A dynamic path $P\left(x_{1}, \ldots, x_{r}\right)$ is called a dynamic cycle if $x_{1}=x_{r}$ and one can traverse this path starting from $x_{1}$ at a time $t$ and returning to $x_{r}=x_{1}$ at the same time $t$. A negative cycle is defined as such a dynamic cycle whose total cost is negative and whose capacity is greater than zero.
It is clear that, if a network, either the original network or the residual network, contains a negative cycle and it can be reached from $s$, then one can continuously travel along this cycle while the cost is decreased unlimited. In such a case, Thus, the problem has no optimal solution.
The original network contains no negative cycle, since all arcs in $A^{+}$ have positive transit times and all arcs in $A^{-}$have zero capacities. Now examine the residual network. Recall that a residual network is generated based on a flow sent from $s$ to $\rho$. For a general flow, the generated residual network may have negative cycles. However, if the flow is sent along a shortest dynamic f-augmenting path, the residual network generated will contain no negative cycle. This can be seen as follows: Suppose that $f$ is a flow sent along a shortest f-augmenting path $P$ and $C$ is a negative cycle in the residual network $N^{\prime}$ generated by $f$. Clearly, $C$ and $P$ must have common sections. Consider the case where they have one common section. Note that this common section will have opposite directions (refer to Sections 3, Chapter 3 and Section 2, Chapter 4). Denote $S_{c}$ and $S_{p}$ as the sections in cycle $C$ and in path $P$, respectively, then we have $\zeta\left(S_{c}\right)=-\zeta\left(S_{p}\right)$. Let $S^{\prime}$ be the remaining section of $C$. Note that the cost of the cycle $\zeta(C)=\zeta\left(S^{\prime}\right)+\zeta\left(S_{c}\right)<0$, i.e., $\zeta\left(S^{\prime}\right)<-\zeta\left(S_{c}\right)=\zeta\left(S_{p}\right)$. Therefore, replacing $S_{p}$ by $S^{\prime}$ in $P$, we can generate a path that is shorter than $P$. This contradicts the assumption that $P$ is a shortest dynamic f-augmenting path.
The analysis above is summarized in the following property.
数学代写|运筹学作业代写OPERATIONS RESEARCH代考|time-varying network
Property 4.1 If a time-varying network $N$ contains no negative cycle, then the residual network generated based on a shortest dynamic $f$-augmenting path contains no negative cycle.
Proof. Let $N^{\prime}$ be the residual network generated based on a flow $f$ sent along a shortest dynamic f-augmenting path $P$ in $N$. Suppose that $N^{\prime}$ contains negative cycles with $C=\left(x_{0}, y_{1}, \ldots, y_{l}, x_{r}, x_{r-1}, x_{1}, x_{0}\right)$ as the one with the minimum cost. Since $C$ is generated based on $f, C$ and $P$ must have common sections.
First, we consider the case where $P$ and $C$ have one common section. The dotted line in Figure $4.1$ represents the shortest dynamic f-augmenting path $P(s, \rho)=\left(s, \ldots, x_{0}, x_{1}, \ldots, x_{r}, \ldots, \rho\right)$, where $S_{p}=\left(x_{0}\right.$ $\left.x_{1}, \ldots, x_{r}\right)$ is a section of $P . S_{c}=\left(x_{r}, x_{r-1}, \ldots, x_{0}\right)$ is the section of $C$ in $N^{\prime}$ (see the solid line in Figure 4.1). Since $S_{p}$ and $S_{c}$ have opposite directions, we have $\zeta\left(S_{c}\right)=-\zeta\left(S_{p}\right)$. Let $S^{\prime}$ be the section $\left(x_{0}, y_{1}, \ldots, y_{l}, x_{r}\right)$. Since $\zeta(C)=\zeta\left(S^{\prime}\right)+\zeta\left(S_{c}\right)<0$, we have $-\zeta\left(S_{c}\right)>\zeta\left(S^{\prime}\right)$. On the other hand, since $\zeta\left(S_{p}\right)=-\zeta\left(S_{c}\right)$, we have $\zeta\left(S_{p}\right)>\zeta\left(S^{\prime}\right)$. Noting that $P(s, \rho)$ is the shortest f-augmenting path from $s$ to $\rho$ in $N^{\prime}$ and both $S_{p}$ and $S^{\prime}$ exist in $N$, we can use $S^{\prime}$ to replace $S_{p}$ in $P(s, \rho)$ to obtain another path $P^{\prime}(s, \rho)$, with $\zeta\left(P^{\prime}\right)<\zeta(P)$. This is a contradiction to the assumption that $P$ is the shortest dynamic f-augmenting path. Therefore, $N^{\prime}$ can not contain any negative cycles.
Now, we prove the claim that if $P(s, \rho)$ and $C$ have more than one common section, then there must exist another cycle in $N^{\prime}$, say $C^{\prime}$, which has only one common section with $P$, and $C$ and $C^{\prime}$ have the same cost. We will prove the case where $P$ and $C$ have two common sections (other cases can be dealt with in a similar way).
Suppose that $P$ and $C$ have two common sections $\left(x_{0}, \ldots, x_{i}\right)$ and $\left(x_{k}, \ldots, x_{r}\right)$. Then, there is another cycle $C^{\prime \prime}=\left(x_{i}, \ldots, x_{p}, \ldots, x_{k}\right.$, $\left.\ldots x_{i}\right)$ in $N^{\prime}$ (see Figure 4.2). Let $S_{1}=\left(x_{i}, \ldots, x_{p}, \ldots, x_{k}\right)$ and $S_{2}=$ $\left(x_{k}, \ldots, x_{q}, \ldots, x_{i}\right)$. Since both sections $S_{1}$ and $S_{2}$ exist in $N, C^{\prime \prime}$ exists in $N$ too. Thus, $C^{\prime \prime}$ must have a non-negative cost, i.e., $\zeta\left(C^{\prime \prime}\right)=$ $\zeta\left(S_{1}\right)+\zeta\left(S_{2}\right) \geq 0$. If $\zeta\left(C^{\prime \prime}\right)>0$, we have $-\zeta\left(S_{1}\right)<\zeta\left(S_{2}\right)$. Then, replacing $S_{2}$ by the section $\left(x_{k}, \ldots, x_{p}, \ldots, x_{i}\right)$ in $C$ will create a new cycle, with cost less than $\zeta(C)$. However, we have assumed that $C$ is the minimal one among all negative cycles. Thus we must have $\zeta\left(C^{\prime \prime}\right)=0$, i.e., $-\zeta\left(S_{1}\right)=\zeta\left(S_{2}\right)$. Then, we can use the section $\left(x_{k}, \ldots, x_{p}, \ldots, x_{i}\right)$ to replace section $S_{2}$ in $C$ and form a new cycle $C^{\prime}$. Note that $C^{\prime}$ and $P$ have one common section only, a case we have proved in the above.
运筹学代考
数学代写|运筹学作业代写OPERATIONS RESEARCH代考|DYNAMIC PATH
正如我们上面提到的，我们将通过在残差网络中依次搜索最短的动态f-augmenting路径来解决这个问题。为了确保算法可以在有限步内完成，我们需要一个条件，即网络不应包含任何“负循环”。以下定义介绍了时变网络中负循环的概念。
定义 4.2 动态路径磷(X1,…,Xr)称为动态循环，如果X1=Xr并且可以从X1一次吨并返回Xr=X1同时吨. 负循环被定义为总成本为负且容量大于零的动态循环。
很明显，如果一个网络，无论是原始网络还是残差网络，都包含一个负循环，并且可以从s，则可以沿着这个循环不断地旅行，同时成本无限地降低。在这种情况下，因此，问题没有最优解。
原始网络不包含负循环，因为所有弧一种+有正传输时间和所有弧一种−容量为零。现在检查残差网络。回想一下，残差网络是基于从s到ρ. 对于一般流，生成的残差网络可能具有负循环。但是，如果流是沿着最短的动态 f-augmenting 路径发送的，则生成的残差网络将不包含负循环。这可以看作如下：假设F是沿着最短 f 增广路径发送的流磷和C是残差网络中的负循环ñ′由产生F. 清楚地，C和磷必须有共同的部分。考虑他们有一个共同部分的情况。请注意，此公共部分将具有相反的方向r和F和r吨这小号和C吨一世这ns3,CH一种p吨和r3一种nd小号和C吨一世这n2,CH一种p吨和r4. 表示小号C和小号p作为循环中的部分C并在路径中磷, 那么我们有G(小号C)=−G(小号p). 让小号′成为剩下的部分C. 注意循环的成本G(C)=G(小号′)+G(小号C)<0， IE，G(小号′)<−G(小号C)=G(小号p). 因此，更换小号p经过小号′在磷，我们可以生成一条比磷. 这与以下假设相矛盾磷是一条最短的动态 f 增广路径。
上述分析总结为以下性质。
数学代写|运筹学作业代写OPERATIONS RESEARCH代考|TIME-VARYING NETWORK
性质 4.1 如果一个时变网络ñ不包含负循环，则基于最短动态生成的残差网络F- 增强路径不包含负循环。
证明。让ñ′是基于流生成的残差网络F沿着最短的动态 f-augmenting 路径发送磷在ñ. 假设ñ′包含负循环C=(X0,是1,…,是一世,Xr,Xr−1,X1,X0)作为成本最低的一个。自从C是根据生成的F,C和磷必须有共同的部分。
首先，我们考虑以下情况磷和C有一个共同的部分。图中虚线4.1表示最短的动态f-augmenting路径磷(s,ρ)=(s,…,X0,X1,…,Xr,…,ρ)， 在哪里小号p=(X0 X1,…,Xr)是一部分磷.小号C=(Xr,Xr−1,…,X0)是部分C在ñ′ s和和吨H和s这一世一世d一世一世n和一世nF一世G你r和4.1. 自从小号p和小号C有相反的方向，我们有G(小号C)=−G(小号p). 让小号′成为部分(X0,是1,…,是一世,Xr). 自从G(C)=G(小号′)+G(小号C)<0， 我们有−G(小号C)>G(小号′). 另一方面，由于G(小号p)=−G(小号C)， 我们有G(小号p)>G(小号′). 注意到磷(s,ρ)是从s到ρ在ñ′和两者小号p和小号′存在于ñ， 我们可以用小号′取代小号p在磷(s,ρ)获得另一条路径磷′(s,ρ)， 和G(磷′)<G(磷). 这与假设相矛盾磷是最短的动态 f 增广路径。所以，ñ′不能包含任何负循环。
现在，我们证明如果磷(s,ρ)和C有多个公共部分，则必须存在另一个循环ñ′， 说C′, 它只有一个公共部分磷， 和C和C′有相同的成本。我们将证明磷和C有两个共同的部分这吨H和rC一种s和sC一种nb和d和一种一世吨在一世吨H一世n一种s一世米一世一世一种r在一种是.
假设磷和C有两个共同的部分(X0,…,X一世)和(X到,…,Xr). 然后，又是一个循环C′′=(X一世,…,Xp,…,X到,…X一世)在ñ′ s和和F一世G你r和4.2. 让小号1=(X一世,…,Xp,…,X到)和小号2= (X到,…,Xq,…,X一世). 由于这两个部分小号1和小号2存在于ñ,C′′存在于ñ也。因此，C′′必须有非负成本，即G(C′′)= G(小号1)+G(小号2)≥0. 如果G(C′′)>0， 我们有−G(小号1)<G(小号2). 然后，更换小号2按部分(X到,…,Xp,…,X一世)在C将创建一个新的循环，成本低于G(C). 然而，我们假设C是所有负循环中最小的一个。因此我们必须有G(C′′)=0， IE，−G(小号1)=G(小号2). 然后，我们可以使用该部分(X到,…,Xp,…,X一世)替换部分小号2在C并形成一个新的循环C′. 注意C′和磷只有一个公共部分，我们在上面已经证明了一个案例。
