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## 数学代写|运筹学代写Operations Research代考|CONTINUOUS DEMAND INSTANTANEOUS REPLENISHMENT MODEL

The inventory-time behaviour is given in Figure 13.1. We place order for $Q$ items when the stock reaches zero and the inventory is instantaneously replenished. Inventory is consumed continuously at the rate of $D$ year.
The coefficients are:
Annual demand $=D /$ year
Order cost $=C_o$
Carrying cost $=C_c$
Order quantity $=Q$
Number of orders $/$ year $=\frac{D}{Q}$

Annual order cost $=\frac{D C_o}{Q}$
Average inventory in the system $=\frac{Q}{2}$
Annual inventory carrying cost $=\frac{Q C_c}{2}$
Total cost $(T C)=\frac{D C_o}{Q}+\frac{Q C_c}{2}$
The value of $Q$ that minimizes the total cost is obtained by setting the first derivative to zero. We get
$$\frac{-D C_o}{Q^2}+\frac{C_c}{2}=0$$
from which
$$Q^=\sqrt{\frac{2 D C_o}{C_c}}$$ and substituting $Q^$ in $T C$, we get
$$T C^*=\sqrt{2 D C_o C_c}$$

## 数学代写|运筹学代写Operations Research代考|ONSIDERING BACKORDERING

In this model, we allow a backorder of $s$ units every cycle and as soon as the order quantity $Q$ arrives, we issue the backordered quantity. Figure $13.3$ shows the model. The maximum inventory held is $I_m=Q-s$. There is an inventory period of $T_1$ per cycle and a backorder period of $T_2$ per cycle.

The coefficients are:
Annual demand $=D /$ year
Order cost $=C_o$
Carrying cost $=C_c$
Shortage (backorder) $\operatorname{cost}=C_s$
Order quantity $=Q$
Backorder quantity $=s$
Maximum inventory in a cycle $=I_m$
Number of orders/year $=\frac{D}{Q}$
Annual order cost $=\frac{D C_o}{Q}$
Average inventory in the system $=\frac{I_m}{2}$
Annual inventory carrying cost $=\frac{I_m C_c}{2}$
Average shortage in the system $=\frac{s}{2}$

Annual shortage cost $=\frac{s C_s}{2}$
Total cost $T C=\frac{D C_o}{Q}+\frac{I_m C_c}{2} \times \frac{T_1}{\left(T_1+T_2\right)}+\frac{s C_s}{2} \times \frac{T_2}{\left(T_1+T_2\right)}$
From similar triangles, we get
and
Substituting, we get
$$T C=\frac{D C_o}{Q}+\frac{(Q-s)^2 C_c}{2 Q}+\frac{s^2 C_s}{2 Q}$$
The values of $Q$ and $s$ that minimize the total cost are obtained by setting the first partial derivative with respect to $Q$ and $s$ to zero. Partially differentiating with respect to $s$ and setting to zero, we get
$$s=\frac{Q C_c}{\left(C_c+C_s\right)}$$
Partially differentiating with respect to $Q$ and substituting for $s$, we get
$$Q^*=\sqrt{\frac{2 D C_o\left(C_c+C_s\right)}{C_c C_s}}$$

## 数学代写|运筹学代写Operations Research代考|CONTINUOUS DEMAND INSTANTANEOUS REPLENISHMENT MODEL

$$\frac{-D C_o}{Q^2}+\frac{C_c}{2}=0$$
M中
$$Q^{=} \sqrt{\frac{2 D C_o}{C_c}}$$

$$T C^=\sqrt{2 D C_o C_c}$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|运筹学代写Operations Research代考|ONTINUOUS VARIABLES—HIGHER DEGREE

Minimize $X_1^3-5 X_1^2+8 X_1+X_2^3-2 X_2^2-10 X_2+10$
Subject to
$$\begin{array}{r} X_1+X_2 \leq 4 \ X_1, X_2 \geq 0 \end{array}$$
Stage: Each variable
State: Resource available for allocation
Decision variable: Values of $X_1$ and $X_2$
Criterion of effectiveness: Minimize $Z$
In this problem we first solve for variable $X_1$ and then for variable $X_2$. The objective function is rewritten as:
Minimize $X_2^3-2 X_2^2-10 X_2+X_1^3-5 X_1^2+8 X_1+10$
One more stage to go $n=1$
\begin{aligned} f_1\left(s_1, X_1\right) & =X_1^3-5 X_1^2+8 X_1+10 \ f_1^\left(s_1\right) & =\text { Minimize } X_1^3-5 X_1^2+8 X_1+10 \ 0 & \leq X_1 \leq s_1 \end{aligned} Differentiating with respect to $X_1$ and equating to zero, we get or $$\begin{array}{r} 3 X_1^2-10 X_1+8=0 \ X_1=2 \text { or } X_1=\frac{4}{3} \end{array}$$ Second derivative is $6 X_1-10$ and takes positive value for $X_1=2$ indicating minimum. Therefore, $X_1{ }^=2$ if $s_1 \geq 2$ and $X_1^=s_1$ if $s_1<2$. \begin{aligned} f_1^\left(s_1\right) & =8-20+16+10=14 \text { if } s_1 \geq 2 \ & =s_1^3-5 s_1^2+8 s_1+10 \text { if } 0=s_1<2 \end{aligned}
Since the function is cubic, we also verify the value of the function at $X_1=0$. At $X_1=0$, the value of $f_1\left(s_1\right)=10$ which is less than 14 and since $s_1$ can take only non-negative values we have $X_1^=0$ and $f_1^\left(s_1\right)=10$.

## 数学代写|运筹学代写Operations Research代考|ACTORIZING THE TERMS

Maximize $2 X_1+3 X_2+X_1 X_2$
Subject to
$$\begin{array}{r} X_1+X_2 \leq 2 \ X_1, X_2 \geq 0 \end{array}$$
Stage: Each variable
State: Resource available for allocation
Decision variable: Values of $X_1$ and $X_2$
Criterion of effectiveness: Maximize $Z$

In this example the term $X_1 X_2$ makes it difficult to separate the objective function in terms of separable functions of the variables. We factorize the objective function as:
$$\text { Maximize }\left(X_1+3\right)\left(X_2+2\right)-6$$
We can leave out the constant from the objective function and write the problem as:
Maximize $\left(X_1+3\right)\left(X_2+2\right)$
Subject to
$$\begin{array}{r} X_1+X_2 \leq 2 \ X_1, X_2 \geq 0 \end{array}$$
One more stage to go $n=1$
\begin{aligned} f_1\left(s_1, X_2\right) & =X_2+2 \ f_1^\left(s_1\right) & =\text { Maximize } X_2+2 \end{aligned} Subject to $0 \leq X_2 \leq s_1$ Here, the maximum value is at $X_2^=s_1$ and $f_1^\left(s_1\right)=s_1+2$ Two more stages to go $n=2$ \begin{aligned} f_2\left(2, X_1\right)= & \left(X_1+3\right) f_1^\left(2-X_1\right) \ f_2^\left(s_1\right)= & \text { Maximize }\left(X_1+3\right)\left(2-X_1+2\right) \ & \text { Subject to } 0 \leq X_1 \leq 2 \end{aligned} Maximize $\left(X_1+3\right)\left(4-X_1\right)$ Subject to $0 \leq X_1 \leq 2$ Maximize $-X_1^2+X_1+12$ Differentiating with respect to $X_1$ and equating to zero, we get $X_1=1 / 2$. The second derivative is negative indicating maximum. We have $X_1^=1 / 2, s_1=3 / 2$ and $X_2^*=3 / 2$ with $Z=49 / 4-$ $6=25 / 4$ (for the original problem)

## 数学代写|运筹学代写Operations Research代考|ONTINUOUSVARIABLESHIGHER DEGREE

$$X_1+X_2 \leq 4 X_1, X_2 \geq 0$$

$$f_1\left(s_1, X_1\right)=X_1^3-5 X_1^2+8 X_1+10 f_1^{\left(s_1\right)} \quad=\text { Minimize } X_1^3-5 X_1^2+8 X_1+100 \leq X_1 \leq s_1$$

$$3 X_1^2-10 X_1+8=0 X_1=2 \text { or } X_1=\frac{4}{3}$$

$$f_1^{\left(s_1\right)}=8-20+16+10=14 \text { if } s_1 \geq 2 \quad=s_1^3-5 s_1^2+8 s_1+10 \text { if } 0=s_1<2$$

## 数学代写|运筹学代写Operations Research代考|ACTORIZING THE TERMS

$$X_1+X_2 \leq 2 X_1, X_2 \geq 0$$

$$\text { Maximize }\left(X_1+3\right)\left(X_2+2\right)-6$$

$$X_1+X_2 \leq 2 X_1, X_2 \geq 0$$

$$f_1\left(s_1, X_2\right)=X_2+2 f_1^{\left(s_1\right)} \quad=\text { Maximize } X_2+2$$

$$f_2\left(2, X_1\right)=\left(X_1+3\right) f_1^{\left(2-X_1\right)} f_2^{\left(s_1\right)}=\quad \text { Maximize }\left(X_1+3\right)\left(2-X_1+2\right) \text { Subject to } 0 \leq X_1 \leq 2$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代写|运筹学代写Operations Research代考|Another Formulation for Subtour Elimination

Let us consider another type of subtour elimination constraint of the form:
\begin{aligned} U_i-U_j+n X_{i j} \leq n-1 & & \text { for } i=1,2, \ldots, n-1 \text { and } j=2,3, \ldots, n \ U_j \geq 0 & & \text { (Bellmore and Nemhauser, 1968) } \end{aligned}

For our 5 -city example we will have $(n-1)^2$ constraints. Let us consider an infeasible solution having a subtour (not involving City 1) $X_{45}=X_{54}=1$. The two relevant constraints are:
and
\begin{aligned} & U_4-U_5+5 X_{45} \leq 4 \ & U_5-U_4+5 X_{54} \leq 4 \end{aligned}
This is clearly infeasible because the two constraints when added together gives
$$5 X_{45}+5 X_{54} \leq 8$$
which is infeasible for $X_{45}=X_{54}=1$. Therefore, every subtour not involving City 1 will violate the relevant set of constraints.
Let us consider a subtour involving City 1 given by
$$X_{12}=X_{23}=X_{31}=1$$
We have two constraints:
\begin{aligned} & U_1-U_2+5 X_{12} \leq 4 \ & U_2-U_3+5 X_{23} \leq 4 \end{aligned}
The constraint $U_3-U_1+5 X_{31} \leq 4$ does not exist for $U_j=1$.
Adding the two constraints, we get
$$U_1-U_3+10 \leq 8 \quad \text { for } X_{12}=X_{23}=1$$
It is possible to have values $U_1$ and $U_3$ that satisfy the constraints and, therefore, the constraint $U_i-U_j+n X_{i j} \leq n-1$ is unable to prevent subtours involving City 1 from occurring. However, we realize that for every subtour involving City 1 there has to be a subtour that does not involve City 1 (the subtour $X_{45}=X_{54}=1$ in our example) and the constraints are able to prevent them from happenning. Therefore, the constraints eliminate all subtours. The only requirement is that we define $d_{i j}=\infty$ (or $M$ ) so that singleton subtours are indirectly eliminated.

## 数学代写|运筹学代写Operations Research代考|The TSP and the Theory of NP-Completeness

Any algorithm that solves a problem carries out various steps that can be reduced to additions, subtraction, multiplication and division and other basic operations. Assuming that each of these operations take unit processing time, it is possible to represent the time taken to implement an algorithm in terms parameters such as problem, size, etc. This function can be a polynomial function or an exponential function. If it is a polynomial function, we say that the algorithm has a complexity of $O\left(N^k\right)$ where $N$ is a problem parameter (say, size) and $k$ is the order of the polynomial. The order is the power corresponding to the highest degree in the polynomial and is used because the rate of increase of the polynomial depends on the order of the polynomial. If the function is exponential, we say that the algorithm is exponential. Examples of exponential functions could be $n !, e^n$, etc.

An algorithm is polynomial if the order of complexity is of the form $O\left(N^k\right)$ and the problem is in the category of “easy” problems. Examples of easy problems are matrix inversion, solving linear equations, assignment problem, etc.

A decision problem is one that has a YES/NO answer. NP is a class of problems with the property that for any instance for which the answer is YES, there is a polynomial proof of YES. If two problems are in NP and if an instance of one can be converted in polynomial time to an instance of another, the problems are reducible. $P$ is a class of problems in NP where there exists a polynomial algorithm. $P$ is polynomially reducible to $Q$. NP complete is a subset of $P$, where the problems are reducible to each other. An optimization problem where the decision problem lies in NP complete is called NP hard. The TSP is an important problem in the class of NP complete problems.

A given problem is NP complete if it can be transformed into zero-one integer programming problem in polynomial time and if zero-one integer programming problem can be transformed to it in polynomial time (Bertsimas and Tsitsiklis, 1997). To show that a given problem is NP complete, it is customary to reduce it to a known NP complete problem. There are several instances where a given problem has been shown to be NP complete by showing that it is reducible to the TSP. (For further reading on the theory of NP-completeness, the reader is referred to Garey and Johnson, 1979, Papadimitriou and Steiglitz, 1982 and Wilf, 1975).

## 数学代写|运筹学代写Operations Research代考|Another Formulation for Subtour Elimination

\Begin{aligned} U_i-U_j+n X_{i j} \leq n-1 & & \text { for } i=1,2, \ldots, n-1 \text { and } j=2,3, \ldots, n \j U_j \geq 0 & & \text { （Bellmore and Nemhauser, 1968）}。 \end{aligned}

\begin{aligned} & u_4-u_5+5 x_{45} \leq 4\leq & u_5-u_4+5 x_{54} \x_{54}。 \end{aligned}

$$5 x_{45}+5 x_{54}。\leq 8$$

$$X_{12}=X_{23}=X_{31}=1$$

`begin{aligned} & u_1-u_2+5 x_{12} \leq 4\leq & u_2-u_3+5 x_{23} \u_2-u_3+5 x_{23} leq 4 \end{aligned} 约束$U_3-U_1+5 X_{31} \leq 4$ 对于$U_j=1$不存在。 将这两个约束相加，我们得到 $$U_1-U_3+10leq 8 quadtext { for }. x_{12}=x_{23}=1$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## avatest™帮您通过考试

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## 数学代写|运筹学代写Operations Research代考|THE PRIMAL DUAL ALGORITHM

Consider the Linear Programming illustration given below
ILlUSTRATION $6.5$
Minimize $Z=3 X_1+4 X_2$
Subject to
\begin{aligned} 2 X_1+3 X_2 & \geq 8 \ 5 X_1+2 X_2 & \geq 12 \ X_1, X_2 & \geq 0 \end{aligned}
This can be solved using the two phase method or the Big-M method. We have already seen that both these would require two iterations to get to the optimal solution for this instance. Both the methods would require at least as many iterations as the number of artificial variables to replace them in the basis.

We have also solved this problem using the dual simplex algorithm and this would also require as many iterations as the number of infeasible basic variables in the initial basis. We solve this problem using a new method called the Primal Dual Algorithm.
We add slack variables $X_3$ and $X_4$ and rewrite the problem as
Minimize $Z=3 X_1+4 X_2$
Subject to
\begin{aligned} 2 X_1+3 X_2-X_3 &=8 \ 5 X_1+2 X_2-X_4 &=12 \ X_1, X_2, X_3, X_4 & \geq 0 \end{aligned}
The dual of this problem is
Maximize $8 y_1+12 y_2$
Subject to
$$\begin{array}{r} 2 y_1+5 y_2 \leq 3 \ 3 y_1+2 y_2 \leq 4 \ -y_1 \leq 0 \ -y_2 \leq 0 \end{array}$$
$y_1, y_2$ unrestricted in sign
The dual solution $y=\left[\begin{array}{ll}0 & 0\end{array}\right]$ is feasible to the dual. Also, this satisfies the third and fourth constraint of the dual as an equation.

## 数学代写|运筹学代写Operations Research代考|GOAL PROGRAMMING

So far, in all our problems encountered in linear programming, we considered a single linear objective function and a set of well defined and rigid linear constraints. When even one of the constraints is violated, we encountered an infeasible solution to the problem.

In practice, we could have multiple objectives when addressing real-life situations. Some constraints may be rigid while some may be flexible and would have targets or goals rather than rigid constraints. We model such situations using a technique called goal programming (Charnes et al., 1960).

Goal programming is one of the ways to address multiple objectives. The easier way is to convert a multiple objective problem into a single objective problem by considering weights for each of the objectives. Another way is to rank the objectives and solve it as a sequence of single objective problems, considering the ranked objectives one at a time. This method should ensure that the solution obtained for a particular objective does not worsen the solution (objective function) of an earlier solved higher ranked objective. This is called lexicographic minimization (optimization) and when applied to goal programming, it is called lexicographic goal programming.

In this section, we illustrate some aspects of lexicographic goal programming through few examples and solve them using graphical and simplex algorithms.

## 数学代写|运筹学代写Operations Research代考|THE PRIMAL DUAL ALGORITHM

$$2 X_1+3 X_2 \geq 85 X_1+2 X_2 \quad \geq 12 X_1, X_2 \geq 0$$

$$2 X_1+3 X_2-X_3=85 X_1+2 X_2-X_4 \quad=12 X_1, X_2, X_3, X_4 \geq 0$$

$$2 y_1+5 y_2 \leq 33 y_1+2 y_2 \leq 4-y_1 \leq 0-y_2 \leq 0$$
$y_1, y_2$ 不绶限制的符昊

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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## 数学代考|运筹学代写Operations Research代考|illustRation of duality

Maximize $\mathrm{Z}=2 \mathrm{x}{1}+6 \mathrm{x}{2}+5 \mathrm{x}{3}$ Subject to: $$\begin{gathered} \mathrm{x}{1}+2 \mathrm{x}{2}+2 \mathrm{x}{3} \leq 10 \ 3 \mathrm{x}{1}+3 \mathrm{x}{2}+2 \mathrm{x}{3} \leq 12 \text { with } \mathrm{x}{1}, \mathrm{x}{2} \geq 0 \end{gathered}$$ i. Construct the dual form of given primal problem. ii. Solve only primal by simplex method and identify basic solution at each iteration. Also, identify basic and non-basic variables of dual problem. Solution: i. Dual form would be: Minimize $\mathrm{Z}^{*}=10 \mathrm{y}{1}+12 \mathrm{y}{2}$ Subject to: $$\begin{gathered} 1 \mathrm{y}{1}+3 \mathrm{y}{2} \geq 2 \ 2 \mathrm{y}{1}+3 \mathrm{y}{2} \geq 6 \ 2 \mathrm{y}{1}+2 \mathrm{y}{2} \geq 5 \text { with } \mathrm{y}{1}, \mathrm{y}{2} \geq 0 \end{gathered}$$ ii. Following would be initial models of both primal and dual: Maximize $\mathrm{Z}=2 \mathrm{x}{1}+6 \mathrm{x}{2}+5 \mathrm{x}{3}+0 \mathrm{~s}{1}+0 \mathrm{~s}{2}$
Subject to:
$$\begin{gathered} x_{1}+2 x_{2}+2 x_{3}+1 s_{1}+0 s_{2}=10 \ 3 x_{1}+3 x_{2}+2 x_{3}+0 s_{1}+1 s_{2}=12 \text { with } x_{1}, x_{2} \geq 0 \end{gathered}$$
where $s_{1}, s_{2}$ are slack variables.

## 数学代考|运筹学代写Operations Research代考|SUMMARY

This chapter discussed concepts of sensitivity analysis and duality by illustrating linear programming problems. It was inferred that coefficients and parameters used in formulation of a linear programming model are only estimates, and it is important to apply sensitivity analysis to identify which of these are sensitive and which are insensitive to change. Sensitivity analysis was performed by finding an allowable feasible range of RHS values, which indicate resource availability and also a range of optimality was identified for objective function coefficients. It is important to note that in this chapter such deduction was done by changing only one coefficient, keeping all others as unchanged, though limited analysis is possible by adopting multiple changes.

Another section was devoted to understanding the concept of duality that was helpful in implementing and interpreting sensitivity analysis. A mirror image called dual can be constructed of original problem that is complementary to primal. Solving any one would provide the solution for the other. The benefit of creating and solving dual is that any variation in solution identifies economic contribution by variable that has caused such variation. The formulation of dual problem and its relationship with primal was discussed in detail.

## 数学代考|运筹学代写Operations Research代考|illustRation of duality

$$\mathrm{x} 1+2 \mathrm{x} 2+2 \mathrm{x} 3 \leq 103 \mathrm{x} 1+3 \mathrm{x} 2+2 \mathrm{x} 3 \leq 12 \text { with } \mathrm{x} 1, \mathrm{x} 2 \geq 0$$

$$1 \mathrm{y} 1+3 \mathrm{y} 2 \geq 22 \mathrm{y} 1+3 \mathrm{y} 2 \geq 62 \mathrm{y} 1+2 \mathrm{y} 2 \geq 5 \text { with } \mathrm{y} 1, \mathrm{y} 2 \geq 0$$
ii. 以下是原始模型和对偶模型的初始模型: 最大化 $\mathrm{Z}=2 \mathrm{x} 1+6 \mathrm{x} 2+5 \mathrm{x} 3+0 \mathrm{~s} 1+0 \mathrm{~s} 2$ 受制于:
$$x_{1}+2 x_{2}+2 x_{3}+1 s_{1}+0 s_{2}=103 x_{1}+3 x_{2}+2 x_{3}+0 s_{1}+1 s_{2}=12 \text { with } x_{1}, x_{2} \geq 0$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。