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# 数学代写|运筹学代写Operations Research代考|MATH3830 OPTIMALITY OF THE MODI METHOD

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## 数学代写|运筹学代写Operations Research代考|OPTIMALITY OF THE MODI METHOD

Let us write the primal and dual of the balanced transportation problem. The primal is:
Minimize $\Sigma \Sigma C_{i j} X_{i j}$
Subject to
\begin{aligned} \Sigma X_{i j}=& a_i \forall i=1, \ldots, m \ \Sigma X_{i j}=& b_j \forall j=1, \ldots, n \ & X_{i j} \geq 0 \end{aligned}
Let us introduce dual variables $u_i$ and $v_j$ corresponding to the two sets of constraints. The dual is:
Maximize $\sum_{i=1}^m a_i u_i+\sum_{j=1}^n b_j v_j$
Subject to
$$u_i+v_j \leq C_{i j}$$
$u_i, v_j$ unrestricted in sign
Since the problem is balanced, we know that
$$\sum_{i=1}^m a_i=\sum_{j=1}^n b_j$$
Because of this the primal has a set of linearly dependent constraints and hence $m+n-1$ linearly independent basic variables. Therefore, in the dual also we have $m+n-1$ linearly independent variables. One dual variable has to be given an arbitrary value. By convention we always put $u_1=0$

## 数学代写|运筹学代写Operations Research代考|SOLVING UNBALANCED TRANSPORTATION PROBLEMS

The dual to the unbalanced transportation problem will have variables $u_i \leq 0$ and $v_j \geq 0$. This means that every value of $u_i$ computed should be strictly non-positive and every $v_j$ computed should be non-negative in order to maintain dual feasibility. This will not be easy to achieve.
The easier option is to convert the unbalanced problem to a balanced problem and apply the MODI method as it is to obtain the optimal solution from the starting basic feasible solution.
This is explained by an example. Consider the transportation problem with three supply points and four demand points (in the usual representation) given in Table 4.27.

This problem is unbalanced because the total supply $\left(\Sigma a_i\right)$ is less than the total requirement $\left(\Sigma b_j\right)$. The total supply is 140 and the total demand is 150 . In this case, we won’t be able to meet a demand of 10 units. We convert this unbalanced problem to a balanced problem by adding an extra supply (row) with 10 units (called dummy row). The cost of transportation from this dummy row to all the columns is zero. This is shown in Table 4.28.

## 数学代写|运筹学代写Operations Research代考|OPTIMALITY OF THE MODI METHOD

$$\Sigma X_{i j}=a_i \forall i=1, \ldots, m \Sigma X_{i j}=\quad b_j \forall j=1, \ldots, n X_{i j} \geq 0$$

$$u_i+v_j \leq C_{i j}$$
$u_i, v_j$ unrestricted in sign

$$\sum_{i=1}^m a_i=\sum_{j=1}^n b_j$$

## 数学代写|运筹学代写Operations Research代考|SOLVING UNBALANCED TRANSPORTATION PROBLEMS

10 个单位的需求。我们通过添加具有 10 个单位的额外供应 (行) (称为虚拟行) 将这个不平衡问题转换为平衡问题。从这个虚拟 行到所有列的运输成本为零。如表 $4.28$ 所示。

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