# Fully Interval Integer Transhipment Problem- A Solution Approach

G. Sudha, G. Ramesh, D. Datta, K. Ganesan

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## To cite this version:

G. Sudha, G. Ramesh, D. Datta, K. Ganesan. Fully Interval Integer Transhipment Problem- A Solution Approach. 4th International Conference on Intelligence Science (ICIS), Feb 2021, Durgapur, India. pp.196-206, 10.1007/978-3-030-74826-5_17. hal-03741707

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# Fully interval integer transhipment problem- A solution approach

G. Sudha¹, G. Ramesh², D. Datta³, K. Ganesan⁴

¹, ², ⁴Department of Mathematics, SRM Institute of science and Technology,
Kattankulathur, Tamil Nadu, India.

³Adjunct faculty, Department of Mathematics, SRM Institute of science and
Technology, (former) Health Physics Division, Bhabha Atomic Research Centre,
Trombay, Mumbai-400085, India

E-Mail: sudhag@srmist.edu.in;
rameshapg@gmail.com; dbbrt_datta@yahoo.com; ganesank@srmist.edu.in

## Abstract—

We present an improved interval Vogel's approximation algorithm for the initial feasible solution of a transhipment problem involving fully interval integer numbers in this paper. The interval variant of MODI algorithm is used to measure the optimality of the interval feasible solution. In a transportation problem, shipments are only permitted between source-sink pairs, while in a transhipment problem, shipments can be permitted both between sources and sinks. Furthermore, there may be points where units of a substance may be transferred from a source to a sink. To demonstrate the effectiveness of the proposed algorithm, a numerical example is given.

*Keywords*— Interval Numbers, New interval arithmetic, Ranking, Interval initial feasible solution, Improved interval Vogel's approximation algorithm.

## 1. INTRODUCTION

Goods are transported from specific sources to specific destinations in classic transportation problems (TP). However, in some situations, products can be delivered to a specific location either directly or through one or more intermediary nodes. Either of these nodes has the ability to function as an origin or a destination. Transhipment issues are models with these additional features. Thus, transhipment issues are a type of transportation problem in which merchandise are elated from a source to a destination via various intermediary nodes (sources/destinations), with the possibility of changing modes of transportation or consolidating or de-consolidating shipments. A transhipment problem's main goal is to find the lowest transportation cost while
satisfying supply and demand at each node. These issues have found a
lot of use in the e-commerce era, when the online shopping business is
thriving thanks to its customers' convenience. To solve the tranship-
ment problem, translate it to an equivalent TP first, and then use exist-
ing techniques to find an initial simple feasible solution. Furthermore,
an optimal solution of equivalent TP can be obtained using the modi-
fied distribution (MODI) process. Orden [17] introduced the concept of
transhipment problem in the year 1956. When the decision parameters
are known vaguely, the classical mathematical techniques are success-
fully applied to model and solve the transhipment problems. But actual-
ly there are problems that the charge coefficients and the supply and
demand quantities may be uncertain due to many reasons. Hence we
must deal with vague in order in formulating and solving real earth
troubles. Hurt et.al [5] have developed an alternative formulations of
the transhipment problem.

Ganesan et.al [3] discussed on arithmetic operations of interval num-
bers and some properties of interval matrices. Pandian et.al [11] studied
a new method for finding an optimal solution of fully interval integer
transportation problems. NagoorGani et.al [8] developed a tranship-
ment problem in fuzzy environment. Rajendran et.al [12] discussed
fully interval transhipment problems. Akilbasha et.al [1] formulated an
innovative exact method for solving fully interval integer transportation
problems.

Sophia Porchelvi et.al [14] studied a comparative study of optimum
solution between interval transportation and interval transhipment prob-
lem. In this paper, we propose a new interval arithmetic and an im-
proved interval Vogel's approximation algorithm to solve the problem
of interval transhipment without translating into crisp form.

The structure of this article is given as follows: Section 2 deals with the basic definitions. Section 3 deals with the mathematical formulation of fully interval transhipment problem. Section 4 introduces improved interval Vogel's approximation algorithm under generalized interval arithmetic. Section 5 provides an example to illustrate the theory developed in this paper. Section 6 concludes this manuscript.
## 2. PRELIMINARIES

The purpose of this segment is to provide some observations, ideas and results which are useful in our further consideration.

### 2.1. Interval numbers

Let $\tilde{a}=[a_1, a_2]=[x \in R: a_1 \le x \le a_2$ and $a_1, a_2 \in R]$ be an interval on the real line R. If $\tilde{a}=a_1=a_2=a$, then $\tilde{a}=[a, a]$ is a real number (or a degenerate interval). The terms interval and interval number are used interchangeably. The mid-point and width (or half-width) of an interval number $\tilde{a}=[a_1, a_2]$ are defined as $m(\tilde{a}) = \left(\frac{a_1 + a_2}{2}\right)$ and $w(\tilde{a}) = \left(\frac{a_2 - a_1}{2}\right)$.

The interval number $\tilde{a}$ can also be expressed in terms of its midpoint and width as $\tilde{a}=[a_1, a_2]=\langle m(\tilde{a}), w(\tilde{a}) \rangle$. We use $IR$ to denote the set of all interval numbers defined on the real line R.

### 2.2. Ranking of Interval Numbers

Sengupta and Pal [13] suggested a easy and powerful index to compare any two intervals on $IR$ through the satisfaction of decision-makers.

**Definition 2.2.1.** Let $\leq$ be an extended order relation between the interval numbers $\tilde{a}=[a_1, a_2]$, $\tilde{b}=[b_1, b_2]$ in $IR$, then for $m(\tilde{a})<m(\tilde{b})$, we construct a premise $(\tilde{a} \leq \tilde{b})$ which implies that $\tilde{a}$ is inferior to $\tilde{b}$ (or $\tilde{b}$ is superior to $\tilde{a}$).

An acceptability function $A_{\leq}: IR \times IR \rightarrow [0, \infty)$ is defined as:

$$A_{\leq}(\tilde{a}, \tilde{b}) = A(\tilde{a} \leq \tilde{b}) = \frac{m(\tilde{b}) - m(\tilde{a})}{w(\tilde{b}) + w(\tilde{a})}, \text{ where } w(\tilde{b}) + w(\tilde{a}) \neq 0.$$

$A_{\leq}$ may be interpreted as the grade of acceptability of the first interval number to be inferior to the second interval number. For any two interval numbers $\tilde{a}$ and $\tilde{b}$ in $IR$ either $A(\tilde{a} \leq \tilde{b}) \geq 0$ (or) $A(\tilde{b} \geq \tilde{a}) \geq 0$ (or) $A(\tilde{a} \leq \tilde{b}) = 0$ (or) $A(\tilde{b} \geq \tilde{a}) = 0$ (or) $A(\tilde{a} \leq \tilde{b}) + A(\tilde{b} \leq \tilde{a}) = 0$.

If $A(\tilde{a} \leq \tilde{b}) = 0$ and $A(\tilde{b} \leq \tilde{a}) = 0$, then we say that the interval numbers $\tilde{a}$ and $\tilde{b}$ are equivalent (non-inferior to each other) and we denote it by $\tilde{a} \approx \tilde{b}$. Also if $A(\tilde{a} \leq \tilde{b}) \geq 0$, then $\tilde{a} \leq \tilde{b}$ and if $A(\tilde{b} \leq \tilde{a}) \geq 0$, then $\tilde{b} \leq \tilde{a}$.
## 2.3. A New Interval Arithmetic

Ming Ma et al. [7] suggested a new fuzzy arithmetic based on the index of locations and the index function of fuzziness. The location index number follows the usual arithmetic where as the fuzziness index functions are assumed to obey the lattice law which is the least upper bound and the greatest lower bound. That is for $a, b \in L$ we define $a \lor b = \max\{a, b\}$ and $a \land b = \min\{a, b\}$.

For any two intervals $\tilde{a} = [a_1, a_2]$, $\tilde{b} = [b_1, b_2] \in IR$ and for $* \in \{+, -, ?\}$ the arithmetic operations on $\tilde{a}$ and $\tilde{b}$ are defined as :

$$ \tilde{a} * \tilde{b} = [a_1, a_2] * [b_1, b_2] = \langle m(\tilde{a}), w(\tilde{a}) \rangle * \langle m(\tilde{b}), w(\tilde{b}) \rangle \\ = \langle m(\tilde{a}) * m(\tilde{b}), \max\{w(\tilde{a}), w(\tilde{b})\} \rangle $$

In particular, here we should use following interval arithmetic operations like addition, subtraction, multiplication and division.

# 3. MAIN RESULTS

## 3.1. Mathematical formulation for using fully interval transhipment problem

In the transshipment table $O_1, O_2, O_3, \dots, O_i, \dots, O_m$ are sources from where goods are to be transported to destinations $D_1, D_2, D_3, \dots, D_j, \dots, D_n$. Any of the sources can transport to any of the destinations. $\tilde{c}_{ij}$ is per unit transporting cost of goods from $i^{th}$ source $O_i$ to $j^{th}$ destination $D_j$ for all $i = 1, 2, \dots, m$ and $j = 1, 2, \dots, n$. $\tilde{x}_{ij}$ is the amount of goods transporting from $i^{th}$ source $O_i$ to $j^{th}$ destination $D_j$. $\tilde{a}_i$ be the amount of goods available at the origins $O_i$ and $\tilde{b}_j$ the demand at the destination $D_j$. The mathematical model of fully interval transportation problem is as follows

$$ \begin{array}{ll} \text{Minimize} & \tilde{Z} \approx \sum_{i=1}^{m} \sum_{j=1}^{n} \tilde{c}_{ij} \tilde{x}_{ij} \\ \text{subject to} & \sum_{j=1}^{n} \tilde{x}_{ij} \approx \tilde{a}_i, \quad i = 1, 2, 3, \dots, m \\ & \sum_{i=1}^{m} \tilde{x}_{ij} \approx \tilde{b}_j, \quad j = 1, 2, 3, \dots, n \\ & \sum_{i=1}^{m} \tilde{a}_i \approx \sum_{j=1}^{n} \tilde{b}_j, \quad \text{where } i = 1, 2, 3, \dots, m; j = 1, 2, 3, \dots, n \text{ and} \\ & \tilde{x}_{ij} \succeq \tilde{0} \quad \text{for all } i \text{ and } j \end{array} $$
<table><caption>Table:1 Interval transhipment table</caption><thead><tr><th></th><th></th><th>1</th><th>2</th><th>...</th><th>i</th><th>...</th><th>m</th><th>m+1</th><th>m+2</th><th>...</th><th>m+j</th><th>...</th><th>m+n</th><th>supply</th></tr><tr><th></th><th></th><th>O<sub>1</sub></th><th>O<sub>2</sub></th><th>...</th><th>O<sub>i</sub></th><th>...</th><th>O<sub>m</sub></th><th>D<sub>1</sub></th><th>D<sub>2</sub></th><th>...</th><th>D<sub>j</sub></th><th>...</th><th>D<sub>n</sub></th><th></th></tr></thead><tbody><tr><td>1</td><td>O<sub>1</sub></td><td>x<sub>11</sub></td><td>x<sub>12</sub></td><td>...</td><td>x<sub>1f</sub></td><td>...</td><td>x<sub>1m</sub></td><td>x<sub>1,m+1</sub></td><td>x<sub>1,m+2</sub></td><td>...</td><td>x<sub>1,m+j</sub></td><td>...</td><td>x<sub>1,m+n</sub></td><td>a<sub>1</sub></td></tr><tr><td>2</td><td>O<sub>2</sub></td><td>x<sub>21</sub></td><td>x<sub>22</sub></td><td>...</td><td>x<sub>2f</sub></td><td>...</td><td>x<sub>2m</sub></td><td>x<sub>2,m+1</sub></td><td>x<sub>2,m+2</sub></td><td>...</td><td>x<sub>2,m+j</sub></td><td>...</td><td>x<sub>2,m+n</sub></td><td>a<sub>2</sub></td></tr><tr><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><td>i</td><td>O<sub>i</sub></td><td>x<sub>i1</sub></td><td>x<sub>i2</sub></td><td>...</td><td>x<sub>if</sub></td><td>...</td><td>x<sub>im</sub></td><td>x<sub>i,m+1</sub></td><td>x<sub>i,m+2</sub></td><td>...</td><td>x<sub>i,m+j</sub></td><td>...</td><td>x<sub>i,m+n</sub></td><td>a<sub>i</sub></td></tr><tr><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><td>m</td><td>O<sub>m</sub></td><td>x<sub>m1</sub></td><td>x<sub>m2</sub></td><td>...</td><td>x<sub>mi</sub></td><td>...</td><td>x<sub>mm</sub></td><td>x<sub>m,m+1</sub></td><td>x<sub>m,m+2</sub></td><td>...</td><td>x<sub>m,m+j</sub></td><td>...</td><td>x<sub>m,m+n</sub></td><td>a<sub>m</sub></td></tr><tr><td>m+1</td><td>D<sub>1</sub></td><td>x<sub>m+1,1</sub></td><td>x<sub>m+1,2</sub></td><td>...</td><td>x<sub>m+1,f</sub></td><td>...</td><td>x<sub>m+1,m</sub></td><td>x<sub>m+1,m+1</sub></td><td>x<sub>m+1,m+2</sub></td><td>...</td><td>x<sub>m+1,m+j</sub></td><td>...</td><td>x<sub>m+1,m+n</sub></td><td>...</td></tr><tr><td>m+2</td><td>D<sub>2</sub></td><td>x<sub>m+2,1</sub></td><td>x<sub>m+2,2</sub></td><td>...</td><td>x<sub>m+2,f</sub></td><td>...</td><td>x<sub>m+2,m</sub></td><td>x<sub>m+2,m+1</sub></td><td>x<sub>m+2,m+2</sub></td><td>...</td><td>x<sub>m+2,m+j</sub></td><td>...</td><td>x<sub>m+2,m+n</sub></td><td>...</td></tr><tr><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><td>m+j</td><td>D<sub>j</sub></td><td>x<sub>m+j,1</sub></td><td>x<sub>m+j,2</sub></td><td>...</td><td>x<sub>m+j,f</sub></td><td>...</td><td>x<sub>m+j,m</sub></td><td>x<sub>m+j,m+1</sub></td><td>x<sub>m+j,m+2</sub></td><td>...</td><td>x<sub>m+j,m+j</sub></td><td>...</td><td>x<sub>m+j,m+n</sub></td><td>...</td></tr><tr><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td></tr><tr><td>m+n</td><td>D<sub>n</sub></td><td>x<sub>m+n,1</sub></td><td>x<sub>m+n,2</sub></td><td>...</td><td>x<sub>m+n,f</sub></td><td>...</td><td>x<sub>m+n,m</sub></td><td>x<sub>m+n,m+1</sub></td><td>x<sub>m+n,m+2</sub></td><td>...</td><td>x<sub>m+n,m+j</sub></td><td>...</td><td>x<sub>m+n,m+n</sub></td><td>...</td></tr><tr><td colspan="2">Demand</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>...</td><td>b<sub>1</sub></td><td>b<sub>2</sub></td><td>...</td><td>b<sub>j</sub></td><td>...</td><td>b<sub>n</sub></td><td></td></tr></tbody></table>

Since any source or destination may ship to any other source or destination in a transshipment problem, it would be convenient to number the origins and destinations sequentially, with the origins numbered 1 to m and the destinations numbered m + 1 to m + n.

<figure><img src="image_3.png" /><figcaption>Figure 1</figcaption></figure>
facients are -1. The problem, on the other hand, can be easily trans-
formed into a normal transportation problem.

$$
t_i = \sum_{j=1, j \neq i}^{m+n} \tilde{x}_{ij}, \quad i = 1, 2, 3, \dots, m, \quad t_i + \tilde{x}_{ii} = \sum_{j=1}^{m+n} \tilde{x}_{ji}, \quad i = 1, 2, 3, \dots, m
$$

$$
\text{and } t_j = \sum_{\substack{i=1 \ i \neq j \\ i \neq j}}^{m+n} \tilde{x}_{ji}, \quad j = m+1, m+2, m+3, \dots, m+n
$$

$$
t_i + \tilde{x}_{ij} = \sum_{i=1}^{m+n} \tilde{x}_{ji}, \quad j = m+1, m+2, m+3, \dots, m+n
$$

where $t_i$ represents the total amount of transshipment through the i-th
origin and $t_j$ represents the total amount shipped put from the j-th destina-
tion as transshipment. Let $T>0$ be sufficiently large number so that $t_i \le T$, for all $i$ and $t_j \le T$ for all $j$. We now write $t_i+x_{ii}=T$ then the non
negative slack variable $\tilde{x}_{ii}$ represents the difference between $T$ and the
actual amount of transshipment through the i-th origin. Similarly, if we
let $t_j+x_{jj}=T$ then the non negative slack variable $\tilde{x}_{jj}$ represents the
difference between $T$ and the actual amount of transshipment through
the j-th destination. It is self-evident that the total volume of products
transshipped at any given time cannot exceed the total amount generat-
ed or obtained . and hence $T = \sum_{i=1}^{m} \tilde{a}_i$.

The transshipment problem then reduces to

$$
\text{Minimize} \quad \tilde{Z} \approx \sum_{i=1}^{m+n} \sum_{\substack{j=1 \\ j \neq i}}^{m+n} \tilde{c}_{ij} \tilde{x}_{ij}
$$

$$
\begin{array}{ll}
\text{subject to} & \displaystyle\sum_{j=1}^{m+n} \tilde{x}_{ij} = \tilde{a}_i + T, \quad i=1,2,3,\dots,m \\
& \displaystyle\sum_{j=1}^{m+n} \tilde{x}_{ij} = T, \quad i=m+1,m+2,m+3,\dots,m+n, \\
& \displaystyle\sum_{i=1}^{m+n} \tilde{x}_{ij} = T, \quad j=1,2,\dots,m \\
& \displaystyle\sum_{i=1}^{m+n} \tilde{x}_{ij} = \tilde{b}_j + T, \quad j=m+1,m+2,m+3,\dots,m+n \\
& \text{and } \tilde{x}_{ij} \ge \tilde{0}, i=1,2,\dots,m+n, j=1,2,\dots,m+n. \\
& \text{where } \tilde{C}_{ii} = 0, i=1,2,\dots,m+n.
\end{array}
$$

A typical transportation problem with (m+n) roots and (m+n) destina-
tions is described by the mathematical model above. The problem's so-
lution includes 2m+2n-1 basic variables. However, the remaining buff-
er stock is represented by m+n of these variables appearing in the diagonal cells if they are omitted. Our interest has m+n-1 basic variables.

**4. Improved Interval Vogel's Approximation algorithm**

**Row opportunity interval cost matrix**

Each element of the same row is subtracted from the row's smallest interval expense.

**Column opportunity interval cost matrix**

The smallest cost of each column in the original interval transhipment cost matrix is subtracted from each segment of the same column. (Note: the diagonal zero is ignored when determining the smallest interval cost.) On the TOC matrix, the proposed algorithm is used. The following are the steps in detail:

**Step 1:** If either (full amount supply > full amount demand) or (total supply total demand), balance the given interval transhipment problem.

**Step 2:** Convert the interval transhipment problem in to interval transportation problem.

**Step 3:** state the all interval parameters supply, demand and unit transhipment cost in terms of midpoint and half width form.

**Step 4:** Obtain the ROC, COC and then TOC matrix

**Step 5:** Identify the boxes in each row with the lowest and next-lowest interval transportation costs, and write the difference (penalty) along the table's side against the corresponding row.

**Step 6:** In each column, find the boxes with the lowest and next-lowest interval transportation costs, and write the difference (penalty) against the corresponding column.

**Step 7:** Determine the most severe penalty possible. If it's on the table's side, give the most space to the box with the lowest transportation interval cost in that row. If the box with the lowest interval cost of transhipment in that column is below the table, give it the most money.

**Step 8:** Select the topmost row and the farthest left column if the penalties for two or more rows or columns are equal.

**Step 9:** The satisfied row or column does not need any further thought. If both the row and column are fulfilled at the same time, delete only one of them, and assign zero supply to the remaining row or column (or demand).

**Step 10:** Continue until all of the rows and columns are satisfied.

**Step 11:** Using the original balanced interval transhipment cost matrix, calculate the cumulative interval carrying cost for the possible allocations.
<figure><img src="image_4.png" /><figcaption>Figure 2</figcaption></figure>

## 5. NUMERICAL EXAMPLE

To solve the following fully interval integer transhipment problem with 2 source and 2 destinations. The supply values of the source S1 and S2 are [4,6], [5,7] units respectively. The demand values of destination D1 and D2 are [3,5], [6,8] units respectively. Transportation cost per unit between various defined source and destination are given in the following table [12].

<table><caption>Table 2. Interval transhipment problem</caption><thead><tr><th></th><th>S1</th><th>S2</th><th>D1</th><th>D2</th><th>Supply</th></tr></thead><tbody><tr><th>S1</th><td>-</td><td>[1,3]</td><td>[6,9]</td><td>[2,4]</td><td>[4,6]</td></tr><tr><th>S2</th><td>[1,3]</td><td>-</td><td>[8,10]</td><td>[2,4]</td><td>[5,7]</td></tr><tr><th>D1</th><td>[0,1]</td><td>[7,9]</td><td>-</td><td>[3,5]</td><td>-</td></tr><tr><th>D2</th><td>[1,3]</td><td>[0,2]</td><td>[1,3]</td><td>-</td><td>-</td></tr><tr><th>Demand</th><td>-</td><td>-</td><td>[3,5]</td><td>[6,8]</td><td>-</td></tr></tbody></table>
<table><caption>Table 3. Equivalent interval transportation problem of the interval transshipment problem</caption><thead><tr><th></th><th>S1</th><th>S2</th><th>D1</th><th>D2</th><th>Supply</th></tr></thead><tbody><tr><th>S1</th><td>[0,0]</td><td>[1,3]</td><td>[6,9]</td><td>[2,4]</td><td>[13,19]</td></tr><tr><th>S2</th><td>[1,3]</td><td>[0,0]</td><td>[8,10]</td><td>[2,4]</td><td>[14,20]</td></tr><tr><th>D1</th><td>[0,1]</td><td>[7,9]</td><td>[0,0]</td><td>[3,5]</td><td>[9,13]</td></tr><tr><th>D2</th><td>[1,3]</td><td>[0,2]</td><td>[1,3]</td><td>[0,0]</td><td>[9,13]</td></tr><tr><th>Demand</th><td>[9,13]</td><td>[9,13]</td><td>[12,18]</td><td>[15,21]</td><td>[45,65]</td></tr></tbody></table>

By applying the proposed algorithm, the initial basic feasible solution is Table 4. Initial basic feasible solution

<table><thead><tr><th></th><th>S1</th><th>S2</th><th>D1</th><th>D2</th><th>Supply</th></tr></thead><tbody><tr><th>S1</th><td>&lt;0,0&gt;<br/>&lt;11,2&gt;</td><td>&lt;1,1&gt;</td><td>&lt;11,1.5&gt;</td><td>&lt;1,1&gt;<br/>&lt;5,3&gt;</td><td>&lt;16,3&gt;</td></tr><tr><th>S2</th><td>&lt;1.5,1&gt;</td><td>&lt;0,0&gt;<br/>&lt;11,2&gt;</td><td>&lt;14,1&gt;</td><td>&lt;1,1&gt;<br/>&lt;6,3&gt;</td><td>&lt;17,3&gt;</td></tr><tr><th>D1</th><td>&lt;0,0.5&gt;</td><td>&lt;14.5,1&gt;</td><td>&lt;0,0&gt;<br/>&lt;11,2&gt;</td><td>&lt;4.5,1&gt;</td><td>&lt;11,2&gt;</td></tr><tr><th>D2</th><td>&lt;2.5,1&gt;</td><td>&lt;0,1&gt;</td><td>&lt;1,1&gt;<br/>&lt;4,3&gt;</td><td>&lt;0,0&gt;<br/>&lt;7,3&gt;</td><td>&lt;11,2&gt;</td></tr><tr><th>Demand</th><td>&lt;11,2&gt;</td><td>&lt;11,2&gt;</td><td>&lt;15,3&gt;</td><td>&lt;18,3&gt;</td><td>&lt;55,3&gt;</td></tr></tbody></table>

The route from source to desination is given as S1 to D2 = <5,3>, S2 to D2 = <6,3>, D2 to D1 = <4,3>

Then the initial solution of the given interval transshipment problem is Min z = <3,1><5,3> + <3,1><6,3> + <2,1><4,3> = <41,3> = [38,44]
<figure><img src="image_5.png" /><figcaption>Figure 3</figcaption></figure>

<table><caption>Table 5. Comparison of solutions</caption><thead><tr><td>S.No</td><td>Solution by the proposed method</td><td>Solution by P.Rajendran and P.Pandian [12]</td><td>Solution by R.Sophiaporchelvi and M.Anitha[14]</td></tr></thead><tbody><tr><td>1</td><td>[38,44]</td><td>[20,67]</td><td>[1,126]</td></tr></tbody></table>

## Comparison of Solutions

<figure><img src="image_6.png" /><figcaption>Figure 4</figcaption></figure>

## 6. CONCLUSION

To discover the primary interval possible explanation to the interval shipment problem, we proposed an improved interval Vogel's approximation algorithm. The interval possible solution's optimality has been checked. A mathematical model is solved to demonstrate the suggested technique, and the outcome is compared to the results of other obtainable methods.
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