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\markboth{Ihda Hasbiyati dkk.} %Jika lebih dari dua penulis, tuliskan sebagai Nama Penulis Pertama dkk.
{Simplex Transportation Method For Determining Transshipment Of Clothing Raw Materials}

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\title{SIMPLEX TRANSPORTATION METHOD FOR DETERMINING TRANSSHIPMENT OF CLOTHING RAW MATERIALS}

\author{IHDA HASBIYATI$^{a}$\footnote{penulis korespondensi}, SYAHPUTRI WAHYUNI$^{b}$, MUFTI SYIFA AMINI$^{c}$, AHRIYATI$^{d}$\\}

\address{$^{a}, ^{b}, ^{c}$ Departement of Mathematics, University of Riau, Indonesia,\\
$^{d}$ Departement of Mathematics, University of Palangkaraya, Indonesia.\\
email : \email{ ihdahasbiyati@lecturer.unri.ac.id}}

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\begin{abstract}
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Diterima ..... \quad Direvisi ..... \quad Dipublikasikan ..... %tanggal-tanggal tersebut \textbf{dikosongkan} saja
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\bigskip

\textbf{Abstrak}. %Dalam bahasa Indonesia
Tujuan dari penulisan ini adalah untuk memberikan gambaran bagaimana menyelesaikan masalah \textit{transshipment} pengiriman bahan baku pakaian dengan metode transportasi simpleks. Untuk menentukan solusi optimal dari masalah transshipment dengan memperhatikan koefisien biaya pengiriman. Proses awal penyelesaian masalah ini dengan mengilustrasikan bagaimana mengubah bilangan fuzzy menjadi bilangan tegas dengan menggunakan fungsi \textit{robust ranking}, kemudian untuk mencari basis awal menggunakan metode aproksimasi Vogel, selanjutnya menggunakan metode simpleks transportasi untuk menguji basis awal tersebut, sehingga diperoleh hasil yang optimal. Sehingga dapat diambil kesimpulan bahwa unit pengiriman barang dari satu sumber ke sumber lain dengan menggunakan metode simpleks transportasi merupakan solusi optimal untuk mencari nilai minimum dari masalah \textit{transshipment}.

\bigskip

\textbf{Abstract}. % Dalam bahasa Inggris
The purpose of this paper is to provide an idea of how to solve the transshipment problem of shipping raw materials of clothing with method transportation simplex. To determine the optimal solution of the transshipment problem by taking into account the freight cost coefficient. The initial process of solving this problem by illustrating how to convert fuzzy numbers into strict numbers using the robust ranking function, then to find the initial basis uses the Vogel approximation method, then uses the transportation simplex method to test the initial basis, so that optimal results are obtained. So the conclusion can be drawn that the unit of shipping goods from one source to another using the transportation simplex method is the optimal solution to find the minimum value of the transshipment problem.

\end{abstract}

\keywords{Robust function ranking, transportation simplex method, transshipment problem, trapezoidal fuzzy number, Vogel approximation method}

\section{Introduction}
Operations research solves problems by producing the optimum value of a decision problem in conditions of limited resources. Transportation is a process man or goods move of a place go to other place by use of a tool help land vehicle, oceanic vehicle, and also air vehicle, well common and also person by use of machine or not utilize machine \cite{Hasbiyati}. Hasbiyati et al. have previously conducted research on transportation problems in various problems \cite{Hasbiyati2}\cite{Hasbiyati3}\cite{Putri}. Part of the transportation problem, namely the route and scheduling problem can be seen in the research conducted by Setiowati et al. \cite{Setiowati}. The transportation problem refers to a special form of linear programming dealing with shipping from various sources to various destinations. The transportation problem is a classic problem in operations research that involves finding the optimal way to move goods from one place to another \cite{Malacky}. Transshipment problem in the standard form is basically a linear min-cost network flow problem and for such types of optimization problems \cite{Ellaimony}. The transportation problem seeks to lower the price of a specific good from a number of suppliers to a variety of destinations \cite{K}. Transshipment problems can exist in every source and destination when shipping from source to destination makes it possible to tranship, that is, goods are sent from several sources to several destinations through other sources and destinations \cite{Okiemute}. The transshipment technique is used to find the shortest route from one point in a network to another \cite{Khurana}.

Several types of uncertainties are encountered in formulating transshipment problem mathematically due to factors like lack of exact information, unobtainable information, frequent changes in rate of fuels, traffic jams or wheather conditions \cite{Kumar}. The transportation problem is basically a linear program that can be solved using the simplex method, but due to its special structure, the simplex transportation method is used, which is computationally much more efficient. This is similar to the discussion, namely the transshipment problem with the transportation simplex method. The initial basic solution used is the Vogel approximation method and the optimal solution used the transportation simplex method.


\section{Methods}
The following will present some methods of supporting the problem of transshipment of raw materials. The study materials such as Transshipment Problem, Trapezoidal Fuzzy Number, Robust Ranking, Vogel Approximation Method, and Simplex Transportation Method.

\subsection{Transshipment Problem}
The explained in the transshipment problem that all sources and destinations can give or receive goods from other sources and destinations. In its application, transshipment can take place at the customer's location (other than the origin and destination of the request) or at a dedicated transshipment location, or both \cite{Wolfinger}. Each point has the potential to become a transit point for goods or a transshipment point. In the transshipment problem, for example $i=1,2,\ldots,m$ as the source of an item to be transported to destination $j=m+1,m+2,…,m+n$. If a sender and receiver have the same point, the sending unit is zero. The coefficient $c_{ij}$ is the cost unit of goods from source $i$ to destination $j$ and  $x_{ij}$ the amount of goods to be transported from source $i$ to destination $j$ while $s_i$ is the amount supply at sources $i$ and $d_j$ is a demand at destination $j$.

In the transshipment problem, each source and destination is seen as a point that can accommodate units of goods distributed for supply or demand. At the source and destination requires buffer stock ($L$) which serves to help ensure the availability of goods and anticipate running out of goods to be given to consumers. Each unit of goods moved to each point must be large enough for all transshipments. Units of goods must not exceed the amount produced or received. The total amount of inventory and the total amount of demand are equal to $L$ and $L$ is added at each source and destination point.
\begin{equation*}
L = \sum_{i=1}^m s_i = \sum_{j=1}^n d_j = \text{buffer stock}
\end{equation*}
Furthermore, the mathematical model of the transshipment problem can be reduced to such as
\begin{equation*} 
\begin{array}{ll@{}ll}
\text{min}& \displaystyle z=\sum_{i=1}^{m+n} \sum_{j=1,i\neq j}^{m+n} c_{ij}x_{ij}, \\
\text{kendala}& \displaystyle\sum_{j=1}^{m+n} x_{ij} = s_i + L,     \quad i=1, 2, \ldots, m,\\
              & \displaystyle\sum_{j=1}^{m+n} x_{ij} = L,     \quad i=m+1, m+2, \ldots, m+n,\\
							& \displaystyle\sum_{i=1}^{m+n} x_{ij} = L,     \quad j=1, 2, \ldots, m,\\
							& \displaystyle\sum_{i=1}^{m+n} x_{ij} = d_j + L,     \quad j=m+1, m+2, \ldots, m+n,\\
							& x_{ij}\geq 0; i,j=1,2,\ldots,m+n; j=1,2,\ldots,m+n, c_{ii}=0.
\end{array}
\end{equation*}

\subsection{Trapezoidal Fuzzy Number}
Fuzzy number can be used in solving many problems such as the transshipment problem which constrains cost uncertainty.  The existence of uncertain parameters gives rise to a new problem model called the fuzzy transportation problem \cite{Bisht}.
\begin{definition} \cite{Panda} \label{def1}
A fuzzy number $A$ is a subset of real line $R$, with the membership function $\mu_A$ satisfying the following properties:
\begin{enumerate}
\item[(i)] 	$\mu_A(x)$ is piecewise continuous in its domain.
\item[(ii)] $A$ is normal, i.e., there is a $x_0 \in A$ such that $\mu_A (x_0)=1$.
\item[(iii)] 	$A$ is convex, i.e., $\mu_A (\lambda x_1+(1-\lambda) x_2 )\geq(\mu_A (x_1 ),\mu_A (x_2 )) \forall x_1,x_2 \in X$.
\end{enumerate}
\end{definition}

\begin{definition} \cite{Savarimuthu} \label{def2}
Trapezoidal fuzzy number $\~{A}$ is a fuzzy number ($a,b,c,d$) shown in Figure which has the membership function $\mu_A (x)$ as following:\\
\begin{equation*}
	\mu_{\tilde{A}}(x)=
	\begin{cases}
0, \;&\mbox{for}\;\;x<a,\\
\dfrac{x-a}{b-a},\;&\mbox{for}\;\;a\leq{x}\leq{b},\\
1, \;&\mbox{for}\;\;b\leq{x}\leq{c},\\
\dfrac{d-x}{d-c},\;&\mbox{for}\;\;c\leq{x}\leq{d},\\
0, \;&\mbox{for}\;\;x>d.
\end{cases}
\end{equation*}
\end{definition}

\vspace{5cm}
\begin{figure}[htbp] \hspace{1.5cm}
\includegraphics[scale=0.13]{kurva.jpg}
\caption{Trapezoidal curve}
\label{f1}
\end{figure}

\subsection{Robust Ranking}
Assume $\~{A} = (a,b,c,d)$ is a trapezoidal fuzzy number. Given the following $\alpha$-cut: (${\alpha_\alpha}^L, {\alpha_\alpha}^U$)=[$(b-a)\alpha+a  ,d-(d-c)\alpha$], which $\alpha \in [0,1]$ . In the decision making process, the ordering of fuzzy number is very important. Robust ranking method is given to determine the rank of trapezoidal fuzzy number. Robust ranking method is transforming trapezoidal fuzzy number into crisp number. If  $\~{A} = (a,b,c,d)$ is a trapezoidal fuzzy number, then the robust ranking is as follow \cite{Ahmed}:
\[R(\~{A}) = \int_0^1{(0.5)({\alpha_\alpha}^L, {\alpha_\alpha}^U)} d\alpha \]

where (${\alpha_\alpha}^L$, ${\alpha_\alpha}^U$) is the $\alpha$-cut of the trapezoidal fuzzy number. In the case of trapezoidal fuzzy number, the robust ranking becomes as follow \cite{Ahmed}:
\begin{equation}
R(\~{A}) = \frac{(a+b+c+d)}{4}
\label{eq1}
\end{equation}

\subsection{Vogel Approximation Method}
The Vogel approximation method deals with finding an optimal solution taking into account the relationships of price indices. This method compares always the two lowest  price indices in both a column and a row \cite{Hlatka}. Vogel’s Approximation Method is known as the best algorithm for generating an efficient initial feasible solution to the transportation problem \cite{Babu}. To determine the optimal solution of a transportation problem, the first step is to determine the initial basis solution. The initial basis solution can be determine by various methods, one of which is the Vogel approach. The steps of the Vogel approach method, namely:\hspace{-2cm}
\begin{itemize}
\item[(i)] For each row or column, determine the difference by subtracting elements smallest cost in a row or column.
\item[(ii)]	Then allocate supply with as much demand as possible to the variable with the smallest cost in the selected row and column. Then readjust supply and demand again, and mark the rows or columns that have been fulfilled. If any row and column are filled simultaneously, select one to be marked.
\item[(iii)] \begin{enumerate} 
\item[a.] If a row or column with zero supply or demand is not marked.
\item[b.] If the row or column with supply and demand is not marked then determine the basis variable in the row or column with the smallest cost.
\item[c.] If all rows and columns that are not marked have supply and demand equal to zero then determine the basic variable that is zero by choosing the lowest cost and stopping. 
\end{enumerate}
\end{itemize}
\vspace{-1.5cm}
\subsection{Simplex Transportation Method}
The transportation simplex method is used because transportation problems are linear programming problems that tend to require a large number of constraints and variables \cite{Haryono}. The computational complexity in the simplex method depends on the number of variables and the number of constraints and is directly proportional to both \cite{Sarode}. The initial basis obtained from solving a problem is not necessarily an optimal solution. The optimal solution is determine by various method to test and optimize the solution. One way to improve the solution of the transportation problem is to use the simplex transportation method or abbreviated as MST.

Let $u_i$ be the dual supply variable and $v_j$ the demand dual variable, with $i=1,2,…,m$ and $j=1,2,…,n$. The dual from the transportation problem is if
\[ \sum_{i=1}^m s_i = \sum_{j=1}^n d_j\]
then it is said to be balanced. In a balanced transportation problem, the dual from can therefore be written:
\[\text{min}\; z' = s_1 u_1+s_2 u_2+\cdots+s_m u_m+d_1 v_1+d_2 v_2+\cdots+d_n v_n,\]
constraint
\begin{equation*}
%\vspace{-0.2cm}
\begin{array}{rcccccccl}
  u_1 &  & &+\;v_1& & & & \leq&c_{11}\\
  u_1 & & & &+\; v_2& & & \leq&c_{12}\\
	\;\vdots & & & & &\ddots & & \leq&\; \vdots\\
  u_1 &  & &&&&+\; v_n  & \leq& c_{1n}\\
	u_2 & &&+\;v_1& & & & \leq& c_{21}\\
  u_2 & & & &+\; v_2& & &\leq& c_{22}\\
	\;\vdots & & & & &\ddots & & \leq&\; \vdots\\
  u_2 & & & & &&+\; v_n  & \leq& c_{2n}\\
	\vdots  & & && &\;\ddots&&\leq&\; \vdots\\
  u_m &&&+\;v_1& & &&\leq& c_{m1}\\
  u_m & &&&+\; v_2&& & \leq& c_{m2}\\
	\vdots &&&& &\;\ddots &&\leq&\; \vdots\\
  u_m &&& && &+\; v_n  & \leq& c_{mn},
\end{array}
\end{equation*}
$u_i$ and $v_j$ independent of sign, $i=1,2,…,m$ and $j=1,2,…,n$.

The steps of the simplex transportation method:
\begin{enumerate}
\item[(i)]	If the question is not balanced, balance it first.
\[\sum_{i=1}^m s_i = \sum_{j=1}^n d_j\]
\item[(ii)] Use one of the methods to find  the initial basic feasible solution.
\item[(iii)] Use the fact that $u_1=0$ for all basic variable to find \[[u_1,u_2,\ldots,u_m,v_1,v_2,\ldots,v_n]\] with the equation
\[u_i + v_j = c_i\]                
\item[(iv)] Determine the possible costs for each unbasic variable using the equation
 \[\={c}_{ij}=u_i+v_j-c_ij\leq 0,\]               
with $\bar{c}_{ij}$ is a possible fee. 
\item[(v)] There are several cases, namely:
\begin{itemize}
\item[a.] If $\={c}_{ij}$ is negative then the solution obtained previously is the optimal solution.
\item[b.]	If $\={c}_{ij}$ is positive then the solution obtained is not optimal. Refine the solution by finding variables that will enter the basis (input variables) and variables that will leave basis (outgoing variables). The input variable is a variable that corresponds to a value of most positive. To find the input and output variables, we construct a closed loop that starts and ends at the input variable. The loop consists of successive vertical and horizontal segments in an undefined direction, meaning they can be clockwise or counterclockwise. The exit variable is determine by selecting the smallest basis variable that is in the constructed loop. Furthermore, the incoming variable is filled in as much as the outgoing variable. For example, if $x_ij$ variable is the input variable, when $x_ij$ is increased by 1 unit to maintain the solution, the base variable in the loop is adjusted according to the amount of demand and supply, namely bi increasing and decreasing the base variable in the loop. This process is summarized in the table with $(+)$ and $(-)$ signs in the corresponding cells. The changes obtained will maintain supply constraint and demand constraint to remain fulfilled. Then return to Step iii.
\end{itemize}
\end{enumerate}

\section{Discussion and Result}
A clothing tailor entrepreneur in Rokan Hilir started his business in 1990. There are several clothing materials that can selected according to customer requests, namely materials Roberto, Maksmara, Songket, and brocade. As time went by the entrepreneur grew greatly rapidly and opened branches (C) with each branch producing similar goods according to customer requests. If customer demand exceeds capacity usually produced every day, then other branches can help produce goods according to demand are still lacking. Branch 1 and branch 2 have Production of 30 pcs, in branch 3, branch 4, and branch 5 requires 25 pcs, 15 pcs, and 20 additional pcs.
	The cost of clothing is not yet known for certain depending on the materials needed, because the cost of clothing depends on the materials required by the customer. More good The price of the materials used is getting more expensive, and vice versa for materials If standard is used then the price is also standard. Transportation cost table with two sources and three destinations can be seen in Table 1.
Then to guarantee each point capable of according the total units of goods to be distributed needs to be increased even L to the point of supply and demand. Next determine the amount supply and demand for buffer stok (L).
\begin{equation*}
L = \sum_{i=1}^m s_i = \sum_{j=1}^n d_j = \text{Buffer Stock}\\
L = \sum_{i=1}^2 s_i = \sum_{j=3}^5 d_j = 60.
\end{equation*}
Estimation of shipping costs for transshipment problem with cost coefficient trapezoidal fuzzy can be seen in Table 1.
\vspace{-0.7cm}
\begin{table}[htbp]
\caption{Transshipment Problem with Cost Coefficient Fuzzy}
\label{tab1}
\begin{tabular}{ccccccc}
\hline
       & C1        & C2 & C3 & C4 & C5 & Supply\\ \hline
C1     & (0;0;0;0) & (1;3;3;5)&(2;3;5;6)&(3;5;6;10)&(5;7;8;12)&30+ L\\
C2     & (1;2;2;3) & (0;0;0;0)&(2;5;8;13)&(1;2;6;11)&(4;6;10;20)&30+ L\\
C3     & (2;3;6;9) & (6;7;9;14)&(0;0;0;0)&(1;2;4;5)&(1;2;5;7)&L\\
C4     & (6;10;13;15) & (4;5;7;8)&(5;8;9;10)&(0;0;0;0)&(1;1;2;4)&L\\
C5     & (2;4;4;6) & (1;1;3;3)&(2;4;5;9)&(9;10;13;16)&(0;0;0;0)&L\\ \hline
Demand & L & L&25+ L&15+ L&20+ L\\ \hline
\end{tabular}
\end{table}

Then substituting L=60 into Table 1, we get the table transportation with trapezoidal fuzzy number cost in Table 2,
\vspace{-0.7cm}
\begin{table}[htbp]
\caption{Transshipment Problem with Cost Coefficient Fuzzy with L=60}
\label{tab2}
\begin{tabular}{ccccccc}
\hline
       & C1        & C2 & C3 & C4 & C5 & Supply\\ \hline
C1     & (0;0;0;0) & (1;3;3;5)&(2;3;5;6)&(3;5;6;10)&(5;7;8;12)&90\\
C2     & (1;2;2;3) & (0;0;0;0)&(2;5;8;13)&(1;2;6;11)&(4;6;10;20)&90\\
C3     & (2;3;6;9) & (6;7;9;14)&(0;0;0;0)&(1;2;4;5)&(1;2;5;7)&60\\
C4     & (6;10;13;15) & (4;5;7;8)&(5;8;9;10)&(0;0;0;0)&(1;1;2;4)&60\\
C5     & (2;4;4;6) & (1;1;3;3)&(2;4;5;9)&(9;10;13;16)&(0;0;0;0)&60\\ \hline
Demand & 60        & 60       &85       &75          &80 & \\ \hline
\end{tabular}
\end{table}

The mathematical model of the standard transportation problem can be written as follows:
\begin{equation*}
\begin{array}{rl}
\mbox{min}\;z \;=& (0,0,0,0)x_{11}+(1,3,3,5)x_{12}+(2,3,5,6)x_{13}+(3,5,6,10)x_{14}\\ \; &+(5,7,8,12)x_{15}+(1,2,2,3)x_{21}+(0,0,0,0)x_{22}+(2,5,8,13)x_{23} \\ \; &+(1,2,6,11)x_{24}+(4,6,10,20)x_{25}+(2,3,6,9)x_{31}+(6,7,9,14)x_{32} \\ \; &+(0,0,0,0)x_{33}+(1,2,4,5)x_{34} +(1,2,5,7)x_{35}+(6,10,13,15)x_{41}\\ \; &+(4,5,7,8)x_{42}+(5,8,9,10)x_{43}+(0,0,0,0)x_{44}+(1,1,2,4)x_{45}\\ \; &+(2,4,4,6)x_{51}+(1,1,3,3)x_{52}+(2,4,5,9)x_{53}+(9,10,13,16)x_{54}\\ \; &+(0,0,0,0)x_{55}
\end{array}
\end{equation*}
s.t.
\begin{equation*}
x_{11}+x_{12}+x_{13}+x_{14}+x_{15}=90
\end{equation*}
\begin{equation*}
x_{21}+x_{22}+x_{23}+x_{24}+x_{25}=90
\end{equation*}
\begin{equation*}
x_{31}+x_{32}+x_{33}+x_{34}+x_{35}=60
\end{equation*}
\begin{equation*}
x_{41}+x_{42}+x_{43}+x_{44}+x_{45}=60
\end{equation*}
\begin{equation*}
x_{51}+x_{52}+x_{53}+x_{54}+x_{55}=60
\end{equation*}
\begin{equation*}
x_{11}+x_{21}+x_{31}+x_{41}+x_{51}=60
\end{equation*}
\begin{equation*}
x_{12}+x_{22}+x_{32}+x_{42}+x_{52}=60
\end{equation*}
\begin{equation*}
x_{13}+x_{23}+x_{33}+x_{43}+x_{53}=85
\end{equation*}
\begin{equation*}
x_{14}+x_{24}+x_{34}+x_{44}+x_{54}=75
\end{equation*}
\begin{equation*}
x_{15}+x_{25}+x_{35}+x_{45}+x_{55}=80
\end{equation*}
$$x_{ij} \geq 0 \; \mbox{dengan}\; i=1, 2, 3, 4, 5 \;\mbox{dan} \; j=1, 2, 3, 4, 5$$ 
$$\tilde{c}_{ii}=0, i=1, 2, 3, 4, 5.$$

The steps to solve the transshipment problem are as follows:
\begin{enumerate}
\item[(i)] Checking the balance between supply and demand. The amount of inventory can be written as
\[\sum_{i=1}^5 s_i = 90+90+60+60+60=360\]
and the demand amount
\[\sum_{j=1}^5 d_j = 90+90+60+60+60=360\]
because supply and demand have the same value, i.e.
\[\sum_{i=1}^5 s_i = \sum_{j=1}^5 d_j = 360\]
so that the transshipment problem is balanced.
\item[(ii)] 	Changing shipping costs in the from of trapezoidal fuzzy numbers into crisp number. Robust ranking is used to change shipping cost coefficient from fuzzy numbers to crisp number of equation i. The following firm numbers are obtained: 
\[R(A)= \frac{a+b+c+d}{4}\] 

\begin{equation*} 
\begin{array}{lll}
\tilde{c}_{11} = (0,0,0,0),& c_{11} = 0\\
\tilde{c}_{12} = (1,3,3,5),& c_{12} = 3\\
\tilde{c}_{13} = (2,3,5,6),& c_{13}= 4\\
\tilde{c}_{14} = (3,5,6,10),& c_{14}= 6\\
\tilde{c}_{15} = (5,7,8,12),& c_{15}= 8\\
\tilde{c}_{21} = (1,2,2,3),& c_{21} = 2\\
\tilde{c}_{22} = (0,0,0,0),& c_{22} = 0\\
\tilde{c}_{23} = (2,5,8,13),& c_{23} = 7\\
\tilde{c}_{24} = (1,2,6,11),& c_{24} = 5 \\
\tilde{c}_{25} = (4,6,10,20),& c_{25}= 10\\
\tilde{c}_{31} = (2,3,6,9),& c_{31} = 5\\
\tilde{c}_{32} = (6,7,9,14),& c_{32} = 9\\
\tilde{c}_{33} = (0,0,0,0),& c_{33} = 0\\
\tilde{c}_{34} = (1,2,4,5),& c_{34} = 3\\
\tilde{c}_{35} = (1,2,5,7),& c_{35} = 4\\
\tilde{c}_{41} = (6,10,13,15),& c_{41} = 11\\
\tilde{c}_{42} = (4,5,7,8),& c_{42} =6\\
\tilde{c}_{43} = (5,8,9,10),& c_{43} = 8\\
\tilde{c}_{44} = (0,0,0,0),&c_{44} = 0\\
\tilde{c}_{45} = (1,1,2,4),&c_{45} = 2\\
\tilde{c}_{51} = (2,4,4,6),& c_{51}= 4\\
\tilde{c}_{52} = (1,1,3,3),& c_{52}= 2\\
\tilde{c}_{53} = (2,4,5,9),& c_{53}= 5\\
\tilde{c}_{54} = (9,10,13,16),& c_{54}= 12\\
\tilde{c}_{55} = (0,0,0,0),&c_{55} =0.\\
\end{array}
\end{equation*}
Changes in shipping costs in the from crisp number can be seen in Table 3. 
\newcommand{\bottombox}[1]{\makebox[1em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
\newcommand{\innerbox}[2]{%}
    \begin{tabular}[b]{c|c}
       \rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
       \multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    \end{tabular}}
\renewcommand{\arraystretch}{1.25}
	\vspace{-0.15cm} 
\begin{table}[htbp]
  \begin{center}
	\caption{Shipping Costs in Crisp Number} \label{kuri4}
	\begin{small}
	\begin{spacing}
	\hspace*{0.01cm} 
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
              & \hspace*{0.3cm}C 1 \hspace*{0.3cm} & \hspace*{0.3cm}C 2  \hspace*{0.3cm} &\hspace*{0.3cm} C 3 \hspace*{0.3cm} &\hspace*{0.3cm} C 4 \hspace*{0.3cm} &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 & \hspace*{-1.6cm}\innerbox{ }{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$3$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{6} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{8} \hspace*{-1.6cm}& {$90$} \\\hline
	C 2 & \hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$0$} \hspace*{-1.6cm} &\hspace*{-1.6cm} \innerbox{ }{$7$} \hspace*{-1.6cm}& \hspace*{-1.6cm}  \innerbox{ }{$5$} \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{ }{10}\hspace*{-1.6cm}&  $90$ \\\hline
	C 3 & \hspace*{-1.6cm}\innerbox{ }{$5$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$9$} \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ }{0} \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ }{3} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{4} \hspace*{-1.6cm} &  $60$ \\\hline
  C 4 &\hspace*{-1.6cm}\innerbox{ }{$11$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$6$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$8$ } \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{ }{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{2} \hspace*{-1.6cm} &  $60$ \\\hline
  C 5 &\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm} &\hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{ }{$5$ } \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ }{12} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{ }{$0$} \hspace*{-1.6cm}&  $60$ \\\hline
  Demand & \hspace*{0.4cm}\bottombox{$60$} & \hspace*{0.4cm}\bottombox{$60$} &\hspace*{0.4cm} \bottombox{$85$} &\hspace*{0.4cm} \bottombox{$75$} &\hspace*{0.4cm} \bottombox{$80$} &    \\\hline	%\end{document}
	\end{tabular}
	%\end{center}
	\end{spacing}
	\end{small}
	\end{center}
	\end{table}

\item[(iii)] Then using the Vogel approach method to find the initial base solution, 
\begin{enumerate}
\renewcommand{\theenumi}{\roman{enumi}}
\item[a.] Investigate the two smallest values in each row and column. If there are the same values, you can choose one of them. Each row and column is obtained as follows
\begin{equation*} 
\begin{array}{lll}
\text{row} \;1 &=& {3-0} = 3\\
\text{row}\; 2 &=& {2-0} = 2\\
\text{row} \;3 &=& {3-0} = 3\\
\text{row} \;4 &=& {2-0} = 2\\
\text{row} \;5 &=& {2-0} = 2\\
\text{column} \;1 &=& {2-0} = 2\\
\text{column} \;2 &=& {2-0} = 2\\
\text{column} \;3 &=& {4-0} = 4 \\
\text{column} \;4 &=& {3-0} = 3 \\
\text{column} \;5 &=& {2-0} = 2. \\
\end{array}
\end{equation*}
Based on the results of each row and column, select the largest value in column 3 which is 4.

\item[b.] Allocate as much demand and inventory as possible to the variables with the smallest cost in the rows and columns. Then delete rows and columns that have been exhausted when allocated. The smallest cost element in column 3 is selected, which is 0 in row 3. Allocated demand and inventory in column 3 row 3, which is 60 pcs. Rows or columns that have been selected are then deleted if they have no inventory or demand. Cost changes can be seen in Table 4. 
%\newcommand{\bottombox}[1]{\makebox[1em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
%\newcommand{\innerbox}[2]{%}
 %   \begin{tabular}[b]{c|c}
  %     \rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
   %    \multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    %\end{tabular}}
\renewcommand{\arraystretch}{1.25}
	\vspace{-0.15cm} 
\begin{table}[htbp]
  \begin{center}
	\caption{Iteration 1 Vogel Approximation Method} \label{kuri5}
	\begin{small}
	\begin{spacing}
	\hspace*{0.01cm} 
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
              & \hspace*{0.3cm}C 1 \hspace*{0.3cm} & \hspace*{0.3cm}C 2  \hspace*{0.3cm} &\hspace*{0.3cm} C 3 \hspace*{0.3cm} &\hspace*{0.3cm} C 4 \hspace*{0.3cm} &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 & \hspace*{-1.6cm}\innerbox{ }{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$3$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{6} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{8} \hspace*{-1.6cm}& {$90$} \\\hline
	C 2 & \hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$0$} \hspace*{-1.6cm} &\hspace*{-1.6cm} \innerbox{ }{$7$} \hspace*{-1.6cm}& \hspace*{-1.6cm}  \innerbox{ }{$5$} \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{ }{10}\hspace*{-1.6cm}&  $90$ \\\hline
	C 3 & \hspace*{-1.6cm}\innerbox{ }{$5$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$9$} \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{60}{0} \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ }{3} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{4} \hspace*{-1.6cm} &  ${}$ \\\hline
  C 4 &\hspace*{-1.6cm}\innerbox{ }{$11$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$6$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$8$ } \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{ }{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{2} \hspace*{-1.6cm} &  $60$ \\\hline
  C 5 &\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm} &\hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{ }{$5$ } \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ }{12} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{ }{$0$} \hspace*{-1.6cm}&  $60$ \\\hline
  Demand & \hspace*{0.4cm}\bottombox{$60$} & \hspace*{0.4cm}\bottombox{$60$} &\hspace*{0.4cm} \bottombox{$25$} &\hspace*{0.4cm} \bottombox{$75$} &\hspace*{0.4cm} \bottombox{$80$} &    \\\hline	%\end{document}
	\end{tabular}
	%\end{center}
	\end{spacing}
	\end{small}
	\end{center}
	\end{table}
	
	\item[c.] Re-investigate each row and column. Then allocate the supply and demand until they are all used up. \newpage
%\newcolumntype{C}{@{}c@{}}
%\newcommand{\bottombox}[1]{\makebox[2em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
%\newcommand{\innerbox}[2]{%}
    %\begin{tabular}[b]{c|c}
       %\rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
       %\multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    %\end{tabular}}
\renewcommand{\arraystretch}{1.25}
	\vspace{-0.15cm} 
\begin{table}[htbp]
  \begin{center}
	\caption{Iteration 2 Vogel Approximation Method} \label{ulya1}
	\begin{small}
	\begin{spacing}
	\hspace*{0.01cm} 
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
              & \hspace*{0.3cm}C 1 \hspace*{0.3cm} & \hspace*{0.3cm}C 2  \hspace*{0.3cm} &\hspace*{0.3cm} C 3 \hspace*{0.3cm} &\hspace*{0.3cm} C 4 \hspace*{0.3cm} &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 & \hspace*{-1.6cm}\innerbox{ }{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$3$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{6} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{8} \hspace*{-1.6cm}& {$90$} \\\hline
	C 2 & \hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$0$} \hspace*{-1.6cm} &\hspace*{-1.6cm} \innerbox{ }{$7$} \hspace*{-1.6cm}& \hspace*{-1.6cm}  \innerbox{ }{$5$} \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{ }{10}\hspace*{-1.6cm}&  $90$ \\\hline
	C 4 &\hspace*{-1.6cm}\innerbox{ }{$11$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$6$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$8$ } \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{60}{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{2} \hspace*{-1.6cm} &  ${}$ \\\hline
  C 5 &\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm} &\hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{ }{$5$ } \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ }{12} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{ }{$0$} \hspace*{-1.6cm}&  $60$ \\\hline
  Demand & \hspace*{0.4cm}\bottombox{$60$} & \hspace*{0.4cm}\bottombox{$60$} &\hspace*{0.4cm} \bottombox{$25$} &\hspace*{0.4cm} \bottombox{$15$} &\hspace*{0.4cm} \bottombox{$80$} &    \\\hline	%\end{document}
	\end{tabular}
	%\end{center}
	\end{spacing}
	\end{small}
	\end{center}
	\end{table}\\
	
In Table 5, each row and column value is obtained, i.e.:\vspace{-0.45cm}
\begin{equation*} 
\begin{array}{lll}
\text{row} \;1 &=& {3-0} = 3\\
\text{row}\; 2 &=& {2-0} = 2\\
\text{row} \;4 &=& {2-0} = 2\\
\text{row} \;5 &=& {2-0} = 2\\
\text{column} \;1 &=& {2-0} = 2\\
\text{column} \;2 &=& {2-0} = 2\\
\text{column} \;3 &=& {5-4} = 1\\
\text{column} \;4 &=& {5-0} = 5\\
\text{column} \;5 &=& {2-0} = 2.
\end{array}
\end{equation*}
Based on the results of each row and column, select the largest value in column 4, which is 5. The largest cost element in column 4 is worth 5. in column 4, select the smallest cost, which is 0 in row 4. Then allocate demand and inventory in column 4 and row 4 as much as 60 pcs. column 4 and row 4 as many as 60 pcs. Cost changes can be seen in Table 6.\\

%\newcolumntype{C}{@{}c@{}}
%\newcommand{\bottombox}[1]{\makebox[2em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
%\newcommand{\innerbox}[2]{%}
    %\begin{tabular}[b]{c|c}
       %\rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
       %\multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    %\end{tabular}}
\renewcommand{\arraystretch}{1.25}
	\vspace{-0.15cm} 
\begin{table}[htbp]
  \begin{center}
	\caption{Iteration 3 Vogel Approximation Method} \label{ulya2}
	\begin{small}
	\begin{spacing}
	\hspace*{0.01cm} 
	
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
              & \hspace*{0.3cm}C 1 \hspace*{0.3cm} & \hspace*{0.3cm}C 2  \hspace*{0.3cm} &\hspace*{0.3cm} C 3 \hspace*{0.3cm} &\hspace*{0.3cm} C 4 \hspace*{0.3cm} &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 & \hspace*{-1.6cm}\innerbox{ }{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$3$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{6} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{}{8} \hspace*{-1.6cm}& {$90$} \\\hline
	C 2 & \hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$0$} \hspace*{-1.6cm} &\hspace*{-1.6cm} \innerbox{ }{$7$} \hspace*{-1.6cm}& \hspace*{-1.6cm}  \innerbox{ }{$5$} \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{ }{10}\hspace*{-1.6cm}&  $90$ \\\hline
  C 5 &\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm} &\hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{ }{$5$ } \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ }{12} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{60}{$0$} \hspace*{-1.6cm}&  ${}$ \\\hline
  Demand & \hspace*{0.4cm}\bottombox{$60$} & \hspace*{0.4cm}\bottombox{$60$} &\hspace*{0.4cm} \bottombox{$25$} &\hspace*{0.4cm} \bottombox{$15$} &\hspace*{0.4cm} \bottombox{$20$} &    \\\hline	%\end{document}
	\end{tabular}
	%\end{center}
	\end{spacing}
	\end{small}
	\end{center}
	\end{table}
	
	In Table 6, each row and column value is obtained, i.e.:
	\begin{equation*} 
\begin{array}{lll}
\text{row} \;1 &=& {3-0} = 3\\
\text{row}\; 2 &=& {2-0} = 2\\
\text{row} \;5 &=& {2-0} = 2\\
\text{column} \;1 &=& {2-0} = 2\\
\text{column} \;2 &=& {2-0} = 2\\
\text{column} \;3 &=& {5-4} = 1\\
\text{column} \;4 &=& {6-5} = 1 \\
\text{column} \;5 &=& {8-0} = 8. \\
\end{array}
\end{equation*}

Based on the results of each row and column, select the largest value in column 5, which is worth 8. The cost element in column 5 is selected the smallest cost, which is worth 0, in row 1. Then allocate demand and inventory in column 5 and row 1 as many as 10 pcs. Cost changes can be seen in Table 7.
%\newcolumntype{C}{@{}c@{}}
%\newcommand{\bottombox}[1]{\makebox[2em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
%\newcommand{\innerbox}[2]{%}
    %\begin{tabular}[b]{c|c}
       %\rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
       %\multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    %\end{tabular}}
\renewcommand{\arraystretch}{1.25}
%\begin{document}
\begin{table}[htbp]
 \begin{center}	
\caption{Iteration 4 Vogel Approximation Method}\label{ulya4}
%\begin{small}
\begin{spacing}
\hspace*{0.1cm} 
\scriptsize{
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
 & \hspace*{0.3cm}C 1  \hspace*{0.3cm} & \hspace*{0.3cm}C 2  \hspace*{0.3cm} &\hspace*{0.3cm} C 3 \hspace*{0.3cm} &\hspace*{0.3cm} C 4 \hspace*{0.3cm} &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 & \hspace*{-1.6cm}\innerbox{60}{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$3$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{6} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{8} \hspace*{-1.6cm}& {$30$} \\\hline
	C 2 & \hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$0$} \hspace*{-1.6cm} &\hspace*{-1.6cm} \innerbox{ }{$7$} \hspace*{-1.6cm}& \hspace*{-1.6cm}  \innerbox{ }{$5$} \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{ }{10}\hspace*{-1.6cm}&  $90$ \\\hline
   Demand & \hspace*{0.4cm}\bottombox{${}$} & \hspace*{0.4cm}\bottombox{$60$} &\hspace*{0.4cm} \bottombox{$25$} &\hspace*{0.4cm} \bottombox{$15$} &\hspace*{0.4cm} \bottombox{$20$} &    \\\hline	\end{tabular}}
%\end{center}
\end{spacing}
%\end{small}
\end{center}
\end{table}
In Table 7, each row and column value is obtained, i.e.:
\begin{equation*} 
\begin{array}{lll}
\text{row} \;1 &=& {3-0} = 3\\
\text{row}\; 2 &=& {2-0} = 2\\
\text{column} \;1 &=& {2-0} = 2\\
\text{column} \;2 &=& {3-0} = 3\\
\text{column} \;3 &=& {7-4} = 3 \\
\text{column} \;4 &=& {6-5} = 1 \\
\text{column} \;5 &=& {10-8} = 2. \\
\end{array}
\end{equation*}
Based on the results of each row and column, select the largest value in row 1, column 2, and column 3, because it has 3 equal values, select one of them, namely in row 1 which is worth 3. The cost element in row 1 is selected the smallest cost, which is worth 0 in column 1. Then allocate demand and inventory in row 1 and column 1 as many as 60 pcs. Cost changes can be seen in Table 8.
%\newcolumntype{C}{@{}c@{}}
%\newcommand{\bottombox}[1]{\makebox[2em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
%\newcommand{\innerbox}[2]{%}
    %\begin{tabular}[b]{c|c}
       %\rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
       %\multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    %\end{tabular}}
\renewcommand{\arraystretch}{1.25}
%\begin{document}
\begin{table}[htbp]
 \begin{center}	
\caption{Iteration 5 Vogel Approximation Method}\label{ulya6}
%\begin{small}
\begin{spacing}
\scriptsize{
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
 & \hspace*{0.3cm}C 2  \hspace*{0.3cm} &\hspace*{0.3cm} C 3 \hspace*{0.3cm} &\hspace*{0.3cm} C 4 \hspace*{0.3cm} &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 &  \hspace*{-1.6cm}\innerbox{ }{$3$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{6} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{8} \hspace*{-1.6cm}& {$30$} \\\hline
	C 2 & \hspace*{-1.6cm}\innerbox{60}{$0$} \hspace*{-1.6cm} &\hspace*{-1.6cm} \innerbox{ }{$7$} \hspace*{-1.6cm}& \hspace*{-1.6cm}  \innerbox{}{$5$} \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{ }{10}\hspace*{-1.6cm}&  $30$ \\\hline
   Demand &\hspace*{0.4cm}\bottombox{$ $} &\hspace*{0.4cm} \bottombox{$25$} &\hspace*{0.4cm} \bottombox{$15$} &\hspace*{0.4cm} \bottombox{$20$} &    \\\hline
		\end{tabular}}
	%\end{center}
	\end{spacing}
	%\end{small}
	\end{center}
	\end{table}
In Table 8, each row and column value is obtained, i.e.:
\begin{equation*} 
\begin{array}{lll}
\text{row} \;1 &=& {4-3} = 1\\
\text{row}\; 2 &=& {5-0} = 5\\
\text{column} \;2 &=& {3-0} = 3\\
\text{column} \;3 &=& {7-4} = 3 \\
\text{column} \;4 &=& {6-5} = 1 \\
\text{column} \;5 &=& {10-8} = 2.\\
\end{array}
\end{equation*}
Based on the results of each row and column, select the largest value in row 2 which is worth 5. The cost element in row 2 is selected the smallest cost which is worth 0 in column 2. Then allocate demand and inventory in row 2 and column 2 as many as 60 pcs. Cost changes can be seen in Table 9.\newpage
%\newcolumntype{C}{@{}c@{}}
%\newcommand{\bottombox}[1]{\makebox[2em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
%\newcommand{\innerbox}[2]{%}
    %\begin{tabular}[b]{c|c}
       %\rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
       %\multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    %\end{tabular}}
%
\renewcommand{\arraystretch}{1.25}
%\begin{document}
\begin{table}[htbp]
 \begin{center}
\caption{Iteration 6 Vogel Approximation Method}\label{yaya1}
%\begin{small}
\begin{spacing}
\hspace*{0.2cm} 
\scriptsize{
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
 &\hspace*{0.3cm} C 3 \hspace*{0.3cm} &\hspace*{0.3cm} C 4 \hspace*{0.3cm} &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 &\hspace*{-1.6cm}\innerbox{25}{$4$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{6} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{8} \hspace*{-1.6cm}& {$5$} \\\hline
	C 2 &\hspace*{-1.6cm} \innerbox{ }{$7$} \hspace*{-1.6cm}& \hspace*{-1.6cm}  \innerbox{}{$5$} \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{ }{10}\hspace*{-1.6cm}&  $30$ \\\hline
   Demand &\hspace*{0.4cm} \bottombox{$ $} &\hspace*{0.4cm} \bottombox{$15$} &\hspace*{0.4cm} \bottombox{$20$} &    \\\hline
		\end{tabular}}
%\end{center}
\end{spacing}
%\end{small}
\end{center}
\end{table}
In Table 9, each row and column value is obtained, i.e.:
\begin{equation*} 
\begin{array}{lll}
\text{row} \;1 &=& {6-4} = 2\\
\text{row} \;4 &=& {7-5} = 2\\
\text{column} \;3 &=& {7-4} = 3\\
\text{column} \;4 &=& {6-5} = 1\\
\text{column} \;5 &=& {10-8} = 2. \\
\end{array}
\end{equation*}
Based on the results of each row and column, select the largest value in column 3 which is worth 3. The cost element in column 3 is selected the smallest cost which is worth 4 in row 1. Then allocate demand and inventory in column 3 and row 1 as many as 25 pcs. Cost changes can be seen in Table 10.
%\newcolumntype{C}{@{}c@{}}
%\newcommand{\bottombox}[1]{\makebox[2em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
%\newcommand{\innerbox}[2]{%}
    %\begin{tabular}[b]{c|c}
       %\rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
       %\multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    %\end{tabular}}
\renewcommand{\arraystretch}{1.25}
%\begin{document}
\begin{table}[htbp]
 \begin{center}	
\caption{Iteration 7 Vogel Approximation Method}\label{yaya2}
%\begin{small}
\begin{spacing}
\hspace*{0.2cm} 
\scriptsize{
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
 &\hspace*{0.3cm} C 4 \hspace*{0.3cm} &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 & \hspace*{-1.6cm} \innerbox{ }{6} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{8} \hspace*{-1.6cm}& {$5$} \\\hline
	C 2 & \hspace*{-1.6cm}  \innerbox{15}{$5$} \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{ }{10}\hspace*{-1.6cm}&  $15$ \\\hline
   Demand &\hspace*{0.4cm} \bottombox{${}$} &\hspace*{0.4cm} \bottombox{$20$} &    \\\hline
	\end{tabular}}
%	\end{center}
	\end{spacing}
	%\end{small}
	\end{center}
	\end{table}
	
	
 In Table 10, each row and column value is obtained, i.e.:
 \begin{equation*} 
\begin{array}{lll}
\text{row} \;1 &=& {8-6} = 2\\
\text{row} \;2 &=& {10-5} = 5\\
\text{column} \;4 &=& {6-5} = 1\\
\text{column} \;5 &=& {10-8} = 2. \\
\end{array}
\end{equation*}
Based on the results of each row and column, select the largest value in row 2 which is worth 5. The cost element in row 2 is selected the smallest cost which is worth 5 in column 4. Then allocate demand and inventory in row 2 and column 4 as many as 15 pcs. Cost changes can be seen in Table 11.\\
%\newcolumntype{C}{@{}c@{}}
%\newcommand{\bottombox}[1]{\makebox[2em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
%\newcommand{\innerbox}[2]{%}
    %\begin{tabular}[b]{c|c}
       %\rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
       %\multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    %\end{tabular}}
\renewcommand{\arraystretch}{1.25}
%\begin{document}
\begin{table}[htbp]
 \begin{center}	
\caption{Iteration 8 Vogel Approximation Method}\label {yaya4}
%\begin{small}
\begin{spacing}
\hspace*{0.2cm} 
\scriptsize{
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
 &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 & \hspace*{-1.6cm}\innerbox{}{8} \hspace*{-1.6cm}& {$5$} \\\hline
	C 2 &  \hspace*{-1.6cm}\innerbox{15}{10}\hspace*{-1.6cm}&  ${}$ \\\hline
   Demand &\hspace*{0.4cm} \bottombox{$5$} &    \\\hline
\end{tabular}}
%\end{center}
\end{spacing}
%\end{small}
\end{center}
\end{table}\\
In Table 11, each row and column value is obtained, i.e.:
\begin{equation*} 
\begin{array}{lll}
\text{row} \;1 &=& {8} = 8\\
\text{row} \;2 &=& {10} =10\\
\text{column} \;5 &=& {10-8} = 2.\\
\end{array}
\end{equation*}
Based on the results of each row and column, select the largest value in row 2 which is worth 10. The cost element in row 2 which is worth 10 is in column 5. Then allocate demand and inventory in row 2 and column 5 as many as 15 pcs. The cost changes can be seen in Table 12.
%\newcolumntype{C}{@{}c@{}}
%\newcommand{\bottombox}[1]{\makebox[2em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
%\newcommand{\innerbox}[2]{%}
    %\begin{tabular}[b]{c|c}
       %\rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
       %\multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    %\end{tabular}}
\renewcommand{\arraystretch}{1.25}
%\begin{document}
\begin{table}[htbp]
 \begin{center}	
\caption{Iteration 9 Vogel Approximation Method}\label {yaya6}
%\begin{small}
\begin{spacing}
\hspace*{0.2cm} 
\scriptsize{
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
 &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 & \hspace*{-1.6cm}\innerbox{5}{8} \hspace*{-1.6cm}& {$5$} \\\hline
   Demand &\hspace*{0.4cm} \bottombox{$5$} &    \\\hline
\end{tabular}}
%\end{center}
\end{spacing}
%\end{small}
\end{center}
\end{table}


In Table 12, each row and column value is obtained, i.e.:
\begin{equation*} 
\begin{array}{lll}
\text{row} \;1 &=& {8}\\
\text{column} \;5 &=& {8}. \\
\end{array}
\end{equation*}
Based on the remaining rows and columns, namely row 1 and column 5, the demand and inventory of 5 units of goods can be directly allocated.


\item[d.] Then the initial base solution of the transshipment problem is obtained. The initial solution obtained is $x_11=60, x_13=25,x_15=5,x_22=60,x_24=15,x_25=15,x_33=60,x_44= 60$, and $x_55  = 60$ with total transportation costs are (150, 230, 405, 640). The results of this calculation can be seen in Table 13.\newpage
%\newcolumntype{C}{@{}c@{}}
%\newcommand{\bottombox}[1]{\makebox[2em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
%\newcommand{\innerbox}[2]{%}
    %\begin{tabular}[b]{c|c}
       %\rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
       %\multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    %\end{tabular}
\renewcommand{\arraystretch}{1.25}
%\begin{document}
\begin{table}[htbp]
  \begin{center}
	\caption{Cost Allocation Vogel Approach Method}\label{yaya7}
		%\begin{small}
	\begin{spacing}
	\hspace*{0.01cm} 
	\scriptsize{
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
                & \hspace*{0.3cm}C 1 \hspace*{0.3cm} & \hspace*{0.3cm}C 2  \hspace*{0.3cm} &\hspace*{0.3cm} C 3 \hspace*{0.3cm} &\hspace*{0.3cm} C 4 \hspace*{0.3cm} &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 & \hspace*{-1.6cm}\innerbox{60}{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$3$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{25 }{$4$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{6} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{5}{8} \hspace*{-1.6cm}& {$90$} \\\hline
	C 2 & \hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ 60}{$0$} \hspace*{-1.6cm} &\hspace*{-1.6cm} \innerbox{ }{$7$} \hspace*{-1.6cm}& \hspace*{-1.6cm}  \innerbox{15}{$5$} \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{15}{10}\hspace*{-1.6cm}&  $90$ \\\hline
	C 3 & \hspace*{-1.6cm}\innerbox{ }{$5$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$9$} \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ 60}{0} \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ }{3} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{4} \hspace*{-1.6cm} &  $60$ \\\hline
  C 4 &\hspace*{-1.6cm}\innerbox{ }{$11$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$6$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$8$ } \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{60 }{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{2} \hspace*{-1.6cm} &  $60$ \\\hline
  C 5 &\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm} &\hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{ }{$5$ } \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ }{12} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{60 }{$0$} \hspace*{-1.6cm}&  $60$ \\\hline
  Demand & \hspace*{0.4cm}\bottombox{$60$} & \hspace*{0.4cm}\bottombox{$60$} &\hspace*{0.4cm} \bottombox{$85$} &\hspace*{0.4cm} \bottombox{$75$} &\hspace*{0.4cm} \bottombox{$80$} &    \\\hline	%\end{document}
	\end{tabular}}
	\end{spacing}
	%\end{small}
	\end{center}
	\end{table}
 \end{enumerate}
 \item[(iv)] Next, the optimality of the initial base solution is checked using the simplex transport method. The values of $u_i$ and $v_j$ are determined through each base variable
 \begin{equation*}
						u_i+v_j=c_{ij},
\end{equation*}
\begin{equation*} 
					\begin{array}{lll}
					\qquad u_1=0, &\mbox{initial} &\mbox{value}\\
					u_1+v_1=0 & \rightarrow  & v_1=0\\
					u_1+v_3=4 & \rightarrow  & v_3=4\\
					u_1+v_5=8 & \rightarrow  & v_5=8\\
					u_2+v_5=10 & \rightarrow & u_2=2\\
					u_2+v_4=5 & \rightarrow  & v_4=3\\
					u_2+v_2=0 & \rightarrow  & v_2=-2\\
					u_3+v_3=0 & \rightarrow  & u_3=-4\\
					u_4+v_4=0 & \rightarrow  & u_4=-3\\					
					u_5+v_5=0 & \rightarrow  & u_5=-8\\
\end{array}
\end{equation*}

     Furthermore, for each nonbasis, $\bar{c}_{ij}$ is determined, namely $\bar{c}_{12}=-5$, $\bar{c}_{14}=-1$, $\bar{c}_{21}=0$, $\bar{c}_{23}=-1$, $\bar{c}_{31}=-9$, $\bar{c}_{32}=-15$, $\bar{c}_{34}=-2$, $\bar{c}_{35}=0$, $\bar{c}_{41}=-14$, $\bar{c}_{42}=-11$, $\bar{c}_{43}=-7$, $\bar{c}_{45}=3$, $\bar{c}_{51}=-12$, $\bar{c}_{52}=-12$, $\bar{c}_{53}=-9$, and $\bar{c}_{54}=-15$. It can be seen that the non-base value of $\bar{c}_{45}=3$ is greater than zero or positive, which means that the initial base solution is not optimal. Furthermore, $x_{45}$ will become the base by forming a loop that starts and ends at $x_{45}$ namely  $x_{45}(+)\to x_{44}(-)\to x_{24}(+)\to x_{25}(-)\to x_{45}(+)$. The exit variable is determined by selecting the smallest cost in the loop which is min$\{x_{25},x_{44}\}= $min$\{15,60\}=15$. Then obtained element changes are $x_{45}=15$, $x_{44}=45$, $x_{24}=30$, and $x_{25}=0$. Cost reduction can be seen in Table \ref{yaya9}.
%\newcolumntype{C}{@{}c@{}}
%\newcommand{\bottombox}[1]{\makebox[2em][r]{#1}\hspace*{\tabcolsep}\hspace*{2em}}%
%\newcommand{\innerbox}[2]{%}
    %\begin{tabular}[b]{c|c}
       %\rule{2em}{0pt}\rule[-2ex]{0pt}{5ex} & \makebox[2em]{#2} \\\cline{2-2}
       %\multicolumn{2}{r}{{#1}\hspace*{1.5\tabcolsep}\hspace*{2em}\rule[-2ex]{0pt}{5ex}}
    %\end{tabular}
\renewcommand{\arraystretch}{1.25}
%\begin{document}
\begin{table}[htbp]
  \begin{center}
	\caption{Vogel Approach Method Cost Reduction}\label{yaya9}
		%\begin{small}
	\begin{spacing}
	\hspace*{0.01cm} 
	\scriptsize{
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
                & \hspace*{0.3cm}C 1 \hspace*{0.3cm} & \hspace*{0.3cm}C 2  \hspace*{0.3cm} &\hspace*{0.3cm} C 3 \hspace*{0.3cm} &\hspace*{0.3cm} C 4 \hspace*{0.3cm} &\hspace*{0.3cm}C 5 \hspace*{0.3cm}&Supplay   \\\hline
  C 1 & \hspace*{-1.6cm}\innerbox{$60$}{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{$3$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{25 }{$4$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{ }{6} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{5}{8} \hspace*{-1.6cm}& {$90$} \\\hline
	C 2 & \hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{60}{$0$} \hspace*{-1.6cm} &\hspace*{-1.6cm} \innerbox{ }{$7$} \hspace*{-1.6cm}& \hspace*{-1.6cm}  \innerbox{$\rightarrow$ $15$}{$5$} \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{$\downarrow$ $15$}{10}\hspace*{-1.6cm}& $90$ \\\hline
	C 3 & \hspace*{-1.6cm}\innerbox{ }{$5$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$9$} \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ 60}{0} \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ }{3} \hspace*{-1.6cm}& \hspace*{-1.6cm}\innerbox{ }{4} \hspace*{-1.6cm} &  $60$ \\\hline
  C 4 &\hspace*{-1.6cm}\innerbox{ }{$11$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$6$} \hspace*{-1.6cm}&\hspace*{-1.6cm}\innerbox{ }{$8$ } \hspace*{-1.6cm} & \hspace*{-1.6cm}\innerbox{$\uparrow$ $60$ }{$0$} \hspace*{-1.6cm}& \hspace*{-1.6cm} \innerbox{$\leftarrow$}{2} \hspace*{-1.6cm} &  $60$ \\\hline
  C 5 &\hspace*{-1.6cm}\innerbox{ }{$4$} \hspace*{-1.6cm} &\hspace*{-1.6cm}\innerbox{ }{$2$} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{ }{$5$ } \hspace*{-1.6cm}&\hspace*{-1.6cm} \innerbox{ }{12} \hspace*{-1.6cm} & \hspace*{-1.6cm} \innerbox{60 }{$0$} \hspace*{-1.6cm}&  $60$ \\\hline
  Demand & \hspace*{0.4cm}\bottombox{$60$} & \hspace*{0.4cm}\bottombox{$60$} &\hspace*{0.4cm} \bottombox{$85$} &\hspace*{0.4cm} \bottombox{$75$} &\hspace*{0.4cm} \bottombox{$80$} &    \\\hline	%\end{document}
	\end{tabular}}
	\end{spacing}
	%\end{small}
	\end{center}
	\end{table}

Furthermore, for each nonbasis, $\bar{c}_{ij}$ is determined, namely $\bar{c}_{12}=-2$, $\bar{c}_{15}=-13$, $\bar{c}_{21}=-15$, $\bar{c}_{24}=-5$, $\bar{c}_{25}=-13$, $\bar{c}_{31}=-13$, $\bar{c}_{32}=-7$, $\bar{c}_{34}=-13$, $\bar{c}_{35}=-21$, $\bar{c}_{41}=-6$, $\bar{c}_{42}=-7$, $\bar{c}_{43}=-3$, $\bar{c}_{51}=-13$, $\bar{c}_{52}=-11$, $\bar{c}_{53}=-5$, and $\bar{c}_{54}=-3$. It can be seen that all nonbases are smaller than zero or negative. The optimization requirements have been met so that the optimal solution to the transshipment problem is obtained, namely ${x_{13}=25}$, ${x_{15}=5}$, ${x_{24}=30}$, and ${x_{45}=15}$. The optimal solution to the transshipment problem can be written in the form of a route can be seen in Figure \ref{fig2}.
\vspace{5cm}
\begin{figure}[htbp]\hspace{0.3cm}
\includegraphics[scale=0.16]{figure.jpg}
\caption{Transshipment Problem Representation}
\label{fig2}
\end{figure}\\

 The total minimum cost of the transshipment problem is
 \begin{equation*}
\begin{array}{lrl}
\mbox{min}\;z &=& \tilde{c}_{13}x_{13}+\tilde{c}_{15}x_{15}+\tilde{c}_{24}x_{24}+\tilde{c}_{45}x_{45}\\
&=& (2,3,5,6)(25)+(5,7,8,12)(5)+(1,2,6,11)(30)+(1,1,2,4)(15)\\
&=& (50,75,125,150)+(25,35, 40,60)+(30,60,180,330)+\\&&(15,15,30,60)\\
&=& (50+25+30+15,75+35+60+15,125+40+180+30,\\&&(150+60+330+60)\\
&=& (120, 185, 375, 600).
\end{array}
\end{equation*}

	Based on the discussion of the fuzzy cost coefficient transshipment problem with the Vogel approach method, the initial base solution is produced. Then to get the optimal solution, the initial base solution optimization test is carried out with the transportation simplex method. The minimum total cost is (120, 185, 375, 600) per pcs.
\end{enumerate}

\section{Conclusion}
Solving the transshipment problem with the transportation simplex method requires several steps to achieve optimal results. The initial solution first converts the cost coefficients in the form of trapezoidal fuzzy numbers into firm numbers using the robust ranking method. After getting a firm number, then determine the initial base solution using the Vogel approach method. The Vogel approach method is the simplest method from other methods. Then test the optimality of the initial base solution and find the optimal solution using the simplex transportation method.
In the transportation simplex method, the minimum cost is obtained which is the optimal route determination in the transshipment problem, namely : 
from the source supplier 1 to distributor 3, then from the source supplier 1 to distributor 5, then the source distributor 2 to distributor 4 to distributor 5, and transshipment is at the distributor 4. The minimum cost obtained from the transshipment problem on the specified route is (120,185,375,600)/unit.

\section{Acknowledgement}
We acknowledge the support from the Department of Mathematics University of Riau and Department of Mathematics University of Palangkaraya.

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\end{document}
