<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">AJCM</journal-id><journal-title-group><journal-title>American Journal of Computational Mathematics</journal-title></journal-title-group><issn pub-type="epub">2161-1203</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/ajcm.2023.134033</article-id><article-id pub-id-type="publisher-id">AJCM-129772</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Physics&amp;Mathematics</subject></subj-group></article-categories><title-group><article-title>
 
 
  Analysis of the Impact of Optimal Solutions to the Transportation Problems for Variations in Cost Using Two Reliable Approaches
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Abdur</surname><given-names>Rashid</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Md.</surname><given-names>Amirul Islam</given-names></name><xref ref-type="aff" rid="aff2"><sup>2</sup></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib></contrib-group><aff id="aff2"><addr-line>Department of Mathematics, Uttara University, Dhaka, Bangladesh</addr-line></aff><aff id="aff1"><addr-line>Department of Mathematics, Jahangirnagar University, Dhaka, Bangladesh</addr-line></aff><pub-date pub-type="epub"><day>13</day><month>10</month><year>2023</year></pub-date><volume>13</volume><issue>04</issue><fpage>607</fpage><lpage>618</lpage><history><date date-type="received"><day>24,</day>	<month>October</month>	<year>2023</year></date><date date-type="rev-recd"><day>10,</day>	<month>December</month>	<year>2023</year>	</date><date date-type="accepted"><day>13,</day>	<month>December</month>	<year>2023</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  In this paper, we have used two reliable approaches (theorems) to find the optimal solutions to transportation problems, using variations in costs. In real-life scenarios, transportation costs can fluctuate due to different factors. Finding optimal solutions to the transportation problem in the context of variations in cost is vital for ensuring cost efficiency, resource allocation, customer satisfaction, competitive advantage, environmental responsibility, risk mitigation, and operational fortitude in practical situations. This paper opens up new directions for the solution of transportation problems by introducing two key theorems. By using these theorems, we can develop an algorithm for identifying the optimal solution attributes and permitting accurate quantification of changes in overall transportation costs through the addition or subtraction of constants to specific rows or columns, as well as multiplication by constants inside the cost matrix. It is anticipated that the two reliable techniques presented in this study will provide theoretical insights and practical solutions to enhance the efficiency and cost-effectiveness of transportation systems. Finally, numerical illustrations are presented to verify the proposed approaches.
 
</p></abstract><kwd-group><kwd>Transportation Problem</kwd><kwd> Initial Basic Feasible Solution</kwd><kwd> Optimal Solution</kwd><kwd> Two Reliable Approaches (theorems) and Numerical Illustrations</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>One of the first uses of linear programming is the transportation problem (TP). It handles the movement of products between many supply points (sources) and several demand locations (destinations). It is possible to construct a wide range of decision issues as TPs, including production distribution, job scheduling, inventory control, and investment analysis. Determining the delivery plan that minimizes overall shipping costs while meeting supply and demand restrictions is the primary goal of the transportation challenge. In operations management, the application of transportation problems for variations in cost is crucial for optimizing supply chain logistics and distribution networks. Let’s consider a manufacturing company that sources raw materials from multiple suppliers and distributes finished products to diverse markets. Fluctuations in fuel prices, labor costs, or other variables can significantly impact transportation expenses. By employing transportation optimization models, operations managers can design efficient delivery routes, select appropriate carriers, and allocate resources effectively. In the face of cost variations, the system can dynamically adjust to minimize transportation expenses. For instance, if fuel costs rise unexpectedly, the optimization model might recommend consolidating shipments, re-routing deliveries, or even exploring alternative transportation modes to maintain cost efficiency. This application of transportation problems in operations management ensures that the supply chain remains agile and responsive to changing cost dynamics. It not only helps control operational expenses but also contributes to overall efficiency and customer satisfaction by ensuring timely and cost-effective deliveries. This approach is particularly relevant in industries where transportation costs form a significant portion of the overall operational expenses, such as manufacturing, retail, and distribution. The foundational transportation model, initially proposed by Hitchcock [<xref ref-type="bibr" rid="scirp.129772-ref1">1</xref>] , underwent advancements by Koppmans [<xref ref-type="bibr" rid="scirp.129772-ref2">2</xref>] and Danzig [<xref ref-type="bibr" rid="scirp.129772-ref3">3</xref>] . Several heuristic techniques, including the North-West corner rule, least cost approach, Vogel’s approximation method (VAM) [<xref ref-type="bibr" rid="scirp.129772-ref4">4</xref>] , Russell’s approximation method [<xref ref-type="bibr" rid="scirp.129772-ref5">5</xref>] , and other renowned heuristics, were employed to derive a simple and practical solution. Kirca and Satir [<xref ref-type="bibr" rid="scirp.129772-ref6">6</xref>] introduced the total opportunity cost approach (TOM) as a heuristic for obtaining the initial viable solution, later expanded by Mathirajan and Meenakshi [<xref ref-type="bibr" rid="scirp.129772-ref7">7</xref>] using VAM. Korukoglu and Balli [<xref ref-type="bibr" rid="scirp.129772-ref8">8</xref>] proposed an enhanced VAM based on total opportunity cost (TOC). Charnes and Cooper [<xref ref-type="bibr" rid="scirp.129772-ref9">9</xref>] devised the stepping stone method, while Dantzig [<xref ref-type="bibr" rid="scirp.129772-ref10">10</xref>] introduced the modified distribution (MODI) approach for verifying optimality in the initial fundamental solution. Numerous scholars, including Amaliah et al. [<xref ref-type="bibr" rid="scirp.129772-ref11">11</xref>] , Hosseini [<xref ref-type="bibr" rid="scirp.129772-ref12">12</xref>] , Jude et al. [<xref ref-type="bibr" rid="scirp.129772-ref13">13</xref>] , Juman and Hoque [<xref ref-type="bibr" rid="scirp.129772-ref14">14</xref>] , Karagul and Sahin [<xref ref-type="bibr" rid="scirp.129772-ref15">15</xref>] , and Uddin and Khan [<xref ref-type="bibr" rid="scirp.129772-ref16">16</xref>] , have delved into the exploration of Initial Basic Feasible Solution (IBFS) as a means to directly attain the optimal solution for Transportation Problems (TP) without relying on Stepping Stone or Modified Distribution (MODI) methods. Rashid [<xref ref-type="bibr" rid="scirp.129772-ref17">17</xref>] contributed a theorem in the context of resolving transportation issues. Ahmed et al. [<xref ref-type="bibr" rid="scirp.129772-ref18">18</xref>] introduced an incessant allocation method to analyze and minimize transportation cost. Taha [<xref ref-type="bibr" rid="scirp.129772-ref19">19</xref>] , studied for finding optimal solution of transportation problems. Amaliah et al. [<xref ref-type="bibr" rid="scirp.129772-ref20">20</xref>] , introduced a heuristic method to find the initial basic feasible solution of transportation problem (TP).</p><p>In the existing literature, the authors obtain that the optimal solution of Transportation Problems (TP) without incorporating changes to the Transportation cost. In general, various communication systems in real-world applications demand representation through variations in Transportation cost, particularly for multiservice facilities catering to customers. This investigation addressed in this paper by integrating the variation of Transportation cost. The presented models emerge as a more reliable framework when compared to the existing models in the literature. An efficient transportation plan not only reduces costs but also improves overall operational efficiency. This can lead to reduced lead times, faster order fulfillment, and enhanced customer satisfaction. The main objective of this paper is to examine the impact of fluctuations in transportation costs on reliability. Significant changes in costs can disrupt the optimal transportation plan and impact various aspects of an organization’s operations. Adapting to these changes effectively through cost management strategies, contingency planning, and informed decision-making is crucial for maintaining efficiency, competitiveness, and financial stability in dynamic business environments. These two suggested theorems have the following benefits:</p><p>1) Analyzing the effect on transportation expenses of possible disruptions or modifications in the availability of resources.</p><p>2) The two suggested theorems in the business sector can be used to examine how sensitive transportation costs are to modifications in the component parts of the cost matrix.</p><p>3) Quickly modifying transportation expenses without sacrificing the best possible outcome in order to effectively manage resources.</p><p>4) Compared to starting from scratch and solving the problem, finding the best solution for the modified problem typically requires less computing work.</p><p>This paper is organized as follows: Section 2: mathematical formulations; Section 3: description of the models; Section 4: analysis of the proposed theorems; Section 5: numerical examples; Section 6: discussion of results; and the last section: the conclusion of the paper.</p></sec><sec id="s2"><title>2. Mathematical Formulation</title><p>Let m suppliers (sources) and n customers (destinations) be considered. The m suppliers can ship a single homogeneous product to any of the n customers at a shipping cost per unit c i j . Let a i represent the number of supply units needed at source i ( i = 1 , 2 , ⋯ , m ) , b j represent the number of demand units needed at destination j ( j = 1 , 2 , ⋯ , n ) and x i j represent the unknown quantity that needs to be transported from source i to destination j. Consequently, the corresponding linear programming model will be</p><p>Minimize (total cost) z = ∑ i = 1 m ∑ j = 1 n c i j x i j</p><p>Subject to</p><p>∑ j = 1 n x i j = a i     for   i = 1 , 2 , ⋯ , m     ( supply   constraint ) ∑ i = 1 m x i j = b j     for   j = 1 , 2 , ⋯ , n     ( demand   constraint ) x i j ≥ 0.</p><p>In matrix form the model can be formulated as <xref ref-type="table" rid="table1">Table 1</xref>.</p></sec><sec id="s3"><title>3. Description of the Model</title><p>In this section, we have shown two modified cost matrix using the addition or subtraction of constants α i ; i = 1 , 2 , ⋯ , m and β j ; j = 1 , 2 , ⋯ , n to specific rows or columns within the matrix, along with the multiplication of constants k within the same cost matrix. Through these systematic modifications, the impact of changes to individual elements on the overall transportation cost structure can be precisely measured (<xref ref-type="table" rid="table2">Table 2</xref> and <xref ref-type="table" rid="table3">Table 3</xref>).</p><p>The total transportation cost of modified cost matrix for TP is minimized by the following formula for <xref ref-type="table" rid="table2">Table 2</xref>.</p><p>z * = ∑ i = 1 m ∑ j = 1 n ( c i j &#177; α i &#177; β j ) x i j</p><table-wrap id="table1" ><label><xref ref-type="table" rid="table1">Table 1</xref></label><caption><title> Matrix form model</title></caption><table><tbody><thead><tr><th align="center" valign="middle" ></th><th align="center" valign="middle"  colspan="2"  >D<sub>1 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>2 </sub></th><th align="center" valign="middle" >⋯</th><th align="center" valign="middle"  colspan="2"  >D<sub>n</sub><sub> </sub></th><th align="center" valign="middle" >Supply</th></tr></thead><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>1</sub><sub> </sub></td><td align="center" valign="middle" >x<sub>11 </sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >x<sub>12</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >⋯</td><td align="center" valign="middle" >x<sub>1n</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >a<sub>1 </sub></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >c<sub>11 </sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c<sub>12</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c<sub>1n</sub></td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>2 </sub></td><td align="center" valign="middle" >x<sub>21</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >x<sub>22</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >⋯</td><td align="center" valign="middle" >x<sub>2n</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >a<sub>2 </sub></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >c<sub>21</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c<sub>22</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c<sub>2n</sub></td></tr><tr><td align="center" valign="middle" >⋮</td><td align="center" valign="middle"  colspan="2"  >⋮</td><td align="center" valign="middle"  colspan="2"  >⋮</td><td align="center" valign="middle" >⋯</td><td align="center" valign="middle"  colspan="2"  >⋮</td><td align="center" valign="middle" >⋮</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>m</sub><sub> </sub></td><td align="center" valign="middle" >x<sub>m</sub><sub>1</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >x<sub>m</sub><sub>2</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >⋯</td><td align="center" valign="middle" >x<sub>mn</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >a<sub>m </sub></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >c<sub>m</sub><sub>1</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c<sub>m</sub><sub>2</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c<sub>mn</sub></td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle"  colspan="2"  >b<sub>1 </sub></td><td align="center" valign="middle"  colspan="2"  >b<sub>2 </sub></td><td align="center" valign="middle" >⋯</td><td align="center" valign="middle"  colspan="2"  >b<sub>n </sub></td><td align="center" valign="middle" ></td></tr></tbody></table></table-wrap><table-wrap id="table2" ><label><xref ref-type="table" rid="table2">Table 2</xref></label><caption><title> Modified cost matrix</title></caption><table><tbody><thead><tr><th align="center" valign="middle" ></th><th align="center" valign="middle"  colspan="2"  >D<sub>1 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>2 </sub></th><th align="center" valign="middle" >⋯</th><th align="center" valign="middle"  colspan="2"  >D<sub>n</sub><sub> </sub></th><th align="center" valign="middle" >Supply</th></tr></thead><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>1</sub><sub> </sub></td><td align="center" valign="middle" >x<sub>11 </sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >x<sub>12</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >⋯</td><td align="center" valign="middle" >x<sub>1n</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >a<sub>1 </sub></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >c 11 &#177; α 1 &#177; β 1</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c 12 &#177; α 1 &#177; β 2</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c 1 n &#177; α 1 &#177; β n</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>2 </sub></td><td align="center" valign="middle" >x<sub>21</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >x<sub>22</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >⋯</td><td align="center" valign="middle" >x<sub>2n</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >a<sub>2 </sub></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >c 21 &#177; α 2 &#177; β 1</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c 22 &#177; α 2 &#177; β 2</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c 2 n &#177; α 2 &#177; β n</td></tr><tr><td align="center" valign="middle" >⋮</td><td align="center" valign="middle"  colspan="2"  >⋮</td><td align="center" valign="middle"  colspan="2"  >⋮</td><td align="center" valign="middle" >⋯</td><td align="center" valign="middle"  colspan="2"  >⋮</td><td align="center" valign="middle" >⋮</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>m</sub><sub> </sub></td><td align="center" valign="middle" >x<sub>m</sub><sub>1</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >x<sub>m</sub><sub>2</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >⋯</td><td align="center" valign="middle" >x<sub>mn</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >a<sub>m</sub><sub> </sub></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >c m 1 &#177; α m &#177; β 1</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c m 2 &#177; α m &#177; β 2</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >c m n &#177; α m &#177; β n</td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle"  colspan="2"  >b<sub>1 </sub></td><td align="center" valign="middle"  colspan="2"  >b<sub>2 </sub></td><td align="center" valign="middle" >⋯</td><td align="center" valign="middle"  colspan="2"  >b<sub>n</sub><sub> </sub></td><td align="center" valign="middle" ></td></tr></tbody></table></table-wrap><table-wrap id="table3" ><label><xref ref-type="table" rid="table3">Table 3</xref></label><caption><title> Modified cost matrix</title></caption><table><tbody><thead><tr><th align="center" valign="middle" ></th><th align="center" valign="middle"  colspan="2"  >D<sub>1 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>2 </sub></th><th align="center" valign="middle" >⋯</th><th align="center" valign="middle"  colspan="2"  >D<sub>n</sub><sub> </sub></th><th align="center" valign="middle" >Supply</th></tr></thead><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>1</sub><sub> </sub></td><td align="center" valign="middle" >x<sub>11 </sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >x<sub>12</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >⋯</td><td align="center" valign="middle" >x<sub>1n</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >a<sub>1 </sub></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >kc<sub>11 </sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >kc<sub>12</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >kc<sub>1n</sub></td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>2 </sub></td><td align="center" valign="middle" >x<sub>21</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >x<sub>22</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >⋯</td><td align="center" valign="middle" >x<sub>2n</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >a<sub>2 </sub></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >kc<sub>21</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >kc<sub>22</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >kc<sub>2n</sub></td></tr><tr><td align="center" valign="middle" >⋮</td><td align="center" valign="middle"  colspan="2"  >⋮</td><td align="center" valign="middle"  colspan="2"  >⋮</td><td align="center" valign="middle" >⋯</td><td align="center" valign="middle"  colspan="2"  >⋮</td><td align="center" valign="middle" >⋮</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>m</sub><sub> </sub></td><td align="center" valign="middle" >x<sub>m</sub><sub>1</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >x<sub>m</sub><sub>2</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >⋯</td><td align="center" valign="middle" >x<sub>mn</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >a<sub>m</sub><sub> </sub></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >kc<sub>m</sub><sub>1</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >kc<sub>m</sub><sub>2</sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >kc<sub>mn</sub></td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle"  colspan="2"  >b<sub>1 </sub></td><td align="center" valign="middle"  colspan="2"  >b<sub>2 </sub></td><td align="center" valign="middle" >⋯</td><td align="center" valign="middle"  colspan="2"  >b<sub>n</sub><sub> </sub></td><td align="center" valign="middle" ></td></tr></tbody></table></table-wrap><p>Therefore, the total cost (objective function) becomes for <xref ref-type="table" rid="table3">Table 3</xref>.</p><p>z * = k ∑ i = 1 m ∑ j = 1 n c i j x i j</p></sec><sec id="s4"><title>4. Analysis of the Proposed Approaches (Theorems)</title><p>Different models have different assumptions and approaches. Using two models provides a more robust analysis that accounts for various scenarios and conditions. It helps identify the potential impact of model-specific assumptions on the results. These models help in understanding how variations in transportation costs affect the optimal solution under different modeling frameworks. In this section we have proposed two theorems to find the effect of optimal solution for transportation problem using variation of costs.</p><p>Theorem 4.1: The optimality of the solution to the original problem is maintained for the new problem if there is an addition or subtraction of a constant quantity α i from each element of the ith row and/or a constant quantity β j from each element of the jth column of the transportation cost matrix [ c i j ] . Additionally, the total transportation cost is altered by</p><p>∑ i = 1 m ( &#177; α i ) a i + ∑ j = 1 n ( &#177; β j ) b j .</p><p>Proof: Let { x i j : i = 1 , 2 , ⋯ , m   ; j = 1 , 2 , ⋯ , n } represent the optimal solution in relation to the original cost matrix [ c i j ] . Then, the objective function for total cost is provided by</p><p>z = ∑ i = 1 m ∑ j = 1 n c i j x i j</p><p>If the real constants α i and β j are added (or subtracted) to (or from) ith row and jth column respectively of the cost matrix [ c i j ] and z * indicates the entire cost of the modified cost matrix, [ c i j &#177; α i &#177; β j ] after that</p><p>z * = ∑ i = 1 m ∑ j = 1 n ( c i j &#177; α i &#177; β j ) x i j</p><p>= ∑ i = 1 m ∑ j = 1 n c i j x i j + ∑ i = 1 m ( &#177; α i ) ∑ j = 1 n x i j + ∑ j = 1 n ( &#177; β j ) ∑ i = 1 m x i j = z + ∑ i = 1 m ( &#177; α i ) a i + ∑ j = 1 n ( &#177; β j ) b j</p><p>Since the terms that are added to z to give z * are independent of x i j , it follows that z * is minimized whenever z is minimized or vice versa and the total</p><p>transportation cost is changed by ∑ i = 1 m ( &#177; α i ) a i + ∑ j = 1 n ( &#177; β j ) b j .</p><p>Theorem 4.2: If a constant amount is multiplied by each element in a transportation cost matrix [ c i j ] , then the overall transportation cost is increased by the constant quantity times, and the best solution of the original problem remains optimal for the new problem.</p><p>Proof: Let { x i j : i = 1 , 2 , ⋯ , m   ; j = 1 , 2 , ⋯ , n } represent the optimal solution in relation to the original cost matrix [ c i j ] . Then, the objective function for total cost is provided by</p><p>z = ∑ i = 1 m ∑ j = 1 n c i j x i j</p><p>If each element of the cost matrix [ c i j ] is multiplied by a constant quantity k then the total cost (objective function) becomes</p><p>z * = ∑ i = 1 m ∑ j = 1 n ( k c i j ) x i j = k ∑ i = 1 m ∑ j = 1 n c i j x i j = k z</p><p>Hence if x i j is the optimal solution of the cost matrix [ c i j ] then it is also optimal for the cost matrix [ k c i j ] and the total cost is increased by k times.</p></sec><sec id="s5"><title>5. Numerical Illustrations</title><p>In real-life scenarios, the transportation problem can be found in various contexts such as supply chain management, logistics, and distribution planning, where minimizing transportation costs while meeting supply and demand requirements is a critical objective. These theorems provide practical methods to address these optimization challenges. Subsequently, we provide numerical examples to validate the introduced theorems.</p><p>Example 5.1: We investigate the following transportation problem to demonstrate Theorem 1.</p><p>To find the optimal cost of the TP we have used least cost method and then employing MODI method. The optimal transportation schedule is presented as <xref ref-type="table" rid="table4">Table 4</xref>.</p><p>Consequently, the optimal solution and the aggregate transportation cost are</p><p>x 12 = 130 ,     x 14 = 20 ,     x 21 = 140 ,     x 24 = 120 ,     x 32 = 270 ,     x 33 = 250 z = 25910.</p><p>We observe the following, when we modified the cost matrix of original transportation problem:</p><p>If we add α 1 = 14 , α 2 = 2 , α 3 = 10 , β 1 = 4 , β 3 = 9 and subtract β 2 = 12 , β 4 = 14 to each unit cost in the original transportation problem by [ c i j &#177; α i &#177; β j ] , the modified cost matrix is given below:</p><p>To find the optimal cost of the modified TP we have also used least cost method and then applying MODI method. The optimal transportation schedule is as <xref ref-type="table" rid="table5">Table 5</xref>.</p><table-wrap id="table4" ><label><xref ref-type="table" rid="table4">Table 4</xref></label><caption><title> Optimal solution original TP</title></caption><table><tbody><thead><tr><th align="center" valign="middle" ></th><th align="center" valign="middle"  colspan="2"  >D<sub>1 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>2 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>3 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>4 </sub></th><th align="center" valign="middle" >Supply</th></tr></thead><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>1 </sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >130</td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >20</td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >150</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >40</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >17</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >34</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >27</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>2 </sub></td><td align="center" valign="middle" >140</td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >120</td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >260</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >15</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >37</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >30</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >25</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>3 </sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >270</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >250</td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >520</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >33</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >28</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >42</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >45</td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle"  colspan="2"  >140</td><td align="center" valign="middle"  colspan="2"  >400</td><td align="center" valign="middle"  colspan="2"  >250</td><td align="center" valign="middle"  colspan="2"  >140</td><td align="center" valign="middle" ></td></tr></tbody></table></table-wrap><table-wrap id="table5" ><label><xref ref-type="table" rid="table5">Table 5</xref></label><caption><title> Optimal solution of modified cost of TP</title></caption><table><tbody><thead><tr><th align="center" valign="middle" ></th><th align="center" valign="middle"  colspan="2"  >D<sub>1 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>2 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>3 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>4 </sub></th><th align="center" valign="middle" >Supply</th></tr></thead><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>1 </sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >130</td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >20</td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >150</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >58</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >19</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >57</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >27</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>2 </sub></td><td align="center" valign="middle" >140</td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >120</td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >260</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >21</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >27</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >41</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >13</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>3 </sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >270</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >250</td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >520</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >47</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >26</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >61</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >41</td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle"  colspan="2"  >140</td><td align="center" valign="middle"  colspan="2"  >400</td><td align="center" valign="middle"  colspan="2"  >250</td><td align="center" valign="middle"  colspan="2"  >140</td><td align="center" valign="middle" ></td></tr></tbody></table></table-wrap><p>Consequently, the optimal solution and the aggregate transportation cost are</p><p>x 12 = 130 ,     x 14 = 20 ,     x 21 = 140 ,     x 24 = 120 ,     x 32 = 270 ,     x 33 = 250 z * = 29780.</p><p>Thus we see that after adding or subtracting constant quantities α i and β j , the optimal schedule remains unchanged and the total transportation cost is changed by</p><p>d = ∑ i = 1 3 ( &#177; α i ) a i + ∑ j = 1 4 ( &#177; β j ) b j         = α 1 a 1 + α 2 a 2 + α 3 a 3 + β 1 b 1 − β 2 b 2 + β 3 b 3 − β 4 b 4         = 14 &#215; 150 + 2 &#215; 260 + 10 &#215; 520 + 4 &#215; 140 − 12 &#215; 400 + 9 &#215; 250 − 14 &#215; 140         = 3870 i . e .   ,     d = z * − z = 29780 − 25910 = 3870</p><p><xref ref-type="table" rid="table6">Table 6</xref> shows some random variations in costs and impact on optimal solution and total cost. The results are investigated through TORA optimization software.</p><p>Example 5.2: We investigate the following transportation problem to demonstrate Theorem 2.</p><p>Applying Vogel’s approximation method for initial solution and then implementing MODI method the optimal transportation schedule is as <xref ref-type="table" rid="table7">Table 7</xref>.</p><p>Consequently, the optimal solution and the aggregate transportation cost are</p><p>x 11 = 30 ,     x 12 = 170 ,     x 13 = 50 ,     x 21 = 100 ,     x 33 = 160 z = 3880.</p><p>If we multiply each element of [ c i j ] by a constant quantity k = 5 the modified cost matrix [ k c i j ] is</p><p>Solving the above problem by using Vogel’s approximation method for initial solution and then employing the MODI method, the optimal transportation table is shown in <xref ref-type="table" rid="table8">Table 8</xref>.</p><table-wrap id="table6" ><label><xref ref-type="table" rid="table6">Table 6</xref></label><caption><title> Optimal results in variations of cost of TP</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >Values of parameters</th><th align="center" valign="middle" >Optimal solution</th><th align="center" valign="middle" >Optimal cost</th><th align="center" valign="middle" >Amount of Changing cost d = z * − z</th></tr></thead><tr><td align="center" valign="middle" >α 1 = 15 ,     α 2 = 20 ,     α 3 = − 5 ,     β 1 = − 10 , β 2 = − 12 ,     β 3 = 35 ,     β 4 = 18</td><td align="center" valign="middle" >x 12 = 130 ,     x 14 = 20 ,     x 21 = 140 , x 24 = 120 ,     x 32 = 270 ,     x 33 = 250</td><td align="center" valign="middle" >z * = 35830</td><td align="center" valign="middle" >9920 unit</td></tr><tr><td align="center" valign="middle" >α 1 = − 25 ,     α 2 = − 15 ,     α 3 = − 18 ,     β 1 = 14 , β 2 = 40 ,     β 3 = 32 ,     β 4 = 35</td><td align="center" valign="middle" >x 12 = 130 ,     x 14 = 20 ,     x 21 = 140 , x 24 = 120 ,     x 32 = 270 ,     x 33 = 250</td><td align="center" valign="middle" >z * = 39760</td><td align="center" valign="middle" >13850 unit</td></tr><tr><td align="center" valign="middle" >α 1 = 25 ,     α 2 = 15 ,     α 3 = 18 ,     β 1 = 14 , β 2 = 40 ,     β 3 = 32 ,     β 4 = 35</td><td align="center" valign="middle" >x 12 = 130 ,     x 14 = 20 ,     x 21 = 140 , x 24 = 120 ,     x 32 = 270 ,     x 33 = 250</td><td align="center" valign="middle" >z * = 73780</td><td align="center" valign="middle" >47870 unit</td></tr><tr><td align="center" valign="middle" >α 1 = 25 ,     α 2 = 15 ,     α 3 = 18 ,     β 1 = − 5 , β 2 = − 10 ,     β 3 = − 8 ,     β 4 = − 13</td><td align="center" valign="middle" >x 12 = 130 ,     x 14 = 20 ,     x 21 = 140 , x 24 = 120 ,     x 32 = 270 ,     x 33 = 250</td><td align="center" valign="middle" >z * = 34400</td><td align="center" valign="middle" >8490 unit</td></tr></tbody></table></table-wrap><table-wrap id="table7" ><label><xref ref-type="table" rid="table7">Table 7</xref></label><caption><title> Optimal solution of original TP</title></caption><table><tbody><thead><tr><th align="center" valign="middle" ></th><th align="center" valign="middle"  colspan="2"  >D<sub>1 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>2 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>3 </sub></th><th align="center" valign="middle" >Supply</th></tr></thead><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>1 </sub></td><td align="center" valign="middle" >30</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >170</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >50</td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >250</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >5</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >7</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >6</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>2 </sub></td><td align="center" valign="middle" >100</td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >100</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >8</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >12</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >13</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>3 </sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >160</td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >160</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >20</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >17</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >9</td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle"  colspan="2"  >130</td><td align="center" valign="middle"  colspan="2"  >170</td><td align="center" valign="middle"  colspan="2"  >210</td><td align="center" valign="middle" ></td></tr></tbody></table></table-wrap><table-wrap id="table8" ><label><xref ref-type="table" rid="table8">Table 8</xref></label><caption><title> Optimal solution of modified cost of TP</title></caption><table><tbody><thead><tr><th align="center" valign="middle" ></th><th align="center" valign="middle"  colspan="2"  >D<sub>1 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>2 </sub></th><th align="center" valign="middle"  colspan="2"  >D<sub>3 </sub></th><th align="center" valign="middle" >Supply</th></tr></thead><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>1 </sub></td><td align="center" valign="middle" >30</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >170</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >50</td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >250</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >25</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >35</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >30</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>2 </sub></td><td align="center" valign="middle" >100</td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >100</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >40</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >60</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >65</td></tr><tr><td align="center" valign="middle"  rowspan="2"  >S<sub>3 </sub></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" >160</td><td align="center" valign="middle" ></td><td align="center" valign="middle"  rowspan="2"  >160</td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >100</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >85</td><td align="center" valign="middle" ></td><td align="center" valign="middle" >45</td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle"  colspan="2"  >130</td><td align="center" valign="middle"  colspan="2"  >170</td><td align="center" valign="middle"  colspan="2"  >210</td><td align="center" valign="middle" ></td></tr></tbody></table></table-wrap><p>Consequently, the optimal solution and the aggregate transportation cost are</p><p>x 11 = 30 ,     x 12 = 170 ,     x 13 = 50 ,     x 21 = 100 ,     x 33 = 160 z * = 19400.</p><p>Thus it is observed that if we multiply each element of [ c i j ] by 5, the optimal schedule remains unchanged and the total transportation cost is increased by 5 times i.e. z * = k z = 5 &#215; 3880 = 19400 .</p><p>Following <xref ref-type="table" rid="table9">Table 9</xref> shows some random variations in costs and impact on</p><table-wrap id="table9" ><label><xref ref-type="table" rid="table9">Table 9</xref></label><caption><title> Optimal results in variations of cost of TP</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >Values of parameters</th><th align="center" valign="middle" >Optimal solution</th><th align="center" valign="middle" >Optimal cost</th><th align="center" valign="middle" >Amount of changing cost d = z * − z</th></tr></thead><tr><td align="center" valign="middle" >k = 10</td><td align="center" valign="middle" >x 11 = 30 ,     x 12 = 170 ,     x 13 = 50 ,     x 21 = 100 ,     x 33 = 160</td><td align="center" valign="middle" >z * = 38800</td><td align="center" valign="middle" >34,920 unit</td></tr><tr><td align="center" valign="middle" >k = 15</td><td align="center" valign="middle" >x 11 = 30 ,     x 12 = 170 ,     x 13 = 50 ,     x 21 = 100 ,     x 33 = 160</td><td align="center" valign="middle" >z * = 58200</td><td align="center" valign="middle" >54,320unit</td></tr><tr><td align="center" valign="middle" >k = 20</td><td align="center" valign="middle" >x 11 = 30 ,     x 12 = 170 ,     x 13 = 50 ,     x 21 = 100 ,     x 33 = 160</td><td align="center" valign="middle" >z * = 77600</td><td align="center" valign="middle" >73,720 unit</td></tr></tbody></table></table-wrap><p>optimal solution and total cost. The results are investigated through TORA optimization software.</p></sec><sec id="s6"><title>6. Discussion of the Results</title><p>In the dynamic landscape of operations management, decision-makers face the challenge of navigating uncertainties, especially in the realm of transportation costs. The primary objective is to enhance decision-making by conducting comprehensive sensitivity analyses. We observed from <xref ref-type="table" rid="table6">Table 6</xref> that when we changed the values of parameters α 1 , α 2 , α 3 , β 1 , β 2 , β 3 and β 4 the optimal solution, x 12 = 130 , x 14 = 20 , x 21 = 140 , x 24 = 120 , x 32 = 270 and x 33 = 250 , remained unchanged, but the total transportation cost changed by a significant amount d = z * − z . And also, we observed that from <xref ref-type="table" rid="table9">Table 9</xref>, when we changed the values of parameters k, the optimal solution x 11 = 30 , x 12 = 170 , x 13 = 50 , x 21 = 100 and x 33 = 160 remain unchanged, but total transportation cost is increased by k times i.e., z * = k z . Significant changes in costs can have a profound impact on the optimal transportation plan. When costs decrease significantly, it may open up opportunities for cost savings and efficiency improvements. For example, if transportation costs increase, it may be necessary to allocate more resources or budget to transportation to maintain service levels. Conversely, if costs decrease, resources may be reallocated to other areas of the business. Transportation costs are closely tied to service quality and customer satisfaction. Significant cost changes can affect delivery times, shipping options, and overall service quality. Analyzing cost changes can help organizations align their transportation strategies with sustainability goals. The sensitivity analysis is vital tools in the real world for understanding and managing the impact of cost variations on transportation problems. These procedures enable organizations to make informed decisions, optimize resource allocation, and remain adaptable in a dynamic business environment.</p></sec><sec id="s7"><title>7. Conclusion</title><p>In this paper, we have provided two approaches (theorems) for solving transportation problems and proved them in a simple and easy way. This method provides a detailed understanding of how adjustments to specific parameters influence the financial aspects of transportation, offering valuable insights for strategic decision-making in transportation management. To present a clear overview of the theorems introduced herein, two illustrative numerical examples have been selected to demonstrate the verification of the theorems. Different models may emphasize different aspects of the Transportation Problem. By using two models, analysts can gain a more comprehensive understanding of how variations in transportation costs impact the problem from different angles, considering numerical factors. The model can be developed for further study to effectively deal with unbalanced transportation problems.</p></sec><sec id="s8"><title>Conflicts of Interest</title><p>The authors declare no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s9"><title>Cite this paper</title><p>Rashid, A. and Islam, Md.A. (2023) Analysis of the Impact of Optimal Solutions to the Transportation Problems for Variations in Cost Using Two Reliable Approaches. American Journal of Computational Mathematics, 13, 607-618. https://doi.org/10.4236/ajcm.2023.134033</p></sec></body><back><ref-list><title>References</title><ref id="scirp.129772-ref1"><label>1</label><mixed-citation publication-type="other" xlink:type="simple">Hitchcock, F.L. (1941) Distribution of a Product from Several Sources to Numerous Localities. Journal of Mathematical Physics, 20, 224-230. https://doi.org/10.1002/sapm1941201224</mixed-citation></ref><ref id="scirp.129772-ref2"><label>2</label><mixed-citation publication-type="other" xlink:type="simple">Koopmans, T.C. (1949) Optimum Utilization of the Transportation System. Econometrica, 17, 136-146. https://doi.org/10.2307/1907301</mixed-citation></ref><ref id="scirp.129772-ref3"><label>3</label><mixed-citation publication-type="book" xlink:type="simple">Dantzig, G.B. (1951) Application of the Simplex Method to a Transportation Problem. In: Koopmans, T.C., Ed., Activity Analysis of Production and Allocation, Wiley, New York, 359-373.</mixed-citation></ref><ref id="scirp.129772-ref4"><label>4</label><mixed-citation publication-type="other" xlink:type="simple">Reinfeld, N.V. and Vogel, W.R. (1958) Mathematical Programming. Prentice-Hall, Englewood Cliffs.</mixed-citation></ref><ref id="scirp.129772-ref5"><label>5</label><mixed-citation publication-type="other" xlink:type="simple">Russell, E.J. (1969) Extension of Dantzig’s Algorithm to Finding an Initial Near-Optimal Basis for the Transportation Problem. Operations Research, 17, 187-191. https://doi.org/10.1287/opre.17.1.187</mixed-citation></ref><ref id="scirp.129772-ref6"><label>6</label><mixed-citation publication-type="other" xlink:type="simple">Kirca, O. and Satir, A. (1990) A Heuristic for Obtaining an Initial Solution for the Transportation Problem. Journal of Operational Research Society, 41, 865-871. https://doi.org/10.1057/jors.1990.124</mixed-citation></ref><ref id="scirp.129772-ref7"><label>7</label><mixed-citation publication-type="other" xlink:type="simple">Mathirajan, M. and Meenakshi, B. (2004) Experimental Analysis of Some Variants of Vogel’s Approximation Method. Asia Pacific Journal of Operational Research, 21, 447-462. https://doi.org/10.1142/S0217595904000333</mixed-citation></ref><ref id="scirp.129772-ref8"><label>8</label><mixed-citation publication-type="other" xlink:type="simple">Korukoglu, S. and Balli, S. (2011) An Improved Vogel’s Approximation Method for the Transportation Problem. Mathematical and Computational Applications, 16, 370-381. https://doi.org/10.3390/mca16020370</mixed-citation></ref><ref id="scirp.129772-ref9"><label>9</label><mixed-citation publication-type="other" xlink:type="simple">Charnes, A. and Cooper, W.W. (1954) The Stepping-Stone Method for Explaining Linear Programming Calculations in Transportation Problems. Management Science, 1, 49-69. https://doi.org/10.1287/mnsc.1.1.49</mixed-citation></ref><ref id="scirp.129772-ref10"><label>10</label><mixed-citation publication-type="other" xlink:type="simple">Dantzig, G.B. (1963) Linear Programming and Extensions. Princeton University Press, Princeton. https://doi.org/10.7249/R366</mixed-citation></ref><ref id="scirp.129772-ref11"><label>11</label><mixed-citation publication-type="other" xlink:type="simple">Amaliah, B., Fatichah, C. and Suryani, E. (2019) Total Opportunity Cost Matrix-Minimal Total: A New Approach to Determine Initial Basic Feasible Solution of a Transportation Problem. Egyptian Informatics Journal, 20, 131-141. https://doi.org/10.1016/j.eij.2019.01.002</mixed-citation></ref><ref id="scirp.129772-ref12"><label>12</label><mixed-citation publication-type="other" xlink:type="simple">Hosseini (2017) Three New Methods to Find Initial Basic Feasible Solution of Transportation Problems. Applied Mathematical Sciences, 11, 1803-1814. https://doi.org/10.12988/ams.2017.75178</mixed-citation></ref><ref id="scirp.129772-ref13"><label>13</label><mixed-citation publication-type="other" xlink:type="simple">Jude, O., Ifeanyichukwu, O.B., Ihuoma, I.A. and Akpos, E.P. (2017) A New and Efficient Proposed Approach to Find Initial Basic Feasible Solution of a Transportation Problem. American Journal of Applied Mathematics and Statistics, 5, 54-61.</mixed-citation></ref><ref id="scirp.129772-ref14"><label>14</label><mixed-citation publication-type="other" xlink:type="simple">Juman, Z.A.M.S. and Hoque, M.A. (2015) An Efficient Heuristic to Obtain a Better Initial Feasible Solution to the Transportation Problem. Applied Soft Computing Journal, 34, 813-826. https://doi.org/10.1016/j.asoc.2015.05.009</mixed-citation></ref><ref id="scirp.129772-ref15"><label>15</label><mixed-citation publication-type="other" xlink:type="simple">Karagul, K. and Sahin, Y. (2020) A Novel Approximation Method to Obtain Initial Basic Feasible Solution of Transportation Problem. Journal of King Saud University—Engineering Sciences, 32, 211-218. https://doi.org/10.1016/j.jksues.2019.03.003</mixed-citation></ref><ref id="scirp.129772-ref16"><label>16</label><mixed-citation publication-type="other" xlink:type="simple">Uddin, S. and Khan, A.R. (2016) Improved Least Cost Method to Obtain a Better IBFS to the Transportation Problem. Journal of Applied Mathematics &amp; Bioinformatics, 6, 1-20.</mixed-citation></ref><ref id="scirp.129772-ref17"><label>17</label><mixed-citation publication-type="journal" xlink:type="simple"><name name-style="western"><surname>Rashid</surname><given-names> A. </given-names></name>,<etal>et al</etal>. (<year>2016</year>)<article-title>Development of a Simple Theorem in Solving Transportation Problems</article-title><source> Journal of Physical Sciences</source><volume> 21</volume>,<fpage> 23</fpage>-<lpage>28</lpage>.<pub-id pub-id-type="doi"></pub-id></mixed-citation></ref><ref id="scirp.129772-ref18"><label>18</label><mixed-citation publication-type="other" xlink:type="simple">Ahmed, M.M., Khan, A.R., Ahmed, F. and Uddin, Md.S. (2016) Incessant Allocation Method for Solving Transportation Problems. American Journal of Operations Research, 6, 236-244. https://doi.org/10.4236/ajor.2016.63024</mixed-citation></ref><ref id="scirp.129772-ref19"><label>19</label><mixed-citation publication-type="other" xlink:type="simple">Taha, H.A. (2007) Operations Research: An Introduction. Pearson Prentice Hall, Upper Saddle River.</mixed-citation></ref><ref id="scirp.129772-ref20"><label>20</label><mixed-citation publication-type="other" xlink:type="simple">Amaliah, B., Fatichah, C. and Suryani, E. (2022) A New Heuristic Method of Finding the Initial Basic Feasible Solution to Solve the Transportation Problem. Journal of King Saud University-Computer and Information Sciences, 34, 2298-2307. https://doi.org/10.1016/j.jksuci.2020.07.007</mixed-citation></ref></ref-list></back></article>