<?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">OJAppS</journal-id><journal-title-group><journal-title>Open Journal of Applied Sciences</journal-title></journal-title-group><issn pub-type="epub">2165-3917</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/ojapps.2023.132017</article-id><article-id pub-id-type="publisher-id">OJAppS-123245</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Biomedical&amp;Life Sciences</subject><subject> Chemistry&amp;Materials Science</subject><subject> Computer Science&amp;Communications</subject><subject> Engineering</subject><subject> Physics&amp;Mathematics</subject></subj-group></article-categories><title-group><article-title>
 
 
  New Approach to Find Initial Basic Feasible Solution (IBFS) for Optimal Solution in Transportation Problem
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Shubham</surname><given-names>Raval</given-names></name><xref ref-type="aff" rid="aff1"><sub>1</sub></xref></contrib></contrib-group><aff id="aff1"><label>1</label><addr-line>Department of Industrial Engineering, University of New Haven, West Haven, USA</addr-line></aff><pub-date pub-type="epub"><day>08</day><month>02</month><year>2023</year></pub-date><volume>13</volume><issue>02</issue><fpage>207</fpage><lpage>211</lpage><history><date date-type="received"><day>14,</day>	<month>January</month>	<year>2023</year></date><date date-type="rev-recd"><day>21,</day>	<month>February</month>	<year>2023</year>	</date><date date-type="accepted"><day>24,</day>	<month>February</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>
 
 
  Minimizing transportation time and getting optimal solution
  s
   are always considered as important factor
  s
   while solving transportation problem. This paper shows a new approach for finding initial basic solution for transportation problem which reduces cost of transportation more than any transportation method such as LCM, northwest, Vogel’s approximation and so on. This method has been illustrated by taking an example; afterwards, it compares basic initial feasible solution with other method
  s
   IBF and optimal dictate solutions such as MODI and Steppingstone method.
 
</p></abstract><kwd-group><kwd>Transportation Problem</kwd><kwd> New Approach for Transportation Problem</kwd><kwd> Initial Basic Feasible Solution</kwd><kwd> Minimizing</kwd><kwd> Transportation Time</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>Transportation problem is a special linear programming problem which concerned with minimization of cost and time. Moreover, transportation problem deals with fulfillment of demand by an obstruction of supply accordingly. Transportation problems for minimizing cost have been studied for a long [<xref ref-type="bibr" rid="scirp.123245-ref1">1</xref>] [<xref ref-type="bibr" rid="scirp.123245-ref2">2</xref>] .</p><p>There are many methods for solving transportation problems such as LCM, Vogel’s approximation, Russell’s approximation, northwest corner method, row minima, column minima and so on [<xref ref-type="bibr" rid="scirp.123245-ref3">3</xref>] .</p><p>More than that there are various ways to make transportation cost even more optimal such as MODI method and stepping stone which is known as optimality test.</p><p>The application of Transportation problem solution is not applicable for only medical, food storage and public sector, it also deals with military operations, rescue equipment and operation with deals with limited time. In these all applications more reduction of Transportation time is appreciated as it deals with either emergency or human safety. What if there is a way to reduce the transportation time more efficiently than other transportation methods like Northwest corner method, LCM, Rusell’s approximation and Vogel’s approximation Furthermore, it compares final solution (minimized cost) with others initial basic feasible solution and optimal solution which can be achieved by MODI and steppingstone method.</p><p>This paper also verifies this new method by the number of basic variables (M + N − 1). And number of demands should be equal to number of supplies.</p></sec><sec id="s2"><title>2. Proposed Procedure of Finding an Initial Basic Feasible Solution</title><p>This paper has assumed that readers are well known with Transportation problems. Moreover, for the recognition this paper assigned a name of this new approach as a Raval’s approximation for finding an initial basic feasible solution for minimum cost of transportation is mentioned as below:</p><p>Step-1: The number of supply and demand should be equal, and metric should have same number of row and column, if it is not then adding dummy column or row in it.</p><p>Step-2: find the number of basic variables by M + N − 1 and select lowest number in the cell accordingly.</p><p>Step-3: After selecting a number of basic variables (M + N − 1) and multiply the lowest element from the demand or supply to the corresponding cell.</p><p>Step-4: For balancing the demand and supply, whatever we multiply with the cells that should be substitute with corresponding demand or supply. Further, select the second lowest element from the demand or supply and the process continues further.</p><p>Step-5: The operation continues until we do not reach the number of basic variables and utilization of all demand and supply.</p><p>Step-6: There should not be any negative elements in cells, demand, and supply, or it violates the basic rule of transportation model.</p><p>Step-7: Multiplication and submission of cells with corresponding demands and supply forgetting initial basic feasible solution</p></sec><sec id="s3"><title>3. Illustrative: Example</title><p><xref ref-type="table" rid="table1">Table 1</xref> shows the time required to transport industrial products from origin to destination. Further it requires to find out optimal solution.</p></sec><sec id="s4"><title>4. Initial Basic Feasible Solution</title><p>Step 1: ensuring table is balanced, number of demands is equal to amount of supply. In this case, it is equal hence, table is balanced.</p><table-wrap-group id="1"><label><xref ref-type="table" rid="table1">Table 1</xref></label><caption><title> Data of example 1. (a) Selection of lowest element from the table; (b) utilization of demand and supply accordingly; (c) selection of second lowest element from the table; (d) utilization of all supply and demand</title></caption><table-wrap id="1_1"><caption><title> (b)</title></caption><table><tbody><thead><tr><th align="center" valign="middle"  rowspan="2"  >Origins</th><th align="center" valign="middle"  colspan="4"  >Destinations</th><th align="center" valign="middle"  rowspan="2"  >Supply</th></tr></thead><tr><td align="center" valign="middle" >D1</td><td align="center" valign="middle" >D2</td><td align="center" valign="middle" >D3</td><td align="center" valign="middle" >D4</td></tr><tr><td align="center" valign="middle" >A</td><td align="center" valign="middle" >6</td><td align="center" valign="middle" >3</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >4</td><td align="center" valign="middle" >22</td></tr><tr><td align="center" valign="middle" >B</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >9</td><td align="center" valign="middle" >2</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >15</td></tr><tr><td align="center" valign="middle" >C</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >8</td><td align="center" valign="middle" >6</td><td align="center" valign="middle" >8</td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >12</td><td align="center" valign="middle" >17</td><td align="center" valign="middle" >9</td><td align="center" valign="middle" >45/45</td></tr></tbody></table></table-wrap><table-wrap id="1_2"><caption><title> (c)</title></caption><table><tbody><thead><tr><th align="center" valign="middle"  rowspan="2"  >Origins</th><th align="center" valign="middle"  colspan="4"  >Destinations</th><th align="center" valign="middle"  rowspan="2"  >Supply</th></tr></thead><tr><td align="center" valign="middle" >D1</td><td align="center" valign="middle" >D2</td><td align="center" valign="middle" >D 3</td><td align="center" valign="middle" >D4</td></tr><tr><td align="center" valign="middle" >A</td><td align="center" valign="middle" >6</td><td align="center" valign="middle" >3</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >4</td><td align="center" valign="middle" >22</td></tr><tr><td align="center" valign="middle" >B</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >9</td><td align="center" valign="middle" >2</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >15</td></tr><tr><td align="center" valign="middle" >C</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >8</td><td align="center" valign="middle" >6</td><td align="center" valign="middle" >8</td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >12</td><td align="center" valign="middle" >17</td><td align="center" valign="middle" >9</td><td align="center" valign="middle" >45/45</td></tr></tbody></table></table-wrap><table-wrap id="1_3"><caption><title> (d)</title></caption><table><tbody><thead><tr><th align="center" valign="middle"  rowspan="2"  >Origins</th><th align="center" valign="middle"  colspan="4"  >Destinations</th><th align="center" valign="middle"  rowspan="2"  >Supply</th><th align="center" valign="middle" ></th></tr></thead><tr><td align="center" valign="middle" >D1</td><td align="center" valign="middle" >D2</td><td align="center" valign="middle" >D3</td><td align="center" valign="middle" >D4</td><td align="center" valign="middle" ></td></tr><tr><td align="center" valign="middle" >A</td><td align="center" valign="middle" >6</td><td align="center" valign="middle" >3</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >4</td><td align="center" valign="middle" >22</td><td align="center" valign="middle"  rowspan="3"  >8 − 7 = 1</td></tr><tr><td align="center" valign="middle" >B</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >9</td><td align="center" valign="middle" >2</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >15</td></tr><tr><td align="center" valign="middle" >C</td><td align="center" valign="middle" >5(7)</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >8</td><td align="center" valign="middle" >6</td><td align="center" valign="middle" >8</td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >12</td><td align="center" valign="middle" >17</td><td align="center" valign="middle" >9</td><td align="center" valign="middle" >45/45</td><td align="center" valign="middle" ></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >7 − 7 = 0</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></tr></tbody></table></table-wrap><table-wrap id="1_4"><caption><title></title></caption><table><tbody><thead><tr><th align="center" valign="middle"  rowspan="2"  >Origins</th><th align="center" valign="middle"  colspan="4"  >Destinations</th><th align="center" valign="middle"  rowspan="2"  >Supply</th><th align="center" valign="middle" ></th></tr></thead><tr><td align="center" valign="middle" >D1</td><td align="center" valign="middle" >D2</td><td align="center" valign="middle" >D 3</td><td align="center" valign="middle" >D4</td><td align="center" valign="middle" ></td></tr><tr><td align="center" valign="middle" >A</td><td align="center" valign="middle" >6</td><td align="center" valign="middle" >3</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >4</td><td align="center" valign="middle" >22</td><td align="center" valign="middle" ></td></tr><tr><td align="center" valign="middle" >B</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >9</td><td align="center" valign="middle" >2</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >15</td><td align="center" valign="middle"  rowspan="2"  >8 − 7 = 1</td></tr><tr><td align="center" valign="middle" >C</td><td align="center" valign="middle" >5(7)</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >8</td><td align="center" valign="middle" >6</td><td align="center" valign="middle" >8</td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >12</td><td align="center" valign="middle" >17</td><td align="center" valign="middle" >9</td><td align="center" valign="middle" >45/45</td><td align="center" valign="middle" ></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >7 − 7 = 0</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></tr></tbody></table></table-wrap><table-wrap id="1_5"><caption><title></title></caption><table><tbody><thead><tr><th align="center" valign="middle"  rowspan="2"  >Origins</th><th align="center" valign="middle"  colspan="4"  >Destinations</th><th align="center" valign="middle"  rowspan="2"  >Supply</th><th align="center" valign="middle" ></th></tr></thead><tr><td align="center" valign="middle" >D1</td><td align="center" valign="middle" >D2</td><td align="center" valign="middle" >D3</td><td align="center" valign="middle" >D4</td><td align="center" valign="middle" ></td></tr><tr><td align="center" valign="middle" >A</td><td align="center" valign="middle" >6</td><td align="center" valign="middle" >3(12)</td><td align="center" valign="middle" >5(2)</td><td align="center" valign="middle" >4(8)</td><td align="center" valign="middle" >22</td><td align="center" valign="middle" >22 − 20 = 8</td></tr><tr><td align="center" valign="middle" >B</td><td align="center" valign="middle" >5</td><td align="center" valign="middle" >9</td><td align="center" valign="middle" >2(15)</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >15</td><td align="center" valign="middle" >15 − 15 = 0</td></tr><tr><td align="center" valign="middle" >C</td><td align="center" valign="middle" >5(7)</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >8</td><td align="center" valign="middle" >6(1)</td><td align="center" valign="middle" >8</td><td align="center" valign="middle" >8 − 7 = 1</td></tr><tr><td align="center" valign="middle" >Demand</td><td align="center" valign="middle" >7</td><td align="center" valign="middle" >12</td><td align="center" valign="middle" >17</td><td align="center" valign="middle" >9</td><td align="center" valign="middle" >45/45</td><td align="center" valign="middle" ></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >7 − 7 = 0</td><td align="center" valign="middle" >12 − 12 = 0</td><td align="center" valign="middle" >17 − 15 = 2</td><td align="center" valign="middle" >9 − 8 = 1</td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td></tr><tr><td align="center" valign="middle" ></td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >0</td><td align="center" valign="middle" >2 − 2 = 0</td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td><td align="center" valign="middle" ></td></tr></tbody></table></table-wrap></table-wrap-group><p>Step 2: selecting lowest number from the cells (from top to bottom by m + n − 1).</p><p>Where, m = number of rows and n = number of columns. In this case, it is 6 that we must select, number of cells.</p><p>Step 3: now, the lowest element from the demand and supply is number 7, moreover, which is the lowest destination which corresponds to demand 7. However, there are two origins which have a number 5, because of that, lowest number of supplies should be selected for getting the lowest number as per optimal solution.</p><p>Step 4: selecting second lowest element from demand and supply and repeat the process so on.</p><p>Step 5: Operation continues until we do not reach the number of basic variables and utilization of all demand and supply.</p><p>Step 6 and step 7: From the table, there is no negative number or elements which violate the basic rule of transportation model. Now, multiplication of the demand/ apply with cells respectively, for getting initial basic feasible solution.</p><p>Initial basic feasible solution:</p><p>3(12) + 5(2) + 4(8) + 2(15) + 5(7) + 6(1) = 149</p><p>In this method, the optimal solution as per IBFS is 149, now this paper’s going to compare other transportation method with this solution further this paper also going to compare this initial basic feasible solution of Raval’s approximation with optimality test of other transportation method, for proving efficiency of this new method.</p></sec><sec id="s5"><title>5. Result and Discussion</title><p>As a result, this paper compares IBFS of Raval’s approximation to other transportation methods such as northwest corner method, least cost method, Vogel’s approximation method, row minima method, column minima method and Rusell’s approximation. Not only it compares the initial basic feasible solution of other methods but it also, compare optimal solution of other method with initial basic feasible solution of Raval’s approximation steel result is more efficient than optimal solution of other methods.</p><table-wrap id="table2" ><label><xref ref-type="table" rid="table2">Table 2</xref></label><caption><title> Comparison of different methods</title></caption><table><tbody><thead><tr><th align="center" valign="middle" >METHODS</th><th align="center" valign="middle" >IBFS</th><th align="center" valign="middle" >OPTIMALITYTEST (MODI/STEPPING STONE)</th></tr></thead><tr><td align="center" valign="middle" >LCM NWCM ROW MINIMA COLUMN MINIMA RUSELL’S APPROXIMATION VOGEL APPROXIMATION RAVAL’S APPROXIMATION</td><td align="center" valign="middle" >150 176 149 150 189 149 149</td><td align="center" valign="middle" >149 149 149 149 149 149 149</td></tr></tbody></table></table-wrap></sec><sec id="s6"><title>6. Comparative Study</title><p>As can be seen from <xref ref-type="table" rid="table2">Table 2</xref> that IBFS of other traditional methods varies from 189 to 149. indeed , methods like row minima and Vogel’s approximation has 149 as IBFS which is lowest number can be achieve from the table however these two method are lengthy while solving. On the other side as can be observe that Raval’s approximation is getting 149 as an IBFS which can be seen in optimality test of others’ methods. Moreover, this method requires less number of operation still it gives assuring optimal solution.</p></sec><sec id="s7"><title>7. Conclusion</title><p>In this writing, an algorithm has been discussed to solve time minimizing transportation problems. Efficiency of the developed algorithm has also been justified by solving numerical problems. During this process, a comparative study between the proposed method and the other existing methods is also carried out; where it is observed that the proposed method requires a minimum number of iterations to obtain the optimal transportation time. Therefore, the proposed algorithm claims its wide application in optimization in solving the time minizine transportation problem.</p></sec><sec id="s8"><title>Conflicts of Interest</title><p>The author declares no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s9"><title>Cite this paper</title><p>Raval, S. (2023) New Approach to Find Initial Basic Feasible Solution (IBFS) for Optimal Solution in Transportation Problem. Open Journal of Applied Sciences, 13, 207-211. https://doi.org/10.4236/ojapps.2023.132017<sup> </sup></p></sec></body><back><ref-list><title>References</title><ref id="scirp.123245-ref1"><label>1</label><mixed-citation publication-type="other" xlink:type="simple">Das, U.K., Babu, MdA., et al. (2014) Logical Development of Vogel’s Approximation Method (LD-VAM): An Approach to Find Basic Feasible Solution of Transportation Problem. International Journal of Scientific &amp; Technology Research (IJSTR), 3, 42-48.</mixed-citation></ref><ref id="scirp.123245-ref2"><label>2</label><mixed-citation publication-type="journal" xlink:type="simple"><name name-style="western"><surname>Khan</surname><given-names> A.R. </given-names></name>,<etal>et al</etal>. (<year>2011</year>)<article-title>A Resolution of the Transportation Problem: An Algorithmic Approach</article-title><source> Jahangirnagar University Journal of Science</source><volume> 34</volume>,<fpage> 49</fpage>-<lpage>62</lpage>.<pub-id pub-id-type="doi"></pub-id></mixed-citation></ref><ref id="scirp.123245-ref3"><label>3</label><mixed-citation publication-type="other" xlink:type="simple">Sharma, J.K. (2009) Operation Research Theory and Application—Assignment Problem. Macmillan Publishers India Ltd, 315-321.</mixed-citation></ref></ref-list></back></article>