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<title>Transshipment problem</title>

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<span id="openzim-page-title" class="mw-page-title-main"><span class="mw-page-title-main">Transshipment problem</span></span>

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<p><b>Transshipment problems</b> form a subgroup of transportation problems, where <a href="Transshipment" title="Transshipment">transshipment</a> is allowed. In transshipment, transportation may or must go through intermediate nodes, possibly changing modes of transport.

</p><p>The <b>Transshipment problem</b> has its origins in medieval times when trading started to become a mass phenomenon. Obtaining the minimum-cost route had been the main priority. However, technological development slowly gave priority to minimum-duration transportation problems.

</p>

<div class="mw-heading mw-heading2"><h2 id="Overview">Overview</h2></div>

<p>Transshipment or Transhipment is the <a href="Shipment" class="mw-redirect" title="Shipment">shipment</a> of <a href="Cargo" title="Cargo">goods</a> or <a href="Intermodal_container" title="Intermodal container">containers</a> to an intermediate destination, and then from there to yet another destination. One possible reason is to change the <a href="Means_of_transport" title="Means of transport">means of transport</a> during the journey (for example from <a href="Ship_transport" class="mw-redirect" title="Ship transport">ship transport</a> to <a href="Road_transport" title="Road transport">road transport</a>), known as <a href="Transloading" title="Transloading">transloading</a>. Another reason is to combine small shipments into a large shipment (consolidation), dividing the large shipment at the other end (deconsolidation). Transshipment usually takes place in <a href="Transport_hub" title="Transport hub">transport hubs</a>. Much international transshipment also takes place in designated <a href="Customs_area" class="mw-redirect" title="Customs area">customs areas</a>, thus avoiding the need for customs checks or duties, otherwise a major hindrance for efficient transport.

</p>

<div class="mw-heading mw-heading2"><h2 id="Formulation_of_the_problem">Formulation of the problem</h2></div>

<p>A few initial assumptions are required in order to formulate the transshipment problem completely:

</p>

<ul><li>The system consists of <i>m</i> origins and <i>n</i> destinations, with the following indexing respectively: <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle i=1,\ldots ,m}">

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<annotation encoding="application/x-tex">{\displaystyle i=1,\ldots ,m}</annotation>

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</math></span><img src="./74690f54a3c93a332ecb2935e900178b9a555483.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.671ex; width:12.282ex; height:2.509ex;" alt="{\displaystyle i=1,\ldots ,m}" loading="lazy"></span>, <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle j=1,\ldots ,n}">

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<annotation encoding="application/x-tex">{\displaystyle j=1,\ldots ,n}</annotation>

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</math></span><img src="./c39c89dce6309ea91bb90112a6babe359d6ff7d6.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.671ex; margin-left: -0.027ex; width:11.819ex; height:2.509ex;" alt="{\displaystyle j=1,\ldots ,n}" loading="lazy"></span></li>

<li>One uniform good exists which needs to be shipped</li>

<li>The required amount of good at the destinations equals the produced quantity available at the origins</li>

<li>Transportation simultaneously starts at the origins and is possible from any node to any other (also to an origin and from a destination)</li>

<li>Transportation costs are independent of the shipped amount</li>

<li>The transshipment problem is a unique Linear Programming Problem (LLP) in that it considers the assumption that all sources and sinks can both receive and distribute shipments at the same time (function in both directions)<sup id="cite_ref-1" class="reference"><a href="#cite_note-1"><span class="cite-bracket">[</span>1<span class="cite-bracket">]</span></a></sup></li></ul>

<div class="mw-heading mw-heading2"><h2 id="Notations">Notations</h2></div>

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<annotation encoding="application/x-tex">{\displaystyle t_{r,s}}</annotation>

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</math></span><img src="./7ed1513962dd74112c8b6e24ae3fe03667f87b12.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:3.042ex; height:2.676ex;" alt="{\displaystyle t_{r,s}}" loading="lazy"></span>: time of transportation from node <i>r</i> to node <i>s</i></li>

<msub>

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<annotation encoding="application/x-tex">{\displaystyle a_{i}}</annotation>

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</math></span><img src="./0bc77764b2e74e64a63341054fa90f3e07db275f.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.671ex; width:2.029ex; height:2.009ex;" alt="{\displaystyle a_{i}}" loading="lazy"></span>: goods available at node <i>i</i></li>

<msub>

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<annotation encoding="application/x-tex">{\displaystyle b_{m+j}}</annotation>

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</math></span><img src="./30bbbc921780ce640753d3e553c87845ed1f15a9.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:4.629ex; height:2.843ex;" alt="{\displaystyle b_{m+j}}" loading="lazy"></span>: demand for the good at node <i>(m+j)</i></li>

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<annotation encoding="application/x-tex">{\displaystyle x_{r,s}}</annotation>

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</math></span><img src="./6514652831074499f3571a2bca38a34dc33e9cf6.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:3.532ex; height:2.343ex;" alt="{\displaystyle x_{r,s}}" loading="lazy"></span>: actual amount transported from node <i>r</i> to node <i>s</i></li></ul>

<div class="mw-heading mw-heading2"><h2 id="Mathematical_formulation_of_the_problem">Mathematical formulation of the problem</h2></div>

<p>The goal is to minimize <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \sum \limits _{i=1}^{m}\sum \limits _{j=1}^{n}t_{i,j}x_{i,j}}">

</mrow>

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<annotation encoding="application/x-tex">{\displaystyle \sum \limits _{i=1}^{m}\sum \limits _{j=1}^{n}t_{i,j}x_{i,j}}</annotation>

</semantics>

</math></span><img src="./8ad5fd25d1c8f911d991a2acc6ff007e748258e0.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.338ex; width:13.522ex; height:7.176ex;" alt="{\displaystyle \sum \limits _{i=1}^{m}\sum \limits _{j=1}^{n}t_{i,j}x_{i,j}}" loading="lazy"></span> subject to:

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<msub>

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</msub>

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<annotation encoding="application/x-tex">{\displaystyle x_{r,s}\geq 0}</annotation>

</semantics>

</math></span><img src="./8dacdff046b2b5012541986f080ce0d004879767.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:7.793ex; height:2.843ex;" alt="{\displaystyle x_{r,s}\geq 0}" loading="lazy"></span>; <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \forall r=1\ldots m}">

</mstyle>

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<annotation encoding="application/x-tex">{\displaystyle \forall r=1\ldots m}</annotation>

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</math></span><img src="./0ba851ff1a040ed54c858d43cb5338c491a01c9c.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:12.14ex; height:2.176ex;" alt="{\displaystyle \forall r=1\ldots m}" loading="lazy"></span>, <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle s=1\ldots n}">

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle s=1\ldots n}</annotation>

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</math></span><img src="./69eb53ccaaa75e4829e50a25fa1dd4221accfc00.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:10.244ex; height:2.176ex;" alt="{\displaystyle s=1\ldots n}" loading="lazy"></span></li>

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<annotation encoding="application/x-tex">{\displaystyle \sum _{s=1}^{m+n}{x_{i,s}}-\sum _{r=1}^{m+n}{x_{r,i}}=a_{i}}</annotation>

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</math></span><img src="./66fc057527e30a003ba141e579d6e46927176492.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.005ex; width:22.844ex; height:7.176ex;" alt="{\displaystyle \sum _{s=1}^{m+n}{x_{i,s}}-\sum _{r=1}^{m+n}{x_{r,i}}=a_{i}}" loading="lazy"></span>; <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \forall i=1\ldots m}">

</mstyle>

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<annotation encoding="application/x-tex">{\displaystyle \forall i=1\ldots m}</annotation>

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</math></span><img src="./558c63a4a02db399a953c4acf7ea1dcdaffcc7a4.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:11.894ex; height:2.176ex;" alt="{\displaystyle \forall i=1\ldots m}" loading="lazy"></span></li>

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<annotation encoding="application/x-tex">{\displaystyle \sum _{r=1}^{m+n}{x_{r,m+j}}-\sum _{s=1}^{m+n}{x_{m+j,s}}=b_{m+j}}</annotation>

</semantics>

</math></span><img src="./e7d4ce4ab60c1ec6505e97b2fa8b74c798a8236c.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.005ex; width:31.105ex; height:7.176ex;" alt="{\displaystyle \sum _{r=1}^{m+n}{x_{r,m+j}}-\sum _{s=1}^{m+n}{x_{m+j,s}}=b_{m+j}}" loading="lazy"></span>; <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \forall j=1\ldots n}">

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle \forall j=1\ldots n}</annotation>

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</math></span><img src="./68d1457af65aa8bb9dd205ab814dc1f7adaa510c.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.671ex; width:11.404ex; height:2.509ex;" alt="{\displaystyle \forall j=1\ldots n}" loading="lazy"></span></li>

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<annotation encoding="application/x-tex">{\displaystyle \sum _{i=1}^{m}{a_{i}}=\sum _{j=1}^{n}{b_{m+j}}}</annotation>

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</math></span><img src="./1be5432c5e7aca4c161eac3e7194a975f2578705.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.338ex; width:17.241ex; height:7.176ex;" alt="{\displaystyle \sum _{i=1}^{m}{a_{i}}=\sum _{j=1}^{n}{b_{m+j}}}" loading="lazy"></span></li></ul>

<div class="mw-heading mw-heading2"><h2 id="Solution">Solution</h2></div>

<p>Since in most cases an explicit expression for the objective function does not exist, an alternative method is suggested by Rajeev and <a href="Satya_Prakash_(physicist)" title="Satya Prakash (physicist)">Satya</a>. The method uses two consecutive phases to reveal the minimal durational route from the origins to the destinations. The first phase is willing to solve <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle n\cdot m}">

</mstyle>

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<annotation encoding="application/x-tex">{\displaystyle n\cdot m}</annotation>

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</math></span><img src="./cb720005d9f6eba5279f276a21f249734e824fff.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.338ex; width:5.114ex; height:1.676ex;" alt="{\displaystyle n\cdot m}" loading="lazy"></span> time-minimizing problem, in each case using the remained <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle n+m-2}">

</mstyle>

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<annotation encoding="application/x-tex">{\displaystyle n+m-2}</annotation>

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</math></span><img src="./f8604ae5565d646f51a448ba8f8a6d8723e4e1ce.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.505ex; width:10.278ex; height:2.343ex;" alt="{\displaystyle n+m-2}" loading="lazy"></span> intermediate nodes as transshipment points. This also leads to the minimal-durational transportation between all sources and destinations. During the second phase a standard time-minimizing problem needs to be solved. The solution of the time-minimizing transshipment problem is the joint solution outcome of these two phases.

</p>

<div class="mw-heading mw-heading3"><h3 id="Phase_1">Phase 1</h3></div>

<p>Since costs are independent from the shipped amount, in each individual problem one can normalize the shipped quantity to <i>1</i>. The problem now is simplified to an assignment problem from <i>i</i> to <i>m+j</i>. Let <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle x'_{r,s}=1}">

</mrow>

</msubsup>

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<annotation encoding="application/x-tex">{\displaystyle x'_{r,s}=1}</annotation>

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</math></span><img src="./d027fd00e8c04ada942bc2ee2239b7c91166c3ec.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:7.793ex; height:2.843ex;" alt="{\displaystyle x'_{r,s}=1}" loading="lazy"></span> be <i>1</i> if the edge between nodes <i>r</i> and <i>s</i> is used during the optimization, and <i>0</i> otherwise. Now the goal is to determine all <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle x'_{r,s}}">

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<annotation encoding="application/x-tex">{\displaystyle x'_{r,s}}</annotation>

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</math></span><img src="./002aa5cf13e39364cf3b13b054ba5e43beb39f45.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:3.532ex; height:2.843ex;" alt="{\displaystyle x'_{r,s}}" loading="lazy"></span> which minimize the objective function:

</p><p><span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle T_{i,m+j}=\sum _{r=1}^{m+n}\sum _{s=1}^{m+n}{t_{r,s}\cdot x'_{r,s}}}">

<msub>

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<annotation encoding="application/x-tex">{\displaystyle T_{i,m+j}=\sum _{r=1}^{m+n}\sum _{s=1}^{m+n}{t_{r,s}\cdot x'_{r,s}}}</annotation>

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</math></span><img src="./7f386bcd77697b07b5373a2e757bb49efe485774.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.005ex; width:25.554ex; height:7.176ex;" alt="{\displaystyle T_{i,m+j}=\sum _{r=1}^{m+n}\sum _{s=1}^{m+n}{t_{r,s}\cdot x'_{r,s}}}" loading="lazy"></span>,<br>

</p><p>such that

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<annotation encoding="application/x-tex">{\displaystyle \sum _{s=1}^{m+n}{x'_{r,s}}=1}</annotation>

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</math></span><img src="./2c16ec0085f4ee937048928801b1f04d7e0fc32b.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.005ex; width:11.887ex; height:7.176ex;" alt="{\displaystyle \sum _{s=1}^{m+n}{x'_{r,s}}=1}" loading="lazy"></span></li>

</mrow>

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<annotation encoding="application/x-tex">{\displaystyle \sum _{r=1}^{m+n}{x'_{r,s}}=1}</annotation>

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</math></span><img src="./0fcdbfb359baeef252ca5f7b046cc8f627aeb82e.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.005ex; width:11.887ex; height:7.176ex;" alt="{\displaystyle \sum _{r=1}^{m+n}{x'_{r,s}}=1}" loading="lazy"></span></li>

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<annotation encoding="application/x-tex">{\displaystyle x'_{m+j,i}=1}</annotation>

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</math></span><img src="./6ea006045b928ac83bb6d99ba8e53938855c4ff0.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:10.246ex; height:3.176ex;" alt="{\displaystyle x'_{m+j,i}=1}" loading="lazy"></span></li>

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<annotation encoding="application/x-tex">{\displaystyle x'_{r,s}=0,1}</annotation>

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</math></span><img src="./2ee93310403735ddcb13ab14ba517f309abe2845.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:9.989ex; height:2.843ex;" alt="{\displaystyle x'_{r,s}=0,1}" loading="lazy"></span>.</li></ul>

<div class="mw-heading mw-heading4"><h4 id="Corollary">Corollary</h4></div>

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<annotation encoding="application/x-tex">{\displaystyle x'_{r,r}=1}</annotation>

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</math></span><img src="./91fb1d1c9c6f9d38c01bcba2fb9b2b3c53ab874e.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:7.763ex; height:2.843ex;" alt="{\displaystyle x'_{r,r}=1}" loading="lazy"></span> and <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle x'_{m+j,i}=1}">

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<annotation encoding="application/x-tex">{\displaystyle x'_{m+j,i}=1}</annotation>

</semantics>

</math></span><img src="./6ea006045b928ac83bb6d99ba8e53938855c4ff0.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:10.246ex; height:3.176ex;" alt="{\displaystyle x'_{m+j,i}=1}" loading="lazy"></span> need to be excluded from the model; on the other hand, without the <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle x'_{m+j,i}=1}">

</mrow>

</msubsup>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle x'_{m+j,i}=1}</annotation>

</semantics>

</math></span><img src="./6ea006045b928ac83bb6d99ba8e53938855c4ff0.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:10.246ex; height:3.176ex;" alt="{\displaystyle x'_{m+j,i}=1}" loading="lazy"></span> constraint the optimal path would consist only of <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle x'_{r,r}}">

</mrow>

</msubsup>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle x'_{r,r}}</annotation>

</semantics>

</math></span><img src="./8527deab8d9887645c0fca3bbcbf6d7b8e767471.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:3.502ex; height:2.843ex;" alt="{\displaystyle x'_{r,r}}" loading="lazy"></span>-type loops which obviously can not be a feasible solution.</li>

<li>Instead of <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle x'_{m+j,i}=1}">

</mrow>

</msubsup>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle x'_{m+j,i}=1}</annotation>

</semantics>

</math></span><img src="./6ea006045b928ac83bb6d99ba8e53938855c4ff0.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:10.246ex; height:3.176ex;" alt="{\displaystyle x'_{m+j,i}=1}" loading="lazy"></span>, <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle t_{m+j,i}=-M}">

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle t_{m+j,i}=-M}</annotation>

</semantics>

</math></span><img src="./47019327b29ff63f616a5d161f3c7e1656f53191.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:12.844ex; height:2.843ex;" alt="{\displaystyle t_{m+j,i}=-M}" loading="lazy"></span> can be written, where <i>M</i> is an arbitrarily large positive number. With that modification the formulation above is reduced to the form of a <a href="Assignment_problem" title="Assignment problem">standard assignment problem</a>, possible to solve with the <a href="Hungarian_method" class="mw-redirect" title="Hungarian method">Hungarian method</a>.</li></ul>

<div class="mw-heading mw-heading3"><h3 id="Phase_2">Phase 2</h3></div>

<p>During the second phase, a time minimization problem is solved with <i>m</i> origins and <i>n</i> destinations without transshipment. This phase differs in two main aspects from the original setup:

</p>

<ul><li>Transportation is only possible from an origin to a destination</li>

<li>Transportation time from <i>i</i> to <i>m+j</i> is the sum of durations coming from the optimal route calculated in Phase 1. Worthy to be denoted by <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle t'_{i,m+j}}">

</mrow>

</msubsup>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle t'_{i,m+j}}</annotation>

</semantics>

</math></span><img src="./12aa2e7e070dd27d09cb3bd3294235caf3f9331e.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:5.495ex; height:3.176ex;" alt="{\displaystyle t'_{i,m+j}}" loading="lazy"></span> in order to separate it from the times introduced during the first stage.</li></ul>

<div class="mw-heading mw-heading4"><h4 id="In_mathematical_form">In mathematical form</h4></div>

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle x_{i,m+j}\geq 0}</annotation>

</semantics>

</math></span><img src="./8a40259a0636a0c3d0a05c60412a31a6d2372515.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:10.246ex; height:2.843ex;" alt="{\displaystyle x_{i,m+j}\geq 0}" loading="lazy"></span> which minimize

</p><p><span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle z=max\left\{t'_{i,m+j}:x_{i,m+j}>0\;\;(i=1\ldots m,\;j=1\ldots n)\right\}}">

<mrow>

<mrow>

</mrow>

</msubsup>

<msub>

</mrow>

</msub>

</mrow>

</mrow>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle z=max\left\{t'_{i,m+j}:x_{i,m+j}>0\;\;(i=1\ldots m,\;j=1\ldots n)\right\}}</annotation>

</semantics>

</math></span><img src="./ff17d99807f8a43d0bebc5c3aa3e61778b88c90c.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.838ex; width:55.444ex; height:4.843ex;" alt="{\displaystyle z=max\left\{t'_{i,m+j}:x_{i,m+j}>0\;\;(i=1\ldots m,\;j=1\ldots n)\right\}}" loading="lazy"></span>,<br>

such that

</p>

</mrow>

</mrow>

</munderover>

<msub>

</mrow>

</msub>

</mrow>

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle \sum _{i=1}^{m}{x_{i,m+j}}=a_{i}}</annotation>

</semantics>

</math></span><img src="./5735d099b0dca5067787718ab1a37c3ac1002110.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.005ex; width:14.856ex; height:6.843ex;" alt="{\displaystyle \sum _{i=1}^{m}{x_{i,m+j}}=a_{i}}" loading="lazy"></span></li>

</mrow>

</mrow>

</munderover>

<msub>

</mrow>

</msub>

</mrow>

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle \sum _{j=1}^{n}{x_{i,m+j}}=b_{m+j}}</annotation>

</semantics>

</math></span><img src="./4e14f2695b79e2651548e573cff621b06ac801f7.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.338ex; width:17.455ex; height:7.176ex;" alt="{\displaystyle \sum _{j=1}^{n}{x_{i,m+j}}=b_{m+j}}" loading="lazy"></span></li>

</mrow>

</mrow>

</munderover>

<msub>

</mrow>

</msub>

</mrow>

</mrow>

</mrow>

</munderover>

<msub>

</mrow>

</msub>

</mrow>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle \sum _{i=1}^{m}{a_{i}}=\sum _{j=1}^{n}{b_{m+j}}}</annotation>

</semantics>

</math></span><img src="./1be5432c5e7aca4c161eac3e7194a975f2578705.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.338ex; width:17.241ex; height:7.176ex;" alt="{\displaystyle \sum _{i=1}^{m}{a_{i}}=\sum _{j=1}^{n}{b_{m+j}}}" loading="lazy"></span></li></ul>

<p>This problem is easy to be solved with the method developed by <a href="Satya_Prakash_(physicist)" title="Satya Prakash (physicist)">Prakash</a>. The set <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \left\{t'_{i,m+j},i=1\ldots m,\;j=1\ldots n\right\}}">

<mrow>

<mrow>

</mrow>

</msubsup>

</mrow>

</mrow>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle \left\{t'_{i,m+j},i=1\ldots m,\;j=1\ldots n\right\}}</annotation>

</semantics>

</math></span><img src="./98b3000bb3e3e3a1b49d08cd021f3a14a25d5e94.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.838ex; width:32.021ex; height:4.843ex;" alt="{\displaystyle \left\{t'_{i,m+j},i=1\ldots m,\;j=1\ldots n\right\}}" loading="lazy"></span> needs to be partitioned into subgroups <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle L_{k},k=1\ldots q}">

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle L_{k},k=1\ldots q}</annotation>

</semantics>

</math></span><img src="./c04c40c116e4df4f06b4a09605618070f55968cb.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.671ex; width:13.745ex; height:2.509ex;" alt="{\displaystyle L_{k},k=1\ldots q}" loading="lazy"></span>, where each <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle L_{k}}">

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle L_{k}}</annotation>

</semantics>

</math></span><img src="./939365b07ac6bf3558837d0c457ca855dcb5e7d4.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.671ex; width:2.672ex; height:2.509ex;" alt="{\displaystyle L_{k}}" loading="lazy"></span> contain the <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle t'_{i,m+j}}">

</mrow>

</msubsup>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle t'_{i,m+j}}</annotation>

</semantics>

</math></span><img src="./12aa2e7e070dd27d09cb3bd3294235caf3f9331e.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:5.495ex; height:3.176ex;" alt="{\displaystyle t'_{i,m+j}}" loading="lazy"></span>-s with the same value. The sequence <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle L_{k}}">

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle L_{k}}</annotation>

</semantics>

</math></span><img src="./939365b07ac6bf3558837d0c457ca855dcb5e7d4.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.671ex; width:2.672ex; height:2.509ex;" alt="{\displaystyle L_{k}}" loading="lazy"></span> is organized as <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle L_{1}}">

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle L_{1}}</annotation>

</semantics>

</math></span><img src="./0e79dc1b001f8b923df475ed14de023cbc456013.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.671ex; width:2.637ex; height:2.509ex;" alt="{\displaystyle L_{1}}" loading="lazy"></span> contains the largest valued <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle t'_{i,m+j}}">

</mrow>

</msubsup>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle t'_{i,m+j}}</annotation>

</semantics>

</math></span><img src="./12aa2e7e070dd27d09cb3bd3294235caf3f9331e.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.338ex; width:5.495ex; height:3.176ex;" alt="{\displaystyle t'_{i,m+j}}" loading="lazy"></span>'s <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle L_{2}}">

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle L_{2}}</annotation>

</semantics>

</math></span><img src="./c6a952cfe42c86b7741f55a817da0e251793a358.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.671ex; width:2.637ex; height:2.509ex;" alt="{\displaystyle L_{2}}" loading="lazy"></span> the second largest and so on. Furthermore, <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle M_{k}}">

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle M_{k}}</annotation>

</semantics>

</math></span><img src="./43175c381ee6ba2694520af1f1a26c676a2726ad.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.671ex; width:3.343ex; height:2.509ex;" alt="{\displaystyle M_{k}}" loading="lazy"></span> positive priority factors are assigned to the subgroups <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \sum _{L_{k}}{x_{i,m+j}}}">

<msub>

</mrow>

</msub>

</mrow>

</munder>

<msub>

</mrow>

</msub>

</mrow>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle \sum _{L_{k}}{x_{i,m+j}}}</annotation>

</semantics>

</math></span><img src="./3ed3a0dadde24afe127ca7ba8f9f7d35a80b3304.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.505ex; width:9.728ex; height:6.009ex;" alt="{\displaystyle \sum _{L_{k}}{x_{i,m+j}}}" loading="lazy"></span>, with the following rule:

</p><p><span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \alpha M_{k}-\beta M_{k+1}=\left\{{\begin{array}{cc}-ve,&if\;\alpha <0\\ve,&if\;\alpha >0\end{array}}\right.}">

<msub>

</mrow>

</msub>

<msub>

</mrow>

</msub>

<mrow>

<mtr>

<mtd>

</mtd>

<mtd>

</mtd>

</mtr>

<mtr>

<mtd>

</mtd>

<mtd>

</mtd>

</mtr>

</mtable>

</mrow>

</mrow>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle \alpha M_{k}-\beta M_{k+1}=\left\{{\begin{array}{cc}-ve,&if\;\alpha <0\\ve,&if\;\alpha >0\end{array}}\right.}</annotation>

</semantics>

</math></span><img src="./43af39d3df10bbae475163cce98fab023c759ad7.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -2.505ex; width:35.503ex; height:6.176ex;" alt="{\displaystyle \alpha M_{k}-\beta M_{k+1}=\left\{{\begin{array}{cc}-ve,&if\;\alpha <0\\ve,&if\;\alpha >0\end{array}}\right.}" loading="lazy"></span>

</p><p>for all <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \beta }">

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle \beta }</annotation>

</semantics>

</math></span><img src="./7ed48a5e36207156fb792fa79d29925d2f7901e8.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -0.671ex; width:1.332ex; height:2.509ex;" alt="{\displaystyle \beta }" loading="lazy"></span>. With this notation the goal is to find all <span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle x_{i,m+j}}">

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle x_{i,m+j}}</annotation>

</semantics>

</math></span><img src="./16b0c93ec0914532defadf365a5a9f79e2bfd5ff.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.005ex; width:5.986ex; height:2.343ex;" alt="{\displaystyle x_{i,m+j}}" loading="lazy"></span> which minimize the goal function

</p><p><span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle z_{1}=\sum _{k=1}^{q}{M_{k}}\sum _{L_{k}}{x_{i,m+j}}}">

<msub>

</mrow>

</msub>

</mrow>

</mrow>

</munderover>

<msub>

</mrow>

</msub>

</mrow>

<msub>

</mrow>

</msub>

</mrow>

</munder>

<msub>

</mrow>

</msub>

</mrow>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle z_{1}=\sum _{k=1}^{q}{M_{k}}\sum _{L_{k}}{x_{i,m+j}}}</annotation>

</semantics>

</math></span><img src="./37d3536d9774900b1eb78d7e94ee3f83d20bf9b7.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.505ex; width:22.433ex; height:7.509ex;" alt="{\displaystyle z_{1}=\sum _{k=1}^{q}{M_{k}}\sum _{L_{k}}{x_{i,m+j}}}" loading="lazy"></span>

</p><p>such that

</p>

</mrow>

</mrow>

</munderover>

<msub>

</mrow>

</msub>

</mrow>

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle \sum _{i=1}^{m}{x_{i,m+j}}=a_{i}}</annotation>

</semantics>

</math></span><img src="./5735d099b0dca5067787718ab1a37c3ac1002110.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.005ex; width:14.856ex; height:6.843ex;" alt="{\displaystyle \sum _{i=1}^{m}{x_{i,m+j}}=a_{i}}" loading="lazy"></span></li>

</mrow>

</mrow>

</munderover>

<msub>

</mrow>

</msub>

</mrow>

<msub>

</mrow>

</msub>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle \sum _{j=1}^{n}{x_{i,m+j}}=b_{m+j}}</annotation>

</semantics>

</math></span><img src="./4e14f2695b79e2651548e573cff621b06ac801f7.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.338ex; width:17.455ex; height:7.176ex;" alt="{\displaystyle \sum _{j=1}^{n}{x_{i,m+j}}=b_{m+j}}" loading="lazy"></span></li>

</mrow>

</mrow>

</munderover>

<msub>

</mrow>

</msub>

</mrow>

</mrow>

</mrow>

</munderover>

<msub>

</mrow>

</msub>

</mrow>

</mstyle>

</mrow>

<annotation encoding="application/x-tex">{\displaystyle \sum _{i=1}^{m}{a_{i}}=\sum _{j=1}^{n}{b_{m+j}}}</annotation>

</semantics>

</math></span><img src="./1be5432c5e7aca4c161eac3e7194a975f2578705.svg" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -3.338ex; width:17.241ex; height:7.176ex;" alt="{\displaystyle \sum _{i=1}^{m}{a_{i}}=\sum _{j=1}^{n}{b_{m+j}}}" loading="lazy"></span></li>

<li><span class="mwe-math-element mwe-math-element-inline"><span class="mwe-math-mathml-inline mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" alttext="{\displaystyle \alpha M_{k}-\beta M_{k+1}=\left\{{\begin{array}{cc}-ve,&if\;\alpha <0\\ve,&if\;\alpha >0\end{array}}\right.}">

<msub>

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<div class="mw-heading mw-heading2"><h2 id="Extension">Extension</h2></div>

<p>Some authors such as Das et al (1999) and Malakooti (2013) have considered multi-objective Transshipment problem.

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<div class="mw-heading mw-heading2"><h2 id="References">References</h2></div>

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</style><cite class="citation web cs1"><a rel="nofollow" class="external text" href="https://www.researchgate.net/publication/317012808">"Transshipment Problem and Its Variants: A Review"</a>. <i>ResearchGate</i><span class="reference-accessdate">. Retrieved <span class="nowrap">2020-11-02</span></span>.</cite></span>

</li>

</ol></div>

<ul><li>R.J Aguilar, Systems Analysis and Design. Prentice Hall, Inc. Englewood Cliffs, New Jersey (1973) pp. 209–220</li>

<li>H. L. Bhatia, K. Swarup, M. C. Puri, Indian J. pure appl. Math. 8 (1977) 920-929</li>

<li>R. S. Gartinkel, M. R. Rao, Nav. Res. Log. Quart. 18 (1971) 465-472</li>

<li>G. Hadley, Linear Programming, Addison-Wesley Publishing Company, (1962) pp. 368–373</li>

<li>P. L. Hammer, Nav. Res. Log. Quart. 16 (1969) 345-357</li>

<li>P. L. Hammer, Nav. Res. Log. Quart. 18 (1971) 487-490</li>

<li>A.J.Hughes, D.E.Grawog, Linear Programming: An Emphasis On Decision Making, Addison-Wesley Publishing Company, pp. 300–312</li>

<li>H.W.Kuhn, Nav. Res. Log. Quart. 2 (1955) 83-97</li>

<li>A.Orden, Management Sci, 2 (1956) 276-285</li>

<li>S.Parkash, Proc. Indian Acad. Sci. (Math. Sci.) 91 (1982) 53-57</li>

<li>C.S. Ramakrishnan, OPSEARCH 14 (1977) 207-209</li>

<li>C.R.Seshan, V.G.Tikekar, Proc. Indian Acad. Sci. (Math. Sci.) 89 (1980) 101-102</li>

<li>J.K.Sharma, K.Swarup, Proc. Indian Acad. Sci. (Math. Sci.) 86 (1977) 513-518</li>

<li>W.Szwarc, Nav. Res. Log. Quart. 18 (1971) 473-485</li>

<li>Malakooti, B. (2013). Operations and Production Systems with Multiple Objectives. John Wiley & Sons.</li>

<li>Das, S. K., A. Goswami, and S. S. Alam. “Multiobjective Transportation Problem with Interval Cost, Source and Destination Parameters.” European Journal of Operational Research, Vol. 117, No. 1, 1999, pp. 100–112</li></ul></div><div><div class="zim-footer">

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