<?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">AM</journal-id><journal-title-group><journal-title>Applied Mathematics</journal-title></journal-title-group><issn pub-type="epub">2152-7385</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/am.2014.52030</article-id><article-id pub-id-type="publisher-id">AM-42170</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>
 
 
  The Optimal Inventory Policy for Reusable Items with Random Planning Horizon Considering Present Value
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>hou-Mei</surname><given-names>Su</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>Shy-Der</surname><given-names>Lin</given-names></name><xref ref-type="aff" rid="aff2"><sup>2</sup></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Li-Fen</surname><given-names>Chang</given-names></name><xref ref-type="aff" rid="aff3"><sup>3</sup></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib></contrib-group><aff id="aff1"><addr-line>Department of Banking and Finance, Takming University of Science and Technology, Taiwan</addr-line></aff><aff id="aff3"><addr-line>Department of Applied Mathematics, Chung Yuan Christian University, Taiwan</addr-line></aff><aff id="aff2"><addr-line>Department of Applied Mathematics and Business Administration, Chung Yuan Christian University, Taiwan</addr-line></aff><author-notes><corresp id="cor1">* E-mail:<email>meimei@takming.edu.tw(HS)</email>;<email>shyder@cycu.edu.tw(SL)</email>;<email>shau.tang@msa.hinet.net(LC)</email>;</corresp></author-notes><pub-date pub-type="epub"><day>17</day><month>01</month><year>2014</year></pub-date><volume>05</volume><issue>02</issue><fpage>292</fpage><lpage>299</lpage><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>
 
 
   We discuss five areas of inventory model, including reusable raw material, EPQ model, optimization, random planning horizon and present value. In the traditional EPQ model, the stock-holding cost of raw material was not counted as a part of relevant cost. We explored the possibility of reducing a company’s impact on the environment and increasing their competitiveness by recycling their repair and waste disposal. The products are manufactured with reusable raw material. Our analysis takes into account the time value, and the present value method is applied to determine the optimal inventory policies for reusable items with random planning horizon. Results show how the heuristic approach can achieve global optimum. Numerical examples are given to validate the proposed system. 
 
</p></abstract><kwd-group><kwd>Reusable; EPQ; Optimization; Random Planning Horizon; Present Value</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>Reuse of material and products is not a new subject. Using the repaired and new made products, Richter [<xref ref-type="bibr" rid="scirp.42170-ref1">1</xref>] has created a model for fixed and variable collection time interval to minimize the cost and found the optimal collection intervals. Richter and Dobos [<xref ref-type="bibr" rid="scirp.42170-ref2">2</xref>] extended the model of Richter [<xref ref-type="bibr" rid="scirp.42170-ref3">3</xref>] with integer setup numbers. By applying Pontryagin’s Maximum Principle, Kleber et al. [<xref ref-type="bibr" rid="scirp.42170-ref4">4</xref>] determined the optional production, remanufacturing, and disposal policy for a cost model. Koh et al. [<xref ref-type="bibr" rid="scirp.42170-ref5">5</xref>] found a joint EOQ and EPQ model in which a fixed proportion of the used products is collected from customers and then recovered for reuse. Konstantaras and Papachristos [<xref ref-type="bibr" rid="scirp.42170-ref6">6</xref>] revised Koh et al.’s paper and used a different analysis to obtain closed form expressions for both the optimal number of setup in the recovery and the ordering processes. Karakayal et al. [<xref ref-type="bibr" rid="scirp.42170-ref7">7</xref>] characterized the optimal acquisition price of the used products and the selling price along with recovery quantities of the reusable components. Up to now Salameh &amp; El-Kassar [<xref ref-type="bibr" rid="scirp.42170-ref8">8</xref>] had the paper to establish an EPQ model taking the stock-holding cost of raw material into consideration and found the optimal lot size. El-Kassaret et al. [<xref ref-type="bibr" rid="scirp.42170-ref9">9</xref>] studied an EPQ model for imperfect quality raw material.</p><p>The EOQ (Economic Ordering Quantity) model was first proposed by Harris [<xref ref-type="bibr" rid="scirp.42170-ref10">10</xref>] and later the EPQ model was developed by E. W. Taft [<xref ref-type="bibr" rid="scirp.42170-ref11">11</xref>]. This paper is mainly based on Richter’s ([1,3]) and Moon and Yun’s [<xref ref-type="bibr" rid="scirp.42170-ref12">12</xref>] idea, and followed the research of Trippi [<xref ref-type="bibr" rid="scirp.42170-ref13">13</xref>], Kim, Philippatos and Chung [<xref ref-type="bibr" rid="scirp.42170-ref14">14</xref>], Moon and Yun [<xref ref-type="bibr" rid="scirp.42170-ref12">12</xref>] and Chung and Lin [<xref ref-type="bibr" rid="scirp.42170-ref15">15</xref>] about time value of money. Kim, Philippatos and Chung [<xref ref-type="bibr" rid="scirp.42170-ref14">14</xref>] have showed a method for evaluating investments in inventory. Moon and Yun [<xref ref-type="bibr" rid="scirp.42170-ref12">12</xref>] have justified the optimality of solutions which are derived from the first order conditions in Kim, Philippatos and Chung [<xref ref-type="bibr" rid="scirp.42170-ref14">14</xref>]. Later Chung and Lin [<xref ref-type="bibr" rid="scirp.42170-ref15">15</xref>] have derived the bounds for the optimal cycle following the optimality of solutions. Using the upper and lower bounds, an algorithm of computing the optimal cycle time is developed. These numerical examples are brought into the algorithm to find the different cycle time. We wish this model is appropriate and practical.</p></sec><sec id="s2"><title>2. The Models</title><p>The mathematical models developed in this study are based on the following definitions and assumptions.</p><p>Definitions:</p><p>PVC(T): the present value of the cash flow for the first inventory horizonC(T): the expected present value of the cash flow for the random planning horizonTRC(T): the total relevant cost per unit timeQ: the order sizeS: the cost of placing an orderP: the production rateD: the demand rateC: the purchasing cost per unit item,</p><p><inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\28b0c0f8-a812-4d21-a335-82c5e80b3f34.png" xlink:type="simple"/></inline-formula>: the reusable rateT: the cycle lengthr: the discount rateh: the stock holding cost of raw materials per item per year and the stock holding cost of finished products per item per yearx: the random planning horizon time.</p><p>Assumptions:</p><p>1) Production rate is greater than demand rate.</p><p>2) Production rate and demand rate are known and constant.</p><p>3) Shortage is not allowed.</p><p>4) A single item is considered.</p><p>5) The time horizon is not infinite.</p><p>6) The stock holding cost of raw materials and products are the same.</p><p>7)<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\0d99105e-537d-4d44-99f1-bfb5d2de09a4.png" xlink:type="simple"/></inline-formula>.</p><p>8) At the end of the planning horizon, the raw material is used up and the productions are sold out.</p><p>9) The random planning horizon time x follows an exponential distribution with parameter<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\1edfeaa4-2bb3-4176-948d-e2aa464e4372.png" xlink:type="simple"/></inline-formula>.</p><p>Two models used in this analysis are illustrated below (see <xref ref-type="fig" rid="fig1">Figure 1</xref>). Model 1 presents the annual total relevant cost, TRC(T). Model 2 presents the expected present value of total relevant cost for the random planning horizon, C(T). The analysis proceeds as follows. First, we find the optimal cycle time separately for each model. Then, we use the numerical examples to show how to find the optimal value of cycle time.</p><p>Model 1: We use the annual total relevant cost to find the optimal solution.</p><p>The annual total relevant cost TRC(T) consists of the following elements:</p><p>The ordering cost per unit time =<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\71f656ba-0994-4083-851f-32146850bb2b.png" xlink:type="simple"/></inline-formula>.</p><p>The purchasing cost per order per unit time =<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\9b20bccf-73f6-4f31-aa3f-834ff17ce189.png" xlink:type="simple"/></inline-formula>.</p><p>The stock holding cost of raw material per unit time =<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\5fe3440a-c3d0-47e0-942f-560e854befb1.png" xlink:type="simple"/></inline-formula>.</p><p>The stock holding cost of products per unit time =<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\08621f04-8067-44b6-bf59-1f282b96ad94.png" xlink:type="simple"/></inline-formula>.</p><p>The total relevant cost per unit time can be expressed as TRC(T) = the ordering cost per unit time</p><p>+ the purchasing cost per order per unit time</p><p>+ the stock holding cost of raw material per unit time</p><p>+ the stock holding cost of products per unit time</p><p>= <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\9786f266-002b-4455-ad1e-96c3e26b87b6.png" xlink:type="simple"/></inline-formula></p><p>The first and second derivatives of TRC(T) are</p><p><inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\3990e86c-22f6-4708-8d87-7a94788cf1d4.png" xlink:type="simple"/></inline-formula>and <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\c826715c-e793-4102-9013-2913ee4cffff.png" xlink:type="simple"/></inline-formula> for all T &gt; 0, respectively.</p><p>Set<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\15557e30-e5e8-43ba-ba15-1467f2159c38.png" xlink:type="simple"/></inline-formula>, then we have the following result:</p><p><img src="htmlimages\8-7401970x\174e1505-8645-455b-9c06-16bcb9f33455.png" /></p><p>The unique solution of above equation is <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\90ed7235-af17-4b6c-a571-5c6d733e6a6a.png" xlink:type="simple"/></inline-formula></p><p>At<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\2c0ca26a-03ba-4dd4-bd2e-7b04ed7c1c3f.png" xlink:type="simple"/></inline-formula>, the TRC(T) has a global minimum on (0,<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\f0f96c7b-5122-4899-aad7-15af991af21f.png" xlink:type="simple"/></inline-formula>) since <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\3c6b6319-3715-49fb-8ad1-fe906d1a2044.png" xlink:type="simple"/></inline-formula> for all T &gt; 0.</p><p>Model 2: Using the expected present value of total relevant cost for random planning horizon C(T) to find the optimal solution.</p><p>We assume that the random planning horizon x is located on the (k + 1)th cycle time. The present value of total relevant cost at the first cycle time PVC(T) consists of the following elements:</p><p>The ordering cost = S.</p><p>The purchasing cost =<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\a8e1ffa4-4ad9-4f4f-8047-601ab5658ec3.png" xlink:type="simple"/></inline-formula>.</p><p>The stock holding cost of raw material =<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\733a77a2-320b-4bf1-b37e-990f44efa685.png" xlink:type="simple"/></inline-formula>.</p><p>The stock holding cost of products =<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\dd61f1f8-18ec-4882-9912-c0d66ab404f0.png" xlink:type="simple"/></inline-formula>.</p><p>The present value of total relevant cost at the first cycle time PVC(T) is as follows:</p><p><inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\36f2d0c9-0a4a-45b5-a6bd-2b3b2a485802.png" xlink:type="simple"/></inline-formula>.</p><p>The present value of total relevant cost from the beginning of the first cycle time to the beginning of the (k + 1)th cycle time is as follows:</p><disp-formula id="scirp.42170-formula144851"><label>(1)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\5875db28-921d-4e23-a501-b4a3cc89c748.png"  xlink:type="simple"/></disp-formula><p>The present value (at time of kT) of total relevant cost at the (k + 1)th cycle time consists of the following elements:</p><p>The purchasing cost =<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\ba81f1d4-8168-49d2-a825-9a7ff9ef4002.png" xlink:type="simple"/></inline-formula>.</p><p>The stock holding cost of raw material =<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\1963b1f2-3e28-4352-9158-84d8b26edd8e.png" xlink:type="simple"/></inline-formula>.</p><p>The stock holding cost of products =<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\936f5e7e-163c-44a5-be33-f158d23dbfb7.png" xlink:type="simple"/></inline-formula>.</p><p>The present value of total relevant cost at the (k + 1)th cycle time is as follows:</p><disp-formula id="scirp.42170-formula144852"><label>(2)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\c0d60562-b2e0-4f87-886b-518e61d3142f.png"  xlink:type="simple"/></disp-formula><p>Now, we want to find the expected present value of total relevant cost for random planning horizon C(T).</p><p>The present value of total relevant cost includes:</p><p>(A) The present value of total relevant cost from the beginning of the first cycle time to the beginning of the (k+1)th cycle time.</p><p>(B) The present value of total relevant cost at the (k + 1)th cycle time.</p><p>If we assume that the probability density function of x is<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\97851fa8-5301-48fe-9373-20cf8ddc29f2.png" xlink:type="simple"/></inline-formula>, the expected present value of the total relevant cost from the beginning to the time x is C(T).</p><disp-formula id="scirp.42170-formula144853"><label>(3)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\a774a9c2-9c7f-4e8d-9661-ccde08274044.png"  xlink:type="simple"/></disp-formula><p>Let <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\0b0e3be6-2ac7-4754-8c5b-a8e83e7b2633.png" xlink:type="simple"/></inline-formula> (4)</p><p>where</p><disp-formula id="scirp.42170-formula144854"><label>. (5)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\f328dd89-cd5e-4b4d-b5fd-739e5108908a.png"  xlink:type="simple"/></disp-formula><p>Then we obtain the first derivative of g(T) as follows:</p><disp-formula id="scirp.42170-formula144855"><label>(6)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\5b8b9638-290e-4896-a103-e61fdfe45e84.png"  xlink:type="simple"/></disp-formula><p>Since <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\a8e88b90-5662-4ad3-9770-48a9ab45e369.png" xlink:type="simple"/></inline-formula> and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\c1be8304-614c-4946-b718-122ae992f1f2.png" xlink:type="simple"/></inline-formula>, the equation <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\415a6b20-5930-4607-81ec-26c02b4d16a8.png" xlink:type="simple"/></inline-formula> has a unique solution.</p><p>From the calculations described above, we have proved that<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\934f00d2-05ad-4481-8cd1-a135c03a68a8.png" xlink:type="simple"/></inline-formula>. Thus, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\20067057-58ff-4a8e-b7f5-cf659e571992.png" xlink:type="simple"/></inline-formula>is a strictly increasing function. From the results above, we conclude that <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\26d598b0-dd98-4b04-b7f4-a6563652be78.png" xlink:type="simple"/></inline-formula> has a unique solution <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\e49ba0af-f0e5-42ad-af2f-283e7e018f49.png" xlink:type="simple"/></inline-formula> as shown in the following:</p><disp-formula id="scirp.42170-formula144856"><label>. (7)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\5a19dbcb-c600-430f-a3a8-d93e96fd8b99.png"  xlink:type="simple"/></disp-formula><p>Since<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\d4099894-e7ba-41a0-92e1-2ed8cd7b678e.png" xlink:type="simple"/></inline-formula>, it is therefore implied that there is a unique solution <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\10c8da50-10b7-45b5-93de-e004b4b98575.png" xlink:type="simple"/></inline-formula> as shown in the following:</p><disp-formula id="scirp.42170-formula144857"><label>. (8)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\567261c7-561d-4a4a-99e7-5069ddc58666.png"  xlink:type="simple"/></disp-formula><p>It is not easy to solve <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\bea9d871-6d67-4eab-9e29-ff5f704a4f2d.png" xlink:type="simple"/></inline-formula> out in this case. Our first step is to find an upper and an lower bound of<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\8a19f885-f995-4b51-af75-6bad88628b30.png" xlink:type="simple"/></inline-formula>. Then, we used the Intermediate Value Theorem and the algorithm of bisection method to find the optimal cycle time. We will now show the procedure of finding a lower bound <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\69bfe4c2-1449-4c04-aa1e-4233ef5accb8.png" xlink:type="simple"/></inline-formula> and an upper bound<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\75fb5f92-e30f-4d14-bee9-3e682739fc1c.png" xlink:type="simple"/></inline-formula>.</p><p>We set the lower bound as<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\1144e919-4d51-4167-9686-92a7d1812ce3.png" xlink:type="simple"/></inline-formula>, and then we found an upper bound of<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\a506afd3-2e45-4c6e-8676-5088825079ad.png" xlink:type="simple"/></inline-formula>:</p><disp-formula id="scirp.42170-formula144858"><label>. (9)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\0765e55a-9c4b-4b9e-860c-e77a68b74a9c.png"  xlink:type="simple"/></disp-formula><p>Setting</p><disp-formula id="scirp.42170-formula144859"><label>, (10)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\6c824298-3849-4bef-8fda-477a0c7deb7d.png"  xlink:type="simple"/></disp-formula><disp-formula id="scirp.42170-formula144860"><label>, (11)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\2d3fd09e-34d5-46c0-9f2c-7eac4a9a84e8.png"  xlink:type="simple"/></disp-formula><disp-formula id="scirp.42170-formula144861"><label>. (12)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\78b318d7-0be8-488e-9b6e-9a729bd9a68d.png"  xlink:type="simple"/></disp-formula><p>And from the above, we can now simplify the inequality to</p><disp-formula id="scirp.42170-formula144862"><label>, (13)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\77fb0788-1fc8-42e7-bea1-fa08b1e9b4c7.png"  xlink:type="simple"/></disp-formula><p>The positive solution of <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\485df579-f1f8-4ba9-9221-ef9f38deaca7.png" xlink:type="simple"/></inline-formula> is</p><disp-formula id="scirp.42170-formula144863"><label>(14)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\0e38dc9f-a298-4899-909b-e5fe55f38b7e.png"  xlink:type="simple"/></disp-formula><p>Theorem: <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\0ff81d70-3d5b-4ef8-80c5-ab849c386570.png" xlink:type="simple"/></inline-formula></p><p>Proof: Since <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\c43f51d1-31b1-4ec7-b53c-adbab628b5fc.png" xlink:type="simple"/></inline-formula> is a strictly increasing function, we conclude that <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\a3608861-3217-417d-8231-688ab84b6688.png" xlink:type="simple"/></inline-formula> for all T and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\2ecbd7d6-b5b3-42ce-babe-d2d3a565c363.png" xlink:type="simple"/></inline-formula>.</p><p>By Equation (97),</p><disp-formula id="scirp.42170-formula144864"><label>(15)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\da71205c-ec25-49ec-9cd5-debc2f5d9ddf.png"  xlink:type="simple"/></disp-formula><disp-formula id="scirp.42170-formula144865"><label>(16)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\65315c39-a30e-4c8e-9030-a6add24316fa.png"  xlink:type="simple"/></disp-formula><disp-formula id="scirp.42170-formula144866"><label>(17)</label><graphic position="anchor" xlink:href="htmlimages\8-7401970x\fd93393f-8793-4c1b-ab30-18c68496ae71.png"  xlink:type="simple"/></disp-formula><p>From the above,<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\e4f5e3c2-4efb-49f0-b338-e824c115d29d.png" xlink:type="simple"/></inline-formula> (18)</p><p>Now we can compute the exact optimal cycle length <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\935454cb-3fbf-4e1e-aef9-e6002c8db5b3.png" xlink:type="simple"/></inline-formula> of model 2 by using the logic of the following algorithm. Our method is similar to the one of Chung and Lin (see [<xref ref-type="bibr" rid="scirp.42170-ref15">15</xref>]).</p><p>Step 1: Let <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\7cff3282-5129-4d52-bc3d-926f3a88865f.png" xlink:type="simple"/></inline-formula></p><p>Step 2: Set <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\d4ce5d2c-2720-45de-9514-77916db9dd78.png" xlink:type="simple"/></inline-formula>and <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\9513510d-dab1-4b93-90a1-35faaf5f2a05.png" xlink:type="simple"/></inline-formula></p><p>where<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\4451df96-d3cc-43e9-b250-75c4a5d6a8ac.png" xlink:type="simple"/></inline-formula>,</p><p><img src="htmlimages\8-7401970x\f675725e-d1ff-421f-8d4c-e41a90e6ae80.png" /></p><p><inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\8c2336fc-2f5e-4512-8196-952c86750b12.png" xlink:type="simple"/></inline-formula>.</p><p>Step 3: Set <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\1dc6e9b4-8c46-4427-87ed-987e12453d4a.png" xlink:type="simple"/></inline-formula></p><p>Step 4: If<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\1a6265b5-be39-41de-a1d8-182b88d6e127.png" xlink:type="simple"/></inline-formula>, go to Step 6.</p><p>Otherwise go to Step 5.</p><p>Step 5: If<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\e75406ba-0ba4-4927-aa4e-3f7ded6336b2.png" xlink:type="simple"/></inline-formula>, then we set<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\d64dd92d-61c8-491f-ab3f-f958f73f19d2.png" xlink:type="simple"/></inline-formula>.</p><p>And if<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\67794434-0cb3-4bf2-b7e5-73ac035e707e.png" xlink:type="simple"/></inline-formula>, then we set<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\635c0333-4e73-41b7-80df-d76ba92eb561.png" xlink:type="simple"/></inline-formula>.</p><p>Then go to Step 3.</p><p>Step 6:<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\0697531f-1714-4f85-8cf4-2fb9c45030b8.png" xlink:type="simple"/></inline-formula>Where <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\80e5a338-0624-48b2-a5ca-497683679c4a.png" xlink:type="simple"/></inline-formula> is the optimal cycle length.</p></sec><sec id="s3"><title>3. Numerical Examples</title><p>Example 1: If we set the numbers as S = 1000, P = 2000, D = 1500, c = 10, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\41718e1b-ef47-4f82-9be7-f4b76cfa4974.png" xlink:type="simple"/></inline-formula>, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\6a7d1da4-549a-47a4-848f-46fb2aaf31a5.png" xlink:type="simple"/></inline-formula>, r = 0.15, h = 2 and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\b5e45223-3cb6-4d27-9c74-f070979f9cd8.png" xlink:type="simple"/></inline-formula>, then we have <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\b89c6a1c-5696-4ce2-88ee-339b03002214.png" xlink:type="simple"/></inline-formula> and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\7e6f6f68-9437-47eb-bdca-50236808abc9.png" xlink:type="simple"/></inline-formula>.</p><p>Example 2: If we set the numbers as S = 2000, P = 2000, D = 1500, c = 10, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\11f9722f-4fb2-4195-b2b8-1e7f8b436809.png" xlink:type="simple"/></inline-formula>, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\ad774324-804a-4c7d-996b-62964e27bfea.png" xlink:type="simple"/></inline-formula>, r = 0.15, h = 2 and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\6485e021-6635-44b0-8d0a-9800cc0836f0.png" xlink:type="simple"/></inline-formula>, then we have <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\ad1571b5-aa2c-4c52-9d91-abcdd1b8103f.png" xlink:type="simple"/></inline-formula> and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\ed30e861-fb1f-408f-a9e0-727c59b2f2ea.png" xlink:type="simple"/></inline-formula>.</p><p>Example 3: If we set the numbers as S = 1000, P = 2000, D = 1500, c = 20, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\28d8f024-4674-4a58-ab35-558c39780e61.png" xlink:type="simple"/></inline-formula>, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\ce93fa91-345f-40c0-ae84-965018cad0da.png" xlink:type="simple"/></inline-formula>, r = 0.15, h = 2 and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\84d9edb8-c434-46cc-b493-5b0f272cedda.png" xlink:type="simple"/></inline-formula>, then we have <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\98908dfc-ae1d-4b3f-b54e-6329fe98716a.png" xlink:type="simple"/></inline-formula> and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\4c504d78-e15b-4e74-ba20-7847bfba92d7.png" xlink:type="simple"/></inline-formula>.</p><p>Example 4: If we set the numbers as S = 1000, P = 2000, D = 1500, c = 10, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\d6ef6551-63f6-4952-96bd-cf75b8917222.png" xlink:type="simple"/></inline-formula>, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\73c7454a-4484-4ce1-875c-1218cc09de03.png" xlink:type="simple"/></inline-formula>, r = 0.15, h = 2 and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\84f8c8e9-cb07-493c-9a1f-0988a99e89c4.png" xlink:type="simple"/></inline-formula>, then we have <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\1a887952-75ef-4dd7-81cf-071e07b5dcc9.png" xlink:type="simple"/></inline-formula> and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\17b24fc9-91f5-42ff-b640-cb4d13961dd4.png" xlink:type="simple"/></inline-formula>.</p><p>Example 5: If we set the numbers as S = 1000, P = 2000, D = 1500, c = 10, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\47f07278-321f-4ae3-b8ae-f47eadcd6c01.png" xlink:type="simple"/></inline-formula>, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\683f74ca-6bd6-4a04-b18d-746636bd9b77.png" xlink:type="simple"/></inline-formula>, r = 0.1, h = 2 and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\d543cf81-437e-4e72-8d3f-ed9b7a0fc494.png" xlink:type="simple"/></inline-formula>, then we have <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\8ca5f966-3bd1-4941-b0f1-1f955d8c19d5.png" xlink:type="simple"/></inline-formula> and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\a122691d-308d-488f-8b5a-e4a544173c1b.png" xlink:type="simple"/></inline-formula>.</p><p>Example 6: If we set the numbers as S = 1000, P = 2000, D = 1500, c = 10, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\5589257d-b2f6-44a7-a987-986122867581.png" xlink:type="simple"/></inline-formula>, <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\827b15c3-72a8-4d26-81a2-c5f4aeff1e95.png" xlink:type="simple"/></inline-formula>, r = 0.15, h = 2 and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\ac3c1d74-9500-4a44-a262-abff7ce63b0f.png" xlink:type="simple"/></inline-formula>, then we have <inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\7de23e1f-f02e-4644-9901-ae5512a6c1df.png" xlink:type="simple"/></inline-formula> and<inline-formula><inline-graphic xlink:href="tmlimages\8-7401970x\723585b0-25a7-44e4-ba2c-d905aa0c9279.png" xlink:type="simple"/></inline-formula>.</p></sec><sec id="s4"><title>4. Conclusion</title><p>The planning horizon is random and has an exponential distribution with a parameter that influences the optimal cycle time. The raw material is used up and the products are sold out at the end of the planning horizon. Thus, the results are more economical in this model. The use of reusable raw material is beneficial and worthwhile. The models in this research can be applied in companies that use reusable or new raw material. There are many interesting results derived from the numerical examples that future research should take into consideration. It is not necessary for the stocking holding cost of raw material and the stock holding cost of products to be the same.</p></sec><sec id="s5"><title>REFERENCES</title></sec><sec id="s6"><title>NOTES</title></sec></body><back><ref-list><title>References</title><ref id="scirp.42170-ref1"><label>1</label><mixed-citation publication-type="other" xlink:type="simple">K. Richter, “The Extended EOQ Repair and Wasted Disposal Model with Variable Setup Number,” European Journal of Operational Research, Vol. 96, No. 2, 1996, pp. 313-324. http://dx.doi.org/10.1016/0377-2217(95)00276-6</mixed-citation></ref><ref id="scirp.42170-ref2"><label>2</label><mixed-citation publication-type="other" xlink:type="simple">K. Richter and I. Dobos, “Analysis of the EOQ Repair and Waste Disposal Problem with Integer Setup Numbers,” International Journal of Production Economics, Vol. 45, No. 1-3, 1999, pp. 443-447. http://dx.doi.org/10.1016/0925-5273(95)00143-3</mixed-citation></ref><ref id="scirp.42170-ref3"><label>3</label><mixed-citation publication-type="other" xlink:type="simple">K. Richter, “The Extended EOQ Repair and Wasted Disposal Model,” International Journal of Production Economics, Vol. 59, No. 1-3, 1996, pp. 463-467. http://dx.doi.org/10.1016/S0925-5273(98)00110-8</mixed-citation></ref><ref id="scirp.42170-ref4"><label>4</label><mixed-citation publication-type="other" xlink:type="simple">R. Kleber, S. Ninner and G. P. Kies Muller, “A Continuous Time Inventory Model for a Product Recovery System with Multiple Options,” International Journal of Production Economics, Vol. 79, No. 2, 2002, pp. 121-141. http://dx.doi.org/10.1016/S0925-5273(02)00256-6</mixed-citation></ref><ref id="scirp.42170-ref5"><label>5</label><mixed-citation publication-type="other" xlink:type="simple">S. G. Koh, H. Hwang and C. S. 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