Equipment replacement problem - Operations Research Stack Exchange most recent 30 from or.stackexchange.com 2022-01-20T19:35:49Z https://or.stackexchange.com/feeds/question/3008 https://creativecommons.org/licenses/by-sa/4.0/rdf https://or.stackexchange.com/q/3008 9 Equipment replacement problem Slim Shady https://or.stackexchange.com/users/1117 2019-11-09T12:08:46Z 2019-11-09T16:00:20Z <p>I have a question on the Equipment Replacement Problem, where the following is taken (with some syntactic modifications) from <a href="https://www.coursehero.com/file/9386204/DP1-Shortest-Path-Alpern/" rel="noreferrer">IB2070 Mathematical Programming II (MP2)</a>, Warwick Business School.</p> <blockquote> <p><strong>Equipment Replacement Problem</strong> <span class="math-container">\begin{array}{cc}\hline\text{Age of Car}&amp;\text{Annual Maintenance}&amp;\text{Trade-in Price at the}\\&amp;\text{Cost}&amp;\text{end of the period}\\\hline0&amp;{\it\unicode{xA3}}2,000&amp;{\it\unicode{xA3}}7,000\\1&amp;{\it\unicode{xA3}}4,000&amp;{\it\unicode{xA3}}6,000\\2&amp;{\it\unicode{xA3}}5,000&amp;{\it\unicode{xA3}}2,000\\3&amp;{\it\unicode{xA3}}9,000&amp;{\it\unicode{xA3}}1,000\\4&amp;12,000&amp;0\\\hline\end{array}</span> When should I trade in my car?</p> <p><strong>Arc Length</strong> <span class="math-container">\begin{align}c_{i,j}=&amp;\,\,\text{maintenance cost during years}\,\,i,i+1,\cdots,j-1\\&amp;+\text{cost of purchasing a new car at year}\,\,i\\&amp;-\text{trade-in value received at year}\,\,j\\\\c_{1,2}=&amp;\,\,{\it\unicode{xA3}}2000+{\it\unicode{xA3}}12000-{\it\unicode{xA3}}7000={\it\unicode{xA3}}7000\\\\c_{2,3}=&amp;\,\,{\it\unicode{xA3}}2000+{\it\unicode{xA3}}4000+{\it\unicode{xA3}}12000-{\it\unicode{xA3}}6000={\it\unicode{xA3}}12000\end{align}</span></p> <p><strong>Shortest Path Problem</strong> <span class="math-container">\begin{array}{|c|c|}\hline c_{i,j}&amp;1&amp;2&amp;3&amp;4&amp;5&amp;6\\\hline1&amp;\phantom{25pt}&amp;7,000&amp;12,000&amp;21,000&amp;31,000&amp;44,000\\\hline2&amp;&amp;&amp;7,000&amp;12,000&amp;21,000&amp;31,000\\\hline3&amp;&amp;&amp;&amp;7,000&amp;12,000&amp;21,000\\\hline4&amp;&amp;&amp;&amp;&amp;7,000&amp;12,000\\\hline5&amp;&amp;&amp;&amp;&amp;&amp;7,000\\\hline\end{array}</span></p> </blockquote> <p>Could somebody please explain how in the last diagram, for <span class="math-container">$c_{2,3}$</span> they get <span class="math-container">$7,000$</span>? I only understand how they get <span class="math-container">$c_{1,i}$</span> for <span class="math-container">$i=1,\ldots,6$</span> but that's all I can understand. If someone explains to me how they get it for <span class="math-container">$c_{2,3}$</span> then I'll be able to understand the whole of it.</p> https://or.stackexchange.com/questions/3008/-/3009#3009 8 Answer by RobPratt for Equipment replacement problem RobPratt https://or.stackexchange.com/users/500 2019-11-09T14:20:49Z 2019-11-09T14:20:49Z <p>It is the same calculation as <span class="math-container">$c_{1,2}$</span>. The purchase price is £12,000 no matter when that purchase is made, and the rest of the costs depend on age of the car (number of years since purchase), not the actual year. So each row in the final table is the same as the previous row shifted one place to the right.</p> https://or.stackexchange.com/questions/3008/-/3010#3010 5 Answer by TheSimpliFire for Equipment replacement problem TheSimpliFire https://or.stackexchange.com/users/123 2019-11-09T14:50:59Z 2019-11-09T14:50:59Z <p>The constant car price means that <span class="math-container">$$c_{i,j}=12\,000-t_j+\sum_i^{j-1}m_i$$</span> where <span class="math-container">$t_j$</span> denotes the trade-in price on year <span class="math-container">$j$</span> and <span class="math-container">$m_i$</span> the maintenance cost on year <span class="math-container">$i$</span>. Thus <span class="math-container">$$c_{i,i+1}=12\,000-t_{i+1}+m_i\tag1.$$</span> However, all indices can be shifted up or down (as @RobPratt mentions) as the age of the car from year <span class="math-container">$i$</span> to year <span class="math-container">$i+1$</span> is constant.</p> <p>In particular, shifting <span class="math-container">$(1)$</span> down by <span class="math-container">$i-1$</span> and <span class="math-container">$i-2$</span> yields <span class="math-container">$c_{1,2}=c_{2,3}$</span> respectively.</p> https://or.stackexchange.com/questions/3008/-/3011#3011 4 Answer by Oguz Toragay for Equipment replacement problem Oguz Toragay https://or.stackexchange.com/users/39 2019-11-09T15:05:59Z 2019-11-09T15:11:31Z <p>In addition to @Rob Pratt's answer, consider the following table which helps you understand the shifts in each row of the matrix in the question:</p> <p><span class="math-container">\begin{array}{cccccc}\hline i &amp; j &amp; j-i &amp; \rm{purchase} &amp; \text{trade-in} &amp; \rm{maintenance} &amp; \rm{total}\\ \hline 1 &amp; 2 &amp; 1 &amp; 12000 &amp; 7000 &amp; 2000 &amp; 7000 \\ 2 &amp; 3 &amp; 1 &amp; 12000 &amp; 7000 &amp; 2000 &amp; 7000 \\ 3 &amp; 4 &amp; 1 &amp; 12000 &amp; 7000 &amp; 2000 &amp; 7000 \\ 4 &amp; 5 &amp; 1 &amp; 12000 &amp; 7000 &amp; 2000 &amp; 7000 \\ 5 &amp; 6 &amp; 1 &amp; 12000 &amp; 7000 &amp; 2000 &amp; 7000 \\ \hline \end{array}</span></p> <p>This table summarizes the calculations for <span class="math-container">$c_{12},c_{23},c_{34},c_{45},c_{56}$</span> in each of its rows. All the costs are the same because the age or the value of <span class="math-container">$j-i$</span> is the same for all rows. Three other tables like the one I mentioned, this time with <span class="math-container">$j-i=2,3,4$</span> can be constructed to cover whole cost table in the question. This problem can also be solved by using <em>Dynamic Programming</em>. </p>