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I have a question on the Equipment Replacement Problem, where the following is taken (with some syntactic modifications) from IB2070 Mathematical Programming II (MP2), Warwick Business School.

Equipment Replacement Problem $$\begin{array}{cc}\hline\text{Age of Car}&\text{Annual Maintenance}&\text{Trade-in Price at the}\\&\text{Cost}&\text{end of the period}\\\hline0&{\it\unicode{xA3}}2,000&{\it\unicode{xA3}}7,000\\1&{\it\unicode{xA3}}4,000&{\it\unicode{xA3}}6,000\\2&{\it\unicode{xA3}}5,000&{\it\unicode{xA3}}2,000\\3&{\it\unicode{xA3}}9,000&{\it\unicode{xA3}}1,000\\4&12,000&0\\\hline\end{array}$$ When should I trade in my car?

Arc Length \begin{align}c_{i,j}=&\,\,\text{maintenance cost during years}\,\,i,i+1,\cdots,j-1\\&+\text{cost of purchasing a new car at year}\,\,i\\&-\text{trade-in value received at year}\,\,j\\\\c_{1,2}=&\,\,{\it\unicode{xA3}}2000+{\it\unicode{xA3}}12000-{\it\unicode{xA3}}7000={\it\unicode{xA3}}7000\\\\c_{2,3}=&\,\,{\it\unicode{xA3}}2000+{\it\unicode{xA3}}4000+{\it\unicode{xA3}}12000-{\it\unicode{xA3}}6000={\it\unicode{xA3}}12000\end{align}

Shortest Path Problem $$\begin{array}{|c|c|}\hline c_{i,j}&1&2&3&4&5&6\\\hline1&\phantom{25pt}&7,000&12,000&21,000&31,000&44,000\\\hline2&&&7,000&12,000&21,000&31,000\\\hline3&&&&7,000&12,000&21,000\\\hline4&&&&&7,000&12,000\\\hline5&&&&&&7,000\\\hline\end{array}$$

Could somebody please explain how in the last diagram, for $$c_{2,3}$$ they get $$7,000$$? I only understand how they get $$c_{1,i}$$ for $$i=1,\ldots,6$$ but that's all I can understand. If someone explains to me how they get it for $$c_{2,3}$$ then I'll be able to understand the whole of it.

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TheSimpliFire
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I have a question on the Equipment Replacement Problem, where the following is taken (with some syntactic modifications) from IB2070 Mathematical Programming II (MP2), Warwick Business School.

Equipment Replacement Problem $$\begin{array}{cc}\hline\text{Age of Car}&\text{Annual Maintenance}&\text{Trade-in Price at the}\\&\text{Cost}&\text{end of the period}\\\hline0&{\it\unicode{xA3}}2,000&{\it\unicode{xA3}}7,000\\1&{\it\unicode{xA3}}4,000&{\it\unicode{xA3}}6,000\\2&{\it\unicode{xA3}}5,000&{\it\unicode{xA3}}2,000\\3&{\it\unicode{xA3}}9,000&{\it\unicode{xA3}}1,000\\4&12,000&0\\\hline\end{array}$$ When should I trade in my car?

Arc Length \begin{align}c_{i,j}=&\,\,\text{maintenance cost during years}\,\,i,i+1,\cdots,j-1\\&+\text{cost of purchasing a new car at year}\,\,i\\&-\text{trade-in value received at year}\,\,j\\\\c_{1,2}=&\,\,{\it\unicode{xA3}}2000+{\it\unicode{xA3}}12000-{\it\unicode{xA3}}7000={\it\unicode{xA3}}7000\\\\c_{2,3}=&\,\,{\it\unicode{xA3}}2000+{\it\unicode{xA3}}4000+{\it\unicode{xA3}}12000-{\it\unicode{xA3}}6000={\it\unicode{xA3}}12000\end{align}

Shortest Path Problem $$\begin{array}{|c|c|}\hline c_{ij}&1&2&3&4&5&6\\\hline1&\phantom{25pt}&7,000&12,000&21,000&31,000&44,000\\\hline2&&&7,000&12,000&21,000&31,000\\\hline3&&&&7,000&12,000&21,000\\\hline4&&&&&7,000&12,000\\\hline5&&&&&&7,000\\\hline\end{array}$$$$\begin{array}{|c|c|}\hline c_{i,j}&1&2&3&4&5&6\\\hline1&\phantom{25pt}&7,000&12,000&21,000&31,000&44,000\\\hline2&&&7,000&12,000&21,000&31,000\\\hline3&&&&7,000&12,000&21,000\\\hline4&&&&&7,000&12,000\\\hline5&&&&&&7,000\\\hline\end{array}$$

Could somebody please explain how in the last diagram, for $$c_{2,3}$$ they get $$7,000$$? I only understand how they get $$c_{1,i}$$ for $$i=1,\ldots,6$$ but that's all I can understand. If someone explains to me how they get it for $$c_{2,3}$$ then I'll be able to understand the whole of it.

Improved TeX formatting
TheSimpliFire
• 4.8k
• 4
• 16
• 52

I have a question on the Equipment Replacement Problem, where the following is taken (with some syntactic modifications) from IB2070 Mathematical Programming II (MP2), Warwick Business School.

Equipment Replacement Problem $$\begin{array}{cc}\hline\text{Age of Car}&\text{Annual Maintenance}&\text{Trade-in Price at the}\\&\text{Cost}&\text{end of the period}\\\hline0&{\it\unicode{xA3}}2,000&{\it\unicode{xA3}}7,000\\1&{\it\unicode{xA3}}4,000&{\it\unicode{xA3}}6,000\\2&{\it\unicode{xA3}}5,000&{\it\unicode{xA3}}2,000\\3&{\it\unicode{xA3}}9,000&{\it\unicode{xA3}}1,000\\4&12,000&0\\\hline\end{array}$$ When should I trade in my car?

Arc Length \begin{align}c_{i,j}=&\,\,\text{maintenance cost during years}\,\,i,i+1,\cdots,j-1\\&+\text{cost of purchasing a new car at year}\,\,i\\&-\text{trade-in value received at year}\,\,j\\\\c_{1,2}=&\,\,{\it\unicode{xA3}}2000+{\it\unicode{xA3}}12000-{\it\unicode{xA3}}7000={\it\unicode{xA3}}7000\\\\c_{2,3}=&\,\,{\it\unicode{xA3}}2000+{\it\unicode{xA3}}4000+{\it\unicode{xA3}}12000-{\it\unicode{xA3}}6000={\it\unicode{xA3}}12000\end{align}

Shortest Path Problem $$\begin{array}{|c|c|}\hline c_{ij}&1&2&3&4&5&6\\\hline1&\phantom{25pt}&7,000&12,000&21,000&31,000&44,000\\\hline2&&&7,000&12,000&21,000&31,000\\\hline3&&&&7,000&12,000&21,000\\\hline4&&&&&7,000&12,000\\\hline5&&&&&&7,000\\\hline\end{array}$$

Could somebody please explain how in the last diagram, for $$c_{2,3}$$ they get $$7,000$$  ? I only understand how they get $$c_{1,i}$$ for $$i=1,..,6$$$$i=1,\ldots,6$$ but that's all I can understand. If someone explains to me how they get it for $$c_{2,3}$$ then i'llI'll be able to understand the whole of it.