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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.

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.

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.

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 $i=1,\ldots,6$ 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.

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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enter image description here 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.

enter image description here

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?

enter image description here

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}

enter image description here

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.

enter image description here

enter image description here

enter image description here

enter image description here

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 $i=1,..,6$ but that's all I can understand. If someone explains me how they get it for $c_{2,3}$ then i'll be able to understand the whole of it

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 $i=1,\ldots,6$ 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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