Machines A, B, C, and D have been in use for several years, while machine E is new. The following table provides data on the value of production of each job when performed on a specific machine.
a. Determine the set of assignments that maximizes production value.
b. What is the total production value of your assignments?
c. Which machine should be retired (i.e., gets no assignment)?
d. If they do retire one machine, will they be as profitable without it as with it? Explain. $12729283040230323134463332529263742931242828\begin{array} { | l | l | l | l | l | l | } \hline & { \text { Machine } } \\\hline \text { Job } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \\1 & 27 & 29 & 28 & 30 & 40 \\ 2 & 30 & 32 & 31 & 34 & 46 \\3 & 33 & 25 & 29 & 26 & 37 \\4 & 29 & 31 & 24 & 28 & 28\\\hline\end{array}$

Question

Machines A, B, C, and D have been in use for several years, while machine E is new. The following table provides data on the value of production of each job when performed on a specific machine.
a. Determine the set of assignments that maximizes production value.
b. What is the total production value of your assignments?
c. Which machine should be retired (i.e., gets no assignment)?
d. If they do retire one machine, will they be as profitable without it as with it? Explain.

12729283040230323134463332529263742931242828\begin{array} { | l | l | l | l | l | l | } \hline & { \text { Machine } } \\\hline \text { Job } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \\1 & 27 & 29 & 28 & 30 & 40 \\ 2 & 30 & 32 & 31 & 34 & 46 \\3 & 33 & 25 & 29 & 26 & 37 \\4 & 29 & 31 & 24 & 28 & 28\\\hline\end{array}

Abigail Flores · Accepted Answer

Final Answer :  NOTE: students need to create an opportunity loss table with a dummy Job row. (a) The optimal assignment is Job 1→D, Job 2→E, Job 3→A, and Job 4→B. (b) The cost of this assignment is 140. (c) C gets no assignment, and should be the machine retired.   A ‾ B ‾ C ‾ D ‾ E ‾  Row Total   Job 1  0 0 0 1 0 1  Job 2  0 0 0 0 1 1  Job 3  1 0 0 0 0 1  Job 4  0 1 0 0 0 1  Job 5  0 ‾ 0 ‾ 1 ‾ 0 ‾ 0 ‾ 1 ‾  Column   Total  1 1 1 1 1 5  Total Cost  140 \begin{array} { | l | c | c | c | c | c | c | } \hline  &  \underline { \mathrm { A } }  &  \underline { \mathrm { B } }  &  \underline { \mathrm { C } }  &  \underline { \mathrm { D } }  &   { \underline { E } }  &  	ext { Row Total } \\hline 	ext { Job 1 }  &  0  &  0  &  0  &  1  &  0  &  1 \\hline 	ext { Job 2 }  &  0  &  0  &  0  &  0  &  1  &  1 \\hline 	ext { Job 3 }  &  1  &  0  &  0  &  0  &  0  &  1 \\hline 	ext { Job 4 }  &  0  &  1  &  0  &  0  &  0  &  1 \\hline 	ext { Job 5 }  &  \underline { 0 }  &  \underline { 0 }  &  \underline { 1 }  &  \underline { 0 }  &  \underline { 0 }  &  \underline { 1 } \\hline 	ext { Column }  &   &   &   &   &   &  \	ext { Total }  &  1  &  1  &  1  &  1  &  1  &  5 \\hline 	ext { Total Cost }  &  140  &   &   &   &   &  \\hline  &   &   &   &   &   &  \\hline\end{array}  Job 1   Job 2   Job 3   Job 4   Job 5   Column   Total   Total Cost  ​ A ​ 0 0 1 0 0 ​ 1 140 ​ B ​ 0 0 0 1 0 ​ 1 ​ C ​ 0 0 0 0 1 ​ 1 ​ D ​ 1 0 0 0 0 ​ 1 ​ E ​ 0 1 0 0 0 ​ 1 ​  Row Total  1 1 1 1 1 ​ 5 ​ ​   (d) Taking away what was arguably the least productive machine still reduces the options of the firm, and alters the opportunity costs. They cannot be better off without the machine.    Assignments   Shipments   A   B   D   E   Row   Total   Job 1  0 0 1 0 1  Job 2  0 0 0 1 1  Job 3  1 0 0 0 1  Job 4  0 1 0 0 1  Column   Total  1 1 1 1 4  Total Cost  140 \begin{array} { | l | c | c | c | c | c | } \hline  &   { 	ext { Assignments } } \\hline 	ext { Shipments }  &  	ext { A }  &  	ext { B }  &  	ext { D }  &  	ext { E }  &  \begin{array} { c } 	ext { Row } \	ext { Total }\end{array} \\hline 	ext { Job 1 }  &  0  &  0  &  1  &  0  &  1 \\hline 	ext { Job 2 }  &  0  &  0  &  0  &  1  &  1 \\hline 	ext { Job 3 }  &  1  &  0  &  0  &  0  &  1 \\hline 	ext { Job 4 }  &  0  &  1  &  0  &  0  &  1 \\hline 	ext { Column }  &   &   &   &   &  \\hline 	ext { Total }  &  1  &  1  &  1  &  1  &  4 \\hline 	ext { Total Cost }  &  140  &   &   &   &  \\hline\end{array}  Shipments   Job 1   Job 2   Job 3   Job 4   Column   Total   Total Cost  ​  Assignments   A  0 0 1 0 1 140 ​  B  0 0 0 1 1 ​  D  1 0 0 0 1 ​  E  0 1 0 0 1 ​  Row   Total  ​ 1 1 1 1 4 ​ ​

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