Exercise 1.15

Question

A company produces two kinds of products. A product of the first type requires $1/4$ hours of assembly labor, $1/8$ hours of testing, and \$1.2 worth of raw materials. A product of the second type requires $1/3$ hours of assembly, $1/3$ hours of testing, and 0.9 worth of raw materials. Given the current personnel of the company, there can be at most 90 hours of assembly labor and 80 hours of testing, each day. Products of the first and second type have a market value of \$9 and \$8, respectively.
\begin{enumerate}[label=(\alph*)]
\item Formulate a linear programming problem that can be used to maximize the
daily profit of the company.
\item Consider the following two modifications to the original problem:
   \begin{enumerate}[label=(\roman*)]
   \item Suppose that up to 50 hours of overtime assembly labor can be sched­uled, at a cost of \$7 per hour.
   \item Suppose that the raw material supplier provides a 10\% discount if the daily bill is above \$300.
   \end{enumerate}
Which of the above two elements can be easily incorporated into the lin­ear programming formulation and how? If one or both are not easy to
incorporate, indicate how you might nevertheless solve the problem.
\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(\alph*)]
\item We model the quantity of the first product by $x_{1}$ and the second by $x_{2}$, both nonnegative. Then the hours required to produce both products can be modelled by $\frac{1}{4} x_{1} +\frac{1}{3} x_{2}$ for assembly and $\frac{1}{8} x_{1} +\frac{1}{3} x_{2}$ for testing. The profit per unit of product one is its price reduced by the costs of its production: $\$9 - \$1.2 = \$7.8$; similarly, the profit per unit of the second product is $\$8 - \$0.9 = \$7.1$. This results in the following linear optimization problem:

\begin{equation}
\begin{array}{lll}
\text{maximize} \quad & 7.8 x_{1} + 7.1 x_{2}\
\text{subject to} \quad
&\dfrac{1}{4} x_{1} +\dfrac{1}{3} x_{2} \leq 90\[5pt]
&\dfrac{1}{8} x_{1} +\dfrac{1}{3} x_{2} \leq 80\
& x_{1}, x_{2} \geq 0.
\end{array}
\label{optim1}
\end{equation}

\item Consider the following two modifications to the original problem:
   \begin{enumerate}[label=(\roman*)]
   \item To accommodate for an additional type of constraint (paid assembly hours), we introduce a new variable $y$ to the original problem \eqref{optim1}:

\begin{equation}
\begin{array}{lll}
\text{maximize} \quad & 7.8 x_{1} + 7.1 x_{2} - 7y\
\text{subject to} \quad
&\dfrac{1}{4} x_{1} +\dfrac{1}{3} x_{2} - y \leq 90\[5pt]
&\dfrac{1}{8} x_{1} +\dfrac{1}{3} x_{2} \leq 80\
& y \leq 50\
& x_{1}, x_{2} \geq 0.
\end{array}
\label{optim2}
\end{equation}

\item It is not possible to incorporate this condition without significant restructuring. There is, however, a workaround: we can formulate two separate linear optimization problems to accommodate for two cases when this discount is taken advantage of and when not. The final answer is then the best result among these two cases. \newline

To do so, modify the linear program \eqref{optim1} to include the restriction $1.2x + 0.9y \leq 300$. Under the assumption that the discount is not employed, an optimal solution to this modified LP yields the highest return.\newline

For the second case, introduce the constraint $1.2x + 0.9y \geq 300$ to \eqref{optim1} to obtain the alternative optimization problem. This condition guarantees that the discount is taken advantage of. One also must modify the objective function to reflect the discount:
$$(9 - 0.9 \cdot 1.2)x + (8 - 0.9 \cdot 0.9)y.$$

Now solve both linear optimization problems and maximize both optimal profits.

\end{enumerate}
\end{enumerate}

Exercise 1.15

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Exercise 1.15

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