数学代写|IMSE881 Linear Programming

A steel company must decide how to allocate next week’s time on a rolling mill, which is a machine that takes unfinished slabs of steel as input and can produce either of two semi-finished products: bands and coils. The mill’s two products come off the rolling line at different rates:
$$
\begin{array}{ll}
\text { Bands } & 200 \text { tons } / \mathrm{hr} \
\text { Coils } & 140 \text { tons } / \mathrm{hr} .
\end{array}
$$
They also produce different profits:
$$
\begin{array}{ll}
\text { Bands } \quad \$ 25 / \text { ton } \
\text { Coils } \quad \$ 30 / \text { ton } .
\end{array}
$$
Based on currently booked orders, the following upper bounds are placed on the amount of each product to produce:
$$
\begin{array}{ll}
\text { Bands } & 6000 \text { tons } \
\text { Coils } & 4000 \text { tons . }
\end{array}
$$
Given that there are 40 hours of production time available this week, the problem is to decide how many tons of bands and how many tons of coils should be produced to yield the greatest profit. Formulate this problem as a linear programming problem. Can you solve this problem by inspection?

Suppose that $Y$ is a random variable taking on one of $n$ known values:
$$
a_1, a_2, \ldots, a_n .
$$
Suppose we know that $Y$ either has distribution $p$ given by
$$
\mathbb{P}\left(Y=a_j\right)=p_j
$$
or it has distribution $q$ given by
$$
\mathbb{P}\left(Y=a_j\right)=q_j .
$$
Of course, the numbers $p_j, j=1,2, \ldots, n$ are nonnegative and sum to one. The same is true for the $q_j$ ‘s. Based on a single observation of $Y$, we wish to guess whether it has distribution $p$ or distribution $q$. That is, for each possible outcome $a_j$, we will assert with probability $x_j$ that the distribution is $p$ and with probability $1-x_j$ that the distribution is $q$. We wish to determine the probabilities $x_j, j=1,2, \ldots, n$, such that the probability of saying the distribution is $p$ when in fact it is $q$ has probability no larger than $\beta$, where $\beta$ is some small positive value (such as 0.05 ). Furthermore, given this constraint, we wish to maximize the probability that we say the distribution is $p$ when in fact it is $p$. Formulate this maximization problem as a linear programming problem.

\begin{aligned}
\operatorname{maximize} & 6 x_1+8 x_2+5 x_3+9 x_4 \
\text { subject to } 2 x_1+x_2+x_3+3 x_4 & \leq 5 \
x_1+3 x_2+x_3+2 x_4 & \leq 3 \
x_1, x_2, x_3, x_4 & \geq 0
\end{aligned}

\begin{array}{ll}
\operatorname{maximize} & 2 x_1+x_2 \
\text { subject to } & 2 x_1+x_2 \leq 4 \
2 x_1+3 x_2 \leq 3 \
4 x_1+x_2 \leq 5 \
x_1+5 x_2 & \leq 1 \
x_1, x_2 & \geq 0
\end{array}

\begin{array}{ll}
\operatorname{maximize} & -x_1-3 x_2-x_3 \
\text { subject to } & 2 x_1-5 x_2+x_3 \leq-5 \
& 2 x_1-x_2+2 x_3 \leq 4 \
& x_1, x_2, x_3 \geq 0 .
\end{array}