Mat 540 Week 1 Math Paper

We simply represent all the nodes by $D_i$ for $ 1\le i \le D_\mathcal{D}$ and $Y_k$ refer as the transmission probability gain of the node $D_i$ and $H$ is defined as the total weight. Given the contact rates and the transmission probability gain, $Y_k$ for $ 1\le i \le \mathcal{D}$, and $H$ can readily be computed. As $Y_k$ are continuous-valued real numbers, we need to quantize these transmission probability gain to run the dynamic optimization procedure i.e.,
\begin{equation}
\Delta Y_k = \mathop {\min }\limits_{\forall i,j;i \ne j} \left| {Y_i - Y_j} \right|,
\end{equation}
with the quantization step of ${\rm N} = \left[ {\frac{H}{{\Delta Y_k}}} \right]$. The nodes are divided into ${L}$ steps and we decide whether node $j$ is selected …show more content…

\end{array}
\end{equation}

The dynamic programming within RACT Algorithm can find the optimal solution for the 0/1 knapsack problem (12). Therefore, we need to show the optimal solution of the 0/1 knapsack problem with quantized real-valued weights.

Alternatively, let $\gamma_i = 1$ and $\chi _{c_i, d_i} = 1$, then $\left\{ {w_k^{*}} \right\}_{k = 1}^{N}$ be the optimal solution of the 0/1 knapsack problem of (12). i.e., $\left\{ {w_k^{*}} \right\}_{k = 1}^{N}$ maximize the utility function of $\mathop {\max }\limits_{\forall {c_i} \in {\cal C},\forall {d_i} \in {\cal D}} \sum ({Y_k^{'}}.{\chi _{{c_i},{d_i}}})$ while satisfying the following constraint:

\begin{equation}
\begin{array}{l}
\sum\nolimits_{k = 1}^N {{w_k} \le H, {w_k} > 0}, \\ \\
{r_{{d_i},{c_i}}} \ge {r_{min}}.
\end{array}
\end{equation}

It is worth noted that (12) automatically implies (9) because ${Y_k^{'}} \ge {Y_k}$, $1 \le k \le N$. Now let suppose, $\left\{ {w_k^{*}} \right\}_{k = 1}^{N}$ is not the optimal solution of the 0/1 knapsack problem (12). Besides, the optimal solution of the 0/1 knapsack problem (12) is $\left\{ {w_k^{\circ}} \right\}_{k = 1}^{N}$. Thence, we …show more content…

\end{equation}

(14) and (15) are due to the fact of ${Y_j^{'}} < {Y_i^{'}}$ and $\sum\nolimits_{k = 1}^{N} {w_k^{*}} + {Y_j^{'}}-{Y_i^{'}} > H$. On other side, as $\left\{ {w_k^{\circ}} \right\}_{k = 1}^{N}$ is the optimal solution of the 0/1 knapsack problem (12), we have,

\begin{equation}
\sum\nolimits_{k = 1}^N {w_k^{\circ}} = \sum\nolimits_{k = 1}^{N} {w_k^{*}} + {Y_j^{'}}-{Y_i^{'}} > \sum\nolimits_{k = 1}^{N} {w_k^{*}}.
\end{equation}

This contradicts ${Y_j^{'}} < {Y_i^{'}}$ and therefore, it can be excluded. Furthermore, we have:
\begin{equation}
\sum\nolimits_{k = 1}^{N} {w_k^{\circ}} = \sum\nolimits_{k = 1}^N {w_k^{*}} + {Y_j^{'}}-{Y_i^{'}} \le \sum\nolimits_{k = 1}^{N} {w_k^{*}}.
\end{equation}

(13) and (17) implies that ${Y_j^{'}}-{Y_i^{'}} \le 0$. By considering, (12) and ${Y_j^{'}}-{Y_i^{'}} \le 0$ we have:

\begin{equation}
\sum\nolimits_{k = 1}^{N} {w_k^{*}} + {Y_j^{'}}-{Y_i^{'}} \le H.
\end{equation}

In conclusion, (18) contradicts the results of $\sum\nolimits_{k = 1}^{N} {w_k^{*}} + {Y_j^{'}}-{Y_i^{'}} >