An example of a sensor network of size 10.
All nodes are video sensors (depicted as small discs)
except the 9 one which is the sink (depicted as a rectangle).
-Large lircles represent the maximum
+Large circles represent the maximum
transmission range which is set to 20 in a square region which is
$50 m \times 50 m$.
\end{scriptsize}
In this one,
the nodes, in a set $N$, are sensors, links, or the sink.
Furthermore, there is an edge from $i$ to $j$ if $i$ can
-send a message to $j$, \textit{i. e.}, the distance betwween
+send a message to $j$, \textit{i. e.}, the distance between
$i$ and $j$ is less than a given maximum
transmission range.
All the possible edges are stored into a sequence
Let thus $R_h$, $R_h \geq 0$,
be the encoding rate of video sensor $h$, $h \in V$.
Let $\eta_{hi}$ be the rate inside the node $i$
-of the production that has beeninitiated by $h$. More precisely, we have
+of the production that has been initiated by $h$. More precisely, we have
$ \eta_{hi}$ is equal to $ R_h$ if $i$ is $h$,
is equal to $-R_h$ if $i$ is the sink, and $0$ otherwise.
The encoding power of the $i$ node is $P_{si}$, $P_{si} > 0$.
-The emmission distortion of the $i$ node is
+The emission distortion of the $i$ node is
$\sigma^2 e^{-\gamma . R_i.P_{si}^{}2/3}$
where $\sigma^2$ is the average input variance and
$\gamma$ is the encoding efficiency coefficient.
The transmission consumed power of node $i$ is
$P_{ti} = c_l^s.y_l$ where $c_l^s$ is the transmission energy
consumption cost of link $l$, $l\in L$. This cost is defined
-as foolows: $c_l^s = \alpha +\beta.d_l^{n_p} $ where
+as follows: $c_l^s = \alpha +\beta.d_l^{n_p} $ where
$d_l$ represents the distance of the link $l$,
$\alpha$, $\beta$, and $n_p$ are constant.
The reception consumed power of node $i$ is
\end{array}
$$
and where the following constraint is added
-$$ $q_i > 0, \forall i \in N $$
+$q_i > 0, \forall i \in N$.
to solve such a problem~\cite{}.
They first introduce dual variables
$u_{hi}$, $v_{h}$, $\lambda_{i}$, and $w_l$ for any
-$h \in V$,$ i \in N$, and $l \in L$.
+$h \in V$, $ i \in N$, and $l \in L$.
\begin{equation}
\begin{array}{l}
L(R,x,P_{s},q,u,v,\lambda,w)=\\
-\sum_{i \in N} q_i^2 + q_i. \left(
-\sum_{l \in L } a_{il}w_l^{(k)}-
+\sum_{i \in N} \left( q_i^2 + q_i. \left(
+\sum_{l \in L } a_{il}w_l-
\lambda_iB_i
-\right)\\
-+ \sum_{h \in V}
-v_h.\dfrac{\ln(\sigma^2/D_h)}{\gamma P_{sh} ^{2/3}} + \lambda_h P_{sh}\\
+\right)\right) \\
++ \sum_{h \in V} \left(
+v_h.\dfrac{\ln(\sigma^2/D_h)}{\gamma P_{sh} ^{2/3}} + \lambda_h P_{sh} \right)\\
+ \sum_{h \in V} \sum_{l\in L}
\left(
\delta.x_{hl}^2 \right.\\
\left.\left. u_{hi} a_{il}
\right)
\right)\\
- + \sum_{h \in V}
+ + \sum_{h \in V} \left(
\delta R_{h}^2
--v_h.R_{h} - \sum_{i \in N} u_{hi}\eta_{hi}
+-v_h.R_{h} - \sum_{i \in N} u_{hi}\eta_{hi}\right)
\end{array}
+\label{eq:dualFunction}
\end{equation}
The proposed algorithm iteratively computes the following variables
\item
$\begin{array}{l}
\lambda_{i}^{(k+1)} = \max\left\{0, \lambda_{i}^{(k)} - \theta^{(k)}.\left(
- q^{(k)}.B_i \right. \right.\\
- \qquad\qquad\qquad -\sum_{l \in L}a_{il}^{+}.c^s_l.\left( \sum_{h \in V}x_{hl}^{(k)} \right) \\
- \qquad\qquad\qquad - \left.\left. \sum_{l \in L} a_{il}^{-}.c^r.\left( \sum_{h \in V}x_{hl}^{(k)} \right) - P_{si}^{(k)} \right) \right\}
+ q^{(k)}.B_i - P_{si}^{(k)} \right. \right.\\
+ \qquad\qquad\qquad -\sum_{l \in L}a_{il}^{+}.c^s_l. \sum_{h \in V}x_{hl}^{(k)} \\
+ \qquad\qquad\qquad - \left.\left. c^r.\sum_{l \in L} a_{il}^{-}. \sum_{h \in V}x_{hl}^{(k)} \right) \right\}
\end{array}
$
\item
-$w_l^{(k+1)} = w_l^{(k+1)} + \theta^{(k)}. \left( \sum_{i \in N} a_{il}.q_i^{(k)} \right)$
+$w_l^{(k+1)} = w_l^{(k+1)} + \theta^{(k)}. \sum_{i \in N} a_{il}.q_i^{(k)} $
\item
-$\theta^{(k)} = \omega / t^{1/2}$
+$\theta^{(k)} = \omega / k^{1/2}$
\item
-$q_i^{(k)} = \arg\min_{q_i>0}
+$q_i^{(k+1)} = \arg\min_{q_i>0}
\left(
q_i^2 + q_i.
\left(
\item \label{item:psh}
$
-P_{sh}^{(k)}
+P_{sh}^{(k+1)}
=
\arg \min_{p > 0}
\left(
\item
$
-R_h^{(k)}
+R_h^{(k+1)}
=
\arg \min_{r \geq 0 }
\left(
$
\item
$
-x_{hl}^{(k)} =
+x_{hl}^{(k+1)} =
\begin{array}{l}
\arg \min_{x \geq 0}
\left(
\end{array}
$
\end{enumerate}
-where the first four elements are dual variable and the last four ones are
-primal ones.
\ No newline at end of file
+for any $h \in V$, $i \in N$, and $l \in L$.
