text:
\section{Introduction}

Ever-increasing demand on high data rates inspires and promotes the
development of wireless technologies. In order to achieve the
desired high information rates, many innovative ideas are introduced
into wireless system designs. Referring to cellular networks,
interference becomes the main limitation of performance improvement
and multi-point coordination is a promising and powerful technology
to improve the efficiency and reliability of wireless
communications. In the scenario of multi-cell orthogonal frequency
division multiple access (OFDMA) networks, a large number of users
try to share the same spectrum to carry out wide-band multimedia
communications and thus the performance of wireless networks is
heavily limited by mutual interference. This fact motivates
researchers to design various power control optimization algorithms
to effectively coordinate interference.

In a multi-cell network, generally speaking the design of
interference coordination is very challenging due to many practical
limitations \cite{Boudreau,Fodor,FeiGao}. One of the difficulties
for interference mitigation stems from the competition of utility
benefits among different base stations (BSs). Specifically, the
interactions between different BSs will greatly affect the whole
network performance. Hence, to efficiently schedule the inter-cell
interference the interactions among different BSs and the
characteristics of the BSs' behaviors should be carefully exploited.
In literature a successful model for the problem of interference
coordination is game theory which can effectively analyze the
behaviors of wireless nodes \cite{Lasaulce,Sung,Menon,Liang,Huang}.

In general, the game models exploited for wireless designs can be
classified into two categories: cooperative and non-cooperative
games. For non-cooperative game, it is convenient to devise totally
distributed algorithms, however it may suffer from a significant
performance loss compared with the optimal centralized solution due
to the fact that there is no cooperation among cells. On the other
hand, cooperative game usually suffers a cost of high overhead and
complexity though it has a better performance.

To overcome the inefficiency of the non-cooperative game and high
overhead of the cooperative game, recently pricing mechanism has
been proposed \cite{Schmidt}, which is employed as an effective
means to stimulate cooperation among players. Specifically, pricing
schemes can guide the players' behaviors toward efficient Nash
equilibrium, by introducing a certain degree of coordination in a
non-cooperative game. This approach has been introduced by C. Shi in
\cite{Non_Separable}. An algorithm for allocating power among
multiple interfering transmitters in a wireless network using OFDM
was presented. The algorithm attempts to maximize the sum over user
utilities, where each user's utility is a function of his total
transmission rate. Users exchange interference prices reflecting the
marginal cost of interference on each sub-channel, and then update
their power allocations given the interference prices and their own
channel conditions. Similar works have also been done in
\cite{OFDMA,MIMO,EW,beamforming,Beamformer} with local interference
pricing, and the corresponding algorithms adjusting beamforming
vectors or power allocation schemes to maximize the sum transmission
rate, respectively. It is interesting that though only limited
information is exchanged in these existing algorithms, the
performance of the distributed algorithms proved to be close to the
centralized optimization under a pricing mechanism
\cite{unified_approach}.

Game theory has achieved a great success in modeling the resource
allocation in multi-cell networks. In addition, it has also
introduced a series of successful mechanisms such as pricing
mechanism. Also inspired by the pricing mechanism, we take a further
step to tackle the resource allocation from the viewpoint of
multi-objective optimization theory. In our work, the performance
metrics of both throughput contribution and power consumption are
taken into account. In an interference-dominated network, at high
signal-to-noise ratio (SNR) regime, it is well-known that increasing
transmit power will be useless to improve system performance. On the
other hand, power consumption is also a very important performance
metric which should be carefully addressed in wireless network
designs, as green communications are of great importance in
practical applications. These two design issues are closely related
with each other, while for most of the existing works, they are
investigated separately. To the best of the authors' knowledge,
there are few works jointly considering energy efficiency and
mitigating the inter-cell interference at meantime
\cite{Fapojuwo,Choi}.

In this paper, based on a multi-objective optimization framework an
energy-efficient power optimization for multi-cell networks is
investigated. Using the pricing mechanism, a bi-objective
optimization problem is formulated for each BS, which aims at both
maximizing the throughput contribution of the BS to the network and
minimizing its total power consumption at the same time. The
throughput contribution of the BS is a function of the transmission
rate of the BS and his interference cost. To find out the tradeoff
between the throughput contribution of the BS and its power
consumption, the Pascoletti and Serafini scalarization method
\cite{mopsp} instead of the widely used weighted sum method is first
applied to transform the bi-objective problem into an equivalent
scalar objective problem considering that Pascoletti and Serafini
scalarization method has strengths in overcoming the deficiencies of
the weighted sum method. Then a proposed adaptive parameter control
algorithm is used to adaptively changing some parameters such that
the complete minimal solution set for the bi-objective optimization
problem can be derived. The minimal solutions are those on the
boundary of the set of feasible solutions. These points have a
common distinguishing characteristic. For a given constant power
consumption, the throughput contribution of the BS is the maximum
one and otherwise for a fixed throughput contribution, the consumed
power is the minimum. Finally, the performance advantage of the
proposed algorithm is demonstrated by the simulation results and it
shows that the proposed algorithm can effectively achieve different
tradeoffs among the involved objective functions. In addition
several famous solutions can be covered by our solution, such as the
solution to capacity maximization problem, the solution to
pricing-based utility optimization problem, and the solution to
equal power allocation problem.

This paper is organized as follows. In Section 2, the system model
is introduced and the corresponding multi-objective optimization
problem is formulated. The proposed adaptive parameter control
algorithm is detailed discussed in Section 3, in which we study the
connection between parameters and minimal solutions. Later the
performance of the algorithm is shown by simulations in Section 4.
Finally, conclusions are drawn in Section 5.

\section{System Model and Problem Formulation}

In this paper, we investigate a downlink OFDMA multi-cell network
with $M$ BSs and $N$ subcarriers. The BSs are connected by
high-speed fiber. Each user is served by only one BS. In addition,
each subcarrier is exclusively assigned to one user. As shown in
Fig.\ref{system_model.eps}, because of the openness of wireless
channels the users in one cell will inevitably receive the
interference signals from the neighboring BSs, especially the users
at the edge of a cell. This fact becomes the most distinguished
factor limiting wireless system performance. As a result,
interference coordination is of great importance and attracts a lot
of attention. Simply speaking, reducing transmit power will
definitely reduce mutual interference and save valuable resources to
realize green communications. However, with decreasing transmit
power meanwhile the throughput also decreases. It is not surprising
that in a complicated interference scenario there is hardly
closed-form optimal solutions. Due to the fact that the considered
optimization problem is in nature a multi-objective optimization
problem, it is impossible to define optimality. There definitely
exists some tradeoff between transmit powers at BSs and the whole
system performance within this awkward scenario and this is the
focus of our work. In view of this, relying on multiple-objective
optimization theory we want to formulate an optimization problem
with multiple objective functions and aim to derive a set of the
optimal solutions. The set will reflect different design preferences
and include several well-known special cases.

In our design for each BS $m$ the following two performance utility
functions are considered simultaneously when it optimizes its
resource allocation.

\noindent 1) Power consumption of the BS;

\noindent 2) Throughput contribution of the BS to the whole network.

\noindent It should be highlighted that the second performance
utility function consists of two parts. One is its own throughput
which is naturally a positive contribution to the whole network. The
other is the interference it caused, which is a negative
contribution as interference will decrease other terminal's
throughput. In the following, we will discuss the formulation of the
throughput contribution in detail.

\subsection*{2.1~~Throughput contribution of the BS to the network}

To measure the throughput contribution of the BS $m$ to the network,
pricing mechanism
\cite{MIMO,EW,beamforming,Beamformer,Non_Separable} is adopted in
our work. Specifically, the pricing-based multi-cell power
allocation game can be formulated as
\begin{equation}\label{game}
\mathcal{G}=\{\mathcal{M},\{\bm p_m\}_{m\in
\mathcal{M}},\{\tilde{U}_m\}_{m\in \mathcal{M}}\},
\end{equation}
where the involved elements are defined as follows \\
$\bullet$~\textbf{Player set}: The set of BSs is denoted by
$\mathcal{M}=\{1,2,\ldots,M\}$. \\
$\bullet$~\textbf{Strategy set}: The allocated power vector $\{\bm
p_1,\ldots,\bm p_M\}$ is defined as
\begin{equation}
\bm p_m=\{[p_m^{[1]},\ldots,p_m^{[N]}]^\textrm{T}:\sum_{n\in
\mathcal{N}}p_m^{[n]}\leq P_{\textrm{max}}\}.
\end{equation}\\
In the previous formulation, $p_m^{[n]}$ is the power allocated to
subcarrier $n$. $\mathcal{N}=\{1,2,\ldots,N\}$ is the set of
subcarriers with the
total number $N$, and $P_{\textrm{max}}$ is the maximum transmit power of each BS.\\
$\bullet$~\textbf{Payoff function set}: The payoff function set is
denoted by $\{\tilde{U}_1,\ldots,\tilde{U}_M\}$, where ${\tilde
{U}}_m$ is a function of the power allocation of all involved BSs.
To highlight this fact, in the following, we can write ${\tilde
{U}}_m=\tilde{U}_m(\bm p_m,\bm p_{-m})$ where ${\bm{p}}_m$ is the
power allocation of BS $m$ and ${\bm{p}}_{-m}$ is the power
allocations of the other BSs except BS $m$ with $\bm p_{-m}=\{\bm
p_{1},\ldots,\bm p_{m-1},\bm p_{m+1},\ldots,\bm p_{M}\}$.
Additionally all the power allocation vectors are assumed to be
known by all BSs.

In order to theoretically analyze the throughput contributions to
the whole network, we first investigate the throughput of BS $m$ on
subchannel $n$, which can be expressed as
\begin{equation}
\begin{split}
U_m^{[n]}
&=\log_2\left(1+\frac{{\left|h_{m,m}^{[n]}\right|}^2p_m^{[n]}}{\sigma^2+\underset{j\in
\mathcal{M}\setminus m}\sum {\left|h_{j,m}^{[n]}\right|}^2
p_j^{[n]}}\right)\\
& =\log_2\left(1+\frac{{\left|h_{m,m}^{[n]}\right|}
^2p_m^{[n]}}{\sigma^2+I_m^{[n]}}\right) \label{interference}
\end{split}
\end{equation}
where \DDcom{$h_{m,j}^{[n]},~\forall j,n$ denotes the complex
channel gain between BS $j$ and the user who is served by BS $m$ on
subchannel $n$}, $\sigma^2$ is the noise variance and $I_m^{[n]}$
denotes the interference term.

From (\ref{interference}) it can be seen that the relationship
between $U_m^{[n]}$ and $I_m^{[n]}$ is nonlinear. To overcome this
problem a linear model called pricing mechanism \cite{OFDMA} is
exploited to represent the total cost BS $m$ needs to pay to the
system when it applies $p_m^{[n]}$ on subchannel $n$.
\begin{equation}\label{total_cost}
\sum_{j\in \mathcal{M}\setminus
m}\pi_j^{[n]}{\left|h_{m,j}^{[n]}\right|}^2p_m^{[n]}
\end{equation}where $\pi_m^{[n]}$ is the interference pricing rate and is defined by \cite{OFDMA}
\begin{equation}\label{cost}
\pi_j^{[n]}=-\frac{\partial U_j^{[n]}}{\partial I_j^{[n]}}.
\end{equation}
Need to notice that the pricing rate requires to be updated after a
BS updates its power. Because only the updated pricing rate makes
sense for the neighboring BSs.

Summing up the costs in (\ref{total_cost}) across all the
subchannels $n\in \mathcal{N}$, the first utility function is
formulated as
\begin{equation}\label{utility_cost}
\tilde{U}_m(\bm p_m,\bm p_{-m})=\underset{n\in \mathcal{N}}\sum
            U_m^{[n]}-\underbrace{\underset{n\in
\mathcal{N}}\sum\underset{j\in \mathcal{M}\setminus
m}\sum\pi_j^{[n]}{\left|h_{m,j}^{[n]}\right|}^2p_m^{[n]}}_{\triangleq C(\bm p_m,\bm p_{-m})}\\
\end{equation}
where the term $C(\bm p_m,\bm p_{-m})$ can be interpreted to be the
cost of BS $m$ that should be paid to the system. As a result,
$\tilde{U}_m(\bm p_m,\bm p_{-m})$ is exploited to measure the
throughput contribution of the BS $m$ to the network.

\noindent\textbf{Remark}: The BSs receive channel state information
(CSI) from their serving users and exchange interference price as
well as a small portion of CSI information with neighboring BSs.
Different from the widely used non-linear Gauss-Seidel algorithm,
the BSs can make their decisions simultaneously with the pricing
mechanism.

\subsection*{2.2~~Problem Formulation}
According to the previous discussions, for the resource optimization
the following two objective functions should be minimized
simultaneously
\begin{align}
&f_1(\bm p_m)=-\underset{n\in \mathcal{N}}\sum
U_m^{[n]}+\underset{n\in \mathcal{N}}\sum\underset{j\in
\mathcal{M}\setminus m}\sum \pi_j^{[n]}{\left|h_{m,j}^{[n]}\right|}^2p_m^{[n]},\\
&f_2(\bm p_m)=\underset{n\in \mathcal{N}}\sum p_m^{[n]}
\end{align}based on which the multi-objective optimization problem (MOP) for each BS
$m$ can be formulated as
\begin{equation}\label{MOP}
\begin{split}
\textrm{MOP:}~~~~~&\underset{\bm
p_m}{\mathop{\textmd{min}}}~~~{\bm{f}}(\bm
p_m)=\left[\begin{array}{cccc}
             f_1(\bm p_m)\\
             f_2(\bm p_m)\\
  \end{array}\right]\\
& \textrm{s.t.}~~g_j(\bm p_m)\geq 0,~~j=1,\ldots,(N+1),\\
\end{split}
\end{equation}where the power constraints are defined as
\begin{equation}
\begin{aligned}
&g_j(\bm p_m)=p_m^{[j]},~~~~~~~~~~~~~~~~~j=1,\ldots,N,\\
&g_j(\bm p_m)=P_{\rm{max}}-\underset{n\in \mathcal{N}}\sum p_m^{[n]},~~j=N+1.\\
\end{aligned}
\end{equation}
As each BS solves MOP individually, in the following $\bm p_m$ is
replaced by ${\bm p}$ for notational simplicity.

Generally speaking, MOPs are usually much more difficult to solve
than its single objective counterpart. The difficulty in solving MOP
(\ref{MOP}) comes from its multi-objective functions and a common
logic to remove this difficulty relies on scalarization
\cite{Survey}. It should be pointed out that in the existing
wireless research, a common logic for scalarization is to replace
the original multiple objectives by their linear weighted sum.
Unfortunately, this kind of scalarization has the following
drawbacks. $\\$ \noindent (1) If the considered optimization problem
is nonconvex, the traditional scalarization will incur some losses.
In specifc, the set of final solutions may not be the Pareto optimal
set \cite{mopsp}. $\\$ \noindent (2) When searching the optimal
points, in the traditional scalarization method the adjustment of
weighting factors may make the objective value change nonuniformly.
As the optimal solutions are unknown, if the objective function
changes sharply some important points are not obtained
\cite{drawbacks}. $\\$ \noindent (3) The following operations after
the traditional scalarization are complicated for nonlinear MOP.
There are two reasons. First, linear weighted operation cannot
change the nonlinear nature of the objective function. Second, the
adaptive updating strategy for the weighting factors are largely
open. In most of the works, the weighting vectors are determined by
the importance among these objectives
\cite{weighted}\cite{weighted1}\cite{weighted2}. However, it is
difficult to find reasonable weighting vectors because there are
usually no direct relationship between weighting vectors and
objective functions, especially the objectives have different
physical meanings.$\\$ To overcome such weakness, from the
perspective of operational mathematics, another famous scalarization
technique named Pascoletti and Serafini Scalarization is much
preferred. In the following, a resource allocation algorithm is
proposed based on Pascoletti and Serafini Scalarization. To the best
of our knowledge, it is the first attempt to take advantage of
Pascoletti and Serafini Scalarization to design resource allocation
algorithm for multi-cell networks.

\section{The Proposed Algorithm Based on Multi-Objective Optimization Theory}

As there are two objective functions, the first question is how to
define the optimality or what kind of solutions we want to achieve.
Referring to the optimal solution, the following properties are
desired to be met. For a given constant power consumption, the
throughput contribution of the BS is the maximum one and on the
other hand for a fixed throughput contribution, the consumed power
is the minimum. From the viewpoint of multi-objective optimization,
this kind of solutions are named as K-minimal points \cite{mopsp}.

In Pascoletti and Serafini method, a new set of control parameters
equivalent to the weighting vectors are introduced. By varying the
parameters, the whole solution set of MOP (\ref{MOP}) can be found.
Thus in the following, we will first discuss the relationship
between MOP and the Pascoletti and Serafini scalarization problem
and then study how the new set of parameters affect minimal
solutions in subsection 3.1. And further in subsection 3.2, we will
study how to adaptively control the parameters such that the whole
solution set can be obtained.

\subsection*{3.1~~Pascoletti and Serafini Scalarization}

As discussed previously, we will choose a scalarization scheme named
Pascoletti and Serafini scalarization to transform the considered
multi-objective optimization problem into a single-objective
optimization problem (SP). Using the Pascoletti and Serafini method,
MOP (\ref{MOP}) is equivalent to the following SP,
\begin{equation}\label{(SP(a,r))}
\begin{split}
\textmd{SP:}~~~~~~\underset{t,\bm p\in
\mathcal{D}}{\mathop{\textmd{min}}}~~~&~~~~~~t\\
~~~\textmd{s.t.}~~~&\bm{a}+t\bm{r}-{\bm{f}}(\bm p)\in \textrm{\textbf{K}},\\
&g_j(\bm p)\geq 0,~~~j=1,\ldots,(N+1),\\
\end{split}
\end{equation}where the parameters are defined as $\bm a=[a_1 ~a_2]^\textrm{T} \in \mathbb{R}^2$ and
$\bm r=[r_1~ r_2]^\textrm{T} \in \mathbb{R}_+^2$. In addition,
$\textrm{\textbf{K}}=\mathbb{R}_+^2$ is the closed pointed convex
cone. The parameter $\bm a$ can be interpreted as a reference point
and the parameter $\bm r$ as a direction \cite{Helbig} as shown in
Fig. \ref{explain.eps}. For a reasonable $\bm r$ and $\bm a$ in the
coordinates, it is always possible to find a minimal $t$ ($<0$) and
then a K-minimal solution $\bar{\bm p}$ corresponding to this $(\bm
a,\bm r)$.

It has been proved in \cite{mopsp} that SP has all key properties a
scalarization method should have for determining minimal solutions
of MOP. If $(\bar{t},\bar{\bm p})$ is a minimal solution of SP, then
$\bar{\bm p}$ is a weakly K-minimal solution of MOP at least.
Besides, by varying $(\bm a,\bm r)$ all K-minimal points of MOP can
be found which are also the solutions of SP. It is obvious that
$(\bm{a},\bm{r})$ are the key parameters to SP and different
$(\bm{a},\bm{r})$ leads to different solutions to SP. As there is
more than one parameter in the transformation procedure, the
motivation is how to reduce the number of parameters and simplify
the computation for the optimal solutions.

\noindent \underline{\textbf{Conclusion 1:}} When the parameter $\bm
r$ is fixed, i.e., $\bm r=\bm r^0$, all K-minimal points can still
be found by varying the parameter $\bm a$.

\textsl{Proof:} The proof has been given in Section 6.1.

As a result, our attention is restricted  to the relationships
between parameter $\bm a$ and the minimal solution $\bm p$ as well
as ${\bm{f}}(\bm p)$ assuming $\bm r$ is constant. Regarding this,
we denote the SP w.r.t. the parameter $\bm{a}$ by
$\textrm{SP}(\bm{a})$ and regard $\bm p$ as a function of $\bm a$,
denoted by ${\bm p}(\bm a)$. Considering the perturbed parameters
$\bm a$ as
\begin{equation}\label{aa}
\bm a\approx \bm a^0 +s\bm v
\end{equation}
where $s \in \mathbb{R}_+$ is the distance between $\bm a$ and $\bm
a^0$, and $\bm v\in \mathbb{R}^{2}$ is the direction in which $\bm
a$ changes.

Substituting (\ref{aa}) into (\ref{(SP(a,r))}), the resulting
$\textrm{SP}({\bm a})$ becomes
\begin{equation}\label{(SP_modified)}
\begin{split}
\textrm{SP}({\bm a}):~~~~\underset{t,\bm p}{\mathop{\textmd{min}}}~~~&~~~~~~t\\
~~~\textmd{s.t.}~~~&(\bm a^0+s\bm v)+t\bm r^0-{\bm{f}}({\bm p})\geq \textbf{0}_2,\\
&{\bm{g}}(\bm p)\geq \textbf{0}_{(N+1)}.\\
\end{split}
\end{equation}

To find the whole solution set, we embark on  a reference problem
$\textrm{SP}(\bm{a}^0)$ and assume its minimal solution $(t^0(\bm
a^0),\bm p^0(\bm a^0))$ as well as the Lagrange multipliers
$(\bm{\mu}^0,\bm{\beta}^0)$ have already been obtained, where
$\bm{\mu}^0$ is the Lagrange multipliers to the first constraint of
(\ref{(SP_modified)}) and $\bm{\beta}^0$ to the second. Based on the
reference point, we can find the next solution $\bm p(\bm a)$ and
thus $\bm f(\bm p(\bm a))$ in the neighborhood of $\bm f(\bm p^0)$
with the help of directional derivatives. Then iteratively
implementing the process until the whole solution set is achieved.
To realize this, the relationship between $\bm a$ and $\bm p(\bm a)$
as well as $\bm f(\bm p(\bm a))$ should be investigated first.

Assume that a variation of the parameters $\bm a$ in one direction
only, i.e., $\bm v$ is fixed and the solutions of
(\ref{(SP_modified)}) only depend on the parameter $s$. Then, we can
regard all the terms $(t, \bm p, \bm{\mu}, \bm{\beta})$ as a
function of $s$, i.e., $\left(t(s), \bm p(s), \bm{\mu}(s),
\bm{\beta}(s)\right)$. For clarity, we use $(t, \bm p, \bm{\mu},
\bm{\beta})$ in the following. The Lagrange function of
(\ref{(SP_modified)}) is first given by:
\begin{equation}\label{Lagrange}
\mathcal{L}(t, \bm p, \bm{\mu},
\bm{\beta})=t-\sum_{i=1}^M{\mu}_i\left[(a_i^0+sv_i)+tr_i^0-f_i({\bm
p})\right]-\sum_{j=1}^{N+1}{\beta}_jg_j(\bm{p})
\end{equation}

For a solution in nonlinear programming \textrm{SP}($s$)
(\ref{(SP_modified)}) to be optimal, Karush-Kuhn-Tucker (KKT)
conditions are necessary:
\begin{equation}\label{KKT}
\begin{split}
\nabla_{t}\mathcal{L}(t, \bm p, \bm{\mu}, \bm{\beta})&=1-\sum_{i=1}^M{\mu}_ir_i^0=0\\
\nabla_{{\bm p}}\mathcal{L}(t, \bm p, \bm{\mu},
\bm{\beta})&=\sum_{i=1}^M{\mu}_i\nabla_{{\bm p}}f_i({\bm
p})-\sum_{j=1}^{N+1}{\beta}_j\nabla_{{\bm p}}g_j({\bm
p})=0\\
\mu_i\left((a_i^0+sv_i)+tr_i^0-f_i({\bm p})\right)&=0,~~\mu_i\geq0,~~~~~\forall i\\
\beta_jg_j({\bm p})&=0,~~\beta_j\geq0,~~~~~\forall j
\end{split}
\end{equation}

These nonlinear equations can be solved with the help of directional
derivatives, which is derived in Section 6.2.

With the solution $(\bar{t},\bar{\bm p}, \bar{\bm \mu},\bar{\bm
\beta})$ obtained by solving (\ref{system equations1}), the minimal
solution of \textrm{SP}($s$) can be attained by
\begin{equation}\label{s1}
\left(\begin{array}{cc}
             t \\
             {\bm p} \\
             \bm \mu \\
             \bm \beta \\
  \end{array}\right)\approx\left(\begin{array}{cc}
             t^0\\
             {\bm p}^0\\
             \bm \mu^0\\
             \bm \beta^0\\
  \end{array}\right)+s\cdot\left(\begin{array}{cc}
             \bar{t}\\
             \bar{{\bm p}}\\
             \bar{\bm \mu}\\
             \bar{\bm \beta}\\
  \end{array}\right)
\end{equation}
So far, the relationship between the minimal solution $({t},{\bm p},
{\bm \mu},{\bm \beta})$ and $s$ is obtained. And in Conclusion 2, we
will show how $s$ affect the minimal points $\bm f({\bm p})$.

\noindent \underline{\textbf{Conclusion 2:}} Since $s$ in (\ref{s1})
can be defined very small ($s\rightarrow 0^+$), the minimal points
$\bm f({\bm p})$ can be approximated using the first-order Taylor
approximation
\begin{equation}\label{approximation}
\bm f( {\bm p})\approx \bm f( {\bm p^0})+s\cdot \bm v+ s(\nabla_{\bm
a}\bar{\tau}^{\delta}(\bm a^0)^\textrm{T}\bm v)\bm r,
\end{equation}
where $\nabla_{\bm a}\bar{\tau}^{\delta}(\bm a^0)=-\bm \mu^0$ is the
derivative of the local minimal value function $\bar{\tau}^{\delta}$
in the point $\bm a^0$.

\textsl{Proof:} The proof can be found in Section 6.3.

Thus far, the function relationship among the minimal solution $\bm
p$, the minimal point $\bm f(\bm p)$ and $\bm a$ can be clearly
seen. Based on the results in (\ref{s1}) and (\ref{approximation}),
we then care how to adaptively control the parameter $\bm a$ such
that all the minimal solution $\bm p$ and the minimal point $\bm
f(\bm p)$ can be obtained instead of artificially modifying $\bm a$.

\subsection*{3.2~~Adaptive control of parameter $\bm a$}

In the following, a procedure is developed to achieve the whole
minimal solution set of the MOP by controlling the choice of the
parameter $\bm a$. First, it is necessary to give the set from which
the parameter $\bm a$ is chosen.

\noindent \textbf{\underline{Theorem 1:}} Define a hyperplane
\begin{equation}
\textrm{\textbf{H}}=\{\bm y\in \mathbb{R}^2|\bm b^\mathrm{T}\bm
y=\beta\},
\end{equation}
with $\bm b\in \mathbb{R}^2$, $\beta\in \{0,1\}$, and $\bm
b^\mathrm{T}\bm r\neq 0$. It is sufficient to get the efficient set
by varying the parameter $\bm a\in \textrm{\textbf{H}}$. Further, it
is shown that a subset $\bm a\in \textrm{\textbf{H}}^a\subset
\textrm{\textbf{H}}$ for
\begin{equation}\label{subset}
\textrm{\textbf{H}}^a=\{\bm y\in \mathbb{R}^2|\bm y=\lambda \bar{\bm
a}^1+(1-\lambda)\bar{\bm a}^2,~~\lambda\in[0,1]\},
\end{equation}
is also sufficient. Where $\bar{\bm a}^1\in \textrm{\textbf{H}}$ and
$\bar{\bm a}^2\in \textrm{\textbf{H}}$ are given by
\begin{equation}
\bar{\bm a}^i:=\bm f(\bar{\bm p}^i)-\bar{t}^i\bm r
~~~\textrm{with}~~\bar{t}^i:=\frac{\bm{b}^\textrm{T}\bm f(\bar{\bm
p}^i)-\beta}{\bm b^\mathrm{T} \bm r},~~~i=1,2.
\end{equation}
with $\bar{\bm p}^1$ the minimal solution of the scalar-valued
problem $\underset{\bm p\in \mathcal{D}}{\rm{min}}~~f_1(\bm p)$ and
$\bar{\bm p}^2$ the minimal solution of $\underset{\bm p\in
\mathcal{D}}{\rm{min}}~~f_2(\bm p)$. Without loss of generality, we
assume $\bar{\bm a}^1$ is smaller than $\bar{\bm a}^2$ on the first
dimension, i.e., $\bar{a}_1^1<\bar{a}_1^2$.

\textsl{Proof:} The proof can be found in our work \cite{ours}.

With the stricter set $\textrm{\textbf{H}}^a$ in (\ref{subset}) from
which $\bm a$ should be chosen, we now want to determine the
parameters $\bm a^0, \bm a^1, \bm a^2,\ldots$ adaptively (starting
with $\bm a^0=\bar{\bm a}^1$) such that the related minimal points
$\bm f(\bm p(\bm a^i)),~i=0,1,2,\ldots,$ gained by solving
$\textrm{SP}(\bm a^i)$ for $i=0,1,2,\ldots,$ have the equal distance
$\alpha>0$, i.e.,
\begin{equation}\label{aaaaa}
\|\bm f(\bm p(\bm a^0))-\bm f(\bm p(\bm a^1))\|=\alpha,
\end{equation}
should be satisfied for any neighboring $\bm a^i$ and $\bm a^{i+1}$.
This metric is applied to avoid the loss of some important points
caused by the sharply change of an objective function. In the
following, we replace $\bm f(\bm p(\bm a^0))$ with $\bm f(\bm p^0)$
for clarity. The advantage of choosing a predefined distance
$\alpha>0$ between two neighbor points can be seen from
Fig.\ref{explain.eps}. With the evenly distributed points, more
accurate information about the relationship between the objectives
would be known.

Then, we aim at finding a direction $\bm v$ and a scalar $s$ such
that $\bm a^1:=\bm a^0+s\bm v$ satisfies (\ref{aaaaa}). Based on
(\ref{approximation}), we have
\begin{equation}
\begin{split}
\alpha&=\|\bm f(\bm p^0)-\bm f(\bm p^1)\|\\
&\approx \|\bm f(\bm p^0)-(\bm f(\bm p^0)+s\bm v+s(-(\bm
\mu^0)^\textrm{T}\bm v)\bm
r)\| \\
&=|s|~\|\bm v+(-(\bm \mu^0)^\textrm{T}\bm v)\bm
r)\| .\\
\end{split}
\end{equation}
Then, the stepsize $s$ and direction $\bm v$ of parameter $\bm a$
can be chosen by
\begin{equation}\label{s_v}
\begin{split}
s^0&=\frac{\alpha}{\|\bm v+(-(\bm \mu^0)^\textrm{T}\bm v)\bm
r)\|},\\
\bm v&=\bm{a}^1-\bm{a}^0.\\
\end{split}
\end{equation}
which leads to the generalized formula of $\bm a$,
\begin{equation}\label{a_expression}
\bm a^{i+1}=\bm a^i + s^i\bm v=\bm a^i+\frac{\alpha}{\|\bm v+(-(\bm
\mu^i)^\textrm{T}\bm v)\bm r)\|}\bm v.
\end{equation}

Based on (\ref{a_expression}), an algorithm is developed to
adaptively solve $\textrm{SP}(\bm a^i)$ with $\bar{\bm a}^1\leq \bm
a^i\leq\bar{\bm a}^2$, whose solutions constitute the solution set.
This algorithm is referred to as adaptive parameter control (APC)
algorithm and is shown in \textbf{Algorithm 1}.

\begin{algorithm}
\caption{Adaptive Parameter Control Algorithm}

\noindent \hangafter=1 \setlength{\hangindent}{4em}
\textbf{Input:~~}Choose $\bm r=(r_1,r_2)^\textrm{T}\in
\mathbb{R}_+^2$ with $r_1>0$, and predefine $\alpha>0$ between the
neighboring two points. The hyperplane is chosen
as\\$~~~~~~~~~~~~\textrm{\textbf{H}}=\{\bm y\in \mathbb{R}^2|\bm
b^\textrm{T}\bm
y=\beta\}$, \\
where $\bm b\in \mathbb{R}^2$ and $\bm b^\textrm{T}\bm r\neq 0$,
$\beta \in \{0,1\}$. Given $M^1\in \mathbb{R}$
with\\
$~~~~~~~~M^1> \underset{\bm p\in \mathcal{D}}{\max}~f_2(\bm
p)-\underset{\bm p\in
\mathcal{D}}{\min}~f_1(\bm p)\frac{r_2}{r_1}$.\\

\noindent \hangafter=1 \setlength{\hangindent}{4em}
\textbf{Step~1:~~}Finding the minimal solution $(\tilde{t}^1,{\bm
p}^1)$ and Lagrange multiplier $\bm \mu^1\in \mathbb{R}_+^2$ of
$\textrm{SP}(\tilde{\bm a}^1)$ with $\tilde{\bm
a}^1=(0,M^1)^\textrm{T}$. Calculate\\
$~~~~~~t^1:=\frac{\bm b^\textrm{T}\bm f(\bm p^1)-\beta}{\bm
b^\textrm{T}\bm r}$ and $\bm
a^1:=\bm f(\bm p^1)-t^1\bm r$.\\
Set $l:=1$.\\

\noindent \hangafter=1 \setlength{\hangindent}{4em}
\textbf{Step~2:~~}Finding the minimal solution $\bm p^E$ of
$\underset{\bm p\in \mathcal{D}}{\min}~f_2(\bm
p)$ and set\\
$~~~~~t^E:=\frac{\bm b^\textrm{T}\bm f(\bm p^E)-\beta}{\bm
b^\textrm{T}\bm r}$ and $\bm
a^E:=\bm f(\bm p^E)-t^E\bm r$.\\
Let $\bm v:=\bm a^E-\bm a^1$.

\noindent \hangafter=1 \setlength{\hangindent}{4em}
\textbf{Step~3:~~}Update $\bm a^{l+1}$ by\\
$~~~\bm a^{l+1}:=\bm a^{l}+\frac{\alpha}{\|\bm v+(-(\bm
\mu^l)^\textrm{T}\bm v)\bm r)\|}\cdot \bm v$.

\noindent \hangafter=1 \setlength{\hangindent}{4em}
\textbf{Step~4:~~}Set $l:=l+1$.\\
$\bullet$ \textbf{If} $\bm a^l=\bm a^1+\rho \bm v$ for a $\rho \in
[0,1]$, find minimal solution $(t^l,\bm p^l)$ and Lagrange
multiplier $\bm \mu^l$ by solving $\textrm{SP}(\bm a^l)$, and go to step 3.\\
$\bullet$ \textbf{Else} stop.

\noindent \hangafter=1 \setlength{\hangindent}{4em}
\textbf{Output:~~}Set $\mathbf{\mathcal{A}}=\{\bm p^1,\ldots,\bm
p^{l-1},\bm p^E\}$ is the minimal solution set of MOP. Set
$\mathbf{\mathcal{B}}=\{\bm f(\bm p^1),\ldots,\bm f(\bm p^{l-1}),\bm
f(\bm p^E)\}$ is an approximation of the minimal points.

\end{algorithm}

In the \textbf{Input} Step, we arbitrarily choose a $\tilde{\bm
a}^1=(0,M^1)^\textrm{T}$ with $M^1> \underset{\bm p\in
\mathcal{D}}{\max}~f_2(\bm p)-\underset{\bm p\in
\mathcal{D}}{\min}~f_1(\bm p)\frac{r_2}{r_1}$ at the initial for the
purpose of engineering practice. Actually, the optimization problem
$\textrm{SP}(\tilde{\bm a}^1,\bm r)$ with such a $\tilde{\bm
a}^1=(0,M^1)^\textrm{T}$ in Step 1 is
\begin{equation}\label{(SP(a1,r))}
\begin{split}
&\underset{\bm p\in \mathcal{D}}{\min}~~~~~~~~t~\\
&~\textmd{s.t.}~~~~~~~~~tr_1-f_1(\bm
p)\geq 0,\\
&~~~~~~M^1+tr_2-f_2(\bm p)\geq 0,\\
&~~~~~~~~~~~~~~t\in \mathbb{R}.
\end{split}
\end{equation}
Therefore $t\geq \frac{f_1(\bm p)}{r_1}$ holds for any feasible
point $(t,\bm p)$, and $M^1+tr_2-f_2(\bm p)>(f_2(\bm p)-f_1(\bm
p)\frac{r_2}{r_1})+\frac{f_1(\bm p)}{r_1}r_2-f_2(\bm p)=0$ is also
satisfied. Therefore, (\ref{(SP(a1,r))}) can be replaced by
$\underset{\bm p\in \mathcal{D}}{\min}~~~\frac{f_1(\bm p)}{r_1}$
which equals to $\underset{\bm p\in
\mathcal{D}}{\textrm{min}}{f_1(\bm p)}$. However, there is no need
to find the exact solution of $\underset{\bm p\in
\mathcal{D}}{\textrm{min}}{f_1(\bm p)}$ in practice, because we can
always find the starting point $\bm a^1$ even if the initial
$\tilde{\bm a}^1$ is roughly given. In addition,
$\textrm{SP}(\tilde{\bm a}^1)$ and $\textrm{SP}({\bm a}^1)$ have the
same minimal solution ${\bm p}^1$ and the same Lagrange multiplier
$\bm \mu^1$, which will be demonstrated in \textbf{Theorem 2}.

\noindent \textbf{\underline{Theorem 2:}} Assume $(\bar{t},\bar{\bm
p})$ is a minimal solution of $\textrm{SP}(\bm a,\bm r)$ with
Lagrange multiplier $\bm \mu$ for arbitrary $\bm a\in \mathbb{R}^2$
and $\bm r\in \mathbb{R}^2$ with $\bm b^\mathrm{T}\bm r\neq 0$.
Surely there could be a $\bar{\bm k}\in \textrm{\textbf{K}}$ with
\begin{equation}
\bm a+\bar{t}\bm r-\bm f(\bar{\bm p})=\bar{\bm k}.
\end{equation}
Further there could be a $\bm a' \in \textrm{\textbf{H}}$ and $t'\in
\mathbb{R}$ such that $(t',\bar{\bm p})$ is a minimal solution of
$\textrm{SP}(\bm a',\bm r)$ with
\begin{equation}
\bm a'+t'\bm r-\bm f(\bar{\bm p})=\bm 0_2.
\end{equation}
In addition $\bm \mu$ is also Lagrange multiplier to the point
$(t',\bar{\bm p})$ for $\textrm{SP}({\bm a}',\bm r)$.

\textsl{Proof:} The proof can be found in our work \cite{ours}.

In Step 1 and Step 2, the subset $\textrm{\textbf{H}}^a$ of the
hyperplane and a direction $\bm v$ with $\bm a^1+s\bm v\in
\textrm{\textbf{H}}$ are determined. And then in Step 3 and Step 4,
the APC algorithm producing an approximation of the efficient set is
done.

\section{Simulation Results}

In this section, the performance the proposed APC algorithm is
assessed. A multi-carrier multi-user OFDM-based system is employed
with $M=19$ BSs as shown in Fig.\ref{system_model.eps}. There are 64
users in the simulation model and each user is served using a
randomly chosen subcarrier. The number of subcarriers is $N=64$ with
the system bandwidth 10 MHz. The distance between adjacent BSs is
1000 m. Reyleigh fast fading is considered, and large scale path
loss is modeled as $\textrm{PL} = 128.1 + 37.6\lg(d)$ \cite{3GPP},
where $d$ is the distance in kilometers.

As our work is studied based on the function in
(\ref{utility_cost}), we choose it as a comparison to the proposed
APC algorithm. This algorithm applies pricing mechanism, and aims at
only maximizing the transmission rate contribution of the BS to the
network, thus we denote it by ``pricing mechanism'' in the
simulations. The solutions of the equal power allocation and utility
maximization are also simulated as the reference points. Here,
utility maximization is that each BS tries to maximize his own
transmission rate ignoring the interference it may cause to other
cells and does not care how much power they will use to achieve the
maximum transmission rate. While the equal power allocation scheme
is that the total power needed in the APC method is uniformly
allocated to all subcarriers. For the initialization, we assume that
the total 30 watts energy is evenly distributed across all the
subcarriers.

Fig.\ref{power_throughput} shows the relationship between power
consumption and the throughput contribution of BS 1 to the network,
which is obtained by the APC method. This curve is the boundary of
the efficient set. For each given power consumption, the
corresponding throughput contribution of BS 1 to the network is the
maximum. And for the contribution of BS 1 to the network, the
corresponding power consumption is the minimum be required. When the
consumed power is greater than 20 watts, the increase of the network
throughput brought by the BS is small. Thus the extra 10 watts power
can be viewed as inefficient.

Fig.\ref{energy_efficiency} gives a parabola-like relationship
between energy efficiency and throughput of the system. Energy
efficiency is defined by the fraction where the numerator is the
throughput of the system and the denominator is the total power
consumption. As our goal is to find an energy-efficient power
allocation scheme which can also guarantee a relatively high system
throughput, we pay close attention to the interesting local area. If
the system agrees to reduce its transmission rate from 8.98 Mbps to
8.68 Mbps, energy efficiency will increase sharply to 47 kbps/watt,
which is much greater than the energy efficiency at 8.98 Mbps.

Fig.\ref{result} shows the comparison between different power
allocation schemes. All the curves are obtained by varying the power
consumption of BS 1 and fixing other BSs' power. It can be seen that
the solutions obtained by solving the pricing mechanism-based
problem are always on the APC curves, no matter whether the
interference pricing rate is updated or not. This result verifies
the comprehensiveness of the proposed APC method. Though it seems
that the solution to the utility maximization is better than the
solutions to pricing mechanism-based problem as well as APC problem
before update the pricing rate, its actual outcome is not so.
Because when we recalculate the system throughput after updating the
pricing rates, the solutions obtained by the pricing mechanism-based
scheme and APC method enjoy much more benefits than the greedy
utility maximization scheme. The reason is that the negative
contribution of BS 1 under utility maximization is great for it
introduces more interference to users in neighboring cells. A final
note about this figure is that all the curves are obtained by
varying the power consumption of BS 1 and fixing other BSs' power,
and the power allocation scheme adopted by the neighboring BSs is
EPA. As a result, there are no significant differences among these
considered schemes since only one player changed his strategy during
this process.

\section{Conclusions}

In this paper, from the perspective of multi-objective optimization
theory an energy-efficient power allocation scheme was developed for
interference-limited multi-cell network.  A bi-objective
optimization problem was first formulated based on the pricing
mechanism. Then using the method of Pascoletti and Serafini
scalarization, the relationship between the varying parameters and
minimal solutions has been discovered, and an adaptive algorithm was
developed to achieve tradeoff between the two objectives. As a
result, all the solutions on the boundary of the efficient solution
set can be computed which are best in terms of both energy
efficiency and inter-cell interference mitigation. Finally, the
performance of
the algorithm was demonstrated in the simulations.\\

\section{Acknowledgments}

This work was supported in part by the National Natural Science
Foundation of China under Grant 61101130, and the Excellent Young
Scholar Research Funding of Beijing Institute of Technology under
Grant 2013CX04038.

\section{Appendices}

\subsection*{6.1~~Proof of Conclusion 1}

Let $\bar{\bm p}$ be a K-minimal solution of the MOP, and set $\bm
a=\bm f(\bar{\bm p})$ and choose $\bm r\in
\textrm{\textbf{K}}\setminus \{\bm 0_2\}$ arbitrarily. Then the
point $(0,\bar{\bm p})$ is a feasible and also minimal solution for
$\textrm{SP}(\bm a,\bm r)$. Otherwise there will be a $t'<0$ and
$\bm p'\in \mathcal{D}$ with $(t',\bm p')$ feasible for
$\textrm{SP}(\bm a,\bm r)$, and a $\bm k'\in \textrm{\textbf{K}}$
with $\bm a+t'\bm r-\bm f(\bm p')=\bm k'$. With these observations,
we have
\[
\bm f(\bar{\bm p})=\bm f(\bm p')+\bm k'-t'\bm r\in \bm f(\bm
p')+\textrm{\textbf{K}}.\] Because $\bar{\bm p}$ is a K-minimal
solution, it can be concluded that $\bm f(\bar{\bm p})=\bm f(\bm
p')$ and thus $\bm k'=t'\bm r$. Since the cone $\textrm{\textbf{K}}$
is pointed, $\bm k'\in \textrm{\textbf{K}}$ and $t'\bm r\in
-\textrm{\textbf{K}}$, which leads to $\bm k'=t'\bm r=\bm 0_2$.
However, it is contradict to $t'<0$ and $\bm r\neq \bm 0_2$. Thus it
can be concluded that if $\bar{\bm p}$ is a minimal solution of the
MOP, then $(0,\bar{\bm p})$ is a minimal solution of
$\textrm{SP}(\bm a,\bm r)$ for the parameter $\bm a:=\bm f(\bar{\bm
p})$ and for arbitrarily $\bm r\in \textrm{\textbf{K}}\setminus
\{\bm 0_2\}$. Further, it can be inferred that all K-minimal points
can be found by varying the parameter $\bm a$ only.

\subsection*{6.2~~Derivation of the solution to (\ref{KKT})}

For the solution $(t, \bm p, \bm{\mu}, \bm{\beta})$ of
$\textrm{SP}(s)$ which is differentiable, the righthanded
derivatives $(\bar{t},\bar{{\bm p}},\bar{\bm \mu},\bar{{\bm
\beta}})$ can be written as
\begin{equation}\label{sys_equs}
\underset{\alpha\rightarrow 0^+}{\lim}\left(
\begin{array}{cccc}
             \frac{t-t^0}{\alpha}\\
              \frac{{\bm p}-{\bm p}^0}{\alpha}\\
              \frac{\bm \mu-\bm \mu^0}{\alpha}\\
              \frac{{\bm \beta}-{\bm \beta}^0}{\alpha}
\end{array}\right)=\left(
\begin{array}{cccc}
             \bar{t}\\
             \bar{{\bm p}}\\
             \bar{\bm \mu}\\
             \bar{{\bm \beta}}\\
\end{array}\right),
\end{equation}based on which it can also be inferred from Theorem 3 of \cite{131} that for
sufficiently small $\alpha$, there exists a unique continuous
function $(t, \bm p, \bm{\mu}, \bm{\beta})$ as the minimal solution
of $\textrm{SP}(s)$. Then we will find the solution $(t, \bm p,
\bm{\mu}, \bm{\beta})$ with the help of $(\bar{t},\bar{{\bm
p}},\bar{\bm \mu},\bar{{\bm \beta}})$ in (\ref{sys_equs}). Because
the KKT conditions (\ref{KKT}) is always satisfied, it follows that
the derivatives $(\bar{t},\bar{{\bm p}},\bar{\bm \mu},\bar{{\bm
\beta}})$ can be easily obtained by solving the following system of
inequalities and equations,
\begin{equation}\label{system equations1}
\begin{split}
-\sum_{i=1}^{M}\bar{\mu}_ir_i^0&=0,\\
\sum_{i=1}^{M}\mu_i^0\nabla_{{\bm p}}^2f_i({\bm p}^0)\bar{{\bm p}}-\sum_{i=1}^{N+1}\beta_i^0\nabla_{{\bm p}}^2&g_i({\bm p}^0)\bar{{\bm p}}\\
+\sum_{i=1}^{M}\bar{\mu}_i\nabla_{{\bm p}}f_i({\bm p}^0)-\sum_{i=1}^{N+1}\bar{\beta}_i^0\nabla_{{\bm p}}&g_i({\bm p}^0)=\bm 0_{N},\\
r_i^0\bar{t}-\nabla_{{\bm p}}f_i({\bm p}^0)^\mathrm{T}\bar{{\bm p}}&=-\bm v,~~~\forall i\in I^+,\\
r_i^0\bar{t}-\nabla_{{\bm p}}f_i({\bm p}^0)^\mathrm{T}\bar{{\bm p}}&\geq -\bm v,~~~\forall i\in I^0,\\
\bar{\mu}_i&\geq 0,~~~~~~\forall i\in I^0,\\
\bar{\mu}_i\big(r_i^0 \bar{t}-\nabla_{{\bm p}}f_i({\bm p}^0)^\mathrm{T}\bar{{\bm p}}+\bm v\big)&=0,~~~~~~\forall i\in I^0,\\
\bar{\mu}_i&=0,~~~~~~\forall i\in I^-.\\
\end{split}
\end{equation}
\begin{equation}\label{system equations2}
\begin{split}
\nabla_{{\bm p}}g_j({\bm p}^0)^\mathrm{T}\bar{{\bm p}}&=0,~~~~~~\forall j\in J^+,\\
\nabla_{{\bm p}}g_j({\bm p}^0)^\mathrm{T}\bar{{\bm p}}&\geq 0,~~~~~~\forall j\in J^0,\\
\bar{\beta}_j&\geq 0,~~~~~~\forall j\in J^0,\\
\bar{\beta}_j(\nabla_{{\bm p}}g_j({\bm p}^0)^\mathrm{T}\bar{{\bm p}})&= 0,~~~~~~\forall j\in J^0,\\
\bar{\beta}_j&= 0,~~~~~~\forall j\in J^-.\\
\end{split}
\end{equation}
where $I^+$, $I^0$, $I^-$ are the active, non-degenerate and
degenerate constraint sets, respectively
\begin{equation}\label{disjoint}
\begin{split}
&I^+:=\{i \in I\mid a_i^0+t^0r_i^0-f_i({\bm p}^0)=0,~~\mu_i^0>0\},\\
&I^0~:=\{i \in I\mid a_i^0+t^0r_i^0-f_i({\bm p}^0)=0,~~\mu_i^0=0\},\\
&I^-:=\{i \in I\mid a_i^0+t^0r_i^0-f_i({\bm p}^0)>0,~~\mu_i^0=0\},\\
\end{split}
\end{equation}
and $J^+$, $J^0$, $J^-$ are another disjoint sets:
\begin{equation}\label{disjoint}
\begin{split}
&J^+:=\{j \in J\mid g_j({\bm p}^0)=0,~~\beta_j^0>0\},\\
&J^0~:=\{j \in J\mid g_j({\bm p}^0)=0,~~\beta_j^0=0\},\\
&J^-:=\{j \in J\mid g_j({\bm p}^0)>0,~~\beta_j^0=0\}.\\
\end{split}
\end{equation}
With the derivatives $(\bar{t},\bar{{\bm p}},\bar{\bm \mu},\bar{{\bm
\beta}})$ obtained by solving (\ref{system equations1}) and
(\ref{system equations2}), the minimal solution of \textrm{SP}($s$)
can be attained from (\ref{sys_equs}), which equals
\begin{equation}
\left(\begin{array}{cc}
             t \\
             {\bm p} \\
             \bm \mu \\
             \bm \beta \\
  \end{array}\right)\approx\left(\begin{array}{cc}
             t^0\\
             {\bm p}^0\\
             \bm \mu^0\\
             \bm \beta^0\\
  \end{array}\right)+s\cdot\left(\begin{array}{cc}
             \bar{t}\\
             \bar{{\bm p}}\\
             \bar{\bm \mu}\\
             \bar{\bm \beta}\\
  \end{array}\right).
\end{equation}

\subsection*{6.3~~Proof of Conclusion 2}

In order to prove the following equation
\begin{equation}\label{f_prof}
\bm f( {\bm p})\approx \bm f( {\bm p^0})+s\cdot \bm v+ s(\nabla_{\bm
a}\bar{\tau}^{\delta}(\bm a^0)^\textrm{T}\bm v)\bm r,
\end{equation}
the definition of the local minimal value function
$\bar{\tau}^{\delta}$ should be given first, and then the derivative
of the local minimal value function in the point $\bm a^0$ should be
provided as well. The local minimal value function
$\tau^{\delta}:\mathbb{R}^2\rightarrow \mathbb{R}$ is defined by
\begin{equation}
{\tau}^{\delta}(\bm a):=\inf\{t\in \mathbb{R} | (t,\bm p)\in
\Sigma\bm a\cap B_{\delta}(t^0,\bm p^0)\},
\end{equation}
where $B_{\delta}(t^0,\bm p^0)$ is the closed ball with radius
$\delta>0$ around $\bm p^0$ and $\Sigma\bm a$ is the constraint set
of SP depending on $\bm a$:
\begin{equation}
\Sigma\bm a:=\{(t,\bm p)\in \mathbb{R}^{N+1} | \bm a+t\bm r-\bm f(
{\bm p})\geq \textbf{0}\}.
\end{equation}

Then the derivative can be achieved by
\begin{equation}\label{proof1}
\begin{split}
\nabla_{\bm a}&\tau^{\delta}(\bm a)=-\bm \mu(\bm a)\\
&-\sum_{i=1}^2\nabla_{\bm a}\bm \mu_i(\bm a)(a_i+t(\bm a)r_i-f_i(\bm p(\bm a))).\\
\end{split}
\end{equation}
Then replace $\bm \mu(\bm a^0)$ by $\bm \mu^0$, we obtain the
derivative in the point $\bm a^0$
\begin{equation}\label{proof2}
\begin{split}
\nabla_{\bm a}&\tau^{\delta}(\bm a^0)=-\bm \mu^0\\
&-\sum_{i\in I^+\bigcup I^0}\nabla_{\bm a}\bm \mu_i(\bm
a^0)\underbrace{(a_i^0+t^0r_i^0-f_i(\bm p^0))}_{=0} \\
&-\sum_{i\in I^-}\nabla_{\bm a}\bm \mu_i(\bm a^0)(a_i^0+t^0r_i^0-f_i(\bm p^0)). \\
\end{split}
\end{equation}
Due to the definition of $I^-$, $a_i^0+t^0r_i^0-f_i(\bm p^0)>0$ for
$i\in I^-$. Since $a_i+t(\bm a)r_i-f_i(\bm p(\bm a))$ is continuous
in $\bm a$, there exists a neighborhood $\mathcal{N}(\bm a^0)$ of
$\bm a^0$ so that for all $\bm a \in \mathcal{N}(\bm a^0)$ it
holds\\
\begin{equation}
a_i+t(\bm a)r_i-f_i(\bm p(\bm a))>0~~~~~\textrm{for}~~~i\in I^-.
\end{equation}
Then $\mu_i(\bm a)=0$ for all $\bm a\in \mathcal{N}(\bm a^0)$ and
$\nabla_{\bm a}\bm \mu_i(\bm a^0)=0$ for $i \in I^-$ can be
obtained. Thus, we get
\begin{equation}\label{proof3}
\nabla_{\bm a}\tau^{\delta}(\bm a^0)=-\bm \mu^0.
\end{equation}

Already obtained the derivative of the local minimal value function
$\nabla_{\bm a}\tau^{\delta}(\bm a^0)$, we then try to obtain the
result in (\ref{f_prof}) with the help of (\ref{proof3}).

Assume that we have already solved the problem $\textrm{SP}(\bm
a^0)$ for the parameters $\bm a^0$ with a minimal solution $(t^0,\bm
p^0)$ and Lagrange multiplier $\bm \mu^0$. Then by using
${\tau}^{\delta}(\bm a^0)={t}(\bm a^0)=t^0$, a first order Taylor
approximation of the local minimal value function to the
optimization problem $\textrm{SP}(\bm a)$ is derived,
\begin{equation}
{t}(\bm a)\approx t^0+\nabla_{\bm a} {\tau}^{\delta}(\bm
a^0)^\textrm{T}(\bm a-\bm a^0)
\end{equation}
Then the approximation for the K-minimal points of MOP dependent on
the parameter $\bm a$ is launched:
\begin{equation}
\begin{split}
\bm f( {\bm p}(\bm a))&=\bm a+t(\bm a)\bm r\\
&\approx \bm a^0+(\bm a-\bm a^0)+(t^0+\nabla_{\bm a}
{\tau}^{\delta}(\bm a^0)^\textrm{T}(\bm
a-\bm a^0))\bm r\\
&=\bm f(\bm p^0)+(\bm a-\bm a^0)+(\nabla_{\bm a} {\tau}^{\delta}(\bm
a^0)^\textrm{T}(\bm
a-\bm a^0))\bm r\\
&=\bm f(\bm p^0)+s\bm v+s(\nabla_{\bm a}\bar{\tau}^{\delta}(\bm
a^0)^\textrm{T}\bm v)\bm r.\\
\end{split}
\end{equation}

\subsection*{6.4~~Proof of Theorem 1}

First, we will prove that it is sufficient to vary the parameter
$\bm a$ on the hyperplane $\textrm{\textbf{H}}=\{\bm y\in
\mathbb{R}^2|\bm b^\textrm{T}\bm y=\beta\}$. Assume $\bar{\bm p}$ is
K-minimal for MOP. For the case that
\begin{equation}\label{aaa}
\bar{t}=\frac{\bm b^\textrm{T}\bm f(\bar{\bm p})-\beta}{\bm
b^\textrm{T} \bm r}~~~\textrm{and}~~~\bm a=\bm f(\bar{\bm
p})-\bar{t}\bm r
\end{equation}
with arbitrarily $\bm r\in \textrm{\textbf{K}}$ and $\bm
b^\textrm{T}\bm r\neq 0$, we have $\bm a\in \textrm{\textbf{H}}$ and
$(\bar{t},\bar{\bm p})$ is feasible for SP($\bm a$). If
$(\bar{t},\bar{\bm p})$ is not a minimal solution of SP($\bm a$),
then there could be another $t'<\bar{t}$, $\bm p'\in \mathcal{D}$
and $\bm k'\in \textrm{\textbf{K}}$ with
\begin{equation}
\bm a+t'\bm r-\bm f(\bm p')=\bm k'.
\end{equation}
Replace $\bm a$ with (\ref{aaa}),
\begin{equation}
\bm f(\bar{\bm p})=\bm f(\bm p')+\bm
k'+\underset{>0}{\underbrace{(\bar{t}-t')}}\underset{\in
\textrm{\textbf{K}} }{\underbrace{\bm r}},
\end{equation}
then it can be concluded that $\bm f(\bar{\bm p})\in \bm f(\bm
p')+\textrm{\textbf{K}}$ for $\bm p'\in \mathcal{D}$, which is
contradict to the definition of $\bar{\bm p}$ K-minimal. Thus
$(\bar{t},\bar{\bm p})$ is a minimal solution of SP($\bm a$). So
far, we have proved that it is sufficient to vary the parameter $\bm
a$ on the hyperplane $\textrm{\textbf{H}}$. Further, we will show
that a subset $\textrm{\textbf{H}}^a\subset \textrm{\textbf{H}}$ is
also sufficient to get the efficient set.

Assume $\bar{\bm p}^1$ is a minimal solution of (\ref{min_f1})
\begin{equation}\label{min_f1}
\underset{\bm p\in \mathcal{D}}{\rm{min}}~~{\bm l^1}^{\textrm{T}}\bm
f(\bm p)
\end{equation}
and $\bar{\bm p}^2$ is a minimal solution of (\ref{min_f2})
\begin{equation}\label{min_f2}
\underset{\bm p\in \mathcal{D}}{\rm{min}}~~{\bm l^2}^{\textrm{T}}\bm
f(\bm p)
\end{equation}
where $\bm l^1=(1,0)$ and $\bm l^2=(0,1)$. The parameters $\bar{\bm
a}^1\in \textrm{\textbf{H}}$ and $\bar{\bm a}^2\in
\textrm{\textbf{H}}$ are given by
\begin{equation}
\bar{\bm a}^i:=\bm f(\bar{\bm p}^i)-\bar{t}^i\bm r
~~~\textrm{with}~~\bar{t}^i:=\frac{\bm{b}^\textrm{T}\bm f(\bar{\bm
p}^i)-\beta}{\bm b^\mathrm{T} \bm r},~~~i=1,2.
\end{equation}

Then, we consider the parameters $\bm a\in \textrm{\textbf{H}}^a$
with the set $\textrm{\textbf{H}}^a$ given by
\begin{equation}
\textrm{\textbf{H}}^a=\{\bm y\in \textrm{\textbf{H}}|\bm y=\lambda
\bar{\bm a}^1+(1-\lambda)\bar{\bm a}^2,~~\lambda\in[0,1]\}.
\end{equation}
It can be inferred from the assumption that $\bar{\bm a}^1,\bar{\bm
a}^2\in \textrm{\textbf{H}}^a\subset \textrm{\textbf{H}}$. For
simplicity, we assume $\bar{\bm a}^1$ is smaller than $\bar{\bm
a}^2$ on the first dimension, i.e., $\bar{a}_1^1<\bar{a}_1^2$.

For any feasible $\bar{\bm p}$, there exists a parameter $\bm a\in
\textrm{\textbf{H}}$ and a $\bar{t}\in \mathbb{R}$ given by
\begin{equation}
\bar{t}=\frac{\bm b^\textrm{T}\bm f(\bar{\bm p})-\beta}{\bm
b^\textrm{T} \bm r}~~~\textrm{and}~~~\bm a=\bm f(\bar{\bm
p})-\bar{t}\bm r
\end{equation}
so that $(\bar{t},\bar{\bm p})$ is a minimal solution of
$\textrm{SP}(\bm a)$. As $\bar{\bm p}^1$ and $\bar{\bm p}^2$ are
minimal solutions of (\ref{min_f1}) and (\ref{min_f2}), we have for
any feasible $\bar{\bm p}$,
\begin{equation}\label{relation}
{\bm l^1}^\textrm{T}\bm f(\bar{\bm p})\geq {\bm l^1}^\textrm{T}\bm
f(\bar{\bm p}^1)~~\textrm{and}~~{\bm l^2}^\textrm{T}\bm f(\bar{\bm
p})\geq {\bm l^2}^\textrm{T}\bm f(\bar{\bm p}^2).
\end{equation}
Suppose that ${\bm l^1}^\textrm{T}\bm f(\bar{\bm p})\geq {\bm
l^1}^{\textrm{T}}\bm f(\bar{\bm p}^2)$, as (\ref{relation}) always
holds, it can be concluded that $\bm f(\bar{\bm p})-\bm f(\bar{\bm
p}^2)\in \textrm{\textbf{K}}$, which is contradict to $\bar{\bm p}$
K-minimal. Thus, it can be shown that
\begin{equation}\label{relation1}
{\bm l^1}^\textrm{T}\bm f(\bar{\bm p}^1)\leq {\bm l^1}^\textrm{T}\bm
f(\bar{\bm p})\leq {\bm l^1}^\textrm{T}\bm f(\bar{\bm p}^2).
\end{equation}
In the same way, the following relation can also be achieved
\begin{equation}\label{relation2}
{\bm l^2}^\textrm{T}\bm f(\bar{\bm p}^2)\leq {\bm l^2}^\textrm{T}\bm
f(\bar{\bm p})\leq {\bm l^2}^\textrm{T}\bm f(\bar{\bm p}^1).
\end{equation}

With the obtained relations (\ref{relation1}) and (\ref{relation2}),
we then demonstrate that the parameter $\bm a$ lies on the segment
between the point $\bar{\bm a}^1$ and $\bar{\bm a}^2$, i.e., $\bm
a=\lambda \bar{\bm a}^1+(1-\lambda)\bar{\bm a}^2$ for a $\lambda\in
[0,1]$. Using the definition of $\bm a$, $\bar{\bm a}^1$ and
$\bar{\bm a}^2$, the following can be obtained:
\begin{equation}\label{fff}
\bm a=\bm f(\bar{\bm p})-\bar{t}\bm r=\lambda (\bm f(\bar{\bm
p}^1)-\bar{t}^1\bm r)+(1-\lambda)(\bm f(\bar{\bm p}^2)-\bar{t}^2\bm
r).
\end{equation}
reformulate (\ref{fff}) as
\begin{equation}\label{reformf}
\bm f(\bar{\bm p})=\lambda \bm f(\bar{\bm p}^1)+(1-\lambda)\bm
f(\bar{\bm p}^2)+(\bar{t}-\lambda \bar{t}^1-(1-\lambda)\bar{t}^2)\bm
r.
\end{equation}
Then we do a case differentiation for $\bar{t}-\lambda
\bar{t}^1-(1-\lambda)\bar{t}^2\geq 0$ and $\bar{t}-\lambda
\bar{t}^1-(1-\lambda)\bar{t}^2<0$ respectively.

For $\bar{t}-\lambda \bar{t}^1-(1-\lambda)\bar{t}^2\geq 0$, we first
divide the set of $\lambda$ into three parts, i.e., $\lambda<0$,
$0<\lambda<1$ and $\lambda>1$, then start by considering the case
that $\lambda<0$.
\begin{equation}
\begin{split}
&~~~~{\bm l^1}^{\textrm{T}}\bm f(\bar{\bm p})\\
&=\lambda {\bm l^1}^\textrm{T}\bm f(\bar{\bm p}^1)+(1-\lambda){\bm
l^1}^\textrm{T}\bm f(\bar{\bm p}^2)+\underbrace{(\bar{t}-\lambda
\bar{t}^1-(1-\lambda)\bar{t}^2)}_{\geq 0}\underbrace{{\bm
l^1}^\textrm{T}\bm
r}_{\geq 0} \\
&\geq \underbrace{\lambda}_{<0} \underbrace{{\bm l^1}^\textrm{T}\bm
f(\bar{\bm p}^1)}_{<{\bm l^1}^\textrm{T}\bm f(\bar{\bm
p}^2)}+(1-\lambda){\bm
l^1}^\textrm{T}\bm f(\bar{\bm p}^2) \\
&>\lambda {\bm l^1}^\textrm{T}\bm f(\bar{\bm p}^2)+(1-\lambda){\bm
l^1}^\textrm{T}\bm f(\bar{\bm p}^2)\\
&={\bm l^1}^\textrm{T}\bm f(\bar{\bm p}^2).
\end{split}
\end{equation}
If ${\bm l^1}^\textrm{T}\bm f(\bar{\bm p})>{\bm l^1}^\textrm{T}\bm
f(\bar{\bm p}^2)$ is satisfied together with ${\bm
l^2}^\textrm{T}\bm f(\bar{\bm p})>{\bm l^2}^\textrm{T}\bm f(\bar{\bm
p}^2)$, it can be concluded that $\bm f(\bar{\bm p})-\bm f(\bar{\bm
p}^2)\in \mathbf{K}$, which is contradicted to $\bar{\bm p}$
K-minimal. Therefore, (\ref{reformf}) is not satisfied for $\lambda
<0$. Then we consider the case that $\lambda>1$.
\begin{equation}
\begin{split}
&~~~~{\bm l^2}^\textrm{T}\bm f(\bar{\bm p})\\
&=\lambda {\bm l^2}^\textrm{T}\bm f(\bar{\bm p}^1)+(1-\lambda){\bm
l^2}^\textrm{T}\bm f(\bar{\bm p}^2)+\underbrace{(\bar{t}-\lambda
\bar{t}^1-(1-\lambda)\bar{t}^2)}_{\geq 0}\underbrace{{\bm
l^2}^\textrm{T}\bm
r}_{\geq 0} \\
&\geq \lambda{\bm l^2}^\textrm{T}\bm f(\bar{\bm
p}^1)+\underbrace{(1-\lambda)}_{<0} \underbrace{{\bm
l^2}^\textrm{T}f(\bar{\bm
p}^2)}_{<{\bm l^2}^\textrm{T}\bm f(\bar{\bm p}^1)}\\
&>{\bm l^2}^\textrm{T}\bm f(\bar{\bm p}^1).
\end{split}
\end{equation}
If ${\bm l^2}^\textrm{T}\bm f(\bar{\bm p})>{\bm l^2}^\textrm{T}\bm
f(\bar{\bm p}^1)$ is satisfied together with ${\bm
l^1}^\textrm{T}\bm f(\bar{\bm p})>{\bm l^1}^\textrm{T}\bm f(\bar{\bm
p}^1)$, it can be concluded that $\bm f(\bar{\bm p})-\bm f(\bar{\bm
p}^1)\in \mathbf{K}$, which is contradicted to $\bar{\bm p}$
K-minimal. Thus, (\ref{reformf}) is not satisfied for $\lambda
>1$. Therefore, it can be concluded that (\ref{reformf}) for the case $\bar{t}-\lambda \bar{t}^1-(1-\lambda)\bar{t}^2\geq 0$ can only be
satisfied for $\lambda \in[0,1]$.

Then we consider the case $\bar{t}-\lambda
\bar{t}^1-(1-\lambda)\bar{t}^2<0$. We first consider the case
$\lambda>1$,
\begin{equation}\label{lamnda1}
\begin{split}
&~~~~{\bm l^1}^\textrm{T}\bm f(\bar{\bm p})\\
&=\lambda {\bm l^1}^\textrm{T}\bm f(\bar{\bm p}^1)+(1-\lambda){\bm
l^1}^T\bm f(\bar{\bm p}^2)+\underbrace{(\bar{t}-\lambda
\bar{t}^1-(1-\lambda)\bar{t}^2)}_{<0}\underbrace{{\bm
l^1}^\textrm{T}\bm
r}_{\geq 0} \\
&\leq \lambda {\bm l^1}^\textrm{T}\bm f(\bar{\bm
p}^1)+(1-\lambda){\bm
l^1}^\textrm{T}\bm f(\bar{\bm p}^2).\\
\end{split}
\end{equation}
As ${\bm l^1}^\textrm{T}\bm f(\bar{\bm p})>{\bm l^1}^\textrm{T}\bm
f(\bar{\bm p}^1)$, (\ref{lamnda1}) can be reformulated as
\begin{equation}\label{lamnda1re}
\underbrace{(\lambda-1)}_{> 0}\big({\bm l^1}^\textrm{T}f(\bar{\bm
p}^1)-{\bm l^1}^\textrm{T}f(\bar{\bm p}^2)\big)>0,
\end{equation}
which is contradict to (\ref{relation}).

For the case $\lambda<0$, it can be obtained in the same way that
\begin{equation}\label{lamnda2}
\lambda {\bm l^2}^\textrm{T}\bm f(\bar{\bm p}^1)+(1-\lambda){\bm
l^2}^\textrm{T}\bm f(\bar{\bm p}^2)\geq {\bm l^2}^\textrm{T}\bm
f(\bar{\bm p})>{\bm l^2}^\textrm{T}\bm f(\bar{\bm p}^2)
\end{equation}
and further we have
\begin{equation}\label{lamnda2re}
\underbrace{(-\lambda)}_{> 0}\big({\bm l^2}^\textrm{T}\bm f(\bar{\bm
p}^2)-{\bm l^2}^\textrm{T}\bm f(\bar{\bm p}^1)\big)>0,
\end{equation}
which is also contradict to (\ref{relation}). Therefore, it can be
concluded that (\ref{reformf}) for the case $\bar{t}-\lambda
\bar{t}^1-(1-\lambda)\bar{t}^2<0$ can only be satisfied for $\lambda
\in[0,1]$.

Based on the previous results, the following conclusion can be drawn: For any
K-minimal solution $\bm p$ of MOP, there exists a parameter $\bm
a\in \textrm{\textbf{H}}^a$ and some $\bar{t}\in \mathbb{R}$ so that
$(\bar{t},\bar{\bm p})$ is
a minimal solution of $\textrm{SP}(\bm a)$.\\

\subsection*{6.5~~Proof of Theorem 2}

Defining the following auxiliary variables
\begin{align}
& t':=\frac{\bm b^\textrm{T}\bm f(\bar{\bm p})-\beta}{\bm
b^\textrm{T} \bm
r} \\
& \bm a':=\bm a+(\bar{t}-t')\bm r-\bar{\bm k}=\bm f(\bar{\bm
p})-t'\bm r,
\end{align} it is straightforward that $\bm a'\in \textrm{\textbf{H}}$ and $\bm a'+t'\bm r-\bm f(\bar{\bm
p})=\bm 0_2$. The point $(t',\bar{\bm p})$ is feasible for
$\textrm{SP}(\bm a')$ and it is also a minimal solution, because
otherwise there exists a feasible point $(\hat{t},\hat{\bm p})$ of
$\textrm{SP}(\bm a')$ with $\hat{t}<t'$ and some $\hat{k}\in
\mathbf{K}$ with
\begin{equation}
\bm a'+\hat{t}\bm r-\bm f(\hat{\bm p})=\hat{\bm k}
\end{equation}
together with the definition of $\bm a'$, it can be concluded that
\begin{equation}
\bm a+(\bar{t}-t'+\hat{t})\bm r-\bm f(\hat{\bm p})=\hat{\bm
k}+\bar{\bm k}\in \mathbf{K}.
\end{equation}
Hence, $(\bar{t}-t'+\hat{t},\hat{\bm p})$ is feasible for
$\textrm{SP}(\bm a)$ with $\bar{t}-t'+\hat{t}<\bar{t}$, which is in
contradiction to the minimality of $(\bar{t},\bar{\bm p})$ for
$\textrm{SP}(\bm a)$. Thus, $(t',\bar{\bm p})$ is also a minimal
solution of $\textrm{SP}(\bm a')$.

Then we demonstrate that the two scalar problems have the same
Lagrange multiplier $\bm \mu$. The Lagrange function $\mathcal{L}$
to the scalar optimization problem $\textrm{SP}(\bm a)$ related to
the MOP is given by
\begin{equation}
\mathcal{L}(t,\bm p,\bm \mu, \bm \beta, \bm a)=t-\bm
\mu^\textrm{T}(\bm a+t\bm r-\bm f(\bm p))- \bm \beta^\textrm{T}\bm
g(\bm p).
\end{equation}
If $\bm \mu$ is Lagrange multiplier to the point $(\bar{t},\bar{\bm
p})$, then it follows
\begin{equation}
\begin{split}
&\nabla_{(t,\bm p)}\mathcal{L}(\bar{t},\bar{\bm p},\bm \mu, \bm
\beta, \bm a)^\textrm{T}\left(\begin{array}{cc}
             t-\bar{t}\\
             \bm p-\bar{\bm p}\\
  \end{array}\right)\\
&=\left[\left(\begin{array}{cc}
             1\\
             0\\
  \end{array}\right)-\sum_{i=1}^{2}\mu_i\left(\begin{array}{cc}
             r_i\\
             -\nabla_{\bm p}f_i(\bar{\bm p})\\
  \end{array}\right)-\sum_{i=1}^{N+1}\beta_i\left(\begin{array}{cc}
             0\\
             \nabla_{\bm p}g_i(\bar{\bm p})\\
  \end{array}\right)\right]^\textrm{T}\left(\begin{array}{cc}
             t-\bar{t}\\
             \bm p-\bar{\bm p}\\
  \end{array}\right).
\end{split}
\end{equation}
Hence $1-\bm \mu^\textrm{T}\bm r=0$ and $(\bm
\mu^\textrm{T}\nabla_{\bm p}{\bm f(\bar{\bm p})})(\bm p-\bar{\bm
p})\geq 0$. Further we have $\bm \mu^\textrm{T}(\bm a+t\bm r-\bm
f(\bm p))=0$. For the minimal solution $(t',\bar{\bm p})$ of the
problem $\textrm{SP}(\bm a')$, it is
\begin{equation}
\bm a'+t'\bm r-\bm f(\bar{\bm p})=\bm 0_2,
\end{equation}
and thus $\bm \mu^\textrm{T}(\bm a'+t'\bm r-\bm f(\bar{\bm p}))=0$.
Together with the following equality
\begin{equation}
\nabla_{(t,\bm p)}\mathcal{L}(t',\bar{\bm p},\bm \mu, \bm \beta, \bm
a')=\nabla_{(t,\bm p)}\mathcal{L}(\bar{t},\bar{\bm p},\bm \mu, \bm
\beta, \bm a),
\end{equation}
we also have
\begin{equation}
\nabla_{(t,\bm p)}\mathcal{L}(t',\bar{\bm p},\bm \mu, \bm \beta, \bm
a')^\textrm{T}\left(\begin{array}{cc}
             t-t'\\
             \bm p-\bar{\bm p}\\
  \end{array}\right)\geq 0.
\end{equation}
Therefore, $\bm \mu$ is also Lagrange multiplier to the point
$(t',\bar{\bm p})$ for the problem $\textrm{SP}(\bm a')$.

redpajama_set_name: RedPajamaArXiv


text:
Why Queen Elizabeth Is a Duke and Not a Duchess
Updated Dec 26, 2018 @ 6:00 am
If you aren't a queen, king, princess, or prince under the British monarchy, the next best thing is being a duke or duchess. The highly distinguished titles are given hereditarily and outrank earls, barons, marquess, and viscounts alike.
Dukedom is very old (and very gendered), and perhaps the best example of the strange inconsistencies with it comes courtesy of Queen Elizabeth II herself. Despite being a woman, the monarch is technically the Duke of Lancaster. And the reason why is pretty weird.
Credit: Bethany Clarke
Historically speaking, despite the fact that dukes and duchesses are considered titles of the same rank, dukedom has come with more privilege — for example, if a duchess marries, her spouse can't inherit her title, but if a duke marries a woman, she will become a duchess. The reason for Elizabeth's dukedom, though, comes down to land power, as Town & Country clarifies.
RELATED: Forget Kate Middleton Driving — Queen Elizabeth Didn't Even Take a Road Test
In the United Kingdom, there's land called the Duchy of Lancaster, which forms a central part of the reigning monarch's income. The Duchy of Lancaster is held and controlled by (you guessed it) the Duke of Lancaster. And this means that the duke must be whomever is currently on the throne. According to the Duchy of Lancaster's official website, it's been this way since way back in 1399.
VIDEO: Meghan Markle and the Queen Are Clashing Over How to Handle Her Dad
The title itself was specified long ago as duke, not duchess, so it was never changed.
A little odd? Perhaps, but it's still not the strangest royal rule we've ever heard ...

redpajama_set_name: RedPajamaCommonCrawl


A beautiful established woodland village enviably positioned within walking distance of the sea front town of Penarth and the undisputed cultural capital of Wales and 'European Capital of Culture' finalist the city of Cardiff. We're Introducing you to our local community magazine, 'Llandough Life' whos' content is entirely the work of Llandough residents. Distributed to 3000 homes each month in the CF64 region, 'Llandough Life' is the go to hand book for news, trades and local knowledge- don't be without it!
The magazine is run by volunteers from the village and is the only magazine in South Wales written by it's residents. We print stories, photographs and experiences of residents and neighbors. Our mission is to create an inclusive community to live. We accept adverts to contribute to printing and distribution, but only if the service is viable in our area.
Residents: 7th February 2019 is the submission deadline for our Valentines Issue of 'Llandough Life'. Send us your photographs, drawings and writing to be considered for publication. Please use your own words and photos so we don't encounter permission issues. By sending us content you agree to letting the 3,000 residents of the Llandough area and their family and friends see it. We hope you enjoy creating and making stories to share as much as we love reading them. If you aren't selected for publication, don't be disheartened- try again for the next edition due for release in March 2019.


In any industry, planning for maintenance downtime is crucial – especially on a mining site where a few minutes lost can equal large production losses.
Oil-cooled transformers have a considerable amount of maintenance that has to be conducted on a regular basis. This includes the testing of cooling liquids by qualified personnel, as well as the replacement of cooling liquids. Oil-type transformers are also susceptible to explosion during operation – a machine failure that cannot be afforded.
The high maintenance of oil-type transformers, the safety and environmental threat it poses, and losses due to downtime therefore prompts engineers and procurement managers to reconsider the components of their production line.
This particular mine is committed to phasing out all of its 30+ oil-type transformers, and we delivered an alternate solution – Cast Resin Dry-Type Transformers.
Due to its design and ability to withstand the harsh conditions on a mine, cast resin transformers call for a low maintenance schedule and thus also considerably less downtime. It is reliable, safe, and flame retardant – crucial factors on a mining site.
Our technical team provided their full support and dedication, during the commissioning and assembly of these transformers on site.


text:
Webdriver Manager [![CircleCI Status](https://circleci.com/gh/angular/webdriver-manager.svg?style=shield)](https://circleci.com/gh/angular/webdriver-manager) [![Join the chat at https://gitter.im/angular/webdriver-manager](https://badges.gitter.im/Join%20Chat.svg)](https://gitter.im/angular/webdriver-manager?utm_source=badge&utm_medium=badge&utm_campaign=pr-badge&utm_content=badge)
=================

A selenium server and browser driver manager for your end to end tests. This is the same tool as `webdriver-manager` from the [Protractor](https://github.com/angular/protractor) repository.

**Note:** Version 9 and lower please reference [pose/webdriver-manager](https://github.com/pose/webdriver-manager). If there are features that existed in version 9 and lower, please open up an issue with the missing feature or a create a pull request.

Getting Started
---------------

```
npm install -g webdriver-manager
```

Setting up a Selenium Server
----------------------------

Prior to starting the selenium server, download the selenium server jar and driver binaries. By default it will download the selenium server jar and chromedriver binary.

```
webdriver-manager update
```

Starting the Selenium Server
----------------------------

By default, the selenium server will run on `http://localhost:4444/wd/hub`.

```
webdriver-manager start
```

Other useful commands
---------------------

View different versions of server and driver files:

```
webdriver-manager status
```

Clear out the server and driver files. If `webdriver-manager start` does not work, try to clear out the saved files.

```
webdriver-manager clean
```

Other topics:
--------------

- [mobile browser support](docs/mobile.md)
- [protractor support](docs/protractor.md)

redpajama_set_name: RedPajamaGithub


text:
Liverpool: Arsenal clash will be acid test for Reds says Jamie Carragher
By Naveen Ullal
August 18, 2015 11:54 BST
After a good start in the first two weeks of the 2015/16 season, Liverpool will take on Arsenal on 24 August Getty
Jamie Carragher has claimed that Liverpool's clash against Arsenal in the upcoming fixture will be an acid test for his former club.
After back-to-back 1-0 wins in the first two fixtures against Stoke City and Premier League newcomers AFC Bournemouth, Brendan Rodgers will take his men to the Emirates on 24 August.
The Northern Irishman fielded Martin Skrtel and Dejan Lovren in the heart of the back four, while summer signings Nathaniel Clyne and Joe Gomez were the two full-backs in the two league matches.
The former Liverpool centre-back has urged the Anfield club to build on their defensive displays.
"The problem last season, equally with the lack of goals, was conceding too much. And now they've got two clean sheets on the bounce," Carragher told Sky Sports.
"Sometimes you can keep a clean sheet but your 'keeper has made great saves or you've been fortunate, but they haven't given away any chances away in two games.
"That's the thing to build on. The goals will come, for sure, as players get fitter and stronger. But the acid test will come away at Arsenal next week considering what happened there last season," the Anfield legend said.
"It will be a test for the back four next week but if they can come through that, then that will give them confidence that they have more organisation, and that can help massively," he explained.
After a 2-0 defeat to West Ham United in the opening match, Arsenal bounced back with a 2-1 victory over Crystal Palace on Sunday, 16 August. Arsene Wenger's side will be keen to seal their first home win of the new season against the Reds.
The Merseyside club have maintained two clean sheets in 2015/16 season. Clyne is delighted after his new club's defensive record, but admitted Liverpool still have scope to improve their back four.
"There's a lot of work going on, we have to improve on our defensive qualities and we showed today that we can be solid in defence," Clyne told Liverpool's official website.
"If we can get a clean sheet there's always the possibility of picking up maximum points like we did today so long the may the clean sheets continue!"
More about Liverpool
Bournemouth boss Eddie Howe: Benteke goal for Liverpool was clearly offside
Christian Benteke impresses Brendan Rodgers after goal on home Liverpool debut
Liverpool manager Brendan Rodgers dismisses Jose Enrique criticism
Liverpool: Reds reject loan bid from Roma for defender Mamadou Sakho
Related topics : Liverpool FC Brendan Rodgers Arsenal FC Arsene Wenger AFC Bournemouth


For the month of April, join us every Friday for our Family Polo Asado Brunch with a buy one get one complimentary offer. Enjoy sharing style grilled meats, delicious tapas and a fine selection of cheese, all served to your table.
To book your table please contact us on +971 4 361 811 or click here.
Unwind and engage with a a hearty, family-friendly Iftar set in contemporary elegance. Located within the sleek Palermo restaurant, this Iftar is designed for large groups with little ones and is filled with a lavish array of sweet and savoury treats, complemented by the authentic spirit of tradition.
To book your Iftar with us please contact us on +971 4 361 8111 or click here.
This March at The Spa, relax and rejuvenate with our special offer including a Balinese Full Body Massage and The Express Facial. Your journey of relaxation will begin with a calming 60 minute full body massage that combines gentle stretching and aromatherapy. After achieving a deep state of relaxation, our revitalising Express Facial will leave your skin healthy and hydrated.
To book your journey of relaxation please click here.


text:
HomePoliticsC'River State DG Commiserates With Victims, Weeps Over Loss Of Their 4 Children In Landslide
C'River State DG Commiserates With Victims, Weeps Over Loss Of Their 4 Children In Landslide
BHG-InfoDesk Admin July 25, 2022
Director General of the Cross River State Primary Healthcare Development Agency, Dr Janet Ekpenyong on Monday wept profusely when she visited the family of middle-aged and physically challenged Ndifreke Sylvester over the loss of their four children following a landslide in the state.
Sources broke the news of a landslide in the community after a heavy downpour on the 13th of July 2022, a tragedy that consumed four children of the same parents and has now attracted humanitarian responses.
The DG expressed pain over the colossal damage that the landslide caused and the wreckage to Ndifreke's household.
Ekpenyong who represented the state governor, Prof Ben Ayade became emotional when she visited to donate medical and other relief items to Ndifreke and other survivors.
She said the loss has been one of the most devastating news that hit her in recent times.
She promised to keep a day-to-day watch over the survivals and ensure that, best practices are carried out on them in the General Hospital.
She further promised the family, the government's commitment to ensure that their welfare and that of other survivors are well taken care of at all times.
"This is a very sad news to behold; waking up to hear of this quantum loss and damages left me devastated all day.
"It's not easy to lose a loved one let alone having four of your children killed in one day.
"The Governor and his dear wife, Senator Ben and Dr Linda Ayade sympathize with the family. They sent us to visit the family, provide both medical and welfare supports and to see that, the two surviving kids are doing well and responding to treatment."
The government promised to defray all medical expenses and ensure that their house is rebuilt.
The landslide incident occurred at Okong-Mbang village, off Nasarawa Community in Calabar.
Unemployed Ndifreke who is physically challenged, his wife and two other children had escaped while the other four who were having their meals were submerged.
Featured Politics


In A Long Way From Home the Adhikari brothers follow their dreams while re-discovering their relationship with one another. Mitali Meelan weaves a contemporary perspective of love while keeping true to the traditions of an Indian middle-class family.
The Adhikaris are a regular, happy family – or so they will have you believe. But at the dinner table, a silence hangs heavy over the three children, especially Arihant and Ishan.
Twenty-two-year-old Arihant possesses a secret talent and a shattered heart. He yearns for his muse and ex-girlfriend, only to surprise himself when they finally meet after a long and difficult time apart. If only Ishan could share some brotherly advice rather than sneak about, aloof and increasingly suspicious.
Ishan twenty-eight, seems to have the job of his dreams – or at least, his parents' dreams. But how long can he stay torn between two worlds? How long before someone finds out about his evenings: about the place he goes to and the woman he sees.
A novel set in the calm before a storm – after which nothing will be the same for this seemingly happy household.
Mitali Meelan works for Shemaroo Entertainment. Mitali is the author of Coffee and Ordinary Life and The Guest. She was first published by Ravinder Singh under the Black Ink imprint.


I nominate Vinay and Rhyme me a Smile for Perfect Poet Award.
PS: Sorry due to some tight schedule last week I am not able to read and comment but now this time I will read as many poets as possible.. SmileS..
i may forgive but won't forget ever ………………………………….
oh oh.. Good.. but try to forget things..
Next PostNext ! Happy Birthday K S BALAN ! இனிய பிறந்தநாள் வாழ்த்துக்கள் கே எஸ் பாலன் !


This article is focused on main changes introduced by the last novelization of the Czech Insolvency Act, i.e. the Act No. 64/2017 Coll., by which the Act No. 182/2008 Coll., the Insolvency Act ("the IA"), is amended, which became effective as of 1st July 2017.
The aim of this rather large novelization is especially: (i) to reinforce the protection of businessmen against so-called bullying insolvency petitions, (ii) to realize changes in the field of the debt relief, which have turned out to be necessary due to the application practise, and (iii) to increase the transparency of the insolvency proceedings.
The novelization embodies new institute of the preliminary assessment of an insolvency petition submitted by a creditor (Section 100a IA). The insolvency court will be newly obliged to assess the insolvency petition filed by a creditor immediately after its delivery, i.e. before its publication in the Insolvency Register. If the court will see any doubts about the legitimacy of such petition, it will decide that neither the insolvency petition nor any other accompanying documents will be published for a certain period of time. This procedure should eliminate negative impacts caused by abusing insolvency petitions, which are sometimes submitted by entrepreneurs in order to deliberately injure their business competitors by the actual fact that the opening of the insolvency proceedings is published in the Insolvency Register.
The novelization modifies the existing legal definition of bankruptcy in the form of insolvency in relation to businessmen keeping accounts. The novelization introduces new institute of so-called coverage gap, which means the difference between the amount of due monetary debts and the amount of debtor's disposable incomes. According to the new legislation (Section 3 (3) IA) there is the presumption of non-bankruptcy, where the coverage gap reported in the statement on the state of liquidity does not exceed 10 % or where the outlook of liquidity developments certifies that the coverage gap will fall under this limit during the period for which the outlook has been drawn up.
Higher protection against so-called bullying insolvency petitions should be also ensured by implementation of the new obligation of the insolvency petitioner to evidence the existence of a mature claim – and to join the submission of this claim to the insolvency petition at the same time (Section 105 IA). In the case that the debtor is a legal entity, the insolvency petitioner, who is keeping accounts or tax records according to the Income Tax Act, will be obliged to give evidence of his claim by (i) the acknowledgement of debt with verified debtor's signature or (ii) the enforceable decision or (iii) the notarial deed with the consent to enforceability or (iv) the executory deed with the consent to enforceability or (v) the confirmation made by an auditor, a court expert witness or a tax advisor and certifying that the creditor accounts for this claim.
The novelization also increases the upper limit of the fine, which may be imposed upon the insolvency petitioner by the respective insolvency court, if he submits flagrant groundless insolvency petition, i.e. from existing sum of CZK 50,000 to CZK 500,000 (Section 128a (3) IA).
In the field of the debt relief the novelization brings the new legislation in relation to the specification of the authorised persons entitled to submit the insolvency petition, where the bankruptcy will be solved by the debt relief. The debt relief petition can no longer be complied by the debtor itself, but it must be made only by an attorney, a notary, a court executor, an insolvency trustee or by an accredited person – a legal entity, which has obtained an accreditation of the Ministry of Justice for providing services in the area of the debt relief (public beneficial legal entities).
The provision about mandatory representation of a debtor by a professional will not be applied, if the debtor is a natural person and has Master's Law or Economic Education or passes the exam for an insolvency trustee, or if such professional acts on behalf of a debtor, who is a legal entity.
The novelization also limits the amount of the remuneration for writing up and filling the debt relief petition. The authorised professional will be entitled to ask for a remuneration in the maximum amount of CZK 4,000 without VAT and in the case of joint debt relief of a married couple in the maximum amount of CZK 6,000 without VAT (Section 390a (3) IA). The accredited persons will be allowed to provide the services free of charge only.
These changes should help to ensure the satisfactory professionalism, when providing services connected with the institute of the debt relief, and at the same time they should lead to price reduction of these services.
Previous: Tax-Loss Carryfowards of Corporations. What is new? What should be done?


text:
Home » News » Mental health guidance for women impacted by recent high-profile media coverage of sexual violence, harassment, and abuse scandals
Mental health guidance for women impacted by recent high-profile media coverage of sexual violence, harassment, and abuse scandals
In consultation with Rape Crisis England and Wales ahead of International Women's Day, we have released guidance to help those who have found recent reporting around sexual violence in the media re-traumatising.
This follows the recent surge in reporting of scandals featuring sexual violence – including scandals in the entertainment and charity sector and during the US Presidential election - that have the potential to set off traumatic feelings in people who have experienced sexual violence.
Research indicates experiences of sexual violence among women are widespread. 61% of women reported having experienced sexual violence during adulthood, and 10 percent reported having experienced sexual violence in the past year. 1.
The Mental Health Foundation focuses on prevention of mental health problems. The advice is designed to help people at risk of being re-traumatised by giving them techniques to gain some control and headspace after they have seen an upsetting report.
Dr Amy Pollard from the Mental Health Foundation said:
"The recent scandals and the growing #MeToo movement have been an important moment for women to raise their voices against sexual violence and harassment.
"But it can bring up a range of conflicting feelings for those who have experienced sexual violence themselves. Being surrounded by constant reports of sexual violence can be re-traumatising for survivors, especially if they feel the coverage is something they can't escape from.
"It's very hard to know how many survivors are affected this way – but we do know that there are many people who are potentially at risk. Many may have never disclosed their experiences.
"But there are positive things people can do to protect their mental health from repeated trauma. Simple techniques like taking a breath when you are feeling anxious about a report you have seen or writing down your thoughts can make a real difference.
"That is why we have published these guidelines today following consultation with experts on managing trauma from sexual violence."
Recent media reports have been linked to an upsurge of women seeking advice and help. Rape Crisis saw a 28% increase in calls in the two weeks after the Harvey Weinstein scandal broke.
Katie Russell, a spokesperson for Rape Crisis said:
"We routinely see marked spikes in contact from victims and survivors of child sexual abuse, rape and all forms of sexual violence when these topics are high profile in the news or featured in the storylines of popular dramas.
"Survivors frequently tell us about the triggering impact this kind of content can have, and we regularly ask media to be mindful of the very many survivors who will always inevitably be among any given audience, because of the huge prevalence of sexual violence."
While the advice has been released for International Women's Day, the Mental Health Foundation and Rape Crisis England & Wales fully acknowledge that men and boys also experience rape and sexual violence and that the impacts on their lives can be similarly devastating and long lasting. These techniques can be applied by anyone, irrelevant of gender.
Read the guidance
Women, Abuse
Please note that these contact details are for media enquiries only.
If you are a journalist and have a media request, please call 020 7803 1110.
For out of hours urgent press enquiries, please email [email protected] or call 07702 873939. If your call is not answered immediately leave a message and we will get back to you as soon as possible.
Unfortunately, we do not have the capacity to respond to student media requests but we hope the website is useful in helping with your studies.
Other enquiries
For emotional support, please contact Samaritans helpline on 116 123. For anyone seeking information on help and support in their area, contact Mind Infoline on 0300 123 3393 or text 86463.


text:
#include <quic/state/test/AckEventTestUtil.h>

namespace quic::test {

AckEventStreamDetailsMatcherBuilder&&
AckEventStreamDetailsMatcherBuilder::setStreamID(const uint64_t streamIdIn) {
  maybeStreamId = streamIdIn;
  return std::move(*this);
}

AckEventStreamDetailsMatcherBuilder&&
AckEventStreamDetailsMatcherBuilder::setStreamBytesAcked(
    const uint64_t ackedIn) {
  maybeStreamBytesAcked = ackedIn;
  return std::move(*this);
}

AckEventStreamDetailsMatcherBuilder&&
AckEventStreamDetailsMatcherBuilder::setStreamBytesAckedByRetrans(
    const uint64_t ackedByRetransIn) {
  maybeStreamBytesAckedByRetrans = ackedByRetransIn;
  return std::move(*this);
}

AckEventStreamDetailsMatcherBuilder&&
AckEventStreamDetailsMatcherBuilder::setMaybeNewDeliveryOffset(
    const folly::Optional<uint64_t>& maybeNewDeliveryOffsetIn) {
  maybeNewDeliveryOffset = maybeNewDeliveryOffsetIn;
  return std::move(*this);
}

AckEventStreamDetailsMatcherBuilder&&
AckEventStreamDetailsMatcherBuilder::addDupAckedStreamInterval(
    const DupAckedStreamIntervals::interval_type& interval) {
  dupAckedStreamIntervals.insert(interval);
  return std::move(*this);
}

AckEventStreamDetailsMatcherBuilder&&
AckEventStreamDetailsMatcherBuilder::addDupAckedStreamInterval(
    const uint64_t startIn,
    const uint64_t endIn) {
  dupAckedStreamIntervals.insert(startIn, endIn);
  return std::move(*this);
}

AckEventStreamDetailsMatcherBuilder&&
AckEventStreamDetailsMatcherBuilder::clearDupAckedStreamIntervals() {
  dupAckedStreamIntervals.clear();
  return std::move(*this);
}

} // namespace quic::test


text:
ENDANGERED SPECIES SPOTLIGHT: Black Rhinoceros
SPECIES- Black Rhinoceros (Diceros bicornis)
CURRENT RANGE- Southern and East Africa
CURRENT THREAT- Illegal poaching, Habitat loss
CONSERVATION STATUS- Critically Endangered
WHERE YOU CAN SEE THEM- Botswana, Namibia, South Africa, Tanzania, and U.S. zoos including the Lincoln Park and St. Louis Zoos.
"Black rhinoceros" is actually a misnomer, as the animal's skin has more of a light grayish-brown color. Its African counterpart, the white rhino, was originally so named due to a misinterpretation of the Afrikaans word wyd (or wide), which refers to its square upper lip. The black rhino, on the other hand, has a pointed, hooked lip, and the two species are now often referred to as square-lipped and hooked-lipped rhinos. The black rhino averages 52-70 inches in height at the shoulder, and 9-12 feet in length, with adults typically weighing anywhere from 1800 to over 6000 pounds. The two horns on their skulls are made out of keratin (the same substance in human hair and fingernails), can grow up to more than 50 inches in length, and are used for defense and digging up roots and breaking branches during feeding. In addition to the shape of the lips, you can distinguish black from white rhinos by their smaller size and the position of their heads: Where white rhinos are grazers who prefer grass, black rhinos hold their heads higher to browse on leaves and twigs. Black rhinos have poor eyesight, but an excellent sense of hearing and smell, using their large, rotating ears to detect predators.
WHY ARE THEY ENDANGERED?
To put it bluntly, because their horns are worth a ridiculous amount of money due primarily to lies purporting their healing and aphrodisiac properties causing a rapidly increasing demand in the traditional Asian medicinal market. With poaching reaching record levels in 2011 (443 killed last year, up from just 13 in 2007), rhino horns have been valued at up to $45,000 a pound on the black market. Although there is not a single scientific report to back up the claims that rhino horn can cure cancer or impotence, these beautiful creatures continue to be killed at an alarming rate: the last Vietnamese rhino was shot and killed in 2009, and in November of 2011 the Western Black Rhinoceros was declared extinct by the International Union for the Conservation of Nature & Natural Resources (IUCN). At current poaching levels, remarkable conservation efforts will be required to save the other Black Rhino subspecies from meeting a similarly tragic fate.
WHAT'S BEING DONE TO SAVE THEM?
If there's good news for rhinos, it's that they have one of the most outspoken and politically active networks within the wildlife conservation community. Organizations such as Save The Rhino, Saving Rhinos, the Black Rhino Foundation, World Wildlife Fund, and the Sheldrick Wildlife Trust are all extremely active in raising awareness and funds for rhino conservation initiatives. On the ground, the war against poaching has become increasingly violent: Rangers have been authorized to shoot poachers on sight, and in the last two months alone police in South Africa have arrested two Vietnamese nationals in illegal possession of rhino horns and seized the assets (over £4 million) of accused poaching kingpin Dawie Groenewald. Though it seems difficult to understand how African nations such as Botswana, South Africa and Tanzania– who depend so heavily on the revenue generated by ecotourism– could let one of the legendary "Big 5" species teeter so close to the brink of extinction, it appears as if they are finally treating the situation with grave severity it merits. Hopefully, with the help of captive breeding programs and private game reserves, it won't prove too little, too late to save the black rhino. –Bret Love
If you enjoyed our story on the Black Rhinoceros, you might also like:
ENDANGERED SPECIES SPOTLIGHT: Dhole
ENDANGERED SPECIES SPOTLIGHT: Iberian Lynx
26 Weird Animals Around the World
Filed Under: Africa, Wildlife Conservation


Like many biomedical investigators, I've been sweating it over the resubmission of an R01 grant my collaborator and I worked furiously on and submitted on November 1. He's the principal investigator, but I'm a coinvestigator with 25% effort; I also wrote one of the three specific aims and most of another, the justification for animal use, and the IACUC (animal use) protocols for the project. Consequently, I have almost as much invested in the success or failure of this grant proposal as its PI does, although he certainly gets props from me for pulling us two co-investigators together with him to do this project. Getting the grant, of course, would be great. True, it wouldn't be as great as my getting another R01 on which I'm the principal investigator (my surgeon's ego still makes me want to prove that I can get another one for myself, crappy NIH funding situation or not), but in essence it's the next best thing and would give a huge boost to my collaborator's career and a nice smaller boost to mine. Thus, getting this one would make me happy, even aside from the nice, fresh infusion of cash into my tight-budgeted lab it would produce. Our score is excellent, and in years past would have been easily in the fundable range; unfortunately, this year, thanks to the budget woes of the NIH, it's right on the cusp. We're riding the knife edge right now, although, given that the PI is a new investigator (the NIH gives new investigators a couple of percentile points' break on the payline in order to make it a little easier for them to be funded), chances as of today look better than 50-50 that we'll get the grant, and it's a nice big one.
Fortunately for our fragile minds, it turns out that the PI's sources tell him that the paylines for the National Cancer Institute will be set this week; so we shouldn't have to sweat too much longer. In the meantime, he turned me on to a blog that I had previously been shockingly unaware of: Grant Writing, Editing, and Grantsmanship. Although the blogger is anonymous (like me, even though my anonymity seems to be falling apart faster than Britney Spears' psyche and career), he or she seems to be hooked into everything an NIH grant applicant wants to hear about, like paylines.
In the meantime, while the PI and I are still sweating it out over our grant, I'm going to sing this Ode to the NIH Payline. Then, in a little more than two years, I can sing it again when I submit the competitive renewal application for my own R01. Well, in reality, I hope I've submitted other grants before that; even better would be to get one funded, so that I don't have all my eggs in one basket, as I do right now. Otherwise I could be in a world of hurt in early 2010 if I fail to renew.
Can someone remind me why I got into this whole surgeon-scientist thing again?


Less is more: we concentrate on one single product group and in this way we achieve appreciable benefits in quality and performance.
That is why Knipex is the world's leading brand in pliers today.
People in more than 100 countries work with Knipex pliers – and this figure is on the increase.
The performance, service life and ergonomics of our tools are guided by the need to meet the high expectations of professional users.
That is why we use particularly high-quality materials and modern production technology.
The range of around 1,000 articles offers the optimum selection for the respective purpose - for modern applications too, such as in solar technology, wind energy and aviation.
For 130 years we have been an owner-managed family enterprise, which is now in its fourth generation.
We feel a special bond with our employees and the people in our region.
We are guided by values, operate according to the principles of sustainability and we consciously assume responsibility in our environment.
Above all, we are committed to social, educational and cultural issues.


Start early on St. Patrick's Day in Wicker Park with our Kegs & Eggs party. We'll open at 8am to get the party started. Tickets are just $40 and available below.
We'll offer classic Irish food features throughout the week as well.


Buy Alien: Isolation (Nostromo Edition) on Daily Game Sale for just €8.94 EUR.
Below you can find "helpful" reviews from verified owners of Alien: Isolation (Nostromo Edition) - provided by Steam. You can provide a review for this game on Steam, and it may be added here automatically after a short time.
Absolutely fantastic game. Masterpiece game, unique atmosphere. Everyone should check this amazing game.
In this game you're playing as a socially awkward introvert woman who runs away and hides from a lovable creature who only wants to give her a hug (and a kiss on the forhead).
A truly beautiful and terrifying game!
The xenomorph looks incredible and stays true to the movie Alien. This game definitely gives that helpless feel and focuses on hiding and surviving, but also give you a bit of control with bombs that you can craft or the much loved flame thrower!
It's definitely a tense and enjoyable game if you give it a chance.
Would definitely recommend to friends, 10/10!
Never have i ever wanted to not play a game but want to play it as much as this game. This game does EVERYTHING a horror game needs perfectly. I'm mostly proud of myself for even manageing to beat this game. The atmosphere, the immersion, the music, the graphics... Up until this game whenever i heard SEGA i would think of Sonic the hedgehog (how times change)... Sega blew my mind with this game. When i have the guts i do plan on replaying this game again sometime or at least continue the DLC's.
Game is scary, and pays great respect to the original film.
video game titles based of film cinema.... ------> GARBAGE, TRASH.
The best alien game out there. Especially for the fans of the original Alien 1979 movie. With great atmosphere and amazing audio design. One of my favorite horror games. So immersive with great attention to detail. Love this game.
It's like they made a simulator for the original alien movie.
no, that's literally what this game feels like it's supposed to be. "alien simulator" complete with being stalked across sevastopol station by one of the SINGLE most terrifying aliens in gaming.
this right here, this'll make you ♥♥♥♥ yourself in terror.
Just finished this materpiece and I recommend it for sure, amazing game and looks amazing on ultra graphics setting.
One of the most scary games I have played (on par with Outlast although on a different style). You always have that feeling that you are not safe anywhere, always checking your back and hearing wierd noises from everywhere.
Long game also, will get you lots of hours of fun, just have a few spare pants available and you are ready to go.
If you like the original Alien (1979) movie then you will love this. Period.
It's nostalgia at its absolute greatest.
6. Meet the Alien and poop your pants.
Can't wait til I get alzheimer's so I'll be able to play this game for the first time again.
Alien: Isolation - New DLC AvailableAbout the GameHOW WILL YOU SURVIVE? Discover the true meaning of fear in Alien: Isolation, a survival horror set in an atmosphere of constant dread and mortal danger. Fifteen years after the events of Alien™, Ellen Ripley's daughter, Amanda enters a desperate battle for survival, on a mission to unravel the truth behind her mother's disappearance. As Amanda, you will navigate through an increasingly volatile world as you find yourself confronted on all sides by a panicked, desperate population and an unpredictable, ruthless Alien. Underpowered and underprepared, you must scavenge resources, improvise solutions and use your wits, not just to succeed in your mission, but to simply stay alive. Overcome an ever-present deadly threat – Experience persistent fear as a truly dynamic and reactive Alien uses its senses to hunt you down and respond to your every move.
Improvise to survive – Hack systems, scavenge for vital resources and craft items to deal with each situation. Will you evade your enemies, distract them or face them head on? Explore a world of mystery and betrayal – Immerse yourself in the detailed setting of Sevastopol, a decommissioned trading station on the fringes of space. Encounter a rich cast of inhabitants in a world scarred by fear and mistrust. When she left Earth, Ellen Ripley promised her daughter Amanda she would return home for her 11th birthday. Amanda never saw her again.
Fifteen years later, Amanda, now a Weyland-Yutani employee, hears that the flight recorder of her mother's ship, the Nostromo, has been recovered at the remote trading station Sevastopol. The temptation for her to finally understand what happened is too much to resist. When the crew arrive at Sevastopol, they find something is desperately wrong. It all seems to be connected to an unknown menace, stalking and killing deep in the shadows. In order to uncover the truth about her mother, Amanda is forced to confront the same terrifying thing that separated them.