Acceptance for publication in Mathematical and Computer Modelling

這一篇投稿到Mathematical and Computer Modelling的文章，終於在第四次修改後接受了！也不得不佩服這位審稿者很認真，鍥而不捨。短短的五週，我們魚雁往返四次～～以下是審稿者最後來信的內容，蠻有趣的！

First, let me say that I am not trying to make life difficult for the authors. I believe that the paper
should be published. I just want to make sure that everything is as clear and accurate as possible.


校訂稿，真的這家出版社很用心～～哪天你有paper被這期刊接受你就知道

以下是TeX檔


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\begin{document}

\title{\textbf{Optimal lot size for an item with partial backlogging when
demand is stimulated by inventory above a specific level}}
\author{{Tsu-Pang Hsieh\setcounter{footnote}{1}\thanks{%
Graduate School of Management Sciences, Aletheia University, Tamsui 251,
Taipei, Taiwan, R.O.C.}, \ Chung-Yuan Dye\setcounter{footnote}{2}\thanks{%
Department of Business Management, Shu-Te University, Yen Chao, Kaohsiung
824, Taiwan, R.O.C.} \ and Liang-Yuh Ouyang\setcounter{footnote}{3}\thanks{%
Department of Management Sciences and Decision Making, Tamkang University,
Tamsui, Taipei 251, Taiwan, R.O.C.} \setcounter{footnote}{-1}\thanks{%
E-mail address: tsupang@gmail.com (T.-P. Hsieh). } } }
\maketitle

\begin{abstract}
In this paper, a deterministic inventory model with time-dependent
backlogging rate is developed. The demand rate is a power function of the
on-hand inventory until down to a certain stock level, at which the demand
rate becomes a constant. We prove that the optimal replenishment policy not
only exists but also is unique. Furthermore, we provide simple solution
procedures for finding the maximum total profit per unit time. Some
numerical examples are given to illustrate the model and conclude the paper
with suggestions for possible future research.

\bigskip \noindent \textbf{Keywords}: inventory, stock-dependent demand,
partial backlogging
\end{abstract}

\newpage

\setlength {\baselineskip} {1.5 \initiallineskip}

\section{Introduction}

\indent\indent In the last several years, the influence of displayed
stock-level on customers has been recognized by many marketing researchers
and practitioners. Baker and Urban \cite{Baker} considered a power-form
inventory-level-dependent demand rate, which would decline along with the
stock level throughout the entire cycle. Mandal and Phaujdar \cite{Mandal}
then incorporated deteriorating items with linearly stock-dependent demand.
Datta and Pal \cite{Datta1} modified the model of Baker and Urban \cite%
{Baker} by assuming that the stock-dependent demand rate was down to a given
level of inventory, beyond which it is a constant. By their assumption, not
all customers are attracted to purchase goods by the huge stock. When the
stock level declines to a certain level, customers arrive to purchase goods
because of its goodwill, good quality or facilities. Since then, research
articles dealing with stock-dependent demand are Urban \cite{Urban1}, Pal et
al. \cite{Pal}, Goh \cite{Goh}, Padmanabhan and Vrat \cite{PVrat}, Giri et
al. \cite{Giri1}, Ray and Chaudhuri \cite{Ray}, Sarker et al. \cite{Sarker},
Datta and Paul \cite{Datta2}, Balkhi and Benkherouf \cite{Balkhi}, Chang
\cite{Tao1}, Hou and Lin \cite{Hou}, Min and Zhou \cite{Min}, Goyal and
Chang \cite{Tao2} and others.

According to Urban \cite{Urban2}, the demand rate like that of Datta and Pal
\cite{Datta1} is only partially dependent on the instantaneous stock level.
Hence, he extended Datta and Pal's \cite{Datta1} model to allow shortages.
Like Montgomery et al. \cite{Montgomery}, Rosenberg \cite{Rosenberg}, Park
\cite{Park}, and Mak \cite{Mak}, Urban \cite{Urban2} assumed that
unsatisfied demand is backlogged at a fixed fraction of the constant demand
rate. Recently, Paul et al. \cite{Paul} termed the demand rate in Datta and
Pal's \cite{Datta1} model as two-component demand rate, and incorporated the
completely backlogged shortages. However, such a situation can be seen to
occur in a case where customers are loyal enough to wait until a fresh lot
arrives during the stock-out period. Their definitions of backlogging rate
discussed above seem to be inappropriate under some circumstances. In
reality, often some customers are conditioned to a shipping delay, and may
be willing to wait for a short time, while other will leave for another
seller because of urgent need. Therefore, the length of the waiting time
until the next replenishment is the main factor for deciding whether the
backlogging will be accepted or not. To reflect this phenomenon, Abad \cite%
{Abad1, Abad2} discussed a pricing and lot-sizing problem for a product with
a variable rate of deterioration, allowing shortages and partial
backlogging. The backlogging rate depends on the time to replenishment - the
longer customers must wait, the greater the fraction of lost sales. However,
the stockout cost (includes the backorder cost and the lost sales cost) is
not used in his model since it not easy to estimate. It seems to indicate
that the stockout cost is assumed to be zero, and its immediate impact is
that there is a lower service level to customers. Researchers on models with
partial backlogging are continuously developed by with San Jos\' e et al. \citep{San1,San2,San3,San4}, and Pentico and Drake \cite{Pentico}.

During the shortages period with partial backlogging, we must distinguish
between the backorders and lost sales cases. The cost arises from losing
some sales resulting in lost profits and annoyed customers. With a lost
sale, it can be considered as the loss of profit on the sales. Moreover, it
also includes the cost of losing the customer, loss of goodwill, and of
establishing a poor record of service. Therefore, if we omit the lost sales
from the profit function (i.e., set the cost of lost sales equal to zero),
then the profit will be overrated. It is true that the unit cost of stockout
is very difficult to measure. However, this does not mean that the unit does
not have some specific values. In practice, the stockout cost can be easy to
obtain from accounting data.

In this paper, we extend Datta and Pal's \cite{Datta1} model to allow
shortages and partial backlogging. Specifically, the backordering rate
during stockout is described by a decreasing function of the waiting time
instead of a constant rate of partial backlogging. In addition, the stockout
cost which is the sum of the backorder cost and the lost sales cost are
included. We then prove that the optimal replenishment policy not only
exists but also is unique. Moreover, numerical examples are used to
illustrate the proposed model, and concluding remarks are provided.

\section{Notation and Assumptions}

\subsection{Notation}

\indent\indent To develop the mathematical model of inventory replenishment
schedule, the notation adopted in this paper is as below:

\leftmargini=25mm \leftmarginii=35mm \leftmarginiii=51mm

\begin{enumerate}
\item[$A=$] the replenishment cost per order

\item[$c=$] the purchasing cost per unit

\item[$s=$] the selling price per unit, where $s>c$

\item[$Q=$] the ordering quantity per cycle

\item[$I=$] the maximum inventory level per cycle

\item[$c_{1}=$] the holding cost per unit per unit time

\item[$c_{2}=$] the backorder cost per unit per unit time

\item[$c_{3}=$] the opportunity cost (i.e., goodwill cost) per unit

\item[$t_{1}=$] the time at which the inventory level reaches $S_{0}$, where
$S_{0}$ is given

\item[$t_{2}=$] the time at which the inventory level reaches zero

\item[$t_{3}=$] the length of period during which shortages are allowed

\item[$T=$] the length of the inventory cycle, hence $T=t_{2}+t_{3}$

\item[$I_{1}\left( t\right) =$] the level of positive inventory at time $t$,
where $0\leq t\leq t_{1}$

\item[$I_{2}\left( t\right) =$] the level of positive inventory at time $t$,
where $t_{1}\leq t\leq t_{2}$

\item[$I_{3}\left( t\right) =$] the level of negative inventory at time $t$,
where $t_{2}\leq t\leq T$

\item[$\Pi \left( t_{1},t_{3}\right) =$] the total profit per unit time with
two-component demand rate

\item[$P\left( t_{2},t_{3}\right) =$] the total profit per unit time with
constant demand rate
\end{enumerate}

\subsection{Assumptions}

In addition, the following assumptions are imposed:

\leftmargini=5mm

\begin{enumerate}
\item Replenishment rate is infinite, and lead time is zero.

\item The time horizon of the inventory system is infinite.

\item The demand rate is dependent on the on-hand inventory down to a level $%
S_{0}$, where $S_{0}$ is given and fixed, beyond which it is assumed to be a
constant, that is, when the on-hand inventory level is $I\left( t\right) $,
the demand rate $R(I\left( t\right) )$ of the item is considered to be of
the form%
\begin{equation*}
R\left( I\left( t\right) \right) =\left\{
\begin{array}{ll}
\alpha \left[ I\left( t\right) \right] ^{\beta }, & I\left( t\right) \geq
S_{0} \\
\ & \\
D, & 0\leq I\left( t\right)
\end{array}%
\right.
\end{equation*}%
where $\alpha >0$ and $0<\beta <1where $\alpha >0$ and $0<\beta <1where $\alpha >0$ and $0<\beta <1where $\alpha >0$ and $0<\beta <1$>
amp;gt;
gt;
amp;gt;
respectively, $D(>0)$ is a constant such that $D=\alpha S_{0}^{\beta }$ .

\item Shortages are allowed and the demand rate $R(I\left( t\right) )$ is
given by:
\begin{equation*}
R(I\left( t\right) )=D,\text{\ }I\left( t\right) <0.
\end{equation*}%
We adopt the concept used in Abad \cite{Abad1, Abad2}, where some of the
unsatisfied demand is backlogged, and the fraction of shortages backordered
is $1/\left( {1+\delta x}\right) $, where $x$ is the waiting time up to the
next replenishment and $\delta $ is a positive constant.
\end{enumerate}

\section{Mathematical formulation}

\indent\indent In the present model, the parameter $S_{0}$ is exogenous.
Depending on the constant $S_0$ and the maximum inventory level $I$, the
inventory problem here has two situations: (i) $I\geq S_{0}$ and (ii) $I<
S_{0}$.

\subsection{Inventory problem with $I\geq S_{0}$}

\begin{figure}[tph]
\centering \includegraphics[height=4in]{backorder.eps}
\caption{Graphical Representation of Inventory System}
\end{figure}

\indent\indent Using above assumptions, the inventory level follows the
pattern depicted in Figure 1. To establish the total relevant profit
function, we consider the following time intervals separately, $\left[
0,t_{1}\right] $, $\left[ t_{1},t_{2}\right] $, and $\left[ t_{2},T\right] $%
. During the interval $\left[ 0,t_{1}\right] $, the inventory is depleted
due to the effect of demand dependent on the on-hand inventory level and
reaches the level $S_{0}$ at time $t=t_{1}$. Hence, the inventory level is
governed by the following differential equation:%
\begin{equation}
\frac{\text{d}I_{1}(t)}{\text{d}t}=-\alpha \left[ I_{1}\left( t\right) %
\right] ^{\beta },0
\end{equation}%
with the boundary condition $I_{1}\left( t_{1}\right) =S_{0}$. Solving the
differential equation (\ref{di1}), we get the inventory level as:%
\begin{equation}
I_{1}(t)=\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) \left(
t_{1}-t\right) \right] ^{\frac{1}{1-\beta }},0\leq t\leq t_{1}. \label{i1t}
\end{equation}

After the time $t=t_{1}$, the demand rate becomes a constant $D$, and the
inventory level falls to zero at time $t=t_{2}$. During the interval $\left[
t_{1},t_{2}\right] $, the inventory is depleted due to the effect of demand.
Hence, the inventory level is governed by the following differential
equation:%
\begin{equation}
\frac{\text{d}I_{2}(t)}{\text{d}t}=-D,t_{1}
\end{equation}%
with the boundary condition $I_{2}\left( t_{2}\right) =0$. Solving the
differential equation (\ref{di2}), we obtain the inventory level as%
\begin{equation}
I_{2}\left( t\right) =D\left( t_{2}-t\right) ,t_{1}\leq t\leq t_{2}.
\label{i2t}
\end{equation}

Due to the continuity of $I_{1}\left( t\right) $ and $I_{2}\left( t\right) $
at point $t=t_{1}$, it follows from Eqs. (\ref{i1t}) and (\ref{i2t}) that%
\begin{equation}
S_{0}=D\left( t_{2}-t_{1}\right) , \label{S01}
\end{equation}%
which implies%
\begin{equation}
t_{2}=t_{1}+\frac{S_{0}}{D}. \label{t2}
\end{equation}%
Thus, $t_{2}$ is a function of $t_{1}$.

Furthermore, at time $t_{2}$, shortage occurs and the inventory level starts
dropping below $0$. During $\left[ t_{2},T\right] $, the inventory level
only depends on demand, and a fraction $\frac{1}{1+\delta \left( T-t\right) }
$ of the demand is backlogged, where $t\in \left[ t_{2},T\right] $. The
inventory level is governed by the following differential equation:%
\begin{equation}
\frac{\text{d}I_{3}(t)}{\text{d}t}=-\frac{D}{1+\delta \left( T-t\right) },%
\text{ }t_{2}
\end{equation}%
with the boundary condition $I_{3}\left( t_{2}\right) =0$. Solving the
differential equation (\ref{di3}), we obtain the inventory level as
\begin{eqnarray}
I_{3}\left( t\right) &=&-\frac{D}{\delta }\bigg\{\ln \left[ 1+\delta \left(
T-t_{1}-\frac{S_{0}}{D}\right) \right] -\ln \left[ 1+\delta \left(
T-t\right) \right] \bigg\} \notag \\
&=&-\frac{D}{\delta }\bigg\{\ln \left[ 1+\delta \left( T-t_{2}\right) \right]
-\ln \left[ 1+\delta \left( T-t\right) \right] \bigg\},\text{ }t_{2}\leq
t\leq T. \label{i3t}
\end{eqnarray}

Therefore, the ordering quantity over the replenishment cycle can be
determined as
\begin{equation}
Q=I_{1}\left( 0\right) -I_{3}\left( T\right) =\left[ S_{0}^{1-\beta }+\alpha
\left( 1-\beta \right) t_{1}\right] ^{\frac{1}{1-\beta }}+\frac{D\ln \left(
1+\delta t_{3}\right) }{\delta }, \label{Q1}
\end{equation}%
and the maximum inventory level per cycle is%
\begin{equation}
I=I_{1}\left( 0\right) =\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta
\right) t_{1}\right] ^{\frac{1}{1-\beta }}. \label{B1}
\end{equation}

Based on Eqs. (\ref{i1t}), (\ref{i2t}), and (\ref{i3t}), the total profit
per cycle consists of the following elements:

\begin{enumerate}
\item ordering cost per cycle $=A$,

\item holding cost per cycle

$=c_{1}\int_{0}^{t_{1}}I_{1}(t)$d$t+c_{1}\int_{t_{1}}^{t_{2}}I_{2}(t)$d$t$

$=c_{1}\int_{0}^{t_{1}}\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right)
\left( t_{1}-t\right) \right] ^{\frac{1}{1-\beta }}$d$t+c_{1}%
\int_{t_{1}}^{t_{2}}D\left( t_{2}-t\right) $d$t$

$=c_{1}\int_{0}^{t_{1}}\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right)
\left( t_{1}-t\right) \right] ^{\frac{1}{1-\beta }}$d$t+c_{1}%
\int_{t_{1}}^{t_{1}+\frac{S_{0}}{D}}\left[ S_{0}-D\left( t-t_{1}\right) %
\right] $d$t$

$=\frac{c_{1}}{\alpha \left( \beta -2\right) }\left\{ S_{0}^{2-\beta }-\left[
S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{2-\beta
}{1-\beta }}\right\} +\frac{c_{1}S_{0}^{2}}{2D}$,

\item backorder cost per cycle

$=c_{2}\int_{t_{2}}^{T}\left[ -I_{3}\left( t\right) \right] $d$t$

$=c_{2}\int_{t_{1}+\frac{S_{0}}{D}}^{T}\left[ -I_{3}\left( t\right) \right] $%
d$t$

$=\frac{c_{2}D}{\delta ^{2}}\left\{ \delta \left( T-t_{2}\right) -\ln \left[
1+\delta \left( T-t_{2}\right) \right] \right\} $

$=\frac{c_{2}D}{\delta ^{2}}\left[ \delta t_{3}-\ln \left( 1+\delta
t_{3}\right) \right] $,

\item opportunity cost due to lost sales per cycle

$=c_{3}D\int_{t_{2}}^{T}\left[ 1-\frac{1}{1+\delta \left( T-t\right) }\right]
$d$t$

$=c_{3}D\int_{t_{1}+\frac{S_{0}}{D}}^{T}\left[ 1-\frac{1}{1+\delta \left(
T-t\right) }\right] $d$t$

$=\frac{c_{3}D}{\delta }\left[ \delta t_{3}-\ln \left( 1+\delta t_{3}\right) %
\right] $,

\item purchase cost per cycle

$=cQ=c\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right] ^{%
\frac{1}{1-\beta }}+\frac{cD}{\delta }\ln \left( 1+\delta t_{3}\right) $,

\item sales revenue per cycle

$=sQ=s\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right] ^{%
\frac{1}{1-\beta }}+\frac{sD}{\delta }\ln \left( 1+\delta t_{3}\right) .$
\end{enumerate}

Therefore, the total profit per unit time of our model is obtained as
follows:
\begin{eqnarray}
\Pi \left( t_{1},t_{3}\right) &=&\frac{1}{t_{2}+t_{3}}\left\{
\begin{array}{l}
\text{sales revenue}-\text{purchase cost}-\text{ordering cost} \\
-\text{holding cost}-\text{backorder cost}-\text{opportunity cost}%
\end{array}%
\right\} \notag \\
&=&\frac{1}{t_{1}+\frac{S_{0}}{D}+t_{3}}\bigg\{\left( s-c\right) \left[
S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{1}{%
1-\beta }}-A \notag \\
&&+\frac{c_{1}}{\alpha \left( \beta -2\right) }\left[ S_{0}^{1-\beta
}+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{2-\beta }{1-\beta }}
\notag \\
&&-c_{1}S_{0}^{2-\beta }\left[ \frac{1}{\alpha \left( \beta -2\right) }+%
\frac{S_{0}^{\beta }}{2D}\right] +D\left( s-c\right) t_{3} \notag \\
&&-\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}%
\left[ \delta t_{3}-\ln \left( 1+\delta t_{3}\right) \right] \bigg\}.
\label{TP}
\end{eqnarray}

To maximize the total profit per unit time, taking the first derivative of $%
\Pi \left( t_{1},t_{3}\right) $ with respect to $t_{1}$ and $t_{3}$,
respectively, we obtain
\begin{eqnarray}
\frac{\partial \Pi \left( t_{1},t_{3}\right) }{\partial t_{1}} &=&-\frac{1}{%
t_{1}+\frac{S_{0}}{D}+t_{3}}\Bigg\{\Pi \left( t_{1},t_{3}\right) -\alpha
\left( 1-\beta \right) c_{1}\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta
\right) t_{1}\right] ^{\frac{\beta }{1-\beta }} \notag \\
&&\times \left[ \frac{s-c}{c_{1}\left( 1-\beta \right) }-\frac{%
S_{0}^{1-\beta }}{\alpha \left( 1-\beta \right) }-t_{1}\right] \Bigg\},
\label{dPt1}
\end{eqnarray}%
and%
\begin{equation}
\frac{\partial \Pi \left( t_{1},t_{3}\right) }{\partial t_{3}}=-\frac{1}{%
t_{1}+\frac{S_{0}}{D}+t_{3}}\left\{ \Pi \left( t_{1},t_{3}\right) -D\left(
s-c\right) +\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] t_{3}%
}{1+\delta t_{3}}\right\} . \label{dPt3}
\end{equation}%
The optimal solution of $\left( t_{1},t_{3}\right) $ must satisfy the
equations $\frac{\partial \Pi \left( t_{1},t_{3}\right) }{\partial t_{1}}=0$
and $\frac{\partial \Pi \left( t_{1},t_{3}\right) }{\partial t_{3}}=0$,
simultaneously, which implies%
\begin{equation}
\Pi \left( t_{1},t_{3}\right) =\alpha \left( 1-\beta \right) c_{1}\left[
S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{\beta }{%
1-\beta }}\left[ \frac{s-c}{c_{1}\left( 1-\beta \right) }-\frac{%
S_{0}^{1-\beta }}{\alpha \left( 1-\beta \right) }-t_{1}\right] , \label{G81}
\end{equation}%
and%
\begin{equation}
\Pi \left( t_{1},t_{3}\right) =D\left( s-c\right) -\frac{D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] t_{3}}{1+\delta t_{3}},
\label{G82}
\end{equation}%
respectively. Because both the left hand sides in Eqs. (\ref{G81}) and (\ref%
{G82}) are the same, hence the right hand sides in these equations are
equal, that is,
\begin{eqnarray}
\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] t_{3}}{1+\delta
t_{3}} &=&D\left( s-c\right) -\alpha \left( 1-\beta \right) c_{1}\left[
S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{\beta }{%
1-\beta }} \notag \\
&&\times \left[ \frac{s-c}{c_{1}\left( 1-\beta \right) }-\frac{%
S_{0}^{1-\beta }}{\alpha \left( 1-\beta \right) }-t_{1}\right] .
\label{dPt11}
\end{eqnarray}%
On the other hand, we substitute $\Pi \left( t_{1},t_{3}\right) $ in (\ref%
{TP}) into Eq. (\ref{G82}) and obtain%
\begin{eqnarray}
\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] \left( t_{1}+%
\frac{S_{0}}{D}+t_{3}\right) t_{3}}{1+\delta t_{3}} &=&A-\left( s-c\right) %
\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{1%
}{1-\beta }} \notag \\
&&-\frac{c_{1}}{\alpha \left( \beta -2\right) }\left[ S_{0}^{1-\beta
}+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{2-\beta }{1-\beta }}
\notag \\
&&+c_{1}S_{0}^{2-\beta }\left[ \frac{1}{\alpha \left( \beta -2\right) }+%
\frac{S_{0}^{\beta }}{2D}\right] \notag \\
&&+D\left( s-c\right) \left( t_{1}+\frac{S_{0}}{D}\right) \notag \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}
\notag \\
&&\times \left[ \delta t_{3}-\ln \left( 1+\delta t_{3}\right) \right] .
\label{dPt31}
\end{eqnarray}

Now, we want to find the value of $\left( t_{1},t_{3}\right) $ which
satisfies Eqs. (\ref{dPt11}) and (\ref{dPt31}), simultaneously. For
convenience, we first let $K\left( t_{1}\right) $ denote the right hand side
of Eq. (\ref{dPt11}), that is,
\begin{eqnarray}
K\left( t_{1}\right) &=&D\left( s-c\right) -\alpha c_{1}\left( 1-\beta
\right) \left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right] ^{%
\frac{\beta }{1-\beta }} \notag \\
&&\times \left[ \frac{s-c}{c_{1}\left( 1-\beta \right) }-\frac{%
S_{0}^{1-\beta }}{\alpha \left( 1-\beta \right) }-t_{1}\right] ,\text{ }%
t_{1}\geq 0. \label{Kt1}
\end{eqnarray}%
It notes that $K\left( t_{1}\right) $ is a continuous function in $t_{1}\in %
\left[ 0,\infty \right) $. Then Eq. (\ref{dPt11}) becomes%
\begin{equation}
D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] \frac{t_{3}}{1+\delta
t_{3}}=K\left( t_{1}\right) , \label{Kt3}
\end{equation}%
which implies%
\begin{equation}
t_{3}=\frac{K\left( t_{1}\right) }{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] -\delta K\left( t_{1}\right) }. \label{t3}
\end{equation}%
Thus, $t_{3}$ is a function of $t_{1}$, and further we have%
\begin{equation}
\frac{\text{d}t_{3}}{\text{d}t_{1}}=\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\left\{ D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] -\delta K\left( t_{1}\right) \right\} ^{2}}\frac{%
\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}. \label{dt3}
\end{equation}%
Furthermore, motivated by Eq. (\ref{dPt31}), we let
\begin{eqnarray}
G\left( t_{1}\right) &=&A-\left( s-c\right) \left[ S_{0}^{1-\beta }+\alpha
\left( 1-\beta \right) t_{1}\right] ^{\frac{1}{1-\beta }}-\frac{c_{1}}{%
\alpha \left( \beta -2\right) }\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta
\right) t_{1}\right] ^{\frac{2-\beta }{1-\beta }} \notag \\
&&+c_{1}S_{0}^{2-\beta }\left[ \frac{1}{\alpha \left( \beta -2\right) }+%
\frac{S_{0}^{\beta }}{2D}\right] +D\left( s-c\right) \left( t_{1}+\frac{S_{0}%
}{D}\right) \notag \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}%
\left[ \delta t_{3}-\ln \left( 1+\delta t_{3}\right) \right] \notag \\
&&-D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] \frac{\left( t_{1}+%
\frac{S_{0}}{D}+t_{3}\right) t_{3}}{1+\delta t_{3}}, \label{Gt1}
\end{eqnarray}%
where $t_{3}$ is given as in Eq. (\ref{t3}). Taking the derivative of $%
G\left( t_{1}\right) $ with respect to $t_{1}$, and by using the relations
shown in Eqs. (\ref{dPt11}), (\ref{Kt1}) and (\ref{dt3}), we obtain%
\begin{eqnarray}
\frac{\text{d}G\left( t_{1}\right) }{\text{d}t_{1}} &=&-\frac{D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] \left( t_{1}+\frac{S_{0}}{D}%
+t_{3}\right) }{\left( 1+\delta t_{3}\right) ^{2}}\frac{\text{d}t_{3}}{\text{%
d}t_{1}} \notag \\
&=&-\left( t_{1}+\frac{S_{0}}{D}+t_{3}\right) \frac{\text{d}K\left(
t_{1}\right) }{\text{d}t_{1}}. \label{dGt1}
\end{eqnarray}

In order to prove the existence and uniqueness of the optimal solution $%
t_{1}^{\ast }$ which satisfies equation $G\left( t_{1}^{\ast }\right) =0$,
we have to investigate the property of function $K\left( t_{1}\right) $.
Taking the derivative of $K\left( t_{1}\right) $ with respect to $t_{1}$, we
have%
\begin{eqnarray}
\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}} &=&-\alpha
^{2}c_{1}\left( 1-\beta \right) \left[ S_{0}^{1-\beta }+\alpha \left(
1-\beta \right) t_{1}\right] ^{\frac{2\beta -1}{1-\beta }}\left[ \frac{\beta
\left( s-c\right) }{c_{1}\left( 1-\beta \right) }-\frac{S_{0}^{1-\beta }}{%
\alpha \left( 1-\beta \right) }-t_{1}\right] \notag \\
&=&-\alpha ^{2}c_{1}\left( 1-\beta \right) \left[ S_{0}^{1-\beta }+\alpha
\left( 1-\beta \right) t_{1}\right] ^{\frac{2\beta -1}{1-\beta }}\left(
\widetilde{t}_{1}-t_{1}\right) , \label{dKt1}
\end{eqnarray}%
where $\widetilde{t}_{1}=\frac{\beta \left( s-c\right) }{c_{1}\left( 1-\beta
\right) }-\frac{S_{0}^{1-\beta }}{\alpha \left( 1-\beta \right) }$. Thus, if
$\widetilde{t}_{1}>0$ (i.e., $S_{0}<\left[ \frac{\alpha \beta \left(
s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}$), we obtain $\left. \frac{%
\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}\right\vert _{t_{1}=\widetilde{t%
}_{1}}=0$, and%
\begin{eqnarray}
K\left( \widetilde{t}_{1}\right) &=&\alpha \left( s-c\right) S_{0}^{\beta
}-\alpha \left( s-c\right) \left( 1-\beta \right) \left[ \frac{\alpha \beta
\left( s-c\right) }{c_{1}}\right] ^{\frac{\beta }{1-\beta }}\text{(because }%
D=\alpha S_{0}^{\beta }\text{)} \notag \\
&=&\alpha \left( s-c\right) \left\{ S_{0}^{\beta }-\left( 1-\beta \right)
\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{\beta }{%
1-\beta }}\right\} . \label{kt1wave}
\end{eqnarray}%
Because $K\left( 0\right) =D\left( s-c\right) -\alpha \left( s-c\right)
S_{0}^{\beta }+c_{1}S_{0}=c_{1}S_{0}>0$ and it can be shown that $%
\lim_{t_{1}\rightarrow \infty }K\left( t_{1}\right) =\infty $, thus, we have
the following result.

\noindent \textbf{Lemma 1. }\textit{Let }$K\left( t_{1}\right) $\textit{\ be
defined as in Eq. }(\ref{Kt1})\textit{, we have:}

\begin{enumerate}
\item[(a)] \noindent \textit{If }$S_{0}\geq \left[ \frac{\alpha \beta \left(
s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}$\textit{, then }$K\left(
t_{1}\right) $\textit{\ is a strictly increasing function in }$t_{1}\in %
\left[ 0,\infty \right) $\textit{, and the minimum of }$K\left( t_{1}\right)
$ is $K\left( 0\right) =c_{1}S_{0}>0$.

\item[(b)] \textit{If }$S_{0}<\left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}$\textit{, then }$K\left( t_{1}\right) $%
\textit{\ is a strictly decreasing function in }$t_{1}\in \left[ 0,%
\widetilde{t}_{1}\right] $\textit{\ and strictly increasing in }$t_{1}\in %
\left[ \widetilde{t}_{1},\infty \right) $\textit{. And }$K\left(
t_{1}\right) $\textit{\ has a minimum value at point }$t_{1}=\widetilde{t}%
_{1}$\textit{. Furthermore, the minimum value }$K\left( \widetilde{t}%
_{1}\right) \geq 0$\textit{\ if }$S_{0}\geq \left( 1-\beta \right) ^{\frac{1%
}{\beta }}\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{%
\frac{1}{1-\beta }}$\textit{, and }$K\left( \widetilde{t}_{1}\right) <0$%
\textit{\ if }$S_{0}<\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[ \frac{%
\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}$\textit{%
.}
\end{enumerate}

\noindent \textbf{Proof: }See Appendix A.

Now, let us consider the following three cases: (i) $S_{0}\geq \left[ \frac{%
\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}$%
,\linebreak (ii) $\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[ \frac{%
\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}\leq
S_{0}<\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{%
1-\beta }}$ and (iii) $S_{0}<\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[
\frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}$.

\subsubsection{Case 1: $S_{0}\geq \left[ \frac{\alpha\beta %
\left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\protect\beta }}$}

\indent\indent In this case, we further consider the following two subcases:
(1) $K\left( 0\right) =c_{1}S_{0}\geq \frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta }$ and (2) $K\left( 0\right) =c_{1}S_{0}<%
\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta }$.

For convenience, we let $F\left( t_{3}\right) $ denote the left hand side of
Eq. (\ref{Kt3}), that is,%
\begin{equation}
F\left( t_{3}\right) =D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right]
\frac{t_{3}}{1+\delta t_{3}},\text{ }t_{3}\geq 0. \label{ft3}
\end{equation}

\subsubsection*{Case 1.1.}

\indent\indent When $S_{0}\geq \left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}$ and $K\left( 0\right) =c_{1}S_{0}\geq
\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta }$, Eq. (%
\ref{ft3}) becomes
\begin{eqnarray*}
F\left( t_{3}\right) &<&\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) %
\right] }{\delta } \\
&\leq &c_{1}S_{0}=K\left( 0\right) ,\text{ for }t_{3}\in \left[ 0,\infty
\right) \text{.}
\end{eqnarray*}%
By Lemma 1(a), we know that $K\left( t_{1}\right) $ is a strictly increasing
function in $t_{1}\in \left[ 0,\infty \right) $, hence for any given $%
t_{1}\in \left[ 0,\infty \right) $, there does not exist a value $t_{3}\in %
\left[ 0,\infty \right) $ such that $F\left( t_{3}\right) =K\left(
t_{1}\right) $, i.e., for any given $t_{1}\in \left[ 0,\infty \right) $, we
can not find a value $t_{3}$ which satisfies Eq. (\ref{dPt11}). However, for
this situation, from Eqs. (\ref{dPt1}), (\ref{G82}), (\ref{Kt1}) and (\ref%
{ft3}), we have%
\begin{equation*}
\frac{\partial \Pi \left( t_{1},t_{3}\right) }{\partial t_{1}}=\frac{F\left(
t_{3}\right) -K\left( t_{1}\right) }{t_{1}+\frac{S_{0}}{D}+t_{3}}<\frac{%
F\left( t_{3}\right) -K\left( 0\right) }{t_{1}+\frac{S_{0}}{D}+t_{3}}<0\text{%
, for any }t_{1}\in \left( 0,\infty \right) \text{ and }t_{3}\in \left(
0,\infty \right) \text{.}
\end{equation*}
Therefore, when $S_{0}\geq \left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}$ and $K\left( 0\right) =c_{1}S_{0}\geq
\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta }$ hold,
i.e., if $S_{0}\geq \max \left\{ \left[ \frac{\alpha \beta \left( s-c\right)
}{c_{1}}\right] ^{\frac{1}{1-\beta }},\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta c_{1}}\right\} $, the maximum value of $%
\Pi \left( t_{1},t_{3}\right) $ occurs at the boundary point $t_{1}^{\ast
}=0 $.

In the special circumstance that $t_{1}^{\ast }=0$, the optimal value of $%
t_{2}$ (denoted by $t_{2}^{\ast }$) can be obtained by Eq. (\ref{t2}) and is
$t_{2}^{\ast }=\frac{S_{0}}{D}$. Then the total profit per unit time in Eq. (%
\ref{TP}) becomes%
\begin{eqnarray}
\Pi \left( t_{3}\right) \equiv \Pi \left( 0,t_{3}\right) &=&D\left(
s-c\right) -\frac{1}{\frac{S_{0}}{D}+t_{3}}\bigg\{A+\frac{c_{1}S_{0}^{2}}{2D}
\notag \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}%
\left[ \delta t_{3}-\ln \left( 1+\delta t_{3}\right) \right] \bigg\}.
\label{Pit3}
\end{eqnarray}%
The necessary condition to find the optimal solution of $\Pi \left(
t_{3}\right) $ is%
\begin{eqnarray*}
\frac{\text{d}\Pi \left( t_{3}\right) }{\text{d}t_{3}} &=&\frac{1}{\left(
\frac{S_{0}}{D}+t_{3}\right) ^{2}}\bigg\{A+\frac{c_{1}S_{0}^{2}}{2D}-\frac{D%
\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] \left( \frac{S_{0}}{D}%
+t_{3}\right) t_{3}}{1+\delta t_{3}} \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}%
\left[ \delta t_{3}-\ln \left( 1+\delta t_{3}\right) \right] \bigg\}=0,
\end{eqnarray*}%
which implies%
\begin{eqnarray}
&&A+\frac{c_{1}S_{0}^{2}}{2D}-\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] \left( \frac{S_{0}}{D}+t_{3}\right) t_{3}}{1+\delta
t_{3}} \notag \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}%
\left[ \delta t_{3}-\ln \left( 1+\delta t_{3}\right) \right] =0. \label{Zt3}
\end{eqnarray}%
Let%
\begin{eqnarray*}
Z\left( t_{3}\right) &=&A+\frac{c_{1}S_{0}^{2}}{2D}-\frac{D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] \left( \frac{S_{0}}{D}%
+t_{3}\right) t_{3}}{1+\delta t_{3}} \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}%
\left[ \delta t_{3}-\ln \left( 1+\delta t_{3}\right) \right] ,\text{ for }%
t_{3}\geq 0.
\end{eqnarray*}%
The derivative of $Z\left( t_{3}\right) $ with respect to $t_{3}$ is
\begin{equation*}
\frac{\text{d}Z\left( t_{3}\right) }{\text{d}t_{3}}=\frac{-D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] \left( \frac{S_{0}}{D}%
+t_{3}\right) }{\left( 1+\delta t_{3}\right) ^{2}}<0,
\end{equation*}%
thus, $Z\left( t_{3}\right) $ is a strictly decreasing function in $t_{3}\in %
\left[ 0,\infty \right) $. Furthermore, we have $Z\left( 0\right) =A+\frac{%
c_{1}S_{0}^{2}}{2D}>0$ and $\lim_{t_{3}\rightarrow \infty }Z\left(
t_{3}\right) =-\infty <0$.>
exists a unique solution $t_{3}^{\ast }\in \left( 0,\infty \right) $ such
that $Z\left( t_{3}^{\ast }\right) =0$, that is, $t_{3}^{\ast }$ is the
unique value which satisfies Eq. (\ref{Zt3}). Summarize the above arguments,
we obtain the following theorem.

\noindent \textbf{Theorem 1. }\textit{For }$S_{0}\geq \max \left\{ \left[
\frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }},%
\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta c_{1}}%
\right\} $\textit{, the optimal value of }$\left( t_{1},t_{2},t_{3}\right) $%
\textit{\ is given by }$t_{1}^{\ast }$\textit{\ }$=0$\textit{, }$t_{2}^{\ast
}=\frac{S_{0}}{D}$\textit{, and }$t_{3}^{\ast }$\textit{\ is the value which
satisfies Eq.} (\ref{Zt3})\textit{.}

When $t_1^{*}=0$, the inventory problem becomes the regular EOQ with
constant demand rate and partial backordering. Once the optimal value $%
\left( t_{1}^{\ast },t_{3}^{\ast }\right) =\left( 0,t_{3}^{\ast }\right) $
is obtained, by Eqs. (\ref{Pit3}) and (\ref{Zt3}), the maximum total profit
per unit time
\begin{equation}
\Pi \left( t_{3}^{\ast }\right) =D\left( s-c\right) -\frac{D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] t_{3}^{\ast }}{\left( 1+\delta
t_{3}^{\ast }\right) } \label{Pi0t3}
\end{equation}%
follows. Besides, from Eq. (\ref{B1}), the maximum inventory level per cycle
is $I^{\ast }=S_{0}$.

\subsubsection{Case 1.2.}

\indent\indent When $S_{0}\geq \left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}$ and $K\left( 0\right) =c_{1}S_{0}<\frac{%
D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta }$, from Lemma
1(a), $K\left( t_{1}\right) $ is a strictly increasing function of $t_{1}\in %
\left[ 0,\infty \right) $, thus we can find a unique value $\widehat{t}%
_{1}\in \left( 0,\infty \right) $ such that $K\left( \widehat{t}_{1}\right) =%
\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta }$.
Furthermore, for any given $t_{1}\geq \widehat{t}_{1}$, we have%
\begin{eqnarray*}
K\left( t_{1}\right) &\geq &K\left( \widehat{t}_{1}\right) \\
&=&\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta } \\
&>&\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta }-%
\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta }\frac{1%
}{1+\delta t_{3}} \\
&=&\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] t_{3}}{%
1+\delta t_{3}}=F\left( t_{3}\right) .
\end{eqnarray*}%
It implies that we can not find a $t_{3}\in \left[ 0,\infty \right) $ such
that Eq. (\ref{dPt11}) holds. Therefore, the optimal solution of $t_{1}$
which satisfies Eq. (\ref{dPt11}) will occur in the interval $\left( 0,%
\widehat{t}_{1}\right) $.

On the other hand, from the definition of $F\left( t_{3}\right) $ in Eq. (%
\ref{ft3}), it can be shown that $F\left( t_{3}\right) $ is a continuous and
strictly increasing function in $t_{3}\in \left[ 0,\infty \right) $.
Besides, we have $F\left( 0\right) =0$, and
\begin{equation*}
\lim_{t_{3}\rightarrow \infty }F\left( t_{3}\right) =\frac{D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta }=K\left( \widehat{t}%
_{1}\right) >K\left( t_{1}\right) \text{, for any }t_{1}\in \left[ 0,%
\widehat{t}_{1}\right) \text{.}
\end{equation*}%
Thus, for any given $t_{1}\in \left[ 0,\widehat{t}_{1}\right) $, there
exists a unique value $t_{3}$ $\in \left( 0,\infty \right) $ such that $%
F\left( t_{3}\right) =K\left( t_{1}\right) $. Consequently, for any given $%
t_{1}\in \left[ 0,\widehat{t}_{1}\right) $, when $S_{0}\geq \left[ \frac{%
\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}$ and $%
K\left( 0\right) =c_{1}S_{0}<\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta }$, i.e., if $\left[ \frac{\alpha \beta
\left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}\leq S_{0}<\frac{D%
\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta c_{1}}$, we can
find a unique value $t_{3}$ $\in \left( 0,\infty \right) $ satisfying Eq. (%
\ref{dPt11}). Therefore, for $\left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}\leq S_{0}<\frac{D\left[ c_{2}+\delta
\left( s-c+c_{3}\right) \right] }{\delta c_{1}}$, once the optimal value $%
t_{1}^{\ast }\in \left[ 0,\widehat{t}_{1}\right) $ is obtained, the optimal
solutions of $t_{2}$, $t_{3}$ and $T$ (denoted by $t_{2}^{\ast }$, $%
t_{3}^{\ast }$ and $T^{\ast }$, respectively) are as follows%
\begin{equation}
t_{2}^{\ast }=t_{1}^{\ast }+\frac{S_{0}}{D}, \label{G8t2}
\end{equation}
\begin{equation}
t_{3}^{\ast }=\frac{K\left( t_{1}^{\ast }\right) }{D\left[ c_{2}+\delta
\left( s-c+c_{3}\right) \right] -\delta K\left( t_{1}^{\ast }\right) },
\label{G8t3}
\end{equation}%
and%
\begin{equation}
T^{\ast }=t_{2}^{\ast }+t_{3}^{\ast }. \label{G8T}
\end{equation}

Now, we want to prove the existence and uniqueness of $t_{1}^{\ast }$ in $%
\left( 0,\widehat{t}_{1}\right) $. By using \linebreak $\left( t_{1}+\frac{%
S_{0}}{D}+t_{3}\right) >0$ and $\frac{\text{d}K\left( t_{1}\right) }{\text{d}%
t_{1}}>0$ for $t_{1}\in \left( 0,\widehat{t}_{1}\right) $, from Eq. (\ref%
{dGt1}), we obtain $\frac{\text{d}G\left( t_{1}\right) }{\text{d}t_{1}}<0>
Therefore, $G\left( t_{1}\right) $ is a strictly decreasing function in $%
t_{1}\in \left[ 0,\widehat{t}_{1}\right) $. Furthermore, from Eq. (\ref{t3}%
), we have $t_{3}\rightarrow \infty $ as $t_{1}\rightarrow \widehat{t}%
_{1}^{-}$, and%
\begin{eqnarray*}
\lim_{t_{1}\rightarrow \widehat{t}_{1}^{-}}G\left( t_{1}\right) &=&A-\left(
s-c\right) \left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) \widehat{t}%
_{1}\right] ^{\frac{1}{1-\beta }}-\frac{c_{1}}{\alpha \left( \beta -2\right)
}\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) \widehat{t}_{1}\right]
^{\frac{2-\beta }{1-\beta }} \\
&&+c_{1}S_{0}^{2-\beta }\left[ \frac{1}{\alpha \left( \beta -2\right) }+%
\frac{S_{0}^{\beta }}{2D}\right] -\frac{D\left( c_{2}+\delta c_{3}\right) }{%
\delta }\left( \widehat{t}_{1}+\frac{S_{0}}{D}\right) \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}-%
\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}%
\lim_{t_{1}\rightarrow \widehat{t}_{1}^{-}}\ln \left( 1+\delta t_{3}\right)
\\
&=&-\infty ,
\end{eqnarray*}%
and%
\begin{eqnarray}
G\left( 0\right) &=&A+\frac{c_{1}S_{0}^{2}}{2D}-D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] \frac{\left( \frac{S_{0}}{D}+t_{3}\right) t_{3}}{%
1+\delta t_{3}} \notag \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}%
\left[ \delta t_{3}-\ln \left( 1+\delta t_{3}\right) \right] \notag \\
&=&A-\frac{c_{1}S_{0}^{2}}{2D}+\frac{c_{1}S_{0}}{\delta } \notag \\
&&-\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}%
\ln \left\{ \frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{D%
\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] -\delta c_{1}S_{0}}%
\right\} . \label{G0}
\end{eqnarray}%
Note that the value in the brace is well-defined, because we have $\frac{D%
\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta }>K\left(
0\right) =c_{1}S_{0}$. Then we have the following result.

\noindent \textbf{Lemma 2.} \textit{For }$\left[ \frac{\alpha \beta \left(
s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}\leq S_{0}<\frac{D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta c_{1}}$\textit{, we
have:}

\noindent (a) \textit{If }$G\left( 0\right) \leq 0$\textit{, then the
optimal value of }$t_{1}$\textit{\ is }$t_{1}^{\ast }=0$\textit{.}

\noindent (b) \textit{If }$G\left( 0\right) >0$\textit{, then the solution }$%
t_{1}^{\ast }\in \left( 0,\widehat{t}_{1}\right) $\textit{\ which satisfies
Eq. }(\ref{dPt31})\textit{\ not only exists but also is unique.}

\noindent \textbf{Proof: }See Appendix B.

Lemma 2(a) indicates that if $\left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}\leq S_{0}<\frac{D\left[ c_{2}+\delta
\left( s-c+c_{3}\right) \right] }{\delta c_{1}}$ and $G\left( 0\right) \leq
0 $, then the optimal time at which the inventory level reaches $S_{0}$ is $%
t_{1}^{\ast }=0$. It implies the maximum inventory level in this system is $%
I^{\ast }=S_{0}$. The corresponding value of $t_{3}$ can be found from Eq. (%
\ref{G8t3}) and is given by $t_{3}=\frac{K\left( 0\right) }{D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] -\delta K\left( 0\right) }$.
However, since $\Pi \left( 0,\frac{K\left( 0\right) }{D\left[ c_{2}+\delta
\left( s-c+c_{3}\right) \right] -\delta K\left( 0\right) }\right) \leq \Pi
\left( 0,t_{3}^{\ast }\right) =\Pi \left( t_{3}^{\ast }\right) $, where $%
t_{3}^{\ast }$ is the value of $t_{3}$ which satisfies Eq. (\ref{Zt3}).
Thus, $t_{3}^{\ast }$ is the optimal value of $t_{3}$, and the maximum total
profit per unit time $\Pi \left( t_{3}^{\ast }\right) $ is obtained by Eq. (%
\ref{Pi0t3}).

Lemma 2(b) reveals that if $\left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}\leq S_{0}<\frac{D\left[ c_{2}+\delta
\left( s-c+c_{3}\right) \right] }{\delta c_{1}}$ and $G\left( 0\right) >0$,
then $t_{1}^{\ast }\in \left( 0,\widehat{t}_{1}\right) $ and is unique.
Furthermore, the unique solution will be proved to be indeed a global
maximum point by checking the second order optimality conditions, that is,
we have the following main result.

\noindent \textbf{Theorem 2. }\textit{For }$\left[ \frac{\alpha \beta \left(
s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}\leq S_{0}<\frac{D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta c_{1}}$\textit{, if }$%
G\left( 0\right) >0$\textit{, then the point }$\left( t_{1}^{\ast
},t_{3}^{\ast }\right) $\textit{\ which satisfies Eqs. }(\ref{dPt11})\textit{%
\ and }(\ref{dPt31})\textit{\ simultaneously is the global maximum point of
the total profit per unit time.}

\noindent \textbf{Proof:} From Lemma 2(b), the solution $t_{1}^{\ast }\in
\left( 0,\widehat{t}_{1}\right) $ which satisfies Eq. (\ref{dPt31}) not only
exists but also is unique. Hence, the value $t_{3}^{\ast }$ can be uniquely
determined by Eq. (\ref{t3}). Furthermore, we can obtain%
\begin{eqnarray*}
\left. \frac{\partial ^{2}\Pi \left( t_{1},t_{3}\right) }{\partial t_{1}^{2}}%
\right\vert _{\left( t_{1},t_{3}\right) =\left( t_{1}^{\ast },t_{3}^{\ast
}\right) } &=&\frac{1}{\left( t_{1}^{\ast }+\frac{S_{0}}{D}+t_{3}^{\ast
}\right) ^{2}}\bigg[\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) -D\left(
s-c\right) +K\left( t_{1}^{\ast }\right) \bigg] \\
&&-\frac{1}{t_{1}^{\ast }+\frac{S_{0}}{D}+t_{3}^{\ast }}\left. \left( \frac{%
\partial \Pi \left( t_{1},t_{3}\right) }{\partial t_{1}}+\frac{\text{d}%
K\left( t_{1}\right) }{\text{d}t_{1}}\right) \right\vert _{\left(
t_{1},t_{3}\right) =\left( t_{1}^{\ast },t_{3}^{\ast }\right) } \\
&=&-\frac{1}{t_{1}^{\ast }+\frac{S_{0}}{D}+t_{3}^{\ast }}\left. \frac{\text{d%
}K\left( t_{1}\right) }{\text{d}t_{1}}\right\vert _{\left(
t_{1},t_{3}\right) =\left( t_{1}^{\ast },t_{3}^{\ast }\right) }<0,
\end{eqnarray*}%
\begin{eqnarray*}
\left. \frac{\partial ^{2}\Pi \left( t_{1},t_{3}\right) }{\partial t_{3}^{2}}%
\right\vert _{\left( t_{1},t_{3}\right) =\left( t_{1}^{\ast },t_{3}^{\ast
}\right) } &=&\frac{1}{\left( t_{1}^{\ast }+\frac{S_{0}}{D}+t_{3}^{\ast
}\right) ^{2}}\bigg\{\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) -D\left(
s-c\right) \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] t_{3}^{\ast }}{%
1+\delta t_{3}^{\ast }}\bigg\} \\
&&-\frac{1}{t_{1}^{\ast }+\frac{S_{0}}{D}+t_{3}^{\ast }}\bigg\{\left. \frac{%
\partial \Pi \left( t_{1},t_{3}\right) }{\partial t_{3}}\right\vert _{\left(
t_{1},t_{3}\right) =\left( t_{1}^{\ast },t_{3}^{\ast }\right) } \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\left(
1+\delta t_{3}^{\ast }\right) ^{2}}\bigg\} \\
&=&-\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\left(
t_{1}^{\ast }+\frac{S_{0}}{D}+t_{3}^{\ast }\right) \left( 1+\delta
t_{3}^{\ast }\right) ^{2}}<0,
\end{eqnarray*}%
and%
\begin{equation*}
\left. \frac{\partial ^{2}\Pi \left( t_{1},t_{3}\right) }{\partial
t_{1}\partial t_{3}}\right\vert _{\left( t_{1},t_{3}\right) =\left(
t_{1}^{\ast },t_{3}^{\ast }\right) }=0.
\end{equation*}%
Thus, the determinant of the Hessian matrix at the stationary point $\left(
t_{1}^{\ast },t_{3}^{\ast }\right) $ is
\begin{eqnarray}
\left\vert \mathbf{H}\right\vert &=&\left. \frac{1}{t_{1}^{\ast }+\frac{S_{0}%
}{D}+t_{3}^{\ast }}\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}\times
\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\left(
t_{1}^{\ast }+\frac{S_{0}}{D}+t_{3}^{\ast }\right) \left( 1+\delta
t_{3}^{\ast }\right) ^{2}}\right\vert _{\left( t_{1},t_{3}\right) =\left(
t_{1}^{\ast },t_{3}^{\ast }\right) } \notag \\
&=&\left. \frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{%
\left( t_{1}^{\ast }+\frac{S_{0}}{D}+t_{3}^{\ast }\right) ^{2}\left(
1+\delta t_{3}^{\ast }\right) ^{2}}\frac{\text{d}K\left( t_{1}\right) }{%
\text{d}t_{1}}\right\vert _{\left( t_{1},t_{3}\right) =\left( t_{1}^{\ast
},t_{3}^{\ast }\right) }>0. \label{hessian1}
\end{eqnarray}%
Consequently, we can conclude that the stationary point $\left( t_{1}^{\ast
},t_{3}^{\ast }\right) $\ for our optimization problem is a global maximum.
This completes the proof. $\square $

Once the optimal solution $\left( t_{1}^{\ast },t_{3}^{\ast }\right) $ is
obtained, we substitute $\left( t_{1}^{\ast },t_{3}^{\ast }\right) $ into
Eqs. (\ref{Q1}) and (\ref{TP}), the optimal ordering quantity per cycle, $%
Q^{\ast }$, and the maximum total profit per unit time, $\Pi \left(
t_{1}^{\ast },t_{3}^{\ast }\right) $, are as follows:%
\begin{equation*}
Q^{\ast }=\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}^{\ast }%
\right] ^{\frac{1}{1-\beta }}+\frac{D\ln \left( 1+\delta t_{3}^{\ast
}\right) }{\delta },
\end{equation*}%
and%
\begin{equation}
\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) =D\left( s-c\right) -\frac{D%
\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] t_{3}^{\ast }}{1+\delta
t_{3}^{\ast }}. \label{p1tT}
\end{equation}

\subsubsection{Case 2: $\left( 1-\protect\beta \right) ^{\frac{1}{\protect%
\beta }}\left[ \frac{\protect\alpha \protect\beta \left( s-c\right) }{c_{1}}%
\right] ^{\frac{1}{1-\protect\beta }}\leq S_{0}<\left[ \frac{\protect\alpha
\protect\beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\protect\beta }%
} $}

\indent\indent Under the condition $S_{0}<\left[ \frac{\alpha \beta \left(
s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}$, we know that $\widetilde{t%
}_{1}>0$, $K\left( t_{1}\right) $ is strictly decreasing in the interval $%
\left[ 0,\widetilde{t}_{1}\right] $, strictly increasing in the interval $%
\left[ \widetilde{t}_{1},\infty \right) $, and has a minimum at the point $%
t_{1}=\widetilde{t}_{1}$. Thus, we obtain $K\left( \widetilde{t}_{1}\right)

\begin{equation*}
K\left( 0\right) =c_{1}S_{0}<\alpha \beta \left( s-c\right) S_{0}^{\beta
}<\alpha \left( s-c\right) S_{0}^{\beta }=D\left( s-c\right) <\frac{D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta }=K\left( \widehat{t}%
_{1}\right) .
\end{equation*}%
Combine the above results, we obtain $K\left( \widetilde{t}_{1}\right)

\widetilde{t}_{1}<\widehat{t}_{1}$.

First, we consider the interval $\left[ \widehat{t}_{1},\infty \right) $. By
using the similar arguments as Case 1.2, it can be shown that for any $%
t_{1}\in \left[ \widehat{t}_{1},\infty \right) $, there does not exist a $%
t_{3}\in \left[ 0,\infty \right) $ such that $F\left( t_{3}\right) =K\left(
t_{1}\right) $. Therefore, the interval $\left[ \widehat{t}_{1},\infty
\right) $ should be excluded from consideration.

Second, we consider the interval $\left[ \widetilde{t}_{1},\widehat{t}%
_{1}\right) $. When $\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[ \frac{%
\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}\leq
S_{0}\linebreak<\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right]
^{\frac{1}{1-\beta }}$, by Lemma 1(b), we have the minimum value $K\left(
\widetilde{t}_{1}\right) \geq 0$. It implies that $K\left( t_{1}\right) $ is
non-negative and strictly increasing in the interval $\left[ \widetilde{t}%
_{1},\widehat{t}_{1}\right) $. Thus, for any given $t_{1}\in \left[
\widetilde{t}_{1},\widehat{t}_{1}\right) $, from Eq. (\ref{ft3}), we have $%
F\left( 0\right) =0$, and $\lim_{t_{3}\rightarrow \infty }F\left(
t_{3}\right) =K\left( \widehat{t}_{1}\right) >K\left( t_{1}\right) $, which
implies for any given $t_{1}\in \left[ \widetilde{t}_{1},\widehat{t}%
_{1}\right) $, there exists a unique value $t_{3}$ $\in \left[ 0,\infty
\right) $ such that $F\left( t_{3}\right) =K\left( t_{1}\right) $, i.e., for
any given $t_{1}\in \left[ \widetilde{t}_{1},\widehat{t}_{1}\right) $, we
can find a value $t_{3}\in \left[ 0,\infty \right) $ which satisfies Eq. (%
\ref{dPt11}).

Next, we consider the interval $\left[ 0,\widetilde{t}_{1}\right) $. Since $%
K\left( t_{1}\right) $ is strictly decreasing in the interval $\left[ 0,%
\widetilde{t}_{1}\right) $. Therefore, from Eq. (\ref{ft3}), we have $%
F\left( 0\right) =0$, and $\lim_{t_{3}\rightarrow \infty }F\left(
t_{3}\right) =K\left( \widehat{t}_{1}\right) >K\left( 0\right) >K\left(
t_{1}\right) $ for all $t_{1}\in \left[ 0,\widetilde{t}_{1}\right) $, which
implies that for any given $t_{1}\in \left[ 0,\widetilde{t}_{1}\right) $,
there exists a unique value $t_{3}$ $\in \left( 0,\infty \right) $ such that
$F\left( t_{3}\right) =K\left( t_{1}\right) $, i.e., for any given $t_{1}\in %
\left[ 0,\widetilde{t}_{1}\right) $, we can find a value $t_{3}\in \left(
0,\infty \right) $ which satisfies Eq. (\ref{dPt11}). As a result, if there
exists a solution $t_{1}^{\ast }\in \left[ 0,\widetilde{t}_{1}\right) $ such
that $G\left( t_{1}^{\ast }\right) =0$, then the corresponding $t_{3}$
(denoted by $t_{3}^{\ast }$) can be unique determined. However, due to the
fact that $\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}<0fact that $\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}<0\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}<0}<0fact that $\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}<0fact that $\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}<0\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}<0}<0$>
amp;gt;
amp;gt;
amp;gt;
gt;
amp;gt;
amp;gt;
amp;gt;
t_{1}\in \left( 0,\widetilde{t}_{1}\right) $, from Eq. (\ref{hessian1}), the
determinant of the Hessian matrix at the stationary point $\left(
t_{1}^{\ast },t_{3}^{\ast }\right) $ is $\left\vert \mathbf{H}\right\vert <0$%
. This means that the stationary point $\left( t_{1}^{\ast },t_{3}^{\ast
}\right) $ is not the optimal solution for our maximum problem. Hence, to
find the optimal solution of $t_{1}$, the interval $\left[ 0,\widetilde{t}%
_{1}\right) $ should be excluded from consideration.

Consequently, when $\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[ \frac{%
\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}\leq
S_{0}<\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{%
1-\beta }}$, the optimal solution of $t_{1}$ which satisfies $F\left(
t_{3}\right) =K\left( t_{1}\right) $ for any $t_{3}\in \left[ 0,\infty
\right) $ will occur in the interval $\left[ \widetilde{t}_{1},\widehat{t}%
_{1}\right) $. Thus we have the following result.

\noindent \textbf{Lemma 3.} \textit{For }$\left( 1-\beta \right) ^{\frac{1}{%
\beta }}\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1%
}{1-\beta }}\leq S_{0}<\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}%
\right] ^{\frac{1}{1-\beta }}$\textit{, we have:}

\noindent (a) \textit{If }$G\left( \widetilde{t}_{1}\right) \leq 0$\textit{,
then the optimal }$t_{1}^{\ast }=0$.

\noindent (b) \textit{If }$G\left( \widetilde{t}_{1}\right) >0$\textit{,
then the solution }$t_{1}^{\ast }\in \left( \widetilde{t}_{1},\widehat{t}%
_{1}\right) $\textit{\ which satisfies Eq. }(\ref{dPt31})\textit{\ not only
exists but also is unique.}

\noindent \textbf{Proof: }See Appendix C.

Lemma 3(a) indicates that if $\left( 1-\beta \right) ^{\frac{1}{\beta }}%
\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{%
1-\beta }}\leq S_{0}<\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}%
\right] ^{\frac{1}{1-\beta }}$ and $G\left( \widetilde{t}_{1}\right) \leq 0$%
, then the optimal time at which the inventory level reaches $S_{0}$ is $%
t_{1}^{\ast }=0$. By using the similar arguments as Lemma 2(a), the maximum
inventory level in this system is $I^{\ast }=S_{0}$, the optimal value of $%
t_{3}$ can be obtained by using Eq. (\ref{Zt3}), and the maximum total
profit per unit time $\Pi \left( t_{3}^{\ast }\right) $ is obtained by Eq. (%
\ref{Pi0t3}).

Lemma 3(b) reveals that if $\left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}\leq S_{0}<\frac{D\left[ c_{2}+\delta
\left( s-c+c_{3}\right) \right] }{\delta c_{1}}$ and $G\left( \widetilde{t}%
_{1}\right) >0$, then $t_{1}^{\ast }\in \left( \widetilde{t}_{1},\widehat{t}%
_{1}\right) $ and is unique. By using the similar proof as Theorem 2, we can
obtain the following result.

\noindent \textbf{Theorem 3.} \textit{For }$\left( 1-\beta \right) ^{\frac{1%
}{\beta }}\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{%
\frac{1}{1-\beta }}\leq S_{0}<\left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}$\textit{, if }$G\left( \widetilde{t}%
_{1}\right) >0$, \textit{then the point }$\left( t_{1}^{\ast },t_{3}^{\ast
}\right) $\textit{\ which satisfies Eqs. }(\ref{dPt11})\textit{\ and }(\ref%
{dPt31})\textit{\ simultaneously and }$t_{1}^{\ast }\in \left( \widetilde{t}%
_{1},\widehat{t}_{1}\right) $\textit{\ is the global maximum point of the
total profit per unit time.}

\subsubsection{Case 3: $S_{0}<\left( 1-\protect\beta \right) ^{\frac{1}{%
\protect\beta }}\left[ \frac{\protect\alpha \protect\beta \left( s-c\right)
}{c_{1}}\right] ^{\frac{1}{1-\protect\beta }}$}

\indent\indent When $S_{0}<\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[
\frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{\beta }{1-\beta
}}$, by Lemma 1(b), we have $K\left( t_{1}\right) $ is strictly decreasing
in the interval $\left[ 0,\widetilde{t}_{1}\right] $, strictly increasing in
the interval $\left[ \widetilde{t}_{1},\infty \right) $, and the minimum
value $K\left( \widetilde{t}_{1}\right) <0$.>
=c_{1}S_{0}>0$ and $\lim_{t_{1}\rightarrow \infty }K\left( t_{1}\right)
=\infty $, there exists two different points $t_{1}^{\prime }$ and $%
t_{1}^{\prime \prime }$ such that $K\left( t_{1}^{\prime }\right) =K\left(
t_{1}^{\prime \prime }\right) =0$, where $0
_{1}
\widetilde{t}_{1}\right)
, we obtain $K\left( \widetilde{t}_{1}\right)
}\right) =K\left( t_{1}^{\prime \prime }\right)
\widehat{t}_{1}\right) $, which implies $0
_{1}

First, we consider the interval $\left[ \widehat{t}_{1},\infty \right) $. By
using the similar arguments as Case 1.2, it can be shown that for any $%
t_{1}\in \left[ \widehat{t}_{1},\infty \right) $, there does not exist a $%
t_{3}\in \left[ 0,\infty \right) $ such that $F\left( t_{3}\right) =K\left(
t_{1}\right) $. Therefore, the interval $\left[ \widehat{t}_{1},\infty
\right) $ should be excluded from consideration.

Second, we consider the interval $\left[ t_{1}^{\prime \prime },\widehat{t}%
_{1}\right) $. Because $F\left( 0\right) =0$, $\lim_{t_{3}\rightarrow \infty
}F\left( t_{3}\right) =K\left( \widehat{t}_{1}\right) >K\left( t_{1}\right) $
for any $t_{1}\in \left[ t_{1}^{\prime \prime },\widehat{t}_{1}\right) $,
and $F\left( t_{3}\right) $ is strictly increasing in the interval $\left[
0,\infty \right) $, which implies for any given $t_{1}\in \left[
t_{1}^{\prime \prime },\widehat{t}_{1}\right) $, there exists a unique value
$t_{3}$ $\in \left[ 0,\infty \right) $ such that $F\left( t_{3}\right)
=K\left( t_{1}\right) $, i.e., for any given $t_{1}\in \left[ t_{1}^{\prime
\prime },\widehat{t}_{1}\right) $, we can find a value $t_{3}$ which
satisfies Eq. (\ref{dPt11}).

Next, we consider the interval $\left[ \widetilde{t}_{1},t_{1}^{\prime
\prime }\right) $. Because for any $t_{3}\in \left[ 0,\infty \right) $, $%
F\left( t_{3}\right) =\linebreak \frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] t_{3}}{1+\delta t_{3}}\geq 0=K\left( t_{1}^{\prime
\prime }\right) >K\left( t_{1}\right) $ for all $t_{1}\in \left[ \widetilde{t%
}_{1},t_{1}^{\prime \prime }\right) $, which implies for any given $t_{1}\in %
\left[ \widetilde{t}_{1},t_{1}^{\prime \prime }\right) $, there does not
exist a value $t_{3}\in \left[ 0,\infty \right) $ such that $F\left(
t_{3}\right) =K\left( t_{1}\right) $, i.e., we can not find a value $t_{3}$
which satisfies Eq. (\ref{dPt11}). Therefore, the interval $\left[
\widetilde{t}_{1},t_{1}^{\prime \prime }\right) $ should be excluded from
consideration.

Finally, we consider the interval $\left[ 0,\widetilde{t}_{1}\right) $. We
can show that there is no point $t_{1}\in \left[ 0,\widetilde{t}_{1}\right) $
satisfying the sufficient condition for the maximality problem of $\Pi
\left( t_{1},t_{3}\right) $, and hence the interval $\left[ 0,\widetilde{t}%
_{1}\right) $ can be excluded from consideration.

Consequently, when $S_{0}<\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[
\frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}$,
the optimal solution of $t_{1}$ which satisfies $F\left( t_{3}\right)
=K\left( t_{1}\right) $ for any $t_{3}\in \left[ 0,\infty \right) $ will
occur in the interval $\left[ t_{1}^{\prime \prime },\widehat{t}_{1}\right) $%
, where $t_{1}^{\prime \prime }\in \left( \widetilde{t}_{1},\widehat{t}%
_{1}\right) $ and satisfies $K\left( t_{1}^{\prime \prime }\right) =0$. Thus
we have the following result.

\noindent \textbf{Lemma 4.} \textit{For }$S_{0}<\left( 1-\beta \right) ^{%
\frac{1}{\beta }}\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right]
^{\frac{1}{1-\beta }}$\textit{, let }$t_{1}^{\prime \prime }\in \left(
\widetilde{t}_{1},\widehat{t}_{1}\right) $\textit{\ and satisfies }$K\left(
t_{1}^{\prime \prime }\right) =0$\textit{. Then we have:}

\begin{enumerate}
\item[(a)] \textit{If }$G\left( \widetilde{t}_{1}\right) \leq 0$\textit{,
then the optimal }$t_{1}^{\ast }=0$\textit{.}

\item[(b)] \textit{If }$G\left( \widetilde{t}_{1}\right) >G\left(
t_{1}^{\prime \prime }\right) >0$\textit{, then the solution }$t_{1}^{\ast
}\in \left( t_{1}^{\prime \prime },\widehat{t}_{1}\right) $\textit{\ which
satisfies Eq. }(\ref{dPt31})\textit{\ not only exists but also is unique.}

\item[(c)] \textit{If }$G\left( \widetilde{t}_{1}\right) >0\geq G\left(
t_{1}^{\prime \prime }\right) $\textit{, then the model reduces to the model
without shortages.}
\end{enumerate}

\noindent \textbf{Proof:} See Appendix D.

\noindent \textbf{Theorem 4.} \textit{For }$S_{0}<\left( 1-\beta \right) ^{%
\frac{1}{\beta }}\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right]
^{\frac{1}{1-\beta }}$\textit{, if }$G\left( \widetilde{t}_{1}\right)
>G\left( t_{1}^{\prime \prime }\right) \geq 0$\textit{, where }$%
t_{1}^{\prime \prime }\in \left( \widetilde{t}_{1},\widehat{t}_{1}\right) $%
\textit{\ and\ satisfies }$K\left( t_{1}^{\prime \prime }\right) =0$\textit{%
, then the point }$\left( t_{1}^{\ast },t_{3}^{\ast }\right) $\textit{\
which satisfies Eqs. }(\ref{dPt11})\textit{\ and }(\ref{dPt31})\textit{\
simultaneously is the global maximum point of the total profit per unit time.%
} \noindent

Lemma 4(a) indicates that if $S_{0}<\left( 1-\beta \right) ^{\frac{1}{\beta }%
}\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{%
1-\beta }}$ and $G\left( \widetilde{t}_{1}\right) \leq 0$, then the optimal
time at which the inventory level reaches $S_{0}$ is $t_{1}^{\ast }=0$. By
using the similar arguments as Lemma 2(a), the maximum inventory level in
this system is $I^{\ast }=S_{0}$, the optimal value of $t_{3}$ can be
obtained by using Eq. (\ref{Zt3}), and the maximum total profit per unit
time $\Pi \left( t_{3}^{\ast }\right) $ is obtained by Eq. (\ref{Pi0t3}).

Lemma 4(b) reveals that if $S_{0}<\left( 1-\beta \right) ^{\frac{1}{\beta }}%
\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{%
1-\beta }}$ and $G\left( \widetilde{t}_{1}\right) >G\left( t_{1}^{\prime
\prime }\right) >0$, then $t_{1}^{\ast }\in \left( t_{1}^{\prime \prime },%
\widehat{t}_{1}\right) $ and is unique. By using the similar arguments as
Theorem 2, Theorem 4 can be showed that the unique solution $\left(
t_{1}^{\ast },t_{3}^{\ast }\right) $ which satisfies Eqs. (\ref{dPt11})\ and
(\ref{dPt31}) indeed a global maximum point. Hence the maximum total profit
per unit time $\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) $ is obtained
by Eq. (\ref{p1tT}).

Lemma 4(c) has proved that if $S_{0}<\left( 1-\beta \right) ^{\frac{1}{\beta
}}\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{%
1-\beta }}$ and $G\left( \widetilde{t}_{1}\right) >0\geq G\left(
t_{1}^{\prime \prime }\right) $, then the optimal value of $t_{1}$ is $%
t_{1}^{\prime \prime }$ and the corresponding optimal value of $t_{3}$ is $%
t_{3}^{\ast }=0$. As a result, the model reduces to the model without
shortages and its total profit per unit time is $\Pi \left( t_{1}^{\prime
\prime },0\right) =D\left( s-c\right) $.

On the other hand, for the special circumstance that $t_{3}^{\ast }=0$, we
will show that if \linebreak $S_{0}<\left( 1-\beta \right) ^{\frac{1}{\beta }%
}\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{%
1-\beta }}$ and $G\left( \widetilde{t}_{1}\right) >0\geq G\left(
t_{1}^{\prime \prime }\right) $, then there exists another value $%
t_{1}^{\ast }\in \left( \widetilde{t}_{1},t_{1}^{\prime \prime }\right) $
such that $\Pi \left( t_{1}^{\prime \prime },0\right) <\Pi \left(
t_{1}^{\ast },0\right) $. By setting $t_{3}^{\ast }=0$, the total profit per
per unit time in Eq. (\ref{TP}) becomes%
\begin{eqnarray}
\Pi \left( t_{1}\right) \equiv \Pi \left( t_{1},0\right) &=&\frac{1}{t_{1}+%
\frac{S_{0}}{D}}\Bigg\{\left( s-c\right) \left[ S_{0}^{1-\beta }+\alpha
\left( 1-\beta \right) t_{1}\right] ^{\frac{1}{1-\beta }} \notag \\
&&-A+\frac{c_{1}}{\alpha \left( \beta -2\right) }\left[ S_{0}^{1-\beta
}+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{2-\beta }{1-\beta }}
\notag \\
&&-c_{1}S_{0}^{2-\beta }\left[ \frac{1}{\alpha \left( \beta -2\right) }+%
\frac{S_{0}^{\beta }}{2D}\right] \Bigg\}\text{, for }t_{1}\in \left(
\widetilde{t}_{1},t_{1}^{\prime \prime }\right) . \label{Pit1}
\end{eqnarray}%
The necessary condition to find the optimal solution of $\Pi \left(
t_{1}\right) $ is
\begin{eqnarray}
\frac{\text{d}\Pi \left( t_{1}\right) }{\text{d}t_{1}} &=&\frac{1}{\left(
t_{1}+\frac{S_{0}}{D}\right) ^{2}}\Bigg\{-\left( s-c\right) \left[
S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{1}{%
1-\beta }}+A \notag \\
&&-\frac{c_{1}}{\alpha \left( \beta -2\right) }\left[ S_{0}^{1-\beta
}+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{2-\beta }{1-\beta }%
}+c_{1}S_{0}^{2-\beta }\left[ \frac{1}{\alpha \left( \beta -2\right) }+\frac{%
S_{0}^{\beta }}{2D}\right] \notag \\
&&+\alpha \left( s-c\right) \left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta
\right) t_{1}\right] ^{\frac{\beta }{1-\beta }}\left( t_{1}+\frac{S_{0}}{D}%
\right) \notag \\
&&-c_{1}\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right] ^{%
\frac{1}{1-\beta }}\left( t_{1}+\frac{S_{0}}{D}\right) \Bigg\} \notag \\
&=&\frac{1}{\left( t_{1}+\frac{S_{0}}{D}\right) ^{2}}\Bigg\{A-\left(
s-c\right) \left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right]
^{\frac{1}{1-\beta }} \notag \\
&&-\frac{c_{1}}{\alpha \left( \beta -2\right) }\left[ S_{0}^{1-\beta
}+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{2-\beta }{1-\beta }%
}+c_{1}S_{0}^{2-\beta }\left[ \frac{1}{\alpha \left( \beta -2\right) }+\frac{%
S_{0}^{\beta }}{2D}\right] \notag \\
&&+\left[ D\left( s-c\right) -K\left( t_{1}\right) \right] \left( t_{1}+%
\frac{S_{0}}{D}\right) \Bigg\}=0, \label{dPit1}
\end{eqnarray}%
which implies%
\begin{eqnarray*}
&&A-\left( s-c\right) \left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right)
t_{1}\right] ^{\frac{1}{1-\beta }}-\frac{c_{1}}{\alpha \left( \beta
-2\right) }\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right]
^{\frac{2-\beta }{1-\beta }} \\
&&+c_{1}S_{0}^{2-\beta }\left[ \frac{1}{\alpha \left( \beta -2\right) }+%
\frac{S_{0}^{\beta }}{2D}\right] +\left[ D\left( s-c\right) -K\left(
t_{1}\right) \right] \left( t_{1}+\frac{S_{0}}{D}\right) =0.
\end{eqnarray*}%
Let%
\begin{eqnarray}
W\left( t_{1}\right) &=&A-\left( s-c\right) \left[ S_{0}^{1-\beta }+\alpha
\left( 1-\beta \right) t_{1}\right] ^{\frac{1}{1-\beta }} \notag \\
&&-\frac{c_{1}}{\alpha \left( \beta -2\right) }\left[ S_{0}^{1-\beta
}+\alpha \left( 1-\beta \right) t_{1}\right] ^{\frac{2-\beta }{1-\beta }%
}+c_{1}S_{0}^{2-\beta }\left[ \frac{1}{\alpha \left( \beta -2\right) }+\frac{%
S_{0}^{\beta }}{2D}\right] \notag \\
&&+\left[ D\left( s-c\right) -K\left( t_{1}\right) \right] \left( t_{1}+%
\frac{S_{0}}{D}\right) ,\text{ }t_{1}\in \left( \widetilde{t}%
_{1},t_{1}^{\prime \prime }\right) . \label{wt}
\end{eqnarray}%
Applying Eq. (\ref{wt}), we obtain the following result.

\noindent \textbf{Theorem 5.} \textit{For }$S_{0}<\left( 1-\beta \right) ^{%
\frac{1}{\beta }}\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right]
^{\frac{1}{1-\beta }}$\textit{,\ if }$G\left( \widetilde{t}_{1}\right)
>0\geq G\left( t_{1}^{\prime \prime }\right) $\textit{, then the solution }$%
t_{1}^{\ast }\in \left( \widetilde{t}_{1},t_{1}^{\prime \prime }\right) $%
\textit{\ which satisfies Eq. }(\ref{dPit1})\textit{\ is the global maximum
point of the total profit per unit time }$\Pi \left( t_{1}\right) $\textit{.}

\noindent \textbf{Proof: }See Appendix E.

\subsection{Inventory problem with $I<>

\indent\indent When the stock level $S_{0}$ at which the demand rate changes
from being inventory-level dependent to a constant $D$ is relatively high,
an optimal inventory control policy would never order enough to raise $I$ to
$S_{0}$. Under this situation that $I<>
function of the inventory level and is always the constant $D$. That is,
what is being discussed in this section is the same as what was discussed in
the previous section, but with a very large value of $S_{0}$. Hence, the
total profit per unit time in Eq. (\ref{TP}) can be rewritten as
\begin{equation}
P\left( t_{2},t_{3}\right) =D\left( s-c\right) -\frac{A}{t_{2}+t_{3}}-\frac{%
c_{1}Dt_{2}^{2}}{2\left( t_{2}+t_{3}\right) }-\frac{D\left[ c_{2}+\delta
\left( s-c+c_{3}\right) \right] }{\delta ^{2}\left( t_{2}+t_{3}\right) }%
\left[ \delta t_{3}-\ln \left( 1+\delta t_{3}\right) \right] . \label{p2}
\end{equation}%
Solving the necessary conditions: $\frac{\partial P\left( t_{2},t_{3}\right)
}{\partial t_{2}}=0$ and $\frac{\partial P\left( t_{2},t_{3}\right) }{%
\partial t_{3}}=0$ for the maximum value of $P\left( t_{2},t_{3}\right) $,
and after algebraic operation, we get%
\begin{equation}
\frac{\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] t_{3}}{1+\delta
t_{3}}-c_{1}t_{2}=0, \label{dp2t1}
\end{equation}%
and%
\begin{eqnarray}
&&A+\frac{c_{1}Dt_{2}^{2}}{2}+\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta ^{2}}\left[ \delta t_{3}-\ln \left(
1+\delta t_{3}\right) \right] \notag \\
&&-D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] \frac{\left(
t_{2}+t_{3}\right) t_{3}}{1+\delta t_{3}}=0. \label{dp2t2}
\end{eqnarray}%
From Eq. (\ref{dp2t1}), we obtain%
\begin{equation}
t_{3}=\frac{c_{1}t_{2}}{\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right]
-\delta c_{1}t_{2}}. \label{p2t2}
\end{equation}%
Motivated by Eq. (\ref{dp2t2}), let%
\begin{eqnarray}
H\left( t_{2}\right) &=&A+\frac{c_{1}Dt_{2}^{2}}{2}+\frac{D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}\left[ \delta
t_{3}-\ln \left( 1+\delta t_{3}\right) \right] \notag \\
&&-D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] \frac{\left(
t_{2}+t_{3}\right) t_{3}}{1+\delta t_{3}},\text{ }t_{2}\geq 0, \label{Ht2}
\end{eqnarray}%
where $t_{3}$ is given as in Eq. (\ref{p2t2}), then the solution of $t_{2}$
can be found by solving $H\left( t_{2}\right) =0$. By using the similar
arguments as the above section, we can easily obtain the following two
results. The proofs are omitted.

\noindent \textbf{Lemma 5.} \textit{The point }$t_{2}^{\ast \ast }\in \left(
0,\frac{c_{2}+\delta \left( s-c+c_{3}\right) }{\delta c_{1}}\right) $\textit{%
\ which satisfies the equation }$H\left( t_{2}\right) =0$ in (\ref{Ht2})%
\textit{\ not only exists but also is unique.}

\noindent \textbf{Theorem 6.} \textit{The point }$\left( t_{2}^{\ast \ast
},t_{3}^{\ast \ast }\right) $\textit{\ which satisfies Eqs. }(\ref{dp2t1})%
\textit{\ and }(\ref{dp2t2})\textit{\ simultaneously is the global maximum
point of the total profit per unit time }$P\left( t_{2},t_{3}\right) $%
\textit{.}


Next, we want to compare the magnitude of the maximum inventory level $%
I^{\ast \ast }$ with the value $S_{0}$. Let us consider the following two
states: (1) $S_{0}\geq \frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) %
\right] }{\delta c_{1}}$ and (2) $S_{0}<\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta c_{1}}$. For state 1: $S_{0}\geq \frac{D%
\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta c_{1}}$, from
Lemma 5 and Theorem 6, we know that the optimal solution $t_{2}^{\ast \ast
}\in \left( 0,\frac{c_{2}+\delta \left( s-c+c_{3}\right) }{\delta c_{1}}%
\right) $. Consequently, the maximum inventory level per cycle $I^{\ast \ast
}=Dt_{2}^{\ast \ast }<\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) %
\right] }{\delta c_{1}}\leq S_{0}$.

For state 2: $S_{0}<\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) %
\right] }{\delta c_{1}}$, i.e., $\frac{S_{0}}{D}<\frac{c_{2}+\delta \left(
s-c+c_{3}\right) }{\delta c_{1}}$. From Eqs. (\ref{G0}) and (\ref{Ht2}), it
is not difficult to check that $H\left( \frac{S_{0}}{D}\right) =G\left(
0\right) $. Besides, it can be shown that $H\left( t_{2}\right) $ in Eq. (%
\ref{Ht2}) is a strictly decreasing function in $t_{2}\in \left[ 0,\frac{%
c_{2}+\delta \left( s-c+c_{3}\right) }{\delta c_{1}}\right) $ in conjunction
with $H\left( 0\right) =A>0$, and $\lim_{t_{2}\rightarrow \frac{c_{2}+\delta
\left( s-c+c_{3}\right) }{\delta c_{1}}^{-}}H\left( t_{2}\right) <0$.>
we investigate the condition under which $H\left( \frac{S_{0}}{D}\right)
\leq 0$ or $>0$, and the following two cases arise.

\begin{enumerate}
\item[(a)] \noindent If $H\left( \frac{S_{0}}{D}\right) =G\left( 0\right)
\leq 0$, by the property of the function $H\left( t_{2}\right) $ and the
Intermediate Value Theorem, we know that the optimal $t_{2}^{\ast \ast }$
must belong to the interval $\left( 0,\frac{S_{0}}{D}\right] $. It in turn
implies that the maximum inventory level per cycle $I^{\ast \ast
}=Dt_{2}^{\ast \ast }\leq D\times \frac{S_{0}}{D}=S_{0}$.

\item[(b)] \noindent If $H\left( \frac{S_{0}}{D}\right) =G\left( 0\right) >0$%
, we know that the optimal $t_{2}^{\ast \ast }$ belongs to the interval
\linebreak$\left( \frac{S_{0}}{D},\frac{c_{2}+\delta \left( s-c+c_{3}\right)
}{\delta c_{1}}\right) $. Obviously, with the constraint $I=Dt_{2}
optimal value $t_{2}^{\ast \ast }=\frac{S_{0}}{D}$ and $P\left( \frac{S_{0}}{%
D},t_{3}^{\ast \ast }\right) =\Pi \left( 0,t_{3}^{\ast }\right) $, where $%
t_{3}^{\ast \ast }=t_{3}^{\ast }$, and $t_{3}^{\ast }$ can be
found by solving Eq. (\ref{Zt3}).
\end{enumerate}

From the analysis carried out so far, once the optimal solution $\left( t_{2}^{\ast \ast
},t_{3}^{\ast \ast }\right) $ is obtained, the optimal ordering quantity per
cycle (denoted by $Q^{\ast \ast }$), the maximum inventory level per cycle
(denoted by $I^{\ast \ast }$) and the maximum total profit per unit time $%
P\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) $ are as follows:%
\begin{equation*}
Q^{\ast \ast }=Dt_{2}^{\ast \ast }+\frac{D\ln \left( 1+\delta t_{3}^{\ast
\ast }\right) }{\delta },
\end{equation*}%
\begin{equation*}
I^{\ast \ast }=Dt_{2}^{\ast \ast },
\end{equation*}%
and%
\begin{equation}
P\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) =D\left( s-c\right) -%
\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] t_{3}^{\ast \ast }%
}{1+\delta t_{3}^{\ast \ast }}. \label{p2t1t2}
\end{equation}
Then we have the following result.

\noindent \textbf{Property 1.}$\,$

\begin{enumerate}
\item[(a)] \textit{If }$S_{0}\geq \frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta c_{1}}$\textit{, there exists a unique }$%
\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) $\textit{\ satisfying
Eqs. }(\ref{dp2t1})\textit{\ and }(\ref{dp2t2})\textit{\ simultaneously,
where }$0
\delta c_{1}}$\textit{, such that }$P\left( t_{2}^{\ast \ast },t_{3}^{\ast
\ast }\right) $\textit{\ is maximum.}

\item[(b)] \textit{If }$S_{0}<\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta c_{1}}$\textit{\ and }$H\left( \frac{S_{0}%
}{D}\right) \leq 0$\textit{, there exists a unique }$\left( t_{2}^{\ast \ast
},t_{3}^{\ast \ast }\right) $ \textit{satisfying Eqs. }(\ref{dp2t1})\textit{%
\ and }(\ref{dp2t2})\textit{\ simultaneously, where }$0
}\leq \frac{S_{0}}{D}$\textit{, such that }$P\left( t_{2}^{\ast \ast
},t_{3}^{\ast \ast }\right) $\textit{\ is maximum.}

\item[(c)] \textit{If }$S_{0}<\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta c_{1}}$\textit{\ and }$H\left( \frac{S_{0}%
}{D}\right) >0$\textit{, then }$P\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast
}\right) $\textit{\ is maximum, where }$t_{2}^{\ast \ast }=\frac{S_{0}}{D}$%
\textit{, and\ }$t_{3}^{\ast \ast }$\textit{\ satisfies Eq. }(\ref{Zt3})%
\textit{.}
\end{enumerate}

\section{Optimal strategies}

\indent\indent By using the relationships among Property 1 and the results
of three cases discussed above, we can establish the following solution
procedures.

\noindent \textbf{Property 2.} \textit{For }$S_{0}\geq \left[ \frac{\alpha
\beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}$\textit{, we
have:}

\begin{enumerate}
\item[(a)] \textit{If }$S_{0}\geq \frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta c_{1}}$\textit{, the optimal total profit
per unit time is }$P\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) $%
\textit{, where }$\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) $%
\textit{\ satisfies Eqs. }(\ref{dp2t1})\textit{\ and }(\ref{dp2t2})\textit{,
and }$t_{2}^{\ast \ast }\in \left( 0,\frac{c_{2}+\delta \left(
s-c+c_{3}\right) }{\delta c_{1}}\right) $\textit{.}

\item[(b)] \textit{If }$S_{0}<\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta c_{1}}$\textit{\ and }$H\left( \frac{S_{0}%
}{D}\right) =G\left( 0\right) \leq 0$\textit{, the optimal total profit per
unit time is }$P\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) $\textit{%
, where }$\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) $\textit{\
satisfies Eqs. }(\ref{dp2t1})\textit{\ and }(\ref{dp2t2})\textit{, and }$%
t_{2}^{\ast \ast }\in \left( 0,\frac{S_{0}}{D}\right] $\textit{.}

\item[(c)] \textit{If }$S_{0}<\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta c_{1}}$\textit{\ and }$H\left( \frac{S_{0}%
}{D}\right) =G\left( 0\right) >0$\textit{, the optimal total profit per unit
time is }$\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) $\textit{, where }$%
\left( t_{1}^{\ast },t_{3}^{\ast }\right) $\textit{\ satisfies Eqs. }(\ref%
{dPt11})\textit{\ and }(\ref{dPt31})\textit{.}
\end{enumerate}

\noindent \textbf{Proof:} See Appendix F.

\noindent \textbf{Property 3.} \textit{For }$\left( 1-\beta \right) ^{\frac{1%
}{\beta }}\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{%
\frac{1}{1-\beta }}\leq S_{0}<\left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}$\textit{, we have:}

\begin{enumerate}
\item[(a)] \textit{If }$G\left( \widetilde{t}_{1}\right) \leq 0$\textit{,
the optimal total profit per unit time is }$P\left( t_{2}^{\ast \ast
},t_{3}^{\ast \ast }\right) $\textit{, where }$\left( t_{2}^{\ast \ast
},t_{3}^{\ast \ast }\right) $\textit{\ satisfies Eqs. }(\ref{dp2t1})\textit{%
\ and }(\ref{dp2t2})\textit{, and }$t_{2}^{\ast \ast }\in \left( 0,\frac{%
S_{0}}{D}\right) $.\textit{\ }

\item[(b)] \textit{If }$G\left( 0\right) >0$\textit{, the optimal total
profit per unit time is }$\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) $%
\textit{, where }$\left( t_{1}^{\ast },t_{3}^{\ast }\right) $\textit{\
satisfies Eqs. }(\ref{dPt11})\textit{\ and }(\ref{dPt31})\textit{.}

\item[(c)] \textit{If }$G\left( 0\right) \leq 0$\textit{\ and }$G\left(
\widetilde{t}_{1}\right) >0$\textit{, the optimal total profit per unit time
is }%
\begin{equation*}
\max \bigg\{\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) ,P\left(
t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) \bigg\}\mathit{,}
\end{equation*}%
\textit{where }$\left( t_{1}^{\ast },t_{3}^{\ast }\right) $\textit{\
satisfies Eqs. }(\ref{dPt11})\textit{\ and }(\ref{dPt31});\textit{\ }$\left(
t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) $\textit{\ satisfies Eqs. }(\ref%
{dp2t1})\textit{\ and }(\ref{dp2t2})\textit{, and }$t_{2}^{\ast \ast }\in
\left( 0,\frac{S_{0}}{D}\right] $.\textit{\ }
\end{enumerate}

\noindent \textbf{Proof:} See Appendix G.

\noindent \textbf{Property 4.} \textit{For }$S_{0}<\left( 1-\beta \right) ^{%
\frac{1}{\beta }}\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right]
^{\frac{1}{1-\beta }}$\textit{, we have:}

\begin{enumerate}
\item[(a)] \textit{If }$G\left( \widetilde{t}_{1}\right) \leq 0$\textit{,
the optimal total profit per unit time is }$P\left( t_{2}^{\ast \ast
},t_{3}^{\ast \ast }\right) $\textit{, where }$\left( t_{2}^{\ast \ast
},t_{3}^{\ast \ast }\right) $\textit{\ satisfies Eqs. }(\ref{dp2t1})\textit{%
\ and }(\ref{dp2t2})\textit{, and }$t_{2}^{\ast \ast }\in \left( 0,\frac{%
S_{0}}{D}\right) $.\textit{\ }

\item[(b)] \textit{If }$G\left( 0\right) >0$\textit{\ and }$G\left(
t_{1}^{\prime \prime }\right) >0$\textit{, the optimal total profit per unit
time is }$\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) $\textit{, where }$%
\left( t_{1}^{\ast },t_{3}^{\ast }\right) $\textit{\ satisfies Eqs. }(\ref%
{dPt11})\textit{\ and }(\ref{dPt31})\textit{.}

\item[(c)] \textit{If }$G\left( 0\right) \leq 0$\textit{\ and }$G\left(
t_{1}^{\prime \prime }\right) >0$\textit{, the optimal total profit per unit
time is}%
\begin{equation*}
\max \bigg\{\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) ,P\left(
t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) \bigg\}\mathit{,}
\end{equation*}%
\textit{where }$\left( t_{1}^{\ast },t_{3}^{\ast }\right) $\textit{\
satisfies Eqs. }(\ref{dPt11})\textit{\ and }(\ref{dPt31})\textit{; }$\left(
t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) $\textit{\ satisfies Eqs. }(\ref%
{dp2t1})\textit{\ and }(\ref{dp2t2})\textit{, and }$t_{2}^{\ast \ast }\in
\left( 0,\frac{S_{0}}{D}\right] $.\textit{\ }

\item[(d)] \textit{If }$G\left( \widetilde{t}_{1}\right) >0\geq G\left(
t_{1}^{\prime \prime }\right) $\textit{, the optimal total profit per unit
time is }$\Pi \left( t_{1}^{\ast }\right) $\textit{, where }$t_{1}^{\ast }$%
\textit{\ satisfies Eq. }(\ref{dPit1})\textit{.}
\end{enumerate}

\noindent \textbf{Proof:} See Appendix H.

\section{Numerical examples}

\indent\indent To illustrate the above results, we apply the proposed
solution procedure to solve the following numerical examples.

\noindent \textbf{Example 1.} We first consider the same example in Paul et
al. \cite{Paul}: $s=15$, $c=10$, $c_{1}=0.8$, $c_{2}=0.4$, $A=100$, $\alpha
=5$, $\beta =0.4$ and $S_{0}=80$ in appropriate units. Besides, we let $%
c_{3}=7$ and $\delta =1$ adapted to our model. We find that $S_{0}-\left[
\frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }%
}=12.67>0$, $S_{0}-\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right]
}{\delta c_{1}}=-367.23<0$>0$. Following
Property 2(c), the optimal solution is: $t_{1}^{\ast }=0.4432$, $t_{2}^{\ast
}=3.2158$, $t_{3}^{\ast }=0.2240$, $I^{\ast }=93.19$, $Q^{\ast }=99.03$, and
$\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) =78.80$.

\noindent \textbf{Example 2.} We then use the data in Datta and Pal \cite%
{Datta1}: $s=20$, $c=10$, $c_{1}=0.5$, $A=10$, $\alpha =0.5$, $\beta =0.4$
and $S_{0}=6$ in appropriate units. Besides, we let $c_{2}=2$, $c_{3}=7$ and
$\delta =1$ adapted to our model. We find that $S_{0}-\left[ \frac{\alpha
\beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}=-4.08<0$,>
S_{0}-\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[ \frac{\alpha \beta
\left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}=3.19>0$ and $G\left(
0\right) =0.95>0$. Following Property 3(b), the optimal solution is: $%
t_{1}^{\ast }=7.0631$, $t_{2}^{\ast }=12.9234$, $t_{3}^{\ast }=0.1790$, $%
I^{\ast }=14.86$, $Q^{\ast }=15.03$, and $\Pi \left( t_{1}^{\ast
},t_{3}^{\ast }\right) =7.29$.

\noindent \textbf{Example 3.} In this example, the same data in Example 2
are used except putting $\alpha =5$ and $S_{0}=28$. We find that $%
S_{0}-\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[ \frac{\alpha \beta
\left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}=-102.46<0$,>
\widetilde{t}_{1}\right) =803.56>0$, $G\left( t_{1}^{\prime \prime }\right)
=-2550.19$. Following Property 4(d), the optimal solution is: $t_{1}^{\ast
}=17.8925$, $t_{2}^{\ast }=19.3693$, $I^{\ast }=Q^{\ast }=946.85$, and $\Pi
\left( t_{1}^{\ast }\right) =301.90$.

\section{Concluding remarks}

\indent\indent In this paper, we develop an EOQ inventory model with
power-form stock-dependent demand rate followed by a constant demand rate,
permitting shortage and time-proportional backlogging rate. In particular,
the backlogging rate considered to be a decreasing function in the waiting
time until the next replenishment is more realistic. It is true that the
stockout cost is very difficult to measure. However, this does not mean that
the unit does not have some specific values. In practice, we can observe
periodically the proportion of demand which would accept backlogging and the
corresponding waiting time for the next replenishment. Then the statistical
techniques, such as the nonlinear regression method, can be used to estimate
the backlogging rate. Similarly, the stockout cost should be easy to obtain,
perhaps from accounting data. Therefore, while considering time-dependent
backlogging rate, we amend Urban's \cite{Urban2} paper by adding both the
backorder cost and the cost of lost sales into the total profit, and
complement the shortcoming of Paul et al. \cite{Paul} in solving the model.

Furthermore, we also provide some useful properties for finding the optimal
replenishment policies. Since decision variables in our problem cannot be
solved by simple algebraic means, they have to be solved numerically by
using Newton--Raphson Method (or any bisection method). Based on our
argument, in order to find a value $t_{1}^{\ast }$ such that $G\left(
t_{1}^{\ast }\right) =0$, a proper choice of the initial value $t_{1}$ is
very important due to possible local maxima. Without the right choice of
initial value, for example, putting $t_{1}<\widetilde{t}_{1}$, it will
converge to a saddle point.

The proposed model can be extended in several ways. For instance, we may
consider finite rate of replenishment. Also, we could extend the
deterministic demand function to stochastic demand patterns. We note that
the reciprocal of a linear function of the waiting time is one of the
possible time-dependent backordering functions, and it would be of interest
to consider other time-dependent backordering functions for this problem.
Finally, we could generalize the model to allow for permissible delay in
payments.

\section*{Acknowledgements}

\indent \indent The authors would like to thank the editor and anonymous reviewers for their
valuable and constructive comments, which have led to a significant
improvement in the manuscript. This research was partially supported by the National Science
Council of the Republic of China under Grant NSC-97-2221-E-156-005
and NSC-97-2221-E-366-006-MY2.


\section*{Appendix A}

\noindent \textbf{The proof of Lemma 1.}

\noindent (a) If $S_{0}\geq \left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}$, then we have $\frac{\beta \left(
s-c\right) }{c_{1}\left( 1-\beta \right) }-\frac{S_{0}^{1-\beta }}{\alpha
\left( 1-\beta \right) }=\widetilde{t}_{1}\leq 0$. From Eq. (\ref{dKt1}), we
know that $\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}>0$, hence $%
K\left( t_{1}\right) $\ is a strictly increasing function in $t_{1}\in \left[
0,\infty \right) $ and $K\left( 0\right) =c_{1}S_{0}>0$ is its minimum.

\noindent (b) If $S_{0}<\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}%
\right] ^{\frac{1}{1-\beta }}$, then we have $\frac{\beta \left( s-c\right)
}{c_{1}\left( 1-\beta \right) }-\frac{S_{0}^{1-\beta }}{\alpha \left(
1-\beta \right) }=\widetilde{t}_{1}>0$. From Eq. (\ref{dKt1}), we know that $%
\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}<0\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}<0}<0$>
gt;
gt;
0,\widetilde{t}_{1}\right) $ and $\frac{\text{d}K\left( t_{1}\right) }{\text{%
d}t_{1}}>0$ for $t_{1}\in \left( \widetilde{t}_{1},\infty \right) $.
Therefore, $K\left( t_{1}\right) $ is strictly decreasing in the interval $%
\left[ 0,\widetilde{t}_{1}\right] $ and is strictly increasing in the
interval $\left[ \widetilde{t}_{1},\infty \right) $. Besides, by the
continuity of $K\left( t_{1}\right) $ in $t_{1}\in \left[ 0,\infty \right) $%
, we see that $K\left( \widetilde{t}_{1}\right) $ is the minimum value of $%
K\left( t_{1}\right) $. Furthermore, from Eq. (\ref{kt1wave}), we have $%
K\left( \widetilde{t}_{1}\right) \geq 0$ if $S_{0}\geq \left( 1-\beta
\right) ^{\frac{1}{\beta }}\left[ \frac{\alpha \beta \left( s-c\right) }{%
c_{1}}\right] ^{\frac{1}{1-\beta }}$, and $K\left( \widetilde{t}_{1}\right)
\leq 0$ if $S_{0}\leq \left( 1-\beta \right) ^{\frac{1}{\beta }}\left[ \frac{%
\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}$. This
completes the proof. $\square $

\section*{Appendix B}

\setcounter{equation}{0}\renewcommand{\theequation}{B.\arabic{equation}}

\noindent \textbf{The proof of Lemma 2.}

\noindent (a) First, we consider $G\left( 0\right) <0$.>
t_{1}\right) $\ is strictly decreasing in the interval $\left[ 0,\widehat{t}%
_{1}\right) $, we can not find a value $t_{1}\in \left[ 0,\widehat{t}%
_{1}\right) $ such that $G\left( t_{1}\right) =0$. However, by Eqs. (\ref%
{Kt1}), (\ref{Kt3}) and (\ref{Gt1}), Eq. (\ref{dPt1}) becomes
\begin{eqnarray}
\frac{\partial \Pi \left( t_{1},t_{3}\right) }{\partial t_{1}} &=&-\frac{1}{%
t_{1}+\frac{S_{0}}{D}+t_{3}}\bigg\{\Pi \left( t_{1},t_{3}\right) +K\left(
t_{1}\right) -D\left( s-c\right) \notag \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] t_{3}}{%
1+\delta t_{3}}-\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right]
t_{3}}{1+\delta t_{3}}\bigg\} \notag \\
&=&-\frac{1}{t_{1}+\frac{S_{0}}{D}+t_{3}}\left\{ -\frac{G\left( t_{1}\right)
}{t_{1}+\frac{S_{0}}{D}+t_{3}}+K\left( t_{1}\right) -\frac{D\left[
c_{2}+\delta \left( s-c+c_{3}\right) \right] t_{3}}{1+\delta t_{3}}\right\}
\notag \\
&=&\frac{G\left( t_{1}\right) }{\left( t_{1}+\frac{S_{0}}{D}+t_{3}\right)
^{2}}. \label{dPt1Gt1}
\end{eqnarray}%
For this situation that $G\left( 0\right) <0$,>
\left( t_{1},t_{3}\right) }{\partial t_{1}}<0\left( t_{1},t_{3}\right) }{\partial t_{1}}<0\left( t_{1},t_{3}\right) }{\partial t_{1}}<0\left( t_{1},t_{3}\right) }{\partial t_{1}}<0$>
amp;gt;
gt;
amp;gt;
\widehat{t}_{1}\right) $, which implies that for any fixed $t_{3}\in \left[
0,\infty \right) $, a smaller value of $t_{1}$ causes a higher value of $\Pi
\left( t_{1},t_{3}\right) $. Therefore, the maximum value of $\Pi \left(
t_{1},t_{3}\right) $ occurs at the boundary point $t_{1}^{\ast }=0$.

Next, if $G\left( 0\right) =0$, then from the property that $G\left(
t_{1}\right) $ is a strictly decreasing in the interval $\left[ 0,\widehat{t}%
_{1}\right) $, we see that $t_{1}^{\ast }=0$ is the unique value which
satisfies $G\left( t_{1}^{\ast }\right) =0$.

\noindent (b) If $G\left( 0\right) >0$, since $G\left( t_{1}\right) $ is a
strictly decreasing function in $t_{1}\in \left[ 0,\widehat{t}_{1}\right) $%
,\ and \linebreak $\lim_{t_{1}\rightarrow \widehat{t}_{1}^{-}}G\left(
t_{1}\right) =-\infty $, by using the Intermediate Value Theorem, there
exists a unique solution $t_{1}^{\ast }\in \left( 0,\widehat{t}_{1}\right) $
such that $G\left( t_{1}^{\ast }\right) =0$, i.e., $t_{1}^{\ast }$ is the
unique solution which satisfies Eq. (\ref{dPt31}). This completes the proof.
$\square $

\section*{Appendix C}

\noindent \textbf{The proof of Lemma 3.}

\noindent (a) When $\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[ \frac{%
\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}\leq
S_{0}<\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{%
1-\beta }}$, from Lemma 1(b), we have $K\left( t_{1}\right) $ is strictly
decreasing in the interval $\left[ 0,\widetilde{t}_{1}\right] $ and strictly
increasing in the interval $\left[ \widetilde{t}_{1},\widehat{t}_{1}\right] $%
. Hence, by Eq. (\ref{dGt1}), we know that $G\left( t_{1}\right) $ has a
global maximum at $\widetilde{t}_{1}$, if $G\left( \widetilde{t}_{1}\right)
<0$,>
. However, from Eq. (\ref{dPt1Gt1}), we obtain that $\frac{\partial \Pi
\left( t_{1},t_{3}\right) }{\partial t_{1}}=\frac{G\left( t_{1}\right) }{%
\left( t_{1}+\frac{S_{0}}{D}+t_{3}\right) ^{2}}<\frac{G\left( \widetilde{t}%
_{1}\right) }{\left( t_{1}+\frac{S_{0}}{D}+t_{3}\right) ^{2}}<0$.>
the similar arguments in Lemma 2(a), we can obtain that $t_{1}^{\ast }=0$.

Next, if $G\left( \widetilde{t}_{1}\right) =0$, since $G\left( t_{1}\right) $
has a global maximum at $\widetilde{t}_{1}$, $\Pi \left( t_{1},t_{3}\right) $
is strictly decreasing in the interval $(0,\widetilde{t}_{1})$ and $(%
\widetilde{t}_{1},\infty )$, respectively. Obviously, $t_{1}=\widetilde{t}%
_{1}$ is an inflection point and the maximum value of $\Pi \left(
t_{1},t_{3}\right) $ occurs at the boundary point $t_{1}^{\ast }=0$.

\noindent (b) The proof is similar as Lemma 2(b), hence we omit it here.
This completes the proof. $\square $

\section*{Appendix D}

\noindent \textbf{The proof of Lemma 4.}

\noindent (a)-(b) The proofs are similar as Lemma 3, hence we omit them here.

\noindent (c) From discussion in Case 3, we see that $K\left( t_{1}\right) $
is strictly increasing in the interval $\left[ \widetilde{t}%
_{1},t_{1}^{\prime \prime }\right] $ and $K\left( t_{1}\right) <0_{1},t_{1}^{\prime \prime }\right] $ and $K\left( t_{1}\right) <0_{1},t_{1}^{\prime \prime }\right] $ and $K\left( t_{1}\right) <0_{1},t_{1}^{\prime \prime }\right] $ and $K\left( t_{1}\right) <0$>
amp;gt;
gt;
amp;gt;
t_{1}\in \left[ \widetilde{t}_{1},t_{1}^{\prime \prime }\right) $. Hence, by
Eq. (\ref{dGt1}), we have $\frac{\text{d}G\left( t_{1}\right) }{\text{d}t_{1}%
}<0$>
i.e., $G\left( t_{1}\right) $ is strictly decreasing in the interval $\left[
\widetilde{t}_{1},t_{1}^{\prime \prime }\right] $. If $G\left( \widetilde{t}%
_{1}\right) >0>G\left( t_{1}^{\prime \prime }\right) $, by using the
Intermediate Value Theorem, there exists a unique solution $t_{1}^{\ast }\in
\left( \widetilde{t}_{1},t_{1}^{\prime \prime }\right) $ such that $G\left(
t_{1}^{\ast }\right) =0$. However, for any $t_{3}\in \left[ 0,\infty \right)
$, $F\left( t_{3}\right) =\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) %
\right] t_{3}}{1+\delta t_{3}}\geq 0=K\left( t_{1}^{\prime \prime }\right)
>K\left( t_{1}^{\ast }\right) $, which implies we can not find a value $%
t_{3}^{\ast }\in \left[ 0,\infty \right) $ such that $F\left( t_{3}^{\ast
}\right) =K\left( t_{1}^{\ast }\right) $, i.e., for fixed $t_{1}^{\ast }\in
\left( \widetilde{t}_{1},t_{1}^{\prime \prime }\right) $ with $G\left(
t_{1}^{\ast }\right) =0$, there does not exist a value $t_{3}\in \left[
0,\infty \right) $ such that Eq. (\ref{dPt11}) holds.

On the other hand, from Eq. (\ref{dPt1}), we have $\frac{\partial \Pi \left(
t_{1},t_{3}\right) }{\partial t_{1}}=\frac{F\left( t_{3}\right) -K\left(
t_{1}\right) }{t_{1}+\frac{S_{0}}{D}+t_{3}}>0$ for all $t_{1}\in \left(
\widetilde{t}_{1},t_{1}^{\prime \prime }\right) $, which implies that for
fixed $t_{3}\in \left[ 0,\infty \right) $, $\Pi \left( t_{1},t_{3}\right) $
is a strictly increasing function in $t_{1}\in \left[ \widetilde{t}%
_{1},t_{1}^{\prime \prime }\right] $. Therefore, a higher value of $t_{1}$
causes a higher value of $\Pi \left( t_{1},t_{3}\right) $, it yields $%
t_{1}^{\ast }=t_{1}^{\prime \prime }$. For this situation, we can find $%
t_{3}^{\ast }=0$ such that $F\left( t_{3}^{\ast }\right) =0=K\left(
t_{1}^{\prime \prime }\right) $. Hence the model reduces to the model
without shortages.

Next, if $G\left( t_{1}^{\prime \prime }\right) =0$, then from the property
that $G\left( t_{1}\right) $ is a strictly decreasing in the interval $\left[
\widetilde{t}_{1},t_{1}^{\prime \prime }\right] $, we see that $t_{1}^{\ast
}=t_{1}^{\prime \prime }$ is the unique value which satisfies $G\left(
t_{1}^{\ast }\right) =0$, and thus the corresponding $t_{3}^{\ast }=0$. This
completes the proof. $\square $

\section*{Appendix E}

\noindent \textbf{The proof of Theorem 5.}

\setcounter{equation}{0}\renewcommand{\theequation}{E.\arabic{equation}}

Due to the relations shown in Eqs. (\ref{Kt3}) and (\ref{wt}), Eq. (\ref{Gt1}%
) can be written as%
\begin{eqnarray*}
G\left( t_{1}\right) &=&A-\left( s-c\right) \left[ S_{0}^{1-\beta }+\alpha
\left( 1-\beta \right) t_{1}\right] ^{\frac{1}{1-\beta }}-\frac{c_{1}}{%
\alpha \left( \beta -2\right) }\left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta
\right) t_{1}\right] ^{\frac{2-\beta }{1-\beta }} \\
&&+c_{1}S_{0}^{2-\beta }\left[ \frac{1}{\alpha \left( \beta -2\right) }+%
\frac{S_{0}^{\beta }}{2D}\right] +D\left( s-c\right) \left( t_{1}+\frac{S_{0}%
}{D}\right) \\
&&+\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}%
\left[ \delta t_{3}-\ln \left( 1+\delta t_{3}\right) \right] \\
&&-D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] \frac{\left( t_{1}+%
\frac{S_{0}}{D}+t_{3}\right) t_{3}}{1+\delta t_{3}} \\
&=&W\left( t_{1}\right) +K\left( t_{1}\right) \left( t_{1}+\frac{S_{0}}{D}%
\right) +\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta
^{2}}\left[ \frac{\delta t_{3}}{1+\delta t_{3}}-\ln \left( 1+\delta
t_{3}\right) \right] \\
&&-D\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] \frac{t_{3}}{%
1+\delta t_{3}}\left( t_{1}+\frac{S_{0}}{D}\right) \\
&=&W\left( t_{1}\right) +\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) %
\right] }{\delta ^{2}}\left[ \frac{\delta t_{3}}{1+\delta t_{3}}-\ln \left(
1+\delta t_{3}\right) \right] .
\end{eqnarray*}%
Note that the term in the brace, $\frac{\delta t_{3}}{1+\delta t_{3}}-\ln
\left( 1+\delta t_{3}\right) $, where $t_{3}<0\left( 1+\delta t_{3}\right) $, where $t_{3}<0\left( 1+\delta t_{3}\right) $, where $t_{3}<0\left( 1+\delta t_{3}\right) $, where $t_{3}<0$>
amp;gt;
gt;
amp;gt;
fact that $K\left( t_{1}\right) <0fact that $K\left( t_{1}\right) <0fact that $K\left( t_{1}\right) <0fact that $K\left( t_{1}\right) <0$>
amp;gt;
gt;
amp;gt;
_{1},t_{1}^{\prime \prime }\right) $), is a strictly increasing function in $%
t_{1}\in \left( \widetilde{t}_{1},t_{1}^{\prime \prime }\right) $ and it
goes to zero as $t_{1}\rightarrow t_{1}^{\prime \prime }$. Hence, $\frac{%
\delta t_{3}}{1+\delta t_{3}}-\ln \left( 1+\delta t_{3}\right) $ is negative
for $t_{1}\in $ $\left( \widetilde{t}_{1},t_{1}^{\prime \prime }\right) $.

Next, we want to prove the existence of $t_{1}^{\ast }\in \left( \widetilde{t%
}_{1},t_{1}^{\prime \prime }\right) $ which satisfies $W\left( t_{1}^{\ast
}\right) =0$. By using Eqs. (\ref{Kt1}) and (\ref{dKt1}), the derivative of $%
W\left( t_{1}\right) $ with respect to $t_{1}$ yields%
\begin{eqnarray*}
\frac{\text{d}W\left( t_{1}\right) }{\text{d}t_{1}} &=&-\alpha \left(
s-c\right) \left[ S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}\right]
^{\frac{\beta }{1-\beta }}+c_{1}\left[ S_{0}^{1-\beta }+\alpha \left(
1-\beta \right) t_{1}\right] ^{\frac{1}{1-\beta }} \\
&&+\left[ D\left( s-c\right) -K\left( t_{1}\right) \right] -\frac{\text{d}%
K\left( t_{1}\right) }{\text{d}t_{1}}\left( t_{1}+\frac{S_{0}}{D}\right) \\
&=&-\alpha c_{1}\left( 1-\beta \right) \left[ S_{0}^{1-\beta }+\alpha \left(
1-\beta \right) t_{1}\right] ^{\frac{\beta }{1-\beta }}\left[ \frac{s-c}{%
c_{1}\left( 1-\beta \right) }-\frac{S_{0}^{1-\beta }}{\alpha \left( 1-\beta
\right) }-t_{1}\right] \\
&&+\left[ D\left( s-c\right) -K\left( t_{1}\right) \right] -\frac{\text{d}%
K\left( t_{1}\right) }{\text{d}t_{1}}\left( t_{1}+\frac{S_{0}}{D}\right) \\
&=&-\frac{\text{d}K\left( t_{1}\right) }{\text{d}t_{1}}\left( t_{1}+\frac{%
S_{0}}{D}\right) <0.
\end{eqnarray*}%
As a result, $W\left( t_{1}\right) $ is a strictly decreasing function in $%
t_{1}\in \left( \widetilde{t}_{1},t_{1}^{\prime \prime }\right) $. Since
\begin{eqnarray*}
W\left( \widetilde{t}_{1}\right) &>&W\left( \widetilde{t}_{1}\right) +\frac{D%
\left[ c_{2}+\delta \left( s-c+c_{3}\right) \right] }{\delta ^{2}}\left[
\frac{\delta t_{3}}{1+\delta t_{3}}-\ln \left( 1+\delta t_{3}\right) \right]
\\
&=&G\left( \widetilde{t}_{1}\right) \\
&>&0,
\end{eqnarray*}%
and $\lim_{t_{1}\rightarrow t_{1}^{\prime \prime }}W\left( t_{1}^{\prime
\prime }\right) =G\left( t_{1}^{\prime \prime }\right) <0$,>
Intermediate Value Theorem, there exists a unique $t_{1}^{\ast }\in \left(
\widetilde{t}_{1},t_{1}^{\prime \prime }\right) $ such that $W\left(
t_{1}^{\ast }\right) =0$.

Once the optimal solution $t_{1}^{\ast }$ is obtained, the maximum total
profit per unit time $\Pi \left( t_{1}^{\ast }\right) $ is%
\begin{eqnarray}
\Pi \left( t_{1}^{\ast }\right) &=&D\left( s-c\right) -K\left( t_{1}^{\ast
}\right) \notag \\
&=&\alpha c_{1}\left( 1-\beta \right) \left[ S_{0}^{1-\beta }+\alpha \left(
1-\beta \right) t_{1}^{\ast }\right] ^{\frac{\beta }{1-\beta }}\left[ \frac{%
s-c}{c_{1}\left( 1-\beta \right) }-\frac{S_{0}^{1-\beta }}{\alpha \left(
1-\beta \right) }-t_{1}^{\ast }\right] . \label{Pit10}
\end{eqnarray}%
Due to the fact that $K\left( t_{1}\right) <0Due to the fact that $K\left( t_{1}\right) <0Due to the fact that $K\left( t_{1}\right) <0Due to the fact that $K\left( t_{1}\right) <0$>
amp;gt;
gt;
amp;gt;
\widetilde{t}_{1},t_{1}^{\prime \prime }\right) $, from Eq. (\ref{Pit10}),
we obtain%
\begin{eqnarray*}
\Pi \left( t_{1}^{\ast }\right) -\Pi \left( t_{1}^{\prime \prime },0\right)
&=&\alpha c_{1}\left( 1-\beta \right) \left[ S_{0}^{1-\beta }+\alpha \left(
1-\beta \right) t_{1}^{\ast }\right] ^{\frac{\beta }{1-\beta }}\left[ \frac{%
s-c}{c_{1}\left( 1-\beta \right) }-\frac{S_{0}^{1-\beta }}{\alpha \left(
1-\beta \right) }-t_{1}^{\ast }\right] \\
&&-D\left( s-c\right) \\
&=&-K\left( t_{1}^{\ast }\right) >0\text{.}
\end{eqnarray*}%
Consequently, if $S_{0}<\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[
\frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }}$
and $G\left( \widetilde{t}_{1}\right) >0\geq G\left( t_{1}^{\prime \prime
}\right) $, then the optimal ordering quantity per cycle is $Q^{\ast }=\left[
S_{0}^{1-\beta }+\alpha \left( 1-\beta \right) t_{1}^{\ast }\right] ^{\frac{1%
}{1-\beta }}$ and the maximum total profit per unit time is obtained by Eq. (%
\ref{Pit10}). This completes the proof. $\square $

\section*{Appendix F}

\noindent \textbf{The proof of Property 2.}

\noindent (a) By Theorem 1, Property 1(a) and using that fact that $\Pi
\left( 0,t_{3}^{\ast }\right) =P\left( \frac{S_{0}}{D},t_{3}^{\ast \ast
}\right) $, we have $P\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right)
>P\left( \frac{S_{0}}{D},t_{3}^{\ast \ast }\right) =\Pi \left( 0,t_{3}^{\ast
}\right) =\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) .$

\noindent (b) By Lemma 2(a), Property 1(b) and using that fact that $\Pi
\left( 0,t_{3}^{\ast }\right) =P\left( \frac{S_{0}}{D},t_{3}^{\ast \ast
}\right) $, we have $P\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right)
\geq P\left( \frac{S_{0}}{D},t_{3}^{\ast \ast }\right) =\Pi \left(
0,t_{3}^{\ast }\right) =\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) .$

\noindent (c) By Theorem 2, Property 1(c) and using that fact that $\Pi
\left( 0,t_{3}^{\ast }\right) =P\left( \frac{S_{0}}{D},t_{3}^{\ast \ast
}\right) $, we have $\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) >\Pi
\left( 0,t_{3}^{\ast }\right) =P\left( \frac{S_{0}}{D},t_{3}^{\ast \ast
}\right) =P\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) .$ This
completes the proof. $\square $

\section*{Appendix G}

\noindent \textbf{The proof of Property 3.}

\noindent (a) Since $G\left( t_{1}\right) $ is a strictly increasing
function in $t_{1}\in \left[ 0,\widetilde{t}_{1}\right) $, if $G\left(
\widetilde{t}_{1}\right) \leq 0$, we have $G\left( 0\right) <0$.
Furthermore, for any given $\left( 1-\beta \right) ^{\frac{1}{\beta }}\left[
\frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{\frac{1}{1-\beta }%
}\leq S_{0}<\left[ \frac{\alpha \beta \left( s-c\right) }{c_{1}}\right] ^{%
\frac{1}{1-\beta }}$, we have $S_{0}<\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] }{\delta c_{1}}$. Therefore, by Lemma 3(a),
Property 1(b) and using that fact that $\Pi \left( 0,t_{3}^{\ast }\right)
=P\left( \frac{S_{0}}{D},t_{3}^{\ast \ast }\right) $, we have $P\left(
t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) >P\left( \frac{S_{0}}{D}%
,t_{3}^{\ast \ast }\right) =\Pi \left( 0,t_{3}^{\ast }\right) =\Pi \left(
t_{1}^{\ast },t_{3}^{\ast }\right) $.

\noindent (b) From Lemma 1(b) and Eq. (\ref{dGt1}), we know that $\frac{%
\text{d}G\left( t_{1}\right) }{\text{d}t_{1}}>0$ for all $t_{1}\in \left( 0,%
\widetilde{t}_{1}\right) $, i.e., $G\left( t_{1}\right) $ is strictly
increasing in the interval $\left[ 0,\widetilde{t}_{1}\right] $. Therefore,
if $G\left( 0\right) >0$, we have $G\left( \widetilde{t}_{1}\right) >0$.
Furthermore, by Theorem 3, Property 1(c) and using that fact that $\Pi
\left( 0,t_{3}^{\ast }\right) =P\left( \frac{S_{0}}{D},t_{3}^{\ast \ast
}\right) $, we have $\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) >\Pi
\left( 0,t_{3}^{\ast }\right) =P\left( \frac{S_{0}}{D},t_{3}^{\ast \ast
}\right) =P\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) $.

\noindent (c) If $H\left( \frac{S_{0}}{D}\right) =G\left( 0\right) \leq 0$,
from Property 1(b), we know that there exists a point $\left( t_{2}^{\ast
\ast },t_{3}^{\ast \ast }\right) $ which satisfies Eqs. (\ref{dp2t1})\ and (%
\ref{dp2t2}), and $t_{2}^{\ast \ast }\in \left( 0,\frac{S_{0}}{D}\right] $
such that $P\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast }\right) $ is a
maximum. On the other hand, since $G\left( \widetilde{t}_{1}\right) \geq 0$,
from Theorem 3, we know that there exists a point $\left( t_{1}^{\ast
},t_{3}^{\ast }\right) $ which satisfies Eqs. (\ref{dPt11})\ and (\ref{dPt31}%
) such that $\Pi \left( t_{1}^{\ast },t_{3}^{\ast }\right) $ is a maximum.
Consequently, the maximum total profit per unit time is $\max \left\{ \Pi
\left( t_{1}^{\ast },t_{3}^{\ast }\right) ,P\left( t_{2}^{\ast \ast
},t_{3}^{\ast \ast }\right) \right\} $. This completes the proof. $\square $

\section*{Appendix H}

\noindent \textbf{The proof of Property 4.}

\noindent (a)-(c) The proofs are similar as Properties 2 and 3, hence we
omit them here.

\noindent (d) If $H\left( \frac{S_{0}}{D}\right) =G\left( 0\right) >0$, from
Property 1(c), we know that there exists a point $t_{3}^{\ast \ast }$ which
satisfies Eq. (\ref{Zt3}) such that $P\left( \frac{S_{0}}{D},t_{3}^{\ast
\ast }\right) $ is a maximum. On the other hand, since $G\left( \widetilde{t}%
_{1}\right) >0>G\left( t_{1}^{\prime \prime }\right) $, from Theorem 5, we
know that there exists a point $t_{1}^{\ast }$ which satisfies Eq. (\ref%
{dPit1}) such that$\ \Pi \left( t_{1}^{\ast }\right) $ is a maximum. Since $%
P\left( \frac{S_{0}}{D},t_{3}^{\ast \ast }\right) =\Pi \left( t_{3}^{\ast
}\right) $, from Eqs. (\ref{Pi0t3}) and (\ref{Pit10}), we obtain
\begin{eqnarray*}
\Pi \left( t_{1}^{\ast }\right) -P\left( \frac{S_{0}}{D},t_{3}^{\ast \ast
}\right) &=&\Pi \left( t_{1}^{\ast }\right) -\Pi \left( t_{3}^{\ast }\right)
\\
&=&\alpha c_{1}\left( 1-\beta \right) \left[ S_{0}^{1-\beta }+\alpha \left(
1-\beta \right) t_{1}^{\ast }\right] ^{\frac{\beta }{1-\beta }} \\
&&\times \left[ \frac{s-c}{c_{1}\left( 1-\beta \right) }-\frac{%
S_{0}^{1-\beta }}{\alpha \left( 1-\beta \right) }-t_{1}^{\ast }\right] \\
&&-D\left( s-c\right) +\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) %
\right] t_{3}^{\ast }}{\left( 1+\delta t_{3}^{\ast }\right) } \\
&=&-K\left( t_{1}^{\ast }\right) +\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] t_{3}^{\ast }}{1+\delta t_{3}^{\ast }}>0
\end{eqnarray*}%
Consequently, the maximum total profit per unit time is $\Pi \left(
t_{1}^{\ast }\right) $.

Next, if $H\left( \frac{S_{0}}{D}\right) =G\left( 0\right) \leq 0$, from
Property 1(b), we know that there exists a $P\left( t_{2}^{\ast \ast
},t_{3}^{\ast \ast }\right) $. Furthermore, since $G\left( \widetilde{t}%
_{1}\right) >0>G\left( t_{1}^{\prime \prime }\right) $, from Lemma 4(c), we
know that there also exists a$\ \Pi \left( t_{1}^{\ast }\right) $. From Eqs.
(\ref{Pit10}) and (\ref{p2t1t2}), we obtain%
\begin{eqnarray*}
\Pi \left( t_{1}^{\ast }\right) -P\left( t_{2}^{\ast \ast },t_{3}^{\ast \ast
}\right) &=&\alpha c_{1}\left( 1-\beta \right) \left[ S_{0}^{1-\beta
}+\alpha \left( 1-\beta \right) t_{1}^{\ast }\right] ^{\frac{\beta }{1-\beta
}} \\
&&\times \left[ \frac{s-c}{c_{1}\left( 1-\beta \right) }-\frac{%
S_{0}^{1-\beta }}{\alpha \left( 1-\beta \right) }-t_{1}^{\ast }\right] \\
&&-D\left( s-c\right) +\frac{D\left[ c_{2}+\delta \left( s-c+c_{3}\right) %
\right] t_{3}^{\ast \ast }}{1+\delta t_{3}^{\ast \ast }} \\
&=&-K\left( t_{1}^{\ast }\right) +\frac{D\left[ c_{2}+\delta \left(
s-c+c_{3}\right) \right] t_{3}^{\ast \ast }}{1+\delta t_{3}^{\ast \ast }}>0%
\text{.}
\end{eqnarray*}%
Consequently, the maximum total profit per unit time is $\Pi \left(
t_{1}^{\ast }\right) $. This completes the proof. $\square $

\setcounter{equation}{0} \renewcommand{\theequation}{B.\arabic{equation}}





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\end{document}


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