2. Model description

In this section we describe the model in more detail. The inventory level process $\mathbf{V}=\{\mathbf{V}(t)\,:\, t\geq 0\}$ under consideration is a jump-fluid process with state-dependent deterministic release rate: for every state $x>0$ the release rate (due to the fluid demand) between jumps is $\alpha (x)$ . The jumps are all upwards and of size q; they represent the arrival (delivery) of orders. There are two types of orders: normal orders and emergency orders. The lead times for both orders are exponentially distributed. Below we successively discuss the release rate process, the policy for placing normal orders and emergency orders, and the lead times in more detail.

Release rate assumption. We assume that

(2.1) \begin{equation}A(x)\,:\!=\, \int_{0}^{x}\dfrac{1}{\alpha (w)}\,{\mathrm{d}} w<\infty ,\end{equation}

for every $x>0$ . The latter assumption simply says that level 0 can be reached in a finite amount of time, since for every $0\leq x_{1}<x_{2}$ , $A(x_{2})-A(x_{1})$ is the time it takes to go from level $x_{2}$ all the way down to level $x_{1}$ if no jumps (deliveries) occur in between. We refer to Harrison and Resnick [Reference Harrison and Resnick16] for an early study of inventory and storage processes with a state-dependent release rate, and to Boxma et al. [Reference Boxma, Kaspi, Kella and Perry7] for a steady-state analysis of storage processes with state-dependent input and output rates.

The finiteness assumption (2.1) simplifies the analysis of the model, but is not necessary. It is enough to assume that $\int_{a}^{x}{1}/{(\alpha (w))}\,{\mathrm{d}} w<\infty $ . This means that level a can be reached from any level $x>a$ .

Remark 2.1. In practice, the price often cannot be changed continuously over time. Thus there are certain predetermined levels such that each of their downcrossings leads to a change in the price. An example is the case when there are switchover levels $\gamma_{1},\gamma_{2},\ldots,\gamma_{n}$ , such that

\begin{equation*}\alpha (x)=\begin{cases}\alpha _{1}, & 0<x\leq \gamma _{1},\\ \alpha _{2}, & \gamma _{1}\leq x<\gamma _{2}, \\\cdot & \\\cdot & \\\alpha _{n}, & \gamma _{n-1}\leq x<\gamma _{n}.\end{cases}\end{equation*}

That is, the controller switches between n different demand rates for which it is natural to assume that $\alpha_{1}<\alpha_{2}<\cdot \cdot \cdot < \alpha_{n}$ .

The policy for placing normal and emergency orders. The jumps of $\mathbf{V}$ represent the arrival (delivery) of normal and emergency orders. If just before downcrossing level b no order is pending, the controller places a normal order, of size q, immediately after the downcrossing of level b; but if at least one order is pending (a normal order or an emergency order) she will not place a new one. If just before downcrossing level a (recall that $a<b$ ) no emergency order is pending, the controller places an emergency order, also of size q, immediately after the downcrossing. This done even if a normal order is still pending. This policy is natural, since level a can be interpreted as a warning level: the inventory level approaches 0 and the controller would like to avoid the risky situation of an empty system, which is interpreted as a period of unsatisfied demand (during which sales may be lost). If just before downcrossing level a no normal order is pending, then also a normal order is placed. This policy ensures that just after hitting level a exactly two orders are pending, one normal and one emergency. Since the lead times of these orders are independent and both exponentially distributed, $\mathbf{V}$ is a regenerative process with, as regeneration epochs, the successive epochs at which $\mathbf{V}$ hits level a from above. In Lemma 3.1 this will be formally stated and proved. A level below a is considered to be precariously low, so that having two orders underway seems justified.

Lead times assumptions. The lead times for normal orders, and also for emergency orders, are assumed to be independent and identically distributed; normal lead times and emergency lead times are also independent of each other. We suppose that normal lead times are exponentially distributed with rate $\lambda_n$ , while emergency lead times are exponentially distributed with rate $\lambda_e$ . It would be natural to assume that $\lambda_e > \lambda_n$ , as an emergency lead time should typically be stochastically shorter than a normal lead time; however, in our analysis such a condition is not necessary. Thus there are two independent suppliers and from each of them one may make just one order at a time. The lead times cannot be expedited.

Parameter assumptions. In this study we restrict attention to parameters a, b, q satisfying $0<a<b<q$ and $ b<a+q$ (see Figure 1). This ordering seems natural, but one can think of other parameter relations, such as $0<a<q<b$ or $ a+q<b$ . In principle the derivations for such other cases are similar. For example, if $a+q<b$ , it can happen that after the arrival of the last pending order the inventory level is still below b but our policy does not indicate a new order. Therefore one might be inclined to consider other parameter combinations. If one wants to do optimization, then of course all possible orderings have to be taken into account.

Figure 1. The inventory level V(t), with state-dependent release rate, normal and emergency orders and constant order sizes q; $\mathbf{C}_i$ is the length of a regenerative cycle.

3. The regenerative structure of the inventory level process

In this section we discuss the regenerative structure of the inventory level process, with the periods between two successive downcrossings of level a as regeneration cycle.

The process $\mathbf{V}$ is not a Markov process (relative to its natural filtration), even though the lead times are exponential. To see this, consider $\mathbf{V}$ for example at some time instant t at which it hits level b. Then either one order is pending (possibly an emergency order) or no order is pending, in which case a new normal order is issued. Therefore the process continues with one pending order whose residual lead time can be $\exp\! (\lambda_n)$ or $\exp\! (\lambda_e)$ , both cases occurring with positive probability. However, from the evolution of $\mathbf{V}$ before t it might be known that exactly two jumps have happened before the first hitting of b at time t; in this case it is clear that the two jumps are due to the arrivals of the normal and the emergency order which were pending at the start of the current cycle, and now the order which is pending immediately after t must be a normal one. Hence the conditional probability law of the process after t is different from the one only conditional on $\mathbf{V}(t)=b$ . In other words, given $\mathbf{V}(t)$ , the future development from t on depends on the history prior to t. Thus $\mathbf{V}$ is not a Markov process. Note however that the joint process of $\mathbf{V}$ and the numbers of pending normal and emergency orders is a Markov process, whereas $\mathbf{V}$ alone is also not Markovian relative to the filtration generated by the three processes together. Furthermore, the jump process of $\mathbf{V}$ is not a Poisson process, since the arrival times of future deliveries depend on the history of $\mathbf{V}$ . Notwithstanding, as we shall prove below, $\mathbf{V}$ is a regenerative process, governed by three known parameters – the decision variables $(a;\,b;\,q)$ – and the known release rate function $\alpha (x)$ .

We are almost ready to derive the steady-state distribution of $\mathbf{V}$ and close this section with an observation that plays an important role in that derivation, and – first – with a lemma that provides some useful insight. In this second lemma we relate the number of pending orders with certain subsets of the sample space of $\mathbf{V}$ .

4. Steady-state analysis

In this section we determine the steady-state density $f({\cdot})$ of the inventory level process $\mathbf{V}$ . Notice that this density always exists, and has finite support $(0,a+2q)$ . As observed in the previous section, $\mathbf{V}$ is not a Markov process. Moreover, the jump process representing the delivery of orders is not a Poisson process since given the ‘present’ value of $\mathbf{V}$ , future deliveries depend on past deliveries. Thus PASTA (Poisson arrivals see time averages) cannot be applied in order to find the steady-state law by the traditional LCT (level crossing technique: see [Reference Brill8] and [Reference Cohen13]) concept of equating the rate of upcrossings of any fixed level x by $\mathbf{V}$ with the corresponding rate of downcrossings. We obtain $f({\cdot})$ via a (as far as we know) novel application of the LCT. Introduce, for $0<x<a+2q$ ,

\begin{equation*}\theta(x) \,:\!=\, E[{\rm number \,of \,downcrossings \,of \,level} \,x \, {\rm during \,a \,regenerative \,cycle}].\end{equation*}

Henceforth $\mathbf{C}$ denotes the length of an arbitrary regenerative cycle, the time between two successive downcrossings of level a. We relate f(x) to $\theta(x)$ in the following lemma. Subsequently we determine $\theta(x)$ in Theorem 4.1.

Lemma 4.1. We have the following relation between the stationary inventory level density f(x) and the mean number of downcrossings in a cycle, $\theta(x)$ : for all $ x \in [0,a+2q)$ ,

(4.1) \begin{equation}\theta(x) = \dfrac{\alpha(x) f(x)}{\alpha(a) f(a)} ,\end{equation}

so that

(4.2) \begin{equation}f(x) = \dfrac{\alpha(a) f(a)}{\alpha(x)} \theta(x) .\end{equation}

Proof. Since $\theta(x)$ is the mean number of downcrossings per regenerative cycle, and $\mathbb{E}[\mathbf{C}]$ the mean cycle length, the downcrossing rate of level x equals $\theta(x)/\mathbb{E}[\mathbf{C}]$ . On the other hand, the downcrossing rate of level x also equals $\alpha(x) f(x)$ . To prove (4.1), and hence also (4.2), it remains to show that

\begin{equation*}\mathbb{E}[\mathbf{C}] = \dfrac{1}{\alpha(a) f(a)} .\end{equation*}

This relation follows by observing that level a is downcrossed exactly once per cycle, and that the downcrossing rate of level a equals $\alpha(a) f(a)$ .

We now derive some properties of $f({\cdot})$ (under the assumption that $A(x)<\infty $ for all $x>0$ ). First, let $\pi$ be the steady-state probability of having zero inventory. By the normalizing condition we have

\begin{equation*} \int_{0}^{a+2q}f(x)\,{\mathrm{d}} x=1-\pi .\end{equation*}

When $A(x)=\infty$ , as for example in the shot noise case $\alpha(w)=w$ , we have $\pi =0$ and the normalizing condition is

\begin{equation*}\int_{0}^{a+2q} f(x)\,{\mathrm{d}} x=1.\end{equation*}

In addition, we have some continuity properties, because the inventory level process decreases in a continuous way. In particular,

(4.3) \begin{equation}f(x{-}) = f(x{+}) \quad \textrm{for} \, x=a, b, a+q, b+q, 2q.\end{equation}

However, $f({\cdot})$ is not continuous at q, because there is a positive probability of reaching q from 0:

\begin{equation*}\alpha (q)[f(q{-})-f(q{+})]=(\lambda_n +\lambda_e )\pi . \end{equation*}

Finally, LCT does work at level 0, where the upward jumps do occur according to a Poisson process:

(4.4) \begin{equation}\alpha (0)f(0)=(\lambda_n +\lambda_e )\pi .\end{equation}

We are now in a position to introduce the steady-state equations which relate the rate of downcrossings among different levels, and thus to obtain expressions for $\theta(x)$ which, via (4.2), translate into expressions for the content level density f(x). In order to ease the notation we use the abbreviations

\begin{equation*}f_n \,:\!=\, \dfrac{\lambda_n}{\lambda_n+\lambda_e}, \quad f_e \,:\!=\, \dfrac{\lambda_e}{\lambda_n+\lambda_e},\end{equation*}

which are the probabilities that a normal order comes first or that an emergency order comes first, and

\begin{equation*} J_{\sigma}(x;\,w)\,:\!=\, {\mathrm{e}}^{-\sigma [A(x)-A(w)]}.\end{equation*}

Observe that $J_{\sigma}(x;\,w)$ is the probability that an ${\rm exp}(\sigma)$ -distributed random variable exceeds the time to decrease from x to w, if no jumps occur. We shall take $\sigma = \lambda_n,\, \lambda_e,\, \lambda_n + \lambda_e$ below. In particular, if both a normal order and an emergency order are pending when the process is at level x, then $J_{\lambda_n+\lambda_e}(x;\,w)$ is the probability that level w is reached from level x before an order is delivered. In the latter case, we even write $J(x;\,w) \,:\!=\, J_{\lambda_n+\lambda_e}(x;\,w)$ . It will be convenient, in the next theorem, to observe that

\begin{equation*}{\mathrm{d}} J(a;\,w) = \dfrac{\lambda_n+\lambda_e}{\alpha(w)} \exp\biggl({-}(\lambda_n+\lambda_e)\int_w^a \frac{1}{\alpha(x)} \,{\mathrm{d}} x \biggr) \,{\mathrm{d}} w\end{equation*}

represents the probability that the first upward jump of the inventory level process, after a has been downcrossed, occurs when the level is between w and $w + {\mathrm{d}} w$ , while $J(a;\,0)$ , the atom at zero, is the probability that zero has been reached from a before an upward jump has occurred.

Proof. (a) $0\leq x <a$ . In this region the number of downcrossings of level x is at most one. Hence the mean number of downcrossings $\theta(x)$ of level $x \in [0,a)$ equals the probability that x is downcrossed, which is $J(a;\,x) = J_{\lambda_n +\lambda_e }(a;\,x)$ .

(b) $a\leq x <b$ . By Lemma 3.2, in this region exactly one order is pending; it can either be the normal order or the emergency order, but not both of them. Accordingly, to determine $\theta (x)$ we distinguish the events where the emergency delivery occurs first and the normal delivery occurs first. We shall now explain the terms of (4.6).

• The term in line 1 of (4.6). This term corresponds to the case that the emergency delivery comes first. The probability of that event is $f_e = {\lambda_e }/{(\lambda_n +\lambda_e) }$ . After the arrival of the emergency order, only the normal order is pending, and there will be no more emergency order in the rest of the cycle. Immediately after the arrival of the emergency order, $\mathbf{V}>b$ , and level $x \in [a,b)$ will be downcrossed at least once until the end of the cycle. That number of downcrossings $\mathbf{D}_x$ is geometrically distributed,

  (4.14) \begin{equation}\mathbb{P}(\mathbf{D}_x=m) = J_{\lambda_n}(x;\,a) (1 - J_{\lambda_n}(x;\,a))^{m-1}, \quad m=1,2,\dots,\end{equation}
  with mean $\mathbb{E}\mathbf{D}_x = 1/J_{\lambda_n}(x;\,a)$ . Indeed, each time when x has been downcrossed, the probability that level a will be reached (and hence the cycle will be ended) before another normal delivery arrives (in which case b will be upcrossed and later downcrossed again, and subsequently x will be downcrossed again) equals $J_{\lambda_n}(x;\,a)$ .

• The term in lines 2–4 of (4.6). The probability that the normal delivery arrives when the state is in $(w,w+{\mathrm{d}} w)$ (for $0<w<a$ ) and before the emergency delivery is $f_n \,{\mathrm{d}} J(a;\,w)$ . Immediately after the arrival of the normal delivery the state becomes $q+w$ (by definition $q+w>b$ ), and from this moment only the emergency order is pending. If that emergency delivery arrives before level x is downcrossed (which has probability $1-J_{\lambda_e}(q+w;\,x)$ ), the process jumps above b, and at the next downcrossing of b a normal order is pending. Hence one again has $1/J_{\lambda_n}(x;\,a)$ downcrossings of level x. If the emergency delivery arrives after x is downcrossed but before a is downcrossed (probability $J_{\lambda_e}(q+w;\,x) (1- J_{\lambda_e}(x;\,a))$ $=J_{\lambda_e}(q+w;\,x) - J_{\lambda_e}(q+w;\,a)$ ), then x has already been downcrossed once, and the process jumps up by q, subsequently b is downcrossed again and a normal order is pending, and we have on average $1/J_{\lambda_n}(x;\,a)$ more downcrossings of level x. If the emergency delivery does not arrive before a is reached (probability $J_{\lambda_e}(q;\,a)$ ), then we have just one downcrossing of x in the cycle.

At this stage we would like to emphasize that, with probability $J(a;\,0)$ , zero is reached before an order has arrived. Indeed, the atom at zero in the integral in (4.6) gives the contribution

\begin{equation*}f_n J(a;\,0) \biggl[(1 - J_{\lambda_e}(q;\,x))\dfrac{1}{J_{\lambda_n}(b;\,a)} + (J_{\lambda_e}(q;\,x)-J_{\lambda_e}(q;\,a))\biggl(1+ \dfrac{1}{J_{\lambda_n}(b;\,a)}\biggr) + J_{\lambda_e}(q;\,a)\biggr].\end{equation*}

These terms correspond to the case in which no delivery arrives before level 0 is reached, after which a normal delivery arrives first (probability $f_n J(a;\,0)$ ). In each of cases (c)–(f) below, where an integral $\int_{w=0}^a z(w) \,{\mathrm{d}} J(a;\,w)$ appears, there is a similar contribution $z(0) J(a;\,0)$ from the atom at 0 as when w is in the interior; similarly for $\int_{w=0}^{x-q} z(w) \,{\mathrm{d}} J(a;\,w)$ in case (d) below.

(c) $b\leq x <q$ . We shall explain all terms of (4.7) successively.

• The terms in lines 1–3 of (4.7). This concerns a case in which the emergency delivery arrives first, when the state is in $(w,w+{\mathrm{d}} w)$ (or in 0, the contribution from the atom). The process now jumps to $q+w>x$ . From this moment only the normal order is pending. We now consider the various possibilities for the arrival of a normal delivery.

1. (i) It arrives with probability $1-J_{\lambda_n}(q+w;\,x)$ before x is crossed. That leads to a jump in the process, some time later a crossing of x, and no further orders until subsequently b is reached. The mean number of crossings of x is now

   \begin{equation*}1 + \dfrac{1-J_{\lambda_n}(b;\,a)}{J_{\lambda_n}(b;\,a)} = \dfrac{1}{J_{\lambda_n}(b;\,a)}.\end{equation*}

2. (ii) It arrives with probability $J_{\lambda_n}(q+w;\,x) - J_{\lambda_n}(q+w;\,b)$ when x has already been crossed a first time, but before b is reached. After b is reached, there are on average

   \begin{equation*}\dfrac{1}{J_{\lambda_n}(b;\,a)}\end{equation*}
   more crossings of x.

3. (iii) It arrives with probability $J_{\lambda_n}(q+w;\,b)$ not before b is reached. After b is reached, there are on average

   \begin{equation*} \dfrac{1-J_{\lambda_n}(b;\,a)}{J_{\lambda_n}(b;\,a)} \end{equation*}
   more arrivals of normal deliveries and subsequent crossings of x.

• The terms in lines 4–6 of (4.7). This concerns a case in which the normal delivery arrives first, when the state is in $(w,w+{\mathrm{d}} w)$ (or, again, in 0, the contribution from the atom). Lines 4–5 are completely analogous to lines 1–2. Line 6 differs more fundamentally from line 3. We get with probability $J_{\lambda_e}(q+w;\,b)-J_{\lambda_e}(q+w;\,a)$ an emergency delivery between b and a, giving rise to, on average, $1/J_{\lambda_n}(b;\,a)$ more crossings of x. With probability $J_{\lambda_e}(q+w;\,a)$ there is no emergency delivery before a is reached and the cycle is ended; in this case there was just one crossing of x.

(d) $q \leq x < a+q$ . We successively explain all terms of the 11 lines of (4.8).

• Lines 1–3 of (4.8). These lines concern the case in which an emergency delivery occurs first, when the state is in $(w,w+ {\mathrm{d}} w)$ , for $x-q<w<a$ (so now there is no contribution from an atom at 0). Distinguish between the three possibilities in which:

1. (i) the normal delivery occurs before x is crossed,

2. (ii) the normal delivery occurs between the crossings of x and b, and

3. (iii) the normal delivery has not yet arrived when b is crossed.

These three possibilities give rise to, on average,

\begin{equation*}\frac{1}{J_{\lambda_n}(b;\,a)},\quad1+\frac{1}{J_{\lambda_n}(b;\,a)}\quad\text{and}\quad\frac{1}{J_{\lambda_n}(b;\,a)}\end{equation*}

crossings of x.

• Lines 4–5 of (4.8). These lines concern the case in which the jump is from $w \in [0,x-q)$ and hence ends up in $q+w<x$ . Now there are two possibilities:

1. (i) the normal delivery arrives before b is reached,

2. (ii) the normal delivery has not yet arrived before b is reached.

These two possibilities give rise to, on average

\begin{equation*}\frac{1}{J_{\lambda_n}(b;\,a)}\quad\text{and}\quad\frac{1-J_{\lambda_n}(b;\,a)}{J_{\lambda_n}(b;\,a)}\end{equation*}

crossings of x.

• Lines 6–9 of (4.8). The normal delivery arrives first, when the state is in $(w,w+ {\mathrm{d}} w)$ , where $x-q \leq w<a$ (lines 6–9). Now distinguish between the four possibilities in which:

1. (i) the emergency delivery arrives before x is crossed,

2. (ii) it arrives between the crossings of x and b,

3. (iii) it arrives between the crossings of b and a,

4. (iv) it does not arrive before the end of the cycle.

The first three possibilities now give rise to, on average,

\begin{equation*}\frac{1}{J_{\lambda_n}(b;\,a)},\quad1+\frac{1}{J_{\lambda_n}(b;\,a)} \quad\text{and}\quad1+\frac{1}{J_{\lambda_n}(b;\,a)}\end{equation*}

crossings of x (in fact we could have combined possibilities (ii) and (iii)). Possibility (iv)gives exactly one downcrossing of x in the whole cycle.

• Lines 10–11 of (4.8). Let $0 \leq w<x-q$ . Now the first jump ends below x, and hence all the mean numbers of crossings of x are one less then for $x-q \leq w<a$ .

(e) $a+q \leq x < b+q$ . We successively explain both lines of (4.9). For this, we first need to define the following two expectations:

• $\beta_n(x)=$ mean number of downcrossings of x in a cycle, when starting from b with only the normal order pending;

• $\beta_e(x)=$ mean number of downcrossings of x in a cycle, when starting from b with only the emergency order pending.

It is easily seen that $\beta_n(x)$ and $\beta_e(x)$ satisfy the following two equations:

\begin{align*} \beta_n(x) &= (1-J_{\lambda_n}(b;\,x-q))(1+\beta_n(x)) + (J_{\lambda_n}(b;\,x-q) - J_{\lambda_n}(b;\,a)) \beta_n(x),\\ \beta_e(x) &= (1-J_{\lambda_e}(b;\,x-q))(1+\beta_n(x)) + (J_{\lambda_e}(b;\,x-q)- J_{\lambda_e}(b;\,a)) \beta_n(x).\end{align*}

Hence, with $a \leq x-q < b$ , equations (4.10) and (4.11) follow.

• Line 1 of (4.9). After the arrival of the emergency delivery and the corresponding jump to $q+w$ , only the normal order is pending. If it arrives before b is reached, the process jumps up by q and we have one crossing of x, and subsequently the process will reach b and we have on average another $\beta_n(x)$ crossings of x. If it does not arrive before b is reached, we get on average $\beta_n(x)$ crossings of x.

• Line 2 of (4.9). Similar to line 1, but now the normal delivery is the first to arrive when the process goes down from a to 0, and only the emergency order is pending. If it arrives before b is reached, the process jumps up by q and we have one crossing of x, and subsequently the process will reach b and we have on average another $\beta_n(x)$ crossings of x. If it does not arrive before b is reached, we have on average $\beta_e(x)$ crossings of x.

(f) $b+q \leq x < 2q$ . In this range there is at most one downcrossing of x during a cycle. Hence $\theta(x)$ is also the probability that level x is downcrossed during the cycle. The first term in the right-hand side of (4.12) concerns the case in which the emergency delivery arrives first, the process jumps to level $q+w$ , and the normal delivery arrives before level $x-q$ is reached, resulting in a jump to a level above x. From here on no delivery can arrive until level b is reached, and thereafter the process can no longer get higher than $b+q$ in the cycle. The second term does the same for the case in which the normal delivery arrives first.

(g) $2q \leq x < a+2q$ . In this range there is again at most one downcrossing of x during a regenerative cycle. Hence $\theta(x)$ is again the probability that level x is downcrossed during the cycle. The two terms in the right-hand side of (4.13) are the same as the first two terms in the right-hand side of (4.12). It should be noticed that here, unlike cases (b)–(f), there is no contribution from the atom at 0: if 0 is reached in the beginning of the cycle, then the level can no longer reach 2q during this cycle.

Remark 4.1. A few terms in (4.7) can be combined, yielding a slight simplification. Observe that for $w \in [b,q)$ , the term inside the first square brackets for $\theta(x)$ in (4.7) can be compressed into

\begin{equation*}\dfrac{1}{J_{\lambda_n}(b;\,a)} + (J_{\lambda_n}(q+w;\,x) - J_{\lambda_n}(q+w;\,b)).\end{equation*}

A similar remark applies in some other formulas. We can now easily integrate that ${1}/{(J_{\lambda_n}(b;\,a))}$ term (taking the atom at 0 into account), which yields

\begin{equation*}f_e \frac{1}{J_{\lambda_n}(b;\,a)}.\end{equation*}

It is reasonably straightforward to verify that (4.3) holds, i.e. the continuity of f(x), and hence of $\theta(x)$ , in $x=a$ , b, $a+q$ , $b+q$ , 2q.

Theorem 4.1 uniquely determines $\theta(x)$ . However, it follows from (4.2) that the content level density f(x) is thus only determined up to the yet unknown constant f(a). The normalizing condition provides one additional equation, i.e. $\int_{0}^{a+2q}f(x)\,{\mathrm{d}} x=1-\pi $ , but the stationary probability $\pi $ of having zero inventory is also unknown as yet. LCT at level zero, as given in (4.4), gives us the extra equation that we need:

\begin{equation*}(\lambda_n +\lambda_e )\pi =\alpha(0)f(0),\end{equation*}

in which an expression for f(0) (or rather $f(0)/f(a)$ ) is obtained by substituting $x=0$ in (4.5) and (4.2). It should be observed that $\pi$ is also the fraction of time in which there is zero inventory, during which no demand is satisfied.

Remark 4.2. In the region $x\in (0,a)$ we can also use the traditional methodology of LCT to obtain the same solution as given in (4.5). To see this, recall that two orders are pending below level a. Since the two lead times are independent and exponentially distributed, we conclude that, below level a, the arrival process of jumps is a Poisson process with rate $\lambda_n +\lambda_e $ . LCT now gives the balance equation

\begin{equation*}\alpha (x)f(x)=(\lambda_n +\lambda_e ) F(x),\end{equation*}

where the right-hand side, with F(x) the steady-state distribution of the inventory level, gives the rate of upcrossings of level $x \in (0,a)$ . Rewriting this as

\begin{equation*}\dfrac{f(x)}{F(x)} = \dfrac{\lambda_n+\lambda_e}{\alpha(x)}\end{equation*}

and integrating gives

\begin{equation*}F(x) = K \exp\biggl((\lambda_n+\lambda_e) \int_0^x \frac{1}{\alpha(w)}\,{\mathrm{d}} w\biggr) ,\end{equation*}

with K some constant. Differentiation yields

\begin{equation*}f(x) = \dfrac{\alpha(a) f(a)}{\alpha(x)} J(a;\,x) ,\end{equation*}

in agreement with (4.5); the last equality was obtained by substituting $x=a$ .

This approach does not work for $x > a$ since there the arrival process of the jumps is not a Poisson process.

Remark 4.3. For some special choices of the state-dependent release rate function $\alpha({\cdot})$ , the expressions in Theorem 4.1 simplify considerably. We mention two special cases.

Example 1: $\alpha(x) \equiv \alpha$ . In this case $A(x) \equiv {x}/{\alpha}$ and hence

\begin{equation*}J_{\sigma}(x;\,y) = {\mathrm{e}}^{-({\sigma}/{\alpha})(x-y)}.\end{equation*}

In this case all integrals appearing in Theorem 4.1 can be easily evaluated.

Example 2: $\alpha(x) =rx$ . In this case A(x) as defined in (2.1) would diverge. However, we can still determine

\begin{equation*}J_{\sigma}(x;\,y) = \biggl(\dfrac{y}{x}\biggr)^ {{\sigma}/{r}}, \quad x>y>0,\end{equation*}

and it should be observed that $J_{\sigma}(x;\,0)=0$ (indeed, the origin can never be reached). Various terms in (4.5)–(4.13) now disappear, and others become relatively simple. These two special cases are quite natural. In the first example one might try to choose $\alpha$ such that a particular profit function is optimized. In the second example the choice of r has to do with pricing: its value reflects the demand caused by a certain price setting.

5. Rate of emergency deliveries

In this section we determine the rate $\rho_e$ of emergency deliveries. Let $\mathbf{I}$ be the number of emergency deliveries during an arbitrary regenerative cycle. We have $\rho_e = {\mathbb{E}[\mathbf{I}]}/{\mathbb{E}[\mathbf{C}]}$ . In the next theorem $\rho_e$ is obtained by determining $\mathbb{E}[\mathbf{I}]$ .

Theorem 5.1. The rate of emergency deliveries is given by

(5.1) \begin{equation}\rho_e = \alpha(a) f(a) \biggl\{f_e+ f_n \int_{0}^{a}[1-J_{\lambda_e }(q+w;\,a)]\,{\mathrm{d}} J(a;\,w)\biggr\} .\end{equation}

6. Rate of normal deliveries

Let $\rho_n$ be the rate of normal deliveries and $\mathbb{E}[\mathbf{N}]$ the mean number of normal deliveries per cycle. Then

\begin{equation*}\rho_n = \dfrac{\mathbb{E}[\mathbf{N}]}{\mathbb{E}[\mathbf{C}]} = \alpha(a) f(a) \mathbb{E}[\mathbf{N}].\end{equation*}

Below we determine $\mathbb{E}[\mathbf{N}]$ .

We shall repeatedly use the fact that (see (4.14)) the mean number of normal deliveries from the moment b is reached while a normal order is pending is given by

\begin{equation*}\mathbb{E}[\mathbf{D}_b]-1 = \frac{1-J_{\lambda_n}(b;\,a)}{J_{\lambda_n}(b;\,a)}.\end{equation*}

Theorem 6.1. The rate of normal deliveries is given by

(6.1) \begin{align}\rho_n &= \alpha(a) f(a)\biggl\{ f_e \int_{0}^{a}\biggl[(1-J_{\lambda_n }(q+w;\,b)) \dfrac{1}{J_{\lambda_n}(b;\,a)} +J_{\lambda_n }(q+w;\,b)\dfrac{1-J_{\lambda_n}(b;\,a)}{J_{\lambda_n}(b;\,a)}\biggr] \,{\mathrm{d}} J(a;\,w)\notag\\ &\quad +f_n\int_0^a\biggl[1+ (1-J_{\lambda_e }(q+w;\,a))\dfrac{1-J_{\lambda_n}(b;\,a)}{J_{\lambda_n}(b;\,a)}\biggr] \,{\mathrm{d}} J(a;\,w) \biggr\}.\end{align}

7. A model variant

In this section we briefly consider the following model variant. The only change with respect to the model of Section 2 is that now, when level b is downcrossed, a normal order is always placed when it is not already pending, even if an emergency order is already pending. As a result, it becomes less likely that level a is reached without any delivery having arrived.

Lemma 3.1 remains true in this variant: the period between two downcrossings of level a is still a regenerative cycle that starts with two pending orders. Statement (b) of Lemma 2 has to be adapted: for $\mathbf{V} \in [a,b)$ , at least one order is pending, instead of exactly one (and the normal order is always pending in this interval). Lemma 4.1, which relates the inventory level density f(x) and the mean number of downcrossings $\theta(x)$ , remains unchanged. Theorem 4.1 needs to be adapted, but only in the region $x \in [b,b+q)$ , which corresponds to the four cases (b)–(e) of the theorem. The reason is that the only difference from the model treated above occurs once level b is downcrossed. There is no change in the regions $[b+q,2q)$ and $[2q,a+2q)$ (cases (f) and (g)) because levels higher than $b+q$ cannot be reached in this model variant after level b is downcrossed. There is no change in the region [0, a) (case (a)) because the regenerative cycle starts the same in the model variant as in the original model, with both a normal order and an emergency order pending. In $[a,b+q)$ the analysis becomes slightly more complicated than in Theorem 4.1. However, all terms corresponding to cases in which the emergency order comes first in the regenerative cycle are unaltered, because a normal order now becomes pending once b is downcrossed, in both the variant and in the original model.

Below we show how formula (4.6) needs to be adapted; this concerns case (b), i.e. the region [a, b). Similar adaptations have to be made for the region $[b,b+q)$ .

Case (b) in the model variant: $a \leq x < b$ . The mean number of downcrossings of level x in the model variant equals

(7.1) \begin{align}\theta(x) &=f_e \dfrac{1}{J_{\lambda_n}(x;\,a)}\notag\\ &\quad +f_n \int_0^a \biggl[(1 - J_{\lambda_e}(q+w;\,b)) \dfrac{1}{J_{\lambda_n}(x;\,a)} + J_{\lambda_e}(q+w;\,b) g(x)\biggr] \,{\mathrm{d}} J(a;\,w),\end{align}

where g(x) is the mean number of downcrossings of level $x \in [a,b)$ in a cycle, from the moment level b is downcrossed in a cycle, given that an emergency order is pending.

The rationale behind (7.1) is as follows.

• The first term in the right-hand side is the same as the first term in (4.6): this is the case in which the emergency delivery occurs first.

• The second term reflects the case in which the normal delivery occurs at level w and before the emergency delivery. The process jumps to $q+w$ . One possibility now is that the emergency delivery comes before b is downcrossed. From then on, there are on average $1/J_{\lambda_n}(x;\,a)$ downcrossings of x in the remainder of the cycle. The second possibility is that the emergency delivery has not occurred before b is reached. From then on, one has on average g(x) downcrossings of level x.

Here g(x) satisfies the following relation:

(7.2) \begin{align}g(x) &= J_{\lambda_n+\lambda_e}(b;\,a) \notag\\ &\quad +f_e (1 - J(b;\,x)) \dfrac{1}{J_{\lambda_n}(x;\,a)}\notag\\&\quad +f_e (J(b;\,x) - J(b;\,a)) \biggl(1 + \dfrac{1}{J_{\lambda_n}(x;\,a)}\biggr)\notag\\&\quad +f_n \int_{w=x}^b \biggl[(1 - J_{\lambda_e}(q+w;\,b)) \dfrac{1}{J_{\lambda_n}(x;\,a)} + J_{\lambda_e}(q+w;\,b) g(x)\biggr] \,{\mathrm{d}} J(b;\,w)\notag\\ &\quad +f_n \int_{w=a}^x \biggl[(1-J_{\lambda_e}(q+w;\,b)) \biggl(1+ \dfrac{1}{J_{\lambda_n}(x;\,a)}\biggr) + J_{\lambda_e}(q+w;\,b) (1 + g(x))\biggr] \,{\mathrm{d}} J(b;\,w) .\end{align}

g(x) can be trivially solved from this relation; observe that one can add the two terms with g(x) in the right-hand side of (7.2) into the product of g(x) and an integral from a to b that does not involve x.

The rationale behind (7.2) is as follows. At the beginning of the remaining cycle, both a normal order and an emergency order are pending. There are now five cases, corresponding to the five lines in (7.2).

• Case 1: both orders remain pending until a is reached. That gives exactly one downcrossing of x.

• Case 2: an emergency delivery occurs first, and before x is downcrossed. The process jumps up by q and eventually reaches b again, but there will be no more emergency order. We have on average $1/J_{\lambda_n}(x;\,a)$ downcrossings of x.

• Case 3: an emergency delivery occurs first, but between levels x and a. Now x has already been downcrossed once, but thereafter the process jumps up by q, and eventually reaches b again; now see Case 2 above.

• Case 4: a normal delivery occurs first, and before x is downcrossed. The process jumps up by q. Now there are two sub-cases: either an emergency delivery occurs before b is reached, or it does not. In the former case one gets $1/J_{\lambda_n}(x;\,a)$ downcrossings of x, and in the latter case the process reaches b again and we are back in the old situation: one has on average g(x) more downcrossings of x.

• Case 5: a normal delivery occurs first, but between levels x and a. Now x has already been downcrossed once, but thereafter the process jumps up by q. Now there are again two sub-cases, just as for Case 4.

Similar adaptations have to be made in cases (c)–(e) of Theorem 4.1.