Optimizing stock levels for service-differentiated demand classes with inventory rationing and demand lead times

Abstract

In this paper, we study a service parts inventory management system for a single product at a parts distribution center serving two priority-demand classes: critical and non-critical. Distribution center keeps a common inventory pool to serve the two demand classes, and provides differentiated levels of service by means of inventory rationing. We assume a continuous review one-for-one ordering policy with backorders and Poisson demand arrivals. Only one demand class provides advance demand information whose orders are due after a deterministic demand lead time, whereas the orders of the other demand class need to be satisfied immediately. The problem has been studied before, but remained a challenging problem. The quality of the existing heuristic for estimating the critical class service levels can diminish significantly in some settings and the search routine for the service level optimization model relies on a brute force approach. Our contribution to the literature is twofold. For the given class of inventory replenishment and allocation policies, first we determine the form of the optimal solution to the service level optimization model, and then we derive an exact optimization routine to determine the optimal policy parameters provided the steady-state distribution is available. The computation of steady-state probabilities is needed only once. Second, we propose an alternative approach to estimate steady-state probabilities. By analyzing the limiting behavior of transition probabilities during infinitesimal time intervals, we are able to characterize the relationships between the steady-state probabilities, which satisfy nicely formed balance equations under the so-called Independence Assumption. In the numerical study section, we show that our approach provides superior performance in estimating service levels than the existing heuristic for all the examples considered. We also compare the performance of using the critical class service levels computed according to our method against the service levels computed by the existing heuristic, and show that our method can provide inventory savings up to 16.67%.

Appendices

Appendix 1: Proof of Proposition 1

Since a one-for-one policy is implemented, whenever a demand of any type occurs, it triggers a replenishment order of size one, and therefore R is incremented by one. If there is a delivery, number of units in the resupply is decremented by one. If it is a due date of the non-critical order, then R is unaffected and hence remains the same.

The dynamics for $Y^{n}$ are also straightforward. Whenever a non-critical demand occurs, $Y^{n}$ is incremented by one. On the other hand, $Y^{n}$ is decremented by one only if it is a due date of the non-critical order. For all other cases, $Y^{n}$ is unaffected.

For the non-critical backorders, the only situation in which $B^{n}$ can be decremented is with the arrival of a delivery ($E_{m}={\text{``}}v{\text{''}}$) when on-hand inventory prior to the delivery is $S_{c}$ and there is at least one non-critical backorder. If on hand inventory equals $S_{c},$ then, by (1), $S=S_{c}+R_{m-1}-B^{n}_{m-1}-Y^{n}_{m-1}$. From this equation, by rearranging the terms we have $B^{n}_{m-1}=R_{m-1}-Y^{n}_{m-1}-(S-S_{c})$. Since the number of non-critical backorders should be at least one, we also have $R_{m-1}-Y^{n}_{m-1}>S-S_{g}$. On the other hand, when a due date of a non-critical order comes ($E_{m}={\text{``}}y{\text{''}}$) and on-hand inventory just prior to the due date is less than or equal to $S_{c}$, then $B^{n}$ is incremented by one. On-hand inventory in this case is given by $\left[ S-R_{m-1}+B^{n}_{m-1}+Y^{n}_{m-1}\right] ^{+}\le S_{c}$, due to (2); which is also equivalent to stating $S-R_{m-1}+B^{n}_{m-1}+Y^{n}_{m-1}\le S_{c}$. Rearranging the terms, we have $R_{m-1}-Y^{n}_{m-1}\ge S-S_{c}+B^{n}_{m-1}$. But $B^{n}_{m-1}$ is a non-negative variable. Hence we should have $R_{m-1}-Y^{n}_{m-1}\ge S-S_{c}$. $\square$

Appendix 2: Proof of Proposition 2

The result $\beta ^{n}(S,S_{c})\ge \beta ^{n}(S,S^{\,\prime }_{c})$ is immediately follows from (11).

To prove $\beta _{c}(S,S_{c})\ \le \beta _{c}(S,S^{\,\prime }_{c})$, let us consider two systems with identical event sequences $\left\{ (m,T_{m},E_{m}); m=1,2,3,\ldots \right\} .$ In the first system, the policy parameters are $(S,S_{c})$ and the resulting states are given by $\left\{ (R_{m},B^{n}_{m},Y^{n}_{m}); m=1,2,3,\ldots \right\} .$ In the second system, the policy parameters are $(S,S_{c}^{\prime })$ with $S_{c}^{\prime }>S_{c}$ and the resulting states are given by $\left\{ (R_{m}^{\prime },B^{n\prime }_{m},Y^{n\prime }_{m}); m=1,2,3,\ldots \right\} .$ We conjecture that $R_{m}^{\prime }=R_{m},\, Y^{n\prime }_{m}=Y^{n}_{m}$ and $B^{n\prime }_{m}\ge B^{n}_{m}$ for all m. With reference to Proposition 1, we can immediately establish $R_{m}^{\prime }=R_{m}$ and $Y^{n\prime }_{m}=Y^{n}_{m}$ for all m. We will prove $B^{n\prime }_{m}\ge B^{n}_{m}$ by induction. For $m=1,\ldots ,S-S_{c}$, it is clear that $B^{n\prime }_{m}= B^{n}_{m}=0$. For induction, we first assume that $B^{n\prime }_{m}\ge B^{n}_{m}$ is true for some m, and then validate the result for $m+1$. Since $(S-1,S)$ policy is followed, the number of backorders can change at most by one unit per event. Therefore, if $B^{n\prime }_{m}> B^{n}_{m}$, $B^{n\prime }_{m+1}\ge B^{n}_{m+1}$ is immediate. Else, $B^{n\prime }_{m}= B^{n}_{m}$ and it is left to prove $B^{n\prime }_{m+1}\ge B^{n}_{m+1}$. It suffices to show that $B^{n\prime }_{m+1}< B^{n}_{m+1}$ is not possible. First, let us suppose $B^{n}_{m+1}=B^{n}_{m}+1$. This can only happen if $R_{m}-Y^{n}_{m}\ge S-S_{c}$ and $E_{m+1}={\text{``}}y$” (the non-critical due date comes). Since $S_{c}^{\prime }>S_{c},$ we have $R_{m}^{\prime }-Y^{n\prime }_{m}=R_{m}-Y^{n}_{m}\ge S-S_{c}^{\prime }$. Since $E_{m+1}={\text{``}}y$”, we will have $B^{n\prime }_{m+1}=B^{n\prime }_{m}+1=B^{n}_{m}+1=B^{n}_{m+1}$. Now let us suppose $B^{n\prime }_{m+1}=B^{n\prime }_{m}-1$. This is possible only if $B^{n\prime }_{m}=R_{m}^{\prime }-Y^{n\prime }_{m}-(S-S_{c}^{\prime })$ and $E_{m+1}={\text{``}}v$” (a delivery occurs). But this would imply $B^{n}_{m}=B^{n\prime }_{m}=R_{m}^{\prime }-Y^{n\prime }_{m}-(S-S_{c}^{\prime })>R_{m}-Y^{n}_{m}-(S-S_{c})$. However, the condition $B^{n}_{m}>R_{m}-Y^{n}_{m}-(S-S_{c})$ is not possible due to (5). Therefore we should have $B^{n\prime }_{m+1}\ge B^{n}_{m+1}$. Proof by induction is completed. Furthermore, conditioned on $B^{n\prime }_{m}\ge B^{n}_{m}$, by (2), at any point in time on-hand inventory in the second system will always be greater than or equal to the on-hand inventory in the second system. Hence, the critical class fill rate for the second system must be at least as high as for the first system. $\square$

Appendix 3: Proof of Proposition 3

Let us consider two cases with $(S,S_{c})$ and $(S^{\prime },S_{c}^{\prime })$ such that $\Delta =S-S_{c}=S^{\prime }-S_{c}^{\prime }$. For any given sample path $\left\{ (m,T_{m},E_{m});m=1,2,3,\ldots \right\}$, $(R_{m},B^{n}_{m},Y^{n}_{m})$ will always be identical to $(R_{m}^{\prime },B^{n\prime }_{m},Y^{n\prime }_{m})$ for an arbitrary m. This is because, due to Proposition 1 the sample path dynamics depend only on $\Delta ,$ the difference between the target inventory S and the threshold level $S_{c}$. Therefore, at any point t in time, system state $\xi _{t}$ will be identical to $\xi _{t}^{\prime }$. Since this is also true for all sample paths, we have $\pi _{(r,b^{n},y^{n})}(S,S_{c})=\pi _{(r,b^{n},y^{n})}(S^{\prime },S_{c}^{\prime }),\left( r,b^{n},y^{n}\right) \in Z_{0}\times Z_{0}\times Z_{0}$. $\square$

Appendix 4: Proof of Lemma 1

By definition,

$$\begin{aligned} \psi _{u}\left( \Delta +k, k\right)= & \sum _{\begin{array}{c} (r,b^{n},y^{n})\in {\mathbb {F}}_{(\Delta +k, k)} \\ \left( \Delta +k-r+b^{n}+y^{n}\right) ^{+}=0 \\ \left( r-b^{n}-y^{n}-\Delta -k \right) ^{+}=u \end{array}}\pi _{(r,b^{n},y^{n})}(\Delta +k, k) \\= & \sum _{\begin{array}{c} (r,b^{n},y^{n})\in {\mathbb {F}}_{(\Delta , 0)} \\ \left( \Delta +k-r+b^{n}+y^{n}\right) ^{+}=0 \\ \left( r-b^{n}-y^{n}-\Delta -k \right) ^{+}=u \end{array}}\pi _{(r,b^{n},y^{n})}(\Delta , 0) {\text { (by Proposition }}~3), \\= & \sum _{\begin{array}{c} (r,b^{n},y^{n})\in {\mathbb {F}}_{(\Delta , 0)} \\ \left( \Delta -r+b^{n}+y^{n}\right) ^{+}=0 \\ \left( r-b^{n}-y^{n}-\Delta \right) ^{+}=u+k \end{array}}\pi _{(r,b^{n},y^{n})}(\Delta , 0) \\= &\, \psi _{u+k}\left( \Delta , 0\right) . \end{aligned}$$

$\square$

Appendix 5: Proof of Lemma 2

By definition,

$$\begin{aligned} \beta ^{c}(\Delta +k, k)= & \, 1-\varphi _{0}(\Delta +k, k ) {\text { (with reference to Eqs.}} (7) {\text { and }} (8))\\= &\, 1- \sum _{u=0}^{\infty }\psi _{u}(\Delta +k, k ) \\= & \, 1- \sum _{u=0}^{\infty }\psi _{u+k}(\Delta , 0 ) {\text { (by Lemma }}~1),\\= &\, 1- \sum _{u=0}^{\infty }\psi _{u}(\Delta , 0 ) \,+\, \sum _{u=0}^{k-1}\psi _{u}(\Delta , 0)\\= &\, 1-\varphi _{0}(\Delta , 0) \,+\, \sum _{u=0}^{k-1}\psi _{u}(\Delta , 0)\\= &\, \beta ^{c}(\Delta , 0)\,+\, \sum _{u=0}^{k-1}\psi _{u}(\Delta , 0). \end{aligned}$$

$\square$

Appendix 6: Proof of Theorem 2

Starting from the initial state (0, 0, 0) at time $t=0$, let

$$\begin{aligned} P_{({\bar{r}},{\bar{b}}^{n},{\bar{y}}^{n}),(r,b^{n},y^{n})}(t,t')= &\, P\left[ \xi _{t'}=(r,b^{n},y^{n})\, | \, \xi _{0}=(0,0,0),\,\xi _{t}=({\bar{r}},{\bar{b}}^{n},{\bar{y}}^{n})\right] . \end{aligned}$$

One-step transition probabilities will be solved for a general system state $(r,b^{n},y^{n})\in {\mathbb {F}}_{(S,S_{c})}$ with $b^{n}\ge 1$, $y^{n}\ge 1$, and $r-b^{n}-y^{n}>S-S_{c}$ (hence, $OH<S_{c}$). Other system states can be solved similarly. By conditioning on the state of the system at time t, there are five possible ways to reach state $(r,b^{n},y^{n})$ in at most one transition over the next infinitesimal $\tau$ time units: a non-critical demand occurs, a critical demand occurs, a delivery is received from the resupply, a due date of a non-critical order comes, or nothing happens. Probabilities of two or more events happening during $(t,t+\tau ]$ are captured within the term $o(\tau )$.

(a) A non-critical demand occurs:

$$\begin{aligned}&P_{(r-1,b^{n},y^{n}-1),(r,b^{n},y^{n})}(t,t+\tau )\nonumber \\&\quad = P\left[ {\text {only a }} \textit{non-critical} {\text { demand occurs during}} \, (t,t+\tau ];\,\, {\text {all }} \textit{r}-1 {\text { units in }} \right. \nonumber \\&\qquad \left. {\text {the resupply at time }} t {\text { are still in the resupply at time }} t+\tau ; \right. \nonumber \\&\qquad \left. \,\, {\text {all }} y^{n}-1 {\text { orders that are ``not yet due'' at time }} t\right. \nonumber \\&\qquad \left. \, {\text {are still ``not yet due'' at time }} t+\tau \,| \, \xi _t=(r-1,b^{n},y^{n}-1),\, \right. \nonumber \\&\qquad \left. \xi _0=(0,0,0)\right] \,\,+\,o(\tau ). \end{aligned}$$

(29)

By the Independence Assumption and using (22) and (25), right-hand side of (29) becomes

$$\begin{aligned}&P_{(r-1,b^{n},y^{n}-1),(r,b^{n},y^{n})}(t,t+\tau )\nonumber \\&\quad = \lambda ^{n}\tau e^{-\lambda \tau } \ q_{t,\tau }(r-y^{n}\,|\,r-y^{n}) \, \, {\tilde{q}}_{t,\tau }(y^{n}-1\,|\,y^{n}-1) \, + \,o(\tau ) \nonumber \\&\quad = \lambda ^{n}\tau e^{-\lambda \tau } \, \left( {\begin{array}{c}r-y^{n}\\ r-y^{n}\end{array}}\right) \, p(\tau )^{r-y^{n}} \, \left[ 1-p(\tau )\right] ^{0} \nonumber \\&\qquad . \, \left( {\begin{array}{c}y^{n}-1\\ y^{n}-1\end{array}}\right) \, {\tilde{p}}(\tau )^{y^{n}-1} \, \left[ 1-{\tilde{p}}(\tau )\right] ^{0} \, + \,o(\tau ) \nonumber \\&\quad = \lambda ^{n}\tau e^{-\lambda \tau } \, p(\tau )^{r-y^{n}} \, {\tilde{p}}(\tau )^{y^{n}-1} \,+ \, o(\tau ) . \end{aligned}$$

(30)

Note that we used $\ q_{t,\tau }(r-y^{n}\,|\,r-y^{n})$ rather than $\ q_{t,\tau }(r-1\,|\,r-1)$. This is because as discussed earlier, conditioned on being at state $\xi _t=(r,b^{n},y^{n})$, delivery process over the next $\tau$ time units is determined by the elements of the set ${\mathbb {U}}(t){\setminus} {\mathbb {V}}(t)$.

(b) A critical demand occurs:

$$\begin{aligned}&P_{(r-1,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )\nonumber \\&\quad = P\left[ {\text {only a }} {\textit{critical}} {\text { demand occurs during}} \, (t,t+\tau ];\,\, {\text {all }} {\textit{r}}-1 {\text { units in }} \right. \nonumber \\&\qquad \left. {\text {the resupply at time }} t {\text { are still in the resupply at time }} t+\tau ; \right. \nonumber \\&\qquad \left. \,\, {\text {all }} y^{n} {\text { orders that are ``not yet due'' at time }} t\right. \nonumber \\&\qquad \left. \, {\text {are still ``not yet due'' at time }} t+\tau \,| \, \xi _t=(r-1,b^{n},y^{n}),\, \right. \nonumber \\&\qquad \left. \xi _0=(0,0,0)\right] \,\,+\,o(\tau ). \end{aligned}$$

(31)

By the Independence Assumption and using (22) and (25), right-hand side of (31) becomes

$$\begin{aligned}&P_{(r-1,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )\nonumber \\&\quad = \lambda ^{c}\tau e^{-\lambda \tau } \ q_{t,\tau }(r-y^{n}-1\,|\,r-y^{n}-1) \, \, {\tilde{q}}_{t,\tau }(y^{n}\,|\,y^{n}) \, + \,o(\tau ) \nonumber \\&\quad = \lambda ^{c}\tau e^{-\lambda \tau } \, \left( {\begin{array}{c}r-y^{n}-1\\ r-y^{n}-1\end{array}}\right) \, p(\tau )^{r-y^{n}-1} \, \left[ 1-p(\tau )\right] ^{0} \nonumber \\&\qquad . \, \left( {\begin{array}{c}y^{n}\\ y^{n}\end{array}}\right) \, {\tilde{p}}(\tau )^{y^{n}} \, \left[ 1-{\tilde{p}}(\tau )\right] ^{0} \, + \,o(\tau ) \nonumber \\&\quad = \lambda ^{c}\tau e^{-\lambda \tau } \, p(\tau )^{r-y^{n}-1} \, {\tilde{p}}(\tau )^{y^{n}} \,+ \, o(\tau ) . \end{aligned}$$

(32)

(c) The delivery from the resupply:

$$\begin{aligned}&P_{(r+1,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )\nonumber \\&\quad = P\left[ {\text {no demand occurs during}} \, (t,t+\tau ];\,\, {\text {among the }} r+1 {\text { units in }} \right. \nonumber \\&\qquad \left. {\text {the resupply at time }} t, {\text { only one of them is received during}} \right. \nonumber \\&\qquad \left. \, (t,t+\tau ] ;\, {\text {all }} y^{n} {\text { orders that are ``not yet due'' at time }} t \right. \nonumber \\&\qquad \left. \, {\text {are still ``not yet due'' at time }} t+\tau \,| \, \xi _t=(r+1,b^{n},y^{n}),\, \right. \nonumber \\&\qquad \left. \xi _0=(0,0,0)\right] \,\,+\,o(\tau ). \end{aligned}$$

(33)

By the Independence Assumption and using (22) and (25), right-hand side of (33) becomes

$$\begin{aligned}&P_{(r+1,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )\nonumber \\&\quad = e^{-(\lambda ^{n}+\lambda ^{c})\tau } \ q_{t,\tau }(r-y^{n}\,|\,r-y^{n}+1) \, \, {\tilde{q}}_{t,\tau }(y^{n}\,|\,y^{n}) \, + \,o(\tau ) \nonumber \\&\quad = e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, \left( {\begin{array}{c}r-y^{n}+1\\ r-y^{n}\end{array}}\right) \, p(\tau )^{r-y^{n}} \, \left[ 1-p(\tau )\right] \nonumber \\&\qquad . \, \left( {\begin{array}{c}y^{n}\\ y^{n}\end{array}}\right) \, {\tilde{p}}(\tau )^{y^{n}} \, \left[ 1-{\tilde{p}}(\tau )\right] ^{0} \, + \,o(\tau ) \nonumber \\&\quad = e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, (r-y^{n}+1) p(\tau )^{r-y^{n}} \left[ 1-p(\tau )\right] \, {\tilde{p}}(\tau )^{y^{n}} \,+ \, o(\tau ) . \end{aligned}$$

(34)

Note that $b^{n}$ has not changed because $OH(t)<S_{c}$ and therefore none of the existing non-critical backorders are cleared, if any.

(d) The non-critical order is due:

$$\begin{aligned}&P_{(r,b^{n}-1,y^{n}+1),(r,b^{n},y^{n})}(t,t+\tau )\nonumber \\&\quad = P\left[ {\text {no demand occurs during}} \, (t,t+\tau ];\,\, {\text {all }} {\textit{r}} {\text { units in the }} \right. \nonumber \\&\qquad \left. {\text {resupply at time }} t {\text { are still in the resupply at time }} t+\tau ; \right. \nonumber \\&\qquad \left. \, {\text {among the }} y^{n}+1 {\text { orders that are ``not yet due'' at time }} t, \right. \nonumber \\&\qquad \left. \, y^{n} {\text { of them are still ``not yet due'' at }} t+\tau \, \right. \nonumber \\&\qquad \left. \,| \, \xi _t=(r,b^{n}-1,y^{n}+1),\,\xi _0=(0,0,0)\right] \,\,+\,o(\tau ). \end{aligned}$$

(35)

By the Independence Assumption and using (22) and (25), right-hand side of (35) becomes

$$\begin{aligned}&P_{(r,b^{n}-1,y^{n}+1),(r,b^{n},y^{n})}(t,t+\tau )\nonumber \\&\quad = e^{-(\lambda ^{n}+\lambda ^{c})\tau } \ q_{t,\tau }(r-y^{n}-1\,|\,r-y^{n}-1) \, \, {\tilde{q}}_{t,\tau }(y^{n}\,|\,y^{n}+1) \, + \,o(\tau ) \nonumber \\&\quad = e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, \left( {\begin{array}{c}r-y^{n}-1\\ r-y^{n}-1\end{array}}\right) \, p(\tau )^{r-y^{n}-1} \, \left[ 1-p(\tau )\right] ^{0} \nonumber \\&\qquad . \, \left( {\begin{array}{c}y^{n}+1\\ y^{n}\end{array}}\right) \, {\tilde{p}}(\tau )^{y^{n}} \, \left[ 1-{\tilde{p}}(\tau )\right] ^{1} \, + \,o(\tau ) \nonumber \\&\quad = e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, p(\tau )^{r-y^{n}-1} \, (y^{n}+1) {\tilde{p}}(\tau )^{y^{n}} \,\left[ 1-{\tilde{p}}(\tau )\right] + \, o(\tau ) . \end{aligned}$$

(36)

(e) Nothing happens:

$$\begin{aligned}&P_{(r,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )\nonumber \\&\quad = P\left[ {\text {no demand occurs during}} \, (t,t+\tau ];\,\, {\text {all }} {\textit{r}} {\text { units in }} \right. \nonumber \\&\qquad \left. {\text {the resupply at time }} t {\text { are still in the resupply at }} t+\tau \right. \nonumber \\&\left. \qquad \, {\text {units}};\, {\text {all }} y^{n} {\text { orders that are ``not yet due'' at time }} t \right. \nonumber \\&\qquad \left. \, {\text {are still ``not yet due'' at time }} t+\tau \,| \, \xi _t=(r,b^{n},y^{n}),\, \right. \nonumber \\&\qquad \left. \xi _0=(0,0,0)\right] \,\,+\,o(\tau ). \end{aligned}$$

(37)

By the Independence Assumption and using (22) and (25), right-hand side of (37) becomes

$$\begin{aligned} P_{(r,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )= & \, e^{-(\lambda ^{n}+\lambda ^{c})\tau } \ q_{t,\tau }(r-y^{n}\,|\,r-y^{n}) \, \, {\tilde{q}}_{t,\tau }(y^{n}\,|\,y^{n}) \, + \,o(\tau ) \nonumber \\= & \, e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, \left( {\begin{array}{c}r-y^{n}\\ r-y^{n}\end{array}}\right) \, p(\tau )^{r-y^{n}} \, \left[ 1-p(\tau )\right] ^{0} \nonumber \\&\quad . \, \left( {\begin{array}{c}y^{n}\\ y^{n}\end{array}}\right) \, {\tilde{p}}(\tau )^{y^{n}} \, \left[ 1-{\tilde{p}}(\tau )\right] ^{0} \, + \,o(\tau ) \nonumber \\= & \, e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, p(\tau )^{r-y^{n}} \, {\tilde{p}}(\tau )^{y^{n}} \,+ \, o(\tau ). \end{aligned}$$

(38)

By conditioning on the state of the system at time t:

$$\begin{aligned} P_{(0,0,0),(r,b^{n},y^{n})}(0,t')\,=\, \sum _{\begin{array}{c} ({\bar{r}},{\bar{b}}^{n},{\bar{y}}^{n})\in {\mathbb {F}}_{(S,S_{c})} \end{array}}P_{(0,0,0),({\bar{r}},{\bar{b}}^{n},{\bar{y}}^{n})}(0,t) \cdot P_{({\bar{r}},{\bar{b}}^{n},{\bar{y}}^{n}),(r,b^{n},y^{n})}(t,t'). \end{aligned}$$

(39)

Then using (39), the probability of being at system state $(r,b^{n},y^{n})$ at time $t'=t+\tau$ can be written as

$$\begin{aligned}&P_{(0,0,0),(r,b^{n},y^{n})}(0,t+\tau )\nonumber \\&\quad = P_{(0,0,0),(r-1,b^{n},y^{n}-1)}(0,t)\cdot P_{(r-1,b^{n},y^{n}-1),(r,b^{n},y^{n})}(t,t+\tau ) \nonumber \,\\&\qquad + \, P_{(0,0,0),(r-1,b^{n},y^{n})}(0,t)\cdot P_{(r-1,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau ) \nonumber \\&\qquad + \, P_{(0,0,0),(r+1,b^{n},y^{n})}(0,t)\cdot P_{(r+1,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau ) \, \nonumber \\&\qquad + \, P_{(0,0,0),(r,b^{n}-1,y^{n}+1)}(0,t)\cdot P_{(r,b^{n}-1,y^{n}+1),(r,b^{n},y^{n})}(t,t+\tau ) \nonumber \\&\qquad + \, P_{(0,0,0),(r,b^{n},y^{n})}(0,t)\cdot P_{(r,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau ) \nonumber \\&\qquad + \, o(\tau ). \end{aligned}$$

(40)

Under the Independence Assumption, each of the one-step transition probabilities on the right-hand side of (40) can be determined using the results in (30), (32), (34), (36), and (38).

Subtracting $P_{(0,0,0),(r,b^{n},y^{n})}(0,t)$ from both sides and taking the limits as $\tau \rightarrow 0$:

$$\begin{aligned}&\lim _{\tau \rightarrow 0} \frac{P_{(0,0,0),(r,b^{n},y^{n})}(0,t+\tau )-P_{(0,0,0),(r,b^{n},y^{n})}(0,t)}{\tau }\nonumber \\&\quad = P_{(0,0,0),(r-1,b^{n},y^{n}-1)}(0,t)\cdot \lim _{\tau \rightarrow 0} \frac{P_{(r-1,b^{n},y^{n}-1),(r,b^{n},y^{n})}(t,t+\tau )}{\tau } \nonumber \\&\qquad + P_{(0,0,0),(r-1,b^{n},y^{n})}(0,t)\cdot \lim _{\tau \rightarrow 0} \frac{P_{(r-1,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )}{\tau } \nonumber \\&\qquad + \, P_{(0,0,0),(r+1,b^{n},y^{n})}(0,t)\cdot \lim _{\tau \rightarrow 0} \frac{P_{(r+1,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )}{\tau } \, \nonumber \\&\qquad + \, P_{(0,0,0),(r,b^{n}-1,y^{n}+1)}(0,t)\cdot \lim _{\tau \rightarrow 0} \frac{P_{(r,b^{n}-1,y^{n}+1),(r,b^{n},y^{n})}(t,t+\tau )}{\tau } \nonumber \\&\qquad - \, P_{(0,0,0),(r,b^{n},y^{n})}(0,t) \cdot \lim _{\tau \rightarrow 0} \frac{\left( 1-P_{(r,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )\right) }{\tau } \nonumber \\&\qquad + \lim _{\tau \rightarrow 0} \,\frac{ o(\tau )}{\tau }. \end{aligned}$$

(41)

The left-hand side of Eq. (41) is

$$\begin{aligned} \lim _{\tau \rightarrow 0} \frac{P_{(0,0,0),(r,b^{n},y^{n})}(0,t+\tau )-P_{(0,0,0),(r,b^{n},y^{n})}(0,t)}{\tau } = P^{\prime }_{(0,0,0),(r,b^{n},y^{n})}(0,t). \end{aligned}$$

(42)

To determine the right-hand side, first we need to determine the limits as $\tau \rightarrow 0$.

Limits as $\tau \rightarrow 0$:

(a) Using (30),

$$\begin{aligned} \lim _{\tau \rightarrow 0} \frac{P_{(r-1,b^{n},y^{n}-1),(r,b^{n},y^{n})}(t,t+\tau )}{\tau }= & \lim _{\tau \rightarrow 0} \frac{\lambda ^{n}\tau e^{-\lambda \tau } \, p(\tau )^{r-y^{n}} \, {\tilde{p}}(\tau )^{y^{n}-1}}{\tau } \,+ \, \lim _{\tau \rightarrow 0} \frac{o(\tau )}{\tau }, \end{aligned}$$

substituting values of $p(\tau )$ from (21) and ${\tilde{p}}(\tau )$ from (24),

$$\begin{aligned}= &\, \lim _{\tau \rightarrow 0} \frac{\lambda ^{n}\tau e^{-\lambda \tau } \, \left[ 1 - \frac{(\lambda ^{n}+\lambda ^{c})\tau }{\lambda ^{c}L+\lambda ^{n}(L-H)}\right] ^{r-y^{n}} \, \left[ 1 - \frac{\tau }{H}\right] ^{y^{n}-1}}{\tau } \nonumber \\= &\, \lambda ^{n}. \end{aligned}$$

(43)

(b) Using (32),

$$\begin{aligned} \lim _{\tau \rightarrow 0} \frac{P_{(r-1,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )}{\tau }= &\, \lim _{\tau \rightarrow 0} \frac{\lambda ^{c}\tau e^{-\lambda \tau } \, p(\tau )^{r-y^{n}-1} \, {\tilde{p}}(\tau )^{y^{n}}}{\tau } \,+ \, \lim _{\tau \rightarrow 0} \frac{o(\tau )}{\tau }, \end{aligned}$$

substituting values of $p(\tau )$ from (21) and ${\tilde{p}}(\tau )$ from (24),

$$\begin{aligned}= &\, \lim _{\tau \rightarrow 0} \frac{\lambda ^{c}\tau e^{-\lambda \tau } \, \left[ 1 - \frac{(\lambda ^{n}+\lambda ^{c})\tau }{\lambda ^{c}L+\lambda ^{n}(L-H)}\right] ^{r-y^{n}-1} \, \left[ 1 - \frac{\tau }{H}\right] ^{y^{n}}}{\tau } \nonumber \\= &\, \lambda ^{c}. \end{aligned}$$

(44)

(c) Using (34),

$$\begin{aligned}&\lim _{\tau \rightarrow 0} \frac{P_{(r+1,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )}{\tau }\\&\quad = \lim _{\tau \rightarrow 0} \frac{e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, (r-y^{n}+1) p(\tau )^{r-y^{n}} [1-p(\tau )]\, {\tilde{p}}(\tau )^{y^{n}}}{\tau } \,+ \, \lim _{\tau \rightarrow 0} \frac{o(\tau )}{\tau }, \end{aligned}$$

substituting values of $p(\tau )$ from (21) and ${\tilde{p}}(\tau )$ from (24),

$$\begin{aligned}= &\, \lim _{\tau \rightarrow 0} \frac{e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, (r-y^{n}+1) \left[ 1 - \frac{(\lambda ^{n}+\lambda ^{c})\tau }{\lambda ^{c}L+\lambda ^{n}(L-H)}\right] ^{r-y^{n}} \, \left[ \frac{(\lambda ^{n}+\lambda ^{c})\tau }{\lambda ^{c}L+\lambda ^{n}(L-H)}\right] \left[ 1 - \frac{\tau }{H}\right] ^{y^{n}}}{\tau }, \end{aligned}$$

replacing the corresponding terms with $\mu$ from (27) and $\vartheta$ from (28),

$$\begin{aligned}= &\, \lim _{\tau \rightarrow 0} \frac{e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, (r-y^{n}+1) \left[ 1 - \mu \tau \right] ^{r-y^{n}} \mu \tau \, \left[ 1 - \vartheta \tau \right] ^{y^{n}}}{\tau } \nonumber \\= &\, (r-y^{n}+1) \mu \lim _{\tau \rightarrow 0} e^{-(\lambda ^{n}+\lambda ^{c})\tau } \lim _{\tau \rightarrow 0} \left[ 1 - \mu \tau \right] ^{r-y^{n}} \lim _{\tau \rightarrow 0} \left[ 1 - \vartheta \tau \right] ^{y^{n}} \nonumber \\= &\, (r-y^{n}+1) \mu . \end{aligned}$$

(45)

(d) Using (36),

$$\begin{aligned}&\lim _{\tau \rightarrow 0} \frac{P_{(r,b^{n}-1,y^{n}+1),(r,b^{n},y^{n})}(t,t+\tau )}{\tau } \\&\quad = \lim _{\tau \rightarrow 0} \frac{e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, p(\tau )^{r-y^{n}-1} \, (y^{n}+1){\tilde{p}}(\tau )^{y^{n}}\left[ 1-{\tilde{p}}(\tau )\right] }{\tau } \,+ \, \lim _{\tau \rightarrow 0} \frac{o(\tau )}{\tau }, \end{aligned}$$

substituting values of $p(\tau )$ from (21) and ${\tilde{p}}(\tau )$ from (24),

$$\begin{aligned}= &\, \lim _{\tau \rightarrow 0} \frac{e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, \left[ 1 - \frac{(\lambda ^{n}+\lambda ^{c})\tau }{\lambda ^{c}L+\lambda ^{n}(L-H)}\right] ^{r-y^{n}-1} \, (y^{n}+1)\left[ 1 - \frac{\tau }{H}\right] ^{y^{n}}(\frac{\tau }{H})}{\tau }, \end{aligned}$$

replacing the corresponding terms with $\mu$ from (27) and $\vartheta$ from (28),

$$\begin{aligned}= &\, \lim _{\tau \rightarrow 0} \frac{e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, \left[ 1 - \mu \tau \right] ^{r-y^{n}-1} \, (y^{n}+1)\left[ 1 - \vartheta \tau \right] ^{y^{n}}\vartheta \tau }{\tau } \nonumber \\= &\, (y^{n}+1)\vartheta \lim _{\tau \rightarrow 0} e^{-(\lambda ^{n}+\lambda ^{c})\tau } \lim _{\tau \rightarrow 0} \left[ 1 - \mu \tau \right] ^{r-y^{n}-1} \lim _{\tau \rightarrow 0} \left[ 1 - \vartheta \tau \right] ^{y^{n}} \nonumber \\= &\, (y^{n}+1) \vartheta . \end{aligned}$$

(46)

(e) Using (38),

$$\begin{aligned}&\lim _{\tau \rightarrow 0} \frac{\left( 1-P_{(r,b^{n},y^{n}),(r,b^{n},y^{n})}(t,t+\tau )\right) }{\tau }\\&\quad = \lim _{\tau \rightarrow 0} \frac{1-e^{-(\lambda ^{n}+\lambda ^{c})\tau } \, p(\tau )^{r-y^{n}} \, {\tilde{p}}(\tau )^{y^{n}} }{\tau } \,+ \, \lim _{\tau \rightarrow 0} \frac{o(\tau )}{\tau }, \end{aligned}$$

substituting values of $p(\tau )$ from (21) and ${\tilde{p}}(\tau )$ from (24),

$$\begin{aligned}= &\, \lim _{\tau \rightarrow 0} \frac{1-e^{-(\lambda ^{n}+\lambda ^{c})\tau } \,\left[ 1 - \frac{(\lambda ^{n}+\lambda ^{c})\tau }{\lambda ^{c}L+\lambda ^{n}(L-H)}\right] ^{r-y^{n}} \, \left[ 1 - \frac{\tau }{H}\right] ^{y^{n}} }{\tau }, \end{aligned}$$

replacing the corresponding terms with $\mu$ from (27) and $\vartheta$ from (28),

$$\begin{aligned}= &\, \lim _{\tau \rightarrow 0} \frac{1-e^{-(\lambda ^{n}+\lambda ^{c})\tau } \,\left[ 1 - \mu \tau \right] ^{r-y^{n}} \, \left[ 1 - \vartheta \tau \right] ^{y^{n}} }{\tau }. \end{aligned}$$

Since there is no point of discontinuity, we can apply the L’Hôpital’s rule. Then the right-hand side becomes

$$\begin{aligned}= &\, \lim _{\tau \rightarrow 0} (\lambda ^{n}+\lambda ^{c}) e^{-(\lambda ^{n}+\lambda ^{c})\tau } \,\left[ 1 - \mu \tau \right] ^{r-y^{n}} \, \left[ 1 - \vartheta \tau \right] ^{y^{n}} \, \nonumber \\&+ \, \lim _{\tau \rightarrow 0} -e^{-(\lambda ^{n}+\lambda ^{c})\tau } \left\{ (r-y^{n})\left[ 1 - \mu \tau \right] ^{r-y^{n}-1} (-\mu ) \left[ 1 - \vartheta \tau \right] ^{y^{n}} \right. \nonumber \\&\left. + \, (y^{n}) \left[ 1 - \vartheta \tau \right] ^{y^{n}-1} (-\vartheta ) \left[ 1 - \mu \tau \right] ^{r-y^{n}} \right\} \nonumber \\= &\, (\lambda ^{n}+\lambda ^{c}) \, + \, (r-y^{n})\mu \, + \, y^{n}\vartheta . \end{aligned}$$

(47)

Plugging the values of these limits into Eq. (41), we have

$$\begin{aligned} P^{\prime }_{(0,0,0),(r,b^{n},y^{n})}(0,t)= &\, P_{(0,0,0),(r-1,b^{n},y^{n}-1)}(0,t) \, \lambda ^{n} \nonumber \\&+\, P_{(0,0,0),(r-1,b^{n},y^{n})}(0,t) \, \lambda ^{c} \nonumber \\&+ \, P_{(0,0,0),(r+1,b^{n},y^{n})}(0,t) \, (r-y^{n}+1) \mu \, \nonumber \\&+ \, P_{(0,0,0),(r,b^{n}-1,y^{n}+1)}(0,t) \, (y^{n}+1) \vartheta \nonumber \\&- \, P_{(0,0,0),(r,b^{n},y^{n})}(0,t) \, \left[ (\lambda ^{n}+\lambda ^{c}) \, + \, (r-y^{n})\mu \, + \, y^{n}\vartheta \right] . \end{aligned}$$

(48)

Assuming the steady-state distributions exist, taking the limits as $t \rightarrow \infty$:

$$\begin{aligned}&\lim _{t \rightarrow \infty } P^{\prime }_{(0,0,0),(r,b^{n},y^{n})}(0,t)\nonumber \\&\quad = \lim _{t \rightarrow \infty } \left\{ P_{(0,0,0),(r-1,b^{n},y^{n}-1)}(0,t) \, \lambda ^{n} \right. \nonumber \\&\qquad \left. + P_{(0,0,0),(r-1,b^{n},y^{n})}(0,t) \, \lambda ^{c} \right. \nonumber \\&\qquad \left. + \, P_{(0,0,0),(r+1,b^{n},y^{n})}(0,t) \, (r-y^{n}+1) \mu \, \right. \nonumber \\&\qquad \left. + \, P_{(0,0,0),(r,b^{n}-1,y^{n}+1)}(0,t) \, (y^{n}+1) \vartheta \right. \nonumber \\&\qquad \left. - \, P_{(0,0,0),(r,b^{n},y^{n})}(0,t) \, \left[ (\lambda ^{n}+\lambda ^{c}) \, + \, (r-y^{n})\mu \, + \, y^{n}\vartheta \right] \right\} \nonumber \\&\quad = \,\, \pi _{(r-1,b^{n},y^{n}-1)} \, \lambda ^{n} + \, \pi _{(r-1,b^{n},y^{n})} \, \lambda ^{c} \nonumber \\&\qquad \, + \, \, \pi _{(r+1,b^{n},y^{n})}\, (r-y^{n}+1) \mu \, + \, \pi _{(r,b^{n}-1,y^{n}+1)} \, (y^{n}+1)\vartheta \nonumber \\&\qquad \, \, - \, \, \pi _{(r,b^{n},y^{n})} \, \left[ (\lambda ^{n}+\lambda ^{c}) \, + \, (r-y^{n})\mu \, + \, y^{n}\vartheta \right] . \end{aligned}$$

(49)

$P_{(0,0,0),(r,b^{n},y^{n})}(0,t)$ is bounded by 0 and 1 for all t. Hence, if $\lim _{t \rightarrow \infty } P^{\prime }_{(0,0,0),(r,b^{n},y^{n})}(0,t)$ converges, it must converge to 0. But as shown, the right-hand side of (49) has a fixed value. Hence the left-hand side of (49) is zero. Rearranging the terms, we have

$$\begin{aligned} \begin{aligned}&\left[ (\lambda ^{n}+\lambda ^{c}) \, + \, (r-y^{n})\mu \, + \, y^{n}\vartheta \right] \pi _{(r,b^{n},y^{n})}\\&\quad = \lambda ^{n}\pi _{(r-1,b^{n},y^{n}-1)} + \, \lambda ^{c}\pi _{(r-1,b^{n},y^{n})} \\&\qquad + \, (r-y^{n}+1) \mu \pi _{(r+1,b^{n},y^{n})} \, + \, (y^{n}+1)\vartheta \pi _{(r,b^{n}-1,y^{n}+1)}. \\ \end{aligned} \end{aligned}$$

$\square$