A dynamic inventory rationing problem with uncertain demand and production rates

Abstract

We investigate the structural properties of a finite horizon, discrete time single product inventory rationing problem, where we allow random replenishment (production) opportunities. In contrast to the standard models of dynamic capacity control in revenue management or production/inventory systems, we assume that the demand/production rates are not known with certainty but lie in some interval. To address this uncertainty, we formulate a robust stochastic dynamic program and show how the structural properties of the optimal policy propagate to the robust counterpart of the problem. Further, we explore how the optimal policy changes with respect to the uncertainty set. We also show that our results can be extended to certain alternative robust formulations.

Appendices

Appendix 1: A proof of theorem

Before proving Theorem 1 we introduce a simple algorithm that starts with $\varvec{\lambda }$ and ends up with $\varvec{\lambda }'$ (where $\varvec{\lambda }_{t} \succeq \varvec{\lambda }'_{t}$) by a sequence of reallocation of probabilities. By definition, $\sum _{i=1}^{n+1} \lambda _{i} \;\le \;1$ and $\sum _{i=1}^{n+1} \lambda '_{i}\;\le \;1$. Now consider the following sequence of vectors, $\varvec{\lambda }^{(1)}, \varvec{\lambda }^{(2)},...,\varvec{\lambda }^{(n+1)}$. Let $\varvec{\lambda }^{(1)}= \varvec{\lambda }$.

Now let $\epsilon _1=\lambda _{1}-\lambda '_{1}$, we construct $\varvec{\lambda }^{(2)}=\varvec{\lambda }^{(1)}+(-\epsilon _1,\epsilon _1,0,0,...,0)$. Obviously $\epsilon _1\ge 0$. In the next iteration, we let $\epsilon _2=\lambda _{2}+\lambda _{1}$ $-\lambda '_{1}-\lambda '_{2}$, again $\epsilon _2\ge 0$ and $\varvec{\lambda }^{(3)}=\varvec{\lambda }^{(2)}+(0,-\epsilon _2,\epsilon _2,0,...,0)$. We continue to iterate similarly for $n$ steps. At step $n$ we have: $\varvec{\lambda }^{(n)}=\varvec{\lambda }^{(n-1)}+(0,0,...,-\epsilon _{n},\epsilon _{n})$. At step $n$ we have: $\varvec{\lambda }^{(n+1)}=\varvec{\lambda }^{(n-1)}+(0,0,...,-\epsilon _{n+1})$. By construction, we have $\varvec{\lambda }^{(n+1)}=\varvec{\lambda }'$. In addition, the sequence of vectors have the property $\varvec{\lambda }^{(1)} \preceq \varvec{\lambda }^{(2)} \preceq ... \preceq \varvec{\lambda }^{(n+1)}$.

Proof

We prove the desired result in two phases corresponding to stages $t$ and $t+1$. Consider two systems that are identical and substitute $\varvec{\lambda }_{t}$ with $\varvec{\lambda }'_{t}$ and $\mu _t$ with $\mu '_t$. In the first phase, we prove that $\Delta v_t(x) \ge \Delta v'_t(x)$ holds at $t$, then in the second phase we prove that $\Delta v_{t+1}(x) \ge \Delta v'_{t+1}(x)$. Please note that at stage $t$, we have $v_{t-1}(x)=v'_{t-1}(x)$, for all $x$ and $\varvec{\lambda }_{t} \ne \varvec{\lambda }'_{t}$ and $\mu _{t+1}\ne \mu '_{t+1}$ whereas in stage $t+1$, $\varvec{\lambda }_{t+1} = \varvec{\lambda }'_{t+1}$ and $\mu _{t+1}=\mu '_{t+1}$.

Suppose $\Delta v_t(x) \ge \Delta v'_t(x)$ holds, then $v_t(x)- v'_t(x) \ge v_t(x-1)- v'_t(x-1)$. This implies that as we replace the $\varvec{\lambda }_{t}$ with $\varvec{\lambda }'_{t}$ at stage $t$ the loss in $v_t(x)$ is greater than loss in $v_t(x-1)$. Whereas as we replace $\mu _t$ with $\mu '_t$ at stage $t$, associated gain in $v_t(x)$ is less than gain in $v_t(x-1)$. We again use the above algorithm in order to perform such a replacement. Hence, we consecutively decrease the arrival probability of a class $i$ by $\epsilon$ and increase a class $j$ by $\epsilon$ where $i< j$. Then we simply increase the production rate by $\mu '_t-\mu _t$.

$$\begin{aligned} \epsilon \left( R_i-\Delta v_{t-1}(x)\right) ^+-\epsilon \left( R_j-\Delta v_{t-1}(x)\right) ^+&\ge \epsilon \left( R_i-\Delta v_{t-1}(x-1)\right) ^+ \nonumber \\&-\epsilon \left( R_j-\Delta v_{t-1}(x-1)\right) ^+. \end{aligned}$$

(11)

We prove the inequality case by case, note that $A$ stands for admission and $R$ stands for rejection. The non-trivial cases are listed below (we do not present the cases where all classes are accepted or all classes are rejected since these are obvious). Please note that accepting a lower class ($j$) means that a higher class ($i$) is always accepted. Also please note that, if a customer class is accepted at an inventory $x-1$ then it is also accepted at $x$, and if it is rejected at $x$ then it is also rejected at $x-1$.

$R_i$	$- \;R_j$	$\ge$	$R_i$	$-\;R_j$
$-\Delta v_{t-1}(x)$	$-\Delta v_{t-1}(x)$	$\ge$	$-\Delta v_{t-1}(x-1)$	$-\Delta v_{t-1}(x-1)$	Result
(A)	(A)		(A)	(R)	$R_i-R_j \;\ge$
					$R_i-\Delta v_{t-1}(x-1)$
(A)	(R)		(A)	(R)	$R_i- \Delta v_{t-1}(x) \;\ge$
					$R_i-\Delta v_{t-1}(x-1)$
(A)	(A)		(R)	(R)	$R_i-R_j \;\ge \;0$
(A)	(R)		(R)	(R)	$R_i- \Delta v_{t-1}(x)\;\ge \;0$

Except for the case in the first row, all inequalities follow easily by concavity of $v(x)$ (a summary of the result is provided in the last column). Consider the case in the first row: because class $j$ is rejected at $x-1$ for this case, we must have $R_j-\Delta v_{t-1}(x-1) \;\le \;0$. This implies that $R_i-R_j \;\ge \;R_i-\Delta v_{t-1}(x-1)$. Lastly, it is obvious that $(\Delta v_{t-1}(x+1))^+ \le (\Delta v_{t-1}(x))^+$.

Next, we consider the second phase where we need to establish that $\Delta v_{t-1}(x) \ge \Delta v'_{t-1}(x)$. We use a similar approach here, but we consider only one operator $T^i$ (admission decision for a single class) and $T$ (production decision) at a time. The cases related to accept all and reject all for both systems at states $x-1$ and $x$ are obvious. Please note that since $\Delta v_{t}(x) \ge \Delta v'_{t}(x)$, therefore $R_i-\Delta v_{t}(x) \le R_i -\Delta v'_{t}(x)$, which means that any class accepted to the initial system will always be accepted to the second system. Except from the obvious cases (accept all, reject all) there are only four alternatives:

$T^iv_{t-1}(x)$	$- T^iv'_{t-1}(x)$	$\ge$	$T^iv_{t-1}(x-1)$	$-T^iv'_{t-1}(x-1)$	Result
(A)	(A)		(R)	(R)	$R_i+v_{t}(x-1)-R_i-v'_{t}(x-1) \ge$
					$v_{t}(x-1)-v'_{t}(x-1)$
(A)	(A)		(R)	(A)	$R_i+v_{t}(x-1)-R_i-v'_{t}(x-1) \ge$
					$v_{t}(x-1)-R_i-v'_{t}(x-2)$
(R)	(A)		(R)	(A)	$v_{t}(x)-R_i- v'_{t}(x-1) \ge$
					$v_{t}(x-1)-R_i-v'_{t}(x-2)$
(R)	(A)		(R)	(R)	$v_{t}(x)-v'_{t}(x-1)-R_i \ge$
					$v_{t}(x-1)-v'_{t}(x-1)$

The first case is clear. Consider the second case, since it is optimal to accept at $x-1$ for the second system, $R_i+v'_{t}(x-2) \ge v'_{t}(x-1)$. The third case is trivial, the LHS can be easily decreased by replacing the optimal action of the first system and the second case is attained. The last case is clear too, since the optimal action of the first system at state $x$ is rejection, $v_{t}(x) \ge v_{t}(x-1)+R_i$.

For the production operator which we denote by $T$ we need to show that it also preserves the inequality. Since $\Delta v_{t+1}(x) \ge \Delta v'_{t+1}(x)$ if it is not optimal to produce in the original system then it is also not optimal to produce in the perturbed system. By concavity we know that the base stock policy is optimal therefore the cases except from the produce at all of the conditions and not produce at all of the conditions the cases are as follows:

$Tv_{t+1}(x)$	$- Tv'_{t+1}(x)$	$\ge$	$Tv_{t+1}(x-1)$	$-Tv_{t+1}(x-1)$	Result
(NP)	(NP)		(P)	(P)	$v_{t}(x)-v'_{t}(x) \ge$
					$v_{t}(x)-v'_{t}(x)$
(P)	(NP)		(P)	(P)	$v_{t}(x+1)-v'_{t}(x) \ge$
					$v_t(x)-v'_{t}(x)$
(NP)	(NP)		(P)	(NP)	$v_t(x)-v'_t(x) \ge$
					$v_t(x)-v'_t(x-1)$
(P)	(NP)		(P)	(NP)	$v_t(x+1)- v'_{t}(x) \ge$
					$v_t(x)-v'_{t}(x-1)$

The first case is clear. In the second case it is optimal to make a production in the original system therefore $v_t(x+1)\ge v_t(x)$. The third case is similar too, since it is not optimal to produce at the second system we have $v'_t(x) \le v'_t(x-1)$. For the last case $v_t(x+1)- v'_{t}(x) \ge v_t(x)- v'_{t}(x)$ and $v_t(x)- v'_{t}(x) \ge v_t(x)-v'_{t}(x-1)$. These results hold for each operator $T_i$ and $T$, hence any convex combination of them satisfies the inequality. This completes the proof.$\square$

Appendix 2: Definition of $\Delta \mathcal{P}_t$ and $\Delta \mathcal{C}_t$

Here we present the definitions of the transformed uncertainty sets employed in the proofs.

$$\begin{aligned} \Delta \mathcal{P}_t&= \left\{ \varvec{\Delta y}=(\Delta y_0,\ldots ,\Delta y_{n+1}):\;\;\;0\; \; \le \Delta y_{i} \;\le \bar{y}_{i,t}-\underline{y}_{i,t},\right. \\&\left. \quad 0\;\le \;q-\sum _{i=1}^{n}\underline{y}_{i,t}\;\le \;\sum _{i=1}^{n} \Delta y_i\;\le \;1 \right\} ,\\ \Delta \mathcal{C}_t&= \left\{ \varvec{\Delta y}=(\Delta y_1,\ldots ,\Delta y_{n+1}):\;\;\;0\; \; \le \Delta y_{i,t} \;\le \bar{y}_{i}-\underline{y}_{i,t}, \right. \\&\left. \quad \sum _{i=1}^{n} b_i \Delta y_i \ge Q-\sum _{i=1}^{n} b_i\Delta y_i,\; 0\;\le \;q-\sum _{i=1}^{n}\underline{y}_{i,t}\;\le \;\sum _{i=1}^{n} \Delta y_i \le \;1 \right\} , \end{aligned}$$