On computing optimal (Q,r) replenishment policies under quantity discounts

Abstract

This article studies the classical reorder quantity, order point (Q,r) continuous review stochastic inventory model with Poisson arrivals and a fixed lead time. This model has been extensively studied in the literature and its use in practice is widespread. This work extends previous research in this area by providing efficient algorithms for the computation of the optimal (Q ^∗,r ^∗) values when there is a multi-breakpoint discount pricing structure.

Appendix

This appendix summarizes some of the previous work in this area. The proofs are sometimes along different lines.

In this model, the inventory position (defined as the inventory at hand plus outstanding orders) provides a suitable state description variable. This is not the case with the inventory level (defined as inventory at hand or net inventory). Indeed, when there is heavy demand during some cycle resulting in a large number of backorders, then the arrival of outstanding orders might never bring the on hand inventory back up to the reorder point again, and hence another order would never be placed under a (Q,r) system that is based on the inventory at hand. However, when a (Q,r) system is based on the inventory position, the holding costs cannot be computed directly. If during some cycle there is a considerable number of backorders, then a large number of orders will be placed, for the reorder point in terms of the inventory position will be crossed a large number of times. If r is the reorder point in terms of the inventory position, then immediately after an order is placed the inventory position is Q+r. Using the Poisson demand arrival assumption we see that the time evolution of the inventory position can be described by a continuous time Markov Chain with state space $\mathcal{S} =\{r+1 ,\ldots , r + Q\}$ and transition diagram given by Fig. 3.

Fig. 3

Inventory position

Since all rates of this Markov chain are equal it follows that the steady state probabilities of the inventory position, π _ip (x)=lim _t→∞ P(X(t)=x)=1/Q, for all x=r+1,…,r+Q, i.e., in equilibrium the inventory position is uniformly distributed over r+1,…,r+Q.

Recall from Sect. 2 that there is a fixed, positive, procurement lead time τ. Also N _τt denotes the number of arrivals in the time interval (t−τ,t]. Arrivals of customers occur according to a Poisson Process with rate λ, the expected value of arrivals during a unit of time. We denote: p _k =P(N _τt =k)=e ^−λτ(λτ)^k/k! and $P_{j}=\sum_{k=0}^{j}p_{k} =P(N_{\tau t} \le j)$. Further, let X(t), Y(t) denote respectively the inventory position and the inventory level at time t. Note that X(t)=Y(t)+O _t Q, all t≥0, where O _t denotes the number of outstanding orders at time t. Since orders placed after t−τ have not arrived by time t, the following equation holds:

$$Y(t) =Y(t-\tau)+ O_{t-\tau} Q -N_{\tau t}.$$

(16)

Note also that X(t−τ)=Y(t−τ)+O _t−τ Q, hence

$$Y(t) =X(t-\tau) -N_{\tau t} .$$

(17)

From (17) it follows that Y(t) is also a continuous time Markov chain. Its state space is the set {…,−2,−1,0,1,2,…,Q+r} and transition diagram given by Fig. 4.

Fig. 4

Inventory level

Even though, the transition rates and diagram of the inventory level process Y(t) are more complex, (17) above allows the computation of the steady state probabilities π(x)=lim _t→∞ P(Y(t)=x) in terms of those of the probabilities π _ip (x)=1/Q and P _j =P(N _τt ≤ j) as follows.

Lemma A.1

For any integer inventory position x the following are true:

$$\pi(x) =\left \{ \begin{array}{l@{\qquad}l}(P_{r+Q-x}-P_{r-x} )/Q, & \mbox{\textit{for}} -\infty< x \leq r,\\P_{r+Q-x}/Q ,& \mbox{\textit{for} } r+1 \leq x \leq r+Q.\end{array} \right .$$

Proof

Since it is easy to see that both X(t) and Y(t) Markov chains are ergodic, we can assume that for t=∞, P(X(t)=k)=π _ip (k)=1/Q and P(Y(t)=x)=π(x). For −∞<x≤r we have:

where in the above we have used the independence of X(t−τ) and N _τt as well as the observation that conditional on X(t−τ)=j, Y(t)=x if and only if N _τt =j−x. Similarly, for r+1≤x≤r+Q we have:

where the first equality above follows from the observation that P(Y(t)=x|X(t−τ)=j)=0 if j<x, since it is not possible for the inventory level at time t to be Y(t)=x≥r+1, if the inventory position at time t−τ is smaller then x. The proof is now complete. □

Next, we assume for the moment for simplicity, that there are no quantity discounts. Now the expected annual cost function C(Q,r), is written as follows

$$ C(Q,r)=\bar{c}_K\frac{1}{Q}+c_h\sum_{x=0}^{r+Q}x\pi(x)-c_p\sum _{x=-\infty}^{0}x\pi(x) + \bar {c}_{\tilde {p}}\sum _{x=-\infty}^{0}\pi(x).$$

(18)

One can simplify C(Q,r) using the lemma below.

Lemma A.2

The following are true:

• (i) $\sum_{x=0}^{r+Q} x \pi(x) = \sum_{x=r+1}^{r+Q} \sum_{i=0}^{x-1}P_{i}/Q$.

• (ii) $-\sum_{x=-\infty}^{-1}x\pi(x) = (\sum_{x=r+1}^{r+Q} (\sum_{i=0}^{x-1}P_{i}+\lambda\tau-x ) )/Q$.

• (iii) $\sum_{x=-\infty}^{0}\lambda\pi(x) = (\sum_{x=r+1}^{r+Q}\lambda (1-P_{x-1}) )/Q$.

Proof

For (i) we have:

Similarly, for (ii) we use the following property of a Poisson distribution: xp _x =λτp _x−1 for any x=1,2,… . We obtain:

Finally, for (iii) we have

 □

Proof of Theorem 1:

It is important to remember the definition of $\bar{c}_{K}, \bar{c}_{i}$ and $\bar {c}_{\tilde {p}}$ on p. 3 so the factor λ is not repeated separately in the calculations below.

(a) From Lemma A.2 we have:

(b) First, we will establish that x ^G exists. Then the uniqueness of x ^G will be established by showing that G(x) is strictly increasing for x≥x ^G+1.

Simple algebra shows that

$$ G(x+1)-G(x)=(c_h+c_p)P_x-\bar {c}_{\tilde {p}}p_x-c_p.$$

(19)

The above implies the following:

$$G(x+1)-G(x)=\left \{ \begin{array}{l@{\quad}l}\leq 0,& \mbox{iff } (c_h+c_p)P_x-\bar {c}_{\tilde {p}}p_x\leq c_p\\>0, & \mbox{iff } (c_h+c_p)P_x-\bar {c}_{\tilde {p}}p_x>c_p.\end{array} \right .$$

We notice that for every x<0, G(x+1)−G(x)=−c _p ≤0. Thus, x ^G is positive if it exists. Since G(x+1)−G(x)>0 for large x, because then G(x+1)−G(x)≈c _h >0, the x-mode x ^G exists.

First we prove uniqueness when $c_{\tilde {p}}$ is positive. Therefore, we show that G(x) is strictly increasing for x≥x ^G+1, by first finding a lower bound for x ^G, using its definition and (19), as follows

i.e.,

$$x^G >\frac{\lambda\tau(\bar {c}_{\tilde {p}}-c_h-c_p)}{\bar {c}_{\tilde {p}}}.$$

Using this bound in (20), we obtain for G(x ^G+2)−G(x ^G+1):

(20)

Thus, G(x ^G+2)>G(x ^G+1) and by an induction argument on x>x ^G we have G(x+2)>G(x+1)>0 for all x>x ^G and we see that x ^G is the maximal minimizing point. This also implies that G(x) is unimodal.

If $c_{\tilde {p}}=0$ then G(x+1)−G(x)=(c _h +c _p )P _x −c _p . Thus, for x ^G we have:

Since P _x is increasing in x, we have that $P_{x}\leq P_{x^{G}-1}$ for x<x ^G and $P_{x}>P_{x^{G}}$ for x>x ^G, thus it follows that x ^G is the maximum minimizing point of the function G(x), i.e, G(x) is unimodal. □