Production-inventory control policy under warm/cold state-dependent fixed costs and stochastic demand: partial characterization and heuristics

Abstract

The motivation for our study comes from some production and inventory systems in which ordering/producing quantities that exceed certain thresholds in a given period might eliminate some setup activities in the next period. Many examples of such systems have been discussed in prior research but the analysis has been limited to production settings under deterministic demand. In this paper, we consider a periodic-review production-inventory model under stochastic demand and incorporate the following fixed-cost structure into our analysis. When the order quantity in a given period exceeds a specified threshold value, the system is assumed to be in a “warm” state and no fixed cost is incurred in the next period regardless of the order quantity; otherwise the system state is considered “cold” and a positive fixed cost is required to place an order. Assuming that the unsatisfied demand is lost, we develop a dynamic programming formulation of the problem and utilize the concepts of quasi-K-convexity and non-K-decreasing to show some structural results on the optimal cost-to-go functions. This analysis enables us to derive a partial characterization of the optimal policy under the assumption that the demands follow a Pólya or uniform distribution. The optimal policy is defined over multiple decision regions for each system state. We develop heuristic policies that are aimed to address the partially characterized decisions, simplify the ordering policy, and save computational efforts in implementation. The numerical experiments conducted on a large set of test instances including uniform, normal and Poisson demand distributions show that a heuristic policy that is inspired by the optimal policy is able to find the optimal solution in almost all instances, and that a so-called generalized base-stock policy provides quite satisfactory results under reasonable computational efforts. We use our numerical examples to generate insights on the impact of problem parameters. Finally, we extend our analysis into the infinite horizon setting and show that the structure of the optimal policy remains similar.

Appendix

1.1 A.1 Proofs of lemmas

Proof of Lemma 2

We let S ₁ and S ₂ denote the minimizers of J ₁(x) and J ₂(x), respectively. (If they are not unique, we take the largest one.) Then, we have either K ₂+J ₁(S ₁)≤J ₂(S ₂) or K ₂+J ₁(S ₁)>J ₂(S ₂).

Case I: Suppose K ₂+J ₁(S ₁)≤J ₂(S ₂). We prove the quasi-K ₁-convexity of J(x) by showing that the function is quasi-K ₁-convex with a changeover at S ₁−R. First, we analyze the interval [−∞,S ₁−R]. For any x≤S ₁−R, min _y≥x+R {K ₂+J ₁(y)}=K ₂+J ₁(S ₁) and hence, we need to compare K ₂+J ₁(S ₁) and min _x+R>y≥x {K ₂ δ(y−x)+J ₂(y)} to determine J(x). Note that min _x+R>y≥x {K ₂ δ(y−x)+J ₂(y)}>J ₂(S ₂), because S ₂ is a minimizer of J ₂(x). Further, since K ₂+J ₁(S ₁)≤J ₂(S ₂), it follows that J(x)=K ₂+J ₁(S ₁), which is a constant.

Next, we analyze the interval [S ₁−R,∞]. To establish the result that J(x) is non-K ₁-decreasing on [S ₁−R,∞], we show that any x ₁,x ₂∈[S ₁−R,∞] satisfying x ₁≤x ₂ imply J(x ₁)≤J(x ₂)+K ₁. At the point x=x ₂, we have either $J(x_{2})=\min_{y\geq x_{2}+R}\{K_{2}+J_{1}(y)\}$ or $J(x_{2})=\min_{x_{2}+R>y\geq x_{2}}\{K_{2}\delta(y-x_{2})+J_{2}(y)\}$.

In the first subcase, $J(x_{1})\leq\min_{y\geq x_{1}+R}\{K_{2}+J_{1}(y)\}\leq\min_{y\geq x_{2}+R}\{K_{2}+J_{1}(y)\}= J(x_{2})\leq J(x_{2})+K_{1}$, where the first inequality follows from the definition of J(x), the second inequality is due to fewer choices since x ₁≤x ₂, and the third equality holds since $J(x_{2})=\min_{y\geq x_{2}+R}\{K_{2}+J_{1}(y)\}$ in this case.

In the second subcase, $J(x_{2})=\min_{x_{2}+R>y\geq x_{2}}\{K_{2}\delta(y-x_{2})+J_{2}(y)\}$. It suffices to consider the case where x ₂ is the minimizer of J ₂(y) on [x ₂,x ₂+R), since this will imply the result for any other x ^∗ minimizing J ₂(y) in this region. Because x ₂ is the minimizer of J ₂(y) on [x ₂,x ₂+R), we have J(x ₂)=J ₂(x ₂). Given x ₁≤x ₂, we have either x ₁+R≤x ₂ or x ₁+R>x ₂. In the former case, we have $J(x_{1})\leq\min_{y\geq x_{1}+R}\{K_{2}+J_{1}(y)\}\leq K_{2}+J_{1}(x_{2})\leq K_{1}+J_{2}(x_{2})=K_{1}+J(x_{2})$, where the first and second inequalities follow from the definition of J(x) and minimization, respectively, the third inequality follows from the assumptions stated in the lemma, and the last equality holds since J(x ₂)=J ₂(x ₂) in this case. Using similar arguments for the case with x ₁+R>x ₂, we can verify that $J(x_{1})\leq\min_{x_{1}+R> y\geq x_{1}}\{K_{2}\delta(y-x_{1})+J_{2}(y)\}\leq K_{2}+J_{2}(x_{2})\leq K_{1}+J_{2}(x_{2})=K_{1}+J(x_{2})$. Hence, we can conclude that J(x) is quasi-K ₁-convex under Case I.

Case II: Suppose K ₂+J ₁(S ₁)>J ₂(S ₂). Then, it follows that the minimizer of J(x) is S ₂, i.e., J(S ₂)=J ₂(S ₂). In this case, we prove the quasi-K ₁-convexity of J(x) by showing that there exists a changeover point b≤S ₂ such that the function is decreasing (constant) in (−∞,b] and non-K ₁-decreasing in [b,+∞].

We first show that J(x) is non-K ₁-decreasing in [S ₂,+∞). Let x ₁,x ₂∈[S ₂,+∞) such that x ₁≤x ₂. At the point x=x ₂, we have either $J(x_{2})=\min_{x_{2}+R>y\geq x_{2}}\{K_{2}\delta(y-x_{2})+J_{2}(y)\}$ or $J(x_{2})=\min_{y\geq x_{2}+R}\{K_{2}+J_{1}(y)\}$. In the first case, we consider the case where x ₂ is the minimizer of J ₂(y) on [x ₂,x ₂+R) (similar to the analysis in Case I), and find J(x ₂)=J ₂(x ₂). Then, we have J(x ₁)≤J ₂(x ₁)≤J ₂(x ₂)+K ₁=J(x ₂)+K ₁, where the second inequality follows from the quasi-K ₁-convexity of J ₂(x). In the second case, $J(x_{2})=\min_{y\geq x_{2}+R}\{K_{2}+J_{1}(y)\}$. Then, by using the definition of J(x) and the relationship x ₁≤x ₂, we can verify $J(x_{1})\leq \min_{y\geq x_{1}+R}\{K_{2}+J_{1}(y)\}\leq\min_{y\geq x_{2}+R}\{K_{2}+J_{1}(y)\}= J(x_{2})\leq J(x_{2})+K_{1}$, and the non-K ₁-decreasing property holds.

Next, we show the existence of the changeover point b. Let a be the smallest number such that K ₂+J ₁(S ₁)>J ₂(a). The existence of a number a<S ₂ is assured by K ₂+J ₁(S ₁)>J ₂(S ₂). Thus, for any x≤a, K ₂+J ₁(S ₁)≤J ₂(x). We consider two cases: a≤S ₁−R and a>S ₁−R.

If a≤S ₁−R, then we set b=a. Notice that for any x≤b(=a), K ₂+J ₁(S ₁)≤J ₂(x), hence it follows that J(x)=K ₂+J ₁(S ₁), that is, J(x) is a constant in (−∞,b]. Moreover, for any x ₁,x ₂∈[b,S ₂] and x ₁≤x ₂, it follows from the definition of J(x) and the quasi-K ₁-convexity of J ₂(x) that J(x ₁)≤J ₂(x ₁)≤K ₁+J ₂(S ₂)=K ₁+J(S ₂)≤K ₁+J(x ₂), where the last inequality holds, because S ₂ is the minimizer of J(x). Hence, J(x) is non-K ₁-decreasing on [b,S ₂].

If a>S ₁−R, then we set b=S ₁−R. Then, for any x≤b=S ₁−R<a, K ₂+J ₁(S ₁)≤J ₂(x). Hence, it follows that J(x)=K ₂+J ₁(S ₁), i.e., a constant in (−∞,b]. For any x∈[b,S ₂], if x≤S ₂−R, then J(x)≤J ₁(S ₂)+K ₂≤J ₂(S ₂)+K ₁=K ₁+J(S ₂); otherwise, J(x)≤K ₂+J ₂(S ₂)≤J ₂(S ₂)+K ₁=K ₁+J(S ₂). So, J(x) is non-K ₁-decreasing in [b,S ₂].

Given that J(x) is non-K ₁-decreasing on [b,S ₂] and [S ₂,+∞], we can easily show that it is non-K ₁-decreasing on [b,+∞]: for any z ₁∈[b,S ₂] and z ₂∈[S ₂,+∞), we have J(z ₂)+K ₁≥J(S ₂)+K ₁≥J(z ₁), which follows since S ₂ is a minimizer and J(x) is non-K ₁-decreasing.

Thus far, we have proved the quasi-K ₁-convexity. Notice that in all cases, J(x) is a constant on the left of the changeover point and non-K ₁-decreasing on the right. Hence, we can conclude that J(x) is non-K ₁-decreasing in ℝ. □

Proof of Lemma 3

We first prove parts (i) and (ii) by considering two separate cases: (a) f _t,2(x)=min _y≥x+R {g _t,2(y)} and (b) f _t,2(x)=min _x+R>y≥x {g _t,1(y)}.

(i) For case (a), we have f _t,1(x)≤min _y≥x+R {K+g _t,2(y)}=K+f _t,2(x). For case (b), f _t,1(x)≤min _x+R>y≥x {Kδ(y−x)+g _t,1(y)}≤K+min _x+R>y≥x {g _t,1(y)}=K+f _t,2(x). The second inequality in the lemma can be easily verified. From the results of part (i), we can also conclude that g _t,1(x)≥g _t,2(x).

(ii) For case (a), $f_{t,1}(x_{1})\leq\min_{y\geq x_{1}+R}\{K+g_{t,2}(y)\}\leq\min_{y\geq x_{2}+R}\{K+g_{t,2}(y)\}=K+f_{t,2}(x_{2})$, where the second inequality follows from that x ₂>x ₁. For case (b), let f _t,2(x ₂)=g _t,1(y(x ₂)) where x ₂+R>y(x ₂)≥x ₂. If y(x ₂)≤x ₁+R, then f _t,1(x ₁)≤K+g _t,1(y(x ₂))=K+f _t,2(x ₂). If y(x ₂)>x ₁+R, then $\min_{y\geq x_{1}+R}\{K+g_{t,1}(y)\}\leq K+g_{t,1}(y(x_{2}))=K+f_{t,2}(x_{2})$. From part (i), we have g _t,1(x)≥g _t,2(x) which implies that $\min_{y\geq x_{1}+R}\{K+g_{t,2}(y)\}\leq\min_{y\geq x_{1}+R}\{K+g_{t,1}(y)\}\leq K+f_{t,2}(x_{2})$. Thus, we have $f_{t,1}(x_{1})\leq\min_{y\geq x_{1}+R}\{K+g_{t,2}(y)\}\leq K+f_{t,2}(x_{2})$.

(iii) Again, consider two cases: $f_{t,1}(x_{2})=\min_{y>x_{2}+R}\{K+g_{t,2}(y)\}$ and $f_{t,1}(x_{2})=\min_{x_{2}+R\geq y\geq x_{2}}\{K\delta(y-x)+g_{t,1}(y)\}$. In the first case, $f_{t,1}(x_{1})\leq\min_{y>x_{1}+R}\{K+g_{t,2}(y)\}\leq \min_{y>x_{2}+R}\{K+g_{t,2}(y)\}=f_{t,1}(x_{2})$. In the second case, we have

where the second inequality follows due to the fact that g _t,2(x)≤g _t,1(x) from part (i) and since x ₁≤x ₂. Using similar arguments, we can also prove that f _t,2(x ₁)≤K+f _t,2(x ₂). □

Proof of Lemma 4

We prove the result by considering three cases.

Case I

$y^{*}_{t,1}(x_{t})=x_{t}$. Then, $y^{*}_{t,2}(x_{t})\geq x_{t}= y^{*}_{t,1}(x_{t})$.

Case II

$x_{t}<y^{*}_{t,1}(x_{t})<x_{t}+R$. If $y^{*}_{t,2}(x_{t})\geq x_{t}+R$, then the result clearly holds. Referring to (4), the condition $x_{t}<y^{*}_{t,1}(x_{t})<x_{t}+R$ implies that $g_{t,1}(y^{*}_{t,1}(x_{t}))\leq g_{t,1}(y)$, for any x _t ≤y<x _t +R. Therefore, if $y^{*}_{t,2}(x_{t})< x_{t}+R$, it can be seen from (5) that $y^{*}_{t,2}(x_{t})=y^{*}_{t,1}(x_{t})$.

Case III

$y^{*}_{t,1}(x_{t})\geq x_{t}+R$. It implies the following two statements: (i) $g_{t,2}(y^{*}_{t,1}(x_{t}))\leq g_{t,1}(y)$, for any x _t ≤y<x _t +R, and (ii) $g_{t,2}(y^{*}_{t,1}(x_{t}))\leq g_{t,2}(y)$, for any x _t +R≤y. By these two statements, we can obtain $y^{*}_{t,2}(x_{t})=y^{*}_{t,1}(x_{t})$.  □

Proof of Lemma 5

Parts (i) and (ii) follow from the definitions of quasi-K-convexity and the critical points.

(iii) Since L(x) is increasing on x≥S _L , L(x ₂)≥L(x ₁). Then, using Lemma 3(ii), we have g _t,1(x ₁)=L(x ₁)+αE[f _t+1,1((x ₁−D _t )⁺)]≤L(x ₂)+αE[f _t+1,2((x ₂−D _t )⁺)]+K≤g _t,2(x ₂)+K.

(iv) By Lemma 3(i), g _t,2(x)≤g _t,1(x) which implies g _t,2(S _t,2)≤g _t,1(S _t,1), because S _t,2 is a minimizer of g _t,2(x). Thus, we have g _t,1(s _t,3)<K+g _t,2(S _t,2)≤K+g _t,1(S _t,1), which by definition, implies s _t,1≤s _t,3.

By Lemma 3(i), g _t,1(S _t,i )<K+g _t,2(S _t,2). Thus, by the definition, S _t,i >s _t,3 for i=1,2.

If S _L ≥S _t,2, clearly s _t,3<S _t,2≤S _L . Otherwise by Lemma 3(i), g _t,1(S _L )<K+g _t,2(S _t,2), which implies s _t,3<S _L .

By definition, g _t,1(s _t,3)<K+g _t,2(S _t,2)≤K+min _y≥x+R g _t,2(y), which implies s _t,4≤s _t,3.

As g _t,1(S _t,1)<g _t,2(S _t,2) and S _t,2−R∈[S _t,1−R,s _t,1], we have s _t,5>S _t,2−R. □

Proof of Lemma 6

(i) To prove the result, it suffices to show that g _t,i (x)≥g _t,i (S _L ) for all $x\geq \hat{S}_{L}>S_{L}$. Applying Lemma 3(iii), we can obtain

(ii) To verify $s_{t,i}\geq\hat{s}_{L}$, we first prove that for t=1,2,…,T and i=1,2,

• (a) f _t,i (x _t ) is a non-increasing function for x _t ≤S _L ;

• (b) g _t,i (y _t ) is a non-increasing function for y _t ≤S _L ;

• (c) g _t,i (y _t )≥g _t,i (S _t,i )+K for any $y_{t}\leq\hat{s}_{L}$.

We prove these results by induction. It is clear that they hold for t=T. Suppose that the results hold for period t+1. Because f _t+1,i (x _t ) and L(x _t ) are non-increasing for x _t ≤S _L , g _t,i (y _t ) is also non-increasing for y _t ≤S _L ,

This completes the proof. □

1.2 A.2 Feasibility analysis of K ₁-convexity (Sect. 3)

We show by contradiction that K ₁-convexity is not preserved in Lemma 2. Assume that K ₂=K ₁. Suppose J ₁(x) and J ₂(x) have the same minimizer denoted by S. Then, by assumption J ₁(S)<J ₂(S). Let s _t,1=min{x|J ₁(x)=J ₁(S)+K ₁} and s _t,2=min{x|J ₂(x)=J ₂(S)+K ₁}. First, select R such that S−R<min{s _t,1,s _t,2}. Then for any x≤S−R, J(x)=K ₁+J ₁(S). Next, suppose J ₁(x) and J ₂(x) are both strictly increasing in (S,+∞) and K ₁+J ₁(S+R)>J ₂(S), then J(S)=J ₂(S). Finally, note that by construction S−s _t,2<R, and thus J(s _t,2)=min{J ₁(s _t,2+R)+K ₁,K ₁+J ₂(S)}. Since we assumed that J ₁(x) is strictly increasing in (S,+∞), we have J ₁(s _t,2+R)≥J ₂(S). Consequently, J(s _t,2)=K ₁+J ₂(S). Now we can pick three points x=S−R, s _t,2, and S with the values J(x)=K ₁+J ₁(S), K ₁+J ₂(S) and J ₂(S), respectively. Recall that a function f(x) is K ₁-convex if, for any x ₁≤x ₂ and λ∈(0,1], f(λx ₁+(1−λ)x ₂)≤λf(x ₁)+(1−λ)(f(x ₂)+K). For K ₁-convexity to hold in this example, we need to show that for any λ∈(0,1], J(s _t,2)≤λJ(S−R)+(1−λ)(J(S)+K ₁), which implies J ₂(S)≤λJ ₁(S)+(1−λ)J ₂(S). However, since we have J ₁(S)<J ₂(S), we have J ₂(S)>λJ ₁(S)+(1−λ)J ₂(S), which contradicts with K ₁-convexity.

1.3 A.3 Infinite horizon case

Similar to the finite-horizon model, we assume that the demands in different periods are realizations from independent and identical distributions. We show that the T-period optimal cost function f _t,1 and f _t,2 both converge uniformly to finite-valued functions f ₁ and f ₂, respectively. This implies that the optimal policy for the infinite-horizon problem inherits the structure of the T-period problem.

Theorem 3

The optimal cost functions f _t,1 and f _t,2 in the T-period problem converge uniformly to functions f ₁ and f ₂ in any finite interval.

Proof

Denote by y _t (x) the optimal inventory level after ordering when the initial inventory level is x. By Lemma 6, it is easy to prove that there exists a constant M such that y _t (x)∈[0,M] for all x∈[0,M] independent of t.

Note the fact that for any constant c ₁, c ₂, c ₃ and c ₄, we have

$$\bigl|\min\{c_1,c_2\}-\min\{c_3,c_4 \}\bigr|\leq \min\bigl\{|c_1-c_3|,|c_2-c_3|,|c_2-c_3|,|c_2-c_4| \bigr\}. $$

Thus, we can obtain

where the fist inequality follows from the above fact and the second from that y _t (x)∈[0,M] for all x∈[0,M] independent of t.

Since M can be chosen arbitrarily large, we have shown that f _t,i (x) converges monotonically and uniformly for all x in any finite interval. The functions f _t,i (x) are continuous and converge uniformly, thus the limit function f _i (x) is also continuous. □

Let the functions f ₁(x) (f ₂(x)) represent the minimum total expected discounted cost of the infinite horizon when the current system state is cold (warm). It follows from the theory of Markov decision processes that

and

$$f_{2}(x_t)=\min\Bigl\{\min_{y>x+R}\bigl\{L(y)+ \alpha Ef_{2}\bigl((y-D)^+\bigr)\bigr\}, \min_{x+R\geq y \geq x}\bigl \{L(y)+\alpha Ef_{1}\bigl((y-D)^+\bigr)\bigr\}\Bigr\}. $$

The convergence guarantees that the optimal policy for the infinite-horizon case takes the same structure as that of the finite horizon case. While the optimal policy can be typically simplified under an infinite horizon, we are not able to do so in our model because none of the finite-horizon decision regions can be eliminated from the policy.

1.4 A.4 OB heuristic policy

The OB heuristic policy is shown in Table 6.

Table 6 OB heuristic policy