Timing order fulfillment of capital goods under a constrained capacity

Abstract

This paper studies the order-fulfillment process of a supplier producing multiple customized capital goods. The times when orders are confirmed by customers are random. The supplier can only work on one product at any time due to capacity constraints. The supplier must determine the optimal time to start the process for each order so that the total expected cost of having the goods ready before or after their orders are confirmed is minimized. We formulate this problem as a discrete time Markov decision process. The optimal policy is complex in general. It has a threshold-type structure and can be fully characterized only for some special cases. Based on our formulation, we compute the optimal policy and quantify the value of jointly managing the order fulfillment processes of multiple orders and the value of taking into account demand arrival time uncertainty.

Appendix

Proof of Lemma 1

Note first that

By using the following identities:

(6)

(7)

we have

Due to IFR assumption, the expression inside the square bracket is increasing in s. Therefore, f(s)≥0 if and only if s≥s ^∗. □

Proof of Proposition 1

For any s<s ^∗, starting production can’t be optimal because waiting for one period and then starting production costs strictly less than immediately starting production (Lemma 1). We show the optimal policy for s≥s ^∗ by contradiction. Suppose there is an s′≥s ^∗ such that a ^∗(s′)=0.

Suppose a ^∗(s)=0 for all s≥s′. Let

$$c_s=\frac{b_s}{\sum_{i\geq s}b_i}(l+1)\pi. $$

By Lemma 1, for any s≥s′ we have

Therefore immediately starting production is better than waiting until the order is confirmed and hence there must be a finite s>s′ such that a ^∗(s)=1.

We let

$$s''=\min\bigl\{s:a^*(s)=1 \mbox{ and } s>s'\bigr\}. $$

By the definitions of s′ and s″, we have a ^∗(s″−1)=0 and a ^∗(s″)=1. This is a contradiction because we know from Lemma 1 that for any s≥s ^∗, immediately starting production costs less than waiting for one period and then starting production. In conclusion, such s′ does not exist and hence a ^∗(s)=1 for all s≥s ^∗. □

Proof of Lemma 2

Let

$$c_s= \biggl(1-\frac{b_s}{\sum_{i\geq s}b_i} \biggr)\pi_2+ \frac{b_s}{\sum_{i\geq s}b_i} \bigl[(l_1+1)\pi_1+(l_2+1) \pi_2+l_2\pi_1 \bigr]. $$

(a) Note first that

Also, by the definition of u(0,2,s),

Therefore,

The result hence follows.

(b) We show that under the condition, w(0,1,s,1)≥w(0,1,s,2) for all s. Let X _s be the time when the order for product 1 is confirmed conditioning on that it is greater than s; that is, Prob(X _s =i)=Prob(X=i|X≥s) for all i≥s. Then,

The first inequality is because it is feasible but not necessarily optimal to start the production of product 1 at time l ₂. The second and the third inequalities are obvious. The fourth follows from the condition π ₁ l ₂≤π ₂ l ₁. □

Proof of Theorem 2

Because w(0,1,s,1)≥w(0,1,s,2), we can simplify (4) to

$$u(0,1,s)=\min\bigl\{w(0,1,s,0),w(0,1,s,2)\bigr\}. $$

The rest of the proof is similar to the proof of Proposition 1 and hence omitted. □

Proof of Lemma 3

Let $s^{*}_{1}$ be the threshold for product 1 defined in Sect. 3. For $s\leq s^{*}_{1}-l_{2}-2$, following from similar arguments used in the proof of Lemma 2, we have

$$f_2(s)=\frac{b_s}{\sum_{i\geq s}b_i}\Biggl[\pi_1\sum _{i=0}^{l_2}\frac {b_{s+i}}{b_s}+\pi_2\sum _{i\geq0}\frac{b_{s+i}}{b_s}+\pi_2l_1- \pi_1l_2\Biggr]. $$

Because of the ILR assumption, the expression inside the square bracket is decreasing in s.

For s≥s ^∗−l ₂, then,

$$u(0,2,s+l_2+1)=\pi_1\sum_{j=0}^{l_1-1}(l_1-j) \frac{b_{s+1+l_2+j}}{\sum_{i\geq s+1+l_2}b_i}+h_1\sum _{j=l_1+1}^\infty(j-l_1) \frac {b_{s+1+l_2+j}}{\sum_{i\geq s+1+l_2}b_i}, $$

$$u(0,2,s+l_2)=\pi_1\sum_{j=0}^{l_1-1}(l_1-j) \frac{b_{s+l_2+j}}{\sum_{i\geq s+l_2}b_i}+h_1\sum_{j=l_1+1}^\infty(j-l_1) \frac {b_{s+l_2+j}}{\sum_{i\geq s+l_2}b_i}. $$

By using identities (6) and (7), we have

Under the conditions that X has ILR and π ₂≥h ₁, f ₂(s) crosses zero at most once, and, if it does, it is from above.

Finally, we shall show

$$\lim_{s\rightarrow(s^*_1-l_2-1)^{-}}f_2(s)\geq f_2\bigl(s^*_1-l_2-1 \bigr)\geq\lim_{s\rightarrow(s^*_1-l_2-1)^{+}}f_2(s). $$

The first inequality is true because based on the definition of $s^{*}_{1}$, when $s=s^{*}_{1}$, a ^∗(0,2,s)=1, not 0. The second inequality is because when $s=s^{*}_{1}-1$, a ^∗(0,2,s)=0, not 1. In summary, f ₂(s) may cross zero at most once, and, if it does, it is from above. We thus can conclude that f ₂(s)≤0 if and only if $s\geq\bar{s}_{2}$. □

Proof of Lemma 4

We show that under the conditions, f ₁(s) crosses zero at most once when s traverses from zero to infinity, and, if it does, it is from below.

By using identities (6) and (7), we have

$$f_1(s)=\pi_1+\pi_2-(\pi_1+h_1) \frac{\sum_{j\geq s+1+l_1}b_j}{\sum_{j\geq s}b_j}. $$

Part (c) is immediate. Part (a) follows because $\sum_{j\geq s+1+l_{1}}b_{j}/\sum_{j\geq s}b_{j}$ is decreasing in s if X has an IFR. Part (b) follows because f ₁(s)≥f(s) when π and h in f(s) are replaced by π ₁ and h ₁, respectively. □

Proof of Theorem 3

(a) The proof is by contradiction. Suppose there is a $\hat{s}$ such that $a^{*}(0,1,\hat{s})=0$.

We first show that $a^{*}(0,1,\hat{s})=0$ implies a ^∗(0,1,s)=0 for all $s\geq\hat{s}$. This can be showed by contradiction. If there is an $s\geq\hat{s}$ such that a ^∗(0,1,s)=0 and a ^∗(0,1,s+1)=2. This implies f ₂(s)≤0. According to Lemma 3, we have f ₂(s+1)≤0, which means that starting the production of product 2 at s+1 would be more costly than delaying the production until one period later. So a ^∗(0,1,s+1)=0, which is a contradiction. Also, because of part (c) of Lemma 4, if a ^∗(0,1,s)=0, then a ^∗(0,1,s+1)≠1. Therefore, if $a^{*}(0,1,\hat{s})=0$, then a ^∗(0,1,s)=0 for all $s\geq\hat{s}$.

We next show that such $\hat{s}$ does not exist. Suppose it did, then

which is a contradiction. Therefore, such $\hat{s}$ does not exist and “wait” can never be the optimal action.

(b) Let δ=(w(0,1,s,2)−w(0,1,s,1))∑ _i≥s b _i . To show the result, it suffices to show that δ as a function of s crosses zero at most once when s traverses from zero to infinity, and if it does, it does so from below. For $s<s^{*}_{1}-l_{2}$,

For $s\geq s^{*}_{1}-l_{2}$,

which, under ILR assumption, can cross zero at most once and if it does, it does so from below. Clearly, if $s^{*}_{1}\leq l_{2}$, then w(0,1,s,2)−w(0,1,s,2) is increasing in s∈[0,∞) and the result is immediate. So we just need to show the result when $s^{*}_{1}>l_{2}$. Let

$$g_1(s)=(\pi_1l_2-l_1 \pi_2)\sum_{i= s}^\infty b_i-(\pi_1+h_1)\sum _{j=l_1+s+1}^{l_1+s^*_1}\sum_{i= j}^\infty b_i+\pi_1\sum_{j=s+l_2+1}^{s^*_1} \sum_{i= j}^\infty b_i+ \pi_1\sum_{j=s+l_2}^{s^*_1-1}b_j, $$

and

$$g_2(s)=(\pi_1l_2-\pi_2l_1) \sum_{i= s}^\infty b_i-( \pi_1+h_1)\sum_{j=l_1+s+1}^{l_1+l_2+s} \sum_{i= j}^\infty b_i. $$

Both g ₁ and g ₂ are defined in [0,∞) and δ=g ₁(s) for $s<s^{*}_{1}-l_{2}$ and δ=g ₂(s) for $s\geq s^{*}_{1}-l_{2}$. We first show that δ(s) is increasing initially and then decreasing in $s\in(0,s^{*}_{1}-l_{2}]$ (i.e., quasi-concave if s were continuous).

Consider two cases. First, if l ₂−1>l ₁, then,

$$g_1(s)-g_1(s-1)=b_{s-1}\Biggl[-( \pi_1l_2-\pi_2l_1)+ \pi_1\sum_{i=l_1+1}^{l_2-1} \frac{b_{i+s-1}}{b_{s-1}}+h_1\sum_{i=l_1+1}^{\infty } \frac{b_{i+s-1}}{b_{s-1}}\Biggr]. $$

Since the expression inside the square bracket is decreasing in s under ILR assumption, g ₁ is quasi-concave. Second, if l ₂−1≤l ₁, then,

(8)

Because of IFR assumption, we know

where the last inequality is because of the definition of $s^{*}_{1}$. Therefore, the expression within { } in (8) is decreasing in s under ILR assumption and hence g ₁ is also quasi-concave in this case. Finally,

From the previous analysis, we know that if $g_{1}(s^{*}_{1}-l_{2}-1)-g_{1}(s^{*}_{1}-l_{2}-2)\leq0$, then $\delta (s^{*}_{1}-l_{2})-\delta(s^{*}_{1}-l_{2}-1)\leq0$. Therefore, we can conclude that δ is quasi-concave in $(0,s^{*}_{1}-l_{2}]$.

Note then that g ₁(s)≤g ₂(s) for $s<s^{*}_{1}-l_{2}$ because of the threshold structure of the optimal policy for product 1. Consider the following two cases. If $g_{2}(s^{*}_{1}-l_{2})\leq0$, then g ₁(s)≤g ₂(s)≤0 for all $s<s^{*}_{1}-l_{2}$. So in this case, if δ crosses zero at all, it must do so at some $s\in[s^{*}_{1}-l_{2},\infty)$ and from below. Suppose $g_{2}(s^{*}_{1}-l_{2})> 0$. Then because δ(s) is quasi-concave in $(0,s^{*}_{1}-l_{2}]$, if δ crosses zero, then it must do so at some $s\in(0,s^{*}_{1}-l_{2}-1)$ from below. Let $z^{*}_{1}=\inf\{ s: w(0,1,s,2)-w(0,1,s,1)\geq0\}$. Then, a ^∗(0,1,s)=2 for $s<z^{*}_{1}$ and a ^∗(0,1,s)=1 otherwise. □

Proof of Theorem 4

(a) Examining the behavior of $z^{*}_{1}$ is equivalent to examining the behavior of δ; that is, to show the results, it suffices to show that δ is decreasing in π ₂ but increasing in l ₂. This is obviously true for π ₂. For l ₂, denote δ and g _i , where i=1,2, as δ(l ₂) and g _i (s,l ₂), respectively, to make their dependence on l ₂ explicit. Suppose $s_{1}^{*}>s$. Then, for $l_{2}<s_{1}^{*}-s$,

which is obviously increasing in l ₂. For $l_{2}\geq s^{*}_{1}-s$, we consider $l_{2}^{2}>l_{2}^{1}\geq s_{1}^{*}-s$. Then,

From the definition of $s^{*}_{1}$ and the fact that $s^{*}_{1}\leq s+l_{2}^{1}$, we have

$$\frac{\pi_1}{\pi_1+h_1}\geq\frac{\sum_{i= s^*_1+l_1+1}^\infty b_i}{\sum_{i= s^*_1}^\infty b_i}\geq\frac{\sum_{i= s+l_2^1+l_1+1}^\infty b_i}{\sum_{i= s+l_2^1}^\infty b_i}\geq \frac{\sum_{i= s+l_2^1+l_1+1}^\infty b_i}{\sum_{i= s}^\infty b_i}. $$

Therefore $\delta(l^{2}_{2})-\delta(l_{2}^{1})\geq0$ and δ is also increasing in this region. Finally, since

which is positive for $s^{*}_{1}>s$. When $s^{*}_{1}\leq s$, δ(l ₂)=g ₂(s,l ₂) for all l ₂ and hence the result follows easily.

(b) We only provide the proof for the statement about h ₁; results about p ₁ and l ₁ can be shown analogously. To show that $z^{*}_{1}$ is increasing in h ₁, it suffices to show that δ is decreasing in h ₁. Similarly, we denote δ, $s_{1}^{*}$, and g _i , i=1,2, as δ(h ₁), $s^{*}_{1}(h_{1})$, and g _i (s,h ₁), respectively, for ease of presentation. Let

$$s^{-1}=\sup\biggl\{h_1:\frac{\sum_{j\geq s+l_1+1}b_j}{\sum_{j\geq s}b_j}- \frac {\pi_1}{\pi_1+h_1}= 0\biggr\}; $$

that is, s ⁻¹ is the inverse function of $s_{1}^{*}$ and it is increasing. For h ₁≤s ⁻¹(s+l ₂), δ(h ₁)=g ₂(s,h ₁), which is obviously decreasing in h ₁.

For h ₁>s ⁻¹(s+l ₂), δ(h ₁)=g ₁(s,h ₁). Let $h_{1}^{2}>h_{1}^{1}>s^{-1}(s+l_{2})$. Then, $s^{*}_{1}(h_{1}^{2})\geq s^{*}_{1}(h_{1}^{1})$ and

The first inequality above is because of the definition of $s_{1}^{*}$. So δ is also decreasing in this region. Finally,

Since $s^{*}_{1}\rightarrow s+l_{2}+1$ as h ₁ approaches s ⁻¹(s+l ₂) from the right, we therefore have

where the last equality is due to the definition of s ⁻¹. So δ is decreasing in h ₁ in [0,∞) and the result follows. □

Theorem 5

Let u _tb (x,s) be the cost of the threshold-based heuristic. Of all N orders, m of them have not been confirmed. Let δ(k)=(n ₁,n ₂,…,n _k ) represents a permutation of (1,2,…,k). Let Λ(k) be the set of all possible permutations. Then,

$$u(\mathbf{x},s)\leq u_{tb}(\mathbf{x},s)\leq\max_{\delta(N)\in \varLambda(N)} \Biggl[l_{n_1}\pi_{n_1}+(l_{n_1}+l_{n_2}) \pi_{n_2}+\cdots+\sum_{i=1}^{N}l_{n_i} \pi_{n_N}\Biggr]+\sum_{i=1}^{m} \pi_i. $$

Proof

The first inequality is obvious. For the second, suppose (n ₁,n ₂,…,n _N ) represent the order by which the products are produced; that is, product n ₁ is produced first, n ₂, the second, etc. By the heuristic, the supplier may start the production of a product either because an order have been confirmed or because the time has passed a threshold and hence production must be started preemptively.

If product n _i ∈C, then its cost is upper bounded by $(l_{n_{1}}+\cdots+l_{n_{i-1}}+l_{n_{i}})\pi_{n_{i}}$; that is, its cost is highest when its order is confirmed when the production of product n ₁ is started.

Suppose that product $n_{i}\in\bar{C}$. If product n _i is produced after its order is confirmed, then its cost is upper bounded by $(l_{n_{1}}+\cdots+l_{n_{i-1}}+l_{n_{i}})\pi_{n_{i}}$. If product n _i is produced preemptively before its order is confirmed, then its cost is upper bounded by $(l_{n_{i}}+1)\pi_{n_{i}}$ because the strategy of producing it as soon as the order is confirmed but not before is a feasible strategy. But $(l_{n_{i}}+1)\pi_{n_{i}}$ is upper bounded by $(l_{n_{1}}+\cdots+l_{n_{i-1}}+l_{n_{i}})\pi_{n_{i}}+\pi_{n_{i}}$. The total number of products in $\bar{C}$ is m. Therefore, given a sample path (n ₁,n ₂,…,n _N ), the cost of the heuristic is upper bounded by

$$\Biggl[l_{n_1}\pi_{n_1}+(l_{n_1}+l_{n_2}) \pi_{n_2}+\cdots+\sum_{i=1}^{N}l_{n_i} \pi_{n_N}\Biggr]+\sum_{i=1}^{m} \pi_i, $$

and hence the result follows.

The bound is independent of time and distributions of confirmation times. Its computation is straightforward. The cost bound is constructed by computing the maximal cost for each product under the heuristic. Each permutation (n ₁,n ₂,…,n _N ) represents a possible sequence by which the products are produced. For an arbitrary product n _i , its cost can’t be higher than $(l_{n_{1}}+l_{n_{2}}+\cdots +l_{n_{i}}+1)\pi_{n_{i}}$. From the bound of the heuristic, one can compute the maximal cost of not considering the capacity linkage across products. Based on our numerical studies, unfortunately, the bound is not very tight. □