Improved Approximation Algorithms for Inventory Problems

Abstract

We give new approximation algorithms for the submodular joint replenishment problem and the inventory routing problem, using an iterative rounding approach. In both problems, we are given a set of N items and a discrete time horizon of T days in which given demands for the items must be satisfied. Ordering a set of items incurs a cost according to a set function, with properties depending on the problem under consideration. Demand for an item at time t can be satisfied by an order on any day prior to t, but a holding cost is charged for storing the items during the intermediate period; the goal is to minimize the sum of the ordering and holding cost.

Our approximation factor for both problems is $O(\log \log \min (N,T))$; this improves exponentially on the previous best results.

N. Olver—Supported by Dutch Science Foundation (NWO) Vidi grant 016.Vidi.189.087.

A Some Omitted Proofs

Proof

(Theorem 3). Let y be a solution to (2). We will first generate two new instances of the subadditive cover over time problem, one being left aligned and the other right aligned.

Given an interval [s, t], define the right-aligned part R([s, t]) and the left-aligned part L([s, t]) by

$$\begin{aligned} R([s,t]) = [s, k2^i] \quad \text { and } \quad L([s,t]) = [k2^i+1,t] , \end{aligned}$$

where i, k are integers such that $k2^i\in [s,t]$ and i is maximal. If $k2^i = t$, then $L([s,t]) = \emptyset$, and if $k2^i + 1=s$ then $R([s,t]) =\emptyset$ by convention. It is clear from this definition that $\{ L([s,t]): v \in V, [s,t] \in \mathcal {W}_v\}$ forms a left-aligned family, and similarly the right-aligned parts form a right-aligned family.

Any LP solution must cover every item by at least half in either the right-aligned or left-aligned part of its demand window. For each $v\in V$ and demand window $[s,t]\in \mathcal {W}_v$, if L([s, t]) receives half a unit of coverage under y, add L([s, t]) as a time window for v in the left-aligned instance; otherwise put R([s, t]) in the right-aligned instance.

It is immediate from the way in which we constructed the two instances that 2y is a feasible solution to each. Hence the combined cost of the optimal solutions to the LP relaxations of the generated instances is at most 4 times that of the original instance. Furthermore, we can translate integral solutions to the left and right aligned instances back to one for the original instance by adding them together, which does not increase the cost by subadditivity of f.    $\square$

Proof

(Lemma 2). We proceed by induction on j, the number of serviced nodes on the subpath of P from the tail of P until before edge e. The claim is clearly true if $j=1$, since the condition ensures that $v_1$ germinated, in which case all edges from $v_1$ to $v_2$ will be deleted by the procedure. (The claim is trivial if $j=0$, in which case edge e is always deleted).

Suppose $j > 1$, and that that the condition holds. First, by considering level $\ell (v_j)$, it follows that $v_j$ germinated. Next, consider level $i=\ell (v_j)-1$. If none of $v_1, v_2, \ldots , v_j$ have level i or less, then the procedure will clearly remove the edges between $v_j$ and $v_{j+1}$, irrespective of what edges have already been removed from P. Otherwise, let q be chosen maximally from $\{1,2,\ldots , j-1\}$ so that $\ell (v_q) \le i$. Then the condition of the lemma holds for an edge between $v_q$ and $v_{q+1}$; hence by our inductive assumption, these edges were removed by the split-and-shift procedure. So at the point that the current edge e is being considered for removal, the subpath of P remaining contains only the services nodes $v_{q+1}, \ldots , v_k$. Since $v_j$ has the smallest level amongst $v_{q+1}, \ldots , v_j$ and has germinated, e will be removed. This completes the induction.    $\square$

Proof

(Lemma 3). Let $k\in \mathbb {N}$ be such that $\frac{1}{16} \alpha \le \frac{1}{k} \le \frac{1}{8}\alpha$. Note that this implies $k\ge 8$. Let $z = f(L_1(x))$.

Claim

If $2^{1/\alpha }z > \hat{f}(x)$, there exists $m \in \mathbb {N}$ such that

$$\begin{aligned} f(L_{\frac{k-m}{k}}(x)) < 2^m z.\end{aligned}$$

(6)

Before we prove the claim, let’s see that it implies the lemma. Suppose that $2^{1/\alpha } f(L_1(x)) > \hat{f}(x)$, since otherwise we are done. The condition of the claim then holds, so take the smallest m that satisfies (6), and let $\theta = \frac{k-m}{k}$. We claim that

$$\begin{aligned} \frac{\alpha }{32} f(L_\theta (x)) \le \hat{f}(x) - \hat{f}( {x}^{| \theta }) . \end{aligned}$$

To see this we first rewrite the right hand side as an integral.

$$\begin{aligned} \hat{f}(x) - \hat{f}( {x}^{| \theta }) = \int _0^1 f(L_\eta (x)) \,d\eta - \int _0^1 f(L_\eta ( {x}^{| \theta })) \,d\eta \\ \nonumber = \int _0^1 f(L_\eta (x)) \,d\eta - \int _0^\theta f(L_\eta (x)) \,d\eta = \int _\theta ^1 f(L_\eta (x)) \,d\eta . \end{aligned}$$

(7)

Recall that $f(L_\eta (x))$ is monotonically decreasing and that $m\ge 1$ so that $\theta +\frac{1}{k} = \frac{k-m+1}{k} \le 1$. Then

$$\begin{aligned} \int _\theta ^1 f(L_\eta (x)) \,d\eta \ge \int _\theta ^{\theta + \frac{1}{k}} f(L_\eta (x)) \,d\eta \ge \frac{1}{k} f(L_{\theta +\frac{1}{k} }(x)). \end{aligned}$$

(8)

Finally, we use the fact that m is minimal, which implies that $f(L_{\frac{k-m+1}{k}}(x)) \ge 2^{m-1}z$, together with (7) and (8):

$$\begin{aligned} \hat{f}(x) - \hat{f}( {x}^{| \theta }) \ge \tfrac{1}{k} 2^{m-1}z = \tfrac{1}{2k} 2^m z > \tfrac{\alpha }{32} f(L_{\theta }(x)). \end{aligned}$$

(9)

In the final inequality of (9) we use that the fact that we chose m to satisfy $2^m z > f(L_{\frac{k-m}{k}}(x))$ and $\frac{1}{k} \ge \frac{1}{16} \alpha$.

Now we proceed to prove the claim. Suppose for contradiction that the condition of the claim holds but no m satisfies inequality (6). Then, in particular it must hold that $f(L_{\frac{1}{k}}(x)) \ge 2^{k-1}z$ and therefore we obtain

$$ \hat{f}(x) \ge \int _0^{\frac{1}{k}} f(L_\eta (x)) \,d\eta \ge \frac{1}{k} f(L_{\frac{1}{k}}(x)) \ge \frac{1}{k} 2^{k-1} z. $$

Since $k \ge 8$, $\tfrac{1}{k} 2^{k-1}z \ge 2^{k/2}z$. Since also $\frac{1}{k} \le \frac{1}{8} \alpha$, we deduce

$$ \hat{f}(x) \ge 2^{k/2}z \ge 2^{4/\alpha } z \ge 2^{1/\alpha } z, $$

contradicting that $2^{1/\alpha }z > \hat{f}(x)$. This proves the claim, and hence the lemma.

   $\square$