A Stochastic and Dynamic Vehicle Routing Problem with Time Windows and Customer Impatience

Abstract

In this paper, we study the problem of designing motion strategies for a team of mobile agents, required to fulfill request for on-site service in a given planar region. In our model, each service request is generated by a spatio-temporal stochastic process; once a service request has been generated, it remains active for a certain deterministic amount of time, and then expires. An active service request is fulfilled when one of the mobile agents visits the location of the request. Specific problems we investigate are the following: what is the minimum number of mobile agents needed to ensure that a certain fraction of service requests is fulfilled before expiration? What strategy should they use to ensure that this objective is attained? This problem can be viewed as the stochastic and dynamic version of the well-known vehicle routing problem with time windows. We also extend our analysis to the case in which the time service requests remain active is itself a random variable, describing customer impatience. The customers’ impatience is only known to the mobile agents via prior statistics. In this case, it is desired to minimize the fraction of service requests missed because of impatience. Finally, we show how the routing strategies presented in the paper can be executed in a distributed fashion.

Appendix

In this appendix, we restate and prove Lemma 4.3.

Lemma 4.3

Assume

$$ \beta \sqrt{\frac{2}{T}} < \lim_{k \to \infty} \frac{m(k)}{\sqrt{kl}} < + \infty. $$

Then, at each epoch r ≥ 1:

$$ \begin{array}{ll} \text{\rm(1)} \quad &\lim_{k \to \infty} n_r^k \overset{a.s}{=} \infty;\\ \text{\rm(2)} \quad &\lim_{k \to \infty} \frac{C_r^k - c}{\sqrt{n_r^k}}\sqrt{m(k)} \overset{a.s}{=} \beta;\\ \text{\rm(3)} \quad &\lim_{k \to \infty} \frac{n_{r}^k}{C_{r-1}^k }\frac{1}{kl/m(k)} \overset{a.s}{=} 1. \end{array} $$

Proof

The arrival rate in a service region is kl/m(k). Consider an arbitrary deterministic time interval c > 0 and let n ^k(c) be the number of Poisson arrivals, with rate kl/m(k), in such time interval; we start by proving that $\lim_{k \to \infty} n^k(c) \overset{\text{a.s.}}{=} \infty$. From Section 2, we have

$$ \lim_{k \to \infty} n^k(c) \overset{\text{a.s.}}{=} \infty \Leftrightarrow \forall N > 0 \lim_{k \to \infty} \mathbb{P}\left[{\bigcup_{p=k}^\infty [ n^p(c) < N ]}\right] = 0. $$

Therefore, we want to show that

$$ \ \forall \, \varepsilon >0 \quad \exists \, \bar{k} : \forall k > \bar{k} \quad \mathbb{P}\left[{\bigcup_{p=k}^\infty [ n^p(c) < N ]}\right] < \varepsilon. $$

(18)

By assumption

$$ L< \lim_{k \to \infty} \frac{m(k)}{\sqrt{kl}} < U, $$

where L and U are two positive constants (their values are of no concern here). Thus, there exists k ₁ > 0 such that for all k ≥ k ₁

$$ L< \frac{m(k)}{\sqrt{kl}} < U. $$

Let k ₂ be the smallest integer k such that $\sqrt{kl}c/L>1$. Now, by using the union bound and assuming k >  max (k ₁,k ₂), we have

$$\begin{array}{rcl} \mathbb{P}\left[{\bigcup_{p=k}^\infty \left[ n^p(c) < N \right]}\right] &\leqslant& \sum\limits_{p=k}^{\infty} \mathbb{P}\left[{n^p(c) < N}\right]\nonumber\\ &=& \sum\limits_{p=k}^{\infty} \sum\limits_{n=0}^{N-1} e^{-\frac{plc}{m(p)}}\frac{\big(plc/m(p) \big)^n}{n!}\nonumber\\ &\leq&\sum\limits_{p=k}^{\infty} \sum\limits_{n=0}^{N-1} e^{-\frac{\sqrt{pl}c}{U}} \frac{\left(\sqrt{pl}c/L \right)^n}{n!}\nonumber\\ &\leq&N \sum\limits_{p=k}^{\infty} e^{-\frac{\sqrt{pl}c}{U}} \big(\sqrt{pl}c/L\big)^{N-1}. \end{array}$$

The series $\sum_{p=0}^{\infty} e^{-\frac{\sqrt{pl}c}{U}} \big(\sqrt{pl}c/L\big)^{N-1}$ is convergent (as it can be easily verified with the comparison test); therefore, $\lim_{k \to \infty } \sum_{p=k}^{\infty} e^{-\frac{\sqrt{pl}c}{U}} \big(\sqrt{pl}c/L\big)^{N-1} = 0$. Let k ₃ be the smallest integer such that, for all k > k ₃, $\sum_{p=k}^{\infty} e^{-\frac{\sqrt{pl}c}{U}} \big(\sqrt{pl}c/L\big)^{N-1} < \varepsilon/N$. Then, by letting $\bar{k} \doteq \max(k_1,k_2,k_3)$, we prove Eq. 18.

Now, the time interval between epochs r − 1 and r (call it τ _r − 1) is at least as large as the computation time c > 0. Thus, if Ω is the set of sample functions ω for which $\lim_{k \to \infty} n^k(c) = \infty$, we have $\lim_{k \to \infty} n^k(\tau_{r-1}) = \infty$ for all ω in Ω. Since ℙ[Ω] = 1 (and $n^k(\tau_{r-1}) = n^k_r$ by definition), part (1) is proven.

We now prove part (2). By Eq. 4, we know that, given a set D _n of n points that are independent and uniformly distributed in a region of unit area, we have $\lim_{n \to \infty} \ensuremath{\operatorname{TSP}}(D_n)/\sqrt{n} \overset{a.s.}{=} \beta$. From part (1) of this Lemma, we also have that $\lim_{k \to \infty} n_r^k \overset{a.s}{=} \infty$. Assume, now, that we scale by a factor $\sqrt{m(k)}$ the coordinates of the demands that arrive in between epochs r − 1 and r, when the overall arrival rate is kl. Let $F_r^k$ be the length of the tour through such scaled demands (the scaled demands are uniformly distributed in a region with unit area). Thus, for any sample function (except possibly for a set of probability zero), $F_r^k/\sqrt{n^k_r}$ runs through the same sequence of values with increasing k as $\ensuremath{\operatorname{TSP}}(D_n)/\sqrt{n}$ runs through with increasing n. Thus if Ω is the set of sample functions ω for which both $\lim_{n \to \infty}\ensuremath{\operatorname{TSP}}(D_n)/\sqrt{n}= \beta$ and $\lim_{k \to \infty} n^k_r \overset{a.s.}{=} \infty$, we have $\lim_{k \to \infty} F^k_r/\sqrt{n^k_r} = \beta$ for all sample functions in Ω. By Eq. 4 and part (1) of the Lemma we have ℙ[Ω] = 1. Thus

$$ \lim_{k \to \infty} \frac{F_r^k}{\sqrt{n^k_r}} \overset{a.s.}{=}\beta. $$

By scaling, we have $F_r^k = (C_r^k - c) \sqrt{m(k)}$, and thus we get the limit in part (2).

Finally, we prove part (3). The number of arrivals in between epochs r − 1 and r is $N^{\frac{kl}{m(k)}}(C_{r-1}^k)$, where $\{N^{\frac{kl}{m(k)}}(t);\, t~\geqslant~0\}$ is a Poisson process with intensity k l/m(k). By the strong law of large numbers for renewal processes (see, for example, [31]) we have

$$ \lim_{t \to \infty} N^1(t)/t \overset{a.s.}{=} 1. $$

For every k, consider the time scaling:

$$ \tau \doteq \frac{kl}{m(k)}t. $$

Notice that in the new time scale the arrival rate is λ = 1 for every k. Let $\tilde{C}_{r-1}^k$ be the length of the time interval between epochs r − 1 and r in the new time scale, i.e. $\tilde{C}_{r-1}^k = \frac{kl}{m(k)}C_{r-1}^k$. Since, by definition, $C_{r-1}^k\geq c>0$, and since by assumption lim _{k → ∞} kl/m(k) = ∞, we have

$$ \lim_{k \to \infty} \tilde{C}_{r-1}^k \overset{a.s.}{=} \infty. $$

Therefore, with similar arguments as before, we obtain

$$ \lim_{k \to \infty} \frac{N^1\big(\tilde{C}_{r-1}^k\big)}{\tilde{C}_{r-1}^k } \overset{a.s.}{=} 1. $$

(19)

By scaling, we have $N^1\big(\tilde{C}_{r-1}^k\big) \equiv N^{\frac{kl}{m(k)}}\big(C_{r-1}^k\big)$, therefore Eq. 19 is equivalent to

$$ \lim_{k \to \infty} \frac{N^{\frac{kl}{m(k)}}\big(C_{r-1}^k\big)}{C_{r-1}^k } \frac{1}{kl/m(k)} \overset{a.s.}{=} 1. $$

Since, by definition, $n_r^k = N^{\frac{kl}{m(k)}}\big(C_{r-1}^k\big)$ , we obtain the claim. □