The Team Surviving Orienteers problem: routing teams of robots in uncertain environments with survival constraints

Abstract

We study the following multi-robot coordination problem: given a graph, where each edge is weighted by the probability of surviving while traversing it, find a set of paths for K robots that maximizes the expected number of nodes collectively visited, subject to constraints on the probabilities that each robot survives to its destination. We call this the Team Surviving Orienteers (TSO) problem, which is motivated by scenarios where a team of robots must traverse a dangerous environment, such as aid delivery after disasters. We present the TSO problem formally along with several variants, which represent “survivability-aware” counterparts for a wide range of multi-robot coordination problems such as vehicle routing, patrolling, and informative path planning. We propose an approximate greedy approach for selecting paths, and prove that the value of its output is within a factor $1-e^{-p_s/\lambda }$ of the optimum where $p_s$ is the per-robot survival probability threshold, and $1/\lambda \le 1$ is the approximation factor of an oracle routine for the well-known orienteering problem. We also formalize an on-line update version of the TSO problem, and a generalization to heterogeneous teams where both robot types and paths are selected. We provide numerical simulations which verify our theoretical findings, apply our approach to real-world scenarios, and demonstrate its effectiveness in large-scale problems with the aid of a heuristic for the orienteering problem.

Appendix

In the following we prove the technical lemma stated in the background section. We start the proof by considering a sequence of Poisson binomial distributions $f_0,f_1,\dots$. The parameters of the nth distribution are denoted as $\{p_{n,k}\}_{k=1}^K$, with $p_{n,1} \le p_{n,2}\le \dots \le p_{n,K}$. The parameters of the $n+1$st distribution are

$$\begin{aligned} \{p_{n+1,k}\}_{k=1}^K = \left\{ \frac{p_{n,1}+p_{n,K}}{2}, \{p_{n,k}\}_{k=2}^{K-1},\frac{p_{n,1}+p_{n,K}}{2}\right\} , \end{aligned}$$

that is, the largest and smallest event probabilities of the nth distribution are averaged to form the $n+1$st distribution. Note that we re-sort the parameters after constructing them from the nth distribution, so it is still true that $p_{n+1,j}\le p_{n+1,k}$ for $j\le k$.

It is easy to verify that this sequence converges to the binomial distribution with parameters K and $p = \frac{1}{K} \sum _{k=1}^K p_k$. We are interested in showing that the tails of the sequence become heavier as n increases. We begin by making some basic observations:

Lemma 9

Define

$$\begin{aligned} \varepsilon _n := \frac{1}{2}\left( p_{n,K}-p_{n,1}\right) \end{aligned}$$

and

$$\begin{aligned} \bar{p}_n := \frac{1}{2}\left( p_{n,K}+p_{n,1}\right) . \end{aligned}$$

Then

$$\begin{aligned} p_{n,1}p_{n,K} = \bar{p}_{n}^2 - \varepsilon _{n}^2, \end{aligned}$$

$$\begin{aligned} (1-p_{n,1})(1-p_{n,K}) = (1-\bar{p}_{n})^2 - \varepsilon _{n}^2, \end{aligned}$$

and

$$\begin{aligned} p_{n,1}(1-p_{n,K}) + p_{n,K}(1-p_{n,1}) = 2\bar{p}_{n}(1-\bar{p}_{n}) + 2\varepsilon _{n}^2. \end{aligned}$$

Proof

Each of these statements follows from straightforward algebra:

$$\begin{aligned} \bar{p}_n^2 - \varepsilon _n^2&= \frac{1}{4} \left( p_{n,1}^2 +2p_{n,1}p_{n,K} + p_{n,K}^2\right) \\&\quad - \frac{1}{4}\left( (p_{n,K}^2 -2p_{n,1}p_{n,K} + p_{n,1}^2) \right) \\&= \frac{1}{4}(4 p_{n,1}p_{n,K}) = p_{n,1}p_{n,K}\\ (1-\bar{p}_{n})^2 - \varepsilon _{n}^2&= 1 -2\bar{p}_n + \bar{p}_n^2 - \varepsilon _n^2\\&= 1 - (p_{n,1}+p_{n,K}) + p_{n,1}p_{n,K}\\&= (1-p_{n,1})(1-p_{n,K})\\ 2\bar{p}_{n}(1-\bar{p}_{n}) + 2\varepsilon _{n}^2&= -2(\bar{p}_n^2 - \varepsilon ^2_n) + 2\bar{p}_n\\&= -2(p_{n,1}p_{n,K}) + (p_{n,1} + p_{n,K}) \\&= p_{n,1}(1-p_{n,K}) + p{n,K}(1-p_{n,1}) \end{aligned}$$

$\square$

It is useful to define an auxiliary sequence of Poisson binomial probability mass functions

$$\begin{aligned} g_n(m) = f(m; \{p_{n,k}\}_{k=2}^{K-1}), \end{aligned}$$

which correspond to the probabilities of m successes excluding the most and least likely “events” of the nth distribution. Note that by definition $g_n(m) = 0$ if $m < 0$ or $m > K-2$. We also define notation for the first and second-order finite difference of $g_n(m)$, which are crucial quantities in our inequalities below.

$$\begin{aligned} \Delta _{1,n}(m)&:= g_n(m) - g_n(m-1),\\&\text {and}\\ \Delta _{2,n}(m)&:= \Delta _{1,n}(m) - \Delta _{1,n}(m-1)\\&=g_n(m) - 2g_n(m-1) + g_n(m-2). \end{aligned}$$

Using these relationships, we can form a succinct recursive description of $f_n(m)$:

Lemma 10

For the sequence of probability mass functions above, we have

$$\begin{aligned} f_{n}(m) = f_{n+1}(m) - \varepsilon ^2_{n} \Delta _{2,n}(m), \end{aligned}$$

and for $F_n(m') := \sum _{m=0}^{m'}f_n(m)$,

$$\begin{aligned} F_n(m') = F_{n+1}(m') - \varepsilon ^2_n \Delta _{1,n}(m'). \end{aligned}$$

Proof

By definition of the probability mass function and $g_n(m)$,

$$\begin{aligned} f_{n}(m)&= p_{n,1}p_{n,K}g_n(m)\\&\quad + (p_{n,1}(1-p_{n,K}) + p_{n,K}(1-p_{n,1}))g_n(m-1)\\&\quad + (1-p_{n,1})(1-p_{n,K})g_n(m-2)\\&= \bar{p}_n^2 g_n(m) + \bar{p}_n(1-\bar{p}_n) g_n(m-1) + (1-\bar{p}_n)^2 g_n(m-2)\\&\qquad - \varepsilon ^2_n ( g_n(m) - 2g_n(m-1) + g_n(m-2))\\&= f_{n+1}(m)-\epsilon ^2_n \Delta _{2,n}(m). \end{aligned}$$

The second equality follows from the identities in Lemma 9, and the second equality follows by definition of $f_{n+1}(m)$ and $\Delta _{2,n}$.

Now taking the summation gives us the second statement:

$$\begin{aligned} F_n(m')&= \sum _{m=0}^{m'} f_{n+1}(m) - \varepsilon ^2_n(g_n(m) - 2g_n(m-1) + g_n(m-2))\\&= F_{n+1}(m') \\&\quad -\,\varepsilon ^2_n \left( \sum _{m=0}^{m'}g_n(m) - 2\sum _{m=0}^{m'-1}g_n(m) + \sum _{m=0}^{m'-2}g_n(m)\right) \\&= F_{n+1}(m') - \varepsilon ^2_n (g_n(m') - g_n(m'-1))\\&= F_{n+1}(m') - \varepsilon ^2_n \Delta _{1,n}(m'). \end{aligned}$$

$\square$

This lemma gives us an exact characterization of the difference between successive distributions in our sequence. Specifically, $f_n(m) \le f_{n+1}(m)$ if and only if the second order finite difference, $\Delta _{2,n}(m)$, is non-negative, and $F_n(m') \le F_{n+1}(m')$ if and only if the first order finite difference, $\Delta _{1,n}(m')$ is non-negative. In the following lemma, we give a sufficient condition on m to ensure that $\Delta _{1,n}(m) \ge 0$.

Lemma 11

Let $\{p_{n,k}\}_{k=1}^K$ and $\mu$ be defined as in Lemma 1, and $\Delta _{1,n}(m)$ defined as the first order finite difference of $g_n(m)$, the Poisson binomial distribution with parameters $\{p_{n,k}\}_{k=2}^{K-1}$. Then for $m \le (1-p_{1,K})(K-2)\frac{\mu }{1-\mu } + p_{1,K}$, $\Delta _{1,n}(m) \ge 0$.

Proof

We start by expressing $g_n(m)$ using the recursive characterization of the Poisson binomial probability mass function given by Chen et al. (1994):

$$\begin{aligned} g_n(m) = \frac{1}{m} \sum _{i=1}^m (-1)^{i-1} g_n(m-i) T_n(i), \end{aligned}$$

where $T_n(i) = \sum _{k=2}^{K-1} \left( \frac{p_{n,k}}{1-p_{n,k}}\right) ^i$. Note that for $i \ge 2$, we have $T_n(i) \le T_n(i-1) \frac{p_{1,K}}{1-p_{1,K}}$. The case $\Delta _{1,n}(m+1) \ge 0$ is equivalent to saying that $\frac{g_n(m+1)}{g_n(m)} \ge 1$. Using the recursive expression above,

$$\begin{aligned} \frac{g_{n}(m+1)}{g_{n}(m)}&= \frac{T_n(1)}{m+1} - \frac{\sum _{i=1}^{m}(-1)^{i-1} g_n(m-i)T_n(i+1) }{(m+1) g_n(m)}\\&\ge \frac{T_n(1)}{m+1}-\frac{\sum _{i=1}^m (-1)^{i-1} g_n(m-i)T_n(i)\left( \frac{p_{1,K}}{1-p_{1,K}}\right) }{(m+1)g_{n}(m)}\\&\ge \frac{T_n(1)}{m+1} - \frac{m}{m+1}\left( \frac{p_{1,K}}{1-p_{1,K}}\right) \\&\ge \frac{K-2}{m+1} \frac{\mu }{1-\mu } - \frac{m}{m+1}\frac{p_{1,K}}{1-p_{1,K}}. \end{aligned}$$

Solving for $m+1$ we have that $\Delta _{1,n}(m+1) \ge 0$ if

$$\begin{aligned} m+1 \le (1-p_{1,K})\left( (K-2)\frac{\mu }{1-\mu }\right) + p_{1,K} \end{aligned}$$

$\square$

Combining Lemmas 10 and 11 completes the proof for Lemma 1.