Robust Task Allocation for Multiple Cooperative Robotic Vehicles Considering Node Position Uncertainty

Abstract

In order for an effective and efficient collaboration among multiple vehicles to be possible, task allocation is essential. In general, task allocation algorithms assume constant costs and attempt to minimize or maximize an expected objective value. However, the optimality and reliability of the results may be greatly affected by uncertainties, such as those of the position of the tasks and the agents. This study addresses such challenges in multi-agent task allocation and proposes a robust task allocation algorithm under the uncertainty in the position of the nodes. To quantify the robustness of the solution, the uncertainty of the cost matrix, and the sensitivity of the solution are considered and evaluated. In order to model the uncertainty, Gaussian approximations are used and the Interval algorithm is employed to investigate the sensitivity of the solution to the uncertainty. For performance evaluation with respect to the travel distance and the value at risk of the proposed methodology, numerical simulations of multi-objective task allocation were carried out.

Appendix A: Validation of Subtour Free Upper Bound

The initial deactivated edge in Eq. 9 guarantees a subtour-free solution. This means that the graph G does not have Hamilton cycles between the intermediate task nodes. This property is demonstrated using the graph theory theorem [49].

Theorem 1

An n × n permutation matrix is the adjacency matrix of some Hamiltonian cyclic graph on n vertices if and only if its characteristic polynomial is $\det (\lambda I_{n} - A_{n}) = \lambda ^{n} - 1$.

In accordance with this theorem, the adjacency matrix for some Hamilton cycles must have the characteristic polynomial $\det (I_{n} - A_{n}) = 0$ with λ = 1. Based on this, the feasibility of the upper bound solution is shown in the following lemma.

Lemma 1

The adjacency matrix A_n = (a_ij)_n×n of a graph G associated solely with tasks is defined as follows:

$${a_{ij}}: = \left\{ {\begin{array}{*{20}{c}} 1 \ or \ 0, & {if \ i < j, \forall i,j \in \mathcal{V}_{T}} \\ 0, & {otherwise} \end{array}} \right.$$

(A1)

The graph G does not have any Hamilton cycle.

Proof

By induction. For n = 2, the adjacency matrix is

$${A_{2}} = \left({\begin{array}{cc} 0 & 0 \\ \mathbf{1} & 0 \end{array}} \right) or {A_{2}} = \left({\begin{array}{cc} 0 & 0 \\ \mathbf{0} & 0 \end{array}} \right).$$

(A2)

For both forms of A₂, the determinant of $A_{2}^{\prime } = I_{2} - A_{2}$ is not zero, where I_n denotes a n × n identity matrix. Thus, the graph G has no Hamilton cycle. If n = k, where k > 2, the determinant of $A_{k}^{\prime }$ has been expanded by cofactors along the first row.

$$\det ({A_{k}}^{\prime}) = (1-a_{11})C_{11} + a_{12}C_{12} + ... + a_{1k}C_{1k} = C_{11}.$$

(A3)

In general, the cofactor of a_ij is defined as follows.

$$C_{ij} = {-1}^{i+j}M_{ij},$$

(A4)

where M_ij is the determinant of the submatrix obtained by deleting the i^th row and j^th column of $A_{k}^{\prime }$. Finally, the determinant of the matrix ${A_{k}}^{\prime }$ can be expressed as below,

$$\begin{array}{@{}rcl@{}} \det ({A_{k}}^{\prime})= C_{11} = \det ({A_{k-1}}^{\prime})& = \det ({A_{k-2}}^{\prime}) = {\ldots} \\ & = \det ({A_{2}}^{\prime}) \neq 0. \end{array}$$

(A5)

Therefore, the graph G has no Hamilton cycle.

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