Dynamic modulation of multi-task priority for controlling redundancy insufficient robots

Abstract

Redundant robots are gaining popularity for their agility in service tasks, but they struggle with managing multiple tasks in dynamic and unstructured environments. Research is currently centered around adjusting task priorities to facilitate the robot’s adaptability to different situational demands. This paper addresses the challenge of automated task prioritization in multi-task handling and presents a solution for robots to effectively execute demanding tasks, even when faced with limited redundancy and multiple constraints. We introduce the concept of secondary merged tasks and formulate task merging as a matrix design problem. An iterative updating algorithm based on real-time task status is proposed to enable automatic prioritization and dynamic adjustment of tasks. This methodology ensures appropriate execution of all tasks at the right time. We analyze the convergence of weight transfer between redundancies and task dependencies, ensuring stable task execution. Simulation experiments and real-world experiments using 9-DOF mobile manipulator and 6-DOF fixed manipulator are conducted to validate the proposed method. This research provides a feasible approach for task prioritization in multi-task handling and holds potential applications.

Appendix A. Proof of Theorem 1

Suppose two elementary tasks $f_p$ and $f_q$, where $f_p$ is being (or has been) activated, i.e., $\bar{f}_p \simeq 1$, and by contrast $f_q$ is idle, i.e., $\bar{f}_q \simeq 0$. We aim to prove that the weight transition can be always correctly achieved for both tasks, such that they can be suitably performed in due course. We open up our proof along the redundancy and task space separately.

1.1 Appendix A.1. Weight transition along redundancy

Proof

Assume an i-th redundancy is available for tasking $f_p$ and $f_q$. If in the winner take all process, the winner is $f_p$,

$$\begin{aligned} \Delta \dot{\alpha }_{i-pq}&{:}{=} \dot{\alpha }_{ip}-\dot{\alpha }_{iq} \nonumber \\&= {\textbf{W}(\textbf{P}, \textbf{S}, \textbf{A})_{ip}-\textbf{W}(\textbf{P}, \textbf{S}, \textbf{A})_{iq}}\nonumber \\&\ge 0 \end{aligned}$$

(A1)

The weight will transition from $f_q$ to $f_p$, and vice versa.

If the winner has been born and the maximum update value is still the winner, then the weight of all non-winners is 0, the weight remains stable, and there is no mutual transition (Alg. 1 line 4–5). If there is a weight transition, the below relationship holds for all i that is not the winner.

$$\begin{aligned} \frac{\textbf{W}(\textbf{P}, \textbf{S}, \textbf{A})_{ip}-\textbf{W}(\textbf{P}, \textbf{S}, \textbf{A})_{iq}}{(\textbf{P}\textbf{S})_{ip}-(\textbf{P}\textbf{S})_{iq}} \ge 1 \end{aligned}$$

(A2)

In Alg. 1 lines 7 to 9, only the same item z is subtracted from all elements, and the relative distance between elements remains the same. Since neither $f_q$ nor $f_p$ is winner, there is no action on line 10.

Then the relative updating difference between $f_p$ and $f_q$ is

$$\begin{aligned} \Delta \dot{\alpha }_{i-pq}&{:}{=} \dot{\alpha }_{ip}-\dot{\alpha }_{iq} \\&= {\textbf{W}(\textbf{P}, \textbf{S}, \textbf{A})_{ip}-\textbf{W}(\textbf{P}, \textbf{S}, \textbf{A})_{iq}}\\&\ge (\textbf{P}\textbf{S})_{ip}-(\textbf{P}\textbf{S})_{iq}\\&= \prod _{u=0}^{i-1}(1-\alpha _{up})\prod _{v=0}^{p-1}(1-\alpha _{iv})\prod _{u\ne i}(\gamma -\alpha _{up})\bar{f}_p\\&- \prod _{u=0}^{i-1}(1-\alpha _{uq})\prod _{v=0}^{q-1}(1-\alpha _{iv})\prod _{u\ne i}(\gamma -\alpha _{uq})\bar{f}_q \end{aligned}$$

Specifically, there are four cases:

Case one: Suppose neither of $f_p$,$f_q$ is occupying a redundancy, i.e., $\alpha _{up}\simeq \alpha _{uq} \simeq 0$, $\forall u\ne i$. Then we have

$$\begin{aligned} \begin{aligned} 0<&\prod _{u=0}^{i-1}(1-\alpha _{up})\prod _{u\ne i}(\gamma -\alpha _{up})\\ \simeq&\prod _{u=0}^{i-1}(1-\alpha _{uq})\!\prod _{u\ne i}(\gamma -\alpha _{uq}) \simeq \gamma ^{n-m-1} \end{aligned} \end{aligned}$$

Denote $c=\gamma ^{n-m-1}>0$ (a constant), then we have

$$\begin{aligned} \dot{\alpha }_{i-pq}\!&\ge \! (\textbf{P}\textbf{S})_{ip}\!-\!(\textbf{P}\textbf{S})_{iq}\! \\&\simeq c\left( \!\prod _{v=0}^{p-1}(1-\alpha _{iv})\bar{f}_p\!-\!\prod _{v=0}^{q-1}(1\!-\!\alpha _{iv})\bar{f}_q\right) \end{aligned}$$

(1). If $p<q$,

we have

$$\begin{aligned} \dot{\alpha }_{i-pq}\! \ge c\prod _{v=0}^{p-1}(1-\alpha _{iv})\left( \bar{f}_p-\!\prod _{v=p}^{q-1}(1-\alpha _{iv})\bar{f}_q\right) \end{aligned}$$

which indicates as $\bar{f}_p$ approaches one and $\bar{f}_q$ approaches zero in line with their task status, $\dot{\alpha }_{i-pq}\ge 0$ is guaranteed, i.e the weight of $f_p$ will increase relatively faster and therefore a higher task priority is correctly given to $f_p$.

(2). If $p>q$, similarly, we have

$$\begin{aligned} \dot{\alpha }_{i-pq}\! \ge c\prod _{v=0}^{q-1}(1-\alpha _{iv})\left( \prod _{v=q}^{p-1}(1-\alpha _{iv})\bar{f}_p-\bar{f}_q\right) \end{aligned}$$

which indicates, similarly, a higher weight will be eventually transited to $f_p$, as $\bar{f}_p$ and $\bar{f}_q$ vary in accordance with their task status. It also suggests that, however, since $f_q$ is previous to $f_p$ by index, until $\bar{f}_q=0$, $\dot{\alpha }_{i-pq}\ge 0$ is not guaranteed. That is, the weight of $f_p$ will not be improved as faster as $f_q$ until $f_q$ is competed, since $f_q$ has a higher indexing priority.

Case two: Suppose only $f_p$ is occupying a redundancy, i.e., $\exists u \ne i, \alpha _{up} =\gamma$. Then $\dot{\alpha }_{ip} =0$ and therefore $\dot{\alpha }_{i-pq} = \dot{\alpha }_{ip}-\dot{\alpha }_{iq} = 0 -\dot{\alpha }_{iq}\le 0$. That is, a relatively faster weight increase will be given to $f_q$. This is in compliance with the fact that $\bar{f}_p$ has been allocated with a redundancy and therefore its weight will not increase. A higher weight will be accordingly transited to $f_q$.

Case three: Suppose only $f_q$ is occupying a redundancy, similarly, we can prove $\dot{\alpha }_{i-pq} \ge 0$ holds, which is consistent with the fact a higher weight is supposed to transit to $f_p$.

Case four: Suppose both substasks are holding redundancies. Then $\dot{\alpha }_{ip} =\dot{\alpha }_{iq} = 0$ and therefore $\dot{\alpha }_{i-pq} = 0$, i.e., there is no relative difference between their updating rate, which is consistent with the fact that tasks that have been (being) executed will not compete for redundancy and there is no weight transition between them.

1.2 Appendix A.2.Weight transition along task

Proof

Suppose the task $f_p$ has been allocated to a u-th redundancy, i.e., $\exists u, \alpha _{up} =\gamma$. Then at any other $\omega$-th redundancy, it satisfies $\dot{\alpha }_{\omega p}\le 0, \forall \omega \ne u$. That is, once a task has been allocated at a certain redundancy, the weights of the task at other redundancies will not increase, which exactly meets the constraint that an assigned task should not jump back and forth.

To sum up, our approach can converge the weights along both the redundancy and the task space. Since each task controller is stable in design, the entire system can be executed stably once the convergence is achieved.