Search for Smart Evaders with Swarms of Sweeping Agents- a Resource Allocation Perspective

Abstract

Assume that inside a given planar circular region, there are smart mobile evaders, that we would like to detect using sweeping agents. We assume that the agents total sensing resources are a line sensor of predetermined length, which is divided between the swarm’s agents. We propose procedures for designing cooperative sweeping processes that guarantee the detection of all evaders that were inside the original evader region. The task is accomplished by deriving conditions on the sweeping velocity of the agents and their paths, thus ensuring that evaders with a given limit on their velocity are caught. A simpler task for the swarm is the confinement of evaders to their initial domain. The feasibility of completing these tasks depends on geometric and dynamic constraints that impose a lower bound on the pursuers’ velocities. This lower bound ensures the satisfaction of the confinement task. Increasing the velocity above the lower bound enables the sweeper swarm of agents to complete the search task as well. We present results on the total search time as a function of the velocity of the swarm’s agents given the initial conditions on the size of the search region and the evaders’ maximal velocity under limited sensing capabilities of the swarm. The established results provide insights on the practical tradeoffs in designing a multi-agent system, where the alternatives are to deploy a smaller number of sophisticated and expensive agents, or to deploy a larger number of simple, cost-effective agents.

Appendices

Appendix A

In this appendix the time of inward advancements until the evader region is bounded by a circle with a radius that is smaller or equal to $\frac {r}{n}$ is computed for a swarm that performs the circular sweep process. This time is denoted by $\widetilde {T}_{in}(n)= \sum \limits _{i = 0}^{{N_{n}} - 2} {{T_{i{n_{i}}}}}$. This proof continues the derivation from Section 4. The expression for the term ${\sum \limits _{i = 0}^{{N_{n}} - 2} {{R_{i}}} }$ is derived in Appendix E of [10]. It is given by,

$$ \sum\limits_{i = 0}^{{N_{n}} - 2} {{R_{i}}} = \frac{{{R_{0}} - {c_{2}}{R_{{N_{n}} - 2}} + ({N_{n}} - 2){c_{1}}}}{{1 - {c_{2}}}} $$

(118)

${R_{{N_{n}} - 2}}$ is calculated in Appendix B of [10]. It is given by,

$$ {R_{{N_{n}} - 2}} = \frac{{{c_{1}}}}{{1 - {c_{2}}}} + {c_{2}}^{{N_{n}} - 2}\left( {{R_{0}} - \frac{{{c_{1}}}}{{1 - {c_{2}}}}} \right) $$

(119)

Substituting the coefficients in Eq. 119 yields,

$$ \begin{array}{l} {R_{{N_{n}} - 2}} = \frac{{r{V_{s}}}}{{2\pi {V_{T}}}} + {\left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right)^{{N_{n}} - 2}}\left( {{R_{0}} - \frac{{r{V_{s}}}}{{2\pi {V_{T}}}}} \right) \end{array} $$

(120)

Substituting the coefficients Eq. 11818 yields,

$$ \begin{array}{@{}rcl@{}} \sum\limits_{i = 0}^{{N_{n}} - 2} {{R_{i}}} &=& \frac{{r{V_{s}}n\left( {{V_{s}} + {V_{T}}} \right)}}{{{{\left( {2\pi {V_{T}}} \right)}^{2}}}}\left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right)\\ &&+ {\left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right)^{{N_{n}} - 1}}\left( {{R_{0}} - \frac{{r{V_{s}}}}{{2\pi {V_{T}}}}} \right) \\&&\times\frac{{n\left( {{V_{s}} + {V_{T}}} \right)}}{{2\pi {V_{T}}}}+ \frac{{({N_{n}} - 2)r{V_{s}}}}{{2\pi {V_{T}}}} \\ &&- \frac{{{R_{0}}n\left( {{V_{s}} + {V_{T}}} \right)}}{{2\pi {V_{T}}}} \end{array} $$

(121)

Plugging the expression for ${\sum \limits _{i = 0}^{{N_{n}} - 2} {{R_{i}}} }$ from Eqs. 118 into 27 results in,

$$ \begin{array}{l} \sum\limits_{i = 0}^{{N_{n}} - 2} {{T_{i{n_{i}}}} = } \frac{{\left( {{N_{n}} - 1} \right)r}}{{n\left( {{V_{s}} + {V_{T}}} \right)}} - \frac{{2\pi {V_{T}}}}{{n{V_{s}}\left( {{V_{s}} + {V_{T}}} \right)}}\left( {\frac{{{R_{0}} - {c_{2}}{R_{{N_{n}} - 2}} + ({N_{n}} - 2){c_{1}}}}{{1 - {c_{2}}}}} \right) \end{array} $$

(122)

And substituting the developed coefficients into Eq. 5 yields,

$$ \begin{array}{@{}rcl@{}} \widetilde{T}_{in}(n)&=& \frac{{\left( {{N_{n}} - 1} \right)r}}{{n\left( {{V_{s}} + {V_{T}}} \right)}} - \frac{r}{{2\pi {V_{T}}}}\left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right) \\ &&-{\left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right)^{{N_{n}} - 1}}\left( {\frac{{{R_{0}}}}{{{V_{s}}}} - \frac{r}{{2\pi {V_{T}}}}} \right)\\ && - \frac{{({N_{n}} - 2)r}}{{n\left( {{V_{s}} + {V_{T}}} \right)}} + \frac{{{R_{0}}}}{{{V_{s}}}} \end{array} $$

(123)

Appendix B

In this appendix the time of inward advancements until the evader region is reduced to a circle with a radius that is smaller or equal to $\frac {2r}{n}$ is computed for a swarm that performs the spiral sweep process. This time is denoted by $\widetilde {T}_{in}(n) = \sum \limits _{i = 0}^{{N_{n}} - 2} {{T_{i{n_{i}}}}}$. This proof continues the derivation from section 5. After rearranging terms (87) resolves to,

$$ \begin{array}{l} \sum\limits_{i = 0}^{{N_{n}} - 2} {{T_{i{n_{i}}}}} = \frac{{2r\left( {{N_{n}} - 1} \right) - n\left( {{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}} - 1} \right)\sum\limits_{i = 0}^{{N_{n}} - 2} {\widetilde{R}_{i}} }}{{n\left( {{V_{s}} + {V_{T}}} \right)}} \end{array} $$

(124)

The term $\sum \limits _{i = 0}^{{N_{n}} - 2} {\widetilde {R}_{i}}$ is calculated in appendix E of [10]. It is given by,

$$ \sum\limits_{i = 0}^{{N_{n}} - 2} {\widetilde{R}_{i}} = \frac{{\widetilde{R}_{0} - {c_{2}}\widetilde{R}_{{N_{n}} - 2} + ({N_{n}} - 2){c_{1}}}}{{1 - {c_{2}}}} $$

(125)

Where the term $\widetilde {R}_{{N_{n}} - 2}$ is calculated in Appendix B of [10]. It is given by,

$$ \widetilde{R}_{{N_{n}} - 2} = \frac{{{c_{1}}}}{{1 - {c_{2}}}} + {c_{2}}^{{N_{n}} - 2}\left( {\widetilde{R}_{0} - \frac{{{c_{1}}}}{{1 - {c_{2}}}}} \right) $$

(126)

Substituting the coefficients in Eq. 126 yields,

$$ \begin{array}{l} \widetilde{R}_{{N_{n}} - 2} = - \frac{{2r}}{{n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}} \\ +{\left( {\frac{{{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}}}{{{V_{s}} + {V_{T}}}}} \right)^{{N_{n}} - 2}} \left( {\frac{{\widetilde{{R_{0}}}n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right) + 2r}}{{n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}} \right) \end{array} $$

(127)

Substituting the coefficients in Eq. 125 yields,

$$ \begin{array}{l} \sum\limits_{i = 0}^{{N_{n}} - 2} {{R_{i}}} = \frac{{{R_{0}}\left( {{V_{s}} + {V_{T}}} \right)}}{{{V_{s}}\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}} - \frac{{r\left( {{V_{s}} + {V_{T}}} \right)}}{{n{V_{s}}\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}} \\+ \frac{{2r\left( {{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}{{n{V_{s}}{{\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}^{2}}}}- \left( {{V_{s}} + {V_{T}}} \right){c_{2}}^{{N_{n}} - 1} \\ \times\left( {\frac{{{R_{0}}n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right) + r\left( {1 + {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}{{n{V_{s}}{{\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}^{2}}}}} \right)\\ { - \frac{{r\left( {{V_{s}} + {V_{T}}} \right)\left( {\frac{{{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}}}{{{V_{s}} + {V_{T}}}}} \right)}}{{n{V_{s}}\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}} - \frac{{2r({N_{n}} - 2)}}{{n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}} \end{array} $$

(128)

Plugging the expression for ${\sum \limits _{i = 0}^{{N_{n}} - 2} {{R_{i}}} }$ from Eqs. 128 into 124 results in,

$$ \begin{array}{*{20}{l}} {\widetilde{T}_{in}}(n) = \sum\limits_{i = 0}^{{N_{n}} - 2} {{T_{i{n_{i}}}}} = \frac{{2r\left( {{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}{{n{V_{s}}\left( {{V_{s}} + {V_{T}}} \right)\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}} \\+ \frac{{{R_{0}}}}{{{V_{s}}}} - \frac{r}{{n{V_{s}}}} -{\left( {\frac{{{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}}}{{{V_{s}} + {V_{T}}}}} \right)^{{N_{n}} - 1}} \\ \times\left( {\frac{{{R_{0}}n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right) + r\left( {1 + {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}{{n{V_{s}}\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}} \right) \\ -{\frac{{r\left( {{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}{{n{V_{s}}\left( {{V_{s}} + {V_{T}}} \right)}} + \frac{{2r}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \end{array} $$

(129)

Appendix C

Lemma 1

For a swarm of n agents, where n is even, that performs the circular pincer sweep process, the limit on the time it takes the swarm to clean the entire evader region as ${n \to \infty }$, is given by,

$$ \begin{array}{@{}rcl@{}} \lim\limits_{n \to \infty } T(n) &=& \lim\limits_{n \to \infty } {T_{circular}}(n) + \lim\limits_{n \to \infty } {T_{in}}(n) \\ &=&\frac{{r\left( {{V_{s}} + {V_{T}}} \right)\ln \left( {\frac{{r{V_{s}}}}{{r{V_{s}} - 2\pi {R_{0}}{V_{T}}}}} \right)}}{{2\pi {V_{T}}^{2}}} - \frac{{{R_{0}}}}{{{V_{T}}}} \end{array} $$

(130)

Proof

We have that,

$$ \lim\limits_{n \to \infty } T(n) = \lim\limits_{n \to \infty } {T_{circular}}(n) + \lim\limits_{n \to \infty } {T_{in}}(n) $$

(131)

The circular sweep times, T_circular(n) are given by,

$$ \begin{array}{l} {T_{circular}}(n) = \frac{{r\left( {{V_{s}} + {V_{T}}} \right)}}{{2\pi {V_{T}}^{2}}}\left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right) + \frac{r}{{n{V_{T}}}}\left( {{N_{n}} - 1} \right) \\ - \frac{{{R_{0}}\left( {{V_{s}} + {V_{T}}} \right)}}{{{V_{s}}{V_{T}}}} + \frac{{2\pi r}}{{{n^{2}}{V_{s}}}} \\ +\frac{{n\left( {{V_{s}} + {V_{T}}} \right)}}{{2\pi {V_{T}}}}{\left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right)^{{N_{n}}}}\left( {\frac{{2\pi {R_{0}}}}{{n{V_{s}}}} - \frac{r}{{n{V_{T}}}}} \right) \end{array} $$

(132)

The number of sweeps it takes the sweepers to reduce the evader region to be bounded by a circle with a radius that is less than or equal to $\frac {r}{n}$, N_n, is given by,

$$ {N_{n}} = \left\lceil {\frac{{\ln \left( {\frac{{2\pi {V_{T}}r - r{V_{s}}n}}{{n\left( {2\pi {R_{0}}{V_{T}} - r{V_{s}}} \right)}}} \right)}}{{\ln \left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right)}}} \right\rceil $$

(133)

The limit on N_n as ${n \to \infty }$ yields,

$$ \lim\limits_{n \to \infty } {N_{n}} = \frac{{\ln \left( {\frac{{r{V_{s}}}}{{r{V_{s}} - 2\pi {R_{0}}{V_{T}}}}} \right)}}{{\lim\limits_{n \to \infty } \ln \left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right)}} $$

(134)

We denote by c_up,

$$ {c_{up}} = \ln \left( {\frac{{r{V_{s}}}}{{r{V_{s}} - 2\pi {R_{0}}{V_{T}}}}} \right) $$

(135)

And by c_down,

$$ {c_{down}} = \frac{{2\pi {V_{T}}}}{{{V_{s}} + {V_{T}}}} $$

(136)

The inward advancement times toward the center of the evader region are given by,

$$ \begin{array}{l} {T_{in}}(n) = \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}{\left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right)^{{N_{n}} - 1}}\left( {\frac{{{R_{0}}}}{{{V_{s}}}} - \frac{r}{{2\pi {V_{T}}}}} \right) + \frac{{{R_{0}}}}{{{V_{s}}}} \end{array} $$

(137)

The limit on the first term in (137) can be written as,

$$ \begin{array}{l} \lim\limits_{n \to \infty } {\left( {1 + \frac{{{c_{down}}}}{n}} \right)^{{N_{n}} - 1}} = {\left( {1 + \frac{{{c_{down}}}}{n}} \right)^{\frac{{{c_{up}}}}{{\ln \left( {1 + \frac{{{c_{down}}}}{n}} \right)}}}} = {e^{{c_{up}}}} \end{array} $$

(138)

We therefore have that,

$$ \lim\limits_{n \to \infty } {\left( {1 + \frac{{{c_{down}}}}{n}} \right)^{{N_{n}} - 1}} = \frac{{r{V_{s}}}}{{r{V_{s}} - 2\pi {R_{0}}{V_{T}}}} $$

(139)

And therefore the limit as ${n \to \infty }$ of the first term in T_in(n) yields,

$$ \lim\limits_{n \to \infty } \frac{{\frac{{r{V_{s}}}}{{r{V_{s}} - 2\pi {R_{0}}{V_{T}}}}}}{n} = 0 $$

(140)

Therefore, the limit as ${n \to \infty }$ on the inward advancement times is given by,

$$ \lim\limits_{n \to \infty } {T_{in}}(n) = \frac{{{R_{0}}}}{{{V_{s}}}} $$

(141)

The limit on the numerator of $\frac {{{N_{n}}}}{n}$ yields,

$$ \lim\limits_{n \to \infty } \ln \left( {\frac{{2\pi {V_{T}}r - r{V_{s}}n}}{{n\left( {2\pi {R_{0}}{V_{T}} - r{V_{s}}} \right)}}} \right) = \ln \left( {\frac{{r{V_{s}}}}{{r{V_{s}} - 2\pi {R_{0}}{V_{T}}}}} \right) $$

(142)

The limit on the denominator of $\frac {{{N_{n}}}}{n}$ yields,

$$ \begin{array}{l} \lim\limits_{n \to \infty } n\ln \left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right) \\ =\ln \left( {\lim\limits_{n \to \infty } {{\left( {1 + \frac{{2\pi {V_{T}}}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \right)}^{n}}} \right) = \ln {e^{\frac{{2\pi {V_{T}}}}{{{V_{s}} + {V_{T}}}}}} = \frac{{2\pi {V_{T}}}}{{{V_{s}} + {V_{T}}}} \end{array} $$

(143)

Therefore, the limit on $\frac {{{N_{n}}}}{n}$ yields,

$$ \lim\limits_{n \to \infty } \left( {\frac{{{N_{n}}}}{n}} \right) = \frac{{\left( {{V_{s}} + {V_{T}}} \right)\ln \left( {\frac{{r{V_{s}}}}{{r{V_{s}} - 2\pi {R_{0}}{V_{T}}}}} \right)}}{{2\pi {V_{T}}}} $$

(144)

Substituting the expressions for the limits that are present in the circular sweep times expression we obtain that,

$$ \begin{array}{@{}rcl@{}} \lim\limits_{n \to \infty } {T_{circular}}(n) &=& \frac{{r\left( {{V_{s}} + {V_{T}}} \right)}}{{2\pi {V_{T}}^{2}}}+ \frac{{\left( {{V_{s}} + {V_{T}}} \right)}}{{2\pi {V_{T}}}} \\&&\times\frac{{r{V_{s}}}}{{r{V_{s}} - 2\pi {R_{0}}{V_{T}}}} \left( {\frac{{2\pi {R_{0}}}}{{{V_{s}}}} - \frac{r}{{{V_{T}}}}} \right) \\ &&+ \frac{{r\left( {{V_{s}} + {V_{T}}} \right)\ln \left( {\frac{{r{V_{s}}}}{{r{V_{s}} - 2\pi {R_{0}}{V_{T}}}}} \right)}}{{2\pi {V_{T}}^{2}}}\\ && - \frac{{{R_{0}}\left( {{V_{s}} + {V_{T}}} \right)}}{{{V_{s}}{V_{T}}}} \end{array} $$

(145)

Rearranging terms yields,

$$ \begin{array}{l} \lim\limits_{n \to \infty } {T_{circular}}(n) = \frac{{r\left( {{V_{s}} + {V_{T}}} \right)\ln \left( {\frac{{r{V_{s}}}}{{r{V_{s}} - 2\pi {R_{0}}{V_{T}}}}} \right)}}{{2\pi {V_{T}}^{2}}} - \frac{{{R_{0}}\left( {{V_{s}} + {V_{T}}} \right)}}{{{V_{s}}{V_{T}}}} \end{array} $$

(146)

Therefore, the limit on the time it takes the swarm to clean the entire evader region as ${n \to \infty }$, is calculated by the addition of the terms in Eqs. 141 and 145 and is given by,

$$ \begin{array}{@{}rcl@{}} \lim\limits_{n \to \infty } T(n) &=& \lim\limits_{n \to \infty } {T_{circular}}(n) + \lim\limits_{n \to \infty } {T_{in}}(n) \\ &=& \frac{{r\left( {{V_{s}} + {V_{T}}} \right)\ln \left( {\frac{{r{V_{s}}}}{{r{V_{s}} - 2\pi {R_{0}}{V_{T}}}}} \right)}}{{2\pi {V_{T}}^{2}}} - \frac{{{R_{0}}}}{{{V_{T}}}} \end{array} $$

(147)

□

Appendix D

Lemma 2

For the spiral pincer sweep process employed by n sweepers, the limit on the critical velocity for the confinement task as ${n \to \infty }$, is given by,

$$ \lim\limits_{n \to \infty } {V_{S}} = {V_{T}}\sqrt {{{\left( {\frac{{\pi {R_{0}}}}{r}} \right)}^{2}} + 1} $$

(148)

Proof

We have that,

$$ \lim\limits_{n \to \infty } {V_{S}} = {V_{T}}\sqrt {\frac{{4{\pi^{2}}}}{{{{\left( {\lim\limits_{n \to \infty } n\ln \left( {\frac{{{R_{0}} + \frac{r}{n}}}{{{R_{0}} - \frac{r}{n}}}} \right)} \right)}^{2}}}} + 1} $$

(149)

The limit in the denominator of Eq. 149 is,

$$ \lim\limits_{n \to \infty } \frac{{\ln \left( {\frac{{{R_{0}} + \frac{r}{n}}}{{{R_{0}} - \frac{r}{n}}}} \right)}}{{\frac{1}{n}}} $$

(150)

In order to calculate the limit in Eq. 150 we apply l’hospital’s rule and obtain,

$$ \begin{array}{l} \lim\limits_{n \to \infty } \left( {\frac{{{R_{0}} - \frac{r}{n}}}{{{R_{0}} + \frac{r}{n}}}} \right)\left( {\frac{{\frac{{ - r}}{{{n^{2}}}}\left( {{R_{0}} - \frac{r}{n}} \right) - \left( {{R_{0}} + \frac{r}{n}} \right)\frac{r}{{{n^{2}}}}}}{{{{\left( {{R_{0}} - \frac{r}{n}} \right)}^{2}}}}} \right)\left( { - {n^{2}}} \right) \end{array} $$

(151)

Applying the limit $n \to \infty$ to Eq. 151 yields,

$$ \frac{{r{R_{0}} + r{R_{0}}}}{{{R_{0}}^{2}}} = \frac{{2r}}{{{R_{0}}}} $$

(152)

Plugging the obtained expression for the limit in Eqs. 152 into 149 yields,

$$ \lim\limits_{n \to \infty } {V_{S}} = {V_{T}}\sqrt {\frac{{4{\pi^{2}}{R_{0}}^{2}}}{{4{r^{2}}}} + 1} $$

(153)

And after simplifying terms we have that,

$$ \lim\limits_{n \to \infty } {V_{S}} = {V_{T}}\sqrt {{{\left( {\frac{{\pi {R_{0}}}}{r}} \right)}^{2}} + 1} $$

(154)

□

Appendix E

Lemma 3

For a swarm of n agents, where n is even, that performs the spiral pincer sweep process, the limit on the time it takes the swarm to clean the entire evader region as ${n \to \infty }$, is given by,

$$ \begin{array}{@{}rcl@{}} \lim\limits_{n \to \infty } T(n) &=& \lim\limits_{n \to \infty } {{T_{in}(n)} +\lim\limits_{n \to \infty } {T_{spiral}}(n)} \\ &=& - \frac{{{R_{0}}}}{{{V_{T}}}} + \frac{{r\left( {{V_{s}} + {V_{T}}} \right)\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{\pi {V_{T}}^{2}{V_{s}}}}\\ &&\times\ln \left( {\frac{{r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - \pi {R_{0}}{V_{T}}}}} \right) \end{array} $$

(155)

Proof

We have that,

$$ \lim\limits_{n \to \infty } T(n) = \lim\limits_{n \to \infty } {T_{spiral}}(n) + \lim\limits_{n \to \infty } {T_{in}}(n) $$

(156)

The limit on the inward advancement times is given by,

$$ \lim\limits_{n \to \infty } {T_{in}}(n) = \lim\limits_{n \to \infty } \left( {\widetilde{T}_{in}(n) + {T_{_{in}last}}(n) + \eta {T_{i{n_{f}}}}}(n) \right) $$

(157)

We have that,

$$ \begin{array}{l} {T_{_{in}last}}(n)= - \frac{{2r}}{{n{V_{s}}\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}} + \frac{r}{{n{V_{s}}}} \\ +\frac{{{c_{2}}^{{N_{n}}}}}{{{V_{s}}}}\left( {\frac{{{R_{0}}n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right) + r\left( {1 + {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}{{n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}} \right) \end{array} $$

(158)

We denote by y the following limit,

$$ y = \lim\limits_{n \to \infty } {\left( {\frac{{{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}}}{{{V_{s}} + {V_{T}}}}} \right)^{{N_{n}}}} $$

(159)

We have that,

$$ \lim\limits_{n \to \infty } {N_{n}} = \lim\limits_{n \to \infty } \frac{{\ln \left( {\frac{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - 2\pi {R_{0}}{V_{T}}}}} \right)}}{{\ln \left( {\frac{{{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}}}{{{V_{s}} + {V_{T}}}}} \right)}} = \frac{{{c_{up}}}}{{\ln {c_{2}}}} $$

(160)

Where c_up is given by,

$$ {c_{up}} = \ln \left( {\frac{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - 2\pi {R_{0}}{V_{T}}}}} \right) $$

(161)

and c₂ is given by,

$$ c_{2}={\frac{{{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}}}{{{V_{s}} + {V_{T}}}}} $$

(162)

Therefore, Eq. 159 takes the form of,

$$ y = \lim\limits_{n \to \infty } {c_{2}}^{\frac{{{c_{up}}}}{{\ln {c_{2}}}}} $$

(163)

applying the natural logarithm function to both sides of the equation yields,

$$ \ln y = \ln {c_{2}}^{\frac{{{c_{up}}}}{{\ln {c_{2}}}}} = {c_{up}} $$

(164)

raising both sides of the equation by an exponent and plugging the value for c_up yields,

$$ \begin{array}{l} y = \lim\limits_{n \to \infty } {\left( {\frac{{{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}}}{{{V_{s}} + {V_{T}}}}} \right)^{{N_{n}}}} = \frac{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - 2\pi {R_{0}}{V_{T}}}} \end{array} $$

(165)

Substituting the result from Eqs. 165 to 158 yields,

$$ \begin{array}{@{}rcl@{}} \lim\limits_{n \to \infty } {T_{_{in}last}}(n) &=& \frac{{r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{\pi {V_{s}}{V_{T}}}} \\ &&+ \frac{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - 2\pi {R_{0}}{V_{T}}}}\\ &&\times\frac{{\pi {R_{0}}{V_{T}} - r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{\pi {V_{s}}{V_{T}}}} \end{array} $$

(166)

And after rearranging terms yields that,

$$ \lim\limits_{n \to \infty } {T_{_{in}last}}(n) = 0 $$

(167)

We have that,

$$ \lim\limits_{n \to \infty } {T_{i{n_{f}}}}(n) = \lim\limits_{n \to \infty } \frac{r}{{n{V_{s}}}}\left( {{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}} - 1} \right) = 0 $$

(168)

$\widetilde {T}_{in}(n)$ is given by,

$$ \begin{array}{*{20}{l}} \widetilde{T}_{in}(n) = \sum\limits_{i = 0}^{{N_{n}} - 2} {{T_{i{n_{i}}}}} = \frac{{{R_{0}}}}{{{V_{s}}}} - \frac{r}{{n{V_{s}}}} + \frac{{2r{c_{2}}}}{{n{V_{s}}\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}} \\- { {c_{2}}^{{N_{n}} - 1}\left( {\frac{{{R_{0}}n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right) + r\left( {1 + {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}{{n{V_{s}}\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}} \right)}\\ { - \frac{{r\left( {{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}{{n{V_{s}}\left( {{V_{s}} + {V_{T}}} \right)}} + \frac{{2r}}{{n\left( {{V_{s}} + {V_{T}}} \right)}}} \end{array} $$

(169)

Taking the limit as ${n \to \infty }$ on $\widetilde {T}_{in}(n)$ yields,

$$ \begin{array}{@{}rcl@{}} \lim\limits_{n \to \infty } \widetilde{T}_{in}(n) &=& \sum\limits_{i = 0}^{{N_{n}} - 2} {{T_{i{n_{i}}}}} = \frac{{{R_{0}}}}{{{V_{s}}}} - \frac{{r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{\pi {V_{s}}{V_{T}}}} \\ &&- \frac{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - 2\pi {R_{0}}{V_{T}}}}\\ &&\times\frac{{\pi {R_{0}}{V_{T}} - r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{\pi {V_{s}}{V_{T}}}} \end{array} $$

(170)

Therefore we have that,

$$ \lim\limits_{n \to \infty } \widetilde{T}_{in}(n) = \frac{{{R_{0}}}}{{{V_{s}}}} $$

(171)

Combining the results from Eqs. 167, 168 and 171 the total time of inward advancements is given by,

$$ \lim\limits_{n \to \infty } {T_{in}}(n) = \frac{{{R_{0}}}}{{{V_{s}}}} $$

(172)

The limit on the spiral sweep times is calculated by,

$$ \begin{array}{l} \lim\limits_{n \to \infty } {T_{spiral}}(n) = \lim\limits_{n \to \infty } \left( {\widetilde{T}_{spiral}(n) + {T_{last}}(n) + \eta {T_{l}}}(n) \right) \end{array} $$

(173)

We have that,

$$ \begin{array}{l} \lim\limits_{n \to \infty } \widetilde{T}_{spiral}(n) \\= - \frac{{\left( {{R_{0}} - \frac{r}{n}} \right)\left( {{V_{s}} + {V_{T}}} \right)}}{{{V_{s}}{V_{T}}}} - \frac{{2r\left( {{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}{{n{V_{T}}{V_{s}}\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}} \\- {c_{4}}\frac{{\left( {{R_{0}}n\left( {{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}} - 1} \right) - r\left( {{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}} + 1} \right)} \right)}}{{n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}} + \frac{{2r\left( {{N_{n}} - 1} \right)}}{{n{V_{T}}}} \end{array} $$

(174)

Where c₄ is given by,

$$ {c_{4}} = \frac{{{V_{s}} + {V_{T}}}}{{{V_{T}}{V_{s}}}}{\left( {\frac{{{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}}}{{{V_{s}} + {V_{T}}}}} \right)^{{N_{n}}}} $$

(175)

The limit on $\frac {{{N_{n}}}}{n}$ is,

$$ \begin{array}{l} \lim\limits_{n \to \infty } \frac{{{N_{n}}}}{n} = \lim\limits_{n \to \infty } \frac{{\ln \left( {\frac{{r\left( {3 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}{{{R_{0}}n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right) + r\left( {1 + {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}}} \right)}}{{n\ln \left( {\frac{{{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}}}{{{V_{s}} + {V_{T}}}}} \right)}} \end{array} $$

(176)

We have that the following limit that is present in the numerator of $\frac {{{N_{n}}}}{n}$ yields,

$$ \lim\limits_{n \to \infty } n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right) = - \frac{{2\pi {V_{T}}}}{{\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }} $$

(177)

Therefore, the limit on the numerator of $\frac {{{N_{n}}}}{n}$ yields,

$$ \ln \left( {\frac{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - 2\pi {R_{0}}{V_{T}}}}} \right) $$

(178)

The limit on the denominator of $\frac {{{N_{n}}}}{n}$ is given by,

$$ {\lim\limits_{n \to \infty } \frac{{\ln \left( {\frac{{{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}}}{{{V_{s}} + {V_{T}}}}} \right)}}{{\frac{1}{n}}}} $$

(179)

In order to calculate the limit in Eq. 179 we apply l’hospital’s rule and obtain,

$$ {\lim\limits_{n \to \infty } \frac{{ - \frac{{2\pi {V_{s}}{V_{T}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}}}{{{n^{2}}\left( {{V_{T}} + {V_{s}}{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}{{ - \frac{1}{{{n^{2}}}}}}} $$

(180)

Simplifying the expression in Eq. 180 yields,

$$ {\frac{{2\pi {V_{s}}{V_{T}}}}{{\left( {{V_{s}} + {V_{T}}} \right)\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}} $$

(181)

We therefore have that the limit on $\frac {{{N_{n}}}}{n}$ as ${n \to \infty }$, is given by,

$$ \begin{array}{l} \lim\limits_{n \to \infty } \frac{{{N_{n}}}}{n} = \frac{{\left( {{V_{s}} + {V_{T}}} \right)\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{2\pi {V_{s}}{V_{T}}}}\ln \left( {\frac{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - 2\pi {R_{0}}{V_{T}}}}} \right) \end{array} $$

(182)

Using the result of the limit from (60) we have that,

$$ \begin{array}{l} \lim\limits_{n \to \infty } \frac{{\left( {{R_{0}}n\left( {{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}} - 1} \right) - r\left( {{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}} + 1} \right)} \right)}}{{n\left( {1 - {e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}}} \right)}} \\=\frac{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - 2\pi {R_{0}}{V_{T}}}}{{2\pi {V_{T}}}} \end{array} $$

(183)

And therefore $\lim \limits _{n \to \infty } \widetilde {T}_{spiral}(n)$ is given by,

$$ \begin{array}{@{}rcl@{}} \lim\limits_{n \to \infty } \widetilde{T}_{spiral}(n) &=& - \frac{{{R_{0}}\left( {{V_{s}} + {V_{T}}} \right)}}{{{V_{s}}{V_{T}}}}\\ && + \frac{{r\left( {{V_{T}} + {V_{s}}} \right)\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{\pi {V_{T}}^{2}{V_{s}}}} \\ &&- \frac{{r\left( {{V_{s}} + {V_{T}}} \right)\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{\pi {V_{T}}^{2}{V_{s}}}} \\&&+ \frac{{2r}}{{{V_{T}}}}\frac{{\left( {{V_{s}} + {V_{T}}} \right)\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{2\pi {V_{s}}{V_{T}}}}\\ &&\times\ln\! \left( {\frac{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - 2\pi {R_{0}}{V_{T}}}}} \right) \end{array} $$

(184)

Simplifying expressions yields,

$$ \begin{array}{@{}rcl@{}} \lim\limits_{n \to \infty } \widetilde{T}_{spiral}(n) &=& - \frac{{{R_{0}}\left( {{V_{s}} + {V_{T}}} \right)}}{{{V_{s}}{V_{T}}}} \\ &&+\frac{{r\left( {{V_{s}} + {V_{T}}} \right)\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{\pi {V_{T}}^{2}{V_{s}}}}\\ &&\times\ln\! \left( {\frac{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{2r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - 2\pi {R_{0}}{V_{T}}}}} \right) \end{array} $$

(185)

We have that,

$$ \lim\limits_{n \to \infty } {T_{last}}(n) = \lim\limits_{n \to \infty } \frac{{2\pi r}}{{{n^{2}}{V_{s}}}} = 0 $$

(186)

and that,

$$ \lim\limits_{n \to \infty } {T_{l}}(n) = \lim\limits_{n \to \infty } \frac{{r\left( {{e^{\frac{{2\pi {V_{T}}}}{{n\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}}} - 1} \right)}}{{n{V_{T}}}} = 0 $$

(187)

and therefore,

$$ \lim\limits_{n \to \infty } {T_{spiral}}(n)=\lim\limits_{n \to \infty } \widetilde{T}_{spiral}(n) $$

(188)

Therefore, the limit on the time it takes the swarm to clean the entire evader region as ${n \to \infty }$, is given by,

$$ \begin{array}{@{}rcl@{}} \lim\limits_{n \to \infty } T(n) &=& \lim\limits_{n \to \infty } \left( {{T_{in}(n)} + {T_{spiral}}(n)} \right) \\ &=&- \frac{{{R_{0}}}}{{{V_{T}}}} + \frac{{r\left( {{V_{s}} + {V_{T}}} \right)\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{\pi {V_{T}}^{2}{V_{s}}}}\\ &&\times\ln \left( {\frac{{r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} }}{{r\sqrt {{V_{s}}^{2} - {V_{T}}^{2}} - \pi {R_{0}}{V_{T}}}}} \right) \end{array} $$

(189)

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