Sensor allocation in diverse environments

Abstract

Sensor coverage varies with location due to factors such as weather, terrain, and obstacles. If a field can be partitioned into zones of homogeneous sensing areas, then the area covered by a random deployment of sensors can be optimized by controlling the number of sensors deployed in each zone. This paper provides formulas to directly calculate the optimal sensor partition in runtime asymptotically equal to the number of zones; to determine the minimum sensor count required to achieve a specific coverage threshold; and to bound the maximum increase in coverage over a strategy oblivious to differences in sensing areas. Results show that this bound is no greater than 13% for a field with two zones. While the analytical solutions assume that each zone is covered independently, sensors are allowed to affect neighboring zones in simulations. Nevertheless, the simulation results support the optimality of the solutions.

Appendices

Appendix 1: Proof of Proposition 5.1

This is evident when i = 1 by setting C ₁ = 1 and $\alpha_1^{\prime}=\alpha_1$. For i ≥ 2 zones, begin with the assumption that the optimal coverage of $Z_{i-1}^{\prime}$ by n sensors is equivalent to Eq. 15, such that the expected coverage under a partition $\beta_i^{\prime}$ is formulated as

$$ \sum_{j=1}^{i-1} \gamma_j \left( 1 - C_{i-1} e^{-\frac{\alpha_{i-1}^{\prime} S}{\sum_{j=1}^{i-1} \gamma_j A} (1 - \beta_i^{\prime}) n} \right) + \gamma_i \left(1 - e^{-\frac{\alpha_i S}{\gamma_i A} \beta_i^{\prime} n} \right) $$

(25)

Taking the second derivative of Eq. 25 with respect to $\beta_i^{\prime}$ shows that it is concave. Therefore, there is one maximum for any value of $\beta_i^{\prime}$. A sketch of the derivation of the optimal value of $\beta_i^{\prime}$ is as follows. Set the first derivative of Eq. 25 with respect to $\beta_i^{\prime}$ equal to 0 and solve for $\beta_i^{\prime}$.

$$ \beta_i^{\prime} = \frac{\alpha_{i-1}^{\prime} \gamma_i}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j} \left( \frac{\sum_{j=1}^{i-1} \gamma_j A}{\alpha_{i-1}^{\prime} S n} \ln \frac{\alpha_i}{\alpha_{i-1}^{\prime} C_{i-1}} + 1 \right) $$

(26)

Substitute this expression into Eq. 25 and reduce to derive an expression for coverage equivalent to Eq. 15

$$ \sum_{j=1}^{i} \gamma_j \left( 1 - \frac{\sum_{j=1}^{i-1} \gamma_j \alpha_i + \alpha_{i-1}^{\prime} \gamma_i}{\sum_{j=1}^{i} \gamma_j \alpha_{i-1}^{\prime}} \cdot \left( \frac{\alpha_{i-1}^{\prime} C_{i-1}}{\alpha_i} \right)^{\frac{\alpha_i \sum_{j=1}^{i-1} \gamma_j}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}} e^{-\frac{\alpha_i \alpha_{i-1}^{\prime} \sum_{j=1}^{i} \gamma_j}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j} \frac{S n}{\sum_{j=1}^{i} \gamma_j A}} \right) $$

(27)

in which $\alpha_i^{\prime}$ and C _i are defined recursively as follows:

$$ \begin{aligned} \alpha_1^{\prime} & = 1 \\ \alpha_{i \ge 2}^{\prime} & = \frac{\alpha_i \alpha_{i-1}^{\prime} \sum_{j=1}^{i} \gamma_j}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j} \\ C_1 & = 1\\ C_{i \ge 2} & = \frac{\sum_{j=1}^{i-1} \gamma_j \alpha_i + \alpha_{i-1}^{\prime} \gamma_i}{\sum_{j=1}^{i} \gamma_j \alpha_{i-1}^{\prime}} \left( \frac{\alpha_{i-1}^{\prime} C_{i-1}}{\alpha_i} \right)^{\frac{\alpha_i \sum_{j=1}^{i-1} \gamma_j}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}} \end{aligned} $$

The derivation of Eqs. 26 and 27 can be found in "Appendix 2". $\square$

Appendix 2: Derivations

1.1 β _opt , Eq. 6 in Sect. 4

Take the derivative of Eq. 5 with respect to β, set the derivative equal to zero, rearrange terms, and eliminate like terms:

$$ \alpha_2 e^{-\frac{\alpha_2 S}{\gamma_2 A} \beta n} = \alpha_1 e^{-\frac{\alpha_1 S}{\gamma_1 A} (1 - \beta) n} $$

Take the log of both sides of the equation and solve for β:

$$ \begin{aligned} \frac{S n}{A} \left( \frac{\alpha_1}{\gamma_1} + \frac{\alpha_2}{\gamma_2} \right) \beta & = \ln \frac{\alpha_2}{\alpha_1} + \frac{\alpha_1 S}{\gamma_1 A} n \\ \frac{\gamma_1 \gamma_2}{\alpha_1 \alpha_2} \frac{S n}{A} \left( \frac{\alpha_1}{\gamma_1} + \frac{\alpha_2}{\gamma_2} \right) \beta & = \frac{\gamma_1 \gamma_2}{\alpha_1 \alpha_2} \left( \ln \frac{\alpha_2}{\alpha_1} + \frac{\alpha_1 S}{\gamma_1 A} n \right)\\ \frac{S n}{A} \left( \frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2} \right) \beta & = \frac{\gamma_2}{\alpha_2} \left( \frac{\gamma_1}{\alpha_1} \ln \frac{\alpha_2}{\alpha_1} + \frac{S}{A} n \right) \\ \beta & = \frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}} \frac{\gamma_2}{\alpha_2} \left( \frac{A}{S n} \frac{\gamma_1}{\alpha_1} \ln \frac{\alpha_2}{\alpha_1} + 1 \right) \\ \end{aligned} $$

Note that α₁ = 1 by definition, so $\ln \frac{\alpha_2}{\alpha_1}=\ln \alpha_2$.

1.2 Two zone optimal coverage, Eq. 8 in Sect. 4

When $n \le -\frac{A}{S} \frac{\gamma_1}{\alpha_1} \ln \alpha_2,$ all sensors are assigned to Z ₁, so the expected coverage is:

$$ \gamma_1 \left(1 - e^{-\frac{\alpha_1 S}{\gamma_1 A} n} \right) $$

To derive an expression for the coverage when $n > -\frac{A}{S} \frac{\gamma_1}{\alpha_1} \ln \alpha_2,$ substitute β _opt into Eq. 5 and simplify

$$ \begin{aligned} & \quad 1 - \gamma_1 e^{-\frac{\alpha_1 S}{\gamma_1 A} \left(1 - \frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}} \, \frac{\gamma_2}{\alpha_2} \,\left( \frac{A}{S n} \frac{\gamma_1}{\alpha_1} \ln \alpha_2 + 1 \right) \right) n} \\ & \qquad - \gamma_2 e^{-\frac{\alpha_2 S}{\gamma_2 A} \, \frac{1}{\frac{\gamma_1}{\alpha_1} + \,\frac{\gamma_2}{\alpha_2}} \, \frac{\gamma_2}{\alpha_2} \left( \frac{A}{S n}\, \frac{\gamma_1}{\alpha_1}\, \ln \alpha_2 + 1 \right) n} \\ & = 1 - \left( \gamma_1 e^{\left( -\frac{\alpha_1}{\gamma_1} \left( \frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2} \right) + 1 + \frac{\alpha_1}{\gamma_1} \,\frac{\gamma_2}{\alpha_2} \right) \frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}}\, \frac{S n}{A}} e^{\frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}} \left( \frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2} \right) \ln \alpha_2} \right. \\ & \qquad \left. + \gamma_2 \right) e^{-\frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}}\, \frac{\gamma_1}{\alpha_1} \ln \alpha_2} e^{-\frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}} \, \frac{S n}{A}} \\ & = 1 - \left( \gamma_1 \alpha_2 + \gamma_2 \right) \alpha_2^{-\frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}} \,\frac{\gamma_1}{\alpha_1}} e^{-\frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}}\, \frac{S n}{A}} \\ & = 1 - \left( \frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2} \right) \left( \alpha_2^\frac{\gamma_2}{\alpha_2} \right)^{\frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}}} e^{\frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}} \,\frac{S n}{A}} \\ \end{aligned} $$

1.3 Two zone minimum sensor count, Eq. 7 in Sect. 4

Just as the formula for C is conditional on whether $n \le -\frac{A}{S} \frac{\gamma_1}{\alpha_1} \ln \alpha_2$ or not, the formula for the minimum number of sensors n to achieve an expected level of coverage C depends on whether or not C is less than or equal to the expected coverage when $n = -\frac{A}{S} \frac{\gamma_1}{\alpha_1} \ln \alpha_2$. The expected coverage for this value of n is:

$$ \gamma_1 \left(1 - e^{\frac{\alpha_1S}{\gamma_1 A}\, \frac{A}{S} \frac{\gamma_1}{\alpha_1} \ln \alpha_2}\right) = \gamma_1 (1 - \alpha_2) $$

To derive a formula for n when $C = \gamma_1 \left(1 - e^{-\frac{\alpha_1 S}{\gamma_1 A} n} \right),$ rearrange the terms, take the log of both sides and solve for n:

$$ n = -\frac{\gamma_1 A}{\alpha_1 S} \ln \frac{\gamma_1 - C}{\gamma_1} $$

To derive a formula for n when $C = 1 - \left( \frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2} \right) \left( \alpha_2^\frac{\gamma_2}{\alpha_2} \right)^{\frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}}} e^{\frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}}\, \frac{S n}{A}}$ rearrange the terms, take the log of both sides and solve for n:

$$ n = \frac{A}{S} \left( \frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2} \right) \left( \ln (1 - C) - \ln \left( \frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2} \right) - \frac{1}{\frac{\gamma_1}{\alpha_1} + \frac{\gamma_2}{\alpha_2}} \frac{\gamma_2}{\alpha_2} \ln \alpha_2 \right) $$

1.4 $\beta_{opt_i}^{\prime},$ Eq. 18 in Sect. 5

Take the derivative of Eq. 25 from "Appendix 1" with respect to $\beta_{opt_i}^{\prime},$ set it equal to 0, rearrange the terms, and eliminate like terms:

$$ \alpha_i e^{-\frac{\alpha_i S}{\gamma_i A} \beta_i^{\prime} n} = \alpha_{i-1}^{\prime} C_{i-1} e^{-\frac{\alpha_{i-1}^{\prime} S}{\sum_{j=1}^{i-1} \gamma_j A} (1 - \beta_i^{\prime}) n} $$

Take the log of both sides and solve for $\beta_i^{\prime}$

$$ \begin{aligned} \frac{\alpha_{i-1}^{\prime} S}{\sum_{j=1}^{i-1} \gamma_j A} \beta_i^{\prime} n + \frac{\alpha_i S}{\gamma_i A} \beta_i^{\prime} n &= \ln \alpha_i - \ln (\alpha_{i-1}^{\prime} C_{i-1}) + \frac{\alpha_{i-1}^{\prime} S}{\sum_{j=1}^{i-1} \gamma_j A} n\\ \frac{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}{\gamma_i \sum_{j=1}^{i-1} \gamma_j} \frac{S n}{A} \beta_i^{\prime} &= \frac{\alpha_{i-1}^{\prime} S n}{\sum_{j=1}^{i-1} \gamma_j A} \left( \frac{\sum_{j=1}^{i-1} \gamma_j A}{\alpha_{i-1}^{\prime} S n} \ln \frac{\alpha_i}{\alpha_{i-1}^{\prime} C_{i-1}} + 1 \right)\\ \beta_i^{\prime} &= \frac{\alpha_{i-1}^{\prime} \gamma_i}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j} \left( \frac{\sum_{j=1}^{i-1} \gamma_j A}{\alpha_{i-1}^{\prime} S n} \ln \frac{\alpha_i}{\alpha_{i-1}^{\prime} C_{i-1}} + 1 \right) \end{aligned} $$

To solve for C _i and $\alpha_i^{\prime},$ first expand $(1 - \beta_i^{\prime})$:

$$ \begin{aligned} & {{\frac{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}} - {\frac{\alpha_{i-1}^{\prime} \gamma_i}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}}\, {\frac{\sum_{j=1}^{i-1} \gamma_jA}{\alpha_{i-1}^{\prime}Sn}} \ln {\frac{\alpha_i}{\alpha_{i-1}^{\prime} C_{i-1}}}} \\ & {\qquad - \frac{\alpha_{i-1}^{\prime} \gamma_i}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}} \\ & {= {\frac{\alpha_i \sum_{j=1}^{i-1} \gamma_j}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}} - {\frac{\alpha_{i-1}^{\prime} \gamma_i}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}}\, {\frac{\sum_{j=1}^{i-1} \gamma_j A}{\alpha_{i-1}^{\prime}Sn}} \ln {\frac{\alpha_i}{\alpha_{i-1}^{\prime} C_{i-1}}}} \end{aligned} $$

Now substitute $1 - \beta_i^{\prime}$ and $\beta_i^{\prime}$ into Eq. 25 and reduce.

$$ \begin{aligned} & \quad \sum_{j=1}^{i} \gamma_j - \sum_{j=1}^{i-1} \gamma_j C_{i-1} e^{-\frac{\alpha_i \alpha_{i-1}^{\prime} S n}{A \left(\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j\right)}} e^{\frac{\alpha_{i-1}^{\prime} \gamma_i}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j} \ln \frac{\alpha_i}{\alpha_{i-1}^{\prime} C_{i-1}}} \\ & - \gamma_i e^{-\frac{\alpha_i \sum_{j=1}^{i-1} \gamma_j}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j} \ln \frac{\alpha_i}{\alpha_{i-1}^{\prime} C_{i-1}}} e^{-\frac{S n}{A}\, \frac{\alpha_i \alpha_{i-1}^{\prime}}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}} \\ & = \sum_{j=1}^{i} \gamma_j - \left( \sum_{j=1}^{i-1} \gamma_j C_{i-1} \frac{\alpha_i}{\alpha_{i-1}^{\prime} C_{i-1}} + \gamma_i \right) \\ & \qquad \cdot \left( \frac{\alpha_{i-1}^{\prime} C_{i-1}}{\alpha_i} \right)^{\frac{\alpha_i \sum_{j=1}^{i-1} \gamma_j}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}} e^{-\frac{\alpha_i \alpha_{i-1}^{\prime} \sum_{j=1}^{i} \gamma_j}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}\, \frac{S n}{\sum_{j=1}^{i} \gamma_j A}} \\ & = \sum_{j=1}^{i} \gamma_j \left( 1 - \frac{\sum_{j=1}^{i-1} \gamma_j \alpha_i + \alpha_{i-1}^{\prime} \gamma_i}{\sum_{j=1}^{i} \gamma_j \alpha_{i-1}^{\prime}} \right. \\ & \qquad \left. \cdot \left( \frac{\alpha_{i-1}^{\prime} C_{i-1}}{\alpha_i} \right)^{\frac{\alpha_i \sum_{j=1}^{i-1} \gamma_j}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j}} e^{-\frac{\alpha_i \alpha_{i-1}^{\prime} \sum_{j=1}^{i} \gamma_j}{\alpha_{i-1}^{\prime} \gamma_i + \alpha_i \sum_{j=1}^{i-1} \gamma_j} \,\frac{S n}{\sum_{j=1}^{i} \gamma_j A}} \right) \end{aligned} $$

Recursive expressions for α _i and C _i can be found in "Appendix 1". Proofs by induction establish that Eqs. 16 and 17 are closed form expressions for α _i and C _i , respectively, which can be substituted in the expression above for $\beta_i^{\prime}$ to yield Eq. 18.

1.5 $\beta_{opt_i},$ Eq. 19 in Sect. 5

This is derived using the method of Lagrange multipliers. A sketch of the derivation is as follows. Set the Lagrange function to

$$ 1 - \sum_{i=1}^k \gamma_i e^{-\frac{\alpha_i S}{\gamma_i A} \beta_i n} - \lambda \left(\sum_{i=1}^k \beta_i - 1\right) $$

(28)

Set the partial derivative with respect to each β _i equal to 0 and solve for each β _i , such that, for all 1 ≤ i ≤ k,

$$ \beta_i = - \frac{\gamma_i A}{\alpha_i S n} \left( \ln \frac{\lambda A}{S n} - \ln \alpha_i \right) $$

(29)

Take the partial derivative of the Lagrangian with respect to λ and set it equal to 0, substitute Eq. 29 for each β _i , and solve for $\ln \frac{\lambda A}{S n}.$

$$ \ln \frac{\lambda A}{S n} = - \frac{1}{\sum_{j=1}^k \frac{\gamma_j}{\alpha_j}} \left( \frac{S n}{A} - \sum_{j=1}^{k} \frac{\gamma_j}{\alpha_j} \ln \alpha_j \right) $$

(30)

Substitute this back into Eq. 29 and simplify to get Eq. 19. Note that k should be set as large as possible such that β _i  > 0, for all 1 ≤ i ≤ k.