Algorithms for sensor-based redistribution of nitrogen fertilizer in winter wheat

Abstract

Several methods were developed for the redistribution of nitrogen (N) fertilizer within fields with winter wheat (Triticum aestivum L.) based on plant and soil sensors, and topographical information. The methods were based on data from nine field experiments in nine different fields for a 3-year period. Each field was divided into 80 or more subplots fertilized with 60, 120, 180 or 240 kg N ha⁻¹. The relationships between plot yield, N application rate, sensor measurements and the interaction between N application and sensor measurements were investigated. Based on the established relations, several sensor-based methods for within-field redistribution of N were developed. It was shown that plant sensors predicted yield at harvest better than soil sensors and topographical indices. The methods based on plant sensors showed that N fertilizer should be moved from areas with low and high sensor measurements to areas with medium values.

The theoretical increase in yield and N uptake, and the reduced variation in grain protein content resulting from the application of the above methods were estimated. However, the estimated increases in crop yield, N-uptake and reduced variation in grain protein content were small.

Appendix A

This appendix describes how to maximize yield (Eqs. 5, 6) given a fixed N application rate (Eq. 7). From Eq. 7 it can be seen that the N _ij ’s depend on each other. Thus, simultaneous maximization of Eqs. 6, 7 is not immediately possible. However, Eq. (6) can be rearranged to

$$ N_{ik_i } = k_i N_i^* - \sum\limits_{j = 1}^{k_i - 1} {N_{ij} } . $$

(13)

Eqs. 5, 6 can now be reduced to

$$ \begin{aligned}{} Y_i^* \; = \sum\limits_{j = 1}^{k_i - 1} {Y_{ij} + Y_{ik_i } } \\ \; = \sum\limits_{j = 1}^{k_i - 1} {\left( {a_i N_{ij}^2 + \left( {b_i + f_i S_{ij} + g_i S_{ij}^2 + h_i S_{ij}^3 } \right)N_{ij} + c_i + d_i S_{ij} + e_i S_{ij}^2 } \right)} + \\ \quad + \left( {a_i \left( {k_i N_i^* - \sum\limits_{j = 1}^{k_i - 1} {N_{ij} } } \right)^2 + \left( {b_i + f_i S_{ik_i } + g_i S_{ik_i }^2 + h_i S_{ik_i }^3 } \right)\left( {k_i N_i^* - \sum\limits_{j = 1}^{k_i - 1} {N_{ij} } } \right) + c_i + d_i S_{ik_i } + e_i S_{ik_i }^2 } \right). \\ \end{aligned} $$

(14)

This equation is simplified by setting $\alpha _{ij} = \left( {b_i + f_i S_{ij} + g_i S_{ij}^2 + h_i S_{ij}^3 } \right)$:

$$ \begin{aligned}{} Y_i^* \; = \;\sum\limits_{j = 1}^{k_i - 1} {\left( {a_i N_{ij}^2 + \alpha _{ij} N_{ij} + c_i + d_i S_{ij} + e_i S_{ij}^2 } \right)} + \\ \quad + \left( {a_i \left( {k_i N_i^* - \sum\limits_{j = 1}^{k_i - 1} {N_{ij} } } \right)^2 + \alpha _{ik_i } \left( {k_i N_i^* - \sum\limits_{j = 1}^{k_i - 1} {N_{ij} } } \right) + c_i + d_i S_{ik_i } + e_i S_{ik_i }^2 } \right). \\ \end{aligned} $$

(15)

In this equation the N _ij ’s are independent and $Y_i^*$ can be maximized by setting the partial derivative of $Y_i^*$ with respect to N _ij to zero:

$$ \frac{{\partial Y_i^* }} {{\partial N_{ij} }} = 0. $$

(16)

Eq. (15) is now differentiated and inserted into Eq. (16):

$$ 2a_i N_{ij} + \alpha _{ij} - 2a_i \left( {k_i N_i^* - \sum\limits_{j = 1}^{k_i - 1} {N_{ij} } } \right) - \alpha _{ik_i } = 0. $$

(17)

This equation is solved with respect to N _ij using Eq. (13).

$$ N_{ij} = \frac{{ - 1}} {{2a_i }}\left( {\alpha _{ij} - 2a_i N_{ik_i } - \alpha _{ik_i } } \right). $$

(18)

Combining this equation with Eq. (13) gives

$$ \begin{aligned}{} N_{ik_i } & = k_i N_i^* - \sum\limits_{j = 1}^{k_i - 1} {N_{ij} } = k_i N_i^* - \sum\limits_{j = 1}^{k_i - 1} {\frac{{ - 1}} {{2a_i }}\left( {\alpha _{ij} - 2a_i N_{ik_i } - \alpha _{ik_i } } \right)} \\ & = k_i N_i^* - (k_i - 1)N_{ik_i } + \sum\limits_{j = 1}^{k_i - 1} {\left( {\frac{{\alpha _{ij} - \alpha _{ik_i } }} {{2a_i }}} \right)} . \\ \end{aligned} $$

(19)

This equation is then solved with respect to $N_{ik_i }$

$$ N_{ik_i } = \frac{1} {{k_i }}\left( {k_i N_i^* + \sum\limits_{j = 1}^{k_i - 1} {\left( {\frac{{\alpha _{ij} - \alpha _{ik_i } }} {{2a_i }}} \right)} } \right) = N_i^* + \sum\limits_{j = 1}^{k_i - 1} {\left( {\frac{{\alpha _{ij} - \alpha _{ik_i } }} {{2a_i k_i }}} \right)} . $$

(20)

Finally, this is inserted into Eq. (18)

$$ \begin{aligned}{} N_{ij} & = N_i^* + \sum\limits_{j = 1}^{k_i - 1} {\left( {\frac{{\alpha _{ij} - \alpha _{ik_i } }} {{2a_i k_i }}} \right)} - \frac{{\alpha _{ij} - \alpha _{ik_i } }} {{2a_i }} \\ & = N_i^* + \frac{1} {{2a_i }}\left( {\sum\limits_{j = 1}^{k_i - 1} {\left( {\frac{{\alpha _{ij} - \alpha _{ik_i } }} {{k_i }}} \right)} - \alpha _{ij} + \alpha _{ik_i } } \right) \\ & = N_i^* + \frac{1} {{2a_i }}\left( { - \alpha _{ij} + \sum\limits_{j = 1}^{k_i } {\frac{{\alpha _{ij} }} {{k_i }}} } \right). \\ \end{aligned} $$

(21)

Expanding $\alpha _{ij}$ gives

$$ N_{ij} = N_i^* + \frac{1} {{2a_i }}\left( { - \left( {b_i + f_i S_{ij} + g_i S_{ij}^2 + h_i S_{ij}^3 } \right) + \sum\limits_{j = 1}^{k_i } {\frac{{\left( {b_i + f_i S_{ij} + g_i S_{ij}^2 + h_i S_{ij}^3 } \right)}} {{k_i }}} } \right). $$

(22)

To obtain a global model of how to redistribute N based on a sensor measurement (S _ij ), it is evident from the above equation that a _i , f _i , g _i and h _i cannot be estimated for each field but must be determined as common factors for all experiments. It is also noted that b _i and the sum are constants within a single field, which implies that Eq. (22) can be rewritten as

$$ N_{ij} = N_i^* + \frac{{f_i S_{ij} + g_i S_{ij}^2 + h_i S_{ij}^3 }} {{ - 2a_i }} + \gamma _i . $$

(23)

Under field conditions $\gamma _i$ can be determined by noting that

$$ \begin{aligned}{} \sum\limits_{J = 1}^{k_i } {N_{ij} } & = \sum\limits_{J = 1}^{k_i } {\left( {N_i^* + \frac{{f_i S_{ij} + g_i S_{ij}^2 + h_i S_{ij}^3 }} {{ - 2a_i }} + \gamma _i } \right)} \\ & = k_i N_i^* + k_i \gamma _i + \sum\limits_{J = 1}^{k_i } {\left( {\frac{{f_i S_{ij} + g_i S_{ij}^2 + h_i S_{ij}^3 }} {{ - 2a_i }}} \right)} , \\ \end{aligned} $$

(24)

and by combining Eqs. 6, 22 the following expression is obtained:

$$ \gamma _i = \frac{{ - 1}} {{k_i }}\sum\limits_{j = 1}^{k_i } {\left( {\frac{{f_i S_{ij} + g_i S_{ij}^2 + h_i S_{ij}^3 }} {{ - 2a_i }}} \right)} . $$

(25)