### Preliminaries

First, the optimization of the grid network is studied by matrix completion, and its network topology is shown in Fig. 1. This network can be widely used in precision agriculture, precision industry, personalized healthcare, and precision medicine, where smart sensor nodes deployed in these applications to detect various physical phenomena. When an event or physical phenomenon occurs, sensor nodes generate alert and alarm to achieve the goal of smart monitoring. For agricultural planting plants, factory production lines, hospital beds, etc., these monitored objects are all regular, so sensor nodes are deployed in an equidistant grid. Therefore, in this paper, we first study the grid network and then generalize it to the general sensor nodes randomly deployed planar network.

We consider a sensor network composed of *N* nodes, and each node generates a data packet to transmit to sink in a round of transmission, and a total of *T* rounds of data are transmitted. The sink will receive *N* × *T* packets, and these data packets can be represented using a matrix *X* (*X* ∈ *R*^{N × T}).

With LRDC, each sensor node will transmit *T* packets to the sink. The minimum amount of data required for the matrix completion technique is called the basic data set, while the excessive part of the data is called backup data set. Therefore, a matrix *Q* is defined as:

$$ {Q}_{i,j}=\left\{\begin{array}{c}\ 0,{X}_{i,j}\in \mathrm{the}\ \mathrm{backup}\ \mathrm{data}\ \mathrm{set}\\ {}1,{X}_{i,j}\in \mathrm{the}\ \mathrm{basic}\ \mathrm{data}\ \mathrm{set}\end{array}\right. $$

where *i*, *j* represents the packet transmitted by the sensor node numbered *i* in the *j*th round.

So the data matrix finally received by sink can be given as:

where ^{′}. ∗ ^{′} represent a scalar product of two matrices.

### LRDC scheme in grid network

The main idea of the LRDC scheme is illustrated in Fig. 2. First, it is necessary to determine which data belongs to basic data set and which data belongs to backup data set before data collection. Then, the basic data transmitted to the sink and the backup data transmitted to its storage location close to the sink in case of retransmission. When the basic data expected to be collected is lost, the sink can send a signal to notify the nodes with the backup data to transmit supplementary data. Therefore, the number of data required for matrix completion technology is satisfied. Finally, the matrix completion technology is used to recover the data that is not transferred to the sink.

The backup data consists of redundant data and the basic data consists of non-redundant data. To determine the locations for storing redundant data, the energy consumption of bottleneck node is need to know. Therefore, in what follows, we will analyze the energy consumption of the bottleneck node.

Energy consumption of bottleneck node: In the LRDC strategy, the redundant packets do not pass through the bottleneck node, so the energy consumption of bottleneck nodes is only related to non-redundant packets.

Therefore, the energy consumption of the bottleneck node can be calculated. According to Eqs. (3), (4), (5), and (6), considering that the transmission radius of each node is *d*, the energy consumption of a packet transmitted by MQAM modulation technique is:

$$ {E}_b=\frac{4}{3}\left(1+\beta \right){N}_f{\sigma}^2\left(M-1\right)\ln \left(\frac{4\left(1-\frac{1}{\sqrt{M}}\right)}{b{P}_e}\right.B{T}_{\mathrm{on}}{G}_d+\left({P}_{\mathrm{circuit}}\right){T}_{\mathrm{on}}+2{P}_{syn}{T}_{\mathrm{start}} $$

(13)

where *P*_{e} is the BER, and the reliability of one hop is as Eq. (14).

$$ \mu ={\left(1-{P}_e\right)}^L $$

(14)

The retransmission mechanism is used to guarantee the success rate of transmission, so the maximum number of retransmissions of a *k*-hop node is given in Theorem 1.

Theorem 1: To guarantee the probability of successful transmission to the destination node is at least *δ*, the maximum number of retransmissions that reach the destination node after *k*-hop is:

$$ {\varsigma}_{k\_\mathit{\max}}=\left\lceil \frac{\log \left(1-\delta \right)}{\log \left(1-{\mu}^k\right)}\right\rceil $$

(15)

Proof: The probability of a node retransmitting *ς*_{k _ max} times still fail is \( {\left(1-{\mu}^k\right)}^{\varsigma_{k\_\max }} \). Therefore, the probability of successful transmission within *ς*_{k _ max} times retransmission is \( 1-{\left(1-{\mu}^k\right)}^{\varsigma_{k\_\max }} \). The probability of successful transmission is required more than *δ*, that is

$$ 1-{\left(1-{\mu}^k\right)}^{\varsigma_{k\_\max }}>\delta $$

Then, we can have

$$ {\varsigma}_{k\_\max }>\frac{\log \left(1-\delta \right)}{\log \left(1-{\mu}^k\right)} $$

Round up the right side of the formula, we can get the formula (15).

Theorem 1 gives the maximum number of retransmissions when the successful transmission rate is at least *δ*. Therefore, the number of expected retransmissions that reach the destination node via *k*-hop can be calculated:

$$ {\varsigma}_k={\sum}_{i=1}^{\varsigma_{k\_\max }}i\bullet {\mu}^k\bullet {\left(1-{\mu}^k\right)}^{i-1} $$

(16)

Figure 3 shows the expectation of retransmissions times. It can be seen that the more hops required to be transferred to destination node, the more expected number of retransmissions. Therefore, nodes near the destination can have reduced number of retransmissions.

Since the number of retransmissions is different for each node, the number of retransmission times can be calculated into the number of packets that each node needs to send, such as:

$$ {m}_{i,j}={x}_{i,j}{\varsigma}_k $$

(17)

where *k* is the number of hops from *S*_{i, j} to sink, and *x*_{i, j} is the number of packets that need to send at *S*_{i, j}.

Therefore, the number of packets forwarded by each node is shown in Theorem 2.

Theorem 2: For grid network, in *T*-round data collection, the amount of data forwarding for each node in the network can be calculated as follows:

$$ \left\{\begin{array}{c}{D}_{a,b}={\sum}_{i=a}^n{\sum}_{j=b}^n\frac{C_{i+j-a-b}^{i-a}}{2^{i+j-a-b}}{m}_{i,j}\ 1<a,b\le n\ \\ {}{D}_{a,b}={\sum}_{k=b}^n\left({m}_{1,k}+\frac{1}{2}{\sum}_{i=2}^n{\sum}_{j=k}^n\frac{C_{i+j-a-b}^{i-a}}{2^{i+j-a-b}}{m}_{i,j}\right)a=1\\ {}{D}_{a,b}={\sum}_{k=a}^n\left({m}_{k,1}+\frac{1}{2}{\sum}_{i=k}^n{\sum}_{j=2}^n\frac{C_{i+j-a-b}^{i-\mathrm{a}}}{2^{i+j-a-b}}{m}_{i,j}\right)b=1\end{array}\right. $$

(18)

where *m*_{i, j} is the amount of data that *S*_{i, j} needs to send in *T*-round data collection.

Proof: First, the node that is not in the first row or first column is analyzed, as shown in Fig. 1. *S*_{a, b} is any node in gird network that is not in the first row or first column. Since the nodes in grid network only transmit data to the left or down, the node that contributes the data of *S*_{a, b} can only be from the upper right of *S*_{a, b}.

The nodes with the same number of hops to *S*_{a, b} belong to the same layer. For any node *S*_{i, j}, the number of hops to reach *S*_{a, b} is *i* + *j* − *a* − *b*, because there are only two options for each hop in grid network. It can only be transmitted down or left, therefore, the node *S*_{i, j} has 2^{i + j − a − b} transmission paths in total.

In 2^{i + j − a − b} transmission paths to *S*_{a, b}, it is easy to know that the packet only needs to be transmitted down *i* − *a* times and to the left *j* − *b* times. Therefore, the path that can reach *S*_{a, b} has \( {C}_{i+j-a-b}^{i-a} \).

Therefore, the probability that the data of *S*_{i, j} can be transmitted to *S*_{a, b} is \( \frac{C_{i+j-a-b}^{i-a}}{2^{i+j-a-b}} \) and the number of data that *S*_{a, b} forward *S*_{i, j} is \( \frac{C_{i+j-a-b}^{i-a}}{2^{i+j-a-b}}{m}_{i,j} \).

Then, since the nodes of the first row can only be transmitted in one direction, the amount of data forwarded by the node *S*_{1, b} in first row is \( \sum \limits_{i=b}^n{m}_{1,i} \). Before that, the number of data forwarding of all nodes except the first row and the first column is calculated. Therefore, considering the second row of nodes, there is \( \frac{1}{2} \) probability that the packets will be transmitted to the first row, and the amount of data forwarded is \( \sum \limits_{k=b}^n{m}_{1,k}+\frac{1}{2}\sum \limits_{i=2}^n\sum \limits_{j=k}^n\frac{C_{i+j-a-b}^{i-a}}{2^{i+j-a-b}}{m}_{i,j} \). Then, Eq. (18) can be obtained.

Theorem 2 gives the amount of data forwarding of all nodes in Grid network, and combined with Eq. (13), the energy consumption of the bottleneck node can be obtained.

The position of the redundant packets: In the above, the energy consumption of the bottleneck node has been obtained. Therefore, the location of redundant packets for each node as in Theorem 3.

Theorem 3: For grid network, using the residual energy of nodes in the network to transmit redundant packets, the storage layer of redundant packets is as follows:

$$ \left\{\begin{array}{c}Y={i}_1+{j}_1-1\\ {}s.t.{D}_{i_1,{j}_1}+\frac{C_{a+b-{i}_1-{j}_1}^{a-{i}_1}}{2^{a+b-{i}_1-{j}_1}}{m}_{a,b}<{D}_{\mathrm{max}}\\ {}{D}_{i_2,{j}_2}+\frac{C_{a+b-{i}_2-{j}_2}^{a-{i}_2}}{2^{a+b-{i}_2-{j}_2}}{m}_{a,b}>{D}_{\mathrm{max}}\mid {i}_2+{j}_2-1<1\\ {}{i}_1+{j}_1={i}_2+{j}_2-1\end{array}\right. $$

(19)

where *m*_{a, b} is the number of packets that *S*_{a, b} needs to send.

Proof: It is easy to know in the grid network that the layer of the node is equal to (*i* + *j* − 1) (The node with the same number of hops are in the same layer). The closer the node is to sink, the lower layer which is located. Therefore, when we require that redundant data be transmitted to nodes as close as possible to the sink, it is required to transmit to the node with the smallest sum of node number.

When the number of non-redundant packets per node is known, according to Eq. (18), the data forwarding amount of each node can be calculated; thus, the maximum amount of data forwarded by the node in network is equal to *D*_{max}. Therefore, it is only necessary to ensure that the amount of data caused by transmitting redundant data does not exceed *D*_{max} and will not reduce the network lifetime.

For the node \( {S}_{i_1,{j}_1} \) in (*i*_{1} + *j*_{1} − 1)^{th} layer, if the redundant data of *S*_{a, b} can be transmitted to *S*_{i, j}, the following conditions will inevitably be satisfied

$$ {D}_{i_1,{j}_1}+\frac{C_{a+b-{i}_1-{j}_1}^{a-{i}_1}}{2^{a+b-{i}_1-{j}_1}}{m}_{a,b}<{D}_{\mathrm{max}} $$

In order to ensure the transmit to the node as close to the sink as possible, it also needs to satisfy the following:

$$ {D}_{i_2,{j}_2}+\frac{C_{a+b-{i}_2-{j}_2}^{a-{i}_2}}{2^{a+b-{i}_2-{j}_2}}{m}_{a,b}>{D}_{\mathrm{max}} $$

Or the next layer of the node has already reached the sink, that is:

Reorganizing the above, we can get (19).

Theorem 3 gives the storage position of redundant data. Therefore, the supplementary data send position in the supplementary data stage also can be known, so the final energy consumption is obtained. However, to obtain the storage location of redundant data, it is necessary to obtain the energy consumption of the bottleneck node. The energy consumption of the bottleneck node is related to the distribution of non-redundant data, so the distribution of non-redundant data needs to be known.

The distribution of redundant data: Since the matrix completion technique cannot recover data matrices in the case of empty rows or empty columns, there exist certain requirements on the number of redundant packets on each sensor.

In ref. [54], the authors have proved that when the collected data obeys Bernoulli distribution, the matrix with missing data can be recovered using matrix completion techniques. We also use the Bernoulli distribution model to determine the number of redundant data packet per node. Theorem 4 gives the number of redundant packets per node under Bernoulli distribution.

Theorem 4: In Bernoulli distribution model, the network collects *T* rounds of data, and the expected number of non-redundant packets that each node needs to send is:

$$ {m}_i=\frac{NT-m}{N} $$

(20)

where *N* is the number of nodes.

Proof: In Bernoulli distribution model, the probability that each packet is a redundant packet is the same, and in the transmission of *T*-round data, the total number of packets that need to be collected is *NT*. Therefore, the probability that a given packet is redundant is \( \frac{1}{NT} \).

Thus, the number of redundant packets per node is \( \frac{1}{NT}\bullet T\bullet m=\frac{m}{N} \).

In the Bernoulli model, the expected number of non-redundant packets for each node is

$$ {m}_i=\frac{NT-m}{N} $$

Theorem 4 gives the number of non-redundant packets per node in the Bernoulli distribution model, which can also get the amount of data forwarded by the bottleneck node.

Since the number of retransmissions of nodes at each layer is different, the data each node sends has different contributions to the data amount of the bottleneck node. In addition, the distribution of redundant data is uniform under the Bernoulli distribution; therefore, a method of unbalanced distribution of redundant data can be used to reduce the data amount of the bottleneck node.

In order to satisfy the requirement for the matrix completion technique to recover the matrix, the probability that empty rows or empty columns are required to be less than the Bernoulli distribution. Then, the minimum amount of non-redundant data per node is given in Theorem 5.

Theorem 5: For a data matrix, when the matrix can be recovered, the minimum amount of data that each sensor node needs to collect is:

$$ x>\frac{mT+m}{NT+m} $$

(21)

Proof: Define event *F* is an event that does not collect data packets for a column, and *x* is the packet that is collected in a sensor node (a row in the matrix).

In the Bernoulli distribution model, the probability that a column has not collected any packets is

$$ {P}_B={\left(1-\frac{m}{NT}\right)}^N={\left(\frac{NT-m}{NT}\right)}^N $$

When there are *x* packets in a row, the probability that a column has no data packets is

$$ P={\left(1-\frac{C_T^{x-1}}{C_T^x}\right)}^N={\left(\frac{T-2x+1}{T-x+1}\right)}^N $$

To make *P* < *P*_{B}, we have

$$ \frac{T-2x+1}{T-x+1}<\frac{NT-m}{NT} $$

Therefore, we can get

$$ x>\frac{mT+m}{NT+m} $$

Theorem 6: In *T*-round data collection, the expected value of each node’s redundant data packets under the Bernoulli distribution model is less than the maximum number of redundant data packets that the matrix can recover.

$$ T-\frac{mT+m}{NT+m}>\frac{NT-m}{N} $$

Proof: According to Theorem 5, under the condition of recovery by matrix completion technology, the minimum number of non-redundant data packets per node is \( \frac{mT+m}{NT+m} \), so the number of redundant packets is

$$ {R}_{\mathrm{max}}=T-\frac{mT+m}{NT+m} $$

According to Theorem 3, the number of redundant data packets per node in Bernoulli distribution model is

$$ {R}_{Ber}=\frac{NT-m}{N} $$

Let \( f(T)={R}_{\mathrm{max}}-{R}_{Ber}=\frac{(NT)^2- mN- NT+m}{N\left( NT+m\right)} \).

Obviously, *N*(*NT* + *m*) must be greater than 0, so only need to prove (*NT*)^{2} − *mN* − *NT* + *m*, and let *g*(*T*) = (*NT*)^{2} − *mN* − *NT* + *m*.

The derivative of *g*(*T*) is

$$ {g}^{\prime }(T)=2{N}^2T-N=N\left(2 NT-1\right) $$

Therefore, *g*(*T*) is a monotonically increasing function in [1, +∞), and it is obviously that *g*(1) is greater than 0, so it can be concluded that *g*(*T*) is greater than 0, so *R*_{max} > *R*_{Ber}.

According to Theorem 5, each sensor node needs to successfully send at least *x* data packets to sink to ensure that the probability of empty rows or empty columns is less than the Bernoulli distribution. Theorem 6 gives proof that the minimum required data in each row for matrix completion is less than the data of each row on Bernoulli distribution model. Therefore, the unbalanced distribution of redundant data packets of nodes can also reduce the maximum energy consumption.

Therefore, a new scheme of unbalanced of redundant data (UORD) distribution is proposed, and in the UORD distribution, the number of redundant data packets per node is determined by the amount of data forwarded by each node, as given in Algorithm 1.

Figure 4 shows the number of non-redundant packets per node with the LRDC scheme. The number of nodes in the network is 100, and the sum of non-redundant packets at the same hop to the sink is counted. It can be clearly seen that with the Bernoulli distribution, the amount of non-redundant data reduced by each node is uniform. However, with the UORD distribution, the non-redundant data for the node with a small number of hops from the sink will not be decreased, but the non-redundant data for the node with a large number of hops from the sink will be greatly reduced. This is because in the grid network, there is only one node that has only one hop from the sink. In other words, the data packets of all the nodes will be forwarded through this node. Therefore, the data load of this node is the largest, and thus, it is better to reduce the non-redundant data of the nodes with more hops because the number of retransmission times of a data packet sent by the nodes is large.

Figure 5 shows the storage location of redundant data. It can be seen that if matrix completion technique is not used, supplementary data only can be sent from itself. In the network with LRDC scheme, redundant data can be transmitted to the node with only two hops from sink when the redundant data meets the Bernoulli distribution. In the UORD distribution, redundant data packets are also transmitted to nodes that are only two hops from the sink. But, since the node with a small hop does not have redundant data packets, these nodes only can send supplementary data from itself.

Since the success rate of the transmission does not reach 100% when the non-redundant data is transmitted, there is still some data loss. Therefore, sink broadcast informs the corresponding node to send data again. However, In LRDC scheme, we can directly transmit the redundant data stored in the nodes near to sink as supplementary data, so the energy consumption of each node under the LRDC scheme is as theorem 7.

Theorem 7: For grid network under the LRDC scheme, when the network transmits *T*-round of data, the energy consumption of each node is:

$$ {E}_{i,j}=\left\{\begin{array}{c}{E}_b\left({D}_{i,j}^{non}+{D}_{i,j}^r+{D}_{i,j}^s\right)\ i+j-1\ge Y\\ {}{E}_b\left({D}_{i,j}^{non}+{D}_{i,j}^s\right)\ i+j-1<Y\end{array}\right. $$

(22)

where *Y* is the storage location of *S*_{i, j} redundant data.

Proof: In the transmission of *T*-round of data, the energy consumption of the LRDC scheme may consist of three parts, including the energy consumption for transmitting non-redundant data packets, the energy consumption for transmitting redundant data packets, and the energy consumption for supplementary data transmission.

It is easy to know that the number of non-redundant packets need to send is *x*_{i, j}. According to formula (17), adopting retransmission mechanism, the number of data packets that each node needs to send is \( {m}_{i,j}^{non} \), so from Theorem 1, the amount of data forwarded by each node is \( {D}_{i,j}^{non} \).

The transmission of redundant data packets also adopts the retransmission mechanism. Therefore, the number of redundant data packets that each node needs to send is \( {m}_{i,j}^r \). However, the transmission of redundant data packets cannot pass through the bottleneck node. Thus, for node with *i* + *j* − 1 ≥ *Y*, Theorem 1 can be used to obtain the amount of data forwarded by each node\( {D}_{i,j}^r \), and for the remaining nodes, the amount of data forwarded is 0.

For supplementary data transmission, because the retransmission mechanism only guarantees the probability of successful transmitted is *δ*, so the probability of loss when transmitting non-redundant packets is (1 − *δ*); therefore, the number of supplementary data packets of nodes need to send is ⌈(1 − *δ*)*x*_{i, j}⌉. The node with redundant data send supplementary data from nodes that store redundant data, and the node without redundant data send supplementary data from itself. Considering node \( {S}_{i^{\prime },{j}^{\prime }} \) has stored redundant data for node *S*_{i, j}, so the number of packets that node needs to send is \( {m}_{i^{\prime },{j}^{\prime}}^s \), then according to Theorem 1 can get the amount of data forwarded by each node is\( {D}_{i,j}^s \).

Reorganizing the above, we can get the energy consumption of the network as below

$$ {E}_{i,j}=\left\{\begin{array}{c}{E}_b\left({D}_{i,j}^{non}+{D}_{i,j}^r+{D}_{i,j}^s\right)\ i+j-1\ge Y\\ {}{E}_b\left({D}_{i,j}^{non}+{D}_{i,j}^s\right)\ i+j-1<Y\end{array}\right. $$

Theorem 7 gives the energy consumption of each node in grid network. Therefore, the maximum energy consumption also can be obtained.

Because in the LRDC scheme, redundant data is transmitted to nodes near the sink, so the number of hops to transmit supplementary data is greatly reduced, thereby reducing the delay of supplementary data transmission. Under this condition, the delay of each node in the grid network is calculated as Theorem 8.

Theorem 8: For grid network, with the LRDC scheme, the delay for each node is:

$$ {\varphi}_{i,j}={m}_{i,j}{\varsigma}_{k_1}+\left(T-{m}_{i,j}\right){\varsigma}_{k_2}+\left\lceil \left(1-\delta \right){m}_{i,j}\right\rceil {\varsigma}_{k_3} $$

(23)

Proof: In grid network, considering the delay of *S*_{i, j}, the delay also can be divided into three parts, including the delay caused by transmitting non-redundant data packets, the delay caused by transmitting redundant data packets, and the delay caused by supplementary data transmission.

For the delay caused by the transmission of non-redundant data packets, it is considered that *S*_{i, j} transmitted to the sink through *k*_{1} hops. Therefore, under the retransmission mechanism, the delay for transmitting this part of data is \( {m}_{i,j}{\varsigma}_{k_1} \).

When transmitting redundant data, it only need to ensure that redundant data is transmitted to the node where it is stored, and the storage location can be obtained by Theorem 3. Therefore, the number of hops from *S*_{i, j} to the storage location, which is considered *k*_{2}, so the delay of this part is \( \left(T-{m}_{i,j}\right){\varsigma}_{k_2} \).

Finally, in the LRDC scheme, the supplemental transmission process requires the transmission of ⌈(1 − *δ*)*m*_{i, j}⌉ supplementary data packets, and transmission can start from the node that stores the redundant data packet. Considering that the number of hops from node for storing redundant packets to sink is *k*_{3}, the delay is \( \left\lceil \left(1-\delta \right)\left(T-{m}_i\right)\right\rceil {\varsigma}_{k_3} \).

Then, we can have

$$ {\varphi}_{i,j}={m}_{i,j}{\varsigma}_{k_1}+\left(T-{m}_{i,j}\right){\varsigma}_{k_2}+\left\lceil \left(1-\delta \right){m}_{i,j}\right\rceil {\varsigma}_{k_3} $$

### LRDC scheme in grid network

In the following, a more common planar random network model [16] will be studied, as shown in Fig. 6.

The sensor nodes in the network are randomly and evenly deployed in a circle with radius *R*, and the transmission radius of each node is considered to be *d*. From Fig. 6, it can be seen that the backup data will only be transmitted to its storage location and will not pass through the bottleneck node. Only when the sink sends a signal to transmit supplementary data, a small amount of backup data will be transmitted to the sink as supplementary data.

Similarly, the energy consumption of the bottleneck node in LRDC scheme needs to be analyzed. By using the shortest path method [16] to transmit data, at a distance of *x* meters from the sink, the number of data forwarding is:

$$ {D}_x=\left(z+1\right)+\frac{z\left(1+z\right)d}{2x} $$

(24)

where *z* is an integer that makes *x* + *zd* just smaller than *R*.

Energy consumption of bottleneck node: According to Eq. (24), when each node sends a data packet, the number of forwarding data of each node of the network can be calculated. However, since the number of data that each node needs to send may not be the same, the number of data that each node forwards needs to be recalculated as Corollary 1.

Corollary 1: In the *T*-round of data collection, the number of data forwarded by the node with the distance x from sink is:

$$ {D}_x={\lambda}_x+\frac{\lambda_{x+d}\left(x+\mathrm{d}\right)+\cdots +{\lambda}_{x+ zd}\left(x+ zd\right)}{x} $$

(25)

where *λ*_{x + kd} represents the data amount of the node in the region between *x* + *kd* and *x* + (*k* + 1)*d*.

Proof: As shown in Fig. 7, considering the node *n*_{x} is located in *S*_{i, k}, and the width of the ring is *d*_{x}. When *d*_{x} is very small, the area where *n*_{x} is located can be approximated as a rectangle. At the same time, due to *d*_{x} is very small, the amount of data that the nodes forward in the same rectangle can be considered the same. Therefore, the area where *n*_{x} is located is

$$ {S}_{i,k}=\alpha x{d}_x $$

And the number of nodes in this area is

$$ {N}_x={S}_{i,k}\rho =\alpha x{d}_x\rho $$

This area needs to transmit *λ*_{x} data in *T*-round data collection, so the amount of data in *T*-round data collection is

$$ {D}_x={\lambda}_x{N}_x $$

It can be seen in Fig. 7 that the area where *n*_{x} is located will forward the data in *S*_{x + r, k}, *S*_{x + 2r, k}, ⋯, *S*_{x + zr, k}(*z* is the maximum value that satisfy *x* + *zr* ≤ *R*), so the total amount of data forwarded is

$$ {\lambda}_x{N}_x+{\lambda}_{x+r}{N}_{x+r}+\cdots +{\lambda}_{x+ zr}{N}_{x+ zr} $$

Therefore, the amount of forwarding data for each fan-shaped ring where *n*_{x} is located is

$$ {D}_x=\left({\lambda}_x{N}_x+{\lambda}_{x+r}{N}_{x+r}+\cdots +{\lambda}_{x+ zr}{N}_{x+ zr}\right)/{N}_x={\lambda}_x+\frac{\lambda_{x+r}\left(x+r\right)+\cdots +{\lambda}_{x+ zr}\left(x+ zr\right)}{x} $$

Corollary 1 gives the amount of data to be forwarded by each node when each node sends different amounts of data in the planar network. Therefore, the energy consumption of the bottleneck node in LRDC scheme can be obtained.

The position of the bottleneck node: Similar to the grid network, considering that nodes with the same hops are in the same layer, the position of redundant data storage also can be obtained.

Theorem 9: In the planar network, for a node with distance *x* from sink, it stores the redundant data in the *Y*th layer, and *Y* is calculated as follows:

$$ \left\{\begin{array}{c}Y=\frac{x- zd-1}{d}+1\\ {}s.t.{D}_{x- zd}+\frac{\lambda_xx}{x- zd}<{D}_{\mathrm{max}}\\ {}{D}_{x-\left(z+1\right)d}+\frac{\lambda_xx}{x-\left(z+1\right)d}>{D}_{\mathrm{max}}\mid x-\left(z+1\right)d\le 0\end{array}\right. $$

(26)

Proof: When the amount of non-redundant data of each node in the network is known, the amount of data *D*_{x} forwarded by each node can be obtained by Corollary 1 so as to obtain the maximum amount of data *D*_{max} to be forwarded.

Considering that node A located at the distance of sink (*x* − *zd*) meter, and node B located at the distance of sink *x* meter. By Corollary 1, it can be obtained that the node A will forward the amount of data transmitted by node B is \( \frac{\lambda_xx}{x- zd} \). Therefore, if the redundant data can be transmitted to the node where the sink distance is (*x* − *zd*), must be satisfied

$$ {D}_{x- zd}+\frac{\lambda_xx}{x- zd}<{D}_{\mathrm{max}} $$

In addition, the redundant data required to be transmitted to the area closest to the sink. Therefore, to satisfy the above conditions, the following should hold:

$$ {D}_{x-\left(z+1\right)d}+\frac{\lambda_xx}{x-\left(z+1\right)d}>{D}_{\mathrm{max}} $$

Or the next hop is arrived at the sink, that is

$$ x-\left(z+1\right)d\le 0 $$

Therefore, the nearest location to which redundant data packets can be transmitted is obtained, so it is possible to calculate how many hops are stored in the node that is away from sink.

Theorem 9 shows the storage location of redundant data. It also needs to know the distribution of redundant data packets to get the storage position of redundant data.

In the planar network, similar to the grid network, according to Theorem 4, redundant data packets can be evenly distributed on each sensor node. Similarly, according to Theorem 5 and Theorem 6, the minimum required non-redundant data for each node can be known and it is less than that in the Bernoulli distribution. Therefore, under UORD distribution, the energy consumption of the bottleneck node is smaller. How to distribute the redundant data packets is shown in Algorithm 2.

Figure 8 shows the sum of the number of non-redundant packets of nodes with the same number of hops in a planar network. It can be seen that the area of each layer gradually increases, and the number of nodes in each layer gradually increases. Therefore, when matrix completion technique is not used, the total number of data packets to be sent per layer increases linearly. Because in the Bernoulli distribution, redundant data packets are uniformly distributed on each node, the total number of data packets to be sent per layer is also linearly increasing. In UORD distribution, redundant data is mainly distributed in high-layer nodes due to the higher number of retransmissions in high-layer node.

Figure 9 shows the location of redundant data packets stored in each node of planar network in LRDC scheme. It can be seen that in a network without using matrix completion technique, each node does not have redundant data, so it is necessary to send data from the original node. In the network using the LRDC scheme, it can be seen that under the Bernoulli distribution, almost all redundant data of nodes are stored at the node that is from one hop from away from sink. Additionally, in the UORD distribution, some nodes still need to send data from itself, because they do not have redundant data packets.

Therefore, the storage location of redundant data has been obtained, and the amount of data forwarded by each node can be obtained, so the energy consumption of each node can also be obtained as in Theorem 10.

Theorem 10: For a planar network, using the LRDC scheme, the energy consumption of each node in the network is

$$ {E}_x=\left\{\begin{array}{c}{E}_b\left({D}_x^{non}+{D}_x^r+{D}_x^s\right)\ \frac{x-1}{d}+1\ge Y\\ {}{E}_b\left({D}_x^{non}+{D}_x^s\right)\ \frac{x-1}{d}+1<Y\end{array}\right. $$

(27)

where *Y* is the location pf redundant packets for the node that distance from sink is *x*.

Proof: Similar to the grid network, the energy consumption of transmission can also be divided into three parts.

For the energy consumption for transmitting non-redundant data, considering that the node’s non-redundant data is *m*_{x}, Similar to Theorem 7, the amount of data that uses the retransmission mechanism can be obtained, and the amount of data forwarded by each node can be obtained according to (27) as \( {D}_x^{non} \).

For redundant data packets, according to Theorem 8, the location to store redundant data packets can be obtained. The amount of redundant data of the node is *T* − *m*_{x}. Similar to the grid network, when the number of layers (\( \frac{x-1}{d}+1 \)) of the node is greater than the number of layers of the bottleneck node, according to Eq. (27), the amount of data forwarded by the node is \( {D}_x^r \), while for the remaining nodes, the amount of data forwarded by the node is 0.

The supplementary data transmission is the same as in grid network. Theorem 8 can obtain the location of redundant data. When the redundant data of nodes is stored by other nodes, the supplementary data of the nodes is sent by other nodes. When the node’s redundant data is still in the original node or there is no redundant data, the supplementary is sent by itself. Therefore, it can also obtain the amount of data \( {D}_x^s \) by each node according to Corollary 1.

Reorganized the above, the energy consumption can be given as:

$$ {E}_x=\left\{\begin{array}{c}{E}_b\left({D}_x^{non}+{D}_x^r+{D}_x^s\right)\ \frac{x-1}{d}+1\ge Y\\ {}{E}_b\left({D}_x^{non}+{D}_x^s\right)\ \frac{x-1}{d}+1<Y\end{array}\right. $$

Theorem 10 gives the energy consumption of various places in the planar network under LRDC scheme. With this theorem, the maximum energy consumption of the network under different distributions of redundant data can be obtained.

Similarly, the LRDC scheme stores redundant data to nodes near the sink, so it also reduces the delay of network and the delay is as given in Theorem 11.

Theorem 11: For a planar network, the delay of each node under the LRDC scheme is as follows:

$$ {\varphi}_x={m}_x{\varsigma}_{k_1}+\left(T-{m}_x\right){\varsigma}_{k_2}+\left\lceil \left(1-\delta \right){m}_x\right\rceil {\varsigma}_{k_3} $$

(28)

Proof: For a planar network, considering the node with the distance x from the sink, similar to grid network, the delay is also divided into three parts.

For the delay caused by the transmission of non-redundant data packets, it is considered that the nodes in the *S*_{x} area need to pass *k*_{1} hops to sink. Therefore, under the retransmission mechanism, the delay for transmitting this part of the data is \( {m}_x{\varsigma}_{k_1} \).

When transmitting redundant data, it is sufficient to ensure that the redundant data arrives at the node where the redundant data is stored, and the storage location can be obtained by Theorem 8. Therefore, the number of hops from the *S*_{x} area to the storage location can be obtained. Considering that is *k*_{2}, this part of the delay is \( \left(T-{m}_x\right){\varsigma}_{k_2} \).

Finally, in the LRDC scheme, the *S*_{x} area needs to be supplemented with ⌈(1 − *δ*)*m*_{x}⌉ data packets in the supplementary transmission, and the transmission can be started directly from the node where redundant data packet is stored. Supposing that the number of hops from the node that stores the redundant data packet to sink is *k*_{3}, the delay is \( \left\lceil \left(1-\delta \right){m}_x\right\rceil {\varsigma}_{k_3} \).