Adaptive duty cycle control–based opportunistic routing scheme to reduce delay in cyber physical systems

Abstract

Wireless sensor networks are the foundation of cyber physical systems. In loss and delay sensitive wireless sensor networks, there is much room for improving the performance of opportunistic routing scheme. In this article, an adaptive duty cycle control–based opportunistic routing scheme was put forward to improve the performance of the opportunistic routing scheme. Our main work is as follows: (a) the proposed duty cycle control–based opportunistic routing scheme implements the opportunistic routing in the duty cycle–based wireless sensor networks. Compared with the opportunistic routing scheme in the non-duty cycle mode, the duty cycle control–based opportunistic routing scheme can effectively reduce the energy consumption by 55.64%, although the performance in the delay is degraded. (b) Next, adaptive duty cycle control–based opportunistic routing scheme is proposed to reduce the delay of nodes in duty cycle control–based opportunistic routing. In adaptive duty cycle control–based opportunistic routing scheme, the node’s duty cycle is increased in the area where the energy has remaining, so that the network delay can be reduced without reducing the network lifetime. Compared with duty cycle control–based opportunistic routing scheme, the maximum end-to-end delay of the node in the network can be reduced by 69.83%.

Introduction

Cyber physical systems (CPS)^1,2 is a complex system that integrates social information based on sensing, embedded computing, network communication, and network control.^3–5 As the basis of the whole system, the Internet of Things (IoT) is an important part of CPS.^6–8 There are many studies on IoT,^3,7,9,10 among which wireless sensor networks (WSNs) are a hot spot.^11,12 The performance of WSNs directly affects the performance of CPS.

With the development of micro-processing technology, the processing capacity of sensor nodes is getting stronger, smaller, and cheaper.^13–16 Compared with ten years ago, the IPhone’s processors have more computing and storage capacity than personal computers and are more powerful but cheaper, according to research. ¹⁷ As a result, various sensor-based devices have seen explosive growth.^18–20 Because the sensor network has fast deployment, flexibility, low price, and various sensing devices, it can realize the universal perception of the objective world. Therefore, WSNs are widely used in many fields, such as medical, military, environmental monitoring.^21–29

According to research, ³⁰ the loss ratio of wireless communication is several orders of magnitude different from that of wired communication. The loss ratio of wireless transmission is even as high as 20% or more. ³¹ Therefore, in order to ensure the reliable transmission of WSN data, researchers have conducted a lot of related research.^32–34

The first to use is the send–wait (SW) protocol. ³⁵ In such a protocol, each time the sender sends data to the receiver, it waits for the receiver to return an ACK message, which means the receiver has successfully received the packet. If the sender successfully receives the ACK message within the specified time, it indicates that the data transmission was successful. Otherwise, the data transmission is failed, and the same data are sent again in the next clock cycle. The above process ends until the packet is successfully sent or the number of retransmissions reaches the maximum number of retransmissions. Obviously, in this method, a data packet may need to be retransmitted multiple times because of the high packet loss ratio. On one hand, it consumes a lot of energy; on the other hand, it brings a lot of delay to data transmission.^34,36,37

In order to improve the deficiencies of the above methods, researchers have proposed various methods. For example, researchers have proposed variants of the SW scheme, such as Goback-N (GBN) protocol, selective repeat (SR).^38,39 Kim proposes a send–wait one data multi-ACK (SW-ODMA) automatic repeat-request (ARQ) protocol to improve the reliability of the energy efficiency transmission. ⁴⁰ In SW-ODMA, every time a receiver receives a packet, the sender returns multiple ACKs. This prevents the ACK from being lost during transmission, resulting in a retransmission.

Opportunity routing (OR) scheme^19,41,42 is another routing protocol that takes full advantage of the broadcast function of wireless channels to guarantee transmission reliability. The OR scheme has changed the one-to-one data transmission mode in traditional data transmission. Since wireless communication has broadcast function, when the sender sends packet, all nodes in the range of sender communication can receive data. According to the research, ⁴³ when broadcasting data packets, it is more appropriate to select a node with a receiving rate between 0.1 and 0.9 as a potential receiver. Therefore, when the sender sends data in the OR scheme, multiple nodes receive data at the same time. As long as one receiver receives successfully, the data transfer is successful.

The OR scheme has attracted the attention of researchers,^19,42 and some researches on the routing protocol of WSNs based on the OR scheme have been proposed. According to our research, these OR schemes are mainly based on WSNs whose nodes are always awake. ⁴¹ Since the sensor node is small and powered by batteries, and its battery capacity is limited due to the cost. Therefore, in order to reduce the energy consumption of nodes, extend lifetime as far as possible, a classic practice is to control the duty cycle of sensor nodes.^18,44,45 That is, the sensor node will be awake/sleep periodically. Since the energy consumed by the node in the sleep state is much less than the energy consumed in the awake state, energy saving is achieved. However, WSNs generally adopt asynchronous clock working mode, and asynchronous mode has a disastrous effect on OR scheme. For example, when the sender sends data, there may not be a receiver in the awake state, or there may be few nodes in the awake state. Therefore, at this time, the OR scheme is degraded to a single-sender and single-receiver transmission mode, or the number of receivers is difficult to reach the number required by the OR scheme, making the OR scheme difficult to apply in the duty cycle–based WSNs.

Therefore, this article first proposes a duty cycle control–based opportunistic routing (DCCOR) scheme. Then an improved DCCOR scheme is proposed, namely adaptive duty cycle control–based opportunistic routing (ADCCOR) scheme. The contribution of the main work is as follows:

1. A DCCOR scheme is proposed in this article, which can make the OR scheme suitable for duty cycle–based WSNs and maintain high performance. The main innovation of DCCOR scheme is that when the sender needs to send data, it calculates the number of nodes in its forwarding set that are awake. If the number of relay sets in the awake state reaches a predetermined number, the data broadcast is started. If the number of nodes in the awake state is less than the predetermined threshold, the sender waits and the wake-up receivers continue to remain awake.

2. An ADCCOR scheme is proposed which can further improve the performance of the DCCOR scheme. The main improvement of ADCCOR is based on the energy consumption characteristics of the WSN. The node with the remaining energy at the far sink area increases its duty cycle, so that when the sender wants send data, the number of awakened nodes increases, or it takes less time to wait for the remaining candidate nodes to awaken. That is, the delay is reduced. Therefore, ADCCOR scheme dynamically adjusts the sensor node’s duty cycle according to the energy consumption of the node, so that the delay can be reduced without changing the lifetime of the network.

3. A lot of theoretical analysis and experimental results have confirmed the effectiveness of the proposed scheme. The DCCOR can be applied to duty cycle–based WSNs, and its energy consumption is reduced by 55.64% compared with WSNs without duty cycle. The proposed ADCCOR scheme reduces the delay by 69.83% compare with the DCCOR scheme, without reducing the network lifetime.

The rest of this article is organized as follows: Section “Related work” mainly introduced the research associated with this article. Section “System model and problem statements” mainly introduced the related model used in this article. Section “The design of ADCCOR scheme” gives the DCCOR scheme and ADCCOR scheme of related specifications. In section “Performance analysis and comparison,” we investigated the OR scheme, the DCCOR scheme and the ADCCOR scheme from the angle of energy consumption and delay both analyzed. Section “Conclusion” gives a summary of this article.

Related work

The most important function of WSNs is to sense data and collect data.^46–50 Collecting data is mainly through the multi-hop routing of the self-organized network between sensor nodes to route data from the source node to the sink. Routing strategies have been fully studied on WSNs. ⁴¹ Therefore, there are many studies related to it. The main factors that are considered in the routing scheme are as follows:

• 1. The routing scheme is energy efficient and can keep the network with a high lifetime.^40,51–55

In WSNs, the energy consumption of sensor nodes mainly has the following parts: ⁵⁶ (a) energy consumed due to transmitting data; (b) energy consumed due to receiving data; (c) energy consumed due to idle waiting; (d) energy consumed due to keeping sleep. According to research, up to 70% of the energy consumed in WSNs is used for data operations, of which data sending and receiving consumes the most energy. ¹⁵ There are many studies showing: The energy consumption of the node in idle waiting time is almost equal to the energy consumption of the received data. ¹⁵ Therefore, to save the energy of nodes, the nodes should be in a sleep state as much as possible.

Therefore, the researchers proposed the duty cycle based on WSNs.^44,45 In such a network, the node will periodically undergo awake/sleep rotation. When the node is awake, it will be able to perform data operation, while there will be no communication during sleep.

• 2. Delay issue for data transmission.^44,56–58

Obviously, the routing scheme needs to make the delay of the data reaching the sink as small as possible. In WSNs, there are many factors influencing delay, among which one of the most important factors influencing delay is the reliability of communication channel.^41,59 If the wireless channel is reliable, the packet can be received by the receiver with only one transmission. If the channel quality is not high, it takes many times to be received by the receiver. Obviously, in a network with a bad wireless communication link, its delay will be significantly higher than that of a good communication network. In addition, the delay is also related to the routing scheme.

• 3. The routing scheme needs to ensure the reliability of the transmission.^60,61

The reliability of data transmission refers to the probability that the packet sent by the source node is successfully received by sink after multi-hop routing. Obviously, the reliability of data transmission is an important performance for WSNs. If the data cannot be collected reliably, it will damage the basic functions of the WSN.

There is a much higher loss ratio in wireless communication than in wired communication.^37,40,44 In addition, there may be attacks in WSNs,^51,52 where malicious nodes interfere with data transmission and damage the reliability of data transmission. Therefore, security is also a problem worthy of attention.^7,53

Research on energy efficiency and lifetime of routing

In order to reduce the energy consumption of nodes and prolong the lifetime of the network, researchers have proposed many routing schemes from different perspectives,^54,55 which are summarized as follows:

• 1. A class of routing strategies based on multi-hop routing.

The first one proposed is the shortest routing scheme. This is the most used routing scheme for WSNs, and the network requirements are relatively simple, which is very suitable for this kind of self-organizing WSN. The main idea of the shortest routing scheme is by continuously iterating, find a path for the node to let its data packet reach to the sink with minimum number of hops.

The standard of the shortest routing scheme to select next hop is the number of hops of nodes from sink. In fact, the standard for selecting next hop in a sensor network with GPS can be the distance of the node from the sink. ⁶²

There is no energy consideration in the shortest routing scheme. Therefore, subsequent researchers proposed an energy-balanced routing scheme. ¹⁷ The main idea of the routing scheme for energy balance is as follows: The set of nodes in the sender’s range that are closer to the sink than the sender is called the sender’s forwarding set. Obviously, in the energy balance routing scheme, the node closest to the sink and with the maximum remaining energy is selected as the relay node in the forwarding set. The energy balance routing scheme is often a multi-objective optimization problem. The solution is to give a certain weight to each of the two target values, then combine the weighted average into one target value, and then select the node with the largest target value as the relay node. ⁵⁰

• 2. Clustering routing scheme.

Different from the above routing scheme, this scheme first divides the network into several clusters. Each cluster selects one node as cluster head, while other nodes in the cluster are called cluster members. When data are routed, all member nodes send their own data to the cluster head of the cluster. The cluster head receives data from all nodes in the cluster for data fusion and then passes through multi-hop between Cluster heads or directly sends it to sink. In fact, if you only see the cluster head node as the node to be routed, the two routing strategies above are actually the same.

Research on delay optimization

In addition to energy consumption, there is another issue that people pay attention to in WSN, which is the delay of data transmission.^44,57,58 The delay of data transmission can be divided into two categories. The first type is one-hop delay, which refers to the delay caused by packet transmission in a single hop. The second kind of delay is the end-to-end delay of data transmission, which refers to the time required for data packets to be received by the sink node from generation. Delay is an important indicator in WSNs, and it has received wide attention from academia.

1. In ORW, ¹⁹ the researchers defined a variable EDC that represents the time from when a node starts transmitting a packet to when the packet is successfully received by the relay node. The EDC is added for each hop to calculate the time required for the packet to be received by the sink node. Forwarding according to the path with the smallest accumulated time, the resulting delay is minimal.

2. Liu et al. ⁵⁶ proposed an FFSC strategy. This is also proposed for networks that use opportunistic routing.

When the sender needs to send data, if there is a node in the forwarding set that is awake and closest to the sink, the sender can get the best relay. At this time, the sender can immediately transmit the data with the farthest distance. However, if the node in the awake state is not the node closest to the sink, then the sender has two options: (a) select the node that is closest to the sink from all nodes that is in the awake state as the relay, and immediately transmit the packet. At this time, the one-hop delay is small, but the number of hops of the route will be more, and the end-to-end delay may be larger. (b) Wait for a while, until the node closest to sink in all forwarding nodes wakes up. At this time, the sender needs to wait for a period to send data, and its one-hop delay is large. However, the sender chooses the node closest to sink as the relay, so the number of route hops is relatively small in this way, and the delay can be reduced.

In order to find the best solution, the authors combine the two problems of minimizing the number of transmission hops and minimizing delays, making them a dual-objective optimization problem and obtaining the final optimized routing.

System model and problem statements

Network model

The network is a circular network with sink node as the center. The nodes in the network are evenly distributed in the network with density $\rho$ . ⁶³ They are responsible for collecting the surrounding data and sending it to the sink node through the relay node after collating and merging. ⁶⁴ The sink node is responsible for receiving packets sent by all nodes in the network and responding accordingly. All sensor nodes in the network have the same initial energy and performance in all aspects. For example, the rate at which a node transmits a data packet is the same. For another example, if the transmission power and the transmission radius are the same, the receiving rate of the node is the same.

The network is a type of WSNs based on duty cycle. Nodes in the network work periodically using sleep/awake rotation. The length of the cycle is $T$ (in seconds). Each cycle is further divided into $\tau$ slots. Each node awakes in only one or several slots and sleeps in other slots. When the node is in the awake state, it can perform all data operations and communication. In the sleep state, data operation and communication cannot be performed.

Reliability model

In this article, the reliability model of the node we refer to the model given by Joo. ⁵⁹ Since the transmission of data packets between nodes is unreliable, the data packets are lost with a certain probability during the transmission between nodes. During the transmission of the node, the propagation path loss can be expressed as

$$\mathrm{PL}\left(d\right)=\mathrm{PL}\left(d_0\right)+10n·\mathrm{lo}{\mathrm{g}}_{10}\left(\frac{d}{d_0}\right)+X_{\sigma }$$

(1)

where $d$ is the distance between two nodes; $d_0$ represents the reference distance; in this article, the reference distance is $1\mathrm{m}$ ; $\mathrm{PL}(d_0)$ represents the path loss of the node at the reference position; $n$ represents the path loss index, in this article, $n=2$ ; $X_{\sigma }$ is a random process, which is a value that follows a Gaussian distribution with a mean and standard deviation of zero. In general, $X_{\sigma }$ is a time-dependent random number. In this article, we use a constant to represent the calculation.

Suppose the node’s transmission power is $P_t$ , additive white Gaussian noise (AWGN) power is $P_n$ ; then the signal-to-noise ratio (SNR) of the receiver node can be expressed as

$$\gamma {\left(d\right)}_{\mathrm{dB}}=P_{\mathrm{tdB}}-\mathrm{PL}{\left(d\right)}_{\mathrm{dB}}-{P_n}_{\mathrm{dB}}$$

(2)

$P_n$ of the node can be expressed by the following formula ⁵⁹

$$P_n=(F+1)k{T}_{0}B$$

(3)

where $F$ represents noise factor; $k$ represents Boltzmann constant; $T_0$ represents the environment temperature; and $B$ represents the equivalent bandwidth. When the ambient temperature is 27°C and there is no external interference, the background noise of this article is $P_n=-115\mathrm{dBm}$ .

Then, according to the communication link model, the success rate of transmitting data packets between nodes can be expressed as ⁵⁹

$$\mathrm{pa}{r}_{\left(d\right)}={(1-\frac{1}{2}\mathrm{ex}{p}^{-\frac{\mathrm{SN}{R}_{\left(d\right)}}{2}\times \frac{B_N}{R_D}})}^{8f}$$

(4)

where $d$ is the distance between two nodes; $\mathrm{SN}{R}_{(d)}$ represents the SNR of the node at this distance; $B_N$ represents the noise bandwidth, in this article, $B_N=30\mathrm{kHz}$ ; $R_D$ indicates the rate at which the node transmits data packets, in this article, $R_D=19.2\mathrm{kbps}$ ; $f$ indicates the size of the packet, because it is in bytes, and it needs to be converted into bits in the specific calculation.

Energy consumption model

For the energy consumption model of the node, we can refer to Joo. ⁵⁵ In the process of transmission, we mainly have the following three types of data packets, which are the data packet containing the information collected by the node, the ACK packet, and CTS packet responsible for the communication between the secondary nodes.^65,66

Since the ACK packet and the CTS packet itself are small, and the energy consumed by the node when receiving these packets is much less than the energy of the transmitted packets. When analyzing the ACK packet and the CTS packet, we only calculate the energy consumed to send these packets, regardless of the energy consumed to receive them.

The energy consumed by the node in sending (receiving) data packet is calculated by the following formula

$$E=\mathrm{Pt}$$

(5)

where $P$ represents the power at which the node sends (receives) the data packet, and $t$ represents the time at which the data packet is sent (received). At the same transmission rate, the time at which the node sends (receives) the data packet is

$$t=\frac{S}{R_D}$$

(6)

where $S$ is the size of the packet (in bits) and $R_D$ indicates the rate at which the node transmits data packets.

In order to calculate the total energy consumed by a node in a data transmission, it is necessary to calculate the number of packets transmitted by the node. We use $N$ to indicate the number of packets and use the superscript $S$ , $R$ , $\mathrm{ACK}$ , and $\mathrm{CTS}$ to indicate that the node sends a packet, receives a packet, sends an $\mathrm{ACK}$ packet, and sends a $\mathrm{CTS}$ packet.

The energy consumed by the current node to send the data packet, the energy consumed by the received data packet, the energy consumed to send the $\mathrm{ACK}$ packet, and the energy consumed to transmit the $\mathrm{CTS}$ packet are

$$E^S=P_t·S^S·N^S/R_D$$

(7)

$$E^R=P_r·S^R·N^R/R_D$$

(8)

$$E^{\mathrm{ACK}}=P_t·S^{\mathrm{ACK}}·N^{\mathrm{ACK}}/R_D$$

(9)

$$E^{\mathrm{CTS}}=P_t·S^{\mathrm{CTS}}·N^{\mathrm{CTS}}/R_D$$

(10)

where $P_r$ represents the received power of the node

$$P_r=P_t-\mathrm{PL}\left(d\right)$$

(11)

The total energy consumed by a node to transmit a packet is

$$E^{\mathrm{data}}=E^S+E^R+E^{\mathrm{ACK}}+E^{\mathrm{CTS}}$$

(12)

Problem statement

The objectives of the DCCOR scheme and the ADCCOR scheme optimization proposed in this article are the same, that is, to design the opportunistic routing scheme for the duty cycle WSN. At the same time, the data transmission reliability of the network meets the requirements of the application, the delay of data transmission should be minimized, and the network lifetime should be maximum. Therefore, the optimization objectives of this article can be expressed as follows:

1. The minimum end-to-end delay, which refers to the time difference between the data packet generated by a node and the successful reception of the data packet by the sink node. We use ${\mathfrak{D}}_{\mathrm{e}2\mathrm{e}}^x$ to represent the end-to-end delay of the node with a distance of $x$ meters from sink. In order to calculate the end-to-end delay of the node, we introduce a new variable ${\mathfrak{D}}_{i\_h2h}^x$ , which indicates the delay of hop $i$ during the transmission of the data packet sent by the node with a distance of $x$ meters from sink. Then, the end-to-end delay of the node is equal to the accumulation of the delay of each hop, that is, ${\mathfrak{D}}_{\mathrm{e}2\mathrm{e}}^x=\sum _{i=1}^k{\mathfrak{D}}_{i\_h2h}^x$ . Where $k$ represents the total number of hops. In order to minimize the end-to-end delay of the node, we should try to make the hop count $k$ as small as possible, and at the same time, let ${\mathfrak{D}}_{i\_h2h}^x$ as small as possible. In summary, the minimum end-to-end delay can be expressed as

   $$\left(\min \right){\mathfrak{D}}_{e2e}^x=\sum _{i=1}^{min\left(k\right)}min\left(d_i^x\right)$$

   (13)

2. The longest network lifetime. In this article, we define the lifetime of the network as the death time of the first node in the network that died due to energy exhaustion. We suppose that the initial energy of the node with a distance of $x$ meters from sink is $E^x$ . The energy consumed by the node with a distance of $x$ meters from sink in each round is ${\epsilon }^x$ . Then, the lifetime of the network can be expressed as

   $$\left(\max \right){\mathcal{L}}^x=\left(\min \right)\frac{E^x}{\left(\min \right){\epsilon }^x}$$

   (14)

Suppose that the initial energy of all nodes in the network is the same, in order to make the lifetime of the network as long as possible, the energy consumed by the nodes in each round of data transmission should be as small as possible. The node that consumes the most energy for each round of data transmission directly determines the lifetime of the network.

In summary, we can summarize the optimization goals of this article as

$$\{\begin{array}{c}\left(\min \right){\mathfrak{D}}_{\mathrm{e}2\mathrm{e}}^x=\sum _{i=1}^{\min \left(k\right)}\min \left(d_i^x\right)\hfill \\ \multicolumn{1}{c}{\left(\max \right){\mathcal{L}}^x=\left(\min \right)\frac{E^x}{\left(\min \right){\epsilon }^x}}\\ \multicolumn{1}{c}{\mathrm{s}.\mathrm{t}.{\mathfrak{D}}_{e2e}^x\le {\mathfrak{D}}^G,{\mathcal{L}}^x\ge {\mathcal{L}}^G,x\le l^x}\end{array}$$

(15)

where ${\mathfrak{D}}^G$ , ${\mathcal{P}}^G$ , and ${\mathcal{L}}^G$ represent the thresholds that we stipulated. Only when it is greater than or less than this threshold, we believe that the purpose of optimization is achieved. $l^x$ indicates the distance from the node at the edge of the network to the sink.

The design of ADCCOR scheme

In this chapter, we first introduce the DCCOR scheme. Based on this, the mechanism of automatically adjusting the duty cycle is added, and the ADCCOR scheme is proposed.

DCCOR scheme

The main ideas of the DCCOR scheme are as follows: first, sender selects the forwarding set (FS) based on its transmission power. The specific approach will be mentioned below. After sender determines the FS, it listens to the channel if there is a packet to send. In the DCCOR scheme, each node is required to initiate a hello message when it is switched to the awake state. After the sender, which is listening for channel, receives the hello message, it sends an RTR (Request to Receive) message to the node, such as A, indicating that the sender is ready to broadcast data. At the same time, sender starts counting $R\_\mathrm{count}$ ++, $R\_\mathrm{count}$ indicates the number of receivers currently in the awake state. After node A receives the RTR message, it keeps the awake state. If the $R\_\mathrm{count}$ is less than the expected number $n$ , then sender continues to repeat the above action waiting for the node in the FS to wake up. After the number of nodes in the awake state reaches the expected $n$ , the sender performs a data broadcast.

After the broadcast, the relay node that received the packet returns an ACK message to the sender. If the sender receives an ACK indicating that the packet was successfully sent, sender returns ETS (End to Send) message. After receiving the ETS message, the data are transmitted. If the node that successfully received the packet is not the sink, the node becomes the new sender and repeats the above process until the packet is sent to the sink.

The above process can be described by Algorithm 1.

Algorithm 1. DCCOR scheme.
The action of Sender
1: For each node has data to transmit such as node A Do 2:  node A turns awake and listens to the channel; 3:  determine the ${\mathcal{F}}_A$ of node A; 4:  ${\mathcal{R}}_A$  = Null;  // Before the start of data transmission relay set is NULL 5:   $R\_\mathrm{count}=0$ ; // The number of relay nodes is 0 6:  IF node A receives the hello message from node B   in ${\mathcal{F}}_A$ ; 7:   $R\_\mathrm{count}$  ++; // Add 1 to the number of relay nodes 8:  IF $R\_\mathrm{count}<n$ 9:    node A send RTR to node B; 10:    Wait next hello message; 11:    ${\mathcal{R}}_A$  = ${\mathcal{R}}_A\cup \mathrm{B}$ ; 12:   Else // The number of relay nodes has reached $n$ 13:    ${\mathcal{R}}_A$  = ${\mathcal{R}}_A\cup \mathrm{B}$ ; 14:    broadcast data to ${\mathcal{R}}_A$ // Broadcast data 15:  End if 16:  IF node A receives an ACK message from node C   in ${\mathcal{R}}_A$ 17:  node A broadcast ETS messages 18: End if 19: End for
The action of Receiver
1: IF a node such as B awake Do 2:  node B broadcast hello; // As soon as it wakes up 3:  IF node B receives a RTR from a sender 4:   keep awake; 5:  End if 6:  IF node B receives a data packet 7:   broadcast an ACK; 8:  End if 9:  IF node B receives a ETS 10:   end of receiving process; 11:  End if 11: End

In the algorithm, symbol representation and meaning are as follows:

${\mathcal{F}}_A$ : the forwarding set of node A as a sender;

${\mathcal{R}}_A$ : the relay nodes set of node A. Collection of awake nodes selected by the sender A;

hello: the message that node sends when it wakes up;

RTR: the message that the sender returns to the receiver’s request to receive data;

ACK: A message sent to the sender after the receiver successfully receives the data packet to inform the sender that the receiver has received the packet successfully;

ETS: indicates that the data were sent successfully.

For example, in Figure 1, when node S wants to send a packet, it begins to monitor the state of the nodes in the FS, that is, the states of nodes A, B, C, D, and E. Assume that the initial threshold number of candidate nodes is 3, FS is empty, and the value of the counter is 0.

Figure 1.

An example of a network using the DCCOR scheme.

When a node in the FS, such as A, wakes up, it sends a hello message to node S. After receiving the message, the node S knows that a new node wakes up, first increments the value of the counter by 1, and then determines whether the value of the counter is less than the threshold. In this example, the value of the counter is 1 and less than the threshold (3). In this case, node S sends an RTR message to node A to keep it in the awake state, while node S continues to listen to the status of other nodes in the FS. When a node such as B wakes up, the value of the counter becomes 2, and the threshold is still not reached, so node B also remains in the awake state. Until a node in the FS, such as C wakes up, the value of the counter is incremented by 1, reaching the threshold. At this point, node S can broadcast the packet.

If node A, B, or C successfully receives the data packet, it will send an ACK message to node S. Once the node S receives the ACK message, it proves that the transmission was successful. In particular, the method by which node A determines its candidate set ${\mathcal{F}}_A$ is as follows: first, the node determines its own coordinates $(x,y)$ and transmit power P and determines the range of regions in which its candidate nodes are located by the transmit power and receive rate ranges ( $r^{0.9}-r^{0.1}$ ). In addition, clear the candidate set.

Subsequently, the node broadcasts a $\mathrm{mes}$ message containing the coordinate information of the node. In order to ensure that this information can be successfully received, we use multiple repeat broadcasts. After node A receives the broadcast message of node B, it obtains the coordinate information of node B and then calculates the distance between nodes A and B ( $l_{\mathrm{AB}}$ ) through the Pythagorean Theorem and its own coordinates. If $r_A^{0.9}\le l_{\mathrm{AB}}\le r_A^{0.1}$ , it indicates that node B is a node in the forwarding set of node A. Therefore, we merge node B into ${\mathcal{F}}_A$ .

Through the above steps, the forwarding set of the node A can be determined. The above method can be represented by Algorithm 2.

Algorithm 2. Determine the ${\mathcal{F}}_A$ of node A.

1: For each node in the network Do 2:   Get $(x,y)$ 3:   Set $P$ 4:    $\mathcal{F}=\mathrm{null}$ 5:   Calculate $r^{0.9}$ and $r^{0.1}$ according to equation (16) 6:   broadcast $\mathrm{mes}$ for three times 7: End if 8: For each node in the network such as node A Do 9:   Get $(x_A,y_A)$ from itself 10:   IF node A receives the $\mathrm{me}{s}_B$ from node B 11:    Get $(x_B,y_B)$ from $\mathrm{me}{s}_B$ 12:    calculate $l_{\mathrm{AB}}=\sqrt{{(x_A-x_B)}^2+{(y_A-y_B)}^2}$ 13:    IF $r_A^{0.9}\le l_{\mathrm{AB}}\le r_A^{0.1}$ 14:      ${\mathcal{F}}_A={\mathcal{F}}_A{\cup }^B$ 15:    End if 16:   End if 17: End for

In the algorithm, symbol representation and meaning are as follows:

• $(x,y)$ : the coordinates of the node;

• $P$ : the transmit power of the node;

• $\mathcal{F}$ : the forwarding set of the node;

• $r^{0.9}$ and $r^{0.1}$ : the minimum and maximum transmission radius of the node under opportunistic routing;

• $\mathrm{mes}$ : a packet broadcast by a node that contains its own coordinate information.

Theorem 1

After the transmission power of the node is selected, the range of the transmission radius of the node can be calculated by the following formula

$$r=d_0{10}^{\frac{P_t-P_n+\frac{2R_D}{B_N}{\log }_e(1-\sqrt[8f]{\mathrm{pa}{r}_{\left(r\right)}})-\mathrm{PL}\left(d_0\right)-X_{\sigma }}{10n}}$$

(16)

Proof

Suppose that the receiving rate of the node whose distance to node S is $r$ is known as $\mathrm{pa}{r}_{(r)}$ , according to equation (4), the $\mathrm{SN}{R}_{(r)}$ of the node S can be calculated as

$$\mathrm{SN}{R}_{\left(r\right)}=-\frac{2R_D}{B_N}{\log }_e(1-\sqrt[8f]{\mathrm{pa}{r}_{\left(r\right)}})$$

(17)

Here, $R_D$ is the data rate, $f$ is the size of a data packet (in bytes), and $B_N$ is the noise bandwidth.

According to equation (2), the path loss of the node is

$$\mathrm{PL}\left(r\right)=P_t-P_n-\mathrm{SN}{R}_{\left(r\right)}$$

(18)

where $P_t$ is the transmission power of node S and $P_n$ is an AWGN power.

Finally, based on equation (1)

$$r=d_0{10}^{\frac{\mathrm{PL}\left(r\right)-\mathrm{PL}\left(d_0\right)-X_{\sigma }}{10n}}$$

(19)

Here, $d_0$ is a reference distance and $\mathrm{PL}(d_0)$ represents the path loss at that distance, $n$ is the path-loss exponent, and $X_{\sigma }$ is a constant random variable to model a specific link with time. Substituting equations (17) and (18) into equation (19)

$$r=d_0{10}^{\frac{P_t-P_n+\frac{2R_D}{B_N}{\log }_e(1-\sqrt[8f]{\mathrm{pa}{r}_{\left(r\right)}})-\mathrm{PL}\left(d_0\right)-X_{\sigma }}{10n}}$$

(20)

In this article, we specify that the packet size is $128\mathrm{bytes}$ , $R_D=19.2\mathrm{kbps}$ , $B_N=30\mathrm{kHz}$ , $P_n=-115\mathrm{dBm}$ , reference distance $d_0=1\mathrm{m}$ , and path loss exponent $n=2$ . Let $\mathrm{pa}{r}_{(r)}$ equal to 0.9 and 0.1, respectively, and you can get the range of the transmission radius of the node. In order to facilitate the calculation, we obtain the final transmission radius by taking the weighted average method.

Theorem 2

According to the range of the transmission radius, the success rate of transmission at different transmission radii, and the number of nodes under the corresponding transmission radius, the average weighted transmission radius of the node is

$$r=\sum _{i=\lceil r_2\rceil }^{\lfloor r_1\rfloor }(\lceil r_2\rceil +i+1)\frac{\mathrm{pa}{r}_i{S}_i^{A\_\mathrm{AREA}}}{\sum _{i=\lceil r_2\rceil }^{\lfloor r_1\rfloor }\mathrm{pa}{r}_i{S}_i^{A\_\mathrm{AREA}}}$$

(21)

where $\lfloor r_1\rfloor$ and $\lceil r_2\rceil$ represent the farthest and closest distance to node S in the transitional area, round down and up, respectively. The values of $r_1$ and $r_2$ can be calculated by nesting $\mathrm{pa}{r}_{\left(r\right)}$ equal to 0.1 and 0.9, respectively, into equation (16). We suppose that all nodes that have a distance to node S from $\lceil r_2\rceil +i$ to $\lceil r_2\rceil +i+1$ meters have the same distance from node S, that is $\lceil r_2\rceil +i+1$ meters. In addition, $S_i^{A\_\mathrm{AREA}}$ represents the area of the ring, and the ring has an inside radius of $\lceil r_2\rceil +i$ meters and a outside radius of $\lceil r_2\rceil +i+1$ meters, the receiving rate of the nodes on the ring is $\mathrm{pa}{r}_i$ .

Proof

Since the probability of successful packet transmission and the number of receiving nodes under different transmission radius are different, it is necessary to weight these values to average.

We suppose that there is a node $A_i$ , which is represented all relay nodes that have a distance to node S from $\lceil r_2\rceil +i$ to $\lceil r_2\rceil +i+1$ meters. The area where these nodes are located is called $A\_\mathrm{ARE}{A}_i$ . Since the nodes in the network are evenly distributed, the number of nodes in the area $A\_\mathrm{ARE}{A}_i$ is positively correlated with the area of the region. We calculate the area of $A\_\mathrm{ARE}{A}_i$ by the following method.

In Figure 2, the distance between the node S and sink is $d$ , with the node S as the center, we get the circle 1 with $r_a$ as the radius. Taking the node S as the center, drawing the circle with $r_b$ as the radius to get the circle 2. Taking sink as the center, draw the circle with the distance between sink and node S as the radius, intersecting the circle 1 and circle 2 at points A, B, C, and D. The overlap between the ring formed by the circle1 and the circle 2 and the sector area OASB is the position of the node in the forwarding set of the node S. In order to facilitate the calculation of the area of the overlap, we made the following auxiliary lines and points in Figure 2. Point E and point F are located on the line connecting the nodes S and sink and intersect at circle 2 and circle 1, respectively; point G located on line BS; respectively connect OA, OS, OB, AS, and DS. OH perpendicular to AS, H is a vertical point, $\angle \mathrm{AOH}={\beta }_1$ .

Figure 2.

Method for calculating the area of a special area.

According to the trigonometric formula

$${\mathrm{\beta }}_1=\mathrm{asin}\frac{\mathrm{AH}}{\mathrm{OA}}=\mathrm{asin}\frac{r_a}{2d}$$

(22)

The acreage of the fan-shaped AOS is

$$S_{S\_\mathrm{AOS}}={\mathrm{\beta }}_1{d}^2$$

(23)

which is

$$S_{S\_\mathrm{AOS}}=\left(\mathrm{asin}\frac{r_a}{2d}\right)d^2$$

(24)

The acreage of △AOS is

$$S_{\_\mathrm{AOS}}=\frac{1}{2}{d}^2\sin \left(2{\beta }_1\right)=\frac{1}{2}{d}^2\sin \left(2\mathrm{asin}\frac{r_a}{2d}\right)$$

(25)

Therefore, the acreage of the irregular area ACS is

$$S_{\mathrm{ACS}}=S_{S\_\mathrm{AOS}}-S_{\mathrm{AOS}}$$

(26)

The fan-shaped ASF center angle is

$$\mathrm{\alpha }=\frac{\pi -2{\mathrm{\beta }}_1}{2}=\frac{\pi -2\mathrm{asin}\frac{r_a}{2d}}{2}$$

(27)

The acreage of the fan-shaped ASF is

$$S_{s\_\mathrm{ASF}}=\frac{\alpha {r_a}^2}{2}=\frac{\pi -2\mathrm{asin}\frac{r_a}{2d}}{4}{r_a}^2$$

(28)

The acreage of the irregular area AFSC is

$$S_{\mathrm{AFSC}}=S_{s\_\mathrm{ASF}}+(S_{S_{\mathrm{AOS}}}-S_{\mathrm{AOS}})$$

(29)

which is

$$\begin{array}{c}\multicolumn{1}{c}{S_{\mathrm{AFSC}}=\frac{\pi -2\mathrm{asin}\frac{r_a}{2d}}{4}{r_a}^2}\\ \multicolumn{1}{c}{+\left(\mathrm{asin}\frac{r_a}{2d}\right)d^2-\frac{r_a}{2}\sqrt{d^2-\frac{{r_a}^2}{4}}}\end{array}$$

(30)

Similarly, the acreage of the irregular area DESG is

$$\begin{array}{c}\multicolumn{1}{c}{S_{\mathrm{DESG}}=\frac{\pi -2\mathrm{asin}\frac{r_b}{2d}}{4}{r_b}^2}\\ \multicolumn{1}{c}{+\left(\mathrm{asin}\frac{r_b}{2d}\right)d^2-\frac{r_b}{2}\sqrt{d^2-\frac{{r_b}^2}{4}}}\end{array}$$

(31)

According to equations (30) and (31), we can give the irregular area of the overlap

$$\begin{array}{c}\multicolumn{1}{c}{S_{\mathrm{fs}}=2(S_{\mathrm{AFSC}}-S_{\mathrm{DESG}})}\\ \multicolumn{1}{c}{=\frac{\pi ({r_a}^2-{r_b}^2)}{2}+\left(\mathrm{asin}\frac{r_b}{2d}\right){r_b}^2-\left(\mathrm{asin}\frac{r_a}{2d}\right){r_a}^2}\\ \multicolumn{1}{c}{+2(\mathrm{asin}\frac{r_a}{2d}-\mathrm{asin}\frac{r_b}{2d})d^2+r_b\sqrt{d^2-\frac{{r_b}^2}{4}}}\\ \multicolumn{1}{c}{-r_a\sqrt{d^2-\frac{{r_a}^2}{4}}}\end{array}$$

(32)

When $r_a=\lceil r_2\rceil +i+1$ , $r_b=\lceil r_2\rceil +i$ , the area of the region $A\_\mathrm{ARE}{A}_i$ is

$$\begin{array}{c}\multicolumn{1}{c}{S_i^{A\_\mathrm{AREA}}=\frac{\pi ({(\lceil r_2\rceil +i+1)}^2-{(\lceil r_2\rceil +i)}^2)}{2}}\\ \multicolumn{1}{c}{+\left(\mathrm{asin}\frac{\lceil r_2\rceil +i}{2d}\right){(\lceil r_2\rceil +i)}^2}\\ \multicolumn{1}{c}{-\left(\mathrm{asin}\frac{\lceil r_2\rceil +i+1}{2d}\right){(\lceil r_2\rceil +i+1)}^2}\\ \multicolumn{1}{c}{+2(\mathrm{asin}\frac{\lceil r_2\rceil +i+1}{2d}-\mathrm{asin}\frac{\lceil r_2\rceil +i}{2d})d^2}\\ \multicolumn{1}{c}{+(\lceil r_2\rceil +i)\sqrt{d^2-\frac{{(\lceil r_2\rceil +i)}^2}{4}}}\\ \multicolumn{1}{c}{-(\lceil r_2\rceil +i+1)\sqrt{d^2-\frac{{(\lceil r_2\rceil +i+1)}^2}{4}}}\end{array}$$

(33)

Simplified to

$$\begin{array}{c}\multicolumn{1}{c}{S_i^{A\_\mathrm{AREA}}=\pi (2r+2i+\frac{1}{2})}\\ \multicolumn{1}{c}{+\left(\mathrm{asin}\frac{\lceil r_2\rceil +i}{2d}\right){(\lceil r_2\rceil +i)}^2}\\ \multicolumn{1}{c}{-\left(\mathrm{asin}\frac{\lceil r_2\rceil +i+1}{2d}\right){(\lceil r_2\rceil +i+1)}^2}\\ \multicolumn{1}{c}{+2(\mathrm{asin}\frac{\lceil r_2\rceil +i+1}{2d}-\mathrm{asin}\frac{\lceil r_2\rceil +i}{2d})d^2}\\ \multicolumn{1}{c}{+(\lceil r_2\rceil +i)\sqrt{d^2-\frac{{(\lceil r_2\rceil +i)}^2}{4}}}\\ \multicolumn{1}{c}{-(\lceil r_2\rceil +i+1)\sqrt{d^2-\frac{{(\lceil r_2\rceil +i+1)}^2}{4}}}\end{array}$$

(34)

Therefore, the transmission radius $r$ of the node S can be expressed by the following weighted average

$$r=\sum _{i=\lceil r_2\rceil }^{\lfloor r_1\rfloor }{k}_i\frac{\mathrm{pa}{r}_i{S}_i^{A\_\mathrm{AREA}}}{\sum _{i=\lceil r_2\rceil }^{\lfloor r_1\rfloor }\mathrm{pa}{r}_i{S}_i^{A\_\mathrm{AREA}}}$$

(35)

Theorem 3

The size of forwarding set of node S can be expressed as

$$\begin{array}{c}\multicolumn{1}{c}{N_S=\rho \{\frac{\pi ({r_1}^2-{r_2}^2)}{2}+\left(\mathrm{asin}\frac{r_2}{2d}\right){r_2}^2}\\ \multicolumn{1}{c}{-\left(\mathrm{asin}\frac{r_1}{2d}\right){r_1}^2+2(\mathrm{asin}\frac{r_1}{2d}-\mathrm{asin}\frac{r_2}{2d})d^2}\\ \multicolumn{1}{c}{+r_2\sqrt{d^2-\frac{{r_2}^2}{4}}-r_1\sqrt{d^2-\frac{{r_1}^2}{4}}\}}\end{array}$$

(36)

where

$${\mathrm{\beta }}_1=\mathrm{asin}\frac{r_1}{2d}$$

(37)

$${\mathrm{\beta }}_2=\mathrm{asin}\frac{r_2}{2d}$$

(38)

Proof

Assuming $r_a=r_1$ , $r_b=r_2$ , then $S_{\mathrm{fs}}$ represents the area of the area where the node candidate set is located. The number of nodes in this part is the size of the forwarding set of node S

$$N_S=\rho S_{\mathrm{fs}}$$

(39)

That is

$$\begin{array}{c}\multicolumn{1}{c}{N_S=\rho \{\frac{\pi ({r_1}^2-{r_2}^2)}{2}+\left(\mathrm{asin}\frac{r_2}{2d}\right){r_2}^2}\\ \multicolumn{1}{c}{-\left(\mathrm{asin}\frac{r_1}{2d}\right){r_1}^2+2(\mathrm{asin}\frac{r_1}{2d}-\mathrm{asin}\frac{r_2}{2d})d^2}\\ \multicolumn{1}{c}{+r_2\sqrt{d^2-\frac{{r_2}^2}{4}}-r_1\sqrt{d^2-\frac{{r_1}^2}{4}}\}}\end{array}$$

(40)

Theorem 4

Assuming that the duty length is $\tau$ , then the duty cycle of each node is $1/\tau$ . The size of forwarding set of node S is $N$ , and the threshold of the candidate(relay) node is $n$ . Then the delay of node S can be expressed as

$$\begin{array}{c}\multicolumn{1}{c}{D_s=(\tau -1)\left(\sum _{m=0}^{n-1}{C}_N^m{P_{u\_(\tau -2)}^1}^m{P_{a\_(\tau -2)}^1}^{N-m}\right)}\\ \multicolumn{1}{c}{+\sum _{k=1}^{\tau -2}\left[k\sum _{m=0}^{n-1}\left(C_N^m{P}_{u\_(k-1)}^m\sum _{j=n-m}^{N-m}{C}_{N-m}^j{P_k^1}^j{P_{a\_k}^1}^{N-m-j}\right)\right]}\end{array}$$

(41)

Proof

Obviously, the delay of node S mainly depends on the time when the nth candidate node wakes up.

Let us first consider one of the $N$ nodes, and since the nodes do not interfere with each other, the time of waking is random. Therefore, the node has the possibility of waking up with $1/\tau$ in each slot, so the probability that the node wakes up in slot $k$ is

$$P_k^1={(1-\frac{1}{\tau })}^k\frac{1}{\tau }$$

(42)

where $k=0,1,\dots ,\tau -2$ .

Since the node must wake up in one of the $\tau$ slots, when $k=\tau -1$ , there are

$$P_{\tau -1}^1={(1-\frac{1}{\tau })}^{\tau -1}$$

(43)

Then, at slot $k$ ( $k=0,1,\dots ,\tau -2,\tau -1$ ), the probability that this node has woken up is

$$P_{u\_k}^1=\sum _{i=0}^k{P}_i^1$$

(44)

The probability that a node wakes up after slot $k$ is

$$P_{a\_k}^1=1-P_{u\_k}^1$$

(45)

When calculating the delay of a single hop of a node, we discuss it in three cases.

Assuming that node $n$ wakes up in the first slot, we can divide $N$ nodes into the following two parts, as shown in Figure 3.

Figure 3.

Node $n$ wakes up in the first slot.

In the first part, there are at least $n$ nodes wake up in the first slot. The remaining nodes belong to the second part; they wake up in the next $\tau -1$ slot. Therefore, the probability that the node $n$ wakes up in the first slot is

$$P_0^n=\sum _{m=n}^N{C}_N^m{P}_{u\_0}^m{P}_{a\_0}^{N-m}$$

(46)

where $C_N^m$ represents the number of combinations, calculated as follows

$$C_N^m=\frac{N!}{(N-m)!m!}$$

(47)

where “!” indicates the factorial.

Assuming that node $n$ wakes up in the last slot, we can divide the $N$ nodes into the following two parts, as shown in Figure 4.

Figure 4.

Node $n$ wakes up in the last slot.

In the first part, at most $n-1$ nodes wake up, and the remaining nodes will wake up in the last slot. Its probability is

$$P_{\tau -1}^n=\sum _{m=0}^{n-1}{C}_N^m{P}_{u\_(\tau -2)}^m{P}_{a\_(\tau -2)}^{1^{N-m}}$$

(48)

Suppose that node $n$ is not awake in the first slot or the last slot, but wakes up at slot $k$ in the middle. We can divide $N$ nodes into three parts, as shown in Figure 5.

Figure 5.

Node $n$ wakes up at slot $k$ .

In part 1, there are $m$ ( $m<n$ ) nodes waking up in any slot from the first slot to slot $k-1$ . In part 2, there are at least $n-m$ nodes waking up at slot $k$ , and the remaining nodes belong to part 3, waking up at slot $k$ or later. Its probability is

$$P_k^n=\sum _{m=0}^{n-1}\left(C_N^m{P}_{u\_(k-1)}^m\sum _{j=n-m}^{N-m}{C}_{N-m}^j{P_k^1}^j{P_{a\_k}^1}^{N-m-j}\right)$$

(49)

In summary, the delay of node S is

$$D_s=0*P_0^n+\sum _{k=1}^{\tau -2}k*P_k^n+(\tau -1)*P_{\tau -1}^n$$

(50)

Substituting the above equations (46), (48), and (49) into equation (50)

$$\begin{array}{c}\multicolumn{1}{c}{D_s=(\tau -1)\left(\sum _{m=0}^{n-1}{C}_N^m{P_{u\_(\tau -2)}^1}^m{P_{a\_(\tau -2)}^1}^{N-m}\right)}\\ \multicolumn{1}{c}{+\sum _{k=1}^{\tau -2}\left[k\sum _{m=0}^{n-1}\left(C_N^m{P}_{u\_(k-1)}^m\sum _{j=n-m}^{N-m}{C}_{N-m}^j{P_k^1}^j{P_{a\_k}^1}^{N-m-j}\right)\right]}\end{array}$$

(51)

The energy consumed by the node can be divided into the following two parts. One part is the energy consumption generated by the node when transmitting data, the other part is the energy consumed when the node keeps listening. We first calculate the energy consumption in the first part.

We have made the following division of the network with radius $R$ . We take the sink node as the center of the WSNs. Starting from the edge of the WSNs, we divide the entire network into a number of rings with $r$ as the ring width, and $r$ is the transmission radius of the nodes in the network. When dividing the WSNs, it may not be possible to divide the network evenly into multiple rings, and the ring closest to the sink (the inner radius is infinitely close to 0) has a ring width of $x\le r$ . First, the nodes located in this ring can communicate directly with sink without using opportunistic routing. Second, since sink is always in working state, there is no need to wait for wakeup, and there is no delay. Therefore, this part is negligible.

We call the ring closest to sink the ring 0, from inside to outside, in order, ring 0, 1, 2, and $m-1$ , and $x+(m-1)r=R$ , $x$ represents the ring radius of ring 0. For the convenience of calculation, we make each ring as simplified as follows.

The inner radius of ring $i$ is $x+(i-1)r$ , and the outer radius is $x+\mathrm{ir}$ . In particular, the inner radius of the ring 0 is infinitely close to zero. For the convenience of calculation, we suppose that all nodes are distributed on the outer radius of the ring.

Theorem 5

According to the above method for dividing the network, the number of nodes on each ring is

$$N_i=\rho \pi r(2x+2\mathrm{ir}-r)$$

(52)

$$N_0=\rho \pi x^2$$

(53)

where $i=1,2,\dots ,m-1$ .

Proof

Since the nodes in the WSN are evenly distributed, as long as the area of each ring is known, the number of nodes on the ring can be easily obtained. The area of ring $i$ is

$$S_i=\pi {[x+\mathrm{ir}]}^2-\pi {[x+(i-1)r]}^2$$

(54)

Simplify the above formula

$$S_i=\pi r(2x+2\mathrm{ir}-r)$$

(55)

In particular, for the ring 0

$$S_0=\pi x^2$$

(56)

The density of nodes in the network is $\rho$ , then the number of nodes on each ring is

$$N_i=\rho \pi r(2x+2\mathrm{ir}-r)$$

(57)

$$N_0=\rho \pi x^2$$

(58)

It should be noted that the calculated value is a decimal, and in reality, the number of nodes must be an integer. Therefore, the calculated value needs to be rounded to approximate.

Consider the packet transmission situation of a certain node separately. When the node is in the edge area of the network, it only sends the data packets collected by itself and does not need to be the relay of other nodes. We suppose that in each round of transmission, each node generates only one data packet. Then, the number of data packets sent by these nodes in the edge area of the network is 1, and the number of received data packets is 0. Correspondingly, the number of ACK packets sent by the node is 0, and the number of sent CTS packets is 1. For nodes in other areas of the network, they need to send their own data packets on one hand and forward the data packets sent by the previous hop node on the other hand.

Theorem 6

Suppose that in a network with radius $R$ , all sensor nodes are evenly distributed. In each round of data transmission, all nodes only generate one data packet, and each node will start its own data transmission process only after all the data packets of its previous hop node have been transmitted. Suppose the candidate number of node is $n$ , the transmission radius is $r$ , and the probability of successful point-to-point transmission is $p$ . According to the previous division of the network, it is supposed that the number of packets received by the node on ring $i$ is $Q_i^R$ , the number of packets sent is $Q_i^S$ , the number of ACK packets and CTS packets sent is $Q_i^{\mathrm{ACK}}$ and $Q_i^{\mathrm{CTS}}$ , respectively

$$Q_i^R=\frac{2x+2\mathrm{ir}+r}{2x+2\mathrm{ir}-r}{Q}_{i+1}^S\mathrm{np}$$

(59)

$$Q_i^S=\frac{2x+2\mathrm{ir}+r}{2x+2\mathrm{ir}-r}{Q}_{i+1}^S(1-{(1-p)}^n)+1$$

(60)

$$\begin{array}{c}\multicolumn{1}{c}{Q_i^{\mathrm{ACK}}=\frac{2x+2\mathrm{ir}+r}{2x+2\mathrm{ir}-r}{Q}_{i+1}^S+Q_i^R\frac{1-{(1-p)}^n}{\mathrm{np}}}\\ \multicolumn{1}{c}{+\frac{2x+2(i-1)r+r}{2x+2(i-1)r-r}{Q}_i^S}\end{array}$$

(61)

$$Q_i^{\mathrm{CTS}}=Q_i^S+Q_i^R(1+\frac{1}{p}-\frac{1-{(1-p)}^n}{n{p}^2})$$

(62)

Proof

For the node on ring $i$ , it receives the packets sent by the node on ring $i+1$ . According to equation (52), the number of nodes on the ring $i+1$ is $N_{i+1}$ , and each node sends $Q_{i+1}^S$ packets, each packet is received $n$ times, and the probability of each successful reception is $p$ , averaged to $N_i$ nodes on ring $i$ , the number of packets received by each node is

$$Q_i^R=\frac{N_{i+1}{Q}_{i+i}^S\mathrm{np}}{N_i}$$

(63)

Simplify the above formula

$$Q_i^R=\frac{2x+2\mathrm{ir}+r}{2x+2\mathrm{ir}-r}{Q}_{i+i}^S\mathrm{np}$$

(64)

The data packet sent by the node on ring $i$ , one is a data packet generated by itself, and the other is to assist the forwarding of the previous hop node. There are a total of $N_{i+1}$ nodes on ring $i+1$ , and each node sends $Q_{i+1}^S$ packets, and the probability of each packet being successfully received is $1-(1-p{)}^n$ , then average to the $N_i$ nodes on ring $i$ . In summary, the number of packets sent by the node on ring $i$ is

$$Q_i^S=\frac{N_{i+1}{Q}_{i+1}^S(1-{(1-p)}^n)}{N_i}+1$$

(65)

Simplify the above formula

$$Q_i^S=\frac{2x+2\mathrm{ir}+r}{2x+2\mathrm{ir}-r}{Q}_{i+1}^S(1-{(1-p)}^n)+1$$

(66)

The ACK packet sent by the node on ring $i$ , one is that when the node wakes up, as the receiver, it informs its state to the next hop. There are $N_{i+1}$ nodes on ring $i+1$ , and each node sends $Q_{i+1}^S$ packets, which need to be received by $N_i$ nodes on ring $i$ . Therefore, the nodes on ring $i$ need to receive $N_{i+1}{Q}_{i+1}^S/N$ packets sent by the previous hop node on average, which requires $N_{i+1}{Q}_{i+1}^S/N$ ACK packets to inform the next hop node that they are waking up. Second, in the process of receiving a data packet, node informs nodes whose priority is lower that it has already received the data packet. Once a node confirms that the packet was successfully received (the node with the higher priority has not received successfully), the node with lower priority will delete it immediately even if it receives the same packet.

The nodes on ring $i$ receive $Q_i^R$ data packets during each round of data transmission, each data packet may be received and forwarded by a node with priority $m$ ( $0\le m\le n$ ), and the corresponding probability is $(1-p{)}^m$ . The third is that when the node acts as the sender, after receiving the ACK packet indicating that the receiver has awake, it sends an ACK to inform the receiver whether to keep the awake state until the sender completes the data transmission, or keep waking up a slot and then directly enter the sleep state. In summary, the number of ACK packets sent by the node is

$$Q_i^{\mathrm{ACK}}=\frac{N_{i+1}{Q}_{i+1}^S}{N_i}+Q_i^R\frac{\sum _{m=0}^{n-1}{(1-p)}^m}{n}+\frac{N_i{Q}_i^S}{N_{i-1}}$$

(67)

Substitute relevant formulas and simplify

$$\begin{array}{c}\multicolumn{1}{c}{Q_i^{\mathrm{ACK}}=\frac{2x+2\mathrm{ir}+r}{2x+2\mathrm{ir}-r}{Q}_{i+1}^S+Q_i^R\frac{1-{(1-p)}^n}{\mathrm{np}}}\\ \multicolumn{1}{c}{+\frac{2x+2(i-1)r+r}{2x+2(i-1)r-r}{Q}_i^S}\end{array}$$

(68)

The CTS packet sent by the node on ring $i$ is divided into three parts. The first is to inform the previous hop node that it is ready to receive the packet. The number of CTS packets sent in this part is equal to the number of packets it receives. The second is to inform the next hop node that it is ready to receive the feedback packet sent by the next hop node. The number of CTS packets sent by this part is equal to the number of data packets sent by it. The third is to inform other candidate nodes that they are ready to receive communication packets of other candidate nodes.

The number of CTS packets sent by the nodes in the third part is related to their priority in the candidate set. Since all sensor nodes are already defined in the WSNs, except for the location and the distance from sink, all other performance and status are the same. Therefore, during a round of data transmission, the probability that a node is at a certain priority is $1/n$ . Suppose that in the process of a certain transmission, the priority of the node is $m$ , and the number of CTS packets it sends has the following cases. For the node $m$ , whether it successfully receives the data packet, it needs to send $m-1$ CTS packets as the response to the previous $m-1$ nodes, and the corresponding probability is $(1-p{)}^{m-1}$ , that is, the previous $m-1$ nodes did not successfully receive the packet. For the $m-1$ nodes in front of the mth node, suppose that the node $k$ receives successfully, the previous $k-1$ nodes all fail to receive, then it needs to send $k$ CTS packets as a response, and the probability at this time is $p(1-p{)}^{m-1}$ . Therefore, the number of CTS packets that the node needs to send is

$$\begin{array}{c}\multicolumn{1}{c}{Q_m^{\mathrm{CTS}-\mathrm{III}}=(m-1){(1-p)}^{m-1}}\\ \multicolumn{1}{c}{+p\sum _{k=1}^{m-1}k{(1-p)}^{k-1}}\end{array}$$

(69)

Let us suppose

$$\mathrm{A}=\sum _{k=1}^{m-1}k{(1-p)}^{k-1}$$

(70)

Multiply both sides by $(1-p)$

$$(1-p)\mathrm{A}=\sum _{k=1}^{m-1}k{(1-p)}^k$$

(71)

Subtract equation (71) from equation (70) to

$$p\mathrm{A}=-(m-1){(1-p)}^{m-1}+\sum _{k=1}^{m-1}{(1-p)}^k$$

(72)

Substituting the above formula into equation (69) is

$$\begin{array}{c}\multicolumn{1}{c}{Q_m^{\mathrm{CTS}-\mathrm{III}}=(m-1){(1-p)}^{m-1}}\\ \multicolumn{1}{c}{-(m-1){(1-p)}^{m-1}+\sum _{k=1}^{m-1}{(1-p)}^k}\end{array}$$

(73)

Simplify the above formula

$$Q_m^{\mathrm{CTS}-\mathrm{III}}=\sum _{k=1}^{m-1}{(1-p)}^k$$

(74)

Summation formula by equal ratio series

$$Q_m^{\mathrm{CTS}-\mathrm{III}}=\frac{1-{(1-p)}^{m-1}}{p}$$

(75)

Since the nodes may be at different priorities in different transmission rounds, we average the number of CTS packets at different priorities

$$Q_i^{\mathrm{CTS}-\mathrm{III}}=\frac{\sum _{m=1}^n{Q}_m^{\mathrm{CTS}-\mathrm{III}}}{n}$$

(76)

Simplify the above formula

$$Q_i^{\mathrm{CTS}-\mathrm{III}}=\frac{1}{p}-\frac{1-{(1-p)}^n}{n{p}^2}$$

(77)

In summary, the number of CTS packets sent by the nodes on ring $i$ is

$$Q_i^{\mathrm{CTS}}=Q_i^R+Q_i^S+Q_i^R(\frac{1}{p}-\frac{1-{(1-p)}^n}{n{p}^2})$$

(78)

Substitute relevant formulas and simplify

$$Q_i^{\mathrm{CTS}}=Q_i^S+Q_i^R\frac{1}{p}(1-\frac{1-{(1-p)}^n}{\mathrm{np}})$$

(79)

Combined with the analysis in the energy consumption model in section “Energy consumption model,” we get the energy consumption when the node transmits the data packets. Next, we consider the energy consumption when the node keeps listening.

Theorem 7

Suppose that the size of forwarding set is $N$ , the node S needs $n$ candidate nodes to wake up to send the data packet, the duty length is $\tau$ , and the listening power is $P_{\mathrm{LPL}}$ . Then, in the process of data transmission, the energy consumed by the node because it remains in the listening state is

$$E_{\mathrm{LPL}}=P_{\mathrm{LPL}}\mathrm{T}=P_{\mathrm{LPL}}\frac{n-1}{N}\sum _{k=0}^{\tau -1}{P}_m^{\mathrm{until}\_k}$$

(80)

Proof

We suppose that the node $n$ wakes up at slot $k$ , and then the previous $n-1$ nodes must wake up before slot $k+1$ . Since the time that the nodes wake up does not affect each other, we can consider the situation of a single node first.

According to equation (42), the probability that a single node wakes up at a certain time in the entire duty length is

$$P_k^1={(1-\frac{1}{\tau })}^k\frac{1}{\tau }$$

(81)

where $k=0,1,\dots ,\tau -2$ .

When $k=\tau -1$

$$P_{\tau -1}^1={(1-\frac{1}{\tau })}^{\tau -1}$$

(82)

The probability that a single node will wake up in or before slot $k$ is

$$P_{u\_k}^1=\sum _{i=0}^k{P}_i^1$$

(83)

Since all nodes in the network will keep awake for a slot, the average time for these $n-1$ candidate nodes to maintain the listening state more than other candidate nodes is

$$T_1^{\mathrm{until}\_k}=\sum _{i=0}^k\frac{P_m^1}{P_{\mathrm{unti}{l}_k}^1}(k-i)$$

(84)

Since node $n$ can wake up at any time of the duty length, the average listening time of each node during a round of data transmission is

$$T=1+\frac{n-1}{N}\sum _{k=0}^{\tau -1}{P}_k^n{T}_1^{\mathrm{until}\_k}$$

(85)

When $k=0$ , the value of $P_k^n$ can be obtained by equation (46); when $k=\tau -1$ , the value of $P_k^n$ can be obtained by equation (48); when k is equal to other values, the value of $P_k^n$ can be obtained by equation (49).

The power of the node to maintain low power listening is $P_{\mathrm{LPL}}$ , so the energy consumed by the node to maintain low power monitoring is

$$E_{\mathrm{LPL}}=P_{\mathrm{LPL}}[1+\frac{n-1}{N}\sum _{k=0}^{\tau -1}{P}_k^n(P_1^{\mathrm{until}\_k}-1)]$$

(86)

ADCCOR scheme

Combined with the analysis in the energy consumption model in section “Energy consumption model,” we get the energy consumption when the node transmits the data packets. Next, we consider the energy consumption when the node keeps listening.

By combining duty cycle control with opportunistic routing, the DCCOR scheme combines the advantages of low power consumption and high reception rate. Due to the characteristics of the wireless network itself, the number of data packets transmitted by the nodes at the edge of the network is much smaller than that of the nodes near the sink, so that the energy consumed by the nodes in the far sink area is much smaller than that of the nodes near the sink area. Thus, when the energy of the node near the sink area is exhausted, the far sink area node still has a large amount of energy remaining. Because the node in the near sink area dies, there is no node in the network that can directly communicate with the sink, that is, the network has also died. Thus, there is a case where a large amount of energy remains in the far sink region node when the network dies.

In order to make full use of this part of the energy, we propose the ADCCOR scheme. That is, based on the DCCOR scheme, the residual energy is used to increase the time that the node keeps listening, to reduce the delay. Except for the first round of data transmission process, which is the same as the DCCOR scheme, in other rounds, it is necessary to adjust the duty cycle continuously.

Theorem 8

We use $E_0^{\mathrm{ALL}\_m}$ and $E_k^{\mathrm{ALL}\_m}$ to represent the total energy consumed by the nodes on ring 0 and ring $k$ , as of round $m$ . $E_{\mathrm{data}}^0$ and $E_{\mathrm{data}}^k$ represent the energy (one round) consumed by the node transmitting packets on ring 0 and ring $k$ , respectively. $E_k^{\mathrm{LPL}\_m}$ indicates the energy consumed by the node on ring $k$ to keep listening in round $m$ . ${\tau }_k^m$ and ${\tau }_k^{m+1}$ represent the duty length of the nodes in round $m$ and round $m+1$ , respectively. Then, the duty length of the node in round $m+1$ can be expressed as

$${\tau }_k^{m+1}=\frac{{\tau }_k^m}{1+M_{m+1}^k}$$

(87)

where $M_{m+1}^k$ represents a multiple of the increase in the listening time of the node on ring $k$ in round $m$ , and

$$M_{m+1}^k=\frac{(E_0^{\mathrm{ALL}\_m}-E_k^{\mathrm{ALL}\_m})+(E_{\mathrm{data}}^0-E_{\mathrm{data}}^k)}{E_k^{\mathrm{LPL}\_m}}$$

(88)

$$E_0^{\mathrm{ALL}\_m}=\sum _{j=1}^m{E}_0^j$$

(89)

$$E_k^{\mathrm{ALL}\_m}=\sum _{j=1}^m{E}_k^j$$

(90)

Proof

When we calculate the duty length of round $m+1$ , we first consider the situation of the previous $m$ rounds of data transfer. Theoretically, after the data transmission of the previous $m$ rounds, the total energy consumption of the node on ring 0 is $E_0^{\mathrm{ALL}\_m}$ , and the total energy consumed by the node on the ring $k$ is $E_k^{\mathrm{ALL}\_m}$

$$E_0^{\mathrm{ALL}\_m}=\sum _{j=1}^m{E}_0^j$$

(91)

$$E_k^{\mathrm{ALL}\_m}=\sum _{j=1}^m{E}_k^j$$

(92)

where $E_i^j$ represents the energy consumed by the node on ring $i$ during the data transmission in round $j$ .

After m rounds, the nodes on ring 0 consume $E_0^{\mathrm{ALL}\_m}-E_k^{\mathrm{ALL}\_m}$ joules more than the nodes on ring $k$ . In round $m+1$ , the energy consumed by the nodes located on ring 0 and ring $k$ due to the transmission of the data packets are $E_{\mathrm{data}}^0$ and $E_{\mathrm{data}}^k$ , respectively. That is, the node on ring 0 consumes $E_{\mathrm{data}}^0-E_{\mathrm{data}}^k$ joules more than the node on ring $k$ . The energy of these two parts can be used to increase the listening time of the nodes on ring $k$ , thereby reducing the delay of the nodes. Thus, the multiple of the increase in listening time can be expressed as

$$M_{m+1}^k=\frac{(E_0^{\mathrm{ALL}\_m}-E_k^{\mathrm{ALL}\_m})+(E_{\mathrm{data}}^0-E_{\mathrm{data}}^k)}{E_k^{\mathrm{LPL}\_m}}$$

(93)

We assume that the length of the entire clock cycle is $T$ , then in round $m$ , the duration of one time slot is $T/{\tau }_k^m$ , then the duration of the time slot in round $m+1$ should be $(1+M_{m+1}^k)T/{\tau }_k^m$ . By dividing the length of the entire clock cycle by the time slot of the node in round $m+1$ , the corresponding duty length can be obtained. At the same time, in order to ensure that the result is an integer, we have to round it up, that is

$${\tau }_k^{m+1}=\frac{{\tau }_k^m}{1+M_{m+1}^k}$$

(94)

In summary, the ADCCOR scheme can be summarized by Algorithm 3.

Algorithm 3. ADCCOR scheme.

1: When the first round of data transmission Do 2:  For each node such as A in network Do 3:    $E_{\mathrm{TOT}}^A=0$ ; 4:   $D{L}_A=D{L}^0$ ; 5:   Calculate $E_{\mathrm{data}}^A$ according to equation (12);    // The energy consumption by transmitting packets 6:   Calculate $E_{\mathrm{LPL}}^A$ according to equation (80);    // The energy consumption by LPL 7:   $E_{\mathrm{TOT}}^A$  = $E_{\mathrm{data}}^A+E_{\mathrm{LPL}}^A$ ; // Total energy consumption 8:  End for 9: End 10: For each round behind Do 11:  For each node such as A Do 12:   Calculate $E_{\mathrm{LPL}}^0$ according to equation (80); 13:   Calculate $E_{\mathrm{LPL}}^A$ according to equation (80); 14:    $M=((E_{\mathrm{TOT}}^0-E_{\mathrm{TOT}}^A)+(E_{\mathrm{LPL}}^0-E_{\mathrm{LPL}}^A))/E_{\mathrm{LPL}}^A$ ; 15:    $M$ round down 16:    $D{L}_A=(1+M)*D{L}_A$ ; //Get a new duty length 17:   Calculate $E_{\mathrm{data}}^A$ according to equation (86); 18:    $E_{\mathrm{TOT}}^A=E_{\mathrm{TOT}}^A+E_{\mathrm{data}}^A+E_{\mathrm{LPL}}^A$ ; //Prepare for the next round 19:  End for 20:   $E_{\mathrm{TOT}}^0=E_{\mathrm{TOT}}^0+E_{\mathrm{data}}^0+E_{\mathrm{LPL}}^0$ ; 21: End for

The meaning of the symbolic representation in the algorithm is as follows:

$D{L}^0$ : the initial duty length of nodes in network;

$D{L}_A$ : the duty length of node A;

$E_{\mathrm{data}}^A$ : the energy consumption of node A due to transfer data packets;

$E_{\mathrm{LPL}}^A$ : the energy consumption of node A due to low power listening;

$E_{\mathrm{TOT}}^A$ : the total energy consumption of node A;

$E_{\mathrm{data}}^0$ : specifically refers to the energy consumed by the node on ring 0 because of the transmission of data packets;

$E_{\mathrm{LPL}}^0$ : specifically refers to the energy consumed by the node on ring 0 because of low power listening;

$E_{\mathrm{TOT}}^0$ : specifically refers to the total energy consumed by the node on ring 0;

$M$ : the multiple of the duty length added of node A.

Performance analysis and comparison

In section “The design of ADCCOR scheme,” we have a detailed description and analysis of DCCOR and found that after the duty cycle control is added to the OR, the delay of the node becomes significantly larger. Therefore, we can reduce the delay of the node by changing the duty cycle of the node without changing the lifetime of the network. In the DCCOR scheme, by combining the duty cycle control and the OR, it has the characteristics of low power consumption and high receiving rate. Due to the characteristics of the WSN itself, the nodes in the far sink area consume far less energy than the nodes in the near sink area. When the nodes in the near sink area die due to energy exhaustion, the nodes in the far sink area have a large amount of energy remaining. Since there are no more nodes in the network that can communicate directly with the sink node, the entire network is in a paralyzed state. Even if there are more energy left in the nodes in the far sink area, it will not help. Therefore, we propose the ADCCOR scheme, which is based on the DCCOR scheme to adjust the duty cycle of the nodes with the remaining energy, to use these energy to reduce the delay.

Below, we will analyze the DCCOR scheme and ADCCOR scheme from the perspective of energy consumption and delay to prove our point of view.

About energy consumption

In section “DCCOR scheme,” we give a detailed method for calculating the energy consumption of nodes in a network using DCCOR scheme and ADCCOR scheme, which will not be repeated here. When calculating the node energy consumption in a network using OR scheme, we can make up the following two parts.

1. The energy consumed by the node due to the listening. We suppose that the power is $P_L$ (in watts) when listening, and the whole duty length is $T$ (in seconds). Since the node is always in the awake state, the listening time is equal to the length of entire duty length. Therefore, the energy consumed due to LPL is

   $$E_{\mathrm{LPL}}=P_L*T$$

   (95)

2. The energy consumed by the node due to the transmission of the packet. Compared with DCCOR scheme and ADCCOR scheme, the number of packets sent and received by nodes in OR scheme are the same, and the number of CTS packets is the same too. Different from DCCOR scheme and ADCCOR scheme, the node in OR scheme does not need to send ACK packets to the upper hop and the next hop, and only needs to send ACK packets to the node whose priority is lower than it. Therefore, the number of packets sent (received) by the node in OR can be expressed as

   $$Q_i^R=\frac{2x+2\mathrm{ir}+r}{2x+2\mathrm{ir}-r}{Q}_{i+1}^S\mathrm{np}$$

   (96)
   $$Q_i^S=\frac{2x+2\mathrm{ir}+r}{2x+2\mathrm{ir}-r}{Q}_{i+1}^S(1-{(1-p)}^n)+1$$

   (97)
   $$Q_i^{\mathrm{ACK}}=Q_i^R\frac{1-{(1-p)}^n}{\mathrm{np}}$$

   (98)
   $$Q_i^{\mathrm{CTS}}=Q_i^S+Q_i^R(1+\frac{1}{p}-\frac{1-{(1-p)}^n}{n{p}^2})$$

   (99)

The definition of the above variables is the same as in section “DCCOR scheme.” Then, according to equations (7)–(10), the energy consumed by the node to send data packets, receive data packets, send ACK packets and CTS packets during a round of data transmission can be obtained; adding those energy together is the total energy consumed due to the transmission of the packet. By adding up the energy of the above two parts, we can get the total energy consumed in the OR scheme.

We suppose that there is a network with a network radius of $R=1000\mathrm{m}$ , and the sensor nodes are evenly distributed in the network with a density of $\rho =0.001$ . The transmission power of the node is $\mathrm{Pt}=-3\mathrm{dBm}$ . The transmission radius takes the weighted average radius $r=63.4351\mathrm{m}$ . At this distance, the receiving rate of a single node is $\mathrm{par}=0.636735$ , and the number of candidate nodes is $n=3$ . The success rate of one-hop is 0.95. In addition, we suppose that the power $P_{\mathrm{LPL}}=-6\mathrm{dBm}$ when the node is in the listening state, and the duty length $\tau =20$ of all nodes in the network. In the above network, the OR scheme, DCCOR scheme, and ADCCOR scheme are respectively used to obtain the energy consumption of the packets transmitted in the network, as shown in Figure 6.

Figure 6.

Energy consumption due to transfer packets in three different OR schemes.

Since the nodes in the DCCOR scheme and the ADCCOR scheme transmit more ACK packets relative to the nodes in the OR scheme, the energy consumed is slightly higher than the OR scheme.

In the ADCCOR scheme, the duty cycle of the nodes in different rounds may be different, and the energy consumed to keep listening is different too. In Figure 7, we give the energy consumed by the nodes in the first five rounds.

Figure 7.

Energy consumption due to listening in ADCCOR scheme.

In order to facilitate comparison with the OR scheme and the DCCOR scheme, we use the average of the energy consumed by listening in the previous rounds. Since the energy consumed by the nodes in the OR scheme and DCCOR scheme to keep listening in different rounds is the same, it is only expressed once, as shown in Figure 8.

Figure 8.

Average energy consumption due to listening in three different OR schemes.

In terms of the energy consumed by listening, the nodes in the OR scheme consume the most because they are always in the awake state. The duty cycle of the node in the DCCOR scheme is always at the initial value, so it is the smallest. The duty cycle in the ADCCOR scheme is always greater than or equal to the DCCOR scheme and is not always equal to 1 as in the OR scheme, so the amount of energy consumed is between the OR scheme and the DCCOR scheme.

By summing the energy consumed by the node’s transmitted data packet with the energy consumed by the listening, the total energy consumed by the node can be obtained. Considering that the total energy consumed by the nodes in the ADCCOR scheme may be different in each round of data transmission, we also use the average value, as shown in Figure 9.

Figure 9.

Average energy consumption in three different OR schemes.

From Figure 9, we can draw the following conclusions. First, by increasing the duty cycle control, the energy consumed by the nodes in the DCCOR scheme is greatly reduced compared to the OR scheme. The lifetime of the node closest to the sink determines the lifetime of the network. From this perspective, the energy consumed by the node in the DCCOR scheme is reduced by 55.64%. With the same initial energy, the lifetime of the nodes in the DCCOR scheme is extended by 125.43%.

Second, by adjusting the duty cycle of the node, that is, using the ADCCOR scheme, the energy consumed by other nodes increases with the number of transmission rounds except for the node closest to the sink and is always no larger than the node closest to the sink. This means that dynamically adjusting the duty cycle does not affect the life of the network. Below, we will prove that by adjusting the duty cycle, the purpose of reducing the delay is indeed achieved.

About delay

Obviously, by increasing the duty cycle control, the node is periodically in a sleep state, and the purpose of reducing energy consumption is achieved. Since the relay node may be in a sleep state, data transmission is interrupted, resulting in a delay. To reduce the delay, we used a method of dynamically adjusting the duty cycle.

In Figure 10, we give the one-hop delay of the node in the DCCOR scheme and the ADCCOR scheme. Since the duty cycle varies in different rounds in the ADCCOR scheme, the size of a time slot is different too. Therefore, we measure the delay in seconds (s).

Figure 10.

Average one-hop delay in DCCOR scheme and ADCCOR scheme.

As can be seen from Figure 10, after increasing the method of adjusting the duty cycle, the one-hop delay of the node becomes smaller and smaller with the increase of the transmission round. In particular, since the sink node is always in the awake state, the delay of the node that can directly communicate with the sink is zero. As of the 25th round, the one-hop delay in the ADCCOR scheme is reduced from 44.52% to 75.50% relative to the nodes in the DCCOR scheme.

The above one-hop delay only reflects the impact of the ADCCOR scheme on the delay of a single node. In order to reflect its impact on the delay of the entire network, the end-to-end delay needs to be analyzed. In Figure 11, we give the end-to-end delay of the nodes in the DCCOR scheme and the ADCCOR scheme.

Figure 11.

Average end-to-end delay in DCCOR scheme and ADCCOR scheme.

Obviously, the end-to-end delay of the nodes in the ADCCOR scheme is significantly smaller than the nodes in the DCCOR scheme, and the farther away from the sink, the smaller the delay. And the more rounds of transmission, the smaller the delay. For the node farthest from the sink node, the maximum end-to-end delay in the network is reduced by 69.83% after 25 rounds. That is to say, after using the ADCCOR scheme, the time that the sink node collects all the packets in the network is shortened by about two-thirds.

Conclusion

As an important part of CPS, WSNs are the cornerstone of the whole system. The performance of WSNs greatly affects the performance of CPS. To this end, our goal is to find better routing solutions, optimize the performance of WSNs, and contribute to the development of CPS.

Related scholars have proposed an OR scheme to improve the reception rate of nodes in WSNs and achieve the goal. However, the OR scheme only focuses on the receiving rate of the node, completely ignoring the energy consumption of the node. In WSNs, in order to reduce the energy consumption of nodes, our most common method is to add duty cycle control. Therefore, in this article, we propose an opportunistic routing that can be applied to the duty cycle–based WSNs, namely DCCOR scheme. According to our research, this article is the first time to apply the opportunistic routing method to the duty cycle WSNs.

Subsequently, in order to make full use of the surplus in the network, we improved the DCCOR scheme and proposed the ADCCOR scheme. The main idea of the ADCCOR scheme is to use the residual energy of the nodes to increase the duty cycle, so that the ADCCOR scheme has a lower delay than DCCOR scheme. The central idea of the ADCCOR scheme is to appropriately increase the duty length of the node (except for the nodes on ring 0) to reduce the delay. Through analysis and calculation, we find that the ADCCOR scheme achieves the goal of reducing node delay without changing the lifetime of the network. This improvement has greatly improved the performance of DCCOR scheme and in the latter comparison with OR scheme, the performance has also been greatly improved.

In summary, we have successfully proposed a DCCOR scheme for use in the duty cycle–based network and improved it. This greatly reduces the energy consumption of the node, and the lifetime of the network is greatly increased, which has a good meaning. Although the delay of DCCOR scheme and ADCCOR in the duty cycle network is larger than that in the non-duty cycle network, this is not the delay caused by the strategy itself, but the problem that all the duty cycle–based networks must exist.