CN107872270A - A kind of relay node selecting method based on optimal threshold transmitting and scheduling in rayleigh fading channel - Google Patents

A kind of relay node selecting method based on optimal threshold transmitting and scheduling in rayleigh fading channel Download PDF

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CN107872270A
CN107872270A CN201710919230.1A CN201710919230A CN107872270A CN 107872270 A CN107872270 A CN 107872270A CN 201710919230 A CN201710919230 A CN 201710919230A CN 107872270 A CN107872270 A CN 107872270A
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CN107872270B (en
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黄亮
岳梦娟
钱丽萍
吴远
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Zhejiang University of Technology ZJUT
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    • H—ELECTRICITY
    • H04—ELECTRIC COMMUNICATION TECHNIQUE
    • H04B—TRANSMISSION
    • H04B7/00—Radio transmission systems, i.e. using radiation field
    • H04B7/14—Relay systems
    • H04B7/15—Active relay systems
    • H04B7/155—Ground-based stations
    • H04B7/15528—Control of operation parameters of a relay station to exploit the physical medium
    • H—ELECTRICITY
    • H04—ELECTRIC COMMUNICATION TECHNIQUE
    • H04B—TRANSMISSION
    • H04B17/00—Monitoring; Testing
    • H04B17/30—Monitoring; Testing of propagation channels
    • H04B17/309—Measuring or estimating channel quality parameters
    • H04B17/336—Signal-to-interference ratio [SIR] or carrier-to-interference ratio [CIR]
    • H—ELECTRICITY
    • H04—ELECTRIC COMMUNICATION TECHNIQUE
    • H04W—WIRELESS COMMUNICATION NETWORKS
    • H04W40/00—Communication routing or communication path finding
    • H04W40/02—Communication route or path selection, e.g. power-based or shortest path routing
    • H04W40/22—Communication route or path selection, e.g. power-based or shortest path routing using selective relaying for reaching a BTS [Base Transceiver Station] or an access point

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Abstract

A kind of relay node selecting method based on optimal threshold transmitting and scheduling in rayleigh fading channel, comprise the following steps：1) channel probability density function and cumulative probability density function after each channel minor sort are calculated；2) each via node channel perception information current time slots fading information, energy and itself energy storage required for via node forwarding information simultaneously calculate average transmission rate notice source node；3) via node determines the minimal disruption probability of each of the links foundation according to the energy condition met required for channel information and forwarding and notifies source node, the repeating power that source node is provided according to link establishment outage probability and via node itself energy storage, determines all qualified via nodes；4) it is all it is eligible in select the node of it is expected that transmission rate is maximum, as optimal via node.When source node transmission power is constant in rayleigh fading channel, end-to-end node selection via node is transmitted the present invention with greatest hope transmission rate.

Description

A kind of via node based on optimal threshold transmitting and scheduling in rayleigh fading channel System of selection
Technical field
The invention belongs to the communications field, more particularly, to the Turbo Detection for Cooperative Communication with energy acquisition and optimizes system It is expected the relay node selecting method of transmission rate.
Background technology
With the broad development of wireless communication technology, wireless communication technology has penetrated into the every aspect of human lives. Meanwhile gradually cause scientific research and engineering field the problem of the communication energy consumption caused by communications industry fast development increases rapidly Pay attention to.Power consumption issues caused by how effectively reducing communications industry are that current scientific research and engineering field are all urgently to be resolved hurrily Problem.Energy acquisition relay network system has merged the advantage of energy acquisition and junction network, is that communication system further develops Extremely promising solution.Energy acquisition junction network utilize from surrounding environment collecting energy (such as solar energy, wind energy, RF etc.) information that via node is sent to by source node is forwarded, on the basis of communication quality is ensured, the wasting of resources is reduced, from And reach energy-saving and emission-reduction, reduce the purpose of communication energy consumption.Energy acquisition part in most of energy acquisition junction network model All it is to be completed by via node, and source node and destination node are still then using traditional power supply mode, the model of communications portion is all It is by single or multiple via node forwarding informations.For energy acquisition junction network, each time slot of different via nodes is received The energy for collecting and storing is different, and the energy for forwarding identical data packet to be consumed is also different, then selection is suitable Via node, then it can lift the capacity usage ratio and transmission rate of whole system.Therefore, when transmission power is limited and constant, How research makes full use of limited energy storage, and suitable via node is selected from energy acquisition junction network so that system reaches Optimal transmission speed is extremely significant.
The content of the invention
In order to overcome the via node in existing energy acquisition junction network to cause forwarding data because collecting energy is limited It is limited, the present invention provide it is a kind of in decoding forwards energy acquisition junction network, for exist in the channel of Rayleigh fading how Via node is selected to maximize the relay node selecting method of system transfer rate with limited collecting energy.
The technical scheme adopted by the invention to solve the technical problem is that：
A kind of relay node selecting method of optimal threshold transmitting and scheduling in rayleigh fading channel, methods described include Following steps：
1) source node to via node, each signal of via node to destination node envelope γ Rayleigh distributeds, And signal to noise ratio γ is obeyedExponential distribution, wherein γ0For average signal-to-noise ratio.Via node instantaneous signal-to-noise ratio γ1,γ2,…,γKBy being γ after ascending order rearrangement(1)≤γ(2)≤…≤γ(k)≤γ(k+1)≤…≤γ(K), γ(k)Claim k-th of order statistic, its probability density function f(k)(γ) is：
Cumulative probability density function F(k)(γ) is：
Wherein, each parameter definition is as follows in formula：
f(k)(γ)：γ(k)Probability density function；
F(k)(γ)：γ(k)Cumulative probability density function；
F(γ)：Signal to noise ratio γ cumulative probability density function, wherein
K：Via node number；
γ0：Average signal-to-noise ratio；
γk：Via node k instantaneous signal-to-noise ratio；
γ(k)：After via node instantaneous signal-to-noise ratio according to sorting from small to large, k-th of order system of signal to noise ratio
Metering
2) each via node is perceived from source node to itself, and itself arrives the letter of destination node
Road information, then by the energy required for forwarding information and itself storage energy notice source node；
The letter that the channel information and via node and destination node perceived by via node receives
Road information, obtain system average signal-to-noise ratio E [γ(k)] as follows：
Average transmission rate E [C(k)] as follows：
Wherein, each parameter definition is as follows in formula：
B：Channel width
Ps：Source node transmission power
3) optimal threshold transmission dispatching method selection via node then stores energy whether to judge link using via node Establish, obtaining outage probability expression formula is：
Wherein, each parameter definition is as follows in formula：
Pk=2Emax/T：Via node k repeating power；EmaxVia node energy storage maximum capacity；T information is from source node The time being sent to required for destination node, information source node to via node and the time phase from via node to destination node Together；
4) system selects Largest Mean average transmission rate E [C from qualified via node(k)], during selection is optimal After node k*Meet following formula：
Constraints is：E[O(k)]≤Emax
For the constraints in formula (7), minimum power and maximum energy storage relation are converted to, obtains following formula：
Constraints：Pk≥2Emax/T
According to step 2), 3) Mean Speed defines in, and obtaining transmission rate when threshold value is fixed is：
Wherein, E [C(0)] represent no data transfer, E [C(0)]=0
According to identical in channel fading, in transmission time identical channel, the time required for information is identical, asks system to pass Defeated expectation energy expenditure：
Energy it is expected that consumption is：
Wherein, E [O(0)] energy expenditure when indicating no information transfer, E [O(0)]=0
According to step 2), 4) transmission rate for obtaining threshold value scheduled transmission algorithm is：
Energy it is expected consumption：
Wherein,
Energy it is expected that consumption is：
The present invention technical concept be：The repeating power of via nodes different first is different, because of each not phase of channel It is different with the energy required for causing synchronization to forward same packet, therefore via node energy storage can be said as can be with The Internet resources of control, realize transmission rate maximum.In other words, it is desirable to judge source node to purpose section by via node energy storage Whether the link of point can be established, and the maximum via node of transmission rate can be reached by being picked out from qualified via node.
Beneficial effects of the present invention are mainly manifested in：For whole energy acquisition junction network, during selection is suitable Not only can be effectively using the energy collected, and the long-term average benefit of system can be increased after node.
Brief description of the drawings
Fig. 1 is the network system schematic diagram for having K via node.
Fig. 2 is the flow chart for selecting suitable via node.
Embodiment
The present invention is described in further detail below in conjunction with the accompanying drawings.
Referring to Figures 1 and 2, a kind of trunk node selection based on optimal threshold transmitting and scheduling in rayleigh fading channel Method, the energy collected can be made full use of by carrying out this method, increase system average transmission rate.The present invention is based on having K The energy acquisition relay network system (as shown in Figure 1) of via node.In energy acquisition junction network, source node is with constant work( Information is sent a via node by rate, and via node will according to the node that therefrom selection transmission rate is maximum and itself energy storage is enough Information is transmitted to destination node.Invention selects suitable via node to reach average transmission rate for energy acquisition junction network Maximum method has following steps (as shown in Figure 2)：
1) source node to via node, each signal of via node to destination node envelope γ Rayleigh distributeds, And signal to noise ratio γ is obeyedExponential distribution, wherein γ0For average signal-to-noise ratio.Via node instantaneous signal-to-noise ratio γ1,γ2,…,γKBy being γ after ascending order rearrangement(1)≤γ(2)≤…≤γ(k)≤γ(k+1)≤…≤γ(K), γ(k)Claim k-th of order statistic, its probability density function f(k)(γ) is：
Cumulative probability density function F(k)(γ) is：
Wherein, each parameter definition is as follows in formula：
f(k)(γ)：γ(k)Probability density function；
F(k)(γ)：γ(k)Cumulative probability density function；
F(γ)：Signal to noise ratio γ cumulative probability density function, wherein
K：Via node number；
γ0：Average signal-to-noise ratio；
γk：Via node k instantaneous signal-to-noise ratio；
γ(k)：After via node instantaneous signal-to-noise ratio according to sorting from small to large, k-th of order statistic of signal to noise ratio；
2) each via node is perceived from source node to itself, and itself arrives the channel information of destination node, then will Energy and itself storage energy notice source node required for forwarding information；
The channel information that the channel information and via node and destination node perceived by via node receives, is obtained To system average signal-to-noise ratio E [γ(k)] as follows：
Average transmission rate E [C(k)] as follows：
Wherein, each parameter definition is as follows in formula：
B：Channel width；
Ps：Source node transmission power；
3) optimal threshold transmission dispatching method selection via node then stores energy whether to judge link using via node Establish, obtaining outage probability expression formula is：
Wherein, each parameter definition is as follows in formula：
Pk=2Emax/T：Via node k repeating power；
EmaxVia node energy storage maximum capacity；
T information be sent to destination node from source node required for time, information source node to via node with from relaying The time of node to destination node is identical, is T/2；
4) system selects Largest Mean average transmission rate from qualified via nodeChoosing Select optimal via node k* and meet following formula：
Constraints is：E[O(k)]≤Emax
Step 4.1：For the constraints in formula (19), minimum power and maximum energy storage relation are converted to, is obtained down Formula：
Constraints：Pk≥2Emax/T
Step 4.2：For the constraints in formula (21), 3) outage probability expression is used instead, i.e., when outage probability is less than Threshold value then link establishment, obtains following formula：
UtilizeOptimal via node k is selected when trying to achieve greatest hope transmission rate*；
Step 4.3：The energy that via node consumed is sent to from source node be equal to it from via node according to information The energy that destination node is consumed is forwarded to, tries to achieve system average energy consumption E [O(k)]：
By formula (23) abbreviation, obtaining the average power consumption after order statistics of via node transmission rate is：
Formula (14) is substituted into formula (24), by integral formula Obtain：
System is obtained by formula (18) (25) and it is expected energy expenditure

Claims (1)

1. a kind of relay node selecting method based on optimal threshold transmitting and scheduling in rayleigh fading channel, its feature exists In：It the described method comprises the following steps：
1) source node to via node, each signal of via node to destination node envelope γ Rayleigh distributeds, and believe Make an uproar and obeyed than γExponential distribution, wherein γ0For average signal-to-noise ratio, via node instantaneous signal-to-noise ratio γ1, γ2,…,γKBy being γ after ascending order rearrangement(1)≤γ(2)≤…≤γ(k)≤γ(k+1)≤…≤γ(K), γ(k) Claim k-th of order statistic, its probability density function f(k)(γ) is：
<mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>f</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mi>K</mi> <mo>!</mo> </mrow> <mrow> <mo>(</mo> <mi>k</mi> <mo>-</mo> <mn>1</mn> <mo>)</mo> <mo>!</mo> <mo>(</mo> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>)</mo> <mo>!</mo> </mrow> </mfrac> <msup> <mi>F</mi> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <msup> <mrow> <mo>&amp;lsqb;</mo> <mn>1</mn> <mo>-</mo> <mi>F</mi> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> <mrow> <mi>K</mi> <mo>-</mo> <mi>k</mi> </mrow> </msup> <mi>f</mi> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>K</mi> </mtd> </mtr> <mtr> <mtd> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </mtd> </mtr> </mtable> </mfenced> <mfrac> <mrow> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> <msub> <mi>&amp;gamma;</mi> <mn>0</mn> </msub> </mfrac> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </msubsup> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>i</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mrow> <mo>(</mo> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mfrac> <mi>&amp;gamma;</mi> <msub> <mi>&amp;gamma;</mi> <mn>0</mn> </msub> </mfrac> </mrow> </msup> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow>
Cumulative probability density function F(k)(γ) is：
<mrow> <mtable> <mtr> <mtd> <mrow> <msub> <mi>F</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mo>=</mo> <mi>Pr</mi> <mi>o</mi> <mi>b</mi> <mo>{</mo> <msub> <mi>&amp;gamma;</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;le;</mo> <mi>&amp;gamma;</mi> <mo>}</mo> <mo>=</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mi>k</mi> </mrow> <mi>K</mi> </msubsup> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>K</mi> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mi>F</mi> <mi>i</mi> </msup> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <msup> <mrow> <mo>&amp;lsqb;</mo> <mn>1</mn> <mo>-</mo> <mi>F</mi> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> <mrow> <mi>K</mi> <mo>-</mo> <mi>i</mi> </mrow> </msup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mi>k</mi> </mrow> <mi>K</mi> </msubsup> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>K</mi> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>j</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>i</mi> </msubsup> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> <mtr> <mtd> <mi>j</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>j</mi> </msup> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mrow> <mo>(</mo> <mi>K</mi> <mo>-</mo> <mi>i</mi> <mo>+</mo> <mi>j</mi> <mo>)</mo> </mrow> <mfrac> <mi>&amp;gamma;</mi> <msub> <mi>&amp;gamma;</mi> <mn>0</mn> </msub> </mfrac> </mrow> </msup> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow>
Wherein, each parameter definition is as follows in formula：
f(k)(γ)：γ(k)Probability density function；
F(k)(γ)：γ(k)Cumulative probability density function；
F(γ)：Signal to noise ratio γ cumulative probability density function, wherein
K：Via node number；
γ0：Average signal-to-noise ratio；
γk:Via node k instantaneous signal-to-noise ratio；
γ(k)：After via node instantaneous signal-to-noise ratio according to sorting from small to large, k-th of order statistic of signal to noise ratio；
2) each via node is perceived from source node to itself, and itself arrives the channel information of destination node, then will forwarding Energy and itself storage energy notice source node required for information；The channel information that is perceived by via node and in After the channel information that node and destination node receive, system average signal-to-noise ratio E [γ are obtained(k)] as follows：
<mrow> <mtable> <mtr> <mtd> <mrow> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>&amp;gamma;</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mi>&amp;gamma;</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>&amp;infin;</mi> </msubsup> <msub> <mi>&amp;gamma;dF</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mi>&amp;gamma;</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>&amp;infin;</mi> </msubsup> <msub> <mi>&amp;gamma;f</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mi>d</mi> <mi>&amp;gamma;</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <msub> <mi>&amp;gamma;</mi> <mn>0</mn> </msub> <mrow> <mo>(</mo> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>K</mi> </mtd> </mtr> <mtr> <mtd> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mtd> </mtr> </mtable> </mfenced> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </msubsup> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>i</mi> </msup> <mfrac> <mn>1</mn> <msup> <mrow> <mo>(</mo> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mfrac> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>3</mn> <mo>)</mo> </mrow> </mrow>
Average transmission rate E [C(k)] as follows：
<mrow> <mtable> <mtr> <mtd> <mrow> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>C</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mi>&amp;gamma;</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>&amp;infin;</mi> </msubsup> <msub> <mi>Blog</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>P</mi> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <msub> <mi>dF</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mi>&amp;gamma;</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>&amp;infin;</mi> </msubsup> <msub> <mi>Blog</mi> <mn>2</mn> </msub> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>P</mi> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <msub> <mi>f</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mi>d</mi> <mi>&amp;gamma;</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mfrac> <mrow> <mi>B</mi> <mrow> <mo>(</mo> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mi>l</mi> <mi>n</mi> <mn>2</mn> </mrow> </mfrac> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>K</mi> </mtd> </mtr> <mtr> <mtd> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mtd> </mtr> </mtable> </mfenced> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </msubsup> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <mfrac> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>i</mi> </msup> <mrow> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> </mfrac> <msup> <mi>e</mi> <mfrac> <mrow> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> <mrow> <msub> <mi>P</mi> <mi>s</mi> </msub> <msub> <mi>&amp;gamma;</mi> <mn>0</mn> </msub> </mrow> </mfrac> </msup> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mfrac> <mrow> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> <mrow> <msub> <mi>P</mi> <mi>s</mi> </msub> <msub> <mi>&amp;gamma;</mi> <mn>0</mn> </msub> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow>
Wherein, each parameter definition is as follows in formula：
B：Channel width；
Ps：Source node transmission power；
3) optimal threshold transmission dispatching method selection via node then stores energy to judge whether link is established using via node, Obtaining outage probability expression formula is：
<mrow> <mtable> <mtr> <mtd> <mrow> <msubsup> <mi>P</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> <mrow> <mi>o</mi> <mi>u</mi> <mi>t</mi> </mrow> </msubsup> <mo>=</mo> <msubsup> <mo>&amp;Integral;</mo> <mn>0</mn> <mi>&amp;infin;</mi> </msubsup> <mi>F</mi> <mrow> <mo>(</mo> <mfrac> <msub> <mi>P</mi> <mi>s</mi> </msub> <msub> <mi>P</mi> <mi>k</mi> </msub> </mfrac> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <msub> <mi>f</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mi>d</mi> <mi>&amp;gamma;</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>K</mi> </mtd> </mtr> <mtr> <mtd> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mtd> </mtr> </mtable> </mfenced> <mrow> <mo>(</mo> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </msubsup> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>i</mi> </msup> <mo>&amp;lsqb;</mo> <mfrac> <mn>1</mn> <mrow> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> </mfrac> <mo>-</mo> <mfrac> <mn>1</mn> <mrow> <mo>&amp;lsqb;</mo> <mfrac> <msub> <mi>P</mi> <mi>s</mi> </msub> <msub> <mi>P</mi> <mi>k</mi> </msub> </mfrac> <mo>+</mo> <mrow> <mo>(</mo> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>&amp;rsqb;</mo> </mrow> </mfrac> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow>
Wherein, each parameter definition is as follows in formula：
Pk=2Emax/T：Via node k repeating power；
EmaxVia node energy storage maximum capacity；
T information be sent to destination node from source node required for time, information source node to via node with from via node Time to destination node is identical, is T/2；
4) system selects Largest Mean average transmission rate E [C from qualified via node(k)], select optimal relaying section Point k*Meet following formula：
<mrow> <mi>max</mi> <mi>E</mi> <mo>&amp;lsqb;</mo> <mo>&amp;lsqb;</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>K</mi> </msubsup> <msub> <mi>C</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mo>&amp;rsqb;</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>6</mn> <mo>)</mo> </mrow> </mrow>
Constraints is：E[O(k)]≤Emax
Step 4.1：For the constraints in formula (6), minimum power and maximum energy storage relation are converted to, obtains following formula：
<mrow> <mi>max</mi> <mi> </mi> <mi>E</mi> <mo>&amp;lsqb;</mo> <mo>&amp;lsqb;</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>K</mi> </msubsup> <msub> <mi>C</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mo>&amp;rsqb;</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>7</mn> <mo>)</mo> </mrow> </mrow>
Constraints：Pk≥2Emax/T
Step 4.2：For the constraints in formula, the outage probability for using step 3) instead represents, i.e., when outage probability is less than threshold It is worth then link establishment, obtains following formula：
<mrow> <mtable> <mtr> <mtd> <mrow> <mi>E</mi> <mo>&amp;lsqb;</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>K</mi> </msubsup> <msub> <mi>C</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mo>=</mo> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>C</mi> <mrow> <mo>(</mo> <mi>K</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msubsup> <mi>P</mi> <mrow> <mo>(</mo> <mi>K</mi> <mo>)</mo> </mrow> <mrow> <mi>o</mi> <mi>p</mi> <mi>t</mi> </mrow> </msubsup> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>K</mi> <mo>-</mo> <mn>1</mn> </mrow> </msubsup> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>C</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mo>(</mo> <mn>1</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>-</mo> <msubsup> <mi>P</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> <mrow> <mi>o</mi> <mi>p</mi> <mi>t</mi> </mrow> </msubsup> <mo>)</mo> <msubsup> <mi>&amp;Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> <mi>K</mi> </msubsup> <msubsup> <mi>P</mi> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> <mrow> <mi>o</mi> <mi>p</mi> <mi>t</mi> </mrow> </msubsup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <msubsup> <mi>&amp;Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>K</mi> </msubsup> <msubsup> <mi>P</mi> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> <mrow> <mi>o</mi> <mi>p</mi> <mi>t</mi> </mrow> </msubsup> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>C</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>C</mi> <mrow> <mo>(</mo> <mi>K</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msubsup> <mi>P</mi> <mrow> <mo>(</mo> <mi>K</mi> <mo>)</mo> </mrow> <mrow> <mi>o</mi> <mi>p</mi> <mi>t</mi> </mrow> </msubsup> <mo>)</mo> </mrow> <mo>+</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>K</mi> <mo>-</mo> <mn>1</mn> </mrow> </msubsup> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>C</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <msubsup> <mi>P</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> <mrow> <mi>o</mi> <mi>p</mi> <mi>t</mi> </mrow> </msubsup> <mo>)</mo> </mrow> <msubsup> <mi>&amp;Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> <mi>K</mi> </msubsup> <msubsup> <mi>P</mi> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> <mrow> <mi>o</mi> <mi>p</mi> <mi>t</mi> </mrow> </msubsup> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow>
UtilizeOptimal via node k is selected when trying to achieve greatest hope transmission rate*；
Step 4.3：From source node the energy that via node consumed is sent to according to information is equal to it forwarded from via node The energy consumed to destination node, try to achieve system average energy consumption E [O(k)]：
<mrow> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>O</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mo>=</mo> <mi>T</mi> <mo>/</mo> <mn>2</mn> <msubsup> <mo>&amp;Integral;</mo> <mrow> <msub> <mi>&amp;gamma;</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>=</mo> <mn>0</mn> </mrow> <mi>&amp;infin;</mi> </msubsup> <msubsup> <mo>&amp;Integral;</mo> <mrow> <msubsup> <mi>&amp;gamma;</mi> <mi>k</mi> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <mfrac> <msub> <mi>P</mi> <mi>s</mi> </msub> <msub> <mi>P</mi> <mi>k</mi> </msub> </mfrac> <msub> <mi>&amp;gamma;</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> </mrow> <mi>&amp;infin;</mi> </msubsup> <mfrac> <msub> <mi>&amp;gamma;</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <msubsup> <mi>&amp;gamma;</mi> <mi>k</mi> <mo>&amp;prime;</mo> </msubsup> </mfrac> <msub> <mi>P</mi> <mi>s</mi> </msub> <msub> <mi>dF</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mi>d</mi> <mi>F</mi> <mrow> <mo>(</mo> <msubsup> <mi>&amp;gamma;</mi> <mi>k</mi> <mo>&amp;prime;</mo> </msubsup> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow>
Abbreviation formula (9), obtaining the average power consumption after order statistics of via node transmission rate is：
<mfenced open = "" close = ""> <mtable> <mtr> <mtd> <mrow> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>O</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mo>=</mo> <mi>T</mi> <mo>/</mo> <mn>2</mn> <msubsup> <mo>&amp;Integral;</mo> <mrow> <msub> <mi>&amp;gamma;</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>=</mo> <mn>0</mn> </mrow> <mi>&amp;infin;</mi> </msubsup> <msubsup> <mo>&amp;Integral;</mo> <mrow> <msubsup> <mi>&amp;gamma;</mi> <mi>k</mi> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <mfrac> <msub> <mi>P</mi> <mi>s</mi> </msub> <msub> <mi>P</mi> <mi>k</mi> </msub> </mfrac> <msub> <mi>&amp;gamma;</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> </mrow> <mi>&amp;infin;</mi> </msubsup> <mfrac> <msub> <mi>&amp;gamma;</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <msubsup> <mi>&amp;gamma;</mi> <mi>k</mi> <mo>&amp;prime;</mo> </msubsup> </mfrac> <msub> <mi>P</mi> <mi>s</mi> </msub> <msub> <mi>dF</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mi>d</mi> <mi>F</mi> <mrow> <mo>(</mo> <msubsup> <mi>&amp;gamma;</mi> <mi>k</mi> <mo>&amp;prime;</mo> </msubsup> <mo>)</mo> </mrow> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mfrac> <mi>T</mi> <mn>2</mn> </mfrac> <msub> <mi>P</mi> <mi>s</mi> </msub> <msubsup> <mo>&amp;Integral;</mo> <mrow> <msub> <mi>&amp;gamma;</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>=</mo> <mn>0</mn> </mrow> <mi>&amp;infin;</mi> </msubsup> <msub> <mi>&amp;gamma;</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <msubsup> <mo>&amp;Integral;</mo> <mrow> <msubsup> <mi>&amp;gamma;</mi> <mi>k</mi> <mo>&amp;prime;</mo> </msubsup> <mo>=</mo> <mfrac> <msub> <mi>P</mi> <mi>s</mi> </msub> <msub> <mi>P</mi> <mi>k</mi> </msub> </mfrac> <msub> <mi>&amp;gamma;</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> </mrow> <mi>&amp;infin;</mi> </msubsup> <mfrac> <mn>1</mn> <msubsup> <mi>&amp;gamma;</mi> <mi>k</mi> <mo>&amp;prime;</mo> </msubsup> </mfrac> <mi>f</mi> <mrow> <mo>(</mo> <msubsup> <mi>&amp;gamma;</mi> <mi>k</mi> <mo>&amp;prime;</mo> </msubsup> <mo>)</mo> </mrow> <msubsup> <mi>d&amp;gamma;</mi> <mi>k</mi> <mo>&amp;prime;</mo> </msubsup> <msub> <mi>dF</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mo>=</mo> </mrow> </mtd> </mtr> </mtable> </mfenced>
<mrow> <mfrac> <mrow> <msub> <mi>TP</mi> <mi>s</mi> </msub> </mrow> <mrow> <mn>2</mn> <msub> <mi>&amp;gamma;</mi> <mn>0</mn> </msub> </mrow> </mfrac> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mi>&amp;gamma;</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>&amp;infin;</mi> </msubsup> <msub> <mi>&amp;gamma;E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>P</mi> <mi>s</mi> </msub> <mi>&amp;gamma;</mi> </mrow> <mrow> <msub> <mi>P</mi> <mi>k</mi> </msub> <msub> <mi>&amp;gamma;</mi> <mn>0</mn> </msub> </mrow> </mfrac> <mo>)</mo> </mrow> <msub> <mi>f</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mrow> <mo>(</mo> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mi>d</mi> <mi>&amp;gamma;</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow>
Formula (1) is substituted into formula (10), and utilizes integral formula Obtain：
<mrow> <mtable> <mtr> <mtd> <mrow> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>O</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>TP</mi> <mi>s</mi> </msub> </mrow> <mn>2</mn> </mfrac> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>K</mi> </mtd> </mtr> <mtr> <mtd> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mtd> </mtr> </mtable> </mfenced> <mfrac> <mrow> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> <msubsup> <mi>&amp;gamma;</mi> <mn>0</mn> <mn>2</mn> </msubsup> </mfrac> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </msubsup> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>i</mi> </msup> <msubsup> <mo>&amp;Integral;</mo> <mrow> <mi>&amp;gamma;</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>&amp;infin;</mi> </msubsup> <msup> <mi>&amp;gamma;e</mi> <mrow> <mo>-</mo> <mfrac> <mrow> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> <msub> <mi>&amp;gamma;</mi> <mn>0</mn> </msub> </mfrac> <mi>&amp;gamma;</mi> </mrow> </msup> <msub> <mi>E</mi> <mn>1</mn> </msub> <mrow> <mo>(</mo> <mfrac> <msub> <mi>P</mi> <mi>s</mi> </msub> <mrow> <msub> <mi>P</mi> <mi>k</mi> </msub> <msub> <mi>&amp;gamma;</mi> <mn>0</mn> </msub> </mrow> </mfrac> <mi>&amp;gamma;</mi> <mo>)</mo> </mrow> <mi>d</mi> <mi>&amp;gamma;</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mi>TP</mi> <mi>s</mi> </msub> </mrow> <mn>2</mn> </mfrac> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>K</mi> </mtd> </mtr> <mtr> <mtd> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mtd> </mtr> </mtable> </mfenced> <mo>(</mo> <mi>K</mi> <mo>-</mo> <mi>k</mi> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <mn>1</mn> <mo>)</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>0</mn> </mrow> <mrow> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mrow> </msubsup> <mfenced open = "(" close = ")"> <mtable> <mtr> <mtd> <mi>k</mi> <mo>-</mo> <mn>1</mn> </mtd> </mtr> <mtr> <mtd> <mi>i</mi> </mtd> </mtr> </mtable> </mfenced> <msup> <mrow> <mo>(</mo> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mi>i</mi> </msup> <mfrac> <mn>1</mn> <msup> <mrow> <mo>(</mo> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>)</mo> </mrow> <mn>2</mn> </msup> </mfrac> <mo>&amp;lsqb;</mo> <mi>ln</mi> <mo>(</mo> <mn>1</mn> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <mfrac> <mrow> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> <msub> <mi>P</mi> <mi>s</mi> </msub> </mfrac> <msub> <mi>P</mi> <mi>k</mi> </msub> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>-</mo> <mfrac> <mrow> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> </mrow> <mrow> <mi>K</mi> <mo>-</mo> <mi>k</mi> <mo>+</mo> <mi>i</mi> <mo>+</mo> <mn>1</mn> <mo>+</mo> <mfrac> <msub> <mi>P</mi> <mi>s</mi> </msub> <msub> <mi>P</mi> <mi>k</mi> </msub> </mfrac> </mrow> </mfrac> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow>
By formula (5) (11), obtain system and it is expected energy expenditure
<mrow> <mtable> <mtr> <mtd> <mrow> <mi>E</mi> <mo>&amp;lsqb;</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>0</mn> </mrow> <mi>K</mi> </msubsup> <msub> <mi>O</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>O</mi> <mrow> <mo>(</mo> <mi>K</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mo>+</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>K</mi> <mo>-</mo> <mn>1</mn> </mrow> </msubsup> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>O</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <msubsup> <mi>&amp;Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> <mi>K</mi> </msubsup> <msubsup> <mi>P</mi> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> <mrow> <mi>o</mi> <mi>p</mi> <mi>t</mi> </mrow> </msubsup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>+</mo> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>O</mi> <mrow> <mo>(</mo> <mn>0</mn> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <msubsup> <mi>&amp;Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>K</mi> </msubsup> <msubsup> <mi>P</mi> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> <mrow> <mi>o</mi> <mi>p</mi> <mi>t</mi> </mrow> </msubsup> </mrow> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>=</mo> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>O</mi> <mrow> <mo>(</mo> <mi>K</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <mo>+</mo> <msubsup> <mi>&amp;Sigma;</mi> <mrow> <mi>k</mi> <mo>=</mo> <mn>1</mn> </mrow> <mrow> <mi>K</mi> <mo>-</mo> <mn>1</mn> </mrow> </msubsup> <mi>E</mi> <mo>&amp;lsqb;</mo> <msub> <mi>O</mi> <mrow> <mo>(</mo> <mi>k</mi> <mo>)</mo> </mrow> </msub> <mo>&amp;rsqb;</mo> <msubsup> <mi>&amp;Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mi>k</mi> <mo>+</mo> <mn>1</mn> </mrow> <mi>K</mi> </msubsup> <msubsup> <mi>P</mi> <mrow> <mo>(</mo> <mi>i</mi> <mo>)</mo> </mrow> <mrow> <mi>o</mi> <mi>p</mi> <mi>t</mi> </mrow> </msubsup> </mrow> </mtd> </mtr> </mtable> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> <mo>.</mo> </mrow>
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