CN102595458A - Distributed multi-path route-healing method - Google Patents

Abstract

The invention discloses a distributed multi-path route-healing method. The method comprises the following steps: nodes of an RRER (Route Replay) message are received and route research is initiated; an intermediate node judges a received RREQ (Route Request) message; the intermediate node forwards an RREQ message with link parameters; routing selection of the RREQ message with parameter is carried out by a destination node; and the destination node returns RREP (Route Space) messages and sets up and repair the routes. According to the method, aiming at the transmission characteristics of different links in Ad hoc network and combining with QoS requirements of the bearer service, multiple transmission paths which not only satisfy the QoS (Quality of Service) requirements of the bearer service but also have minimum repair expenditure are selected in the multiple selectable repair paths, so as to achieve network repair and effectively improve the repair benefits and the invulnerability of the repaired network.

Description

A kind of distributed multipath method for repairing route

Technical field

The invention belongs to wireless communication technology field, particularly the multipath method for repairing route in the wireless self-organization network.

Background technology

Along with the development of distributed wireless networks technology and communication service, loaded service type and QoS demand are in continuous increase, for example on distributed wireless networks: need the data service of high speed reliable transmission, the video that needs real-time Transmission or audio service etc.In the existing distributed wireless network, Ad hoc network because of its flexibly self-organization just receiving application more and more widely.

On the one hand, Ad hoc network organizing is convenient, flexible.Ad hoc network need not to be provided with any center control nodes, and all node status wherein all are equality.Node in the network not only has the transceive data function of common mobile terminal, but also has the message relay transfer capability.When source node and the destination node of communication was not in direct communications range, they can carry out multi-hop communication through the form of a plurality of intermediate node relayings forwardings.On the other hand, the Radio Link between Ad hoc network node is subject to multiple factor affecting and produces interruption, thereby influences the business transmission of the network carrying.Because Ad hoc network node has characteristics such as removable and energy constraint, owing to move or depleted of energy when causing the network topology structure variation, existing Radio Link will interrupt when node.In addition, the air interference of Ad hoc network vulnerable environment of living in also can make wireless chaining status change or interrupts.

At present adopt dual mode mostly for the reparation of Ad hoc network broken link, a kind of is overall situation reparation, and promptly the source node from communication begins, and searches the route that arrives destination node again.This repair mode can produce bigger time delay and bandwidth cost in the pathfinding process.Particularly when the interruption of communication link is the local interruption that causes because individual node moves, can bring higher reparation expense and the serious system availability that reduces with the method that the overall situation is repaired.A kind of in addition repair mode commonly used is local the reparation, promptly initiates the route repair process from the neighbor node that interrupts node, in interrupting node periphery range of nodes, carries out routing again.The coverage of this reparation is less, and bandwidth of being brought and time overhead overall situation reparation relatively decrease, but in more abominable communication environment, possibly have a plurality of malfunctioning nodes and many broken link in the network, and local reparation may be failed.

The route recovery technology of Ad hoc network routing protocol is adopted in existing Ad hoc network broken link reparation more.Typical A d hoc network routing protocol has with DSDV (Destination-Sequenced Distance-Vector Routing) etc. and is the priori formula Routing Protocol of representative and is the reaction equation Routing Protocol of representative with AODV (Ad hoc On-Demand Distance Vector Routing), AOMDV (Ad hoc On-Demand Multipath Distance Vector Routing) etc.

The DSDV Routing Protocol generates and safeguards route through regularly mutual route message between neighbor node, and its route table items generates prior to data transfer needs.When the network wireless link occurs fault; Fault message is surveyed and is broadcasted by the neighbor node of malfunctioning node; Inundation is carried out in this broadcasting in network; After the source node of communication is received fault message, search in the routing table other alternative routes that arrive destination node, and with this path as repair path.In the more frequent Ad hoc network of change in topology, the priori routing iinformation that the DSDV agreement is safeguarded can lose efficacy very soon, and the success rate that network is repaired is lower.

The AODV agreement belongs to a kind of typical reaction formula Routing Protocol, when source node when destination node is sent bag, source node just can be initiated the route querying process in network.When the Radio Link in the network breaks down; Malfunctioning node sends fault message to the source node direction; This information is transmitted to source node along original route in the other direction; After source node is received fault message, will initiate the route querying process once more, search a new route that arrives destination node as repair path through broadcasting.Though AODV has solved the inefficacy property problem of priori routing iinformation in the DSDV repair process, its repair path is Dan Jing, can not effectively utilize the ability of mulitpath parallel transmission in the Ad hoc network.

The AOMDV agreement has then been introduced multipath thought, realizes the multipath transmission through obtaining many acyclic and the disjoint paths of link, has reduced the dependency degree of network connectivty to individual paths.Promptly when certain the bar link occurs fault in the network, machine-processed according to the AOMDV multipath, the path that source node can be abandoned breaking down, and transmit in other paths of selecting to have set up.

CHAMP (CacHing And Multiple Path) agreement utilization bag caching mechanism and the shortest multipath route pathfinding mechanism are when downstream node breaks down; Through reporting to its upstream node; Message by upstream node utilizes caching mechanism will just transmit is again transmitted according to the backup path of the shortest multipath pathfinding mechanism foundation, reaches the purpose of quick formation repair path.

QRMP (QoS Routing with Mobility Prediction) agreement is then when node motion produces transmission path fault; Send the route request of rebuilding by source node with the limited mode of flooding, and utilize the node mobility forecasting mechanism source, destination node between seek the reparation that failure path is realized in the most steady path of satisfying the QoS demand.

Existing Ad hoc network recovery technique is primarily aimed at the reachability problem of transmission path, the path of breaking off is repaired with the fastest mode, but do not consider the survivability problem of repair path itself.For example: a certain zone of Ad hoc network receives lasting interference, and single footpath repair path that DSDV, AODV and QRMP repair back formation possibly interrupt once more, and this will trigger once new repair process, causes too much reparation expense thus.

Secondly; More existing technology are not considered the QoS demand of Business Stream in repair process;, do not carry out choose reasonable according to demand for services parameters such as the bandwidth of Business Stream, time delay, packet loss though but there are many repair paths in AOMDV and CHAMP agreement.

In addition; Existing recovery technique is not often considered the reparation expense in path; Network node can consume certain resource in repair process, for example: the repair mode that node energy, transmission bandwidth etc., choose reasonable have low overhead value is the effective means that improves network system efficient.AODV and QRMP agreement all are the route repair of being initiated by source node, and its repair process is not optimized processing according to the link repair expense, with consuming too much resource.

Summary of the invention

The objective of the invention is to have proposed a kind of distributed multipath method for repairing route in order to solve the problems referred to above that route repair exists in the existing wireless self-organization network.

Technical scheme of the present invention is: a kind of distributed multipath method for repairing route specifically comprises the steps:

S1: certain node in the middle of a communication link of network breaks down; The upstream node of this malfunctioning node direction towards the source sends the RRER message; Receive the node initiation routing broadcast message RREQ of RRER message, seek the route of the downstream node that arrives the malfunctioning node direction;

S2: the RREQ message to intermediate node is received is judged, if the source address of the message that this RREQ message is a node oneself to send or this RREQ message appears in the historical record of intermediate node with request ID, then abandons this message; If the source address of this RREQ message does not appear in the historical record of this intermediate node with request ID, then this source address is write in the historical record with request ID, search towards the path of this message destination address, change step S3;

S3: if the aim sequence in the path found number is greater than the aim sequence in the RREQ message number; Then show and found a paths; In the reparation routing parameter of RREQ message, fill in the parameter information of the link between this node and its upstream node, transmit the RREQ message to the next-hop node of institute's storage purpose node; This intermediate node extracts the information in the RREQ message simultaneously, is used for making up reverse routing table, and packet can transmit along the rightabout of institute's routing when guaranteeing that this path is selected, waits for regular hour Δ T, if overtime then delete reverse routing table; Said regular hour Δ T should be able to guarantee that the RREQ message can pass whole network and produce a RREP message that sends to source node;

S4: after destination node receives the RREQ packet; Wait for the time of second of Δ T at least; At Δ T in the time of second, the message that never arrives with the path is carried out qos parameter and repairs routing parameter and extract, carry out route selection; Routing iinformation to after selecting makes up corresponding RREP message, last these RREP messages of destination node clean culture; Each RREP message is given source node according to different reverse routing table loopbacks, and is every through an intermediate node, and then hop count adds 1 and write down the information of this message, for each bar Business Stream is set up the forward direction routing table as repairing the routing table that successful posterior nodal point uses;

S5: each RREP message is got back to source node according to the order hop-by-hop of intermediate node, and every the hop count of RREP message adds 1 through an intermediate node, and each intermediate node extracts the relevant information of RREP message simultaneously, makes up the forward direction routing table.

Further, said RREQ message comprises qos parameter and repairs routing parameter.

Further, said qos parameter comprises time delay, bandwidth, the packet loss of current each Business Stream, and said reparation routing parameter is to initiate node to time delay, bandwidth, the packet loss of repairing the link between all adjacent nodes the destination node from repairing.

Step S4 is described, and to carry out the detailed process of route selection following:

S41: destination node compares the qos parameter of each bar link and the qos parameter of Business Stream, selects available stream, makes up the bandwidth traffic allocation matrix $C=\left[\begin{array}{cccc}{c}_{11}& c_{12}& \mathrm{\Λ}& c_{1n}\\ c_{21}& c_{22}& \mathrm{\Λ}& c_{2n}\\ M& M& O& M\\ c_{m1}& c_{m2}& \mathrm{\Λ}& c_{\mathrm{Mn}}\end{array}\right],$ But the number of m representative routing, n represents the number of Business Stream, and each row of matrix is represented a link, and each row is represented a Business Stream, c _IjRepresent i bar chain road direction j row traffic flow assignment cost for bandwidth; If the bandwidth resources of link are greater than the bandwidth demand amount of Business Stream; Then be provided with a virtual service flow absorb more than resource; Its expense is made as 0, for not giving the link bandwidth of traffic flow assignment, its expense is made as infinity;

S42: obtain initial allocative decision: in the bandwidth traffic allocation matrix that constitutes, at first at minimal-overhead expense c with the least member method _IjCarry out bandwidth traffic between link corresponding i and the Business Stream j and distribute, the bandwidth stream value of distribution is got the minimum value of path bandwidth and business demand bandwidth.If the bandwidth traffic in path has assigned and has then deleted the corresponding overhead cost of this row; If the bandwidth of Business Stream has assigned and has then deleted the corresponding overhead cost of these row; In the overhead cost of not scratched, seek minimum value then; And make bandwidth traffic and distribute, finish up to whole assignment of traffic, formed an initial allocative decision this moment;

S43: the check number of asking nonbasic variable with the potential method: the matrix of structure one (m+1) * (n+1) $C^{\′}=\left[\begin{array}{ccccc}{c}_{11}^{\′}& c_{12}^{\′}& \mathrm{\Λ}& c_{1n}^{\′}& u_1\\ c_{21}^{\′}& c_{22}^{\′}& \mathrm{\Λ}& c_{2n}^{\′}& u_2\\ M& M& O& M& M\\ c_{m1}^{\′}& c_{m2}^{\′}& \mathrm{\Λ}& c_{\mathrm{Mn}}^{\′}& u_m\\ v_1& v_2& \mathrm{\Λ}& v_n& 0\end{array}\right],$ Make v _n=0, for each basic lattice c ' is arranged _Ij=u _i+ v _j, wherein

Be expense, utilize initial condition v _n=0, obtain all u _i, v _j, the check number of calculating nonbasic variable

If satisfy σ _Ij>=0, then current allocative decision is exactly an optimal distributing scheme, and calculating stops, otherwise changes step S44;

S44: get the minimum nonbasic variable of a check number and make into basic variable; Its corresponding lattice are for advancing Ji Ge; To advance basic lattice is that starting point is made the loop circuit that all the other summits are basic lattice, on the loop circuit, from the bandwidth traffic of all even number lattice points, selects minimum value θ as adjustment amount, and these lattice are from basic lattice; Corresponding variable is for leaving basic variable, θ=min{x _Ij, (i j) is even number lattice point in the loop circuit };

S45: the bandwidth traffic on the loop circuit is made adjustment: the freight volume of all odd number lattice adds adjustment amount θ, and the freight volume of all even number lattice deducts adjustment amount θ, obtains a new bandwidth traffic allocative decision, changes step S43.

Beneficial effect of the present invention:, the present invention proposes a kind of based on transport service QoS demand and have the minimum multipath method for repairing route of repairing expense to the problem that existing route recovery technique in the Ad hoc network exists.This method is directed against the transmission feature of different links in the Ad hoc network and combines the QoS demand of bearer service; In many optional repair paths; Selection can be satisfied bearer service QoS demand; Have minimum many transmission paths of repairing expense simultaneously and realize the network reparation, effectively promoted the survivability of repairing income and having repaired network.

Description of drawings

Fig. 1 is the inventive method schematic flow sheet.

Fig. 2 is a routing failure RRER message format sketch map.

Fig. 3 is a route requests RREQ message format sketch map.

Fig. 4 is a route replies RREP message format sketch map.

Fig. 5 is a faulty link sketch map among the embodiment.

Fig. 6 is the link sketch map of repair process 1 among the embodiment.

Fig. 7 is the link sketch map of repair process 2 among the embodiment.

Embodiment

Below in conjunction with accompanying drawing and concrete embodiment the present invention is done further elaboration.

At first the basic conception of using is done an explanation:

Basic variable: the place of dispense flow rate in the bandwidth traffic allocation table.

Nonbasic variable: the place that does not have dispense flow rate in the bandwidth traffic allocation table.

Ji Ge: basic variable x _IjCorresponding lattice (i, j).

Loop circuit: if one group of lattice point can be write as following form: (i through after the suitable ordering ₁, j ₁), (i ₁, j ₂), (i ₂, j ₂), (i ₂, j ₃), (i ₃, j ₃) ..., (i _s, j _s), (i _s, j ₁), claim that then this group lattice point constitutes the loop circuit.Wherein the i representative is gone, and j represents row, and its subscript is represented line number or columns.Concerning a loop circuit, having a clockwise direction starting point is exactly No. 1 grid, and then is exactly No. 2 grid, is No. 3 grid then, thereby can define odd number lattice and even number lattice.

The schematic flow sheet of a kind of distributed multipath method for repairing route of the present invention is as shown in Figure 1, specifically comprises the steps:

Receive the node initiation route querying of RRER message: certain node in the middle of a communication link of network breaks down; The upstream node of this malfunctioning node direction towards the source sends the RRER message; RRER message format sketch map is as shown in Figure 2; Receive the node initiation routing broadcast message RREQ of RRER message, RREQ message format sketch map is as shown in Figure 3, seeks the route of the downstream node that arrives the malfunctioning node direction;

Intermediate node is judged the RREQ message of receiving: the RREQ message to intermediate node is received is judged; If the source address of the message that this RREQ message is a node oneself to send or this RREQ message appears in the historical record of intermediate node with request ID, then abandon this message; If the source address of this RREQ message does not appear in the historical record of this intermediate node with request ID, then this source address is write in the historical record with request ID, search towards the path of this message destination address, change step S3;

Intermediate node is transmitted the RREQ message of band chain road parameter: if the aim sequence in the path of finding number is greater than the aim sequence in the RREQ message number; Then show and found a paths; In the reparation routing parameter of RREQ message, fill in the parameter information of the link between this node and its upstream node, transmit the RREQ message to the next-hop node of institute's storage purpose node; This intermediate node extracts the information in the RREQ message simultaneously, is used for making up reverse routing table, waits for regular hour Δ T, if overtime then delete reverse routing table; Said regular hour Δ T should be able to guarantee that the RREQ message can pass whole network and produce a RREP message that sends to source node;

Destination node is carried out Path selection to band parameters R REQ message: after destination node receives the RREQ packet; Wait for the time of second of Δ T at least, at Δ T in the time of second, the message that never arrives with the path is carried out qos parameter and repairs routing parameter and extract; Carry out route selection; Routing iinformation to after selecting makes up corresponding RREP message, and RREP message format sketch map is as shown in Figure 4, last these RREP messages of destination node clean culture; Each RREP message is given source node according to different reverse routing table loopbacks, and is every through an intermediate node, and then hop count adds 1 and write down the information of this message, for each bar Business Stream is set up the forward direction routing table;

Destination node loopback RREP message; Set up and repair route: each RREP message is got back to source node according to the order hop-by-hop of intermediate node, and every through an intermediate node, the hop count of RREP message adds 1; Each intermediate node extracts the relevant information of RREP message simultaneously, makes up the forward direction routing table.

Specifically describe with two embodiment below.

Embodiment one:

Like mistake! Do not find Reference source.In the shown wireless self-organization network, have one from source node through N1, N2, N3, N4 communication link to destination node, on this communication link, have a node N3 to break down.Be the process that the link at the place of breaking down is repaired below.

Step 1: N3 breaks down in the middle of a communication link of network; Send the RRER message by N3 to its upstream node (towards the direction in source); RRER message N2 initiates routing broadcast message RREQ immediately, seeks the route of downstream node (towards the direction of the purpose) N4 that arrives malfunctioning node.In order to limit the broadcasting of message, carry out local route repair, the value that hop count is set here is 2.The qos parameter of message is as shown in table 1; Comprise the qos parameter (time delay, bandwidth, packet loss) of each Business Stream at present; The reparation routing parameter is as shown in table 2, for initiating node to repairing the link QoS information (time delay, bandwidth, packet loss) between all adjacent nodes the destination node from repairing.

Table 1

Table 2

Step 2: intermediate node N5, N6, N7 judge the RREQ message of receiving; The source address of this RREQ message does not appear in the historical record of this intermediate node with request ID; Then this source address is write in the historical record with request ID, searched towards the path of this message destination address;

The aim sequence in the path that step 3:N5, N6, N7 find number is greater than the aim sequence in the RREQ message number; Show and found a paths; In the reparation routing parameter of RREQ message, fill in parameter (bandwidth, time delay, the packet loss) information of the link between this node and its upstream node N2; Hop count adds 1, transmits the RREQ message to next-hop node (the being destination node) N4 of institute's storage purpose node; Simultaneously; N5, N6, N7 node extract the information in the RREQ message; Be used for making up reverse routing table; Wait for regular hour Δ T, said regular hour Δ T should be able to guarantee that the RREQ message can pass whole network and produce a RREP message that sends to source node, if overtime then delete reverse routing table;

After step 4: destination node N4 receives the RREQ packet, wait for the time of second of Δ T at least, at Δ T in the time of second; The message that never arrives with the path is carried out qos parameter and repairs routing parameter and extract; The path that obtains has: N2->N5->N4, N2->N6->N4, N2->N7->N4; N2->N5->N6->N4, N2->N7>N6->N4.

The bandwidth on these 5 roads, time delay, ` packet loss parameter can be represented with matrix X: the qos parameter value on every road of capable vector representation of matrix wherein, and for each row vector, its first row parameter is a bandwidth, the secondary series parameter is that time delay, the 3rd row parameter are packet loss.

Table 3 bandwidth traffic allocation table

Article three, the bandwidth of Business Stream, time delay, packet loss parameter can be represented with matrix Y: the qos parameter value of every Business Stream of capable vector representation of matrix wherein; For each row vector; Its first row parameter is a bandwidth, and the secondary series parameter is that time delay, the 3rd row parameter are packet loss.

Qos parameter (the row vector of Y matrix) that now will every Business Stream compares with the qos parameter (row of X matrix is vectorial) on each bar road, and delay parameter and the packet loss of finding out each bar Business Stream is all more than or equal to all couplings of the relevant parameter on each bar road.Utilize the Optimum Theory method to carry out route selection and (see step 6); Routing iinformation after selecting is passed through to make up a plurality of RREP messages; Source address and source sequence number that the destination address of RREP message and aim sequence number duplicate RREQ respectively; The source sequence number of RREP message is the latest sequence number value of node maintenance, and hop-count value is changed to 0, last these RREP messages of destination node clean culture; Each RREP message is passed to source node according to different paths, and the information of this message of intermediate node record is for each bar Business Stream foundation is transmitted.

Step 5: each RREP message is transmitted back to source node according to different reverse routes, and every the hop count of RREP message adds one through an intermediate node, and each intermediate node extracts the relevant information of RREP message simultaneously, makes up the forward direction routing table.

Step 6: Optimum Theory route selecting method.Concrete grammar is seen S41, S42, S43, S44, the S45 of summary of the invention, can obtain table 4 by S43.

Table 4 potential method is asked the check numerical table

The existing bandwidth value of known five available road A1, A2, A3, A4, A5 is respectively 6,8, and the bandwidth demand value of 10,8,6, three Business Stream B1, B2, B3 is respectively 5,6,3, and the expense of each traffic flow assignment bandwidth is given on each bar road, and is as shown in table 5.

Table 5

Can know that by table 5 the total bandwidth resource on available road is greater than the demand of Business Stream, need Business Stream B4 of imagination to be used for the bandwidth of absorbing redundant now, make its bandwidth demand amount B4=38-14=24, expense c4=0.

Ask initial allocation and transportation scheme with the least member method: at first from overhead cost table (table 6), find minimal-overhead x14=0, mark with containing the underscore font in the table.

Table 6

Give B4 the bandwidth resource allocation of A1 then, the numerical value of distribution is the minimum value of A1 existing resource and the current demand of B4.Table 7 is the assignment of traffic table, contains the adjustment of underscore font representation flow.

Table 7

	B1	B2	B3	B4	Existing resource
						A1				6	6-6
A2					8
						A3					10
A4					8
						A5					6
Demand	5	6	3	24 -6

This moment, the A1 existing resource was 0, had not had resource to divide again, scratched expense that A1 is expert at, and found minimal-overhead in the expense of never scratching again, and is as shown in table 8.

Table 8

	B1	B2	B3	B4	Existing resource
						A1	Scratch	Scratch	Scratch	Scratch	6 -6
A2				0	8
						A3					10

A4					8
						A5					6
Demand	5	6	3	24 -6

Give B4 the A2 resource allocation then, as shown in table 9.

Table 9

	B1	B2	B3	B4	Existing resource
						A1	Scratch	Scratch	Scratch	Scratch	6-6
A2				8	8 -8
						A3					10
A4					8
						A5					6
Demand	5	6	3	24-6 -8

This moment, A2 did not have resource to provide, scratched expense that A2 is expert at, and found minimal-overhead in the expense of never scratching again, and is as shown in table 10.

Table 10

	B1	B2	B3	B4	Existing resource
						A1	Scratch	Scratch	Scratch	Scratch	6-6
A2	Scratch	Scratch	Scratch	Scratch	8-8
						A3				0	10
A4					8
						A5					6
Demand	5	6	3	24-6-8

Give B4 the A3 resource allocation, as shown in table 11.

Table 11

	B1	B2	B3	B4	Existing resource
						A1	Scratch	Scratch	Scratch	Scratch	6-6
A2	Scratch	Scratch	Scratch	Scratch	8-8
						A3				10	10 -10
A4					8
						A5					6

Demand

5

6

3

24-6-8 -10

This moment, the bandwidth demand of B4 was satisfied, scratched B4 place row expense, and A3 does not have resource to reallocate simultaneously, scratches expense that A3 is expert at.Because ranks are scratched simultaneously, this has belonged to degenerate problem, look for need or being listed as from the row scratched this moment any space (X31, X32, X33, X44, X45) as basic lattice, putting its bandwidth traffic is 0.Select X33 as basic lattice now, putting its bandwidth traffic is 0 (table 12), finds minimal-overhead in the expense of never scratching then.

Table 12

	B1	B2	B3	B4	Existing resource
						A1	Scratch	Scratch	Scratch	Scratch	6-6
A2	Scratch	Scratch	Scratch	Scratch	8-8
						A3	Scratch	Scratch	Scratch	Scratch	10-10
A4			10	Scratch	8
						A5				Scratch	6
Demand	5	6	3	24-6-8-10

Give B3 the A4 resource allocation, as shown in table 13.

Table 13

	B1	B2	B3	B4	Existing resource
						A1	Scratch	Scratch	Scratch	Scratch	6-6
A2	Scratch	Scratch	Scratch	Scratch	8-8
						A3	Scratch	Scratch	Scratch	Scratch	10-10
A4			3	Scratch	8 -3
						A5				Scratch	6
Demand	5	6	3 -3	24-6-8-10

This moment, B3 was satisfied, scratched B3 place row expense, found minimal-overhead in the expense of never scratching again, and is as shown in table 14.

Table 14

	B1	B2	B3	B4	Existing resource
						A1	Scratch	Scratch	Scratch	Scratch	6-6
A2	Scratch	Scratch	Scratch	Scratch	8-8
						A3	Scratch	Scratch	Scratch	Scratch	10-10
A4		11	Scratch	Scratch	8-3

A5			Scratch	Scratch	6
						Demand	5	6	3-3	24-6-8-10

Give B2 the A4 resource allocation, as shown in Tble 15.

Table 15

	B1	B2	B3	B4	Existing resource
						A1	Scratch	Scratch	Scratch	Scratch	6-6
A2	Scratch	Scratch	Scratch	Scratch	8-8
						A3	Scratch	Scratch	Scratch	Scratch	10-10
A4		5	Scratch	Scratch	8-3 -5
						A5			Scratch	Scratch	6
Demand	5	6 -5	3-3	24-6-8-10

This moment, A4 did not have resource to distribute, scratched expense that A4 is expert at, and found minimal-overhead in the expense of never scratching again, and is shown in table 16.

Table 16

	B1	B2	B3	B4	Existing resource
						A1	Scratch	Scratch	Scratch	Scratch	6-6
A2	Scratch	Scratch	Scratch	Scratch	8-8
						A3	Scratch	Scratch	Scratch	Scratch	10-10
A4	Scratch	Scratch	Scratch	Scratch	8-3-5
						A5	13		Scratch	Scratch	6
Demand	5	6-5	3-3	24-6-8-10

Give B1 the A5 resource allocation, shown in table 17.

Table 17

	B1	B2	B3	B4	Existing resource
						A1	Scratch	Scratch	Scratch	Scratch	6-6
A2	Scratch	Scratch	Scratch	Scratch	8-8
						A3	Scratch	Scratch	Scratch	Scratch	10-10
A4	Scratch	Scratch	Scratch	Scratch	8-3-5
						A5	5		Scratch	Scratch	6 -5
Demand	5 -5	6-5	3-3	24-6-8-10

This moment, the B1 demand satisfied, and scratched B1 place row expense, found minimal-overhead and dispense flow rate in the expense of never scratching again, and is shown in table 18.

Table 18

	B1	B2	B3	B4	Existing resource
						A1	Scratch	Scratch	Scratch	Scratch	6-6
A2	Scratch	Scratch	Scratch	Scratch	8-8
						A3	Scratch	Scratch	Scratch	Scratch	10-10
A4	Scratch	Scratch	Scratch	Scratch	8-3-5
						A5	Scratch	1	Scratch	Scratch	6-5 -1
Demand	5-5	6-5 -1	3-3	24-6-8-10

So far, the scheme of initially allocating and transporting is come out, and is shown in table 19:

Table 19

	B1	B2	B3	B4	Existing resource
						A1				6	6-6
A2				8	8-8
						A3			0	10	10-10
A4		5	3		8-3-5
						A5	5	1			6-5-1
Demand	5-5	6-5-1	3-3	24-6-8-10

Ask check number to make the table of (M+1) * (N+1) with gesture position method, wherein M delegated path bar is several 5, and N represents professional fluxion 4.Make (M+1) behavior v _jOK, (N+1) classifies u as _iRow, v _N=0.Utilize initial condition v _N=0, and c _Ij=u _i+ v _j(c wherein _IjBe band black underscore font in the expense correspondence table), can obtain all u _i, v _jObtain all unknown quantity u _i, v _jAfterwards, utilize σ _Ij=c _Ij-(u _i+ v _j) obtain the corresponding check number σ of all Fei Jige (containing underscore font value corresponding) _Ij(angle brackets value corresponding), shown in table 20:

Table 20

	B1	B2	B3	B4	Ui
						A1	<-8>; 1	<-8>; 2	<-6>; 3	0	0
A2	<-3>; 6	<-5>; 5	<-5>; 4	0	0
						A3	<-2>; 7	<-2>; 8	9	0	0

A4	<2>; 12	11	?10	<-1>; 0	1
						A5	13	14	?<2>; 15	<-4>; 0	4
Vj	9	10	?9	0

Table 20 does not satisfy all σ Ij>=0 usefulness loop circuit method is adjusted:

The minimum value of selecting in the check number-8 (goes up table and has two-8; Optional one, suppose to select (A1 corresponding with B1-8), send out with the loop circuit and select a loop (following table contains the node that the underscore font marks the loop); get the minimum value 0 of all even number lattice points on the loop circuit; the odd number lattice point on the loop circuit adds adjustment amount 0, the even number lattice point on the loop circuit deducts adjustment amount 0, and is shown in table 21:

Table 21

	B1	B2	B3	B4	Ui
						A1	0+0			6-0
A2				8
						A3			0-0	10+0
A4		5-0	3+0
						A5	5-0	1+0
Vj

Calculate check number, shown in table 22:

Table 22

	B1	B2	B3	B4	Ui
						A1	1	<0>; 2	<2>; 3	0	0
A2	<5>; 6	<3>; 5	<3>; 4	0	0
						A3	<6>; 7	<6>; 8	<8>; 9	0	0
A4	<2>; 12	11	10	<-9>;0	9
						A5	13	14	<2>; 15	<-12>; 0	12
Vj	1	2	1	0

Check number does not meet the demands; Basic variable is left in minimum value-12 conduct of getting in the check number; Select a loop circuit (table 23 contains the node that the underscore font marks the loop), the minimum value 5 of getting all even number lattice points on the loop circuit adds adjustment amount 5 to the odd number lattice point on the loop circuit; Even number lattice point on the loop circuit deducts adjustment amount 5, and is shown in table 23:

Table 23

B1

B2

B3

B4

Ui

A1	0+5			6-5
						A2				8
A3				10
						A4		5	3
A5	5-5	1		0+5
						Vj

Calculate check number, shown in table 24:

Table 24

	B1	B2	B3	B4	Ui
						A1	1	<-12>; 2	<-10>; 3	0	0
A2	<5>; 6	<-9>; 5	<-9>; 4	0	0
						A3	<6>; 7	<-6>; 8	<-4>; 9	0	0
A4	<14>; 12	11	10	<3>; 0	-3
						A5	<12>; 13	14	<2>; 15	0	0
Vj	1	14	13	0

Check number does not meet the demands; Basic variable is left in minimum value-12 conduct of getting in the check number; Select a loop circuit (table 25 contains the node that the underscore font marks the loop), the minimum value 1 of getting all even number lattice points on the loop circuit adds adjustment amount 1 to the odd number lattice point on the loop circuit; Even number lattice point on the loop circuit deducts adjustment amount 1. owing to have unnecessary 1 even number lattice point all to equate (being 1) and all be minimum 1, and is as shown in Tble 25:

Table 25

	B1	B2	B3	B4	Ui
						A1	5	0+1		1-1
A2				8
						A3				10
A4		5	3
						A5		1-1		5+1
Vj

Calculate check number, shown in table 26:

Table 26

B1

B2

B3

B4

Ui

A1	1	2	<2>; 3	0	0
						A2	<5>; 6	<3>; 5	<3>; 4	0	0
A3	<6>; 7	<6>; 8	<8>; 9	0	0
						A4	<2>; 12	11	10	<-9>; 0	9
A5	<12>; 13	<12>; 14	<14>; 15	0	0
						Vj	1	2	1	0

Check number does not meet the demands; Basic variable is left in minimum value-9 conduct of getting in the check number; Select a loop circuit (table 27 contains the node that the underscore font marks the loop); get the minimum value 0 of all even number lattice points on the loop circuit, the odd number lattice point on the loop circuit is added adjustment amount 0, thereby the even number lattice point on the loop circuit deducts adjustment amount 0. x14 for leaving basic lattice.The flow adjustment, shown in table 27:

Table 27

	B1	B2	B3	B4	Ui
						A1	5	1+0		0-0
A2				8
						A3				10
A4		5-0	3	0+0
						A5				6
Vj

Calculate check number, shown in table 28:

Table 28

	B1	B2	B3	B4	Ui
						A1	1	2	<2>; 3	<9>; 0	-9
A2	<-4>; 6	<-6>; 5	<-6>; 4	0	0
						A3	<-3>; 7	<-3>; 8	<-1>; 9	0	0
A4	<2>; 12	11	10	0	0
						A5	<3>; 13	<3>; 14	<5>; 15	0	0
Vj	10	11	10	0

Check number does not meet the demands, and the minimum value of getting in the check number is-6, because two of existence all are-6 lattice, optional x22 is as leaving basic variable.Select a loop circuit (table 29 contains the node that the underscore font marks the loop), the minimum value 5 of getting all even number lattice points on the loop circuit adds adjustment amount 5 to the odd number lattice point on the loop circuit, and the even number lattice point on the loop circuit deducts adjustment amount 5, and is shown in table 29:

Table 29

	B1	B2	B3	B4	Ui
						A1	5	1
A2		0+5		8-5
						A3				10
A4		5-5	3	0+5
						A5				6
Vj

Calculate check number, shown in table 30:

Table 30

	B1	B2	B3	B4	Ui
						A1	1	2	<-4>; 3	<3>; 0	-3
A2	<2>; 6	5	<-6>; 4	0	0
						A3	<3>; 7	<3>; 8	<-1>; 9	0	0
A4	<8>; 12	<6>; 11	10	0	0
						A5	<9>; 13	<9>; 14	<5>; 15	0	0
Vj	4	5	10	0

Check number does not meet the demands, and the minimum value of getting in the check number leaves basic variable for-6 conducts.Select a loop circuit (table 31 contains the node that the underscore font marks the loop), the minimum value 3 of getting all even number lattice points on the loop circuit adds adjustment amount 3 to the odd number lattice point on the loop circuit, and the even number lattice point on the loop circuit deducts adjustment amount 3, and is shown in table 31:

Table 31

	B1	B2	B3	B4	Ui
						A1	5	1
A2		5	+3	3-3
						A3				10
A4			3-3	5+3
						A5				6
Vj

Calculate check number, shown in table 32:

Table 32

	B1	B2	B3	B4	Ui
						A1	1	2	<2>; 3	<3>; 0	-3
A2	<2>; 6	5	4	0	0
						A3	<3>; 7	<3>; 8	<5>; 9	0	0
A4	<8>; 12	<6>; 11	<6>; 10	0	0
						A5	<9>; 13	<8>; 14	<11>; 15	0	0
Vj	4	5	4	0

Check number meets the demands, and the optimum allocation strategy of bandwidth traffic is shown in table 33:

Table 33

	B1	B2	B3	B4	Bandwidth resources
						A1	5	1			6
A2		5	3		8
						A3				10	10
A4				8	8
						A5				6	6
Bandwidth demand	5	6	3	24

Embodiment two

Below in conjunction with mistake! Do not find Reference source.The present invention is done further elaboration.

Step 1: N3 breaks down in the middle of a communication link of network; Send the RRER message by N3 to its upstream node (towards the direction in source); N2 initiates routing broadcast message RREQ immediately, seeks the route of downstream node (towards the direction of the purpose) N4 that arrives malfunctioning node.In order to limit the broadcasting of message, carry out local route repair, the value that hop count is set here is 2.The qos parameter of message comprises the qos parameter (time delay, bandwidth, packet loss) of each Business Stream at present, and repairing routing parameter is to initiate node to repairing the link QoS information (time delay, bandwidth, packet loss) between all adjacent nodes the destination node from repairing.

Step 2: intermediate node N5, N6 judge the RREQ message of receiving; The source address of this RREQ message does not appear in the historical record of this intermediate node with request ID; Then this source address is write in the historical record with request ID, searched towards the path of this message destination address;

The aim sequence in the path that step 3:N5, N6 find number is greater than the aim sequence in the RREQ message number; Show and found a paths; In the reparation routing parameter of RREQ message, fill in parameter (bandwidth, time delay, the packet loss) information of the link between this node and its upstream node N2; Hop count adds 1, transmits the RREQ message to next-hop node (the being destination node) N4 of institute's storage purpose node; Simultaneously, N5, N6 node extract the information in the RREQ message, are used for making up reverse routing table, wait for the regular hour, if overtime then delete reverse routing table;

After step 4: destination node N4 receives the RREQ packet; Wait for the time of second of Δ T at least, at Δ T in the time of second, the message that never arrives with the path is carried out qos parameter and repairs routing parameter and extract; The path that obtains has: N2->N5->N4, N2->N6->N4.

The bandwidth on these 2 roads, time delay, packet loss parameter can be represented with matrix X: the qos parameter value on every road of capable vector representation of matrix wherein, and for each row vector, its first row parameter is a bandwidth, the secondary series parameter is that time delay, the 3rd row parameter are packet loss.

Article three, the bandwidth of Business Stream, time delay, packet loss parameter can be represented with matrix Y: the qos parameter value of every Business Stream of capable vector representation of matrix wherein; For each row vector; Its first row parameter is a bandwidth, and the secondary series parameter is that time delay, the 3rd row parameter are packet loss.

Qos parameter (the row vector of Y matrix) that now will every Business Stream compares with the qos parameter (row of X matrix is vectorial) on each bar road, and delay parameter and the packet loss of finding out each bar Business Stream is all more than or equal to all couplings of the relevant parameter on each bar road.Utilize the Optimum Theory method to carry out route selection and (see step 6); Routing iinformation after selecting is passed through to make up a plurality of RREP messages; Source address and source sequence number that the destination address of RREP message and aim sequence number duplicate RREQ respectively; The source sequence number of RREP message is the latest sequence number value of node maintenance, and hop-count value is changed to 0, last these RREP messages of destination node clean culture; Each RREP message is passed to source node according to different paths, and the information of this message of intermediate node record is for each bar Business Stream foundation is transmitted.

Step 5: each RREP message is transmitted back to source node according to different reverse routes, and every the hop count of RREP message adds one through an intermediate node, and each intermediate node extracts the relevant information of RREP message simultaneously, makes up the forward direction routing table.

Step 6: Optimum Theory route selecting method

Two road A1 that selected, A2 and three Business Stream B1, B2, B3 are arranged now, and the total bandwidth resource on the road of selecting is bigger, shown in table 34 than the total bandwidth resource of three traffic flow requirements:

Table 34

This is the problem of a supply greater than demand, Business Stream B4 of imagination, and demand b4=14-9=5, expense c4=0 asks initial allocation and transportation scheme with the least member method:

From table 35, find minimal-overhead 0, minimal-overhead is with containing the underscore font representation.

Table 35

Give B4 the resource allocation of A1, shown in table 36:

Table 36

	B1	B2	B3	B4	Existing resource
						A1				5	6-5
A2					8
						Demand	4	3	2	5 -5

This moment, B4 was satisfied, and scratched this row expense, found minimal-overhead 1 in the expense of never scratching again, and is shown in table 37:

Table 37

Give B1 the A1 resource allocation, shown in table 38:

Table 38

	B1	B2	B3	B4	Existing resource
						A1	1			5	6-5 -1
A2					8
						Demand	4 -1	3	2	5-5

This moment, A1 did not have resource to provide, scratched this row expense, found minimal-overhead 4 in the expense of never scratching again, and is shown in table 39:

Table 39

Give B3 the A2 resource allocation, shown in table 40:

Table 40

	B1	B2	B3	B4	Existing resource
						A1	1			5	6-5-1
A2			2		8 -2
						Demand	4-1	3	2 -2	5-5

This moment, B3 was satisfied, scratched this row expense, found minimal-overhead 5 in the expense of never scratching again, and is shown in table 41:

Table 41

Give B2 the A2 resource allocation, shown in table 42:

Table 42

	B1	B2	B3	B4	Existing resource
						A1	1			5	6-5-1
A2		3	2		8-2 -3
						Demand	4-1	3 -3	2-2	5-5

This moment, B2 was satisfied, scratched this row expense, found minimal-overhead 6 in the expense of never scratching again, and is shown in table 43:

Table 43

Give B1 the A2 resource allocation, shown in table 44:

Table 44

	B1	B2	B3	B4	Existing resource
						A1	1			5	6-5-1
A2	3	3	2		8-2-3 -3
						Demand	4-1 -3	3-3	2-2	5-5

So far, the scheme of initially allocating and transporting is come out, and is shown in table 45:

Table 45

B1

B2

B3

B4

Existing resource

A1	1			5	6-5-1
						A2	3	3	2		8-2-3-3
Demand	4-1-3	3-3	2-2	5-5

Ask check number with gesture position method, what be with underscore is nonbasic variable, shown in table 46:

Table 46

	B1	B2	B3	B4	Ui
						A1	1	<2>; 2	<4>; 3	0	0
A2	6	5	4	<-5>; 0	5
						Vj	1	0	-1	0

Because check number 2,4 ,-5 is not nonnegative number entirely, so will adjust, get the minimum nonbasic variable-5 of check number as advancing basic variable with the loop circuit method, do the loop circuit, the bandwidth traffic on the loop circuit is adjusted, shown in table 47:

Table 47

	B1	B2	B3	B4	Existing resource
						A1	1+3			5-3	6-5-1
A2	3-3	3	2	0+3	8-2-3-3
						Demand	4-1-3	3-3	2-2	5-5

Ask check number with gesture position method again, what be with underscore is nonbasic variable, shown in table 48:

Table 48

	B1	B2	B3	B4	Ui
						A1	1	<-3>; 2	<-1>; 3	0	0
A2	<5>; 6	5	4	0	0
						Vj	1	5	4	0

Because check number-3 ,-1,5 is not nonnegative number entirely, so will adjust, get the minimum nonbasic variable-3 of check number as advancing basic variable with the loop circuit method, do the loop circuit, the bandwidth traffic on the loop circuit is adjusted, shown in table 49:

Table 49

	B1	B2	B3	B4	Existing resource
						A1	4	0+2		2-2	6-5-1
A2		3-2	2	3+2	8-2-3-3
						Demand	4-1-3	3-3	2-2	5-5

Ask check number with gesture position method again, what be with underscore is nonbasic variable, shown in table 50:

Table 50

	B1	B2	B3	B4	Ui
						A1	1	2	<2>; 3	<3>; 0	-3
A2	<2>; 6	5	4	0	0
						Vj	4	5	4	0

Because check number 2,2,3 is nonnegative number entirely, obtained optimum bandwidth traffic allocative decision, shown in table 51:

Table 51

	B1	B2	B3	B4	Existing resource
						A1	4	2			6-5-1
A2		1	2	5	8-2-3-3
						Demand	4-1-3	3-3	2-2	5-5

The content of not doing in this specification to describe in detail belongs to this area professional and technical personnel's known prior art.The above is merely preferable enforcement of the present invention, and is in order to restriction the present invention, not all within spirit of the present invention and principle, any modification of being done, is equal to replacement, improvement etc., all should be included within protection scope of the present invention.

Claims (4)

1. a distributed multipath method for repairing route specifically comprises the steps:

S1: certain node in the middle of a communication link of network breaks down; The upstream node of this malfunctioning node direction towards the source sends the RRER message; Receive the node initiation routing broadcast message RREQ of RRER message, seek the route of the downstream node that arrives the malfunctioning node direction;

S2: the RREQ message to intermediate node is received is judged, if the source address of the message that this RREQ message is a node oneself to send or this RREQ message appears in the historical record of intermediate node with request ID, then abandons this message; If the source address of this RREQ message does not appear in the historical record of this intermediate node with request ID, then this source address is write in the historical record with request ID, search towards the path of this message destination address, change step S3;

S3: if the aim sequence in the path found number is greater than the aim sequence in the RREQ message number; Then show and found a paths; In the reparation routing parameter of RREQ message, fill in the parameter information of the link between this node and its upstream node, transmit the RREQ message to the next-hop node of institute's storage purpose node; This intermediate node extracts the information in the RREQ message simultaneously, is used for making up reverse routing table, and packet can transmit along the rightabout of institute's routing when guaranteeing that this path is selected, waits for regular hour Δ T, if overtime then delete reverse routing table; Said regular hour Δ T should be able to guarantee that the RREQ message can pass whole network and produce a RREP message that sends to source node;

S4: after destination node receives the RREQ packet; Wait for the time of second of Δ T at least; At Δ T in the time of second, the message that never arrives with the path is carried out qos parameter and repairs routing parameter and extract, carry out route selection; Routing iinformation to after selecting makes up corresponding RREP message, last these RREP messages of destination node clean culture; Each RREP message is given source node according to different reverse routing table loopbacks, and is every through an intermediate node, and then hop count adds 1 and write down the information of this message, for each bar Business Stream is set up the forward direction routing table as repairing the routing table that successful posterior nodal point uses;

S5: each RREP message is got back to source node according to the order hop-by-hop of intermediate node, and every the hop count of RREP message adds 1 through an intermediate node, and each intermediate node extracts the relevant information of RREP message simultaneously, makes up the forward direction routing table.

2. method for repairing route according to claim 1 is characterized in that, said RREQ message comprises qos parameter and repairs routing parameter.

3. method for repairing route according to claim 1; It is characterized in that; Said qos parameter comprises time delay, bandwidth, the packet loss of current each Business Stream, and said reparation routing parameter is to initiate node to time delay, bandwidth, the packet loss of repairing the link between all adjacent nodes the destination node from repairing.

4. according to claim 2 or 3 described method for repairing route, it is characterized in that step S4 is described, and to carry out the detailed process of route selection following:

S41: destination node compares the qos parameter of each bar link and the qos parameter of Business Stream, selects available stream, makes up the bandwidth traffic allocation matrix $C=\left[\begin{array}{cccc}{c}_{11}& c_{12}& \mathrm{\Λ}& c_{1n}\\ c_{21}& c_{22}& \mathrm{\Λ}& c_{2n}\\ M& M& O& M\\ c_{m1}& c_{m2}& \mathrm{\Λ}& c_{\mathrm{Mn}}\end{array}\right],$ But the number of m representative routing, n represents the number of Business Stream, and each row of matrix is represented a link, and each row is represented a Business Stream, c _IjRepresent i bar chain road direction j row traffic flow assignment cost for bandwidth; If the bandwidth resources of link are greater than the bandwidth demand amount of Business Stream; Then be provided with a virtual service flow absorb more than resource; Its expense is made as 0, for not giving the link bandwidth of traffic flow assignment, its expense is made as infinity;

S42: obtain initial allocative decision: in the bandwidth traffic allocation matrix that constitutes, at first at minimal-overhead expense c with the least member method _IjCarry out bandwidth traffic between link corresponding i and the Business Stream j and distribute, the bandwidth stream value of distribution is got the minimum value of path bandwidth and business demand bandwidth.If the bandwidth traffic in path has assigned and has then deleted the corresponding overhead cost of this row; If the bandwidth of Business Stream has assigned and has then deleted the corresponding overhead cost of these row; In the overhead cost of not scratched, seek minimum value then; And make bandwidth traffic and distribute, finish up to whole assignment of traffic, formed an initial allocative decision this moment;

S43: the check number of asking nonbasic variable with the potential method: the matrix of structure one (m+1) * (n+1) $C^{\′}=\left[\begin{array}{ccccc}{c}_{11}^{\′}& c_{12}^{\′}& \mathrm{\Λ}& c_{1n}^{\′}& u_1\\ c_{21}^{\′}& c_{22}^{\′}& \mathrm{\Λ}& c_{2n}^{\′}& u_2\\ M& M& O& M& M\\ c_{m1}^{\′}& c_{m2}^{\′}& \mathrm{\Λ}& c_{\mathrm{Mn}}^{\′}& u_m\\ v_1& v_2& \mathrm{\Λ}& v_n& \end{array}\right],$ Make v _n=0, for each basic lattice c ' is arranged _Ij=u _i+ v _j, wherein

Be expense, utilize initial condition v _n=0, obtain all u _i, v _j, the check number of calculating nonbasic variable

If satisfy σ _Ij>=0, then current allocative decision is exactly an optimal distributing scheme, and calculating stops, otherwise changes step S44;

S44: get the minimum nonbasic variable of a check number and make into basic variable; Its corresponding lattice are for advancing Ji Ge; To advance basic lattice is that starting point is made the loop circuit that all the other summits are basic lattice, on the loop circuit, from the bandwidth traffic of all even number lattice points, selects minimum value θ as adjustment amount, and these lattice are from basic lattice; Corresponding variable is for leaving basic variable, θ=min{x _Ij, (i j) is even number lattice point in the loop circuit };

S45: the bandwidth traffic on the loop circuit is made adjustment: the freight volume of all odd number lattice adds adjustment amount θ, and the freight volume of all even number lattice deducts adjustment amount θ, obtains a new bandwidth traffic allocative decision, changes step S43.

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