WO2010148595A1

WO2010148595A1 - Allocating method and apparatus for the network service route

Info

Publication number: WO2010148595A1
Application number: PCT/CN2009/074079
Authority: WO
Inventors: 鲁春怀; 欧雪刚
Original assignee: 中兴通讯股份有限公司
Priority date: 2009-06-23
Filing date: 2009-09-21
Publication date: 2010-12-29
Also published as: CN101651621B; BRPI0924503A2; CN101651621A

Abstract

An allocating method and apparatus for the network service route are provided. In the method, the services to be routed are ordered and stored according to a priority order; by searching in the preset equipment database, each node in the network is equipped with an equipment which has the lowest cost; each service is obtained as a current service according to a priority order respectively; a route is allocated for the current service according to the preset rules, the equipping of the node equipment passed by the route is altered according to the requirement of the current service, so that the consuming cost is lowest when the current service passes this route comparing with the multiple selectable routes which can be passed; wherein the cost includes: the cost of the node and link the route passed. According to the technology scheme provided by the invention, a planning result that the whole network cost is better can be obtained, and the nodes which the service routes pass can be equipped with the equipments which can satisfy the requirement.

Description

Sub-method

Wood domain

In the case of the communication domain, it involves the routing methods of each route. The background wood is exchanged from the o a ca S c ed O ca ewo SO ) The large demand for each of the woods needs to be obtained by the user's needs due to the operation of the obstacles. From the pieces to solve. The following basic energy needs to be completed. Each routing city has different requirements for different user results. The cost is not the case for most users. The cost is now the cost of the city. The cost of each cost is that each of the routes can be filled with the most cost of each demand. The method of phase-receiving the sheep is to divide each route into the first-time minimum or the most-to-one route. The cost is to be the most costly in the first place. The scheme has the following woods. First, considering the cost factor, considering the cost in each setting, there is a large bureau, and the cost is the result of the situation.

The result is not used. Content

There is a big cost in the phase, and the purpose of presenting the book is to provide

Each routing scheme is resolved. The interface of this article mentions the routing methods of each of the gods. Each routing sub-method of the present method is arranged in the order of each of the pre-sequenced wells, and each of the pre-sequences in the predetermined ones is the first in each of the pre-sequences. Each of the former is costed by the cost of the most costly package route compared to the extra route. Another aspect of this article mentions the various branching devices of the Shinagawa. Each of the routing devices includes sheep, sheep, sheep, and sheep, and the sheep are stored in the predetermined order.

The most costly sheep are placed in the first order of the sheep, and each of the remaining labors is divided into the former, each of which is divided by the former. The cost. The most routing of the routing values to the front of the algorithm is the most costly route. The bearer of each route is solved by solving the problem that the cost in the phase is large and the full capacity is set to the full cost. As a result, each route can fill the demand. Other specialties of this book will be made clear in the Ming and Qing dynasty. The purpose of this document is not specifically indicated in the description, requirements, and contents. The unexplained step of understanding the cost of the part of the book is intended to be inconsistent. In The current flow of each of the routing methods of the flow 2 of each of the routing methods of the flow of the flow of the flow of the flow of the flow of the flow of the flow of the flow of the flow of the flow of the flow of the flow Routing device

. The rest of the routing parameters of each routing route are all the way to the front of the algorithm. The most expensive way to set the cost is to load the most. The first load is in the present + first. Book each

In the order of the most costly, the first order is the same as the previous ones.

Excess routing costs are the most cost-effective routing

cost. In the case of no love, the special ones in this book. The following conclusions

The Yu Ming solution is not used for the final version. The method first introduces the routing method of each of the gods. 1 The flow of each route is divided into methods. Each of the routing sub-methods includes the following steps S1-step S103) Step S: All the first-ordered sorting wells are stored in each of the first, and each of the first-ordered first-orders is first sorted first according to each of the examples) , each. The results are sorted by the same size in each case. Sort the order of the first photos. Save the order in the form of each list. Step S103, each of the predetermined cost among the predetermined ones is included in each of the predetermined ones

It is set to use the most cost-effective delivery of each of the most costly in each of the predetermined ones. Step S105 firstly separates each of the previous steps S107, and the previous photos are divided into the former each of the required requirements, and the cost of the excess route is compared with the cost of the most cost-routed route.

cost. Then, the routing rules in the 5107 are divided into the following steps: the following steps are required: Step G) Steps. Cost of all cities to the former All the cities in the city will work on each route in the routing of information such as the cost of the following information, the pre-existing, the pre-existing or the redundant routing, and other information. Among them, in the following way, the route that has been found by the 1 table has been found to be the best route to the route that has not been found. If the same number of cases are the smallest, the most versatile, and the most dependent, then the mercury needs to be the same, and the length depends on the nature. step. Steps unresolved in all the roads of the former front phase If the capacities of the two ends of the remaining capacity are not full, the demand is large, or the capacity is larger. The capacity is larger than the inherent maximum capacity.

First of all, whether it has the capacity before the load, otherwise the capacity will be large until the capacity before the load is smaller than the inherent maximum capacity. The annual maximum rate is calculated in the annual load. The capacity is taken in each demand. ST - ST 2 is Its rate is not the maximum rate. In the case of large capacity before the demand, the demand for the front-down phase of the large annual rate ST-year must be to the ST-year needs for the year-end of the two-year phase. Among them, the year of the new year is to consider whether there is ST-rate of fruit, whether it has a straight repair year, or whether the market is pre-positioned with the addition of ST-rate. If you add the ST-rate of the sheep to repair the year, you must replace the media with a higher median rate, and you need to find the full cost of all the media that can meet the requirements. If the respective capacity of the front and the next are full, the results will be more. The step package for paying the sheep first considers the capacity of the large sheep that has been set up, but it is not the best to replace the sheep with the smallest capacity before the cost of the sheep with the highest capacity until the capacity is full or all the sheep have been placed. Both have replaced the largest volume of transactions. If all the sheep are replaced with the maximum capacity, the capacity of the sheep is still not satisfied. If you want to add more cash, the most costly cashmere capacity will be replaced by the replacement of the most costly cashmere. If the maximum capacity of the sheep is still not satisfied with the demand, then the sheep will be added until the capacity is full or the total capacity of all the sheep has been used. If all the users with the largest capacity are still dissatisfied with the demand, they can only meet the needs of the larger capacity. The most cost-effective of all the requirements can be more important. step. The result of the calculation is calculated from the cost of the previous year, the cost of the core is lower than the cost of the previous one, and the cost of the update will be the cost of the previous year before the previous route, that is, the cost of the previous year, that is, the previous years) The cost of each cost before and after the cost. Each of the previous cost-to-sales sub-annual costs costs the cost of the previous cost in the following ways. Each capacity / capacity. Among them, the cost of each cost is replaced by each cost, otherwise the cost is paid. The cost is the same as before. Among them, if there is a replacement of the sheep, the cost of paying the sheep will be paid. The capacity of each capacity / delivery capacity of the sheep will be replaced by the cost of the sheep before the replacement of the cost of the replacement of the sheep. If there is a large capacity in the year, it costs the sheep cost of the year. Each / rate is otherwise used for the cost of each sheep, and the cost of each sheep is increased. The package is not in the year. the way

It depends on the situation. Steps . Whether there is an unreasonable list in the previous phase

If it is step C. Whether or not the existence of the fruit does not exist, the former points are the least cost of the preceding steps or the least costly. The best that has been found by the former. If the previous purpose is obtained, the previous each bearer is obtained by the otherwise step. If you consider the cost of each route, you need to compare the cost of each route to the cost of each item, to the length or the reliability.

If the number is equal, compare the first. If the most cost-effective road is redundant, it will be carried by the excess road. The purpose of each of the preceding and the former before the bearer carries the route in the manner that the destination is traced back. Otherwise, the previous steps will be taken. The routing method of each is not the same as S's non-intelligence. Not in the situation Take this with it. The results of the full cost of each route method of the Shengang are mentioned. Each route can fill the demand. 2 The flow of each route is divided into methods. Each of the routing sub-methods of the present invention includes the following steps S201 to S209, step S201, and the respective arrangements of the respective lists are arranged in the order of the first order, and the first order is the same. In step S203, each step S207 of each list intermediary in which each of the fixed costs is the most expensive and the most costly is not disposed, and each of the sub-costs is divided into the respective requirements. The one in 1 is no longer. Step S209: Whether there are unreasonable fruits in each list is the step S205, otherwise the flow ends. 3 The remainder of the flow is the most costly route. The remaining routes with the most cost are to be packaged in the following steps 5301 - 5173. Step S301 The cost of route 1 to all cities is 0. Put each pre-C ode by 1. Which will be -1 or 0 or 1 is in the middle

The reason for this. Step S303 Year list of C ode phase The year before the year is 1 year, 2 years, 3 years, and the list contains the remaining years. Step S305, in the list, take out the route of the other end of the year, which is outside the C ode, and then the e ode of the e ode. Step 5311 Then Step 5307 Step S307 Can each year go to the next e ode? S309 Otherwise, in step S311, whether each capacity can be annual or not, and whether the capacity of each end of the year can be full or not can not be filled with the maximum annual rate of the large-capacity both ends.

Steps to 1 are no longer. Step S309 calculates the cost e Cos to the e ode in each year. If the e Cos is smaller than the cost before the e ode, the cost of the e ode is updated. e Cos sets the pre-C ode of the e ode to 0 in the year list of the S 311 C ode phase. Whether there is an unreasonable year step S313 whether there is a result of 0 or the end of each route is a result, then step S315 finds that all the costs from 0 are the smallest

If there are more than one. The C ode will be found by 1 If the C ode is not the purpose) Then Step S303 Otherwise Step 5317 Step 5317 Each pre-routing information of each mercury

In the year of each year, each of the capacity update of each of the routes is formed. 4 The flow of each item of the book can be left. Whether each of the 4 items of this item can be packaged in the following steps S401 - Step S421 In step S401, the capacity is in step S403, otherwise step 5407, step 40, whether the capacity has reached the maximum rate, and then the flow cannot be ended. Otherwise, step S405, step 40, the capacity of the large step 407 is large, then step S409, otherwise step 5417 Step 409, the step of step S4, the step of step S4, and whether the result is step 41. Otherwise, step S403, step S41, before the intersection capacity is full, the demand is step S4, otherwise step S415, step 4 is lower. If the result is step S419, otherwise step S421 and step 419 are each step S421. 5 meaning of this. among them

Introduced 5 books

In the year of the year, the years of year 1, year 2, year 3, year 4, year 5 and year 6 were respectively, respectively, C, and . Following knot

. There are 1 A 2 ST - each A 2 ST - 1 Each 3 C ST - each 4 ST - each 5 C ST - is the same for each.

5 years of cost and balance 20 in each of the following packages O, 4 1 rate ST - fixed cost 10 4 sheep cost 20 4 rate ST -4 fixed cost 40 1 sheep cost 10 80 in each

CS pay C4 cross sheep 100 paying cost 100 in each middle and lower e ce maximum rate ST - fixed cost 100 for each 1 15 with 1, 2 for sheep S for 3, 4. e ce2 maximum rate ST -4 fixed cost 200 used each 1 4 with 1, 2, 4, 4 used for sheep S

Each of the routing packets has the following 0. Each sorting is the same as each of the five different sorts. Each of the different sorts is 1 each 2 each 3 each 4 each 5 2) Pre-set the most fixed cost, C, and e Ce S sheep. 3) First, each 1 point is from the market table 1. 5 Before the cost

0 each e ce

Sheep CS

-1 each e ce

Sheep CS

C -1 each e ce

Sheep CS

-1 each e ce

Sheep CS

-1 each e ce

Sheep CS 4) Connected Year 1 year 2. First consider the rate of each year of each route 1 year 1 often ST - 4 city needs to replace e ce replace the cost of e ce2 4 sheep CS sheep cost cost school ce2 fixed cost 4 sheep cost CS sheep cost e ce The fixed cost of S sheep costs the cost of year 1 2). 2 460 of the year 2 of year 1 of the year. The cost of the city C is updated in Table 2.

2

Before the cost

Each e ce2 sheep CS 4

460 each e ce2

Sheep CS 4

C 460 each e ce2

Sheep CS 4

-1 each e ce

Sheep CS

-1 each e ce

Sheep CS 5 ) , , C out route 0 cost minimum

The cost of C is the same.

) 1 from end to 仟 1.

6) Each of the 2 routes is no longer from the result to C year 2. 7) However, each of the 3 is found by City Table 3. 3

Before the cost

-1 each e ce2

Sheep CS 42 each e ce2 sheep CS 4

C -1 each e ce2

Sheep CS 4

-1 each e ce

Sheep CS

-1 each e ce

Sheep CS 8) Find all connected years, year 3, year 4. If each route is 1 year 1 , there is no need to repair the cost of the ST - the cost of each set of fees does not need to repair the city 4 sheep cost each 3 rate each 3 mesh / 4 sheep rate 20 fee delivery cost S sheep cost. Each of the 3 rates each 3 orders / delivery 2) / delivery of the sheep 2 costs of the cost of the same cost of each year to the cost 2 cost of the cost) + annual cost / 4 49. Consider again the rate of 3 years and 3 years. ST - City C needs to re-describe the cost of 3 to C fees per year 2* sheep cost delivery cost + annual cost 204. Update the city, C, and Table 4 from the results of 4 years and 3 years.

4

9) Find the least cost route 1 among all 0s.

10) Phase 1 year 2 of the phase. Year 1 is no longer considered for each year. If each route is 2 years old, it has just been introduced to ST-, so city C does not need to set the cost of 2 to C for each year. Cost Submission Cost Annual Cost / 4 47 C C's delivery cost due to the cost of delivery. Cost from C to C

47 96 is smaller than the cost before C 204. Update Table 5 of C.

5

11) Find the C C with the lowest cost of 0 for each purpose. So each 3 ends by C by year 1 year 2.

12) Each of the 4 routes is 3 years old. 1 year 2 The ST-capacity does not need to be more, and the C's are set to take the year 3, then C needs to be more expensive, so the route of each 4 is C. Year 2 years 4 years.

13) The most 5 looking for by the city table 6. 6

Before the cost

-1 person e ce2

Sheep CS 42

-1 each e ce2

Sheep CS 4

C each e ce2

Sheep CS 4

-1 each e ce

Sheep CS

-1 each e ce

Sheep CS 14) C connected to the year 2, year 3, year 4. If the rate of year 2 is 2 years, the rate of year 2 must be as large as ST - each with ST - rate of sheep, so each 5 cannot be 2 years. If the year is 3, the rate of ST-C is just ST-free and the cost of the sheep is increased. Cost Cost of sheep 2 Cost of payment Annual cost 134. If the light year is 4 years, the rate of 4 must be strong. The ST-4C city must replace the sheep with the cost of 4 sheep. 4 sheep cost sheep cost delivery cost annual cost / 4 69. Update the city's Table 7.

7

15) All the routes that are the least costly by 0 will be route 1. 16) Phase of the year 4 years 5 years 4 is C The year to come is no longer considered. In each of the five cities, the cost of raising the sheep fee is required. 2 The cost of the sheep is the cost of the annual cost of 202. The cost of the cost of each year is 205. 8

Before the cost

-1 each e ce2

Sheep CS 42

134 C each e ce2

Sheep CS 4 O

C each e ce2

Sheep CS 4

69 C each e ce

Sheep CS 4

271 each e ce

Sheep CS O 17) All from 0% of the lowest cost

Route 1. 18) All years of the year 1, year 3 years 6 . If the rate of year 1 is 1 year, the rate must be as large as ST - and each has ST -16 at the rate of year. Therefore, each year 5 cannot be 3 years old. C is the year of the year. The cost of the cost of raising the sheep's fee is the cost of the sheep. The cost of the annual cost is 132. The cost of the cost of each of the six years is 132 266. The cost of the city is smaller than the previous cost. 9

Before the cost

-1 each e ce2

Sheep CS 42

134 C each e ce2 sheep anti CS 4 O

C each e ce2

Sheep CS 4

69 C each e ce

Sheep CS 4 271 each e ce

Sheep CS O 19) All from 0 in the middle of the cost is smaller

Each has been completed by 5 years from C years 3 years. 9 The cost of each obtained 214 years cost 20*6 120 Full cost 2260 The result of each phase is 2, each C, 3 C, 4 C, 5 C, respectively, regardless of the cost of the minimum or the most recent route. However, the results are shown in Table 10. 1

Cost each sheep

540 e ce2 CS O 4 2

630 e ce2 CS O 42 O 1

C 630 e ce2 CS O 42 O 1

290 e ce1 CS O 1

The cost of each of the 290 e ce1 CS O 1 2380 full cost 2500 is 240 times higher than the cost presented here to arrive at the cost of each route proposed by the present invention. The cost of each route in the phase is compared to the cost of each route to the full demand. The method will be full. The device is provided with various routing devices of the Shengang. 6 The meaning of each routing device. 7 The meaning of each routing device. 6 The routing devices of the book are stored in sheep, sheep 2, sheep 3, and sheep 4 below. Save the sheep in the order of the first order, the cost of each of the stocks in the predetermined ones, the most costly, the most costly sheep, and the first part of the sheep. The sheep 4 is divided into the cost of the cost package, which is the cost of the cost of the package. 7 divided sheep 4 steps package 40, 42, 44 46, 48, 50, 52 of which 40% of the cost of all the cities before the first all the city's work before the 42nd to 40th in the middle Each of the pre-phases is unrecognized from 44 to 42. Each of the two ends of each of the required capacity sets is provided, and each of the seven sets of each of the seven sheep is in the form of a step-by-step update 440, an update sub-442, and an updater. 444 where the update sub-440 is in the capacity that the road does not have the capacity before the bearer Until the capacity of each of the large capacity is less than the intrinsic maximum capacity, the updater 442 is in need of the capacity before the capacity is large, and each of the first update sub-444s before the previous demand is full. The situation is more confusing. From 46 to 44, the results of the previous calculations are calculated from the cost of the previous year, and the cost of the core is lower than the cost of the previous one.

The former route 46 is not included in the package, and the costs and costs before the cost of the previous one are accumulated. 48 to 46 44 Is there an unreasonable year in the list of years before the 50th to 48th

52 to 42. The 50th is the least costly or the least costly intermediary. The best results that have been found before are obtained from the previous goals. The results of the various requirements of each route can be set for each of the results of the cost of the cost of the sheep and sheep. Each of the routing branches mentioned in the book can be filled with the results of the full cost. This is a requirement for the SS and so on. However, each step or step of the wood of the domain is to be stored in the device by means of a general device without them being concentrated in the device of the sheep or distributed in a plurality of devices. They are not separately made into integrations or multiple or steps in them are not integrated. This is not intended for a specific piece of knot.

The use of this book is not used to make the wood in this domain. The repairs, equivalent substitutions, etc. made within the spirit of this text are included in this solid.

Claims

Asking for the route of each route, it is special

Will be sorted by the order of each

The most costly in each of the reservations

First, the order is the same as each

In the previous photos, the cost of each of the previous routes is more expensive than the cost of the extra routes. The cost of the cost package routed.

The method of requirement 1 is characterized in that it is divided into the steps of the routing of the previous requirements . All the cities before the cost of each of the first

Steps . Unreasonable in all the previous phases of the former

Step C. Whether the capacity of both ends of the remaining capacity in the unreasoned road is full or not, the demand is large, and the capacity of the two ends is larger than the inherent maximum capacity.

Steps . The result of the calculation is calculated. The cost of the previous route is lower. The cost of the core is lower than the cost of the previous one.

Previous route

Steps . Is there an unreasonable result in the list of prephases? Step C

step. Whether there is a fruit that does not exist The steps are all the least costly or the least costly intermediary. Will

The best by the previous found. The purpose of each of the former is to obtain the previous bearing by the otherwise step. . The method of claim 2 is characterized in that it is determined in step C that whether there is capacity before the load, or whether the capacity is large until the capacity before the load is larger than the inherent maximum capacity.

In the case of a large capacity, each demand is more in front of the lower phase.

Whether the demand is higher or not before the previous capacity is exceeded. . Each of the routing stations of the gods is characterized by

Save the sheep in the order in which they are sorted first.

The most costly

The sheep are in the order of the sheep, and each

In the case of the sheep, the cost of the route is calculated by the cost of the most costly package. . The device of claim 4 is characterized by a sub-pack

The cost of all the cities that are located in the first place

Work before

It is unreasonable in all the roads of the former front phase Each of the two ends of the capacity update with a large demand

The result of the setting is calculated. The cost of the previous year is lower. The cost of the core is lower than the cost of the previous one.

Is there an unreasonable list in the previous phase?

Whether or not there is

It is the least costly or the least costly medium. The best results that have been found are obtained from the previous ones. . The device of claim 5 is characterized in that the capacity of the step package is increased until the capacity of the road has no capacity before the load, until the capacity before the load is larger than the inherent maximum capacity.

The first update is based on the situation where the capacity is large.

The first update is better placed in the case of the demand before the previous capacity is dissatisfied. . The device of requirement 5 is characterized by

Tired of the cost and cost of each cost before the previous cost.