当前位置:网站首页>Hcip OSPF comprehensive experiment
Hcip OSPF comprehensive experiment
2022-07-27 02:23:00 【less than _ ermi】
Experimental content :

Experimental analysis and ideas :
1、 Subnet partition
By region ; The area is divided freely
2、IP Address configuration
3、 Default route +NAT
4、MGRE Environment configuration
5、oSPF agreement
6、 Reissue
7、 Inter domain routing summary and extradomain routing summary
8、 Air interface anti loop routing
9、 Special area
10、 Change the convergence time
11、oSPF authentication
12、 Test the accessibility of the whole network
Experimental process :
Subnet partition
Yes 6 Regions , Variable length subnet mask , borrow 3 Bit Division 8 Subnet
area0 ( There are 3 Net segment 3 Loop back , borrow 3 Bit delimitation 8 Subnet )
172.16.32.0/19
172.16.32.0/22 172.16.48.0/22 172.16.40.0/22 172.16.36.0/22
172.16.56.0/22 172.16.52.0/22 172.16.44.0/22 172.16.60.0/22
area1 ( There are 1 Net segment 3 Loop back , borrow 2 Bitwise speech 4 Subnet )
172.16.64.0/19
172.16.64.0/21 172.16.80.0/21 172.16.72.0/21 172.16.76.0/21
area2 ( There are 2 Net segment 1 Loop back , borrow 2 Bit delimitation 4 Subnet )
172.16.128.0/19
172.16.128.0/21 172.16.144.0/21 172.16.136.0/21 172.16.152.0/21
area3 ( There are 2 Net segment 1 Loop back , borrow 2 Bit delimitation 4 Subnet )
172.16.160.0/19
172.16.160.0/21 172.16.176.0/21 172.16.168.0/21 172.16.184.0/21
area4 ( There are 1 Net segment 2 Loop back , borrow 2 Bit delimitation 4 Subnet )
172.16.192.0/19
172.16.192.0/21 172.16.208.0/21 172.16.200.0/21 172.16.216.0/21
RIP There are 2 Loopback , borrow 1 Bit delimitation 2 Subnet
172.16.96.0/19
172.16.96.0/20 172.16.112.0/20
Topology

IP Address configuration / Yes r3、r5、r6、r7 Configure default routes and NAT( With R3 For example )
[r3]ip route-static 0.0.0.0 0 172.16.44.2
[r3]acl 2000
[r3-acl-basic-2000]rule permit source any
[r3]int s4/0/0
[r3-Serial4/0/0]nat outbound 2000MGRE Environment configuration (r3 Centered )
r3 To configure
[r3]inter t0/0/0
[r3-Tunnel0/0/0]ip add 10.1.1.1 24
[r3-Tunnel0/0/0]tunnel-protocol gre p2mp
[r3-Tunnel0/0/0]source s4/0/0
[r3-Tunnel0/0/0]nhrp entry multicast dynamic r5 To configure (r6\r7 Configuration is the same as r5, Register in the center )
[r5]inter t0/0/0
[r5-Tunnel0/0/0]ip add 10.1.1.2 255.255.255.0
[r5-Tunnel0/0/0]tunnel-protocol gre p2mp
[r5-Tunnel0/0/0]source s4/0/0
[r5-Tunnel0/0/0]nhrp network-id 100
[r5-Tunnel0/0/0]nhrp entry 10.1.1.1 172.16.44.1 registeroSPF agreement ( Declare the division of each router r4 Outside , With r1 For example Pay attention to the announcement area )
[r1]ospf 1 router-id 1.1.1.1
[r1-ospf-1]area 1
[r1-ospf-1-area-0.0.0.1]network 172.168.72.1 0.0.0.0
[r1-ospf-1-area-0.0.0.1]network 172.168.64.1 0.0.0.0Reissue ( stay r9\r12 To configure )
r12( establish rip Then republish )
[r12]rip 1
[r12-rip-1]network 172.16.0.0
[r12]ospf 1
[r12-ospf-1]import-route rip
r9( Create two processes ospf 1 And ospf 2)
[r9]ospf 1
[r9-ospf-1]import-route ospf 2
Inter domain routing summary ( stay r3\r6\r7 On the configuration )
[r3]ospf 1
[r3-ospf-1]area 1
[r3-ospf-1-area-0.0.0.1]abr-summary 172.16.64.0 255.255.224.0Extraterritorial routing summary ( stay r9\r12 On the configuration )
[r9]ospf 1
[r9-ospf-1]asbr-summary 172.16.192.0 255.255.224.0Air interface anti loop routing ( stay r3\r6\r7\r9\r12)
[r3]ip route-static 172.16.64.0 19 NULL 0Special area
1) Complete tip blending area1 Adjust to the complete terminal area
[r1]ospf 1
[r1-ospf-1]arp-ping
[r1-ospf-1]area 1
[r1-ospf-1-area-0.0.0.1]stub[r3]ospf 1
[r3-ospf-1]area 1
[r3-ospf-1-area-0.0.0.1]stub no-summary 2)Area2 Adjust to complete ASSA
[r11]ospf 1
[r11-ospf-1]area 2
[r11-ospf-1-area-0.0.0.2]nssa[r6]ospf 1
[r6-ospf-1]area 2
[r6-ospf-1-area-0.0.0.2]nssa no-summary3)Area3 Adjust to complete ASSA The configuration is the same as above , here r9 appear 3 Class default r9 to r10 Republish a default message
[r9]ospf 2
[r9-ospf-2]default-route-advertisechange hello Time
[r3]interface s4/0/0
[r3-serial4/0/0]ospf timer hello 5 oSPF authentication
[r3]interface serial4/0/0
[r3-serial0/0/1]ospf authentication-mode md5 1 123456Test the accessibility of the whole network (172.16.60.1 by r4 The loopback of )



End of experiment
边栏推荐
- Wechat applet: user wechat login process (attached: flow chart + source code)
- C语言——while语句、dowhile语句、for循环和循环结构、break语句和continue语句
- (超详尽版,不懂随时评论)Codeforces Round #804 (Div. 2)C The Third Problem
- Codeforces Round #810 (Div. 2), problem: (B) Party
- C语言——数据类型、基本数据类型的取值范围
- 多线程中 synchronized 锁升级的原理是什么?
- JVM面试题(面试必备)
- RISC-V工具链编译笔记
- Lecture 3 - GPIO input / output library function usage and related routines
- Esp8266wi fi data communication
猜你喜欢

Esp8266wi fi access cloud platform
npm报错, Error: EPERM: operation not permitted, mkdir

在有序数组找具体某个数字

HCIA Basics (1)

Ogeek meetup phase I, together with cubefs, is hot

Explain exi interrupt through the counting experiment of infrared sensor

数字集成电路:CMOS反相器(一)静态特性

C language - first program, print, variables and constants
![C language implementation of the small game [sanziqi] Notes detailed logic clear, come and have a look!!](/img/b9/ade9a808a3f6d24cd9825dc9b010c1.png)
C language implementation of the small game [sanziqi] Notes detailed logic clear, come and have a look!!

HCIP-第六天-OSPF静态大实验
随机推荐
(super detailed version, don't know to comment at any time) codeforces round 804 (Div. 2) C the third problem
HCIP第一天
(CF1691D) Max GEQ Sum
Lvs+keepalived project practice
(题意+详细思路+加注释代码) Codeforces Round #805 (Div. 3)F. Equate Multisets
NB-IOT接入云平台
Codeforces Round #810 (Div. 2), problem: (B) Party
C语言——while语句、dowhile语句、for循环和循环结构、break语句和continue语句
C语言——数据类型、基本数据类型的取值范围
Lora gateway node converges sensor data
Lesson 5 - key control LED
NAT (network address translation protocol)
怎么判断一个数是奇数还是偶数?
静态路由基本配置 实现全网可达
在有序数组找具体某个数字
C language - assignment operator, compound assignment operator, self increasing and self decreasing operator, comma operator, conditional operator, goto statement, comment
离开页面的提示
Golang - sync包的使用 (WaitGroup, Once, Mutex, RWMutex, Cond, Pool, Map)
(prefix and / thinking) codeforces round 806 (Div. 4) F Yet Another Problem About Pairs Satisfying an Inequality
求解整数的每一位