vlsm technologyytfyuftuvygubiugiugu.pptx

ksuha;uihasogafi;ughfi;h;hfsoi;fbbfsjfsdfnfnnfnnngovand jamal sa abdulraman oisjhofhaof88hadfsdohsdfgsfosahfdaif

1. Soran University Faculty of Science Mr.Ababakr I Rasul Lecture 1 Subnetting and VLSM 1 [email protected]

2. IP address classes Class Start address Finish address A 0.0.0.0 126.255.255.255 B 128.0.0.0 191.255.255.255 C 192.0.0.0 223.255.255.255 D 224.0.0.0 239.255.255.255 E 240.0.0.0 255.255.255.255 2

3. Private IP Addresses These IP addresses are used for internal use by company or home networks that need to use TCP/IP 3 Class Private Start Address Private End Address A 10.0.0.0 10.255.255.255 B 172.16.0.0 172.31.255.255 C 192.168.0.0 192.168.255.255

4. Subnetting Subnetting an IP network is to separate a big network into smaller multiple networks for reorganization and security purposes.

5. Table :Subnet a network Subnet mask 128 192 224 240 248 252 254 255 Bits value 128 64 32 16 8 4 2 1 2 ^ power number 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0 Bits borrowed 1 2 3 4 5 6 - - Subnet prefix Class C Class B Class A /25 /17 /9 /26 /18 /10 /27 /19 /11 /28 /20 /12 /29 /21 /13 /30 /22 /14 - /23 /15 -

6. Subnet mask Table : Default Subnet Masks for Class A, Class B and Class C Networks IP Address Class Total # Of Bits For Network ID / Host ID Default Subnet Mask First Octet Second Octet Third Octet Fourth Octet Class A 8 / 24 11111111 (255) 00000000 (0) 00000000 (0) 00000000 (0) Class B 16 / 16 11111111 (255) 11111111 (255) 00000000 (0) 00000000 (0) Class C 24 / 8 11111111 (255) 11111111 (255) 11111111 (255) 00000000 (0)

7. How to determine host and subnet A- (2 n ) where N is equal to number of bits borrowed. determine number of total subnet. A- (2 h -2) where H equal to number of host bits determine number of valid host per subnet subnet

8. Example Class C: 192.168.100.50 / 26 Subnet mask: 255.255.255.192 Number of subnets: 2 n = 2 2 = 4 ( n = number of borrowed bits ) Number of usable host: 2 h -2 = 2 6 -2 = 64 – 2 = 62 To find network address Convert both IP address and subnet mask to binary number 64 32 16 8 4 2 1 192. 168. 100. 50 11000000. 10101000. 01100100. 00110010 IP address 255. 255. 255. 192 and 11111111. 11111111. 11111111. 11000000 subnet mask ----------------------------------------------------------- 110000000. 10101000. 01100100. 00000000 192 . 168 . 100 . 0 Network address: 192.168.100.0 First usable host: 192.168.100.1 last host =number of host or(valid host) + network address Last host: 192.168.100.62 last host =0 + 62 Broadcast address: 192.168.100.63

9. Example 200.10.5.68/28 what is subnet mask, network, first host, broadcast ,last host ? How many hosts or subnets are valid? 200.10.5.68/28 255.255.255.240 200.10.5.01000100 And 255.255.255.11110000 ----------------------------------- 200.10.5. 64 200.10.5.64 network address 200.10.5.65 first host 200.10.5.78 last host last host =number of host + network address =64+14 200.10.5.79 broadcast 2 n =2 4 = 16 subnet 2 H -2= 2 4 -2=14 hosts

10. Variable-length subnet masking ( VLSM)

11. What Variable-Length Subnet Masking (VLSM)) VLSM amounts to "subnetting subnets” which means that VLSM allows network engineers   to divide an IP address   space into a hierarchy of subnets of different sizes, making it possible to create subnets with very different host counts without wasting large numbers of addresses.

12. VLSM Available subnet - 192.168.2.0/24   Solution:  In this network we have 6 networks (LAN1 – LAN2 – LAN3 – WAN link1 – WAN link2 – WAN link3)  1- Determine the class of this network 192.168.2.0/24  (Class: C , N = 24bits , H = 8bits, Default Mask = 24)  2- Order the networks from the largest size to the smallest:  1) LAN 2 (50 hosts)  2) LAN1 (24 hosts)  3) LAN3 (8 hosts)  4) WAN link 1 - WAN link 2 - WAN link 3 (2 hosts) 

13. VLSM (cont….) 3-Start from the biggest network:  1) LAN 2 (50 hosts):  H = 6 bits -> 2 6 -2 = 62 hosts  S = 2 bits -> 2 2 = 4 subnets  mask = N + S = 24+2 = /26 =(255.255.255.192)  LAN2 will take the subnet ID: 192.168.2.0 /26  Subnet 192.168.2.0/26 00000000 subnetmask 255.255.255.192 11000000 and N etwork address 192.168.2.0 0= 00000000 F irst address 192.168.2.1 L ast address 192.168.2.62 last host =network address +number of host or(valid host)=0+62=62 Bro dcast adress 192.168.2.63 192.168.2.64 N ext network address for another host

14. VLSM (cont….) 2) LAN 1 (24 hosts):  H = 5 bits -> 2 5 -2 = 30 hosts  S = 3 bits -> 2 3 = 8 subnets  /mask = N + S = 24+3 = /27 =(255.255.255.224)  subnet 192.168.2.64/27 01000000 subnet mask 255.255.255.224 11100000 and N etwork address 192.168.2.64 64= 01000000 F irst address 192.168.2.65 L ast address 192.168.2.94 last host =network address +number of host or(valid host)=64+30=94 Bro dcast adress 192.168.2.95 192.168.2.96 N ext network address for another host

15. VLSM (cont….) 3) LAN 3 (8 hosts):  H = 4 bits -> 2 4 -2 = 14 hosts  S = 4 bits -> 2 4 = 16 subnets  /mask = N + S = 24+4 = /28 =(255.255.255.240)  subnet 192.168.2.96/28 01100000 subnet mask 255.255.255.240 11110000 and N etwork address 192.168.2.96 96 = 01100000 F irst address 192.168.2.97 L ast address 192.168.2.110 last host =network address +number of host or(valid host)=96+14=110 Bro dcast adress 192.168.2.111 192.168.2.112 N ext network address for another host

16. VLSM (cont….) 4) WAN Links 1,2,3 (2 hosts):  H = 2 bits -> 2 2 -2 = 2 hosts  S = 6 bits -> 2 6 = 64 subnets  /mask = N + S = 24+6 = /30 =(255.255.255.252)  192.168.2.112/30 subnet 192.168.2.112/30 01110000 subnet mask 255.255.255.252 11111100 and N etwork address 192.168.2.112 112= 01110000 F irst address 192.168.2.113 L ast address 192.168.2.114 last host =network address +number of host or(valid host)=112+2=114 Bro dcast adress 192.168.2.115 192.168.2.116 N ext network address for another host

17. VLSM (cont….) 4) WAN Links 1,2,3 (2 hosts):  H = 2 bits -> 2 2 -2 = 2 hosts  S = 6 bits -> 2 6 = 64 subnets  /mask = N + S = 24+6 = /30 =(255.255.255.252)  192.168.2.116/30 subnet 192.168.2.116/30 01110000 subnet mask 255.255.255.252 11111100 and N etwork address 192.168.2.116 116= 01110000 F irst address 192.168.2.117 L ast address 192.168.2.118 last host =network address +number of host or(valid host)=116+2=118 Bro dcast adress 192.168.2.119 192.168.2.120 N ext network address

18. S ubnetting HomeWork Write of the subnet mask, Network address, first host address, last host address, broadcast address, and number subnet and hosts 192.168.5.100 / 29 192.168.4.87/25 18

19. VLSM Homework Find the (VLSM) and determine the range of Hosts the following term . Available subnet = 192.168.1.0/24