Menu Content/Inhalt

Identification






Mot de passe oublié ?
Solution to Exercise 2 Convertir en PDF Version imprimable Suggérer par mail
Écrit par CCNA   
26-05-2008
  1. Subnet 172.160.0.0/18
    • Step 1 : IP class
         172.160.0.0/18 (classless) --> 172.160.0.0/16 (B- class)
      C class subnetting
    • Step 2 : host part    
      27 26 25 24 23 22 21 20
      1 1 0
      0 0 0 0
               --> 27+26 = 128+64 = 192
    • Step 3 : number of subnets
         22-2 = 2 subnets available
    • Step 4 : number of hosts per subnet
         214-2 = 16382 hosts per subnet
    • Step 5 : block size
         Block size = 256-192= 64
    • Step 6 : subnet mask
         Subnet mask = 255.255.192.0
    • Step 7 : subnets and their characteristics   
      NetID IP range Broadcast
      1 172.160.64.0 64.1-->127.254 172.160.127.255
      2 172.160.128.0 128.1-->191.254 172.160.191.255

       
  2. Subnet 172.160.0.0/20
    • Step 1 : IP class
         172.160.0.0/20 (classless) --> 172.160.0.0/16 (B- class)
      C class subnetting
    • Step 2 : host part    
      27 26 25 24 23 22 21 20
      1 1 1
      1
      0 0 0
               --> 27+26+25+24 = 128+64+32+16 = 240
    • Step 3 : number of subnets
         24-2 = 14 subnets available
    • Step 4 : number of hosts per subnet
         212-2 = 4094 hosts per subnet
    • Step 5 : block size
         Block size = 256-240= 16
    • Step 6 : subnet mask
         Subnet mask = 255.255.255.240
    • Step 7 : subnets and their characteristics   
      NetID IP range Broadcast
      1 172.160.16.0 16.1-->31.254 172.160.31.255
      2 172.160.32.0 32.1-->63.254 172.160.63.255
             
      14 172.160.224.0
      224.1-->239.254
      172.160.239.255

     
  3. Subnet 172.160.0.0/21
    • Step 1 : IP class
         172.160.0.0/21 (classless) --> 172.160.0.0/16 (B- class)
      C class subnetting
    • Step 2 : host part    
      27 26 25 24 23 22 21 20
      1 1 1
      1
      1
      0 0
               --> 27+26+25+24+23 = 128+64 = 248
    • Step 3 : number of subnets
         25-2 = 30 subnets available
    • Step 4 : number of hosts per subnet
         211-2 = 2046 hosts per subnet
    • Step 5 : block size
         Block size = 256-248= 8
    • Step 6 : subnet mask
         Subnet mask = 255.255.248.0
    • Step 7 : subnets and their characteristics   
      NetID IP range Broadcast
      1 172.160.8.0 8.1-->15.254 172.160.15.255
      2 172.160.16.0 16.1-->23.254 172.160.23.255
             
      30 172.160.240.0 240.1-->247.254 172.160.247.255

       
  4. Subnet 172.160.0.0/26
    • Step 1 : IP class
         172.160.0.0/26 (classless) --> 172.160.0.0/16 (B- class)
      C class subnetting
    • Step 2 : host part    
      27 26 25 24 23 22 21 20
      1 1 0
      0 0 0 0
               --> 27+26 = 128+64 = 192
    • Step 3 : number of subnets
         210-2 = 1022 subnets available
    • Step 4 : number of hosts per subnet
         26-2 = 62 hosts per subnet
    • Step 5 : block size
         Block size = 256-192= 64
    • Step 6 : subnet mask
         Subnet mask = 255.255.255.192
    • Step 7 : subnets and their characteristics   
      NetID IP range Broadcast
      1 172.160.0.64 65-->126 172.160.0.127
      2 172.160.0.128 129-->190 172.160.0.191
             
      1021 ? ? ?
      1022 ? ? ?
Dernière mise à jour : ( 06-06-2008 )
 
< Précédent   Suivant >