Sub Neting exercises.

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Presentation transcript:

Sub Neting exercises

Exercise 1 The host has an IP address of 172.16.179.237/19. What are the first address and last addresses in the subnet? Mask : 11111111.11111111. 11100000.00000000 172. 16 . 10110011.0 172. 16 .10100000.0 The first address: 172.16.160.0/19.

Exercise 1 Last address : Mask : 11111111.11111111. 11100000.00000000 172. 16 . 10110011.0 172. 16 .10111111.11111111 172. 16 .191 .255

Exercise 2 The network consists of 525 hosts. What should be the network mask length used to accommodate them? We need 10 bit for host 2^9=512 2^10=1024 The mask will be /22 In binary it looks like this: 11111111.11111111.11111100.00000000. /22 = 255.255.252.0

Exercise 3 Use 192.168.1.0/24 to address the following five subnets. Maximize the network mask length to accommodate the number of hosts specified below: Subnet 1 = 76 hosts,Subnet 2 = 40 hosts ,Subnet 3 = 16 hosts, Subnet 4 = 2 hosts. Subnet 1 = 76 hosts 26 = 64 (62 valid host addresses is not enough) 27 = 128 (126 valid host addresses is the closest number) Result: Subnet 1 mask is /25 or 255.255.255.128 (32 – 7 = 25). Subnet 2 = 40 hosts 26 = 64 (62 valid host addresses) Result: Subnet 2 mask is /26 or 255.255.255.192 (32 – 6 = 26). Subnet 3 = 16 hosts 24 = 16 Result: Subnet 3 mask is /28 or 255.255.255.240 (32 – 4 = 28). Subnet 4 = 2 hosts 22 = 4 (2 valid host addresses) Result: Subnet 4 mask is /30 or 255.255.255.252 (32 – 2 = 30)

Exercise 4

Exercise 5