(6 marks) Consider the network configuration shown below. The customer network i
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Question
(6 marks) Consider the network configuration shown below. The customer network is allocated the address space 140.2.0.0/16. The Border Router A of the customer network advertises an aggregate route for that address range and internally uses subnetting on the entire third octet space to create 255 sub-network addresses, not all of which are actually allocated and assigned since the customer currently does not need so many networks. The ISP router B advertises the default route (i.e. 0.0.0.0/0) back to the customer network. This implies that all traffic destined to target prefixes not located in the customer's network must be sent via the ISP's network. Identify and explain any one specific problem this configuration could lead to and provide a remedial solution to address it. (200 words) e) uses third octet space to create 2sg ISP Network 18.2.2.1 Advertisement from customer: 140.2.0.0/16 18.2.2.2 Advertisement from ISP: 0.0.0.0/o Customer Network Customer networks: 140.2.1.0/24, 140.2.2.0/24, 140.2.5.0/24)Explanation / Answer
1) six subnets.
2) Subnet A should have enough addresses to support 250 interfaces.
Number bits used for identification of host in subnet A
28-2 =256-2=254 addresses are available.(n=8)
2 addresses are used for the broad cast and subnet.
Number of bits for network: 232-n =232-8 =224 hence
Subnet A address: 214.97.254.0./24
Subnet A broad cast address: 214.97.254.255/24
214.97.254.11111111 /24=214.97.254.255/25
Subnet B should have enough addresses to support 120 interfaces. And
Number bits used for identification of host in subnet B
27-2 =128-2=126 addresses are available.(n=7)
2 addresses are reserved for the broad cast and subnet.
Number of bits used for network: 232-n =232-7 =225 hence
Subnet B address: 214.97.255.0/25
Subnet B broad cast address: 214.97.255.127/25
214.97.255.01111111/25=214.97.255.127/25
Subnet C should have enough addresses to support 120 addresses.
Number bits used for identification of host in subnet C
27-2 =128-2=126 addresses are available.(n=7)
2 addresses are reserved for the broad cast and subnet.
Number of bits used for network: 232-n =232-7 =225 hence
Subnet C address: 214.97.255.128/25
Subnet C broad cast address: 214.97.255.255 /25
214.97.255.11111111/25=214.97.254.255/25
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