Question

Consider the topology shown in Figure 2. Denote the three subnets with hosts (starting clockwise at 12:00) as Networks A,B, and C. Denote the subnets without hosts as Networks D,E, and F. Figure 2. Network of Question 2. a) Assign network addresses to each of these six subnets, with the following constraints: - All addresses must be allocated from 214.97.254/23; - Subnet A should have enough addresses to support 250 interfaces; - Subnet B should have enough addresses to support 120 interfaces; - Subnet C should have enough addresses to support 120 interfaces. Subnets D, E and F should each be able to support two interfaces. For each subnet, the assignment should take the form a⋅b⋅c⋅d/x or a⋅b⋅c⋅d/x−e.f.g⋅h/y. b) Using your answer to part (a), provide the forwarding tables (using longest prefix matching) for each of the three routers.

Answer 1

Subnet assignments:

A: 214.97.254.0/24

B: 214.97.254.128/25

C: 214.97.254.192/25

D: 214.97.254.224/30

E: 214.97.254.228/30

F: 214.97.254.232/30

Forwarding tables:

R1: A (214.97.254.0/24), D (214.97.254.224/30)

R2: B (214.97.254.128/25), E (214.97.254.228/30)

R3: C (214.97.254.192/25), F (214.97.254.232/30)

Step-by-step explanation:

In part (a), subnet A is allocated the address range 214.97.254.0/24 to support 250 interfaces. Subnets B and C are given 214.97.254.128/25 and 214.97.254.192/25 respectively, each supporting 120 interfaces. Subnets D, E, and F are assigned 214.97.254.224/30, 214.97.254.228/30, and 214.97.254.232/30, each capable of supporting two interfaces.

In part (b), the forwarding tables for each router are established using longest prefix matching. Router 1 forwards traffic for networks A and D, Router 2 for networks B and E, and Router 3 for networks C and F. This ensures that each router directs packets to the correct subnet based on the longest matching prefix.

Answer 2

To assign network addresses to the six subnets, we start with the network address 214.97.254/23. Subnet A needs a /24 subnet mask to support 250 interfaces, while Subnet B and C need /24 subnet masks to support 120 interfaces each. Subnets D, E, and F can use /30 subnet masks to support 2 interfaces each.

Step-by-step explanation:

To assign network addresses to each of the six subnets in the given topology, we start with the network address 214.97.254/23. This means that the subnet mask is /23, which gives us 2^9 - 2 = 510 possible addresses for each subnet. Subnet A needs to support 250 interfaces, so it requires a /24 subnet mask. Subnet B and C need to support 120 interfaces each, so they also require a /24 subnet mask. Subnets D, E, and F only need to support 2 interfaces each, so they can use a /30 subnet mask.

The assignments are as follows:

Network A: 214.97.254.0/24

Network B: 214.97.255.0/24

Network C: 214.97.0.0/24

Network D: 214.97.1.0/30

Network E: 214.97.2.0/30

Network F: 214.97.3.0/30