Network Layer (Part 2) - Addressing, Routing & IP

1. Forwarding vs. Routing

Forwarding (The “Doer”): Moves packets from a router’s input port to the correct output port. Happens at extremely short timescales (nanoseconds) and is implemented in hardware.
Routing (The “Thinker”): The background algorithms that determine the end-to-end paths packets take. Happens at longer timescales (seconds) and is implemented in software.

2. IP Addressing & CIDR

An IP address (IPv4) is a 32-bit identifier for host/router interfaces.

Note: Switches and Hubs operate at Layer 2/Layer 1, so they do not have IP addresses!
Classful Addressing (The Old Way): Divided IPs into rigid classes (Class A = /8, Class B = /16, Class C = /24). This caused rapid address exhaustion because Class B was too large for most companies, but Class C was too small.
CIDR (Classless Inter-Domain Routing - The New Way): Allows subnet masks of any arbitrary length.
- Format: a.b.c.d/x (where x is the number of bits in the subnet portion).
- Subnet part: The high-order (leftmost) bits.
- Host part: The low-order (rightmost) bits.

The Subnetting Math Rules (Guaranteed Exam Topic)

If you are given 200.23.16.130/23:

Subnet Mask: The first 23 bits are 1s, the remaining 9 bits are 0s. 11111111.11111111.11111110.00000000 = 255.255.254.0
Network Address: Perform a Bitwise AND between the IP and the Subnet Mask. (This zeroes out the host bits).
Total IPs: If there are $h$ host bits (e.g., $32 - 23 = 9$ ), Total IPs = $2^{h}$ (e.g., $2^{9} = 512$ ).
Usable (Valid) Hosts: Always $2^{h} - 2$ . You must subtract 2 because:
- All 0s in the host part = Network Address (identifies the subnet itself).
- All 1s in the host part = Broadcast Address (sends to everyone in the subnet).

3. Longest Prefix Matching

Routers use aggregate forwarding tables to save space. When a packet arrives, the router compares the destination IP to the table.

The Rule: If an IP matches multiple subnet ranges in the table, the router forwards it to the interface with the longest prefix (the highest /x number), because it is the most specific match.

4. DHCP & NAT

DHCP (Dynamic Host Configuration Protocol)

Dynamically assigns IP addresses to hosts when they join a network.

DORA Process: Discover, Offer, Request, ACK.
DHCP provides more than just an IP. It also provides:
1. The Subnet Mask.
2. The First-hop Router (Default Gateway).
3. The DNS Server address.

NAT (Network Address Translation)

Solves IP exhaustion by hiding an entire local network behind one single public IP address.

How it works:
- Outgoing: Router replaces the private source IP and Port with its public NAT IP and a newly generated Port.
- Incoming: Router checks its NAT Table, matches the destination Port, and replaces the public IP/Port back to the private IP/Port.
The Controversy: NAT violates the “end-to-end” principle by interfering with IP addresses at the network edge. It also breaks peer-to-peer apps (like Skype or gaming servers) because external hosts cannot initiate a connection to a hidden private IP (NAT Traversal problem).

5. IP Datagrams & Fragmentation

The IP Header is typically 20 bytes (plus a 20-byte TCP header = 40 bytes of total overhead).

MTU (Maximum Transfer Unit): The largest possible frame a specific link can handle (e.g., Ethernet MTU is usually 1500 bytes).
Fragmentation: If a router receives an IP datagram larger than the next link’s MTU, it must chop the data into smaller fragments.
Crucial Rule: Fragments are only reassembled at the final destination, not by the routers in between.

Fragmentation Math Rules:

If MTU = 1500 bytes, the maximum data payload per fragment is $1500 - 20 (IP Header) = 1480 bytes$ .

MF (More Fragments) Flag: 1 if there are more fragments coming. 0 if it is the absolute last fragment.
Offset: Tells the receiver where this fragment belongs. Calculated as: (Total Data Bytes sent so far) / 8.

6. IPv6 & Tunneling

Created to solve IP exhaustion (128-bit addresses).

Key changes: Fixed 40-byte header, and fragmentation is NOT allowed by routers (must be done by the end host).
Tunneling: How to pass IPv6 packets through older IPv4 networks? The IPv6 packet is encapsulated inside the payload of an IPv4 packet.

7. Routing in the Internet (AS, IGP, EGP)

The internet is divided into Autonomous Systems (ASes) (e.g., ISPs, universities).

Intra-AS Routing (IGP - Interior Gateway Protocols): Routing within a specific AS.
- RIP (Routing Information Protocol): Distance Vector. Metric = Hop count (Max 15). Advertises every 30s. If no ad for 180s, link is dead.
- OSPF (Open Shortest Path First): Link State. Uses Dijkstra’s algorithm. Floods the entire AS.
Inter-AS Routing (EGP - Exterior Gateway Protocols): Routing between different ASes.
- BGP (Border Gateway Protocol): The de facto protocol of the internet.
- Policy-based routing: BGP doesn’t just look at distance; it looks at business rules. For example, a customer network (Dual-homed to two ISPs) will never advertise a transit route between its two ISPs, because it gets no profit for carrying their traffic.

Sample Exam Questions & Solutions

Question 1: Subnetting Math

Given the IP address 200.23.16.130/23: a) How many total IP addresses are in this subnet? b) How many valid IP addresses can be assigned to hosts? c) Find the network (subnet) address.

Solution: a) A /23 mask leaves $32 - 23 = 9$ bits for hosts. Total IPs = $2^{9} = 512$ . b) Valid hosts exclude the network and broadcast addresses. $512 - 2 = 510$ . c) The 23rd bit falls in the 3rd octet.

16 in binary is 00010000. 130 in binary is 10000010.
The subnet mask for /23 is 255.255.254.0 (11111111.11111111.11111110.00000000).
Bitwise AND between IP and Mask zeroes out the host bits.
Network Address = 200.23.16.0.

Question 2: IP Fragmentation Math

A host sends a large IP datagram containing 6000 bytes of data and a 20-byte IP header. It travels over a link with an MTU of 1500 bytes. a) How many fragments are created? b) For the second fragment, what is the Total Length (bytes), the Data Size (bytes), and the MF flag? c) For the last fragment, what is the Total Length (bytes), Data Size (bytes), and the MF flag?

Solution:

Step 1: Calculate Max Payload per fragment. MTU (1500) - Header (20) = 1480 bytes of data per fragment.
Step 2: Split the 6000 bytes of data. $6000 = 1480 + 1480 + 1480 + 1480 + 80$ . a) 5 fragments are created. b) Second fragment: Data Size = 1480 bytes. Total Length = $1480 + 20 = 1500 bytes$ . MF Flag = 1 (more fragments follow). c) Last fragment: Data Size = 80 bytes. Total Length = $80 + 20 = 100 bytes$ . MF Flag = 0 (no more fragments).

Question 3: Longest Prefix Matching

A router uses longest prefix matching with the following table:

Interface 0: 10.1.0.0/16
Interface 1: 10.1.2.0/24
Interface 2: 10.1.2.128/25
Interface 3: Otherwise Which interface will a packet destined for 10.1.2.200 be forwarded to?

Solution: First, check which ranges it fits into:

Matches /16 (Starts with 10.1)
Matches /24 (Starts with 10.1.2)
Matches /25 (Starts with 10.1.2 and the last octet is between 128 and 255. 200 fits here!) Because it matches all three, the router uses the Longest Prefix (the most specific match). /25 is the longest prefix. Answer: It will be forwarded to Interface 2.

Question 4: Network Address Translation (NAT)

A home network uses a NAT router with the public IP 138.76.29.7. An internal host (10.0.0.1) sends a TCP request from source port 3345 to an external web server (128.119.40.186 on port 80). a) What are the Source IP and Source Port on the packet after it leaves the NAT router? b) When the web server replies, what Destination IP and Destination Port will it put on the packet?

Solution: a) The NAT router replaces the internal private IP/Port with its own public IP and a new, unique port (e.g., 5001). Source IP: 138.76.29.7. Source Port: 5001. (The destination IP/Port remain the external server’s). b) The web server replies to the public-facing NAT router. Destination IP: 138.76.29.7. Destination Port: 5001. (When the NAT router receives this, it will look up 5001 in its table and forward it internally back to 10.0.0.1:3345).

Question 5: BGP Routing Policy Theory

Autonomous System X is a customer network connected to two different provider ISPs (AS B and AS C). Explain why AS X will not advertise a route to AS B through AS C (e.g., path C-X-B).

Solution: BGP is a policy-based routing protocol driven by business agreements. AS X is a paying customer of both AS B and AS C. If AS X advertises a route between B and C, it would act as a transit network, carrying traffic between the two providers for free. Since AS X receives no profit for routing this transit traffic (and it would consume X’s bandwidth), its routing policy will explicitly forbid advertising the C-X-B path. An ISP access network only routes traffic that originates from or is destined for its own network.

CITS3002

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Week 7