1. Anjuman College of Engineering and Technology Department of Computer Science and Engineering VII Sem. TCP and IP (Elective I)
Prof. Nazish Khan

2. Syllabus: TCP and IP
Unit I: Network architecture-Standards, TCP/IP Model Overview, Networking Technologies: LANS, WANS, Connecting Devices. Internetworking concept, Internet Backbones, NAP, ISPs, RFCs and Internet Standards.
Unit II: Classful Internet address, CIDR-Subnetting and Supernetting, ARP, RARP, OOTP, DHCP.
Unit III: IP Datagram-IP Package-IP forwarding and routing algorithms, computing paths, RIPOSPF, ICMP, IGMP.
Unit IV: TCP header, services, Connection establishment and termination, Interactive data flow, Bulk data flow, Flow control and Retransmission, TCP timers, Urgent Data processing, Congestion control, Extension headers.
Unit V: Switching technology, MPLS fundamentals, signaling protocols, LDP, IP traffic engineering, ECMP, SBR, Routing extensions for traffic engineering, Traffic engineering limitations and future developments.
Unit VI: IP security protocol-IPv6 addresses, Packet format, Multicast, Anycast, ICMPv6, Interoperation between IPv4 and IPv6-QoS, Auto configuration.

3. Course Outcomes:
Upon completion of this course students will be able to:
differentiate functioning of OSI and TCP/IP Model; explain the basics of networking and Internet.
classify and solve problems on IP Addressing.
analyze the roles, services and features of various Network Layer and Routing protocols.
analyze the roles, services and features of Transport layer Protocols.
explain the concepts of Switching Technologies and Traffic Engineering.
explain security at IP, compare IPv4 and IPv6.

4. CO 3:analyze the roles, services and features of various Network Layer and Routing protocols.
Internet Protocols

5. Introduction:
The Internet Protocol (IP) is the transmission mechanism used by the TCP/IP protocols.
IP is an unreliable and connectionless datagram protocol.
It is also called as best effort delivery protocol.
The term best effort means that it does not provide error checking and tracking.
IP tries its best to get a transmission done through to its destination.
IP does not guarantees the data transmission.
It must be paired with a reliable protocol to provide reliability, such as TCP.
IP is a connectionless protocol.
This means that whatever data transmission takes place between the source and destination, it could arrive out of order, lost or corrupted.

6. Datagram: Packets are called as Datagram.
A datagram is a packet of variable length consisting of two parts:
Header and Data.
The minimum size of an IP datagram is 20 bytes and maximum size is bytes.
The size of the header is 20 to 60 bytes, it contains information which is related to routing and delivery.
Figure in next slide shows the IP Datagram format.
A brief description of each field is also given.

7. Datagram:

8. Datagram:
Version:
Which version of the protocol the datagram belongs to.
The current version number is 4.
Next version: 6
IHL:
The number of 32-bit words in the header
Because this is 4 bits, the max header length is 15 words (i.e. 60 bytes)
The header is at least 20 bytes, but options may make it bigger

9. Datagram: Type of Service:
Contains a 3-bit precedence field (that is ignored today), 4 service bits, and 1 unused bit.
The four service bits can be:
Normal (default)
minimize delay
maximize throughput
maximize reliability
minimize cost
This is a "hint" of what characteristics of the physical layer to use
The Type of Service is not supported in most implementations. However, some implementations have extra fields in the routing table to indicate delay, throughput, reliability, and cost.

10. Datagram: Total Length: total length of the datagram in bytes.
we know where the data starts by the header length
we know the size of the data by computing "total length - header length"
Identification:
Uniquely identifies the datagram.
Usually incremented by 1 each time a datagram is sent.
All fragments of a datagram contain the same identification value.
This allows the destination host to determine which fragment belongs to which datagram.

11. Datagram: Flags and Fragmentation Offset: Used for fragmentation
DF means do not fragment. It is a request to routers not to fragment the datagram since the destination is incapable of putting the pieces back together.
MF means more fragments to follow. All fragments except the last one have this bit set. It is needed to know if all fragments of a datagram have arrived.
Fragment offset:
number of fragment.

12. Datagram: Time to Live: Upper limit of routers
usually set to 32 or 64.
decremented by each router that processes the datagram,
router discards the datagram when TTL reaches 0.
Protocol:
Tells IP where to send the datagram up to.
1 means ICMP
2 means IGMP
6 means TCP
17 means UDP

13. Datagram: Header checksum: Only covers the header, not the data.
Source IP address:
The sender
Destination IP address:
the final destination
Options:
Optional data.
Some examples include having the router put in a IP address of router and a time stamp so the final destination knows how long it took to get to each hop.
The source and destination in the IP header is the original source and the final destination! The physical layer addresses pass the datagram from router to router. So, while the physical layer addresses change from router to router, the source and destination IP addresses in the IP datagram remain constant!

14. Datagram: The checksum: How to compute a checksum?
Put a 0 in the checksum field.
Add each 16-bit value together.
Add in any carry
Inverse the bits and put that in the checksum field.
To check the checksum:
Add each 16-bit value together (including the checksum).
Add in carry.
Inverse the bits.
The result must be 0.
Remember, only the bits in the header are calculated in the IP checksum.

15. Example:
Consider the following IP header, with source IP address of and destination address of All values are given in hex: c
92 cc 00 00
92 95 ba 14
a9 7c 15 95

16. Example:
So, first add all 16-bit values together, adding in the carry each time:
4500
+ 006c
----
456c
+ 92cc
d838
+ 0000
+ 3806
1103e <---But, we have a carry here! So, remove the leftmost bit and add it back in. So, we get: 103e + 1 = 103f.

17. Example:
103f a2d4 + ba14 15ce8 <---Again, we have a carry here! So, remove the leftmost bit and add it back in. So, we get: 5ce8 + 1 = 5ce9. 5ce9 + a97c <---Again, we have a carry here! So, remove the leftmost bit and add it back in. So, we get: = bfb

18. Example: Now we have to inverse the bits. 1bfb = 0001 1011 1111 1011
inverse bits: = e404
So, the checksum is e404. So, the IP header we send looks like: c
92 cc 00 00
38 06 e4 04
92 95 ba 14
a9 7c 15 95
As an exercise, please act as the receiver, compute the checksum on that packet, and make sure the result is 0!

19. IP Fragmentation:
Note: the total_length field in the IP header is 16 bits. that means the max size of of an IP datagram is bytes.
BUT, the physical layer may not allow a packet size of that many bytes (for example, a max ethernet packet is 1500 bytes)
SO, IP must sometimes fragment packets.
When an IP datagram is fragmented, each fragment is treated as a separate datagram.
it is reassembles at the final destination, not at a router!
it does that because the router may have to fragment it again!

20. IP Fragmentation: Each fragment has its own header.
The identification number is copied into each fragment.
One bit in the "flags" field says "more fragments are coming. If that bit is 0, then it signifies this is the last fragment.
The "fragment offset" field contains the offset of the data.
fragment flag of 0 and offset of 0 means the datagram is not fragmented.
fragment offset is measured in units of 8 bytes (64 bits). That is because the fragment offset field is 3 bits shorter than the total length field (and 2^3 is 8).

21. IP Fragmentation: The entire flags field looks like this:
| bit 0 | bit 1 | bit 2 |
bit 0: not used
bit 1: if 1, it means "don't fragment". If IP must fragment the packet and this bit is set, IP throws away the datagram.
bit 2: The fragment flag.

22. Example:
Suppose we have a physical layer that can transmit a maximum of 660 bytes. And, suppose IP wants to send 1460 bytes of data. So, the IP datagram is a total of 1480 bytes, including the 20 byte IP header:
| 20-byte ip header | 1460 bytes of data |
Here is what IP sends:
First packet:
bytes:
| IP header | first 640 bytes of data |

23. Example: In that packet, "fragment flag" is 1, offset is 0.
Second packet:
bytes:
| IP header | second 640 bytes of data |
In that packet, "fragment flag" is 1, offset is 80. The offset is 80 because (80 * 8) is 640, so the offset of that data is 640 byes into the packet.
Note: all other fields of the IP header are identical to the first packet (except the checksum)!
Third packet:
| IP header | third 180 bytes of data |

24. Example:
In that packet, "fragment flag" is 0, offset is 160. The offset is 160 because (160 * 8) is 1280, so the offset of that data is 1280 byes into the packet.
Note: all other fields of the IP header are identical to the first packet except the checksum.
IMPORTANT: The routers see 3 separate packets. The final destination reassembles the packet before passing the packet to the upper layers.