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Lecture 1 1
Lecture 1: TCP/IP
TCP/IP Layer Structure IP TCP UDP
Lecture 1 2
Introduction
To achieve the global connectivity, ideally we have a virtual global network to which every machine can connect to.
However, in the real world, we do not have such global network. Many different networks with different
technologies and protocols are existed all over the world.
We called they are heterogeneous networks.
Lecture 1 3
Introduction
Alternative approach
We establish physical links and routers to connect them together;
we apply the same higher level communication protocol for each machine such that receivers can understand the content of packets sent from senders.
Solution: TCP/IP
Lecture 1 4
TCP/IP Model
TCP/IP = Transmission Control Protocol/Internet Protocol.
Developed in 1970s by the US Department of Defense.
Application
Transport (TCP)
Internet (IP)
Link
Physical
Lecture 1 5
TCP/IP Model
Physical (Chapter 4)
Link (Chapter 5)
Internet (IP = Internet Protocol) specify the format of the packets sent across
the Internet and the mechanisms used to forward packets from a station through one or more routers to the destination station.
Lecture 1 6
TCP/IP Model
Transport (TCP = Transmission Control Protocol) specify an end-to-end protocol for the reliable
transfer of data between two programs.
Application specify how one application uses an Internet.
Lecture 1 7
TCP/IP Model
TH
IH
HH
Data
TCP Data
IP Data
H-to-N Data HT
Source machine
Application
Transport (TCP)
Internet (IP)
Host-to-Network
Destination machine
Application
Transport (TCP)
Internet (IP)
Host-to-Network
Lecture 1 8
IP
Internet Protocol
Main characteristics
Hierarchical addressing: IP address are 32 bits in length and are used in the source and destination address fields of the IP datagram.
Connectionless routing: each data packet is an individual datagram to do the routing.
Lecture 1 9
IP Address
Dotted Decimal Notation
It is difficult for human being to read 32-bit IP addresses in technical documents or through application programs.
Thus IP addresses are written as four decimal integers separated by decimal points, where each integer gives the value of one octet of the IP address (1 octet = 8 bits).
Lecture 1 10
IP Address
Example 1
10000000 00001010 00000010 00011110is written as 128.10.2.30
Example 2
149.8.12.40 is written as10010101 00001000 00001100 00101000
Lecture 1 11
Two components: network id. and host id.
Network id.: identifies the network;
Host id.: identifies the station (or host computer) on that network (as identified by the network id.).
IP Address
IP address
Network id. Host id.
Lecture 1 12
IP Address
Trade off between the size of the network id. field and that of the host id. field. Larger network id. larger possible number of
networks (in the Internet) with each network having smaller number of hosts.
Larger host id. larger number of hosts in a network but the possible number of networks is smaller.
To accommodate networks of different sizes, we classified networks into 5 categories: A, B, C, D and E.
Lecture 1 13
IP Address
0 Network Host
10 Network Host
110 Network Host
1110 Multicast address
11110 Reserved for future use
ClassA
B
C
D
E
32 bits
Lecture 1 14
IP Address
Class A 126 networks (7-bit network id. 2 special cases);
All 0s (0.0.0.0): allow only at system startup and never be a valid destination address. Once a machine learns its correct IP address, all 0s will not be used.
All 1s (127.xx.yy.zz): reserved for loop back test (testing the TCP/IP on a local machine, send a packet from its output port and receive it from its own input port).
16.8 million hosts (24-bit host id. 2 special cases);All 0s (xx.0.0.0): network.All 1s (xx.255.255.255): broadcast within the network.
Lecture 1 15
IP Address
Class B 16,382 networks (14-bit network id. 2); 65534 hosts (16-bit host id. 2);
Class C 2 million networks (21-bit network id. 2); 254 hosts (8-bit host id. 2);
Class D 28 bits to specify a multicast group; can be used only as destination address;
Lecture 1 16
IP Header
Version IHL Type of service Total length
Identification DF
MF Fragment offset
Time to live Protocol Header checksum
Source address
Destination address
Options (0 or more words)
32 bits
Lecture 1 17
IP Header
Total length: the total length of the datagram (including header);
Datagram = IP-PDU, packet routed in IP layer;
Maximum: 65,535 bytes (1 byte = 8 bits);
How to handle a message if its size is larger than the maximum packet size of a physical network? Solution: fragmentation and reassembling.
Lecture 1 18
IP Header: Fragmentation
Fragmentation of IP datagram is allowed because This feature facilitates building an Internet
with components networks accommodating different maximum packet sizes.
IP datagram
Datagram header
DatagramData Area
FrameHeader
Frame Data AreaPhysical Network frame
Lecture 1 19
IP Header: Fragmentation
When an IP datagram is encapsulated by a physical network frame (e.g. Ethernet frame), since the size of a frame is limited (e.g. for fairness), the size of an IP datagram is also limited. Thus we need to apply fragmentation for the transmitted message longer than the limited size.
Packet too long may suffer long transmission delay and even cause network congestion. Thus short packet is preferred.
Lecture 1 20
IP Header: Fragmentation
When the size of a datagram is larger than the maximum one, a router breaks the datagram up into a number of small fragments.
The destination host's IP layer can then reassemble the fragments back to the complete datagram before passing it up to upper layer protocol (say TCP) entity.
Lecture 1 21
IP Header: Fragmentation
Identifier: When a large datagram needs to be fragmented, all its fragments carry the same value in the identifier field.
The destination host can determine which datagram the current fragment belongs to and reassemble the original datagram.
DF: when it sets to 1, it tells the Internet (router) not to fragment the datagram.
Lecture 1 22
IP Header: Fragmentation
MF: when it sets to 1, it stands for 'more fragment'. All fragments of a datagram except the last one
have this bit set.
Fragment offset: it tells where in the containing datagram this fragment belongs. To reassemble, the destination host must obtain
all fragments starting with the fragment that has offset 0 through the fragment with the highest offset.
Lecture 1 23
IP Header
Time to live: it specifies how long, in seconds, the datagram is allowed to remain in the internet system.
Protocol: it tells the network access layer in the destination host, which upper protocol process to give the datagram to. Usually it will be TCP or UDP.
Lecture 1 24
IP Header
Header checksum: a checksum verifying the header only;
Source address;
Destination address;
Options
Lecture 1 25
IP Routing
hosts
message
NetA
NetC
NetBNetD
R(AC)
H1
H2
H4
H3
R(ABD)
Lecture 1 26
IP Routing
Consider H1 would like to send a packet to H3.
H1 is the end station of the network NetA, and H2, H3 and H4 are the end stations of the network NetD.
H1 communicates with other stations by using the native protocol of the network NetA (say PrA, e.g. Ethernet).
Similarly, H2, H3 and H4 communicate with each other with the native protocol of NetD (say PrD, e.g. Token Ring).
Lecture 1 27
IP Routing
It is possible that PrA, PrB , PrC and PrD are not the same.
H1 does so by using IP protocol which H1, R(ABD) and H3 all understand and agree upon.
H1 puts H3's IP address in the destination address and its own IP address in the source address.
H1 also puts the destination address of R(ABD) in the destination address field of the header of PrA-PDU.
Lecture 1 28
IP Routing
When the PrA-PDU is routed by NetA to the destination R(ABD), R(ABD) will extract the IP datagram from the PrA-PDU and look at the destination address and decide that the destination is on H3.
So R(ABD) sends the IP datagram to station H3, this time embedding the datagram in a PrD-PDU.
When H3 receives the PrD-PDU, it will extract the IP datagram and obtain the data.
Lecture 1 29
IP RoutingET = Ethernet Tail
TT = Token Ring Tail
Px(y) = Physical address of y in x
H1 App.
TCP
IP
H-to-N
H3Datagram
IP Header
DataS=H1
D=H3...
IP
H-to-N
App.
TCP
IP
H-to-N
Router R(ABD)
D=PrA(R(ABD))
…ET
Ethernet Header
D=PrD(H3)
…TTDatagram
Token Ring Header
Network NetA Network NetD
Datagram
Lecture 1 30
IP Routing
How does the router make a suitable decision to route the packet to H3? Solution: Routing table.
A network on the Internet is usually designated by the network prefix of its IP address followed by appending 0's to the suffix.
Example: the network 144.214 in the next slide is usually designated 144.214.0.0.
Lecture 1 31
IP Routing
A router is connected to more than one network. Hence it has multiple IP addresses. Example: the router below has two IP
addresses: 144.214.0.15 and 144.120.12.9.
router144.214.0.0
144.120.0.0
144.214.0.15
144.120.12.9
Lecture 1 32
IP Routing
R1 R2
R3
NetA
NetB NetC NetD
NetE
Routing table at R2
Destination Next hop
NetA R1 NetB R1 NetC direct deliver NetD direct deliver NetE R3
Lecture 1 33
IP Routing
In practice, the networks are identified by its IP address.
The router uses a network-id mask (also called a subnet mask) to extract the network id from a (destination station) IP address and search the routing table for a match.
Lecture 1 34
IP Routing
R1 R2
R3
20.0.0.0
40.0.0.0 128.1.0.0 192.4.10.0
144.214.0.0
40.0.0.7
128.1.0.8
144.214.0.5
128.1.0.9
192.4.10.9
20.0.0.7
192.4.10.8
Lecture 1 35
IP Routing
Routing table at R2 would look like
Destination Mask Next hop
20.0.0.0 255.0.0.0 128.1.0.840.0.0.0 255.0.0.0 128.1.0.8128.1.0.0 255.255.0.0 direct deliver192.4.10.0 255.255.255.0 direct deliver144.214.0.0 255.255.0.0 192.4.10.8
Lecture 1 36
IP Routing
Example Datagram P arrive at R2 with destination
address 144.214.10.18. For each entry in the routing table, the
corresponding mask is “anded” with the destination address and the result (144.214.0.0) is compared to the destination (network) field.
If a match is found, it will be sent to the address at the next hop field (192.4.10.8).
Lecture 1 37
Companion IP Protocols
The core IP protocol is for the sending of datagrams between stations across the Internet.
There are a number of companion protocols to handle other functions.
Two important protocols will be described: ICMP (Internet Control Message Protocol) and ARP (Address Resolution Protocol).
Lecture 1 38
ICMP
It is used to communicate control messages between host and router, among routers and between hosts.
ICMP messages are embedded in the data field of a datagram and the protocol type is set to 1.IP
headerIP data field
IP datagram
ICMP message
Protocol = 1
Lecture 1 39
ICMP
Most ICMP messages are for signaling error or unusual situations.
Messages between routers and hosts:'can't reach destination'
'Time-to-live expired'
'illegal parameter''slow down - congestion''there is a better route to send data', …, etc.
Lecture 1 40
ICMP
Messages between hosts:'can't read application''reassembly time expired''strange parameter''slow down - congestion''echo request''echo reply', …, etc.
Lecture 1 41
ARP
When an IP datagram arrives at a destination router, the router will send the datagram to the destination host over the destination network.
Since the format of the physical network address (e.g. Ethernet address) is different from that of the IP address, usually the router has a table to map the destination IP address to its corresponding physical network address.
Lecture 1 42
ARP
The router then sends the datagram to the destination by encapsulating the datagram in the corresponding physical network address.
However, if the router does not know it, how does the router send the datagram? The router may not know the mapping if the
configuration of the physical network is changed, or the station is just joined the network.
Solution: ARP
Lecture 1 43
ARP
R144.214.0.0
144.120.0.0
144.214.01.5
144.120.12.9
Datagram with destination address Ap
Lecture 1 44
ARP
An IP datagram with destination address Ap (e.g. 144.120.60.8) arrives at the (destination) router R.
The router wants to know the Ethernet address of the station with IP address Ap. R broadcasts a request: “Who owns IP address Ap?” on the destination LAN (e.g. 144.120.0.0).
Lecture 1 45
ARP
Only the destination with the IP address Ap will response, giving its physical network address to R (e.g. Ethernet address E).
Then R updates its table and send the datagram to the destination.
Note that a source station can use ARP to find the local network address of the router if necessary.
Lecture 1 46
TCP
Transmission Control Protocol
Functions:
To provide a point-to-point reliable connection oriented service for upper (application) layer entities.
To provide for multiplexing of multiple transport connections over a single network.
Lecture 1 47
TCP Segment
TCP PDUs are called segments.
Fixed size header (20 bytes);
The data field can be up to 216 40 bytes (TCP and IP headers), i.e. 65,495 bytes.
Lecture 1 48
TCP Segment
Data (optional)
Destination port
Acknowledgement number
Options (0 or more 32-bit words)
Source port
Sequence number
TCP headerlength
FIN
RST
PSH
ACK
URG
SYN
Window size
Urgent pointerChecksum
32 bits
Lecture 1 49
TCP Segment
Source port and destination port: TCP port numbers that identify the application programs at the ends of the connection. A port number plus an IP address form an
unique transport service access point (TSAP).
Sequence number (SEQ): identify the position in the sender’s byte stream of the data in the segment.
Lecture 1 50
TCP Segment
Acknowledgement number (ACKN): identifies the number of the octet that the source expects to receive next.
TCP header length: TCP segment’s header (in units of 32-bit words);
Lecture 1 51
TCP Segment
URG When it sets to 1, the urgent pointer is in use. The pointer is used to specify the position in
the segment where urgent data ends. This is used to draw attention of the receiver.
ACK When it sets to 1, the field of the
acknowledgement number (ACKN) is valid.
Lecture 1 52
TCP Segment
PSH When it sets to 1, it indicates to the receiver
that it should deliver the data (and any already buffered) to the application program.
Otherwise, the receiver may buffer (and only deliver when buffer is full) for efficiency.
RST When it sets to 1, reset the connection.
Lecture 1 53
TCP Segment
SYN Used for connection set-up; SYN = 1, ACK = 0 connection set-up
request; SYN = 1, ACK = 1 connection set-up accept;
FIN Used for connection release; When it sets to 1, the sender has reached end
of its byte stream.
Lecture 1 54
TCP Services
Connection set-up three-way handshake
Sender Events
Receiver
Events
Network Message
sSYN(SEQ = x)
SYN(SEQ = y, ACKN = x + 1)
SYN(SEQ = x + 1, ACKN = y + 1)
Lecture 1 55
TCP Services
The advantage of three-way handshake is that it still works even the TCP segment containing the connection-accept segment is lost.
Note that a new set of starting sequence numbers is used on connection set-up. This is to avoid any segment from a previous connection session between the same processes from confusing the current connection.
Lecture 1 56
TCP Services
Connection release
FIN
FIN
ACK
Sender Events
Receiver
Events
Network Message
s
Lecture 1 57
TCP Services
Data transfer
Damaged and lost segments are handled by a positive acknowledgement time-out retransmission mechanism.
Duplicated and out of order segments are detected by use of the sequence number field.
Lecture 1 58
TCP Services
Flow control It is affected by a window mechanism. The send window size can be dynamically changed
by the receiver (based on its buffer condition). Window advertisement (the window size field in
the TCP header) specifies how many octets (1 octet = 8 bits) of data that the receiver is prepared to accept.
Example : maximum segment size of the sender is 1000 octets and maximum window advertisement is 2000 octets.
Lecture 1 59
TCP Services
advertise window = 2500
send data octets 1 - 1000send data octets 1001 - 2000send data octets 2001 - 2500
ack up to 1000, window = 1500ack up to 2000, window = 500ack up to 2500, window = 0
Sender Events
Receiver
Events
Network Message
s
receive ack for 1000receive ack for 2000receive ack for 2500 application reads 2000
octetsack up to 2500, window = 2000
ack up to 3500, window = 1000ack up to 4500, window = 0
application reads 1000 octetsack up to 4500, window = 1000
send data octets 2501 - 3500send data octets 3501 - 4500
receive ack for 4500
receive ack for 3500receive ack for 4500
Lecture 1 60
TCP Services
Congestion control: slow-start algorithm TCP (sender) maintains two windows, a send
window Ws which is set by receiver’s window advertisement, and a congestion window Wc.
The sender uses the smaller of the two for actual transmission.
A threshold, T, is an integer such that the congestion window will increase exponentially before reaching the threshold.
Usually T will be initially set to 64k bytes.
Lecture 1 61
TCP Services
Procedure:1. Wc = 1.
2. When (i) a window is sent, (ii) there is no time-out, and
(iii) Wc is smaller than the threshold,
Wc = min(2 Wc, threshold)
(growth rate is exponential).3. When (i) a window is sent, (ii) there is no time-out,
and(iii) Wc is not smaller than the threshold,
Wc = Wc + 1 (growth rate is linear).
4. When a time-out occurs, T = Wc / 2 and Wc = 1.
Lecture 1 62
TCP Services
Slow-Start
0
10
20
30
40
50
60
70
80
0 5 10 15 20
Transmission numbers
Co
ng
est
ion
win
do
w (
kbyt
es)
Timeout
Threshold
New Threshold
Lecture 1 63
TCP Services
Trans. No. Wc (kbytes) Trans. No. Wc (kbytes)
0 1 13 11 2 14 22 4 15 43 8 16 84 16 17 165 32 18 326 64 19 357 65 20 368 66 21 379 67 22 3810 68 23 3911 69 24 4012
(Timeout)70
Lecture 1 64
TCP Multiplexing
A host use an unique IP address to communicate through the Internet.
Within that machine, there may be multiple application programs requiring remote communication services.
The TCP layer implements multiple transport connections over a single network interface.
Lecture 1 65
TCP Multiplexing
Host A Host B
Internet
TCP
IP
X Y
TCP
IP
M N
144.214.12.38
205.10.11.09
144.214.12.38:23
144.214.12.38:290
205.10.11.09:2529
205.10.11.09:1326
Lecture 1 66
UDP
User Data ProtocolConnectionless transport protocol
suitable for applications requiring short communication exchanges;
packet is up to 64 kbytes.32 bits
Destination portSource port
ChecksumDatagram length
User data
Lecture 1 67
Tutorial 1
1. When an IP datagram is to be routed through a network whose maximum packet size is smaller than that of the datagram, it is fragmented into smaller datagrams. Where do you think is better to reassemble the datagram? At the next router or at the destination host (IP layer)? Explain.
Lecture 1 68
Tutorial 1
2. Most IP datagram reassembly algorithms have a timer to avoid having a lost fragment tie up reassembly buffers forever. Suppose a datagram is fragmented into 4 fragments. The first 3 fragments arrive, but the last one is delayed. Eventually the timer goes off and the three fragments in the receiver’s memory are discarded. A little later, the last fragment stumbles in. What should be done with it?
Lecture 1 69
Tutorial 1
3. How many responses a router expects to get when it broadcasts an ARP request? Why?
4. You have just explained the ARP protocol to a friend. When you are all done, he says: “I have got it. ARP provides a service to the network layer, so it is part of the data link layer.” What do you say to him?
Lecture 1 70
Tutorial 1
5. Write out the following IP address in dotted decimal format:
10010000 11001000 00100101 01000001
6. Is the IP address space efficiently utilized? Explain.Suppose that instead of using 16 bits for the network part of a class B address, 20 bits has been used. How many class B networks would there have been?
Lecture 1 71
Tutorial 1
7. What is the size of the port number space for TCP? What is the maximum size of a TCP segment? Under what condition is this maximum size achievable?
8. Consider a TCP connection over the Internet. When a time-out occurs on the sending of a segment, which is by far most likely the cause: (i) congestion, (ii) error: damaged or lost IP datagram (which encapsulates the TCP segment). Explain.
Lecture 1 72
Tutorial 1
9. Consider the slow start flow control algorithm used in TCP. Suppose the maximum segment size is 1 Kbytes. Suppose the congestion window Wc just before a time-out was 32 Kbytes. What are the congestion window sizes for the first 8 transmissions after the time-out? Assume that there are no time-outs during these 8 transmissions.