Netwerken · Netwerken TCP / IP. 3de jaar Netwerkbeheerder Page: 2 Syntra Antwerpen &...

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TCP / IP



• IPv4 Essentials: Architectural overview.

– The 4-layer model.– ARP– IP– ICMP– IGMP– TCP / UDP

• IPv4 Essentials: addressing / netmasking.

– Classfull IP adressing– Limitations of classfull adressing– Subnetting– Subnetting example

Module 2: IPv4 Essentials



2.1.1 The 4-layer model.

2.1 IP Essentials: Architectural overview.

Layer 5,6,7Layer 5,6,7

Layer 4Layer 4

Layer 3Layer 3

Layer 1,2Layer 1,2

FTPFTPSMTPSMTP TelnetTelnet SNMPSNMP TFTPTFTP NFSNFS

TCPTCP UDPUDP

IPIP ICMPICMP

ARPARP

Media Access (LAN & WAN)Media Access (LAN & WAN)

RARPRARP





• Layer 1, Media access layer (OSI-layers 1 & 2).

– Network interface layer.– Transmitting and receiving of data (frames).

• Layer 2, IP/ARP/ICMP/IGMP (OSI-layer 3)

– Internet layer.– IP (Internet Protocol): routing of data packets.– ARP (Address Resolution Protocol): collects hardware addresses of hosts in the same

physical network.– ICMP (Internet Control Message Protocol): sends messages and creates error messages

when packets are not correctly delivered.– IGMP (Internet Group Management Protocol): used for multicasting.






• Layer 3, TCP / UDP (OSI-layer 4)

– Transport layer.– TCP (Transport Control Protocol): connection-oriented communication, creates a logical link

between sender and receiver reliable communication.– UDP (User Datagram Protocol): connection-less communication. No checking to see if data

arrives correctly at the receiver unreliable communication.

• Layer 4, SMTP, FTP, Telnet, … (OSI-layers 5,6 & 7)

– Application layer.– All different IP applications are situated in this layer :

• SMTP: Mail• FTP: File Transfer Protocol.• …





2.1.2 ARP.

• Sender must know MAC address of receiver in order to communicate.

• Address resolution is the process of linking an IP address to a MAC address.

• Broadcast based.

• How ?





2.1.2 ARP.





2.1.2 ARP.

• ARP-cache used to minimalize number of broadcasts.

• Contains dynamic and static addresses:

– dynamic: addresses are added and removed automatically.– static: address stays in cache till the next reboot of the system.

• ARP-cache contains permanently the broadcast address of the local subnet. It’s never shown if you look at the content of the cache.

• Every entry in the cache has a potential lifetime of 10 minutes.

• If an entry is not used within 2 minutes it is removed.

• If an entry is used within 2 minutes it stays in the cache.

• Every entry is always removed after max 10 minutes, even if it is being used!

• When cache is full FIFO principle (First In, First Out).





2.1.3 IP.

• Connection-less protocol No reliable communication.

• No acknowledgement when packets arrive correctly at the receiver.

• Acknowledgements are handled by the applications in the upper layers.

• Following fields are added into the IP-header when a packet arrives from the transport layer:


Field Function

IP address sender The IP address of the sender (source IP)

IP address receiver The IP address of the receiver (destination IP)

Protocol Informs the IP-layer of the receiver to which protocol the data has to be transferrred: TCP or UDP

Checksum A calculation method to check if the IP packet contains errors

TTL (Time To Live) The number of seconds a data packet can exist. Every time a packet passes a router, TTL = TTL -1. When TTL = 0 the packet is dropped.




2.1.3 IP.

• How does the routing process work?

On the PC:


New IP packet

Localaddress?

yes Packet is sent directlyto receiver

noExisting routeto receiver?

Packet is sent viathis route to receiver

yes

no

Packet is sent tothe default gateway




2.1.3 IP.

On the router:



Incoming IP packet

TTL = 0 ??

TTL = TTL - 1

Drop Packetyesno

Fragment packet ??

Calculate newchecksum

LocalAddress ??

Forward to receiver

Forward todefault router

yes

Create new header for every fragment withfollowing fields:•A flag to indicate that other fragments follow.•A fragment ID.•A fragment offset (tells receiver how many fragments to use for reassembling ip packet.

yes

noExisting routeto receiver?

no

yes

Packet is sent usingthis route

no Fragment packet ??

Calculate newchecksum

Fragmentation if needed.Procedure idem as described above

Fragmentation if needed.Procedure idem as described above



3.1.4 ICMP.

• Doesn’t make IP reliable.

• Tries to report errors and to give feedback on certain conditions.

• ICMP packages can be delayed or can be lost during transmission.

• Unreliable.





2.1.5 IGMP.

• Defined in RFC 1112.

• Only used for multicasting (one to many transmissions).

• IGMP information is sent to all routers supporting multicasting.

• IGMP Packets are sent as datagrams and therefore unreliable.

• PS.: Use following url to look for RFC’s:

http://www.rfc-editor.org/rfcsearch.html









2.1.5 IGMP.

• Defined in RFC 1112.

• Only used for multicasting (one to many transmissions).

• IGMP information is sent to all routers supporting multicasting.

• IGMP Packets are sent as datagrams and therefore unreliable.

• PS.: Use following url to look for RFC’s:










2.1.6 TCP / UDP.

• TCP: Connection oriented service reliable.• UDP: Connection-less service unreliable.





2.1.7 Layered communication.

• How do layers address each other?

At least 3 addresses: IP + Port Number + Protocol type (TCP or +UDP) = Socket


To: 34To: 34From: 2813From: 2813

To: 193.210.191.190To: 193.210.191.190From: 193.210.191.2From: 193.210.191.2

To: 27BACD456FECTo: 27BACD456FECFrom: A460F41BC52DFrom: A460F41BC52D

MACMAC

IPIP

TCPTCP

Applic.Applic.From: Prot + From: Prot + IPIP + + Port Port ((> Socket> Socket))To: Prot + IP + Port (< Socket)To: Prot + IP + Port (< Socket)

From: 2813From: 2813To: 34To: 34

From: 193.210.191.2From: 193.210.191.2To: 193.210.191.190To: 193.210.191.190

From: A460F41BC52DFrom: A460F41BC52DTo: 27BACD456FECTo: 27BACD456FEC

From: Prot + From: Prot + IPIP + + Port Port ((> Socket> Socket))To: Prot + IP + Port (< Socket)To: Prot + IP + Port (< Socket)


From: 34From: 34To: 2813To: 2813

To: 2813To: 2813From: 34From: 34

From: 193.210.191.190From: 193.210.191.190To: 193.210.191.2To: 193.210.191.2

To: 193.210.191.2To: 193.210.191.2From: 193.210.191.190From: 193.210.191.190

From: 27BACD456FECFrom: 27BACD456FECTo: A460F41BC52DTo: A460F41BC52D

To: A460F41BC52DTo: A460F41BC52DFrom: 27BACD456FECFrom: 27BACD456FEC



2.2.1 Classfull IP adressing.

• IP address = 32 bit

• IP address has 2 parts:

– Network ID or network prefix: identifies the network where the system belongs too.– Host ID or host number: identifies the system itself.

2.2 IP Essentials: addressing / netmasking.

Network ID

32 bit IPv4 address

Host ID

• 2 hosts on the same network both have the same network prefix but different host number.

• 2 hosts on a different network both have a different network prefix but could have same host number (not necessary)






Network : “0” + 7 bitsdec. addr. : 1.x.x.x to 126.x.x.xSubnet mask : 255.0.0.0126 networks of 16.777.214 hosts50% of total IPv4 address space

Network :” 10” + 14 bitsdec.addr. : 128.0.x.x to 191.255.x.xSubnet mask : 255.255.0.016.384 networks of 65.534 hosts25% of total IPv4 address space

Network :” 110” + 21 bitsdec.addr.: 192.0.0.x to 223.255.255.xSubnet mask : 255.255.255.02.097.152 networks of 254 hosts.12,5% of total IPv4 address space.

Class A (1-126)

0 + 7bits 24 bits

Network Host

Class B (128-191)

1 0 + 14 bits 16 bits

HostNetwork

Class C (192-223)

1 1 0 + 21 bits 8 bits

HostNetwork






Network : “1110” + 4 bitsdec. addr. : 224.x.x.x to 239.x.x.xSubnet mask : 128.0.0.016 networks of 16.777.214 hosts.6,25% of total IPv4 address space.

Network :” 1111” + 20 bitsdec.addr. : 240.x.x.x to 255.x.x.xSubnet mask : 255.255.255.01.048.576 networks of 254 hosts6,25% of total IPv4 address space.

Class D (224-239)

1 1 1 0 + 4 bits 24 bits

Host

Class E (240-255)

1 1 1 1 + 20 bits 8 bits

HostNetwork

Network

• Class D used for multicasting.

• Class E Experimental use only.




2.2.2 Limitations of classfull IP adressing.

• In the early days of internet IP address space was given to company based on what they were asking, not on what they needed.

• This was allowed because the IPv4 address space seemed inexhaustible infinitesimal big.

• Nobody worried about exhausting the available IPv4 address space.

• By choosing to work with a 32 bit address the available address space is 232 = 4.294.967.296 addresses

• Problem for medium-sized companies :– a class C network = too small– a class B network = too big

• As a result lots of companies received an IP network range which was too big lots of IP addresses were unusable.

• Not a very efficient use of the IPv4 address space!





2.2.3 Subnetting.

• Third level added in the structure of the IP address Subnet number


Network prefix Host number

Classfull hierarchy with 2 levels

Network prefix Host numberSubnet number

Subnet hierarchy with 3 levels

extended network prefix

• Extended network prefix = Network prefix + Subnet number

• Defined by the subnet mask.




2.2.3 Subnetting.

• Bits in subnet mask and the bits in IP address have a 1 to 1 relation.

• For the bits in subnet mask equal to 1 corresponding bits in IP address are part of extended network prefix.

• For the bits in subnet mask equal to 0 corresponding bits in IP address are part of host number


130.5.5.25 (class B) 10000010 . 00000101 . 00000101 . 00011001

255.255.255.0 11111111 . 11111111 . 11111111 . 00000000

Extended Network Prefix

SubnetNumberNetwork Prefix

HostNumber




2.2.3 Subnetting.

• Sometimes one only speaks about the length of the extended network prefix and not about the subnet mask.

• Length of extended network prefix = number of bits equal to 1 in subnet mask.

• Different notation:

– subnet mask 255.255.255.0 /24 (24 bits are equal to 1 in subnet mask)– subnet mask 255.255.0.0 /16 (16 bits are equal to 1 in subnet mask)– subnet mask 255.0.0.0 /8 (8 bits are equal to 1 in subnet mask)


130.5.5.25 10000010 . 00000101 . 00000101 . 00011001

255.255.255.0 11111111 . 11111111 . 11111111 . 00000000

130.5.5.25 / 24 10000010 . 00000101 . 00000101 . 00011001

or




2.2.4 Subnetting example.

1. Facts:

Network number 193.1.1.0 / 24 has been assigned to a company.6 subnets have to be created.The biggest subnet must contain 25 hosts.

1. Questions:

Calculate all possible subnets and give the respective subnet ID’s.Give the broadcast address for every subnet.Calculate and give all host addresses for the second subnet.






1. Answer:

Define subnet mask / extended network prefix.

How many bits do we need to create 6 subnets ?

3 bits (binary system !) 23 = 8 possible subnets

What is the extended network prefix ?

We have received a /24 network ID already 24 bits in subnet mask. We want to subnet this network in 8 different subnets we need the 3 extra bits to do this. In total we need 24 + 3 = 27 bits for the extended network prefix

We have 5 bits left for the host number. 25 = 32 possible combinations the addresses with all bits = 0 (subnet ID) or all bits = 1 (broadcast) are never a host address Maximum number of hosts = 32 – 2 = 30 (still more than is needed)






1. Answer:

Define subnet mask / extended network prefix.


193.1.1.0 11000001 . 00000001 . 00000001 . 000

255.255.255.224 11111111 . 11111111 . 11111111 .

Extended Network Prefix – 27 bits

SubnetNumber

bits

Network Prefix

HostNumberbits

00000

111 00000





1. Answer:

Define different subnet ID’s.


Basic network address : 11000001.00000001.00000001.00000000 = 193.1.1.0 / 24

Subnet #0 : 11000001.00000001.00000001.00000000 = 193.1.1.0 / 27

Subnet #1 : 11000001.00000001.00000001.00100000 = 193.1.1.32 / 27

Subnet #2 : 11000001.00000001.00000001.01000000 = 193.1.1.64 / 27

Subnet #3 : 11000001.00000001.00000001.01100000 = 193.1.1.96 / 27

Subnet #4 : 11000001.00000001.00000001.10000000 = 193.1.1.128 / 27

Subnet #5 : 11000001.00000001.00000001.10100000 = 193.1.1.160 / 27

Subnet #6 : 11000001.00000001.00000001.11000000 = 193.1.1.192 / 27

Subnet #7 : 11000001.00000001.00000001.11100000 = 193.1.1.224 / 27





1. Answer:

Subnet with all bits = 0 and all bits = 1.


Subnet ID : 193.1.1.0 / 27 11000001 . 00000001 . 00000001 . 000

27 bit - prefix

00000

Network ID : 193.1.1.0 / 24 11000001 . 00000001 . 00000001 . 000

24 bit - prefix

00000

Subnet broadcast : 193.1.1.255 / 27 11000001 . 00000001 . 00000001 . 111

27 bit - prefix

11111

Network broadcast : 193.1.1.255 / 24 11000001 . 00000001 . 00000001 . 111

24 bit - prefix

11111





1. Answer:

Define all possible host addresses for subnet #2.


Subnet #2 : 11000001.00000001.00000001.01000000 = 193.1.1.64 / 27

Host #1 : 11000001.00000001.00000001.01000001 = 193.1.1.65 / 27

Host #2 : 11000001.00000001.00000001.01000010 = 193.1.1.66 / 27

Host #3 : 11000001.00000001.00000001.01000011 = 193.1.1.67 / 27

Host #4 : 11000001.00000001.00000001.01000100 = 193.1.1.68 / 27

Host #5 : 11000001.00000001.00000001.01000101 = 193.1.1.69 / 27

Host #6 : 11000001.00000001.00000001.01000110 = 193.1.1.70 / 27

Host #7 : 11000001.00000001.00000001.01000111 = 193.1.1.71 / 27

Host #8 : 11000001.00000001.00000001.01001000 = 193.1.1.72 / 27

Host #9 : 11000001.00000001.00000001.01001001 = 193.1.1.73 / 27

Host #10 : 11000001.00000001.00000001.01001010 = 193.1.1.74 / 27





1. Answer:



Subnet #2 : 11000001.00000001.00000001.01000000 = 193.1.1.64 / 27

Host #11 : 11000001.00000001.00000001.01001011 = 193.1.1.75 / 27

Host #12 : 11000001.00000001.00000001.01001100 = 193.1.1.76 / 27

Host #13 : 11000001.00000001.00000001.01001101 = 193.1.1.77 / 27

Host #14 : 11000001.00000001.00000001.01001110 = 193.1.1.78 / 27

Host #15 : 11000001.00000001.00000001.01001111 = 193.1.1.79 / 27

Host #16 : 11000001.00000001.00000001.01010000 = 193.1.1.80 / 27

Host #17 : 11000001.00000001.00000001.01010001 = 193.1.1.81 / 27

Host #18 : 11000001.00000001.00000001.01010010 = 193.1.1.82 / 27

Host #19 : 11000001.00000001.00000001.01010011 = 193.1.1.83 / 27

Host #20 : 11000001.00000001.00000001.01010100 = 193.1.1.84 / 27





1. Answer:



Subnet #2 : 11000001.00000001.00000001.01000000 = 193.1.1.64 / 27

Host #21 : 11000001.00000001.00000001.01010101 = 193.1.1.85 / 27

Host #22 : 11000001.00000001.00000001.01010110 = 193.1.1.86 / 27

Host #23 : 11000001.00000001.00000001.01010111 = 193.1.1.87 / 27

Host #24 : 11000001.00000001.00000001.01011000 = 193.1.1.88 / 27

Host #25 : 11000001.00000001.00000001.01011001 = 193.1.1.89 / 27

Host #26 : 11000001.00000001.00000001.01011010 = 193.1.1.90 / 27

Host #27 : 11000001.00000001.00000001.01011011 = 193.1.1.91 / 27

Host #28 : 11000001.00000001.00000001.01011100 = 193.1.1.92 / 27

Host #29 : 11000001.00000001.00000001.01011101 = 193.1.1.93 / 27

Host #30 : 11000001.00000001.00000001.01011110 = 193.1.1.94 / 27





1. Answer:

Define broadcast address for every subnet.

Broadcast address = address with all bits of the host number = 1


Basic network address : 11000001.00000001.00000001.00000000 = 193.1.1.0 / 24

Broadcast subnet #0 : 11000001.00000001.00000001.00011111 = 193.1.1.31 / 27









Netwerken · Netwerken TCP / IP. 3de jaar Netwerkbeheerder Page: 2 Syntra Antwerpen &...

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