1. Chapter 5 Addressing
Dr. Clincy
Lecture

2. Example 4 Solution The mask is 11111111 11111111 11111111 11000000 or
A company is granted the site address (class B). The company needs 1000 subnets. Design the subnets.
Solution
The number of 1s in the default mask is 16 (class B).
The company needs 1000 subnets. This number is not a power of 2. The next number that is a power of 2 is 1024 (210). We need 10 more 1s in the subnet mask.
The total number of 1s in the subnet mask is 26 ( ).
The total number of 0s is 6 ( ).
The mask is or
The number of subnets is 1024.
The number of addresses in each subnet is 26 (6 is the number of 0s) or 64.
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Lecture

3. Example 4
Subtract 63 from 255 to get 192
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Lecture

4. SUPERNETTING
Although class A and B addresses are dwindling – there are plenty of class C addresses
The problem with C addresses is, they only have 256 hostids – not enough for any midsize to large size organization – especially if you plan to give every computer, printer, scanner, etc. multiple IP addresses
Supernetting allows an organization the ability to combine several class C blocks in creating a larger range of addresses
Note: breaking up a network = subnetting
Note: combining Class-C networks = supernetting
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Lecture

5. Assigning or Choosing Class C Blocks
When assigning class C blocks, there are two approaches: (1) random and (2) superblock
Random Approach: the routers will see each block as a separate network and therefore, for each block there would be an entry in the routing table – a router contains an entry for each destination network
Superblock Approach: instead of multiple routing table entries, there would be a single entry. However, the choices of blocks need to follow a set of rules:
#1 – the # of blocks must be a power of 2 (ie. 1, 2, 4, 8 …)
#2 – blocks must be contiguous (no gaps between blocks)
#3 – the 3rd byte of the first address in the superblock must be evenly divisible by the number of blocks – ie. if the # of blocks is N, the 3rd byte must be divisible by N
Number of 1s removed from Default mask is dictated by the number of C blocks combined (ie 1 for 2, 2 for 4, 3 for 8, etc)
Dr. Clincy
Lecture

6. Example 5
A company needs 600 addresses. Which of the following set of class C blocks can be used to form a supernet for this company?
Solution
1: No, there are only three blocks. Must be a power of 2
2: No, the blocks are not contiguous.
3: No, 31 in the first block is not divisible by 4.
4: Yes, all three requirements are fulfilled. (1. Power of 2, 2. Contiguous and 3. 3rd byte of 1st address is divisible by 4: 32/4=8)
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Lecture

7. Example 8
A supernet has a first address of and a supernet mask of How many blocks are in this supernet and what is the range of addresses?
Solution
The default mask has 24 1s because is a class C.
Because the supernet mask is , the supernet has 21 1s.
Since the difference between the default and supernet masks is 3, there are 23 or 8 blocks in this supernet.
Because the blocks start with and must be contiguous, the blocks are , , ………
The first address is The last address is
The total number of addresses is 8 x 256 = 2048
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Lecture

8. Explain Supernetting Conceptually
Back out this bit from netid into host id
Causes these 2 blocks to combine as a single block
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Lecture

9. Variable-length subnetting
Suppose you were granted a Class C address – this mean you would have 8 bits to play with
Also, suppose you needed 5 subnets consisting of the following # of hosts: 60, 60, 60, 30 and 30
If you used a 2 bit subnet mask – can get 4 subnets with 64 stations each (too big)
If you used a 3 bit subnet mask – can get 8 subnets with 32 stations each (too small)
What’s the solution ?
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Lecture

10. Variable-length Subnetting
Solution: used 2 subnet masks – one applied after the other
Could use a 2 bit subnet mask and get 4 subnets with 64 stations each - this would satisfy the three 60-host subnet requirement – therefore the subnet mask would be (192)
We could then further divide one of the 64-host subnets into two 32-host subnets by applying this mask (224) after this mask of (192) is used
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Lecture

11. Ch 5 Classless Addressing
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Lecture

12. Classful Addressing is Obsolete
Guess What ?
Classful Addressing is Obsolete
However, understanding the classful approach will help you easily understand the classless approach
Quickly explain classless vs classful (leave address aggregation for the routing topics)
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Lecture

13. CLASSLESS ADDRESSING
Recall the problems with Classful addressing – you have to get a predefined block of addresses – in most cases, the block is either too large or too small
In the 1990’s, ISP came into prominence – they provide Internet access for individuals to midsize organizations that don’t want sponsor their own Internet service (ie. , etc).
The ISP’s are granted several B and C blocks of addresses and they subdivide their address space into groups of 2, 4, 8, 16, etc.. – blocks can be variable length
Because of the up rise of ISP’s, in 1996, the Internet Authorities announced a new architecture called Classless Addressing (making classful addressing obsolete)
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Lecture

14. Number of Addresses in a Classless Block
There are two conditions
Condition 1: the number of addresses in a block; it must be a power of 2 (2, 4, 8, . . .). A household may be given a block of 2 addresses. A small business may be given 16 addresses. A large organization may be given 1024 addresses.
Another Condition:
The beginning address must be evenly divisible by the number of addresses.
For example, if a block contains 4 addresses, the beginning address must be divisible by 4. If the block has less than 256 addresses, we need to check only the rightmost byte. If it has less than 65,536 addresses, we need to check only the two rightmost bytes, and so on.
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Lecture

15. Classless Subnet Illustration
Netid subnetid 1
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Lecture

16. Example 9
Which of the following can be the beginning address of a block that contains 16 addresses?
Solution
The address is eligible because 32 is divisible by 16. The address is eligible because 80 is divisible by 16.
Dr. Clincy
Lecture

17. Example 10
Which of the following can be the beginning address of a block that contains 1024 addresses?
Solution
To be divisible by 1024, the rightmost byte of an address should be 0 because any value in that first byte will be a fraction of 1024 (ie. 0 to 255).
To be divisible by 1024, the rightmost byte should be 0 and the second rightmost byte must be divisible by 4 because for every unique number in the second byte position, there exist 256 addresses in the first byte position that maps to it. To get 1024 addresses overall, you will need an increment of 4 in the 2nd byte position.
Therefore, the 2nd byte needs to be divisible by 4.
Only the address meets this condition.
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Lecture

18. Mask
Recall the Classful approach, only given an IP – the user defined their mask
For the Classless approach, when an org is given a block, it’s given both the starting address and the mask – these two pieces of info defines the entire block
For classless case, instead of writing out the full mask, we just specify the number of 1’s in the mask and append it to the address – this is called slash notation or CIDR (classless interdomain routing) notation
For classless addressing, the prefix refers to the common part of the address (ie. network portion)
For classless addressing, the suffix refers to the varying part of the address (ie. host portion)
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Lecture

19. A block in classes A, B, and C can easily be represented in slash notation as A.B.C.D/ n where n is either 8 (class A), 16 (class B), or (class C).
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Lecture

20. There are only 8 addresses in this block.
Example 11
A small organization is given a block with the beginning address and the prefix length /29 (in slash notation). What is the range of the block?
Solution
The beginning address is To find the last address we keep the first 29 bits and change the last 3 bits to 1s.
Beginning:
Ending :
There are only 8 addresses in this block.
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Lecture

21. Example 13
What is the network address if one of the addresses is /27?
Solution
The prefix length is 27, which means that we must keep the first 27 bits as is and change the remaining bits (5) to 0s. The 5 bits affect only the last byte. The last byte is Changing the last 5 bits to 0s, we get or 64. The network address is /27.
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Lecture

22. Example 14
An organization is granted the block /26. The organization needs to have four subnets. What are the subnet addresses and the range of addresses for each subnet?
Solution
The suffix length is 6. This means the total number of addresses in the block is 64 (26). If we create four subnets, each subnet will have 16 addresses.
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Lecture

23. Chapter 7: The Infamous IP
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Lecture

24. Position of IP in TCP/IP protocol suite
Packets in the IP layer are called datagrams
IP is an unreliable and connectionless datagram protocol
To make IP reliable, TCP protocol is added
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Lecture

25. IP datagram
IP datagram is variable length consisting of two parts (header, data)
Header is bytes & contains routing and deliver info
Ver – version of IP
HLEN – header length – total length of the header field (in 4-byte words or units)
Service type – now called Differentiated Services – tells the service type (ie. ftp, dns, telnet, etc..) – will come back to this
Total length – defines the total length of the datagram including the header – need this to determine if padding is needed – recall Ethernet frame can range bytes – so if the IP datagram is less than 46 bytes (need padding)
Identification – used for fragmentation – networks that are not able to encapsulate the full IP datagram will need to fragment – will come back to this
Flags – used for fragmentation – will come back to this
Fragmentation offset – used for fragmentation – will come back to
Time to live – datagram life time as it travels – used to control the number of hops (routers) a datagram can traverse – fix infinite loop problems
Protocol – defines the higher level protocol (ie. TCP, UDP, ICMP, ICMP, etc..) that’s using the service of the IP layer – since the IP Muxes data from the Transport layer – this field is used to demux
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Lecture

26. IP datagram Header cont…
Header Checksum – error checking (will cover later)
Source Address – IP address of the source (remain unchanged as data traverses)
Destination Address - IP address of the destination (remain unchanged as data traverses)
Option – are not required for every datagram – used for network testing and debugging – will cover in more detail later
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Lecture

27. FRAGMENTATION
Recall we stated that networks that are not able to encapsulate the full IP datagram will need to fragment
As the datagrams travel through the network hitting various Routers –
the router “decapsulates” the IP datagram from the frame
The router then processes it
Then the router encapsulates it in another frame
This is how routers are able to communicate with various networks
Router 1
Router 2
Network 1
Network 2
Network 3
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Lecture

28. MTU
Each Data Link Protocol has it own frame format – one field defines the max size of the data field – when datagram is encapsulated, the total size of the datagram must not exceed that max size (why ??? - HW/SW limitations of the physical network)
That value is called a MTU (maximum transfer unit)
The largest possible MTU is 65,535 and if this is used – it makes the IP protocol independent of the underlying physical network
If any other MTU is used, there will be cases possibly where the datagram needs to be fragmented in order to pass through that network
As it passes through the network, a previous fragment can be fragmented again if that physical network has a smaller MTU
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Lecture

29. Flag field
Fields related to the fragmentation are the ID field, flags field and fragmentation offset field
Id – combo of the Id and source Ip address (IP protocol used a counter to label datagram)
Flags: 1st reserved, if D set, can’t fragment (must drop if can’t pass), if D=0, can fragment. If M is set, means more fragments exist
Fragment offset – shows relative position of the fragment with respect to the whole datagram
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Lecture

30. Fragmentation example
Take a datagram of original size 4000 bytes (byte 0 to 3999) and fragment it into 3 fragments
The fragment offset is measured in units of 8 bytes. So the first offset would be 0/8=0 since the starting byte position is 0
The second starting byte position is 1400 and therefore the offset is 1400/8= 175
The third starting byte position is 2800 and therefore the offset equals 2800/8=350
This is done to ensure the offset can fit in the 13-bit field
Routers/Hosts that fragment must pick a size of each fragment so that the 1st byte is divisible by 8 (ie. 0, 8, 16, 24 ……696 …… 1400 …… ……… 2800 … etc)
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Lecture

31. Total Length Id isn’t changing
Detailed
example
Total Length Id isn’t changing
Allow “more” fragmentation
XDM
D=1, can’t frag
D=0, can frag
M=1, more frag exist
M=0, no more frag exist
offset
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Lecture

32. Re-assembly
Even if the fragments arrived to the destination out-of-order, the destination host could reassemble by:
The 1st fragment always has an offset of zero
If the 1st fragment’s length is divided by 8, it equals to the 2nd fragment’s offset
If the 1st and 2nd fragments’ total length are divided by 8, it equals to the 3rd fragment’s offset
Continue …
The last fragment’s “more” bit should be set to 0 – meaning no more fragments remaining
Dr. Clincy
Lecture

33. Recall - IP datagram
IP datagram is variable length consisting of two parts (header, data)
Header is bytes & contains routing and deliver info
Haven’t covered options yet
Option – are not required for every datagram – used for network testing and debugging – will cover in more detail later
Dr. Clincy
Lecture

34. Option format
Composed of a 1-byte code field, a 1-byte length field and a variable-sized data field
Length field defines the total length of the option (including the code field)
Data field contains the data of the specific option – some option types don’t require data
Code field is 8-bits long and contains 3 subfields: copy, class and number
Copy: controls presence of option. If 0, means copy options to the first fragment only; if 1, means copy option to all fragments
Class: defines general purpose of options. If 00, options is used for datagram control; if 10, options used for management and debugging.
Number: defines the type of option. As of now, only 6 types defined
Dr. Clincy
Lecture

35. Regarding the Number field
Number: defines the type of option. As of now, only 6 types defined
2 of the option types are 1-byte in size (doesn’t need length and data fields)
4 of the options are multiple-byte and require the length and data fields
Used as a filler between options (using a 16-bit or 32-bit boundary) – know the starting point of the next option
Used at the end of the last option for padding
Record the Internet routers that can handle the datagram ( can list up to 9 router IP addresses)
Used by the source to predetermine a route for a datagram as it traverses
Used by the source to predetermine a route too (but more relaxed than the Strict Source Route Option)
Record the time the datagram is processed by a router
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Lecture

36. Regarding the Record route option
The Tx creates a placeholder for the visited routers to fill in their IP addresses
The pointer field is used to point to the first empty entry so the router knows where to enter it’s outgoing IP address (address the datagram is leaving)
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Lecture

37. Record route concept Can have only 3 IP addresses because of 12+3=15
Outgoing IP address
Pointer field value of 4 when starting out
Increment pointer
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Lecture

38. Regarding the Strict source route option
Option used by the source to predetermine a route for the datagram as it traverses the Internet
In this case, the routers are specified up front in dictating the specific route. All routers MUST be visited – if other routers are visited, the datagram is dropped) – if all of the listed routers are not visited, the datagram is dropped
Routers are entered by the sender
Why: security, distinguish among different networks, don’t want certain traffic to leave your network, etc.
Dr. Clincy
Lecture

39. Loose source route option
Similar to the Strict Source Route Option but more relaxed
In this case, the routers are specified up front and all MUST be visited ( however, other routers can be visited too)
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Lecture

40. Timestamp option
Used to record the time of datagram processing by a router (expressed in milliseconds from midnight)
Use this to track the routers’ behavior – time from one router to the next
O-flow: # of routers that could not add their timestamp
Flags: dictates what the router should do (ie. add timestamp, add timestamp & IP address, etc..)
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Lecture

41. CHECKSUM
The error detection method used by most TCP/IP protocols is called checksum
The checksum protects against bit corruption that could possibly occur during transmission
Checksum calculated at the Tx and is appended with the sent data
The Rx repeats the calculation in determining if the data is correct or not
Give them an analogy in base-10
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Lecture

42. To create the checksum the sender does the following:
1. The packet is divided into k sections, each of n bits (usually 16)
2. All sections are added together using one’s complement arithmetic.
3. The final result is complemented to make the checksum.
Checksum process at the receiver is as follows:
The received packet is divided into k sections
All sections are added together
3. The final result is complemented and should equal zero if correct
NOTE: value + (-value) = 0
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Lecture

43. When to apply the checksum
For IP datagram, Checksum is used on the header only (and not the data)
The header needs to be check because it’s changing router-to-router (the data itself is static)
Recall that the higher-level protocols encapsulate data into the datagram and uses their own checksum
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Lecture

44. Recall Binary Addition
1010 (neg 5)
+0010 (pos 2)
1100 (neg 3)
1101 (neg 2)
+0111 (pos 7)
10100 (overflow – add the 1 back)
0101 (pos 5)
Recall complement
0011
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Lecture

45. ROUTING IP OVER ATM
The IP packet is encapsulated in cells (not just one). An ATM network has its own definition for the physical address of a device. Binding between an IP address and a physical address is attained through a protocol called ATMARP.
Each Router has an IP address which associates with the packet-switch side of the network (Internet)
The ATM side of the router uses its own 20-byte physical ATM address
And in guiding the cells across the ATM network, Virtual Circuit Identifiers are used
In a LAN case, broadcasting is used by ARP – in a ATM case, broadcasting can’t be used – another approach is needed - ATMARP
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Lecture

46. Next Slides - Go Over Final Project
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Lecture

47. CS4622 Final Project
Objective: Research & Teach a “CS4622 Related “ Subject at a high-level using existing knowledge
Teams will teach the class about a specific topic at a “high-level” relating to topics covered in CS4622 – no paper required – only power point slides required – you can scan images into the slides
Due July 16th by 5pm – upload ppt slide presentation to D2L – docked 1% per each hour late
Presentations will occur on July 17th (odd-numbered teams) and July 19th (even-numbered teams) - no teams will have to sit through 2 evenings of presentations – you can show up on your presentation day only.
Your team has 20 minutes for the presentation – must be in powerpoint form
All team members must present
Docked for going over or below time limit of 20 minutes (5% per minute)
Graded on (1) lecture clarity, flow and thoroughness (65%), (2) slide quality (15%), and (3) perceived team work (20%)
Your presentation should flow as one lecture and not disjointed (can’t divide the subject – everyone must understand the entire subject – if not, will lose points on flow)
Slides sent on July 16th by 5pm will be the final presentation used – no need to bring slides – can not change slides after July 16th (5 pm) – can only use the slides posted
You should list the various sources you referenced in your lecture
Team decision and strategy in determining “how” to cover the topic in 20 minutes.
Team members are responsible for researching (and learning) together and organizing the lecture as a team – teamwork – I can sense teamwork or not before and during the presentation
The Presentation topics will not be covered on any exam
Will lose points for reading from cue cards or any source (including the actual ppt slides) - seeking for you to understand the topic at a high-level well enough to teach and instruct - not read. Will lose points for very “wordy” slides.
Team numbers don’t indicate presentation order.
NOTE: it take approximately 10 to 14 power point slides to conduct a 20 minute talk – your team is responsible for making sure the talk is timed for 20 minutes
Words of advice: (1) don’t wait until the last minute to get started and (2) if you have problems with a team mate, let me know ASAP – don’t wait 1 or 2 weeks before the presentation (or the day of the presentation) – if so, it’s too late to take action

48. Topics Team 1 – Dark Web/I2P/Deep Web Team 2 – SCTP
Team 3 – Network Security: DNS Spoofing, ARP Poisoning
Team 4 – Network Security: DoS Attack, Smurf Attack
Team 5 – Internet of Things (IoT)
Team 6 – Mobile IP
NOTE 1: You are responsible for contacting your team members via initially
NOTE 2: Will lose points if you use Wikipedia (non-credible source)
Odd-numbered teams present July 17th (Teams 1, 3 and 5)
Even-numbered teams present July 19th (Teams 2, 4 and 6)
Only need to come on your team’s presentation day
NOTE: Online lectures #24 and #25 will be posted on July 17th and July 19th, respectively

49. You only need to come on your team’s presentation day
Presentations will occur on: July 17th (odd-numbered teams: 1, 3 and 5) July 19th (even-numbered teams: 2, 4 and 6) -
You only need to come on your team’s presentation day
TEAM
FULL_NAME 1
Cardwell, Jon
Fredricks, Mark
Leiper, Jake
Vina, Ryan 2
Ahmed, Nawal
Clifford, Harrison
Kimani, Frank 3
Crews, Randy
Hoffler, Stephen
Natic, Alex 4
Countess, Raymond
Fuentes, Ernesto
Negahdar, Arash
Nunnelley, Ben 5
Farris, Jacques
Lee, Joshua
Pellegrini, Adam 6
Hunsinger, Clayton
Jiang, Xiaoju
Taylor, Jeremy
Dr. Clincy
Lecture