This document discusses IPv4 addressing and subnetting. It begins by explaining that an IPv4 address is a 32-bit logical address written in dotted decimal notation, with each octet ranging from 0-255. It then covers the different IPv4 address classes (A, B, C, D, E), their subnet masks, and address ranges. Classless Inter-Domain Routing (CIDR) is introduced as eliminating classful networking. The document also includes problems about configuring forwarding tables and longest prefix matching.

1. Computer Networks (CIE 447)
Problem set 3 – Chapter 4
Dr. / Samy Soliman
T. A. :Eng. Menna Mohamed
:Eng. Nourhan Tarek
CIE 447- SPRING 2020

2. IPv4 addresses
IPv4 address is the logical address of a network device, which is composed of a 32-bits
address.
It is written as 4 octets, that are separated by a dot.
Ex.: 172.16.254.1
Each octet in the IPv4 address has a value range from 0 to 255 , where each octet is a 4-bit
long.
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3. The subnet
Class A
Class A addresses were originally designed for networks with a very large number of hosts.
The default subnet mask is 255.0.0.0 or /8.
Some of this class addresses are reserved or invalid to be assigned to hosts.
0.0.0.1 to 0.255.255.255 are invalid host addresses.
127.0.0.1 to 127.255.255.255 are invalid host addresses, However, they are used for testing the
local computer
127.0.0.1 is used for testing the TCP/IP.
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4. The subnet
Classes B & C
Class B
Class B addresses were originally designed for medium-sized to large-sized networks.
The default subnet mask is 255.255.0.0 or /16.
The range of valid addresses are from 128.0.0.0 to 191.255.0.0 / 16.
Class C
Class C addresses were designed for small networks.
The default subnet mask is 255.255.255.0 or /24.
The range of valid addresses are from 192.0.0.0 to 123.255.255.0 / 24.
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5. The subnet
Classes D & E
Class D
Class D addresses are reserved for multicasting IP addresses.
The addresses are not assigned to hosts, which means there is no default subnet mask.
 the range of valid addresses is from 224.0.0.0 to 239.255.255.255
Class E
Class E addresses are experimental and reserved for future use.
These addresses are not assigned to hosts and there is no default subnet mask.
The valid range of addresses is from 240.0.0.0 to 255.255.255.255
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6. Classless Interdomain routing (CIDR)
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Class
Size
of network
number bit
field
Size of rest bit
field
Start address End address
Default subnet
mask
CIDR notation
Class A
(unicast)
8 24 0.0.0.0
127.255.255.2
55
255.0.0.0 /8
Class B
(unicast)
16 16 128.0.0.0
191.255.255.2
55
255.255.0.0 /16
Class C
(unicast)
24 8 192.0.0.0
223.255.255.2
55
255.255.255.0 /24
Class D
(multicast)
not defined not defined 224.0.0.0
239.255.255.2
55
not defined not defined
Class E
(reserved)
not defined not defined 240.0.0.0
255.255.255.2
55
not defined not defined

7. Classless Interdomain routing (CIDR)
Classless
Completely eliminates traditional concepts of Class A, B and C addresses.
Network prefix based:
 Routers do not make any assumption on the basis of the three leading bits.
They require an explicit network prefix to determine dividing point between
network_id (prefix) and host_id.
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8. Classless Interdomain routing (CIDR)
Example:
Subnet 1: 213.2.96.0 → 11010101.00000010.01100000.00000000
Subnet 2: 213.2.97.0 → 11010101.00000010.01100001.00000000
Subnet 3: 213.2.98.0 → 11010101.00000010.01100010.00000000
Subnet 4: 213.2.99.0 → 11010101.00000010.01100011.00000000
Supernet mask: 255.255.252.0
Supernet address: 213.2.96.0/22 → 11010101 . 00000010 . 011000 00 . 00000000
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9. Longest prefix match
Matching the longest prefix of an IP destination address to the entries in a routing table, to
determine the appropriate outgoing link.
Subnet 1’s IP addresses are from 213.2.96.0 → 11010101.00000010.01100000.00000000
to 213.2.96.23 → 11010101.00000010.01100000.00110111
The longest prefix match that is used in the forwarding table of the router is:
11010101.00000010.01100000.00
Subnet 2’s IP addresses are from 213.2.97.0 → 11010101.00000010.01100001.00000000
to 213.2.97.226 → 11010101.00000010.01100001.11100010
The longest prefix match that is used in the forwarding table of the router is:
11010101.00000010.01100001
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10. Problem 4
 Consider the network below.
a. Suppose that this network is a datagram network.
Show the forwarding table in router A, such that all
traffic destined to host H3 is forwarded through interface
3.
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11. Problem 4
 Consider the network below.
a. Suppose that this network is a datagram network.
Show the forwarding table in router A, such that all
traffic destined to host H3 is forwarded through interface
3.
Solution:
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Destination address Interface link
H3 3

12. Problem 4
 Consider the network below.
b. Suppose that this network is a datagram network. Can
you write down a forwarding table in router A, such that
all traffic from H1 destined to host H3 is forwarded
through interface 3, while all traffic from H2 destined to
host H3 is forwarded through interface 4? (Hint: this is a
trick question.)
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13. Problem 4
 Consider the network below.
b. Suppose that this network is a datagram network. Can
you write down a forwarding table in router A, such that
all traffic from H1 destined to host H3 is forwarded
through interface 3, while all traffic from H2 destined to
host H3 is forwarded through interface 4? (Hint: this is a
trick question.)
Solution:
No, because the forwarding table is based on the
destination address, not the source address
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Destination address Interface link
H3 3
H3 4

14. Problem 4
 Consider the network below.
c. Now suppose that this network is a virtual circuit
network and that there is one ongoing call between H1
and H3, and another ongoing call between H2 and H3.
Write down a forwarding table in router A, such that all
traffic from H1 destined to host H3 is forwarded through
interface 3, while all traffic from H2 destined to host H3
is forwarded through interface 4.
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15. Problem 4
 Consider the network below.
c. Now suppose that this network is a virtual circuit
network and that there is one ongoing call between H1
and H3, and another ongoing call between H2 and H3.
Write down a forwarding table in router A, such that all
traffic from H1 destined to host H3 is forwarded through
interface 3, while all traffic from H2 destined to host H3
is forwarded through interface 4.
Solution: the forwarding table at router A
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14
24
Incoming
interface
Incoming VC Outgoing
interface
Outgoing VC
1 11 3 12
2 21 4 22

16. Problem 4
 Consider the network below.
d. Assuming the same scenario as (c), write down the
forwarding tables in nodes B, C, and D.
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17. 14
24
Problem 4
 Consider the network below.
d. Assuming the same scenario as (c), write down the
forwarding tables in nodes B, C, and D.
Solution:
The forwarding table at router B:
The forwarding table at router C:
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Incoming
interface
Incoming
VC
Outgoing
interface
Outgoing
VC
1 12 2 13
Incoming
interface
Incoming
VC
Outgoing
interface
Outgoing
VC
1 22 2 23
The forwarding table at router D:
Incoming
interface
Incoming
VC
Outgoing
interface
Outgoing
VC
1 13 3 14
2 23 3 24

18. Problem 5(*)
Consider a VC network with a 2-bit field for the VC
number. Suppose that the network wants to set up
a virtual circuit over four links: link A, link B, link C,
and link D.
Suppose that each of these links is currently carrying
two other virtual circuits, and the VC numbers of
these other VCs are as follows:
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In answering the following questions, keep in mind that each of the existing VCs may only be traversing
one of the four links.
a. If each VC is required to use the same VC number on all links along its path, what VC number could be
assigned to the new VC?

19. Problem 5(*)
Consider a VC network with a 2-bit field for the VC
number. Suppose that the network wants to set up
a virtual circuit over four links: link A, link B, link C,
and link D.
Suppose that each of these links is currently carrying
two other virtual circuits, and the VC numbers of
these other VCs are as follows:
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In answering the following questions, keep in mind that each of the existing VCs may only be traversing
one of the four links.
a. If each VC is required to use the same VC number on all links along its path, what VC number could be
assigned to the new VC?
Solution:
A new VC can’t be established, as the VC number is 2 bits.
∴ The max. number of links that could be established 22 = 4 links

20. Problem 5(*)
Consider a VC network with a 2-bit field for the VC
number. Suppose that the network wants to set up
a virtual circuit over four links: link A, link B, link C,
and link D.
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b. If each VC is permitted to have different VC numbers in the different links along its path (so that
forwarding tables must perform VC number translation), how many different combinations of four VC
numbers (one for each of the four links) could be used?

21. Problem 5(*)
Consider a VC network with a 2-bit field for the VC
number. Suppose that the network wants to set up
a virtual circuit over four links: link A, link B, link C,
and link D.
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b. If each VC is permitted to have different VC numbers in the different links along its path (so that
forwarding tables must perform VC number translation), how many different combinations of four VC
numbers (one for each of the four links) could be used?
Solution:
The number of link = 4 and each link has 2 VC numbers
∴ The total combinations of the 4 VC links = 24 = 16 combination
(10, 00, 00, 10) or (00, 10, 00, 10)

22. Problem 10
Consider a datagram network using 32-bit host
addresses. Suppose a router has four links,
numbered 0 through 3, and packets are to be
forwarded to the link interfaces as follows:
a. Provide a forwarding table that has five entries,
uses longest prefix matching, and forwards
packets to the correct link interfaces.
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Destination Address Range Link
Interface
11100000 00000000 00000000 00000000
Through
11100000 00111111 11111111 11111111
0
11100000 01000000 00000000 00000000
Through
11100000 01000000 11111111 11111111
1
11100000 01000001 00000000 00000000
Through
11100001 01111111 11111111 11111111
2
Otherwise 3

23. Problem 10
Consider a datagram network using 32-bit host
addresses. Suppose a router has four links,
numbered 0 through 3, and packets are to be
forwarded to the link interfaces as follows:
a. Provide a forwarding table that has five entries,
uses longest prefix matching, and forwards
packets to the correct link interfaces.
Solution:
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Destination Address Range Link
Interface
11100000 00000000 00000000 00000000
Through
11100000 00111111 11111111 11111111
0
11100000 01000000 00000000 00000000
Through
11100000 01000000 11111111 11111111
1
11100000 01000001 00000000 00000000
Through
11100001 01111111 11111111 11111111
2
Otherwise 3
Entry no. Prefix match Link interface
1 11100000 00 0
2 11100000 01000000 1
3 1110000 2
4 11100001 1 3
5 Otherwise 3

24. Problem 10
Consider a datagram network using 32-bit host addresses. Suppose a router has four
links, numbered 0 through 3, and packets are to be forwarded to the link interfaces as
follows:
b. Describe how your forwarding table determines the appropriate link interface for
datagrams with destination addresses:
11001000 10010001 01010001 01010101
11100001 01000000 11000011 00111100
11100001 10000000 00010001 01110111
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25. Problem 10
Consider a datagram network using 32-bit host
addresses. Suppose a router has four links, numbered 0
through 3, and packets are to be forwarded to the link
interfaces as follows:
b. Describe how your forwarding table determines the
appropriate link interface for datagrams with destination
addresses:
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Entry no. Prefix match Link interface
1 11100000 00 0
2 11100000 01000000 1
3 1110000 2
4 11100001 1 3
5 Otherwise 3
Solution:
11001000 10010001 01010001 01010101 → it matches 5𝑡ℎ entry: otherwise, it will be forwarded to link 3
11100001 01000000 11000011 00111100 → it matches 3𝑡ℎ entry, it will be forwarded to link 2
11100001 10000000 00010001 01110111 → it matches 4𝑡ℎ
entry, it will be forwarded to link 3

26. Problem 11
Consider a datagram network using 8-bit host addresses. Suppose a router uses longest prefix
matching and has the following forwarding table:
For each of the four interfaces, give the associated range of destination host addresses and the
number of addresses in the range.
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27. Problem 11
Consider a datagram network using 8-bit host addresses. Suppose a router uses longest prefix matching and has
the following forwarding table:
For each of the four interfaces, give the associated range of destination host addresses and the number of
addresses in the range.
Solution:
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Destination Address Range Link
Interface
No. of
addresses in a
link
00000000 Through 00111111 0 26 = 64
01000000 Through 01011111 1 25 = 32
01100000 Through 01111111 2 25 + 26
= 32 + 64
= 96
10000000 Through 10111111 2
11000000 Through 11111111 3 26
= 64

28. Problem 12(*)
Consider a datagram network using 8-bit host addresses. Suppose a router uses longest prefix
matching and has the following forwarding table:
For each of the four interfaces, give the associated range of destination host addresses and the
number of addresses in the range.
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29. Problem 12(*)
Consider a datagram network using 8-bit host addresses. Suppose a router uses longest prefix
matching and has the following forwarding table:
For each of the four interfaces, give the associated range of destination host addresses and the
number of addresses in the range.
Solution:
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Destination Address Range Link
Interface
No. of
addresses in a
link
10000000 Through 11111111 0 27 = 128
10000000 Through 10111111 1 26 = 64
11100000 Through 11111111 2 25 =32
00000000 Through 01111111 3 27
= 128

30. Problem 13
Consider a router that interconnects three subnets: Subnet 1, Subnet 2, and Subnet 3.
Suppose all of the interfaces in each of these three subnets are required to have the prefix
223.1.17/24. Also suppose that Subnet 1 is required to support at least 60 interfaces, Subnet 2 is
to support at least 90 interfaces, and Subnet 3 is to support at least 12 interfaces.
Provide three network addresses (of the form a.b.c.d/x) that satisfy these constraints.
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31. Problem 13
Consider a router that interconnects three subnets: Subnet 1, Subnet 2, and Subnet 3. Suppose all of the interfaces in each
of these three subnets are required to have the prefix 223.1.17/24. Also suppose that Subnet 1 is required to support at least
60 interfaces, Subnet 2 is to support at least 90 interfaces, and Subnet 3 is to support at least 12 interfaces.
Solution:
The total no. of addresses = 28
= 256 and by mapping the CIDR into binary address
The IP address: 223.1.17.0 → network address, can’t be used as an interface and the IP address: 223.1.17.225 → broadcast
address
For subnet 1: it needs 60 interfaces +2 (network and broadcast) = 26
= 64 ∴ The subnet 1 address: 223.1.17.0/25
For subnet 2: it needs 90 interfaces +2 (network and broadcast) = 27
= 128 ∴ The subnet 2 address: 223.1.17.64/24
For subnet 3: it needs 12 interfaces +2 (network and broadcast) = 24
= 16 ∴ The subnet 3 address: 223.1.17.192/28
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223 1 17 x
11011111 00000001 00010001 xxxxxxxx

32. Problem 17
Consider the topology shown in Figure 4.17. Denote the three
subnets with hosts (starting clockwise at 12:00) as Networks A,
B, and C. Denote the subnets without hosts as Networks D, E,
and F.
a. Assign network addresses to each of these six subnets, with
the following constraints:
All addresses must be allocated from 214.97.254/23;
Subnet A should have enough addresses to support 250 interfaces;
Subnet B should have enough addresses to support 120 interfaces;
Subnet C should have enough addresses to support 120 interfaces.
subnets D, E and F should each be able to support two interfaces.
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33. Problem 17
Consider the topology shown in Figure 4.17. Denote the three
subnets with hosts (starting clockwise at 12:00) as Networks A,
B, and C. Denote the subnets without hosts as Networks D, E,
and F.
a. Assign network addresses to each of these six subnets, with
the following constraints:
All addresses must be allocated from 214.97.254/23;
∴ The total no. of available addresses = 29 = 512
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214 97 254 x
11010110 01100001 11111110 xxxxxxxx

34. Problem 17
a. Assign network addresses to each of these six subnets, with the
following constraints:
Subnet A should have enough addresses to support 250 interfaces;
∴ it needs 250 interfaces +2 (network and broadcast) = 28
= 256
Subnet B should have enough addresses to support 120 interfaces;
∴ it needs 120 interfaces +2 (network and broadcast) = 27
= 128
Subnet C should have enough addresses to support 120 interfaces.
∴ it needs 120 interfaces +2 (network and broadcast) = 27 = 128
subnets D, E and F should each be able to support two interfaces.
∴ Each needs 2 interfaces = 22
= 4
∴ The total no. of the designed addresses = 516, which exceeds the
available number of addresses
Sol. : Deduct a number of addresses from one of the subnets → Subnet B
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35. Problem 17
a. Assign network addresses to each of these six subnets, with the following
constraints:
Subnet A should have enough addresses to support 250 interfaces;
∴ it needs 250 interfaces +2 (network and broadcast) = 28
= 256
The range of addresses are 214.97.255.00000000 → 214.97.255. 00000001
The range of addresses are 214.97.255.0 → 214.97.255.254
Subnet A’s network address: 214.97.255/24
Subnet B should have enough addresses to support 120 interfaces;
∴ it needs 120 interfaces +2 (network and broadcast) = 27
= 128
Only 120 addresses will be used. ∴ a set of IP addresses will be deducted.
The range of addresses are 214.97.254.00000000 → 214.97.255. 01111111
∵ The required no. of addresses to be deducted = 8
∴ The deducted range of addresses: 214.97.254.00000000 → 214.97.254.00000111
CIE 447- SPRING 2020 35
∴ Subnet B’s network address : 214.97.254.0/25 - 214.97.254.0/29

36. Problem 17
a. Assign network addresses to each of these six subnets, with the
following constraints:
Subnet C should have enough addresses to support 120 interfaces.
∴ it needs 120 interfaces +2 (network and broadcast) = 27
= 128
The range of addresses are 214.97.254.10000000 → 214.97.254. 11110111
The range of addresses are 214.97.254.128 → 214.97.254.247
Subnet C’s network address: 214.97.254.128/25
subnets D, E and F should each be able to support two interfaces.
Subnet D has 2 addresses : 214.97.254.0/31
Subnet E has 2 addresses : 214.97.254.2/31
Subnet F has 4 addresses : 214.97.254.4/30
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37. Problem 17
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38. Problem 17
Consider the topology shown in Figure 4.17. Denote the three subnets with hosts (starting
clockwise at 12:00) as Networks A, B, and C. Denote the subnets without hosts as Networks D, E,
and F.
b. Using your answer to part (a), provide the forwarding tables (using longest prefix matching) for
each of the three routers.
Router 1:
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Longest prefix match Link interface
11010110 01100001 11111111 A
11010110 01100001 11111110 0000000 D
11010110 01100001 11111110 000001 F

39. Problem 17
Consider the topology shown in Figure 4.17. Denote the three subnets with hosts (starting
clockwise at 12:00) as Networks A, B, and C. Denote the subnets without hosts as Networks D, E,
and F.
b. Using your answer to part (a), provide the forwarding tables (using longest prefix matching) for
each of the three routers.
Router 2:
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Longest prefix match Link interface
11010110 01100001 11111110 1 C
11010110 01100001 11111110 0000001 E
11010110 01100001 11111110 000001 F

40. Problem 17
Consider the topology shown in Figure 4.17. Denote the three subnets with hosts (starting
clockwise at 12:00) as Networks A, B, and C. Denote the subnets without hosts as Networks D, E,
and F.
b. Using your answer to part (a), provide the forwarding tables (using longest prefix matching) for
each of the three routers.
Router 3:
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Longest prefix match Link interface
11010110 01100001 11111110 0 B
11010110 01100001 11111110 0000000 D
11010110 01100001 11111110 0000001 E

41. Homework
Solve problems 14, 15 and 16.
Due date: 5/4/2020
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