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CHAPTER

PRACTICE SET
Questions
Q4-1.

The transport layer communication is between two ports; the network layer
communication is between two hosts. This means that each layer has a different source/destination address pair; each layer needs a different header to
accommodate these pair of addresses. In addition, there are other pieces of
information that need to be separately added to the corresponding header.

Q4-2.

Routing cannot be done at the transport layer, because the communication at
the transport layer is one single logical path between the source port and the
destination port. Routing cannot be done at the data-link layer because the
communication at the data-link layer is between two nodes (one single path);
there is no need for routing. On the other hand, there are several possible paths
for a packet between the source host and destination host at the network layer.
Routing is the job of selecting one of these paths for the packet.

Q4-3.

Forwarding is delivery to the next node. A router uses its forwarding table to
send a packet out of one of its interfaces and to make it to reach to the next
node. In other words, forwarding is the decision a router makes to send a
packet out of one of its interfaces. Routing, on the other hand, is an end-to-end
delivery resulting in a path from the source to the destination for each packet.
This means a routing process is a series of forwarding processes. To enable
each router to perform its forwarding duty, routing protocols need to be running all of the time to provide updated information for forwarding tables.
Although forwarding is something we can see in the foreground, in the background, routing provides help to the routers to do forwarding.

Q4-4.
a. In the datagram approach, the forwarding decision is made based on the

destination address in the packet header.
b. In the virtual-circuit approach, the forwarding decision is based on the

label in the packet header.
Q4-5.

The number of virtual circuits is 28 = 256.

Q4-6.

The three phases are setup phase, data transfer, and teardown phase.

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2

Q4-7.

None of these services are implemented for the IP protocol in order to make it
simple.

Q4-8.

Four types of delays are transmission delay, propagation delay, processing
delay, and queuing delay.

Q4-9.

The throughput is the smallest transmission rate, or 140 Kbps. The bottleneck
is now the link between the source host and R1.

Q4-10.

The minimum length of the IPv4 header is 20 bytes and the maximum is 60
bytes. The value of the header length field defines the header length in multiples of four bytes, which means that HLEN can be between 5 and 15. It cannot
be less than 5 and it cannot be greater than 15. It is exactly 5 when there is no
option.

Q4-11.

The identification numbers need to be contiguous. The identification number
of the last datagram should be 1024  100  1 = 1123.

Q4-12.

Since the fragmentation offset field shows the offset from the beginning of the
original datagram in multiples of 8 bytes, an offset of 100 indicates that the
first byte in this fragment is numbered 800, which means bytes numbered 0 to
799 (for a total of 800 bytes) were sent before.

Q4-13.

If the first and the last addresses are known, the block is fully defined. We can
first find the number of addresses in the block (N) and then find the prefix
length (n).
N  (last address)  (first address)  1
n  32  log2N
Block: (first address)/n

Q4-14.

If the first and the number of addresses (N) are known, the block is fully
defined. We can find the prefix length (n) using the number of addresses.
n  32  log2N
Block: (first address)/n

Q4-15.

Many blocks can have the same prefix length. The prefix length only determines the number of addresses in the block, not the block itself. Two blocks
can have the same prefix length but start in two different points in the address
space. For example, the following two blocks have the same prefix length, but
they are definitely two different blocks. The length of the blocks is the same,
but the blocks are different.
127.15.12.32/27

Q4-16.

174.18.19.64/27

We cannot find the prefix length because we don’t know the length of the
block. The second given address can be an address in the middle. We need the
last address, or the length of the block to find the prefix length.

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Q4-17.

There is no need for a router and subnetting. Each customer can be directly
connected to the ISP server. In this case, the set of addresses assigned to customers can be thought of as belonging to a single block with the prefix length
n (the prefix length assigned to the ISP).

Q4-18.

The three auxiliary protocols are ICMP, IGMP, and ARP.

Q4-19.

The header length is 6  4  24. The option length is then 24  20 = 4 bytes.

Q4-20.

It can be 23 or 1. It cannot be 0 because it means the packet cannot travel at
all. It cannot be 301, because the length of the value field is 8 bits, which
means the maximum value is 255.

Q4-21.

The protocol field and the port numbers both have the same functionality:
multiplexing and demultiplexing. Port numbers are used to do these tasks at
the transport layer; the protocol field is used to do the same at the network
layer. We need only one protocol field at the network layer because payload
taken from a protocol at the source should be delivered to the same protocol at
the destination. The client and server processes, on the other hand, normally
have different port numbers (ephemeral and well-known), which means we
need two port numbers to define the processes. The size of the protocol field
defines the total number of different protocols that use the service of the network layer, which is a small number (eight bits is enough for this purpose). On
the other hand, many new applications may by added every day that needs a
larger size of the port number field (sixteen bits is assigned).

Q4-22.

Two fields, source IP address and the identification, are needed to uniquely
define fragments belonging to the same datagram. The value of the identification field is not enough because two sources may start with the same identification number.

Q4-23.

Each datagram should have a unique identification number that distinguishes
it from other datagrams sent by the same source. The identification number is
copied into all fragments. In other words, the identification number glues all
fragments belonging to the same datagram together.

Q4-24.

MPLS adds an extra header to an IP datagram. This means that MPLS implicitly creates a new layer in which a datagram is encapsulated. This layer is
between the network layer and the data-link layer.

Q4-25.

If this happens, we may enter a loop, a vicious circle. The first datagram is in
error; the second datagram reports error in the first. If the second datagram is
also in error, the third datagram will be carrying error information about the
second, and so.

Q4-26.

The source IP address is the IP address of the router interface from which the
original IP datagram is received. The destination IP address is the IP address
of the original source host that sent the original datagram. In other words, the

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reporting router in this case acts as a source host. This proves that a router
needs an IP address for each of its interfaces.
Q4-27.

According to the principle we mentioned in the text, the shortest path is the
inverse of the original one. The shortest path is G  E  B  A.

Q4-28.

According to the principle we mention in the text, the shortest path from A to
N can be found in two steps. We first use the shortest path from A to H to
move to node H. We then use the shortest path from node H to N. The result is
A  B  H  G  N.

Q4-29.

Link-state routing uses Dijkstra’s algorithm to first create the shortest-path
tree before creating the forwarding table. The algorithm needs to have the
complete LSDB to start.

Q4-30.

The path-vector routing algorithm is actually distance-vector routing using the
best path instead of the shortest distance as the metric. Each node first creates
a forwarding table, assuming it can only reach immediate neighbors. The forwarding table is gradually improved as path vectors arrive from the immediate
neighbors.

Q4-31.

The three ASs described in the text are stub, multihomed, and transient. The
first two do not allow transient traffic; the third does. The stub and multihomed ASs are similar in that they are either the sink or source of traffic; the
first is connected to only one other AS, but the second is connected to more
than one ASs.

Q4-32.

We can say that a number of hops in RIP is the number of networks a packet
travels to reach its final destination. The first network, in which the original
host is located, is normally not counted in this calculation because the source
host does not take part in routing. To reduce the traffic of exchanging routing
updates, the hosts in the Internet do not take part in this process. This is done
because the number of hosts in the Internet is much larger than the number of
routers. Including hosts in this process makes the routing-update traffic
unbearable.

Q4-33.

The source and destination IP addresses in datagrams carrying payloads
between the hosts are the IP addresses of the hosts; the IP addresses carrying
routing update packets between routers are IP addresses of the routing interfaces from which the packets are sent or received. This shows that a router
needs as many IP addresses as it has interfaces.

Q4-34.

Each datagram has a different source IP address: the IP address of the interface from which it is sent out (a router can have only one immediate neighbor
on each interface). Each datagram also has a different destination IP address:
the IP address of the router interface at which it arrives.

Q4-35.

Although RIP is running as a process using the service of the UDP, the process
is called a daemon because it is running all the time in the background. Each

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router acts both as a client and a server; it acts as a client when there is a message to send; it acts as a server when a message arrives.
Q4-36.

RIP messages are short with clear message boundaries. It is not efficient to use
the service of TCP with all of the connection establishment and connection
teardown overhead.

Q4-37.

OSPF divides an AS into areas, in which routing in each area is independent
from the others; the areas only exchange a summary of routing information
between them. RIP, on the other hand, considers the whole AS as one single
entity.

Q4-38.

If the AS is small, it is normally recommended to consider it as only one area
(the backbone area) to reduce the overhead of information exchange between
areas.

Q4-39.

In RIP, each router just needs to share its distance vector with its neighbor.
Since each router has one type of distance vector, we need only one update
message. In OSPF, each router needs to share the state of its links with every
other router. Since a router can have several types of links (a router link, a network link, ), we need several update messages.

Q4-40.

We need to have OSPF processes that run all the time because we never know
when an OSPF message will arrive. These processes are running at the network layer, not at the application layer. They are normally referred to as daemons.

Q4-41.

The type of payload can be determined from the value of the protocol field.
The protocol field value for ICMP is 01; for OSPF, it is 89.

Q4-42.
a. router link

b. router link

c. network link

Q4-43.

It cannot. A link needs to be advertised in a router link LSP; a network needs
to be advertised in a network link LSP.

Q4-44.

Each AS is independent, which means that it can run one of the two common
intradomain routing protocols (RIP or OSPF). On the other hand, the whole
Internet is considered as one entity, which means that we must run only one
interdomain routing protocol (the common one is BGP).

Q4-45.

BGP is designed to create semi-permanent communication between two BGP
speakers; this requires the service of TCP. A connection is made between the
two speakers and remains open, while the messages are exchanged between
them. UDP cannot provide such a service.

Q4-46.

The intradomain routing routes the packet inside an autonomous system that is
totally in the control of the organization. On the other hand, the interdomain
routing routes the packet through an autonomous system that is out of the con-

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trol of the organization; the organization needs to apply a policy to decide
through which AS the packet should pass.
Q4-47.

The following shows the use of each attribute:
a. The LOCAL-PREF is used to implement the organization policy.
b. The AS-PATH defines the list of autonomous systems through which the

destination can be reached.
c. The NEXT-HOP defines the next router to which the data packet should be

forwarded.
Q4-48.

In multicasting, the sender host sends only one copy of the message, but it is
multiplied at the routers if needed; all multiplied copies have the same destination address. In multiple-unicasting, the sender host sends one copy for each
destination; each copy has its own destination address.

Q4-49.

Sending a multiple-recipient e-mail is a case of multiple unicasting. An email
message needs recipient addresses at the application layer, which cannot be
translated to a multicast address at the network layer. The recipients of an
email address do not necessarily belong to the same group. In other words, we
a one-to-many communication at the application layer, which should not be
confused to one-to-many communication at the network layer.

Q4-50.

The multicast address block is 224.0.0.0/4. In other words, a multicast address
is between 224.0.0.0 and 239.255.255.255. Based on this criteria we have
a. A multicast

Q4-51.

b. A multicast

c. Not a multicast

In each case, we find the corresponding block to be able to find the group
a. 224.0.1.7 belongs to the block 224.0.1.0/24; it belongs to the internetwork

control block.
b. 232.7.14.8 belongs the block 232.0.0.0/8; it belongs to the SSM block.
c. 239.14.10.12 belongs the block 239.0.0.0/8; it belongs to the administra-

tively scoped block.
Q4-52.

If a host is a member of N multicast group, it will have N multicast addresses.

Q4-53.

The group list is the union of the individual lists; it is {G1, G2, G3, G4}.

Q4-54.
a. In unicast communication, the destination is only one of the leaves of the

tree in each transmission.
b. In multicast communication, the destination may be one or more leaves of

the tree in each transmission.

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Q4-55.
a. In the source-based tree approach, we need 20  4 = 80 shortest-path trees.
b. In the group-shared tree, we need only 4 shortest-path trees, one for each

group.
Q4-56.

DVMRP allows a router to create the shortest path-tree whenever it receives a
multicast packet (on demand). The number of shortest-path trees in DVMRP
that use the source-based approach is huge. This means if each router created
all of the required multicast shortest-path trees, it would be a huge overhead.

Q4-57.

Each router using DVMRP creates the shortest-path three in three steps:
a. In the first step, the router uses the RPF algorithm to keep only packets that

have arrived from the source using the shortest-path three. In other words,
the first part of the tree is made using the RPF algorithm.
b. In the second step, the router uses the RPB algorithm to create a broadcast

tree.
c. In the third step, the router use the RPM algorithm to change the broadcast

tree created in the second step to a multicast tree.
Q4-58.

MOSPF uses Dijkstra's algorithm to create the whole broadcast path tree in
one shot, but DVMRP needs to use three steps because it does not have the
LSDB to use Dijkstra's algorithm.

Q4-59.

Every multicast routing algorithm needs to somehow use a unicast protocol in
its operation. For example, DVMRP needs to use RIP and MOSPF needs to
use OSPF. Although PIM also needs to use a unicast protocol, the protocol can
be either RIP or OSPF.

Q4-60.

PIM-DM is very similar to DVMRP, but it does not care about controlling the
broadcast step of DVMRP because it assumes that most networks have a loyal
member in each group. It only uses the first step (RPF) and the third step
(RPM).

Q4-61.

In PIM-DM, it is assumed that most networks have a loyal member in each
group, so it does not matter if the first packet reaches all networks. In PIMSM, it is assumed that a few networks has a loyal member in each group, so
broadcasting is wasting the bandwidth.

Q4-62.

We can say (1) larger address space, (2) better header format, (3) new options,
(4) allowance for extension, (5) support for resource allocation, and (6) support for more security.

Q4-63.

The flow field can be used in several ways. It allows IPv6 to be used as a connection-oriented protocol. It also allows IPv6 to give priority to different payloads, such as giving high priority to real-time multimedia applications.

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Q4-64.

A compatible address is an address of 96 bits of 0s followed by 32 bits of an
IPv4 address. A mapped address is an address of 80 bits of 0s followed by 16
bits of 1s and followed by 32 bits of an IPv4 address. A compatible address is
used when a computer using IPv6 wants to send a packet to another computer
using IPv6. A mapped address is used when a computer using IPv6 wants to
send a packet to a computer still using IPv4.

Q4-65.

The three protocols IGMP, ICMP, and ARP in IPv4 have been combined into a
single protocol, ICMPv6.

Q4-66.

The IP header is included because it contains the IP address of the original
source. The first 8 bytes of the data are included because they contain the first
section of the TCP or UDP header which contains information about the port
numbers (TCP and UDP) and sequence number (TCP). This information
allows the source to direct the ICMP message to the correct application.

Problems
P4-1.

The total length of the datagram is (00A0)16  160 bytes. The header length is
5  4 = 20. The size of the payload is then 160  20  140. The efficiency 
140 / 160  87.5%.

P4-2.

We analyze each byte or group of bytes to answer the questions:
a. The second hex digit in the first byte is 5 (HLEN), which means that the

header length is only 5  4  20 bytes.

b. There are no options because the header size is only 20 bytes.
c. The total length of the packet is (0054)16 or 84 bytes. Since the header is 20

bytes, it means the packet is carrying 64 bytes of data.

d. Since the flags field fragmentation offset bit is all 0s, the packet is not frag-

mented.
e. The value of the TTL field is (20)16 or 32 in decimal, which means the

packet may visit up to 32 more routers.

f. The value of the protocol field is 6, which means that the packet is carrying

a segment from the TCP protocol.
P4-3.

The following fields can be changed from one router to another:
a. HLEN: If there is option change
b. Total length: If fragmented or options change
c. Flags: If fragmented
d. Fragmentation Offset: If fragmented
e. Time-to-Live; Decremented at each router
f. Header Checksum: Need to change because of other changes

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P4-4.

In each case, we first need to think about the value of M and then the value of
the offset:
a. Since M = 1, it means there are more fragments and this is the first or mid-

dle; since the offset field is zero, it means this is the first fragment.
b. Since M = 1, it means there are more fragments and this the first or middle;

since the offset field is nonzero, it means this is a middle fragment.
P4-5.

Let us discuss each case separately:
a. Packet sniffing can be defeated if the datagram is encrypted at the source

and decrypted at the destination using an unbreakable scheme.
b. Packet modification can be defeated using a strong message integrity

scheme.
c. IP spoofing can be defeated using a strong entity authentication scheme.
P4-6.

The size of the address in each case is the base to the power of the number of
digits:
a. The size of the address space is 216  65,536.
b. The size of the address space is 166 = 16,777,216.
c. The size of the address space is 84 = 4096.

P4-7.

We change each byte to the corresponding binary representation:
a. 01101110 00001011 00000101 01011000
b. 00001100 01001010 00010000 00010010
c. 11001001 00011000 00101100 00100000

P4-8.

We change each 8-bit section to the corresponding decimal value and insert
dots between the bytes.
a. 94.176.117.21

P4-9.

b. 137.142.208.49

c. 87.132.55.15

The class can be defined by looking at the first byte (see figure 4.31):
a. Since the first byte is between 128 and 191, the class is B.
b. Since the first byte is between 192 and 223, the class is C.
c. Since the first byte is between 240 and 255, the class is E.

P4-10.

The class can be defined by checking the first few bits (see figure 4.31). We
need to stop checking if we find a 0 bit or four bits have already been checked.
a. Since the first bit is 0, the Class is A.
b. Since the first four bits are 1110, the class is D.
c. Since the first three bits are 110, the class is C.

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P4-11.

P4-12.

The whole block can be represented as 0.0.0.0/0. The first address in the class
is 0.0.0.0. The prefix is 0 because no bits define the block; all bits define the
address itself. Another test to prove that the prefix is 0 is that the number of
addresses in the block can be found as 232n. The value of n should be zero to
make the number of addresses N  232.
The size of the block can be found as N = 232n:
a. 2320  4,294,967,296

P4-13.

c. 23232  1

The prefix can be found as n  32  log2N:
a. n  32  log2 1  32

P4-14.

b. 23214  262,144

b. n  32  log2 1024  22

c. n  32  log2232  0

We can first write the prefix in binary and then change each 8-bit chunk to
decimal:
 mask: 0.0.0.0
 mask: 255.252.0.0
c. 11111111 11111111 11111111 11111100  mask: 255.252.255.252
a. 00000000 00000000 00000000 00000000

b. 11111111 11111100 00000000 00000000

P4-15.

We first write the mask in binary notation and then count the number of leftmost 1s.
 n: 11
b. 11111111 11110000 00000000 00000000  n: 12
c. 11111111 11111111 11111111 10000000  n: 25
a. 11111111 11100000 00000000 00000000

P4-16.

We first write each potential mask in binary notation and then check if it has a
contiguous number of 1s from the left followed by 0s.
 Not a mask
 A mask
c. 11111111 11111111 11111111 00000110  Not a mask
a. 11111111 11100001 00000000 00000000
b. 11111111 11000000 00000000 00000000

P4-17.

We can write the address in binary. Set the last 32  n bits to 0s to get the first
address; set the last 32  n bits to 1s to get the last address. You can use one of
the applets at the book website to check the result.
a.
Given:
First:
Last:

00001110 00001100 01001000 00001000 14.12.72.8/24
00001110 00001100 01001000 00000000 14.12.72.0/24
00001110 00001100 01001000 11111111 14.12.72.255/24

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b.
Given:
First:
Last:

11001000 01101011 00010000 00010001 200.107.16.17/18
11001000 01101011 00000000 00000000 200.107.0.0/18
11001000 01101011 00111111 11111111 200.107.63.255/18

Given:
First:
Last:

01000110 01101110 00010011 00010001 70.110.19.17/16
01000110 01101110 00000000 00000000 70.110.0.0/16
01000110 01101110 11111111 11111111 70.110.255.255/16

c.

P4-18.

We write the address in binary and then keep only the leftmost n bits.
a. 10101010 00101000 00001011
b. 01101110 00101000 111100
c. 01000110 00001110 00

P4-19.

The administration can use DHCP to dynamically assign addresses when a
host needs to access the Internet. This is possible because, in most organizations, not all of the hosts need to access the Internet at the same time.

P4-20.

Both NAT and DHCP can be used for this purpose. DHCP dynamically
assigns one of the assigned addresses when a host needs to access the Internet;
NAT permanently assigns a set of private addresses to the host, but maps the
private address to the global address when a host needs to use the Internet.

P4-21.

The total number of addresses is 28 = 256. This means we have 64 addresses
for each network. We can divide the whole address space into four blocks
(blocks 0 to 3), each of 64 addresses. The addresses in each block are allocated as (0 to 63), (64 to 127), (128 to 191), and (192 to 255). It can be
checked that each block is allocated according to the two restrictions needed
for the proper operation of CIDR. First, the number of addresses in each block
is a power of 2. Second, the first address in each block is divisible by the number of addresses in the block, as shown below:
Block 0: 0/64 = 0

Block 1: 64/64 = 1 Block 2: 128/64 = 2 Block 3: 192/64 = 3

The prefix length for each group is ni  8  log264 = 2. We can then write the
ranges in binary to find the prefix for each block.
Block
0
1
2
3

Range
0 to 63
64 to 127
128 to 191
192 to 255

Range in binary
00000000 to 00111111
01000000 to 01111111
10000000 to 10111111
11000000 to 11111111

n
2
2
2
2

Prefix
00
01
10
11

12

The following shows the outline and the forwarding table. Note that each
interface can use one of the addresses in the corresponding block.

Addresses:
0 to 63

0/2

P4-22.

Addresses:
64 to 127

Addresses:
128 191

128/2

64/2

m1

m2

m0

m3

Addresses:
192 to 255

192/2

Forwarding table
Prefix Interface
m0
00
m1
01
m2
10
m3
11

The total number of addresses is 212 = 4096. This means that there are 512
addresses for each network. We can divide the whole address space into eight
blocks (blocks 0 to 7), each of 512 addresses. The addresses in each block are
allocated as (0 to 511), (512 to 1023), (1024 to 1535), (1536 to 2047), ,
(3584 to 4095). It can be checked that each block is allocated according to the
two restrictions needed for the proper operation of CIDR. First, the number of
addresses in each block is a power of 2. Second, the first address is divisible
by the number of addresses as shown below:
Block 0: 0 / 512 = 0

Block 1: 512 / 512 = 1

Block 2: 1024 / 512 = 2

The prefix length for each group is ni  12  log2512 = 3. We can then write
the ranges in binary to find the prefix for each block.
Block
Range
0
0 to 511

Range in binary
000000000000 to 000111111111

1

512 to 1023

001000000000 to 001111111111

2

1024 to 1535

010000000000 to 010111111111

3

1536 to 2047

011000000000 to 011111111111

4

2048 to 2559

100000000000 to 100111111111

5

2560 to 3071

101000000000 to 101111111111

6

3072 to 3583

110000000000 to 110111111111

7

3584 to 4095

111000000000 to 111110000000

n

3
3
3
3
3
3
3
3

Prefix
000
001
010
011
100
101
110
111

The following figure shows the outline and the forwarding table. Note that
each interface can use one of the addresses in the corresponding block. The
addresses are written in decimal (not dotted-decimal) because of the address
space size.

13

Addresses:
0 to 512

0/3

Addresses:
512 to 1023

1024/3

512/3

m0
m7

m1
m6

Addresses:
2048 to 2559

m2
m5

2560/3

2048/3

P4-23.

Addresses:
1024 to 1535

1536/3

m3
m4
3072/3

Addresses:
2560 to 3071

Addresses:
1536 to 2047

Addresses:
3072 to 3583

3584/3

Addresses:
3584 to 4095

Forwarding table
Prefix Interface
m0
000
m1
001
m2
010
m3
011
m4
000
m5
000
m6
000
m7
000

The total number of addresses is 29 = 512. We need, however, to check
whether address allocation is done according to the restrictions for CIDR’s
proper operation. The address allocations to the networks are (N0: 0 to 63),
(N1: 64 to 255), and (N2: 256 to 511). Each range is a power of 2, which
means that the first restriction is fulfilled. The second restriction (the first
address in the block should divide the number of addresses in the block) is fulfilled for N0 and N2, but not for N1:
N0: 0 / 64 = 0

N1: 64 / 256 = 0.25

N2: 256 / 256 = 1

One solution would be to think of the addresses in N1 as the aggregation of
two contiguous blocks (64 to 127) and (128 to 256) connected to the same
interface. We call these blocks 1-1 and 1-2. The prefixes for blocks are
n0  9  log264 = 3
n1-2  9  log2128  2

Block
Range
0
0 to 63

n1-1  9  log264  3
n2  9  log2256  1

Range in binary
000000000  000111111

1-1

64 to 127

001000000  001111111

1-2

128 to 255

010000000  011111111

2

256 to 511

100000000  111111111

n

3
3
2
1

Prefix
000
001
01
1

Based on the above table, we can show the outline of the internet and
addresses and the forwarding table, as shown below. Note that the address
aggregation in N1 is transparent to the user as long as the router forwards the
packet according to its forwarding table. If we need to be fair, we should say
that N1 actually has two network addresses because it is made of two blocks.
The administration can easily divide the block into two subblocks with a
router.

14

Addresses:
0 to 63

Addresses:
64 to 255
N0

Addresses:
256 to 511
N2

N1

0/3

64/3

256/1

128/2

m1
m2

m0

P4-24.

Forwarding table
Prefix Interface
m0
000
m1
001
m1
01
m2
1

One way to do this is to first find the size of each block. We can then add the
size to the first address in the block to find the last address. Next, we can put
the blocks together to find whether they can be combined into a larger block.
Block
Size
a
N  23226 64
b
c

N  23226 64

N  23225 128

First address

Last address

16.27.24.0/26
16.27.24.64/26
16.27.24.128/25

 16.27.24.63/26
 16.27.24.127/26
 16.27.24.255/26

Original
blocks

n:26
N:64

n:26
N:64

Block a

Block b

16.27.24.255/24

16.27.24.128/25

16.27.24.127/26

16.27.24.64/26

16.27.24.0/26

16.27.24.63/26

Since the blocks are contiguous, we can combine the three blocks into a larger
one. The new block has 256 addresses and n  32 log2256 = 24.

n:25
N:128
Block c

n:24

New block
P4-25.

16.27.24.0/24

16.27.24.255/24

N:256

The organization is granted 23221  211  2048 addresses. The medium-size
company has 23222  210  1024 addresses. Each small organization has 232
23  29  512 addresses. We can plot the range of addresses for each organization as shown below:
Large organization:
Medium organization:
Small organization 1:
Small organization 2:

12.44.184.0/21
12.44.184.0/22
12.44.188.0/23
12.44.190.0/23

to
to
to
to

12.44.191.255/21
12.44.187.255/22
12.44.189.255/23
12.44.191.255/23

The company install a router whose forwarding table is based on the longest-

15

prefix match first principle as shown below.
Network address /mask
00001100 00101100 1011110
00001100 00101100 1011111
00001100 00101100 101110

Next hop


Interface
Small organization 1
Small organization 2
Medium organization

Let us use three cases to show that the packets are forwarded correctly.
a. Assume a packet with the destination address 12.44.185.110 is arrived. The

router first extracts the first 23 bits (00001100 00101100 1011100) and
check to see if it matches with the first row of the table, which does not. It
then checks with the second row, which does not match either. The router
next extracts the first 22 bits (00001100 00101100 101110), which matches
with the last entry. The packet is correctly forwarded to the interface of the
medium organization.
b. Assume a packet with the destination address 12.44.190.25 is arrived. The
router first extracts the first 23 bits (00001100 00101100 1011111) and
check to see if it matches with the first row of the table, which does not. It
then checks with the second row, which does. The packet is correctly forwarded to the interface of second small organization.
c. Assume a packet with the destination address 12.44.189.24 is arrived. The
router first extracts the first 23 bits (00001100 00101100 1011110) and
check to see if it matches with the first row of the table, which does The
packet is correctly forwarded to the interface of first small organization.
P4-26.
a. The number of addresses in the ISP block is N  232204096. We can

add 4095 (which is N  1) to the first address to find the last one (note that
the addition can be done in base 256, as described in Appendix B. In base
256, 4095 is (15.255). We have
First address: 16.12.64.0/20

Last address: 16.12.79.255/20

The prefix length for each organization is ni  32  log2 256 = 24. We
assume that the addresses are allocated from the beginning of the ISP block
with each organization consuming 256 addresses. The following shows
how addresses are allocated. Note that the prefix for each block is 24 bits.
Block
1
2
3
4
5
6
7
8
Unassigned

First address
16.12.64.0/24
16.12.65.0/24
16.12.66.0/24
16.12.67.0/24
16.12.68.0/24
16.12.69.0/24
16.12.70.0/24
16.12.71.0/24
16.12.72.0/21









Last address
16.12.64.255/24
16.12.65.255/24
16.12.66.255/24
16.12.67.255/24
16.12.68.255/24
16.12.69.255/24
16.12.70.255/24
16.12.71.255/24
16.12.79.255/21

n
24
24
24
24
24
24
24
24
21

16

The unallocated addresses, which can be reserved for the future use of the
ISP, are 16.12.72.0/21 to 16.12.79.255/21, for a total of 2048 addresses.
b. The simplified outline is given below. Note that packets having destination

addresses with the last prefix in the figure are discarded until these
addresses are assigned.

Org.: Organization
I.: Interface
Dis.: Discard

Internet

ISP

Org. 1

m2

Default

Org. 2

0001000

00001100 01000000

m3

Org. 3

0001000

00001100 01000001

0001000

00001100 01000010

0001000

00001100 01000011

Org. 5

0001000

00001100 01000100

Org. 6

0001000

00001100 01000101

m7

Org. 7

0001000

00001100 01000110

m8

Org. 8

0001000

00001100 01000111

0001000

00001100 01001

m4

m0

m5
m6
Dis.

Forwarding table
Prefix

m1

Org. 4

Interface
m0
m1
m2
m3
m4
m5
m6
m7
m8
Dis.

P4-27.
a. The number of addresses in the ISP block is N  232212048. We can

add 2047 (which is N  1) to the first address to find the last one (note that
the addition can be done in base 256, as described in Appendix B. In base
256, 2047 is (7.255). We have
First address: 80.70.56.0/21

Last address: 80.70.63.255/21

b. To make the number of addresses in each block a power of 2 (first CIDR

restriction), we assign the following ranges to each organization. The prefix length for each organization is ni = 32 log2Ni where Ni is the number
of addresses assigned to that organization. Note that the unused addresses
cannot fit in a single block (second CIDR restriction).
Block

Size

1

512

80.70.56.0/23

First address

2

512

80.70.58.0/23

3

256

80.70.60.0/24

4

256

80.70.61.0/24

5

64

80.70.62.0/26

6

64

80.70.62.64/26

7

64

80.70.62.128/26

Unused

320

80.70.62.192

Last address

 80.70.57.255/23
 80.70.59.255/23





80.70.60.255/24
80.70.61.255/24
80.70.62.63/26
80.70.62.127/26
80.70.62.191/24

n

23
23
24
24
26
26
26

80.70.63.255

c. The simplified outline is given below. Note that to make the forwarding

table operable, we need to divide the unused addresses into two blocks.

17

Packets with destination addresses matching the last two prefixes are discarded by the router.

Forwarding table
Prefix
Org.: Organization
I.: Interface
Dis.: Discard

Internet

ISP

m0

P4-28.

Org. 1

m2

01010000 01000110 0011100

Org. 2

01010000 01000110 0011101

m3

Org. 3

m4

Org. 4

01010000 01000110 00111100
01010000 01000110 00111101

m5
m6
Dis.

Default

m1

m7

Org. 5
Org. 6
Org. 7

01010000 01000110 00111110 00
01010000 01000110 00111110 01
01010000 01000110 00111110 10
01010000 01000110 00111110 11
01010000 01000110 00111111

I.
m0
m1
m2
m3
m4
m5
m6
m7
Dis.

The total number of addresses in the organization is N  23216 65,536.
a. Each subnet can have Nsub = 65,536 /1024 = 64 addresses.
b. The subnet prefix for each subnet is nsub  32  log2Nsub 32  6  26.
c. Now we can calculate the first and the last address in the first subnet. The

first address is the beginning address of the block; the last address is the
first address plus 63.
First address: 130.56.0.0/26

Last address: 130.56.0.63/26

d. To find the first address in subnet 1024, we need to add 65,472 (1023 64)

in base 256 (0.0.255.192) to the first address in subnet 1. The last address
can then be found by adding 63 to the first. 
First address: 130. 56.255.192/26
P4-29.

Last address: 130. 56.255.255/26

Router R1 has four interfaces. Let us investigate the possibility of a packet
with destination 140.24.7.194 from each of these interfaces and see how it is
routed.
a. The packet can arrive from one of the interfaces m0, m1, and m2, because

one of the computers in organization 1, 2, or 3 could have sent this packet.
The prefix /26 is applied to the address, resulting in the network address
140.24.7.192/26. Since none of the network addresses/masks matches this
result, the packet is sent to the default router R2.
b. The packet cannot arrive at router R1 from interface m3 because this

means that the packet must have arrived from interface m0 of router R2,
which is impossible because if this packet arrives at router R2 (from any
interface), the prefix length /26 is applied to the destination address, resulting in the network address/mask of 140.24.7.192/26. The packet is sent out
from interface m1 and directed to organization 4 and never reaches router
R1.

18

P4-30.

The packet is sent to router R1 and eventually to organization 1 as shown
below:
a. Router R2 applies the mask /26 to the address (or it extracts the leftmost 26

bits) resulting in the network address/mask of 140.24.7.0/26, which does
not match with the first entry in the forwarding table.
b. Router R2 applies the mask /24 to the address (or it extracts the leftmost 24

bits) resulting in the network address/mask of 140.24.7.0/24, which
matches with the second entry in the forwarding table. The packet is sent
out from interface m0 to router R1.
c. Router R1 applies the mask /26 to the address (or it extracts the leftmost 26

bits) resulting in the network address/mask of 140.24.7.0/26, which
matches with the first entry in the forwarding table. The packet is sent out
from interface m0 to organization 1.
P4-31.

We have
Dxy min {(cxa + Day), (cxb + Dby), (cxc + Dcy), (cxd + Ddy)}
Dxy min {(2 + 5), (1 + 6), (3 + 4), (1 + 3)} = min {7, 7, 7, 4} = 4

P4-32.

At time t1, we have one periodic timer, ten expiration timers, and no garbage
collection timer. An expiration timer becomes invalid after 180 seconds. This
means, at time t2, we have one periodic timer, nine expiration timers, and one
garbage collection timer (for the one which has become invalid).

P4-33.
a. The hello message (type 1) is used by a router to introduce itself to neigh-

boring routers and to introduce already-known neighboring routers to other
neighbors.
b. The data description message (type 2) is sent in response to a hello mes-

sage. A router sends its full LSDB to the newly joined router.
c. The link-state request message (type 3) is sent by a router that needs infor-

mation about a specific LS.
d. The link-state update message (type 4) is sent by a router to other routers

for building the LSDB. There are five different versions of this message to
announce different link states.
e. The link-state acknowledge message (type 5) is sent by a router to

announce the receiving of a link-state update message. This message is
used to provide reliability for the main message used in OSFP.

19

The following shows the initialization and two rounds of updates. Although
the process is asynchronous, which means that a node can initialize itself and
fire updates to its neighbors at the any time, we have assumed the updates
takes place in an orderly way (A, B, C, D). After all nodes has sent their
updates, a new round starts and those nodes that have seen any change, will
fire updates again. The result should be the same using any other order. The
process should stop after there is no change in any node.

Note: Colored boxes show the changes in the metric.

B

C

D

3
0
2
5
2

A
B 2
C 0
D 4
3

6
5
4
0
4

6
5
4
0

A
B
C
D

A
B
C
D

A 0
B 3
C
D 6

A
B
C
D

+3
3
0
2
5

A
B 2
C 0
D 4

A
B
C
D

A updates
B and D.

+6
8

8

A 0
After each node B 3
initializes itself. C
D 6
1

8

A

8

P4-34.

B updates
A, C, and D.

A
B
C
D

+3
0
3
5
6
5

A
B
C
D

3
0
2
5

A
B
C
D

+2
5
2
0
4
6

A
B
C
D

+5
6
5
4
0

C updates
B and D.

A
B
C
D

0
3
5
6

A
B
C
D

+2
3
0
2
5

A
B
C
D

5
2
0
4

A
B
C
D

+4
6
5
4
0

A
B
C
D

+5
5
2
0
4

A
B
C
D

6
5
4
0

D updates
A, B, and C.

+6
A 0
B 3
C 5
D 6

+4
A 3
B 0
C 2
D 5

A updates
B and D.

A
B
C
D

0
3
5
6

A
B
C
D

3
0
2
5

A
B
C
D

5
2
0
4

+6
A 6
B 5
C 4
D 0

C updates
B and D..

A
B
C
D

0
3
5
6

+2
A 3
B 0
C 2
D 5

A
B
C
D

5
2
0
4

A
B
C
D

+3

+4

No more updates need to be sent; the system is stable.

6
5
4
0

20

P4-35.

Two nodes, A and D, see the changes (see Table 4.4, line 16). These two nodes
update their vectors immediately. We assume that changes in each round are
fired in the order A, B, C, D. The following shows that the internet is actually
stable after two rounds of updates, but more updates are fired to assure the
system is stable. We have shown only three columns for each forwarding
table, but RIP usually uses more than columns. Also note that we have used
the yellow color to show the changed in cost field, which triggers updates. The
cost and the next hop fields participate in updating.

Cost

Change in cost

A

Next hop

B

Changes A 0 A
occured B 3 A
5 B
in the links. C
D 1 D
1

A
B
C
D

3
0
2
5

C

B
B
B
B

A
B
C
D

5
2
0
4

D

B
C
C
C

A
B
C
D

0
3
5
1

B
C
C
C

A
B
C
D

3
0
2
4

0
3
5
1

B
C
C
C

A
B
C
D

5
2
0
4

+5

+1
D sends A
updates to B
A, B, and C. C
D

B
B
B
A

A
B
C
D

3
0
2
4

B
B
B
A

D
D
D
D
2

+1

+3
A sends A
updates to B
C
B and D. D

1
5
4
0

B
C
C
C

A
B
C
D

1
4
4
0

D
A
D
D

A
B
C
D

1
4
4
0

D
A
D
D

+4
A
B
C
D

5
2
0
4

B
C
C
C

21

The following shows how the forwarding tables will be changed.

Unstable Cost

A

B

C

0
3
5
6

A
A
Β
A

A ? ?
B
C
C 0 C
D 4 C
3

Next hop

D
D
D
D
D

A
B
C
D

6
5
4
0

D
D
D
D

4

B
B
A
C

A ? ?
B
C
C 0 C
D 4 C

A
B
C
D

+6
6
5
4
0

A
B
C
D

A
A
Β
A

A
B
C
D

B updates
A and D,

+3
A 0 A
B 3 A
C
Β
D 6 A

A
B
C
D

3
0
8
5

B
B
A
C

A ? ?
B
C
C 0 C
D 4 C

A
B
C
D

+5
6
5
4
0

D
D
D
D

A 0 A
C updates B 3 A
D.
C
Β
D 6 A

A
B
C
D

3
0
8
5

B
B
A
C

A ? ?
B
C
C 0 C
D 4 C

A
B
C
D

+4
6
5
4
0

D
D
D
D

+6
A
A
Β
A

A
B
C
D

3
0
8
5

+5
B
B
A
C

A
B
C
D

10
9
0
4

+4
D
C
C
C

A
B
C
D

6
5
4
0

D
D
D
D

0
3
10
6

A
A
Β
A

A
B
C
D

3
0
13
5

+3
B
B
A
C

A
B
C
D

10
9
0
4

Β
C
C
C

A
B
C
D

6
5
4
0

+6
D
D
D
D

0
3
10
6

+3
A
A
Β
A

A
B
C
D

3
0
13
5

B
B
A
C

A
B
C
D

10
9
0
4

Β
C
C
C
7

A
B
C
D

6
5
4
0

+5
D
D
D
D

A 0
C updates B 3
C 10
D.
D 6

A
A
Β
A

A
B
C
D

3
0
13
5

B
B
A
C

A
B
C
D

10
9
0
4

Β
C
C
C

A
B
C
D

6
5
4
0

+4
D
D
D
D

A 0
D updates B 3
A, B, and C. C 10
D 6

+6
A
A
Β
A

3
0
9
5

+5
B
B
A
C

10
9
0
4

+4
D
C
C
C

A
B
C
D

A
B updates B
A, B, and C. C
D

0
3
10
6

8

8

0
3
5
6

8

A updates
B and D.

+3
3
0
8
5

8

1

A 3 B
B 0 B
C
B
D 5 D
2

8

A
B
C
D

Change in cost

8

Changes
occurred
in the links.

Cost

8

P4-36.

5
A updates
B and D.

A
B
C
D

6
B updates
A and D.

A
B
C
D

A
B
C
D

A
B
C
D

8
6
5
4
0

D
D
D
D

No more updates need to be sent; the system is stable.

Note that there are some unstable cost values that are not finalized. These
wrong pieces of information may create looping in the system; the packet may
bound back and forth until the system becomes stable. Eight updates are
needed to stabilize the system.

22

P4-37.

The number of operations in each iteration of the algorithm is n, in which n is
the number of nodes in the network. In computer science, this complexity is
written as O(n) and is referred to as Big-O notation.

P4-38.

The following shows the advertisement in each case (a triplet defines the destination, cost, and the next hop):
a. From A to B: (A, 0, A), (B, ∞, A), (C, 4, A), (D, ∞, B).
b. From C to D: (A, 4, C), (B, ∞, D), (C, 0, C), (D, ∞, C).
c. From D to B: (A, ∞, B), (B, ∞, D), (C, 6, D), (D, 0, D).
d. From C to A: (A, ∞, C), (B, 8, D), (C, 0, C), (D, 6, C).

P4-39.

The following shows the advertisement in each case (a triplet defines the destination, cost, and the next hop):
a. From A to B: (A, 0, A), (C, 4, A).
b. From C to D: (A, 4, C), (C, 0, C).
c. From D to B: (C, 6, D), (D, 0, D).
d. From C to A: (B, 8, D), (C, 0, C), (D, 6, C).

P4-40.

P4-41.

We can guess the new routing table because the only way each node can reach
node E is via D. The following shows the new network and the forwarding
tables. Note that we only add an entry to each forwarding table for nodes A, B,
C, and D. The forwarding table for node E is totally new.

A
B
C
D
E

0
5
4
7 B
8 B

A
B
C
D
E

4
8 D
0
6
7 D

5

A

A
B
C
D
E

B

5
0
8 D
2
3 D

2

4
C

D

6
A
B
C
D
E

1

E

7 Β
2
6
0
1

A
B
C
D
E

8 D
3 D
7 D
1
0

The forwarding table for node A can be made using the least-cost tree, as
shown below:

Destination
A
B
C
D
E
F
G

Cost
0
2
7
3
6
8
9

Next hop

B
B
B
B

Forwarding table for node A

23

B

C

8

8

3

A

8

8

The following shows the steps to create the shortest-path tree for node G and
the forwarding table for this node.

3

A

B

C

G
E

F

D

E

F

1

8

8

G

8

8

D

1

Iteration 1

7

3

A

B

C

8

Initialization

8

P4-42.

7

3

A

B

C

G
D

E

F

8

3

1

G
D

E

F

8

3

1

Iteration 3

Iteration 2

9

7

3

A

B

C

9

7

3

A

B

C

G
D

E

F

8

3

1

G
D

E

F

8

3
Iteration 5

1

Iteration 4

9

7

3

A

B

C
G

D

E

8

3

F
1
Iteration 6

Destination
A
B
C
D
E
F
G

Cost
9
7
3
8
3
1
0

Next hop
F
F
F
F

Forwarding table for node G

24

The following shows the steps to create the shortest path tree for node B and
the forwarding table for this node.

2

0

5

2

0

5

A

B

C

A

B

C

G
E

F

D

E

4

5

4
Iteration 1

Initialization

F
8

D

8

8

8

G

8
2

0

5

2

0

5

A

B

C

A

B

C
G

G

8

8

P4-43.

D

E

F

D

E

F

5

4
Iteration 2

6

5

4
Iteration 3

6

2

0

5

2

0

5

C

A

B

C

A

B

G

G
D

E

5

4

F

8

6

D

E

F

5

4

6

Iteration 4

8

Iteration 5

2

0

5

A

B

C
G

D

E

F

5

4
Iteration 6

6

7

Destination
A
B
C
D
E
F
G

Cost
2
0
5
5
4
6
8

Next hop

A
E
E

Forwarding table for node B

25

P4-44.

The following shows the shortest-path tree and the forwarding table for node
A.

A

4

B

A

C

G

5
F

F

D

3

E

A

G

5

D

3

4

B
C

4

B

E

A

B

4

8
C

8
C

G

5

5
F

D

3

E

7

F

D

3

6

4

A

G

E

4

A

B

7

6

B

8
C

8
C

G

5

5
F

D

3

E

7

3

4
B

8
C

G

5
F

D

3

E

6

7

F

D

6

A

G

E

Destination
A
B
C
D
E
F
G

7

6

Cost
0
4
5
3
6
7
8

Next hop

D
B
B

Forwarding table for node A

P4-45.

The number of searches in each iteration of Dijkstra’s algorithm is different.
In the first iteration, we need n number of searches, in the second iteration, we
need (n  1), and finally in the last iteration, we need only one. In other words,
the total number of searches for each node to find its own shortest-path tree is
Number of searches n  n 1  n 2  n  3    3  2  1 n (n 1) 2

The series can be calculated if it is written twice: once in order and once in the
reverse order. We then have n items, each of value (n + 1), which results in n(n
+ 1). However, we need to divide the result by 2. In computer science, this
complexity is written as O(n2) and is referred to as Big-O notation.

26

P4-46.

The following shows the initialization and updates.

A

C

B

A A
B A, B
C
D A, D
E
1

A
B
C
D
E

A A
B A, B
C
D A, D
E

A
B
C
D
E

A
B
C
D
E

A
A, B
A, B, C
A, D
A, B, E

A
B
C
D
E

A
A, B
A, B, C
A, D
A, B, E

B, A
B
B, C

D

E

A
B C, B
C C
D
E C, E

A D, A
B
C
D D
E

A
B E, B
C E, C
D
E E

B, A
B
B, C
B, A, D
B, E

A
B C, B
C C
D
E C, E

A D, A
B D, A, B
C
D D
E

A
B E, B
C E, C
D
E E

A
B
C
D
E

B, A
B
B, C
B, A, D
B, E

A
B
C
D
E

C, B, A
C, B
C

A
B
C
D
E

E, B, A
E, B
E, C

C, E

A D, A
B D, A, B
C
D D
E

A
B
C
D
E

B, A
B
B, C
B, A, D
B, E

A
B
C
D
E

C, B, A
C, B
C
C, B, A, D
C, E

A
B
C
D
E

A
B
C
D
E

E, B, A
E, B
E, C
E, B, A, D
E

B, E
2

D, A
D, A, B
D, A, B, C
D
D, A, B, E

E

P4-47.

Router R1, using its OSPF forwarding table, knows how to forward a packet
destined for N4. R1 announces this reachability to R5 using an eBGP session.
R5 adds an entry to its RIP forwarding table that shows R1 as the next router
for any packet destined for N4.

P4-48.

Router R9 knows how to reach N13 through its RIP forwarding table. R9
advertises this reachability to R4 using an eBGP session. R4 in turn advertises
this reachability to R2 using an iBGP session. R2 then advertises this reachability to R6 using an eBGP session. R6 advertises its reachability to R8. R8
adds an entry to its RIP forwarding table to show that any packet destined for
N13 should be forwarded to R6.

P4-49.

The packet that has arrived through m2 should be forwarded because if R3
wants to send a packet to the source, S, in the reverse path, it will send it
through the interface m2.

P4-50.

This is where IGMP can help. Using IGMP, the router should collect information about the interests of networks. When a packet with a multicast destination address G arrives, the router should send the packet only from those
interfaces connected to the interested networks.

27

P4-51.

The following shows the two cases. In the first case, router R is the root of the
unicast communication. In the second case, router S is the root of the multicast
communication.

S

S

R

R
a. OSPF: R is the root.

P4-52.

b. MOSPF: S is the root.

See Figure 4.72.
a. The size of a RIP message that advertises a single network is 24 [4  (20 

1)] bytes.
b. The size of a RIP message that advertises n networks is [4  (20  n)]

bytes.
The following shows the contents of the RIP message (See Figure 4.72).

net 1
All 0s
All 0s
4
All 0s

Family: 2

net 2
All 0s
All 0s

Header
Network 1

Reserved
All 0s

Network 2

Version
Family: 2

2
All 0s

Family: 2

net 3
All 0s
All 0s

Network 3

2

1
All 0s

Family: 2

net 4
All 0s
All 0s

5

Network 4

P4-53.

28

P4-54.

The following table shows the comparison:
Field
VER
HLEN
Service (or traffic class)
Flow label
Total length
Payload length
Identification
Flags
Flag offset
TTL (or hop limit)
Protocol
Checksum
Source Address
Destination Address

IPv4


IPv6











P4-55.
a.
b.
c.
d.

0000:0000:0000:0000:5555:5555:5555:5555
0000:0000:0000:0000:AAAA:AAAA:AAAA:AAAA
5555:5555:5555:5555:5555:5555:5555:5555
7777:7777:7777:7777:7777:7777:7777:7777

a.
b.
c.
d.

0:FFFF:FFFF::
1234:2346:3456::FFFF
0:1::FFFF:1200:1000
::FFFF:FFFF:24.123.12.6

P4-56.

P4-57.
a. 0000:0000:0000:0000:0000:0000:0000:2222
b. 1111:0000:0000:0000:0000:0000:0000:0000
c. 000B:000A:00CC:0000:0000:0000:1234:000A
P4-58.
a. 0000 :0000:0000:0000:0000:0000:0000:0002
b. 0000:0023:0000:0000:0000:0000:0000:0000
c. 0000:000A:0000:0000:0000:0000:0000:0003


29

P4-59.

One way to solve the problem is to write the binary notation in each case and
keep the block prefix:
a. 1111 1110 1000 0000  0001 0010
b. 1111 1101 0010 0011  0000 0000
c. 0011 0010 0000 0000  0000 0000

P4-60.

Link local
Unique local unicast
Global unicast

Both subnets keep the given global routing prefix. Each subnet adds a 16-bit
subnet identifier. We assume the subnet identifiers start from (0001)16, but
they can also start from (0000)16.
a. The first subnet block is 2000:1234:1423:0001/64.
b. The second subnet block is 2000:1234:1423:0002/64.