CCNA Exploration Chapter 6 VLSM and CIDR. Study questions.
6.1
What is classful IP addressing?
An addressing system based on five fixed classes. The class is determined by
the first four bits of the first octet of the IP address. Classes A, B and C are
used for addressing hosts on networks. Class D is reserved for multicasting
and Class E is reserved for experimental use and future use.
What range of addresses are in class A?
0.0.0.0 to 127.255.255.255
This is everything starting with binary 0.
Which class A addresses are reserved for special purposes and therefore
cannot be used for networks and hosts?
0.0.0.0/8 cannot be used because 0.0.0.0 is used to specify a default route.
127.0.0.0/8 cannot be use because 127.0.0.1 is used for loopback testing.
How many usable class A networks exist?
126
What is the subnet mask for class A (in both dotted decimal form and prefix
form)?
255.0.0.0 or /8
How many host addresses does a class A network provide? And how do you
work this out?
There are 24 host bits. 224 = 16,777,216. Take off two for network and
broadcast addresses. 16,777,214 host addresses.
What range of addresses are in class B?
128.0.0.0 to 191.255.255.255
This is everything starting with binary 10.
How many usable class B networks exist? And how do you work it out?
First octet range is 128 to 191. That makes 64 different first octets. Second
octet is also for network addressing. 8 bits. 28 = 256.
Each first octet can have 256 different second octets. 256 x 64 = 16384.
Therefore there are 16384 class B networks.
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What is the subnet mask for class B (in both dotted decimal form and prefix
form)?
255.255.0.0 or /16
How many host addresses does a class B network provide? And how do you
work this out?
There are 16 host bits. 216 = 65536. Take off two for network and broadcast
addresses. 65534 host addresses.
What range of addresses are in class C?
192.0.0.0 to 223.255.255.255
This is everything starting with binary 110.
How many usable class C networks exist? And how do you work it out?
First octet range is 192 to 223. That makes 32 different first octets. Second
and third octets are also for network addressing. 16 bits. 2 16 = 65536.
Each first octet can have 65536 different network addresses.
65536 x 32 = 2,097,152.
Therefore there are 2,097,152 class C networks.
What is the subnet mask for class C (in both dotted decimal form and prefix
form)?
255.255.255.0 or /24
How many host addresses does a class C network provide? And how do you
work this out?
There are 8 host bits. 28 = 256. Take off two for network and broadcast
addresses. 254 host addresses.
What developments have allowed us to carry on using IP version 4 without
completely running out of addresses?
VLSM and CIDR to avoid wasting too many addresses.
NAT and private addresses. LANs can be given private addresses (which may
be duplicated without restriction) for internal use and use NAT in order to
share a few public addresses for external use.
What are the Class D addresses, 224.0.0.0 to 239.255.255.255 used for?
Multicast addressing. A multicast address is used rather than a broadcast
address to communicate with a group of devices, e.g. routers running RIPv2.
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Why did early routing protocols such as RIPv1 not include subnet mask
information in routing updates?
It was not necessary. Classful addressing was in use. The router could find
out the class and therefore the subnet mask by inspecting the first four bits of
the IP address.
An administrator needs to address the following networks:
Site A: 400 hosts
Site B: 350 hosts
Site C: 300 hosts
Site D: 300 hosts
Site E: 250 hosts
And 6 point-to-point links.
Classful subnetting is used, with Class B private address 172.16.0.0.
All subnets must have the same mask.
The subnets should be as small as possible.
How many bits are borrowed?
What is the subnet mask?
How many subnets would this give?
How many hosts on each?
How many addresses are wasted on point-to-point links?
Borrow 7 bits
Subnet mask 255.255.254.0 or /23
128 subnets
510 hosts on each. (512 addresses but 2 used for network and broadcast)
On each point to point link there are 2 hosts, so 508 unused addresses. This
gives 508 x 6 = 3048 wasted addresses on the links.
Why was CIDR introduced?
To give a less wasteful use of public address space. Instead of allocating a
classful fixed size address space of 256 (class C) or 65,534 (class B) to each
organisation, it was possible to be more flexible and allocate blocks based on
any subnet mask. The number of addresses could be matched more closely
to the requirements.
CIDR also allows addresses to be aggregated and summarised so that routing
tables can be smaller.
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What address could be used to summarise the following group of addresses?
194.3.0.0/21
194.3.8.0/22
194.3.12.0/22
194.3.16.0/22
194.3.20.0/22
194.3.0.0/19
What address could be used to summarise the following group of addresses?
170.16.0.0/16
170.17.0.0/17
170.17.128.0/17
170.16.0.0/15
What information do classless routing protocols need to include in their routing
updates that classful protocols do not?
Subnet mask for each network.
Which routing protocols support CIDR and VLSM?
RIPv2, EIGRP, OSPF, IS-IS, BGP.
What is a supernet?
An group of networks that are aggregated and given a summary address.
Subnet 172.16.0.0/16 by borrowing 4 bits.
Network address
and mask
172.16.0.0/20
172.16.16.0/20
172.16.32.0/20
172.16.48.0/20
172.16.64.0/20
172.16.80.0/20
172.16.96.0/20
172.16.112.0 /20
172.16.128.0/20
172.16.144.0/20
172.16.160.0/20
172.16.176.0/20
172.16.192.0/20
172.16.208.0/20
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Host addresses
Broadcast address
172.16.0.1 - 172.16.15.254
172.16.16.1 -172.16.31.254
172.16.32.1 -172.16.47.254
172.16.48.1 -172.16.63.254
172.16.64.1 -172.16.79.254
172.16.80.1 -172.16.95.254
172.16.96.1 -172.16.111.254
172.16.112.1 -172.16.127.254
172.16.128.1 -172.16.143.254
172.16.144.1 -172.16.159.254
172.16.160.1 -172.16.175.254
172.16.176.1 -172.16.191.254
172.16.192.1 -172.16.207.254
172.16.208.1 -172.16.223. 254
172.16.15.255
172.16.31.255
172.16.47.255
172.16.63.255
172.16.79.255
172.16.95.255
172.16.111.255
172.16.127.255
172.16.143.255
172.16.159.255
172.16.175.255
172.16.191.255
172.16.207.255
172.16.223. 255
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172.16.224.0/20
172.16.240.0/20
172.16.224.1 -172.16.239.254
172.16.240.1 -172.16.255.254
172.16.239.255
172.16.255.255
Take the fourth subnet from the previous question and subnet it again by
borrowing an additional two bits.
Network address
and mask
172.16.48.0/22
172.16.52.0/22
172.16.56.0/22
172.16.60.0/22
Host addresses
Broadcast address
172.16.48.1 -172.16.51.254
172.16.52.1 -172.16.55.254
172.16.56.1 -172.16.59.254
172.16.60.1 -172.16.63.254
172.16.51.255
172.16.55.255
172.16.59.255
172.16.63.255
A router has a route to 172.17.0.0/16 via s0/0. It also has a summary route to
172.16.0.0/15 via fa0/0. A packet arrives addressed to 172.17.3.2. Which
route will it take?
It will go on the route to 172.17.0.0/16 via s0/0. This gives a more specific
match.
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