What is the subnetwork number of a host

What is the subnetwork number of a host

What is the maximum number of IP addresses that can be assigned to hosts on a local
subnet that uses the 255.255.255.224 subnet mask?
A. 14
B. 15
C. 16
D. 30
E. 31
F. 62

 

·         D. A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets, each
with 30 hosts. Does it matter if this mask is used with a Class A, B, or C network
address? Not at all. The number of host bits would never change.

 

You have a network that needs 29 subnets while maximizing the number of host addresses
available on each subnet. How many bits must you borrow from the host field to provide
the correct subnet mask?
A. 2
B. 3
C. 4
D. 5
E. 6
F. 7

 

·         D. A 240 mask is 4 subnet bits and provides 16 subnets, each with 14 hosts. We need
more subnets, so let's add subnet bits. One more subnet bit would be a 248 mask. This
provides 5 subnet bits (32 subnets) with 3 host bits (6 hosts per subnet). This is the
best answer.

 

What is the subnetwork address for a host with the IP address 200.10.5.68/28?
A. 200.10.5.56
B. 200.10.5.32
C. 200.10.5.64
D. 200.10.5.0

 

·         C. This is a pretty simple question. A /28 is 255.255.255.240, which means that
our block size is 16 in the fourth octet. 0, 16, 32, 48, 64, 80, etc. The host is in the
64 subnet.

 

The network address of 172.16.0.0/19 provides how many subnets and hosts?
A. 7 subnets, 30 hosts each
B. 7 subnets, 2,046 hosts each
C. 7 subnets, 8,190 hosts each
D. 8 subnets, 30 hosts each
E. 8 subnets, 2,046 hosts each
F. 8 subnets, 8,190 hosts each

 

·         F. A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only
3 subnet bits, but it provides 13 host bits, or 8 subnets, each with 8,190 hosts.

 

Which two statements describe the IP address 10.16.3.65/23? (Choose two.)
A. The subnet address is 10.16.3.0 255.255.254.0.
B. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
C. The last valid host address in the subnet is 10.16.2.254 255.255.254.0.
D. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.
E. The network is not subnetted.

 

·         B,D The mask 255.255.254.0 (/23) used with a Class A address means that there are
15 subnet bits and 9 host bits. The block size in the third octet is 2 (256 - 254). So this
makes the subnets in the interesting octet 0, 2, 4, 6, etc., all the way to 254. The host
10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the
2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254.

 

 

If a host on a network has the address 172.16.45.14/30, what is the subnetwork this
host belongs to?
A. 172.16.45.0
B. 172.16.45.4
C. 172.16.45.8
D. 172.16.45.12
E. 172.16.45.16

 

·         D. A /30, regardless of the class of address, has a 252 in the fourth octet. This means
we have a block size of 4 and our subnets are 0, 4, 8, 12, 16, etc. Address 14 is obviously
in the 12 subnet.

 

 

Which mask should you use on point-to-point WAN links in order to reduce the waste
of IP addresses?
A. /27
B. /28
C. /29
D. /30
E. /31

 

·         D. A point-to-point link uses only two hosts. A /30, or 255.255.255.252, mask provides
two hosts per subnet.

 

What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
A. 172.16.36.0
B. 172.16.48.0
C. 172.16.64.0
D. 172.16.0.0

 

·         C. A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so
we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet
is 72.0, so the broadcast address of the 64 subnet is 71.255.

 

 

You have an interface on a router with the IP address of 192.168.192.10/29. Including
the router interface, how many hosts can have IP addresses on the LAN attached to the
router interface?
A. 6
B. 8
C. 30
D. 62
E. 126

 

·         A /29 (255.255.255.248), regardless of the class of address, has only 3 host bits.
Six hosts are the maximum number of hosts on this LAN, including the router
interface.

 

 

You need to configure a server that is on the subnet 192.168.19.24/29. The router has
the first available host address. Which of the following should you assign to the server?
A. 192.168.19.0 255.255.255.0
B. 192.168.19.33 255.255.255.240
C. 192.168.19.26 255.255.255.248
D. 192.168.19.31 255.255.255.248
E. 192.168.19.34 255.255.255.240

 

·         C. A /29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets
are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet, and since 32 is the next
subnet, the broadcast address for the 24 subnet is 31. 192.168.19.26 is the only correct
answer.

 

You have an interface on a router with the IP address of 192.168.192.10/29. What is
the broadcast address the hosts will use on this LAN?

·        
A. 192.168.192.15
B. 192.168.192.31
C. 192.168.192.63
D. 192.168.192.127
E. 192.168.192.255

·         A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the
subnets are 0, 8, 16, 24, etc. 10 is in the 8 subnet. The next subnet is 16, so 15 is the

·          

·        
broadcast address.

 

You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which
classful subnet mask would you use?
A. 255.255.255.192
B. 255.255.255.224
C. 255.255.255.240
D. 255.255.255.248

 

·         B. You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides
16 subnets with 14 hosts—this will not work. The mask 255.255.255.224 provides
8 subnets, each with 30 hosts. This is the best answer.

 

If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what
would be the valid subnet address of this interface?
A. 172.16.112.0
B. 172.16.0.0
C. 172.16.96.0
D. 172.16.255.0
E. 172.16.128.0

 

·         A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth
octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and
1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off
or on—which is a value of 0 or 128. The host in the question is in the 0 subnet, which
has a broadcast address of 127 since 112.128 is the next subnet.

 

 

Using the illustration from the previous question, what would be the IP address of S0
if you were using the first subnet? The network ID is 192.168.10.0/28 and you need to
use the last available IP address in the range. Again, the zero subnet should not be considered
valid for this question.
A. 192.168.10.24
B. 192.168.10.62
C. 192.168.10.30
D. 192.168.10.127

 

·         C. A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question
stated not to use subnet zero) and the next subnet is 32, so our broadcast address is 31.
This makes our host range 17-30. 30 is the last valid host.

 

Which configuration command must be in effect to allow the use of 8 subnets if the
Class C subnet mask is 255.255.255.224?
A. Router(config)#ip classless
B. Router(config)#ip version 6
C. Router(config)#no ip classful
D. Router(config)#ip unnumbered
E. Router(config)#ip subnet-zero
F. Router(config)#ip all-nets

 

·         E. A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000)
and provides 8 subnets, each with 30 hosts. However, if the command ip subnet-zero
is not used, then only 6 subnets would be available for use.

 

You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
A. 172.16.17.1 255.255.255.252
B. 172.16.0.1 255.255.240.0
C. 172.16.20.1 255.255.254.0
D. 172.16.16.1 255.255.255.240
E. 172.16.18.255 255.255.252.0
F. 172.16.0.1 255.255.255.0

 

·         E. A Class B network ID with a /22 mask is 255.255.252.0, with a block size of 4 in
the third octet. The network address in the question is in subnet 172.16.16.0 with a
broadcast address of 172.16.19.255. Only option E has the correct subnet mask listed,
and 172.16.18.255 is a valid host.

 

. Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following
can be valid host IDs on the LAN interface attached to the router? (Choose two.)
A. 172.16.0.5
B. 172.16.1.100
C. 172.16.1.198
D. 172.16.2.255
E. 172.16.3.0
F. 172.16.3.255

 

·         D,E The router's IP address on the E0 interface is 172.16.2.1/23, which is
255.255.254.0. This makes the third octet a block size of 2. The router's interface is in
the 2.0 subnet, and the broadcast address is 3.255 because the next subnet is 4.0. The
valid host range is 2.1 through 3.254. The router is using the first valid host address in
the range.

 

. To test the IP stack on your local host, which IP address would you ping?
A. 172.0.0.1
B. 1.0.0.127
C. 127.0.0.1
D. 127.255.255.255
E. 255.255.255.255

 

·         C. To test the local stack on your host, ping the loopback interface of 127.0.0.1.