Your network uses the18.104.22.168 class B address. You need to support 459 hosts per subnet, while accommodating the maximum number of subnets. Which mask would you use?
To obtain 459 hosts the number of host bits will be 9. This can support a maximum of
510 hosts. To keep 9 bits for hosts means the last bit in the 3rd octet will be 0. This gives
255.255.254.0 as the subnet mask.
Using a subnet mask of 255.255.255.224, which of the IP addresses below can you
assign to the hosts on this subnet? (Select all that apply)
Answer: C, D, E
Since the subnet mask is 255.255.255.224, the number of network hosts that is available
is 30. Every network boundary will be a multiple of 32. This means that every subnet will
be a multiple (0, 32, 64, 96, 128, 160, 192, 224) and the broadcast address for each of
these subnets will be one less this number (31, 63, 95, 127, 159, 191, 223). Therefore, any
IP address that does not end in one of these numbers will be a valid host IP address.
C. Valid Host in subnetwork 2 (22.214.171.124 to 126.96.36.199)
D. Valid Host in subnetwork 1 (188.8.131.52 to 184.108.40.206)
E. Valid Host in subnetwork 2 (192.168.16.64 to 192.168.16.95)
A. This will be the broadcast address for the 220.127.116.11/27 network.
B. This will be the broadcast address for the 18.104.22.168/27 network
F. This will be the network address for the 22.214.171.124/27 network.
Your ISP has assigned you the following IP address and subnet mask:
IP address: 126.96.36.199
Subnet mask: 255.255.255.240
Which of the following addresses can be allocated to hosts on the resulting subnet?
(Select all that apply)
Answer: A, C, D
IP address = 11001000.10001101.00011011.00000000 = 188.8.131.52
Subnet mask = 11111111.11111111.11111111.11110000 = 255.255.255.240
Subnet # = 11001000.10001101.00011011.00000000 = 184.108.40.206
Broadcast = 11001000.10001101.00011011.00001111 = 220.127.116.11
The valid IP address range = 18.104.22.168 - 22.214.171.124
The IP network 126.96.36.199 is subnetted using a /24 mask. How many usable
networks and host addresses can be obtained from this?
A. 1 network with 254 hosts
B. 4 networks with 128 hosts
C. 2 networks with 24 hosts
D. 6 networks with 64 hosts
E. 8 networks with 36 hosts
A subnet with 24 bits on would be 255.255.255.0. Since this is a class C network, this
subnet can have only 1 network and 254 usable hosts.
Given that you have a class B IP address network range, which of the subnet masks
below will allow for 100 subnets with 500 usable host addresses per subnet?
Using the 2n-2 formula for host addresses, 29-2 = 510 host address, so a 9-bit subnet
mask will provide the required number of host addresses. If these 9 bits are used for the
hosts in a class B network, then the remaining 7 bits are used for the number of networks.
Again using the 2n-2 formula, we have 2n-2 = 126 networks that are available.
A. This will provide for only 1 network with 216-2 = 65534 hosts
B. This will provide for 6 networks with 8190 host addresses.
D. This will provide 254 networks and 254 hosts.
E. This will provide 2046 different networks, but each network will have only 30 hosts.
You have a class C network, and you need to design it for 5 usable subnets with each subnet handling a minimum of 18 hosts each. Which of the following network masks should you use?
The default subnet mask for class C network is 255.255.255.0. If one has to create 5
subnets, then 3 bits are required. With 3 bits we can create 6 subnets. The remaining 5
bits are used for Hosts. One can create 30 hosts using 5 bits in host field. This matches
with the requirement.
A, B. This is an illegal subnet mask for a class C network, as the third octet can not be
divided when using a class C network.
C. This is the default subnet mask for a class C network. It provides for one network, with
254 usable host IP addresses.
E. This subnet mask will provide for 14 separate networks with 14 hosts each. This does
not meet the requirement of a minimum of 18 hosts.
The 188.8.131.52 network was subnetted using a /28 subnet mask. How many usable subnets and host addresses per subnet were created as a result of this?
A. 2 networks with 62 hosts
B. 6 networks with 30 hosts
C. 16 networks and 16 hosts
D. 62 networks and 2 hosts
E. 14 networks and 14 hosts
F. None of the above
A class C subnet with a 28 bit mask requires 4 bits for the network address, leaving 4 bits
for host addresses. Using the 2n-2 formula (24-2 in this case) we have 14 host addresses
and 14 network addresses.
A. This would be the result of a /26 network mask
B. This would be the result of a /27 network mask
C. Remember we need to always subtract two for the network and broadcast addresses, so
this answer is incorrect.
D. This would be the result of a /30 network mask.
The 184.108.40.206 network is subnetted using a /26 mask. How many networks and
IP hosts per network exists using this subnet mask?
A. 4 networks with 64 hosts
B. 64 networks and 4 hosts
C. 2 networks and 62 hosts
D. 62 networks and 2 hosts
E. 6 network and 30 hosts
A class C network with a 26 bit mask requires 2 bits for the network address, leaving 6
bits for host addresses. Using the 2n-2 formula (22-2 for the network and 26-2 for hosts)
we have 2 network addresses and 62 host addresses.
A, B. This is not a possible combination. No network mask will provide for 64 usable
hosts, because we must always subtract 2 for the network and broadcast address.
D. This would be the result of a /30 mask.
E. This would be the result of a /27 network mask.
You have a class B network with a 255.255.255.0 mask. Which of the statements
below are true of this network? (Select all valid answers)
A. There are 254 usable subnets.
B. There are 256 usable hosts per subnet.
C. There are 50 usable subnets.
D. There are 254 usable hosts per subnet.
E. There are 24 usable hosts per subnet.
F. There is one usable network.
Answer: A, D
The default subnet mask for Class B is 255.255.0.0. Thus an extra 8 bits have been used
for the network portion, leaving 8 for hosts. The 2n - 2 formula (28 - 2 in this case for
both the network and IP hosts) gives us 254 networks and 254 hosts per network.
B. We must remember to always subtract 2 (one for the network, and one for the
broadcast) so the result is 254, not 256.
C, E. No possible network mask would give us this exact number of subnets or hosts.
F. This would be true if this were a class C network, not a class B.