Subnetting Questions and Answers

Q: Which one of the following CIDR prefixes exactly represents the range of IP addresses 10.12.2.0 to 10.12.3.255?

The prefix is 10.12.2.0/23.The range 10.12.2.0 to 10.12.3.255 holds \(2 \times 256 = 512 = 2^{9}\) addresses, so 9 host bits and a prefix length of \(32 - 9 = 23\). A /23 block must start on an even third octet; 2 is even (binary 0000 0010), so 10.12.2.0/23 spans exactly 10.12.2.0 to 10.12.3.255.10.12.2.0/24 covers only 10.12.2.0 to 10.12.2.255. 10.12.0.0/22 covers 10.12.0.0 to 10.12.3.255, twice the range. 10.12.2.0/22 is not a valid block start, since a /22 must begin at a third octet divisible by 4; it denotes the same 10.12.0.0 to 10.12.3.255 block.

Q: Assuming classbased addressing scheme, in an university has 35 (thirty five) departments and uses a class B address. If use 6 (six)-bit subnet number and 10-bit host number then how many Ethernet segments and maximum hosts in each Ethernet segment are possible ?

The answer is 64, 1022.A class B address has 16 host bits, split here into a 6-bit subnet number and a 10-bit host number:\[ \begin{aligned} \text{subnets} &= 2^{6} = 64 \\ \text{hosts} &= 2^{10} - 2 = 1022 \end{aligned} \]Two host addresses in every subnet are reserved, the all-zeros host part for the network and the all-ones host part for broadcast, so each Ethernet segment can have 1022 hosts. 64 segments are enough for the 35 departments. 1024 forgets the two reserved addresses, while 256 and 254 are the figures for an 8-bit host field.

Q: The network address of 172.16.0.0/19 provides how many subnets and hosts?

Correct answer: 8 subnets, 8,190 hosts each172.16.0.0 is a class B network with a natural /16, so a /19 borrows three bits and leaves thirteen: \[ 2^{3} = 8\ \mathrm{subnets}, \qquad 2^{13}-2 = 8190\ \mathrm{hosts}. \] All eight subnets are usable on any modern IOS, and the answers offering seven subnets come from the obsolete rule that excluded the all-zeros subnet.

Q: To test the IP stack on your local host, which IP address would you ping?

Correct answer: 127.0.0.1127.0.0.1 is the loopback address, and pinging it exercises the local TCP/IP stack without touching any network hardware - a reply proves the software is installed and running. 127.0.0.0 is the network address of the loopback block and 127.0.0.255 a broadcast within it; neither is the host address to test.

Which one of the following CIDR prefixes exactly represents the range of IP addresses 10.12.2.0 to 10.12.3.255?

A.10.12.2.0/23
B.10.12.2.0/24
C.10.12.0.0/22
D.10.12.2.0/22

Exam Year(s)

GATE CS - 2024,

Answer: A

The prefix is 10.12.2.0/23.

The range 10.12.2.0 to 10.12.3.255 holds \(2 \times 256 = 512 = 2^{9}\) addresses, so 9 host bits and a prefix length of \(32 - 9 = 23\). A /23 block must start on an even third octet; 2 is even (binary 0000 0010), so 10.12.2.0/23 spans exactly 10.12.2.0 to 10.12.3.255.

10.12.2.0/24 covers only 10.12.2.0 to 10.12.2.255. 10.12.0.0/22 covers 10.12.0.0 to 10.12.3.255, twice the range. 10.12.2.0/22 is not a valid block start, since a /22 must begin at a third octet divisible by 4; it denotes the same 10.12.0.0 to 10.12.3.255 block.

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Assuming classbased addressing scheme, in an university has 35 (thirty five) departments and uses a class B address. If use 6 (six)-bit subnet number and 10-bit host number then how many Ethernet segments and maximum hosts in each Ethernet segment are possible ?

A.64, 256
B.64, 254
C.64, 1024
D.64, 1022

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The answer is 64, 1022.

A class B address has 16 host bits, split here into a 6-bit subnet number and a 10-bit host number:

\[ \begin{aligned} \text{subnets} &= 2^{6} = 64 \\ \text{hosts} &= 2^{10} - 2 = 1022 \end{aligned} \]

Two host addresses in every subnet are reserved, the all-zeros host part for the network and the all-ones host part for broadcast, so each Ethernet segment can have 1022 hosts. 64 segments are enough for the 35 departments. 1024 forgets the two reserved addresses, while 256 and 254 are the figures for an 8-bit host field.

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The network address of 172.16.0.0/19 provides how many subnets and hosts?

A.7 subnets, 30 hosts each
B.8 subnets, 8,190 hosts each
C.8 subnets, 2,046 hosts each
D.7 subnets, 2,046 hosts each

Answer: B

Correct answer: 8 subnets, 8,190 hosts each

172.16.0.0 is a class B network with a natural /16, so a /19 borrows three bits and leaves thirteen: \[ 2^{3} = 8\ \mathrm{subnets}, \qquad 2^{13}-2 = 8190\ \mathrm{hosts}. \] All eight subnets are usable on any modern IOS, and the answers offering seven subnets come from the obsolete rule that excluded the all-zeros subnet.

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To test the IP stack on your local host, which IP address would you ping?

A.127.0.0.0
B.1.0.0.127
C.127.0.0.1
D.127.0.0.255

Answer: C

Correct answer: 127.0.0.1

127.0.0.1 is the loopback address, and pinging it exercises the local TCP/IP stack without touching any network hardware - a reply proves the software is installed and running. 127.0.0.0 is the network address of the loopback block and 127.0.0.255 a broadcast within it; neither is the host address to test.

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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 host?

A.172.16.112.0
B.172.16.0.0
C.172.16.96.0
D.172.16.255.0

Answer: A

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 128 is the next subnet.

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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

Answer: 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.

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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

Answer: 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.

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Using the following illustration, what would be the IP address of E0 if you were using the eighth subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. The zero subnet should not be considered valid for this question.

A.192.168.10.142
B.192.168.10.66
C.192.168.100.254
D.192.168.10.143

Answer: A

A /28 is a 255.255.255.240 mask. Let's count to the ninth subnet (we need to find the broadcast address of the eighth subnet, so we need to count to the ninth subnet). Starting at 16 (remember, the question stated that we will not use subnet zero, so we start at 16, not 0), 16, 32, 48, 64, 80, 96, 112, 128, 144. The eighth subnet is 128 and the next subnet is 144, so our broadcast address of the 128 subnet is 143. This makes the host range 129-142. 142 is the last valid host.

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A network administrator is connecting hosts A and B directly through their Ethernet interfaces, as shown in the illustration. Ping attempts between the hosts are unsuccessful. What can be done to provide connectivity between the hosts?

1.A crossover cable should be used in place of the straight-through cable.

2.A rollover cable should be used in place of the straight-through cable.

3.The subnet masks should be set to 255.255.255.192.

4.A default gateway needs to be set on each host.

5.The subnet masks should be set to 255.255.255.0.

A.1 only
B.2 only
C.3 and 4 only
D.1 and 5 only
E.2 and 5 only

Answer: D

First, if you have two hosts directly connected, as shown in the graphic, then you need a crossover cable. A straight-through cable won't work. Second, the hosts have different masks, which puts them in different subnets. The easy solution is just to set both masks to 255.255.255.0 (/24).

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10.

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

Answer: 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.

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