IP Addressing

Question 1

In the IPv4 addressing format, the number of networks allowed under Class C addresses is

• A. 2^14
• B. 2^7
• C. 2^21
• D. 2^24

Answer: (C)

IP Addressing IP Addressing
Discuss it

In class C, 8 bits are reserved for Host Id and 24 bits are reserved for Network Id. Out of these 24 Network Id bits, the leading 3 bits are fixed as 110. So remaining 21 bits can be used for different networks. See this for more details.

Question 2

In the IPv4 addressing format, the number of networks allowed under Class C addresses is

• A. 2^14
• B. 2^7
• C. 2^21
• D. 2^24

Answer: (C)

IP Addressing IP Addressing
Discuss it

In class C, 8 bits are reserved for Host Id and 24 bits are reserved for Network Id. Out of these 24 Network Id bits, the leading 3 bits are fixed as 110. So remaining 21 bits can be used for different networks. See this for more details.

Question 3

A n Internet Service Provider (ISP) has the following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20. The ISP wants to give half of this chunk of addresses to Organization A, and a quarter to Organization B, while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?

• A. 245.248.136.0/21 and 245.248.128.0/22
• B. 245.248.128.0/21 and 245.248.128.0/22
• C. 245.248.132.0/22 and 245.248.132.0/21
• D. 245.248.136.0/24 and 245.248.132.0/21

Answer: (A)

GATE CS 2012 IP Addressing
Discuss it

See Question 4 of http://www.geeksforgeeks.org/computer-networks-set-1/

Question 4

Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same netmask N. Which of the values of N given below should not be used if A and B should belong to the same network?

• A. 255.255.255.0
• B. 255.255.255.128
• C. 255.255.255.192
• D. 255.255.255.224

Answer: (D)

GATE CS 2010 IP Addressing
Discuss it

See question 2 http://www.geeksforgeeks.org/computer-networks-set-4-2/

Question 5

If a class B network on the Internet has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet?

• A. 1022
• B. 1023
• C. 2046
• D. 2047

Answer: (C)

GATE CS 2008 IP Addressing
Discuss it

See Question 3 of http://www.geeksforgeeks.org/computer-networks-set-7/

Question 6

The address of a class B host is to be split into subnets with a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet?

• A. 62 subnets and 262142 hosts.
• B. 64 subnets and 262142 hosts.
• C. 62 subnets and 1022 hosts.
• D. 64 subnets and 1024 hosts.

Answer: (C)

GATE-CS-2007 IP Addressing
Discuss it

See question 1 of http://www.geeksforgeeks.org/computer-networks-set-10/

Question 7

The subnet mask for a particular network is 255.255.31.0. Which of the following pairs of IP addresses could belong to this network?

• A. 172.57.88.62 and 172.56.87.233
• B. 10.35.28.2 and 10.35.29.4
• C. 191.203.31.87 and 191.234.31.88
• D. 128.8.129.43 and 128.8.161.55

Answer: (D)

GATE-CS-2003 IP Addressing
Discuss it

  Background required - Subnetting and SubNet Mask Answer - D Explanation - Suppose we have a host A with IP Address IP_A and Subnet mask of the subnet of which A is a part is M_s. Now when A wants to send packet to a host B with an IP address say IP_B, then A will first AND the subnet-Mask M_s with IP_A to find out the subnet-id ID_N of the subnet in order to identify whether IP_B belongs to the same network. Now if B also belongs to the same network, then IP_B BITWISE-AND M_s must be equal to ID_N which A has calculated. If IP_A and IP_B belongs to same network, we must have

	IPA BITWISE-AND Ms == IPBBITWISE-AND Ms

Out of the 4 options, only option D matches.

128.8.129.43 BITWISE-AND 255.255.31.0 = 128.8.1.0
128.8.161.55 BITWISE-AND 255.255.31.0 = 128.8.1.0

You can check rest of the options. This explanation has been provided by Pranjul Ahuja.

Question 8

Consider the following routing table at an IP router For each IP address in Group-I identify the correct choice of the next hop from Group-II using the entries from the routing table above.

   List-I               List-II
A. 128.96.171.92     1. Interface 0
B. 128.96.167.151    2. Interface 1
C. 128.96.163.121    3. R2
D. 128.96.165.121    4. R3
                     5. R4
Codes:
     A B C D
(a)  1 3 5 4
(b)  1 4 2 5
(c)  2 3 4 5
(d)  2 3 5 4

• A. a
• B. b
• C. c
• D. d

Answer: (A)

GATE-CS-2015 (Set 2) IP Addressing 50 Computer Networks MCQs with Answers
Discuss it

Taking the 1st IP Address: 128.96.171.92

Bitwise AND between 128.96.171.92 and 255.255.254.0 we get the subnet ID as follows:25525511111110012896101010119212896101010100

∴ Subnet ID = 128.96.170.0
∴ 128.96.171.92 will forward to interface 0

Taking the 2nd IP Address: 128.96.167.151

Bitwise AND between 128.96.167.151 and 255.255.254.0 we get, 255255111111100128961010011115112896101001100

∴ Subnet ID = 128.96.166.0
∴ 128.96.167.151 will forward to interface R2

Taking the 3rd IP Address: 128.96.163.151

Bitwise AND between 128.96.167.151 and 255.255.254.0 we get, 255255111111100128961010001115112896101000100

∴ Subnet ID = 128.96.162.0 (Doesn’t match with any given interface)

Now, Bitwise AND between 128.96.167.151 and 255.255.252.0 we get, 255255111111000128961010001115112896101000000

∴ Subnet ID = 128.96.160.0 (Doesn’t match with any given interface)
∴ 128.96.163.151 will forward to default interface R4

Taking the last IP Address: 128.96.164.121

Bitwise AND between 128.96.164.121 and 255.255.254.0 we get, 255255111111100128961010010012112896101001000

∴ Subnet ID = 128.96.164.0
∴ 128.96.167.151 will forward to interface R3

∴ Option (A) is the correct answer.

Question 9

Which of the following IP address can be used in WAN?

• A. 10.0.0.1
• B. 172.16.0.10
• C. 15.1.5.6
• D. None

Answer: (C)

IP Addressing 50 Computer Networks MCQs with Answers
Discuss it

10.0.0.1  and 172.16.0.10 are private IP address. Thus cannot be used in WAN.  15.1.5.6 is a public IP address.

Question 10

Which of the following can be used as both Source and Destination IP?

• A. 198.168.1.255
• B. 10.0.0.1
• C. 127.0.0.1
• D. 255.255.255.255

Answer: (B)

IP Addressing
Discuss it

192.168.1.255 is a Direct Broadcast Address (DBA),  DBA  can't be used as Source Address. 127.0.0.1 Loopback address again can't be used as the source address. 255.255.255.255 is Limited Broadcast Address, which cannot be used as DBA. 10.0.0.1 can be both used as both source and destination IP address.