What is the maximum number of host IP addresses that can exist in a class B network group of answer choices?

Following questions have been asked in GATE CS 2007 exam. 

1) 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) 
Maximum number of subnets = 2^6-2 =62. 
Note that 2 is subtracted from 2^6. The RFC 950 specification reserves the subnet values consisting of all zeros (see above) and all ones (broadcast), reducing the number of available subnets by two. 

Maximum number of hosts is 2^10-2 = 1022. 
2 is subtracted for Number of hosts is also. The address with all bits as 1 is reserved as broadcast address and address with all host id bits as 0 is used as network address of subnet. 
In general, the number of addresses usable for addressing specific hosts in each network is always 2^N – 2 where N is the number of bits for host id. 
See this for details 

2) The message 11001001 is to be transmitted using the CRC polynomial x^3 + 1 to protect it from errors. The message that should be transmitted is: 
(A) 11001001000 
(B) 11001001011 
(C) 11001010 
(D) 110010010011 

Answer (B) 
The polynomial x^3+1 corresponds to divisor is 1001. 
 

11001001 000 <--- input right padded by 3 bits 1001 <--- divisor 01011001 000 <---- XOR of the above 2 1001 <--- divisor 00010001 000 1001 00000011 000 10 01 00000001 010 1 001 00000000 011 <------- remainder (3 bits)

See this for division process. 
After dividing the given message 11001001 by 1001, we get the remainder as 011 which is the CRC. The transmitted data is, message + CRC which is 11001001 011. 

3) The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is: 
 

Answer (C) 
 

Distance between stations = L KM Propagation delay per KM = t seconds Total propagation delay = Lt seconds Frame size = k bits Channel capacity = R bits/second Transmission Time = k/R Let n be the window size. UtiliZation = n/(1+2a) where a = Propagation time / transmission time = n/[1 + 2LtR/k] = nk/(2LtR+k) For maximum utilization: nk = 2LtR + k Therefore, n = (2LtR+k)/k Number of bits needed for n frames is Logn.

See this for details. 

4) Match the following: 

(P) SMTP (1) Application layer (Q) BGP (2) Transport layer (R) TCP (3) Data link layer (S) PPP (4) Network layer (5) Physical layer

(A) P – 2 Q – 1 R – 3 S – 5 
(B) P – 1 Q – 4 R – 2 S – 3 
(C) P – 1 Q – 4 R – 2 S – 5 
(D) P – 2 Q – 4 R – 1 S – 3 

Answer (B) 
SMTP is an application layer protocol used for e-mail transmission. 
TCP is a core transport layer protocol. 
BGP is a network layer protocol backing the core routing decisions on the Internet 
PPP is a data link layer protocol commonly used in establishing a direct connection between two networking nodes. 

Please see GATE Corner for all previous year paper/solutions/explanations, syllabus, important dates, notes, etc. 

Please write comments if you find any of the answers/explanations incorrect, or you want to share more information about the topics discussed above
 

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A B and C Classes of Networks

Internet addresses are allocated by the InterNIC (//www.internic.net ), the organization that administers the Internet. These IP addresses are divided into classes. The most common of these are classes A, B, and C. Classes D and E exist, but are not generally used by end users. Each of the address classes has a different default subnet mask. You can identify the class of an IP address by looking at its first octet. Following are the ranges of Class A, B, and C Internet addresses, each with an example address:

  • Class A networks use a default subnet mask of 255.0.0.0 and have 0-127 as their first octet. The address 10.52.36.11 is a class A address. Its first octet is 10, which is between 1 and 126, inclusive.
  • Class B networks use a default subnet mask of 255.255.0.0 and have 128-191 as their first octet. The address 172.16.52.63 is a class B address. Its first octet is 172, which is between 128 and 191, inclusive.
  • Class C networks use a default subnet mask of 255.255.255.0 and have 192-223 as their first octet. The address 192.168.123.132 is a class C address. Its first octet is 192, which is between 192 and 223, inclusive.

In some scenarios, the default subnet mask values do not fit the needs of the organization, because of the physical topology of the network, or because the numbers of networks (or hosts) do not fit within the default subnet mask restrictions. The next section explains how networks can be divided using subnet masks.

Class A Network (/ 8 Prefixes)

This network is 8-bit network prefix. Its highest bit is set to 0, and contains a 7-bit network number and a 24-bit host number.

A maximum of 126, which is (2 7 -2,) networks can be defined; two is subtracted because all an (0 and 1) subnet cannot be used in certain routers using RIP-1 Protocol. Each network supports a maximum of 16,777,214 (2 24 -2) hosts per network. You must subtract two because the base network represents host “0”, and the last host on the network is actually used for 1s (“broadcast”) and may not be assigned to any host.

The class A network address block contains 2 31 power (2,147,483,648) individual addresses. The IPv4 address space contains a maximum of 2 32 power (4,294,967,296) addresses, which mean that a class A network address space is 50% of the total IPv4 unicast, address space.

Class B Networks (/16 Prefixes)

This network is a 16-bit network prefix; its highest bit order is set to 1-0. It is a 14-bit network number with a 16-bit host number.

This class defines 16,384 (2 14 ) /16 networks, and supports a maximum of 65,534 (2 16 -2) hosts per network. Class B /16 block address is (1,073,741,824) = 2 30; therefore it represent 25% of the total IPV4.

Class C Networks (/24 Prefixes)

This is a 24-bit network prefix; it has a 3 bit set to the highest order 1-1-0. It is a 21-bit network number with 8-bit host number.

This class defines a maximum of 2,097,152 (2 21 ) /24 networks. And each network supports up to 254 (2 8 -2) hosts. The entire class C network represents 2 29 (536,870,912) addresses; therefore it is only 12.5 % of the total IPv4.

Other Networks

There are two other networks that are not commonly used, class D and Class E.
Class D has its highest bit order set to 1-1-1-0 it is used to support multicasting.
Class E has its highest bit order set to 1-1-1-1 which is reserved for experimental use.

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What is the maximum number of usable hosts with a class B network address?

Class B Networks (/16 Prefixes) It is a 14-bit network number with a 16-bit host number. This class defines 16,384 (2 14 ) /16 networks, and supports a maximum of 65,534 (2 16 -2) hosts per network. Class B /16 block address is (1,073,741,824) = 2 30; therefore it represent 25% of the total IPV4.

What is the maximum number of host IP addresses?

i.e 2048-2 = 2046 are the maximum number of host the given range of IP address can have. Show activity on this post.

What is the maximum number of hosts per subnet in Class B?

The power of 2 minus 2 =maximum number of host per subnet Which would leave 11 bits for host addressing, hence 2 ^ 11, or 2048.

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