1. While opening a TCP connection, the initial sequence number is to be derivedusing a time-of-day (ToD) clock that keeps running even when the host is down.The low order 32 bits of the counter of the ToD clock is to be used for the initialsequence numbers. The clock counter increments once per millisecond. Themaximum packet lifetime is given to be 64s.Which one of the choices given below is closest to the minimum permissible rateat which sequence numbers used for packets of a connection can increase?(A) 0.015/s (B) 0.064/s (C) 0.135/s (D) 0.327/s

ANSWER – (B) 0.064/sEXPLANATION - Sequencer number: Frame from a sending station are numbered sequentially we need to set a limit

2. Let G(x) be the generator polynomial used for CRC checking. What is thecondition that should be satisfied by G(x) to detect odd number of bits in error?(A) G(x) contains more than two terms(B) G(x) does not divide 1+xk, for any k not exceeding the frame length(C) 1+x is a factor of G(x)(D) G(x) has an odd number of terms.

Answer: (C) 1+x is a factor of G(x)

Explanation:

When the polynomial code method is employed the sender and receiver must agree upon a generator polynomial G(x). Both the high and low order bit of the generator must be 1. So 1 + x is a factor of G(x)

Alternatively: For odd number of bits error, the error function will be having factor of (1 + x2 + x4 + ....) which can not be divided by generator (1 + x) as a factor

3. In Ethernet when Manchester encoding is used, the bit rate is:(A) Half the baud rate. (B) Twice the baud rate.(C) Same as the baud rate. (D) None of the above

Answer (A) Half the baud rate

Explanation : In Manchester encoding, the bitrate is half of the baud rate.

4. Which one of the following uses UDP as the transport protocol?(A) HTTP (B) Telnet (C) DNS (D) SMTP

ANSWER – (C) DNS

EXPLANATION :

DNS primarily uses User Datagram Protocol (UDP) on port number 53 to serve requests. DNS queries consist of a single UDP request from the client followed by a single UDP reply from the server.

5. There are n stations in a slotted LAN. Each station attempts to transmit with aprobability p in each time slot. What is the probability that ONLY one stationtransmits in a given time slot?

(A) np(1-p)n-1 (B) (1-p)n-1 (C) p(1-p)n-1 (D) 1-(1-p)n-1

Answer - (A) np(1-p)n-1

Explanation : The probability that a particular station transmits and no body else transmits = p*(1-p)^(n-1)The probability that any station can transmit = n*(probability that a particular station transmits) = n*p*(1-p)^(n-1).

6. In a token ring network the transmission speed is 10^7 bps and the propagation speed is 200 metres/micro second. The 1-bit delay in this network is equivalent to:

(A) 500 metres of cable.(B) 200 metres of cable.(C) 20 metres of cable.(D) 50 metres of cable.

Answer (C) 20 metres of cable.

Explanation : Transmission delay for 1 bit t = 1/(10^7) = 0.1 micro seconds.200 meters can be traveled in 1 micro second. Therefore, in 0.1 micro seconds, 20 meters can be traveled.

7. The address of a class B host is to be split into subnets with a 6-bit subnetnumber. What is the maximum number of subnets and the maximum number ofhosts 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 host

Answer (C) 62 subnets and 1022 hosts

Explanation : 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

8. The message 11001001 is to be transmitted using the CRC polynomial x3 1 toprotect it from errors. The message that should be transmitted is:

(A) 11001001000 (B) 11001001011(C) 11001010 (D) 110010010011

Answer (B) 11001001011

Explanation : The polynomial x^3+1 corresponds to divisor is 1001.

11001001 000 <--- input right padded by 3 bits1001 <--- divisor01011001 000 <---- XOR of the above 2 1001 <--- divisor00010001 000 100100000011 000 10 0100000001 010 1 00100000000 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.

9. 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) P – 1 Q – 4 R – 2 S – 3

Explanation : 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 InternetPPP is a data link layer protocol commonly used in establishing a direct connection between two networking nodes