Question 1
A TCP server application is programmed to listen on port number P on host S. A TCP client is connected to the TCP server over the network. Consider that while the TCP connection was active, the server machine S crashed and rebooted. Assume that the client does not use the TCP keepalive timer. Which of the following behaviours is/are possible?
If the client was waiting to receive a packet, it may wait indefinitely.
If the client sends a packet after the server reboot, it will receive a RST segment.
The TCP server application on S can listen on P after reboot.
If the client sends a packet after the server reboot, it will receive a FIN segment.
Question 1 Explanation: 
  1. True
    Since broken connections can only be detected by sending data, the receiving side will wait forever. This scenario is called a “half-open connection” because one side realizes the connection was lost but the other side believes it is still active.
  2. True
    The situation resolves itself when client tries to send data to server over the dead connection, and server replies with an RST packet (not FIN).
  3. True
    Yes, a TCP Server can listen to the same port number even after reboot.  For example, the SMTP service application usually listens on TCP port 25 for incoming requests. So, even after reboot the port 25 is assigned to SMTP.
  4. False
    The situation resolves itself when client tries to send data to server over the dead connection, and server replies with an RST packet (not FIN), causing client to finally to close the connection forcibly.
    FIN is used to close TCP connections gracefully in each direction (normal close of connection), while TCP RST is used in a scenario where TCP connections cannot recover from errors and the connection needs to reset forcibly.
Question 2
Consider the sliding window flow-control protocol operating between a sender and a receiver over a full-duplex error-free link. Assume the following:
  • The time taken for processing the data frame by the receiver is negligible.
  • The time taken for processing the acknowledgement frame by the sender is negligible.
  • The sender has an infinite number of frames available for transmission.
  • The size of the data frame is 2,000 bits and the size of the acknowledgment frame is 10 bits.
  • The link data rate in each direction is 1 Mbps (=106bits per second).
  • One way propagation delay of the link is 100 milliseconds.
The minimum value of the sender’s window size in terms of the number of frames, (rounded to the nearest integer) needed to achieve a link utilization of 50% is ______.
Question 2 Explanation: 

Tt(packet) = L / B.W => 2000 bits / 10^6 bps = 2  x 10^-3 sec = 2 millisec

Tt(Ack) = L / B.W. => 10 bits / 10^6 bps = 10^-5 sec = 10^-2 millisec = 0.01 millisec

Tp = 100 millisec

Total time = Tt(packet) + 2 x Tp + Tt(Ack)

=> 2 + 2 x 100 + 0.01 = 202.01 millisec

Efficiency = 50 % = ½

Efficiency = Useful Time /  Total time

½ = n x Tt / Total time 


=> 2 x n x Tt =  Total time

=>2 x n x 2 = 202.01  

=> n = 202.01 / 4 => 50.50


For minimum, we have to take ceil, Hence size of window = 51

Question 3
Consider two hosts P and Q connected through a router R. The maximum transfer unit (MTU) value of the link between P and R is 1500 bytes, and between R and Q is 820 bytes. A TCP segment of size 1400 bytes was transferred from P to Q through R, with IP identification value as 0x1234. Assume that the IP header size is 20 bytes. Further, the packet is allowed to be fragmented, i.e., Don’t Fragment (DF) flag in the IP header is not set by P. Which of the following statements is/are correct?
If the second fragment is lost, R will resend the fragment with the IP identification value 0x1234.
If the second fragment is lost, P is required to resend the whole TCP segment.
Two fragments are created at R and the IP datagram size carrying the second fragment is 620 bytes.
TCP destination port can be determined by analysing only the second fragment.
Question 3 Explanation: 
Question 4
Consider the following two statements.
S1: Destination MAC address of an ARP reply is a broadcast address.
S2: Destination MAC address of an ARP request is a broadcast address.
Which of the following choices is correct?
Both S1and S2are true.
S1is true and S2is false.
S1is false and S2is true.
Both S1and S2are false.
Question 4 Explanation: 
ARP request is Broadcasting. ARP reply is unicasting.
Question 5

What is the distance of the following code 000000, 010101, 000111, 011001, 111111?

Question 5 Explanation: 
Distance = Minimum hamming distance = 2
010101 ⊕ 011001 = 001100
Question 6

Assume that you have made a request for a web page through your web browser to a web server. Initially the browser cache is empty. Further, the browser is configured to send HTTP requests in non-persistent mode. The web page contains text and five very small images. The minimum number of TCP connections required to display the web page completely in your browser is ______.

Question 6 Explanation: 
In non-persistent HTTP connection for every object, there is a TCP connection established. Therefore, 1 TCP connection for text and 5 TCP connections for images required.
Hence, 1 Text + 5 Image = 6 Objects
Question 7

Consider the following statements about the functionality of an IP based router.

    I. A router does not modify the IP packets during forwarding.
    II. It is not necessary for a router to implement any routing protocol.
    III. A router should reassemble IP fragments if the MTU of the outgoing link is larger than the size of the incoming IP packet.

Which of the above statements is/are TRUE?

I and II only
II only
I only
II and III only
Question 7 Explanation: 
I: The packet contains Header and data. The router modifies the header details like TTL.
II: Is True.
III: Reassemble is not necessary at the router.
Question 8

An organization requires a range of IP addresses to assign one to each of its 1500 computers. The organization has approached an Internet Service Provider (ISP) for this task. The ISP uses CIDR and serves the requests from the available IP address space The ISP wants to assign an address space to the organization which will minimize the number of routing entries in the ISP’s router using route aggregation. Which of the following address spaces are potential candidates from which the ISP can allot any one to the organization?

I and II only
III and IV only
II and III only
I and IV only
Question 8 Explanation: 
Given CIDR IP is and for HID 32 - 17 = 15 bits can be used.
And to Assign an IP address for 1500 computer, we require 11 bit from HID part.
So NID + SID = 17 + 4 = 21 bits and HID = 11 bits
202.61.0 0000 000.00000000
So, from the given option, possible IP Address is
I. 84 -> 0 1010 100 (Because in HID bit 1 is not possible)
II. 104 -> 0 1101 000
III. 64 -> 0 1000 000
IV. 144 -> 1 0010 000 (Because in NID bit 1 is not possible )
Question 9

Consider a TCP connection between a client and a server with the following specifications: the round trip time is 6 ms, the size of the receiver advertised window is 50 KB, slow start threshold at the client is 32 KB, and the maximum segment size is 2 KB. The connection is established at time t=0. Assume that there are no timeouts and errors during transmission. Then the size of the congestion window (in KB) at time t+60 ms after all acknowledgements are processed is ______.

Question 9 Explanation: 
Threshold = 32 Kb, MSS = 2KB, RTT = 6ms
Here, t + 60 is nothing but at the 10 RTT (60/6 = 10), but here it’s asking after all acknowledgement are processed it means after the 10th RTT, i.e., at the 11RTT.
1st transmission: 2 KB
2nd transmission: 4 KB
3rd transmission: 8 KB
4th transmission: 16 KB
5th transmission: 32 KB (Threshold reached)
6th transmission: 34 KB
7th transmission: 36 KB
8th transmission: 38 KB
9th transmission: 40 KB
10th transmission: 42 KB
At the completion of 10th transmission RTT = 10*6 = 60 ms
For the 11th transmission, The congestion window size is 44 KB.
Question 10

The protocol data unit (PDU) for the application layer in the Internet stack is

Question 10 Explanation: 
The PDU for Data Link layer, Network layer, Transport layer and Application layer are frame, datagram, segment and message respectively.
Question 11

Which of the following transport layer protocols is used to support electronic mail?

Question 11 Explanation: 
TCP is used in transport layer to carry out mail which is initiated by application layer protocol SMTP.
Question 12

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

Question 12 Explanation: 
For class C address, size of network field is 24 bits. But first 3 bits are fixed as 110; hence total number of networks possible is 221.
Question 13

An Internet Service Provider (ISP) has the following chunk of CIDR-based IP addresses available with it: 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 and
B and
C and
D and
Question 13 Explanation: 
Question 14

Consider a source computer (S) transmitting a file of size 106 bits to a destination computer (D) over a network of two routers (R1 and R2) and three links (L1, L2 and L3). L1 connects S to R1; L2 connects R1 to R2; and L3 connects R2 to D. Let each link be of length 100 km. Assume signals travel over each link at a speed of 108 meters per second. Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D?

1005 ms
1010 ms
3000 ms
3003 ms
Question 14 Explanation: 

Propagation delay = (Distance) / (Velocity) = 3*105/108 = 3ms
Total transmission delay for 1 packet = 3 * L / B = 3*(1000/106) = 3ms. Because at source and 2 routers, we need to transmit the bits.
The first packet will reach destination = Tt + Tp = 6ms.
While the first packet was reaching to D, other packets must have been processing in parallel. So D will receive remaining packets 1 packet per 1 ms from R2. So remaining 999 packets will take 999 ms.
And total time will be 999 + 6 = 1005 ms
Question 15

Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

14 MSS
12 MSS
Question 15 Explanation: 
Given initial threshold = 8
Time = 1 during 1st trans. , window size = 2 (Slow start),
Time = 2 congestion window size = 4 (double the no. of ack.)
Time = 3 congestion window = 8
Time = 4 congestion window size = 9, after threshold, increase by one additive increase.
Time = 5 transmit 10 MSS, but time out occur congestion window size = 10
Hence threshold = (congestion window size)/2 = 10/2 = 5
Time = 6 transmit 2(since in the question, they are saying ss is starting from 2)
Time = 7 transmit 4
Time = 8 transmit 5
Time = 9 transmit 6
Time = 10 transmit 7
Question 16

Which of the following assertions is FALSE about the Internet Protocol (IP)?

It is possible for a computer to have multiple IP addresses
IP packets from the same source to the same destination can take different routes in the network
IP ensures that a packet is discarded if it is unable to reach its destination within a given number of hops
The packet source cannot set the route of an outgoing packets; the route is determined only by the routing tables in the routers on the way
Question 16 Explanation: 
Because in strict source routing or loose source routing path is set by the source not by router and main task of router is to check outgoing path with the help of forwarding table inside it.
Question 17

Which of the following functionalities must be implemented by a transport protocol over and above the network protocol?

Recovery from packet losses
Detection of duplicate packets
Packet delivery in the correct order
End to end connectivity
Question 17 Explanation: 
End to end connectivity is the required functionality provided by Transport protocol.
Question 18

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

A and
B and
C and
D and
Question 18 Explanation: 
To find whether hosts belong to same network or not , we have to find their net id, if net id is same then hosts belong to same network and net id can be find by ANDing subnet mask and IP address. (Bitwise AND) = (Bitwise AND) =
Question 19

A 2 km long broadcast LAN has 107 bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2×108 m/s. What is the minimum packet size that can be used on this network?

50 bytes
100 bytes
200 bytes
None of the above
Question 19 Explanation: 
Minimum packet size for a CSMA/CD LAN is the frame which cover whole RTT(round trip time). i.e. Tt = 2Tp
d= 2 km = 2 x 103 m, v = 2 x 108 m/s, B= 107
Tp = d / v = 2 x 103 /(2 x 108 ) seconds = 10-5 seconds
Let L bits be minimum size of frame, then Tt = t L / B = L / 107 seconds
Now, Tt = 2Tp
L/107 = 2 x 10-5 = 200 bits = (200 / 8) bytes = 25 bytes
Question 20

Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 µs. Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 µs. What is the maximum achievable throughput in this communication?

7.69 × 106 bps
11.11 × 106 bps
12.33 × 106 bps
15.00 × 106 bps
Question 20 Explanation: 
Given, Tt = 50 μs, Tp = 200 μs, L = 1000 bytes, N = 5,
Transmission rate , Tt = L / B.W
Therefore, B.W. = L / Tt = 1000 bytes/ 50 μs = 8000 bits / 50 μs=160 Mbps
Efficiency = N / 1 + 2a, where a = Tp / Tt
Efficiency = 5 * 50 / (50+400) = 250/450 = 5/9
Maximum achievable throughput = Efficiency * B.W = (5/9)*160 Mbps = 88.88 Mbps = = 11.11 x 106 bytes per second
*Actual option should be in bytes per second.
Question 21

A simple and reliable data transfer can be accomplished by using the ‘handshake protocol’. It accomplishes reliable data transfer because for every data item sent by the transmitter __________.

in this case receiver has to respond that receiver can be able to receive the data item.
Question 22

Start and stop bits do not contain an ‘information’ but are used in serial communication for

Error detection
Error correction
Slowing down the communications
Question 22 Explanation: 
The start and stop bits are used to synchronize the serial receivers.
Question 23

Match the pairs in the following questions by writing the corresponding letters only.

(A) IEEE 488    (P) specifies the interface for connecting a single device
(B) IEEE 796    (Q) specifies the bus standard for connecting a computer to 
                    other devices including CPU’s
(C) IEEE 696    (R) specifies the standard for an instrumentation bus
(D) RS232-C     (S) specifies the bus standard for the “backplane” bus called  
Out of syllabus.
Question 24
Consider the three-way handshake mechanism followed during TCP connection established between hosts P and Q. Let X and Y be two random 32-bit starting sequence numbers chosen by P and Q respectively. Suppose P sends a TCP connection request message to Q with a TCP segment having SYN bit = 1, SEQ number = X, and ACK bit = 0. Suppose Q accepts the connection request. Which one of the following choices represents the information present in the TCP segment header that is sent by Q to P?
SYN bit = 1, SEQ number = X+1, ACK bit = 0, ACK number = Y, FIN bit = 0
SYN bit = 0, SEQ number = X+1, ACK bit = 0, ACK number = Y, FIN bit = 1
SYN bit = 1, SEQ number = Y, ACK bit = 1, ACK number = X+1, FIN bit = 0
SYN bit = 1, SEQ number = Y, ACK bit = 1, ACK number = X, FIN bit = 0
Question 24 Explanation: 


Q will send the SYN bit = 1 to the connection establishment.

Q Seq number will be Y different from X

ACK bit = 1 because sending the ACK

ACK number = X+1 (Next seq number id)

FIN bit = 0 (Because establishing the connection) 

Question 25
Consider the cyclic redundancy check (CRC) based error detecting scheme having the generator polynomial X3+X+1. Suppose the message m4m3m2m1m0=11000 is to be transmitted. Check bits c2c1c0are appended at the end of the message by the transmitter using the above CRC scheme. The transmitted bit string is denoted by m4m3m2m1m0c2c1c0. The value of the checkbit sequence c2c1c0is
Question 25 Explanation: 
Question 26
Consider a computer network using the distance vector routing algorithm in its network layer. The partial topology of the network is as shown below.   The objective is to find the shortest-cost path from the router R to routers P and Q. Assume that R does not initially know the shortest routes to P and Q. Assume that R has three neighbouring routers denoted as X, Y and Z. During one iteration, R measures its distance to its neighbours X, Y and Z as 3, 2 and 5, respectively. Router R gets routing vectors from its neighbours that indicate that the distance to router P from routers X, Y and Z are 7, 6 and 5, respectively. The routing vector also indicates that the distanceto router Q from routers X, Y and Z are 4, 6 and 8, respectively. Which of the following statement(s) is/are correct with respect to the new routing table of R, after updation during this iteration
The next hop router for a packet from R to P is Y.
The distance from R to Q will be stored as 7.
The next hop router for a packet from R to Q is Z.
The distance from R to P will be stored as 10.
Question 26 Explanation: 

Given R gets the distance vector (3,2,5)

After the one iteration distance vector from X to P, Y to P,  and Z to P is (7, 6, 5) respectively

The distance vector from R to P via X Y Z is (3+7, 2+6, 5+5) =(10, 8, 10)

So Take minimum distance from R to P which is 8 via Y

After the iteration distance vector from X to Q, Y to Q, Z to Q is ( 4, 6, 8) respectively

The distance vector from R to Q via X Y Z is (3+4, 2+6, 5+8) = (7, 8 13)

So Take minimum distance from R to Q  which is 7 via X.

Question 27
Consider a network using the pure ALOHA medium access control protocol, where each frame is of length 1,000 bits. The channel transmission rate is 1 Mbps (= 106bits per second). The aggregate number of transmissions across all the nodes (including new frame transmissions and retransmitted frames due to collisions) is modelled as a Poisson process with a rate of 1,000 frames per second. Throughput is defined as the average number of frames successfully transmitted per second. The throughput of the network (rounded to the nearest integer) is _________. 
Question 27 Explanation: 

1 frames takes = Tt = L / B.w. => 1000 / 10^6 = 1 millisec

1000 frame Tt = 1000 x 1 millisec = 1 sec

In 1 sec, 1000 frames sends, which is 1 millisec per frame.

So, G = 1


Efficiency of Pure Aloha (η) = G x e-2G

where G = Number of requests per time slot willing to transmit.

e = Mathematical constant approximately equal to 2.718

So, η = 1 x 2.718(-2 x 1) = 0.1353

Therefore, In 1 sec1000 frames = 0.1353 x 1000 = 135.3(closest integer) =>135


Throughput =>  135

Question 28
Consider the string abbccddeee. Each letter in the string must be assigned a binary code satisfying the following properties:
  1. For any two letters, the code assigned to one letter must not be a prefix of the code assigned to the other letter.
  2. For any two letters of the same frequency, the letter which occurs earlier in the dictionary order is assigned a code whose length is at most the length of the code assigned to the other letter.
Among the set of all binary code assignments which satisfy the above two properties, what is the minimum length of the encoded string?
Question 28 Explanation: 

Input String : abbccddeee

The character frequencies are













Binary Code






Question 29

Which one of the following statements is FALSE?

Packet switching leads to better utilization of bandwidth resources than circuit switching.
Packet switching results in less variation in delay than circuit switching.
Packet switching requires more per packet processing than circuit switching.
Packet switching can lead to reordering unlike in circuit switching.
Question 30

Which one of the following statements is FALSE?

TCP guarantees a minimum communication rate
TCP ensures in-order delivery
TCP reacts to congestion by reducing sender window size
TCP employs retransmission to compensate for packet loss
Question 30 Explanation: 
Option B:
Sequence numbers can allow receivers to discard duplicate packets and properly sequence reordered packets.
Option C:
If the congestion is deleted, the transmitter decreases the transmission rate by a multiplicative factor.
Option D:
Acknowledgement allows the sender to determine when to retransmit lost packets.
Question 31

Which one of the following statements is FALSE?

HTTP runs over TCP
HTTP describes the structure of web pages
HTTP allows information to be stored in a URL
HTTP can be used to test the validity of a hypertext link
Question 31 Explanation: 
Note: Out of syllabus.
Question 32

A sender is employing public key cryptography to send a secret message to a receiver. Which one of the following statements is TRUE?

Sender encrypts using receiver’s public key
Sender encrypts using his own public key
Receiver decrypts using sender’s public key
Receiver decrypts using his own public key
Question 32 Explanation: 
Sender can encrypts using the receiver public key and receiver decrypts it using his own private key.
Question 33

A subnet has been assigned a subnet mask of What is the maximum number of hosts that can belong to this subnet?

Question 33 Explanation: 
Maximum no. of hosts = 2(no. of bits in HID) - 2
= 26- 2
= 64 - 2
= 62
Question 34

A host is connected to a Department network which is part of a University network. The University network, in turn, is part of the Internet. The largest network in which the Ethernet address of the host is unique is:

the subnet to which the host belongs
the Department network
the University network
the Internet
Question 34 Explanation: 
The answer is option (D), in a specified LAN technology - Ethernet is mentioned here. So, MAC addresses will be specifically taken as physical address which is unique in the entire world.
Question 35

In TCP, a unique sequence number is assigned to each

Question 35 Explanation: 
In TCP, a unique sequence number is assigned to each byte.
Question 36

Which of the following objects can be used in expressions and scriplets in JSP (Java Server Pages) without explicitly declaring them?

session and request only
request and response only
response and session only
session, request and response
Question 36 Explanation: 
Note: Out of syllabus.
Question 37

Consider the following statements:
I. telnet, ftp and http are application layer protocols.
II.l EJB (Enterprise Java Beans) components can be deployed in a J2EE (Java2 Enterprise Edition) application server.
III. If two languages conform to the Common Language Specification (CLS) of the Microsoft.NET framework, then a class defined in any one of them may be inherited in the other.
Which statements are true?

l and II only
II and III only
l and III only
I, II and III
Question 37 Explanation: 
If two languages conform to the common language specification (CLS) of the Microsoft.NET framework.
Then there are certain compliance rules which may be used for inheritance. So other statement (I) and (II) are True.
Question 38

A serial transmission T1 uses 8 information bits, 2 start bits, 1 stop bit and 1 parity bit for each character. A synchronous transmission T2 uses 3 eight bit sync characters followed by 30 eight bit information characters. If the bit rate is 1200 bits/second in both cases, what are the transfer rates of T1 and T2?

100 characters/sec, 153 characters/sec
80 characters/sec, 136 characters/sec
100 characters/sec, 136 characters/sec
80 characters/sec, 153 characters/sec
Question 38 Explanation: 
T1: 1 char = (8 + 2 + 1 + 1) = 12 bits
Transfer rate = 1200/12 = 100 char/sec
T2: Transfer character in bits = 24 + 240 = 264 bits
In 264 = 30 characters
Then in 1200 = ? 264/30 = 1200/x
x = 136.3 char/sec
So, correct option is (C).
Question 39

In a data link protocol, the frame delimiter flag is given by 0111. Assuming that bit stuffing is employed, the transmitter sends the data sequence 01110110 as

Question 39 Explanation: 
In the data link layer, bits stuffing is employed then bit stuffing is done using the flag delimiter. If there is a flag of n bits then we will compare the data sequence with the flag and for every n-1 bits matched found, a bit 0 is stuffed in the data sequence.
Thus using the above logic,
Delimiter flag: 0111
Data sequence: 01110110
So, for a flag of 4 bits we will compare data sequence with a pattern of 3 bits, i.e., 011.
0 1 1 0 1 0 1 1 0 0
In the above pattern the underlined bits are found matched. Hence, 0 in italics is stuffed. Thus resulting in the data sequence as 0110101100 which is option (D).
Question 40

In a sliding window ARQ scheme, the transmitter's window size is N and the receiver's window size is M. The minimum number of distinct sequence numbers required to ensure correct operation of the ARQ scheme is

min (M,N)
max (M,N)
M + N
Question 40 Explanation: 
For such a scheme to work properly, we will need a total of M+N distinct sequence numbers.
Question 41

Consider a 10 Mbps token ring LAN with a ring latency of 400 µs. A host that needs to transmit seizes the token. Then it sends a frame of 1000 bytes, removes the frame after it has circulated all around the ring, and finally releases the token. This process is repeated for every frame. Assuming that only a single host wishes to transmit, the effective data rate is

Question 41 Explanation: 
Note: Out of syllabus.
Question 42

A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?

5 Kbps
10 Kbps
15 Kbps
20 Kbps
Question 42 Explanation: 
Question 43

Consider a simplified time slotted MAC protocol, where each host always has data to send and transmits with probability p = 0.2 in every slot. There is no backoff and one frame can be transmitted in one slot. If more than one host transmits in the same slot, then the transmissions are unsuccessful due to collision. What is the maximum number of hosts which this protocol can support, if each host has to be provided a minimum through put of 0.16 frames per time slot?

Question 43 Explanation: 
Let there be N such hosts. Then when one host is transmitting then others must be silent for successful transmission. So throughput per host,
0.16 = 0.2 × 0.8N-1
⇒ 0.8 = 0.8N-1
⇒ N = 2
Question 44

In the TCP/IP protocol suite, which one of the following is NOT part of the IP header?

Fragment Offset
Source IP address
Destination IP address
Destination port number
Question 44 Explanation: 
Destination port number is not present at IP header.
Question 45

A TCP message consisting of 2100 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame, excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the total IP overhead in the second network for this transmission?

40 bytes
80 bytes
120 bytes
160 bytes
Question 45 Explanation: 
At Router-1:
2120B reach R1's network layer. It removes original IP header, fragments data part at IP and then appends IP header to all fragments and forwards . So, it divides 2100 Bytes into two fragments of size 1200 and 900. And both fragments are sent to R2.
At Router-2:
Both fragments that reach R2 exceed MTU at R2. So, both are fragmented. First packet of 1200B is fragmented into 3 packets of 400 Bytes each. And second packet of 900B is fragmented into 3 fragments of 400, 400 and 100 Bytes respectively.
So, totally 6 packets reach destinations.
So, total IP overhead = 6 × 20 = 120 Bytes
Question 46

Suppose that the maximum transmit window size for a TCP connection is 12000 bytes. Each packet consists of 2000 bytes. At some point of time, the connection is in slow-start phase with a current transmit window of 4000 bytes. Subsequently, the transmitter receives two acknowledgements. Assume that no packets are lost and there are no time-outs. What is the maximum possible value of the current transmit window?

4000 bytes
8000 bytes
10000 bytes
12000 bytes
Question 46 Explanation: 
Since maximum transmit window size = 12000 B
and packet size =2000 B (or MSS)
Receiver window size = 6 MSS and
Current sender window size = 2 MSS
Slow start threshold = receiver window/2 = 3 MSS
Now current sender window size = 2 MSS <3 MSS,
which implies transmission is in slow start phase.
After receiving first Ack: Current sender window should increase exponentially to 4 MSS but since threshold = 3 MSS, current sender window size goes to threshold which is 3 MSS, then after receiving second Ack: Since now it is in congestion avoidance phase, sender window size increases linearly which makes current sender window
= 4 MSS
= 4 × 2000 B
= 8000 B
Question 47

Traceroute reports a possible route that is taken by packets moving from some host A to some other host B. Which of the following options represents the technique used by traceroute to identify these hosts

By progressively querying routers about the next router on the path to B using ICMP packets, starting with the first router
By requiring each router to append the address to the ICMP packet as it is forwarded to B. The list of all routers en-route to B is returned by B in an ICMP reply packet
By ensuring that an ICMP reply packet is returned to A by each router en-route to B, in the ascending order of their hop distance from A
By locally computing the shortest path from A to B
Question 47 Explanation: 
Traceroute works by sending packets with gradually increasing TTL value, starting with TTL value of 1. The first router receives the packet, decrements the TTL value and drops the packet because it then has TTL value zero. The router sends an ICMP time exceeded message back to the source. The next set of packets are given a TTL value of 2.
So the first router forwards the packets, but the second router drops them and replies with ICMP time exceeded. Proceeding in this way, traceroute uses the returned ICMP time exceeded messages to build a list of routers that packets traverse, until the destination is reached and returns an ICMP echo reply message.
Question 48

Which of the following statements is TRUE about CSMA/CD

IEEE 802.11 wireless LAN runs CSMA/CD protocol
Ethernet is not based on CSMA/CD protocol
CSMA/CD is not suitable for a high propagation delay network like satellite network
There is no contention in a CSMA/CD network
Question 48 Explanation: 
For CSMA/CD requires that sender is to be transmitting atleast till the first bit reaches the receiver. So the collision will be eliminated in case if it is present.
For networks with high propagation delay this time becomes too long hence the minimum packet size required becomes too big to be feasible.
Question 49

Which of the following statements is FALSE regarding a bridge?

Bridge is a layer 2 device
Bridge reduces collision domain
Bridge is used to connect two or more LAN segments
Bridge reduces broadcast domain
Question 49 Explanation: 
Bridge devices works at the data link layer of the open system interconnected (OSI) model, connecting two different networks together and providing communication between them. So, option A, C are true.
The bridge acts as a interface between two networks and speed the traffic between them and there by reduces the collision domain.
So, option B also True.
Question 50

Count to infinity is a problem associated with

link state routing protocol.
distance vector routing protocol.
DNS while resolving host name.
TCP for congestion control.
Question 50 Explanation: 
Distance vector routing protocol uses the Bellman-Ford algorithm and Ford-Fulkerson algorithm.
The Bellman-Ford algorithm does not prevent routing loops from happening and suffers from the count-to-infinity problem.
There are 50 questions to complete.

Access subject wise (1000+) question and answers by becoming as a solutions adda PRO SUBSCRIBER with Ad-Free content

Register Now