Computer-Networks
Question 1 |
Which of the following protocol pairs can be used to send and retrieve e-mails (in that order)?
SMTP, MIME | |
SMTP, POP3 | |
IMAP, POP3 | |
IMAP, SMTP |
POP3: Post Office Protocol (Responsible for retrieve email)
SMTP: Simple Mail Transfer Protocol (Responsible for send Email)
IMAP: Internet Message Access protocol (Responsible for store and view)
MIME: Multi purpose Internet Mail Extensions (For media)
Question 2 |
The value of 351 mod 5 is ______.
3 | |
5 | |
2 | |
1 |
⇒ 31 = 3 ⇒ 3 mod 5 = 3
32 ⇒ 9 mod 5 = 4
33 ⇒ 27 mod 5 = 2
34 ⇒ 81 mod 5 = 1
35 ⇒ 243 mod 5 = 3
For every four numbers sequence is repeating.
So, (51 % 4) = 3
⇒ 33 = 27
⇒ 27 mod 5 = 2
Question 3 |
Consider three machines M, N and P with IP addresses 100.10.5.2, 100.10.5.5, and 100.10.5.6 respectively. The subnet mask is set to 255.255.255.252 for all the three machines. Which one of the following is true?
M, N, and P all belong to the same subnet | |
Only M and N belong to the same subnet | |
M, N and P belong to three different subnets | |
Only N and P belong to the same subnet |
Therefore, N and P belong to the same subnet.
Question 4 |
In an RSA cryptosystem, the value of the public modulus parameter n is 3007. If it is also known that Φ(n) = 2880, where Φ() denotes Euler’s Quotient Function, then the prime factor of n which is greater than 50 is ______.
107 | |
97 | |
45 | |
92 |
n = 3007, fi(n) = 2880 → fi(n) = (p – 1) (q – 1),
where p, q are prime factor of n.
The unit place of n is 7, it is a prime number and factor will be
1.7=7
11*17
21*37
31*47
….
31*97 =>3007
n = 3007 => 31*97
Therefore, 31 & 97 are the two prime numbers, which is satisfying the condition and 97 is greater than 50.
So, 97 is the correct answer.
Other methods:
When ϕ(n) is given when n=pq where p and q are prime numbers, then we have
ϕ(n) = (p−1)(q−1) = pq−(p+q)+1
But pq=n,
therefore, ϕ(n) = n−(p+q)+1 and p+q = n+1−ϕ(n).
Now, p and q are the roots of the equation,
x2 − (p+q)x + pq = (x-p)(x-q)
Substituting for p+q and pq in the above equation
x2 – (n+1-ϕ(n))x + n
Question 5 |
Consider that 15 machines need to be connected in a LAN using 8-port Ethernet switches. Assume that these switches do not have any separate uplink ports. The minimum number of switches needed is _____.
3 | |
7 | |
1 | |
5 |
Therefore, the total required number of the switches = Ceil (15 /7) = 3
Question 6 |
Suppose that in an IP-over-Ethernet network, a machine X wishes to find the MAC address of another machine Y in its subnet. Which one of the following techniques can be used for this?
X sends an ARP request packet to the local gateway’s IP address which then finds the MAC address of Y and sends to X | |
X sends an ARP request packet with broadcast IP address in its local subnet | |
X sends an ARP request packet to the local gateway’s MAC address which then finds the MAC address of Y and sends to X | |
X sends an ARP request packet with broadcast MAC address in its local subnet |
Since both are present in the same subnet thus an ARP request packet can be sent as broadcast MAC address, all will see but the only destination will reply as a unicast reply.
Video Reference :
http://eclassesbyravindra.com/mod/page/view.php?id=147
Question 7 |
Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (= 109 bits/second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.
33 | |
34 | |
35 | |
36 |
The process of using all the sequence number and repeating a previously used sequence number.
The time taken to wrap around is called wrap around time:
Minimum Time = Wrap around time = Total number of bits in sequence number / Bandwidth = 232 * 8 / 109 = 34.35 == 34 (closest integer)
Question 8 |
Match the following:
Field Length in bits P. UDP Header's Port Number I. 48 Q. Ethernet MAC Address II. 8 R. IPv6 Next Header III.32 S. TCP Header's Sequence Number IV. 16
P-III, Q-IV, R-II, S-I | |
P-II, Q-I, R-IV, S-III | |
P-IV, Q-I, R-II, S-III | |
P-IV, Q-I, R-III, S-II |
Q. Ethernet MAC Address – 48 bits
R. IPV6 Next Header – 8 bits
S. TCP Header’s Sequence Number – 32 bits
Question 9 |
(i) The cwnd increase by 2 MSS on every successful acknowledgement.
(ii) The cwnd approximately doubles on every successful acknowledgement.
(iii) The cwnd increase by 1 MSS every round trip time.
(iv) The cwnd approximately doubles every round trip time.
Which one of the following is correct?
Only (ii) and (iii) are true | |
Only (i) and (iii) are true
| |
Only (iv) is true | |
Only (i) and (iv) are true |
Initially, TCP starts with cwnd of 1 MSS. On every ack, it increases cwnd by 1 MSS.
That is, cwnd doubles every RTT.
Initially sends 1 segment. On ack, sends 2 segments.
After these 2 acks come back, sends 4 segments etc.
TCP rate increases exponentially during slow start.
Slow start continues till cwnd reaches threshold.
After threshold is reached, cwnd increases more slowly, by one 1 MSS every RTT.
Question 10 |
Consider a simple communication system where multiple nodes are connected by a shared broadcast medium (like Ethernet or wireless). The nodes in the system use the following carrier-sense based medium access protocol. A node that receives a packet to transmit will carrier-sense the medium for 5 units of time. If the node does not detect any other transmission in this duration, it starts transmitting its packet in the next time unit. If the node detects another transmission, it waits until this other transmission finishes, and then begins to carrier-sense for 5 time units again. Once they start to transmit, nodes do not perform any collision detection and continue transmission even if a collision occurs. All transmissions last for 20 units of time. Assume that the transmission signal travels at the speed of 10 meters per unit time in the medium.
Assume that the system has two nodes P and Q, located at a distance d meters from each other. P starts transmitting a packet at time t=0 after successfully completing its carrier-sense phase. Node Q has a packet to transmit at time t=0 and begins to carrier-sense the medium.
The maximum distance d (in meters, rounded to the closest integer) that allows Q to successfully avoid a collision between its proposed transmission and P’s ongoing transmission is ___________.
50 | |
51 | |
52 | |
53 |
Now signal travels at the speed of 10 meters per unit time.
Therefore, in 5 unit time, it can travel a maximum distance (d) of 50 m (5*10), which allows the receiver (Q) to sense that the channel is busy.
Question 11 |
Consider an IP packet with a length of 4,500 bytes that includes a 20-byte IPv4 header and a 40-byte TCP header. The packet is forwarded to an IPv4 router that supports a Maximum Transmission Unit (MTU) of 600 bytes. Assume that the length of the IP header in all the outgoing fragments of this packet is 20 bytes. Assume that the fragmentation offset value stored in the first fragment is 0.
The fragmentation offset value stored in the third fragment is __________.
144 | |
145 | |
146 | |
147 |
Therefore Payload = 600 – 20 = 580 bytes.
As we know fragment size should be multiple of 8 but 580 bytes is not a multiple of 8, therefore fragment size is 576 bytes.
Offset value of kth fragment = Fragment size *( kth fragment – 1) / scaling factor
Offset value of third fragment = 576 * (3-1) / 8 = 144
Question 12 |
Consider a TCP client and a TCP server running on two different machines. After completing data transfer, the TCP client calls close to terminate the connection and a FIN segment is sent to the TCP server. Server-side TCP responds by sending an ACK, which is received by the client-side TCP. As per the TCP connection state diagram (RFC 793), in which state does the client-side TCP connection wait for the FIN from the server-side TCP?
LAST-ACK | |
TIME-WAIT | |
FIN-WAIT-1 | |
FIN-WAIT-2 |
i.e. waiting for the ACK for own FIN segment.
There are two possibilities here:
I. If Client receives ACK for its FIN then client will move to FIN-WAIT-2 and will wait for matching FIN from server side.
After receiving the FIN from server, client will send ACK and move to TIME-WAIT state.
II. Client has sent FIN segment but didn’t get ACK till the time.
Instead of ACK, client received FIN from server side.
Client will acknowledge this FIN and move to CLOSE state.
Here Client will wait for the ACK for its own FIN.
After receiving ACK, client will move to TIME-WAIT state.
Here we encounter First Case.
So, the solution is (D).
Refer this TCP state transition diagram:
Question 13 |
(I) S can launch a birthday attack to replace m with a fraudulent message.
(II) A third party attacker can launch a birthday attack to replace m with a fraudulent message.
(III) R can launch a birthday attack to replace m with a fraudulent message.
Which of the following are possible security violations?
(I) and (II) only | |
(I) only | |
(II) only | |
(II) and (III) only |
(I) Can the sender replace the message with a fraudulent message?
Yes, definitely because the sender will encrypt the message with its private key.
It can encrypt another message also with its private key.
(II) Can the third party send a fraudulent message?
No, because the third party doesn’t know about the private key of the sender.
(III) Can receiver send the fraudulent message?
No, the receiver also doesn’t know about the Private key of the sender.
So receiver also cannot send the fraudulent message.
Question 14 |
A computer network uses polynomials over GF(2) for error checking with 8 bits as information bits and uses x3 + x + 1 as the generator polynomial to generate the check bits. In this network, the message 01011011 is transmitted as
01011011010 | |
01011011011 | |
01011011101 | |
01011011100 |
= 1∙x3+0∙x2+1∙x1+1∙x0
= 1011
Message = 01011011
So, the message 01011011 is transmitted as
Question 15 |
In a RSA cryptosystem, a participant A uses two prime numbers p = 13 and q = 17 to generate her public and private keys. If the public key of A is 35, then the private key of A is _________.
11 | |
12 | |
13 | |
14 |
Given, p=13, q=17, e=35 (Public key), d=? (Private key)
As per RSA Algorithm; following steps:
Step 1: Find n = p×q = 13×17 = 221
Step 2: Find ∅(n) = (p-1)×(q-1) = 12×16 = 192
Step 3: d×e mod ∅(n) = 1 ⇒ (d = e(-1) mod ∅(n))
or
d×e = 1 mod ∅(n)
⇒ d×35 mod 192 = 1
Question 16 |
The value of parameters for the Stop-and-Wait ARQ protocol are as given below:
Bit rate of the transmission channel = 1 Mbps.
Propagation delay from sender to receiver = 0.75 ms.
Time to process a frame = 0.25 ms.
Number of bytes in the information frame = 1980.
Number of bytes in the acknowledge frame = 20.
Number of overhead bytes in the information frame = 20.
Assume that there are no transmission errors. Then, the transmission efficiency (expressed in percentage) of the Stop-and-Wait ARQ protocol for the above parameters is _________ (correct to 2 decimal places).
89.33% | |
89.34% | |
89.35% | |
89.36% |
B = 1Mbps, L = 1980Bytes, Overhead = 20Bytes
TProc = 0.25ms, LAck = 20Bytes
Tp=0.75ms
Total Data size(L) = (L + overhead) = 1980+20 = 2000Bytes
Efficiency of Stop & Wait ARQ?
Tt = L/B = 2000Bytes/1Mbps = (2000×8bits)/(106 b/s) = 16msec
TAck = LAck/B = (20×8bits)/(106 bits/sec) = 0.16msec
∴ In Stop and Wait ARQ, efficiency
ƞ = Tt/(Tt+TAck+2Tp+TProc) = 16ms/(16+0.16+2×0.75+0.25ms) = 16ms/17.91ms = 0.8933
Question 17 |
Consider the following statements about the routing protocols, Routing Information Protocol (RIP) and Open Shortest Path First (OSPF) in an IPv4 network.
-
- I: RIP uses distance vector routing
-
- II: RIP packets are sent using UDP
-
- III: OSPF packets are sent using TCP
- IV: OSPF operation is based on link-state routing
Which of the statements above are CORRECT?
I and IV only | |
I, II and III only | |
I, II and IV only | |
II, III and IV only |
RIP is one of the oldest DVR protocol which employ the hop count as a routing metric.
II: RIP packets are sent using UDP. “TRUE”
RIP uses the UDP as its transport protocol, and is assigned the reserved port no 520.
III: OSPF packets are sent using TCP. “FASLE”
OSPF encapsulates its data directly into IP Packets and does not use either TCP or UDP.
IV: OSPF operation is based on link state routing. “TRUE”
OSPF is a routing protocol which uses link state routing (LSR) and works within a single autonomous system.
Hence correct is answer “C”.
Question 18 |
Consider socket API on a Linux machine that supports connected UDP sockets. A connected UDP socket is a UDP socket on which connect function has already been called. Which of the following statement is/are CORRECT?
-
I. A connected UDP socket can be used to communicate with multiple peers simultaneously.
II. A process can successfully call connect function again for an already connected UDP socket.
I only | |
II only | |
Both I and II | |
Neither I nor II |
I. A connected UDP socket can be used to communicate with only one peer.
A DNS client can be configured to use one or more servers, normally by listing the IP addresses of the servers in the file /etc/resolv.conf.
If a single server is listed, the client can call connect, but if multiple servers are listed the client cannot call connect.
II. A process with a connected UDP socket can call connect function again for that socket for one of two reasons:
(a) To specify a new IP address and port.
(b) To unconnect the socket.
Hence, the correct answer is (B).
Question 19 |
The maximum number of IPv4 router addresses that can be listed in the record route (RR) option field of an IPv4 header is _________.
9 | |
10 | |
11 | |
12 |
In IPv4 header, 40 bytes are reserved for OPTIONS.
For Record Route to stores, 1 byte is used to store type of option, 1 byte for length and 1 byte for pointer. Out of 40 bytes, 37 bytes are left.
Each IP4 address takes 32 bits or 4 bytes.
Therefore, it can store at most floor (37/4) = 9 router addresses.
Hence correct answer is 9 router address.
Question 20 |
Consider two hosts X and Y, connected by a single direct link of rate 106bits/sec. The distance between the two hosts is 10,000 km and the propagation speed along the link is 2×108m/sec. Host X sends a file of 50,000 bytes as one large message to host Y continuously. Let the transmission and propagation delays be p milliseconds and q milliseconds, respectively. Then the values of p and q are
p=50 and q=100 | |
p=50 and q=400 | |
p=100 and q=50 | |
p=400 and q=50 |
B = 106 bits/sec
L = 50000 Bytes
d = 10000 Km = 107 m
v = 8×108 m/sec
Transmission time,
P = L/B = 50000×8bits/ 106 bits/sec = 0.4sec = 400msec
Propagation time,
q = d/v = 107m/ 2×108 m/s = 0.05sec = 50 msec
Question 21 |
Which one of the following protocols is NOT used to resolve one form of address to another one?
DNS | |
ARP | |
DHCP | |
RARP |
Except DHCP, remaining all the protocols are used to resolve one form of address to another one.
I. DNS is going to convert hostname to IP address.
II. ARP is going to convert IP to MAC.
III. DHCP is going to assign IP dynamically.
IV. RARP is going to convert MAC to IP.
Question 22 |
(i) HTTP
(ii) FTP
(iii) TCP
(iv) POP3
(i) and (ii) only | |
(ii) and (iii) only | |
(ii) and (iv) only | |
(iv) only |
A protocol that requires keeping of the internal state on the server is known as a stateful protocol.
Stateless – HTTP, IP
Stateful – FTP, SMTP, POP3, TCP
TCP is stateful as it maintains connection information across multiple transfers, but TCP is a Transport layer protocol.
FTP and POP3 is stateful Application layer protocol.
Question 23 |
Consider that B wants to send a message m that is digitally signed to A. Let the pair of private and public keys for A and B be denoted by Kx– and Kx+ for x = A,B, respectively. Let Kx(m) represent the operation of encrypting m with a key Kx and H(m) represent the message digest. Which one of the following indicates the CORRECT way of sending the message m along with the digital signature to A?
 | |
 | |
 | |
 |
Message digest is a hash value generated by applying a function on it.
Message digest is encrypted using private key of sender, so it can only be decrypted by public key of sender.
This ensures that the message was sent by the known sender.
Message digest is sent with the original message to the receiving end, where hash function is used on the original message and the value generated by that is matched with the message digest.
This ensures the integrity and thus, that the message was not altered.
Digital signature uses private key of the sender to sign message digest.
Question 24 |
An IP datagram of size 1000 bytes arrives at a router. The router has to forward this packet on a link whose MTU (maximum transmission unit) is 100 bytes. Assume that the size of the IP header is 20 bytes.
The number of fragments that the IP datagram will be divided into for transmission is _________.
13 | |
14 | |
15 | |
16 |
Size of Datagram (L) = 1000 bytes
MTU = 100 bytes
Size of IP header = 20 bytes
Size of Data that can be transmitted in one fragment (payload) = 100 – 20 = 80 bytes
Size of Data to be transmitted = Size of Datagram – size of header = 1000 – 20 = 980 bytes
No. of fragments required = ⌈980/80⌉ = 13
Question 25 |
For a host machine that uses the token bucket algorithm for congestion control, the token bucket has a capacity of 1 megabyte and the maximum output rate is 20 megabytes per second. Tokens arrive at a rate to sustain output at a rate of 10 megabytes per second. The token bucket is currently full and the machine needs to send 12 megabytes of data. The minimum time required to transmit the data is seconds _________.
1.1 sec | |
1.2 sec | |
1.3 sec | |
1.4 sec |
S = C / (M – P)
Where,
M = Maximum output rate,
C = capacity of the bucket,
P = Rate of arrival of a token,
Given, M=20 Mb, C=1Mbps, P=10 Mbps
Therefore, S= 1 Mb / (20-10) Mbps = 1/10 = 0.1 sec
Since, the bucket is initially full, it already has 1 Mb to transmit so it will be transmitted instantly.
So, we are left with only (12 – 1) Mb, i.e. 11 Mb of data to be transmitted.
Therefore, time required to send the 11 MB will be 11 * 0.1 = 1.1 sec
Question 26 |
A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps = 1000 bits/second). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one-way propagation delay is 100 milliseconds.
Assuming no frame is lost, the sender throughput is _________ bytes/second.
2500 | |
2501 | |
2502 | |
2503 |
Frame size (L) =1000 bytes
Sender side bandwidth (BS) = 80 kbps = 10 * 103 bytes/sec
Acknowledgement size (LA) =100 bytes
Receiver side bandwidth (BR) = 8 kbps = 1 * 103 bytes/sec
Propagation delay (Tp) =100 ms
By formula:
Transmission delay (Tt ) = L/BS = 1000 bytes / 10 * 103 bytes/sec = 100 ms
Acknowledge delay (Tack ) = LA / BR = 100 bytes / 1 * 103 bytes/sec = 100 ms
Total cycle time = Tt + 2 * Tp + Tack = 100 ms + 2 * 100 ms + 100 ms = 400 ms
Efficiency (η) = Tt / Total cycle time = 100 ms / 400 ms = 1 / 4 = 0.25
Throughput = Efficiency (η) * Bandwidth (BS) = 0.25 * 10 *103 bytes/s = 2500 bytes/second
Question 27 |
Anarkali digitally signs a message and sends it to Salim. Verification of the signature by Salim requires
A station stops to sense the channel once it starts transmitting a frame. | |
The purpose of the jamming signal is to pad the frames that are smaller than the minimum frame size. | |
A station continues to transmit the packet even after the collision is detected. | |
The exponential backoff mechanism reduces the probability of collision on retransmissions. |
It is a protocol of data link layer and it tells how the data can be formatted to transmit and how to place the data on network for transmission.
Now considering the Ethernet protocol we will discuss all the options one by one
I. In Ethernet the station does not required, stops to sense the channel prior frame transmission.
II. A Jamming signal used to inform all the other devices or stations about collision that has occurred, so that further data transmission is stopped.
III. No, stations sends jamming signal if collusion is detected
IV. To reduce the probability of collision on retransmissions an exponential back off mechanism is used.
This is only True.
Question 28 |
In an Ethernet local area network, which one of the following statements is TRUE?
A station stops to sense the channel once it starts transmitting a frame. | |
The purpose of the jamming signal is to pad the frames that are smaller than the minimum frame size. | |
A station continues to transmit the packet even after the collision is detected. | |
The exponential backoff mechanism reduces the probability of collision on retransmissions. |
It is a protocol of data link layer and it tells how the data can be formatted to transmit and how to place the data on network for transmission.
Now considering the Ethernet protocol we will discuss all the options one by one
I. In Ethernet the station does not required, stops to sense the channel prior frame transmission.
II. A Jamming signal used to inform all the other devices or stations about collision that has occurred, so that further data transmission is stopped.
III. No, stations sends jamming signal if collusion is detected
IV. To reduce the probability of collision on retransmissions an exponential back off mechanism is used.
This is only True.
Question 29 |
Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a webpage from a remote server, assuming that the host has just been restarted.
HTTP GET request, DNS query, TCP SYN | |
DNS query, HTTP GET request, TCP SYN | |
DNS query, TCP SYN, HTTP GET request | |
TCP SYN, DNS query, HTTP GET request |
Question 30 |
A network has a data transmission bandwidth of 20 × 106 bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is _________ bytes.
200 | |
201 | |
202 | |
203 |
Given,
Bandwidth (B) = 20 × 106 bps
TP = 40 μs ⇒ 40 × 10– 6 sec
Suppose minimum frame size is L.
Tt = 2 × TP ⇒ L / B = 2 × TP
⇒ L = 2 × TP × B = 2 × 40 × 10-6 × 20 × 106 = 1600 bits ⇒ 200 bytes
Therefore, L = 200 bytes
Question 31 |
I. At least three non-overlapping channels are available for transmissions.
II. The RTS-CTS mechanism is used for collision detection.
III. Unicast frames are ACKed.
All I, II and III | |
I and III only | |
II and III only | |
II only |
I. This is true, maximum 3 overlapping channels are possible in Wifi.
II. The RTS (Request To Send) and CTS(Clear To Send) are control frames which is used for collision avoidance, not in collision detection, (so, II is False)
III. Every frame in Wifi is Acked, because Wifi stations do not use collusion detection. (True)
Question 32 |
Consider a 128 × 103 bits/ second satellite communication link with one way propagation delay of 150 milliseconds. Selective retransmission (repeat) protocol is used on this link to send data with a frame size of 1 kilobyte. Neglect the transmission time of acknowledgement. The minimum number of bits required for the sequence number field to achieve 100% utilization is __________.
4 | |
5 | |
6 | |
7 |
a = Tp / Tt where Tp is propagation delay, and Tt is transmission delay.
Given, B = 128 kbps, Tp = 150 msec,
L = 1 KB = 1 * 8 * 210 bits
Tt = L / B ⇒ 1 * 8 * 210 bits / 128 * 103 bps ⇒ 0.064 sec = 64 msec
So, a = 150 msec / 64 msec = 2.343
Efficiency (η) = 100 % ⇒ 1 = N/ 1 + 2 * a
So, N = 1 + 2 * a ⇒ 1 + 2 * 2.343 = 5.686
No. of sequence numbers requires in SR is 2*N = 2 *5.686 = 11.375
Minimum No. of bits required in the sequence number = [ log2 (11.375) ] = 4
Question 33 |
Suppose two hosts use a TCP connection to transfer a large file. Which of the following statements is/are False with respect to the TCP connection?
- 1. If the sequence number of a segment is m, then the sequence
number of the subsequent segment is always m+1.
2. If the estimated round trip time at any given point of time is t sec, the value of the retransmission timeout is always set to greater than or equal to t sec.
3. The size of the advertised window never changes during the course of the TCP connection.
4. The number of unacknowledged bytes at the sender is always less than or equal to the advertised window.
III only | |
I and III only | |
I and IV only | |
II and IV only |
If the sequence no. of the segment is m, then the sequence number of the subsequent segment depends on the current segment size.
II. True.
If the estimated RTT at any given point of time is t second, then the value of the re-transmission timeout is always set to greater than or equal to t sec.
III. False.
The size of the advertised window may change during the course of the TCP connection depending on the processing capability at the receiver’s side and the network traffic.
IV. True.
The number of unacknowledged bytes at the sender is always less than or equal to the advertised window, because the sender never sends no. of bytes greater than advertised window.
Question 34 |
Which one of the following fields of an IP header is NOT modified by a typical IP router?
Checksum | |
Source address | |
Time to Live (TTL) | |
Length |
Option A (Checksum) needs to be updated by each visited Router since TTL Value is modified.
Option D (Length) also modified whenever there is a need of performing the fragmentation process.
Option B (Source Address) can’t be modified by an IP router. Only NAT can modify it.
Question 35 |
Consider a LAN with four nodes S1, S2, S3 and S4. Time is divided into fixed-size slots, and a node can begin its transmission only at the beginning of a slot. A collision is said to have occurred if more than one node transmit in the same slot. The probabilities of generation of a frame in a time slot by S1, S2, S3 and S4 are 0.1, 0.2, 0.3 and 0.4, respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is _________.
0.4404 | |
0.463 | |
0.464 | |
0.465 |
S2→0.2
S3→0.3
S4→0.4
The probability of sending a frame without any collision by any of these stations is
Question 36 |
Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is _________.
320 | |
321 | |
322 | |
323 |
Tp = 20 ms
η ≥ 50%
For η≥50% ⇒ L≥BR
⇒ L = 64×103×2×20×10-3
= 2560bits
= 320bytes
Question 37 |
Identify the correct order in which a server process must invoke the function calls accept, bind, listen, and recv according to UNIX socket API.
listen, accept, bind recv | |
bind, listen, accept, recv | |
bind, accept, listen, recv | |
accept, listen, bind recv |
Question 38 |
Which one of the following statements is NOT correct about HTTP cookies?
A cookie is a piece of code that has the potential to compromise the security of an internet user | |
A cookie gains entry to the user’s work area through an HTTP header | |
A cookie has an expiry date and time | |
Cookies can be used to track the browsing pattern of a user at a particular site
|
Cookies are not piece of code, they are just strings typically in the form of key value pairs.
Question 39 |
A link has a transmission speed of 106 bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgement has negligible transmission delay, and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at the nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the one-way propagation delay (in milliseconds) is ___________.
12 | |
13 | |
14 | |
15 |
L = 1000
η = 25%
Tp = ?
In stop-and-wait, η = 1/1 + 2a
⇒1/4 = 1/1 + 2a ⇒ 1 + 2a = 4
2a = 3; a = 32
Tx = L/B = 8×103/106 = 8ms
Tp/Tx = 3/2; 2Tp = 3Tx
2Tp = 24ms
Tp = 12ms
Question 40 |
Host A sends a UDP datagram containing 8880 bytes of user data to host B over an Ethernet LAN. Ethernet frames may carry data up to 1500 bytes (i.e. MTU = 1500 bytes). Size of UDP header is 8 bytes and size of IP heard is 20 bytes. There is no option field in IP header. How many total number of IP fragments will be transmitted and what will be the contents of offset field in the last fragment?
6 and 925 | |
6 and 7400 | |
7 and 1110 | |
7 and 8880 |
Total Size excluding IP Header = 8888 bytes.
Number of fragments = ceil(8880+ UDP or TCP header /1500-IP header)
= ceil(8880+8 /1500-20)
= ceil(8888/1480)
= 7
Offset of last fragment = (MTU-IP header ) *( number of fragments -1) / scaling factor = 1110 (scaling factor of 8 is used in offset field).
= (1500-20)* (7-1)/8
= 1110
Question 41 |
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.
Group I Group II
(i) 128.96.171.92 (a) Interface 0
(ii) 128.96.167.151 (b) Interface 1
(iii) 128.96.163.121 (c) R2
(iv) 128.96.165.121 (d) R3
(e) R4
i-a, ii-c, iii-e, iv-d | |
i-a, ii-d, iii-b, iv-e | |
i-b, ii-c, iii-d, iv-e | |
i-b, ii-c, iii-e, iv-d |
Question 42 |
Consider the following statements.
-
I. TCP connections are full duplex.
II. TCP has no option for selective acknowledgment.
III. TCP connections are message streams.
Only I is correct | |
Only I and III are correct | |
Only II and III are correct | |
All of I, II and III are correct |
Question 43 |
Consider a CSMA/CD network that transmits data at a rate of 100 Mbps (108 bits second) over a 1 km (kilometer) cable with no repeaters. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable?
8000 | |
10000 | |
16000 | |
20000 |
B = 100 Mbps
d = 1 km
v = ?
In CSMA/CD, L = 2×d/v×B
⇒ v = 2dB/L = 2×103×108/104
⇒ v = 20,000 km/sec
Question 44 |
Consider a network connected two systems located 8000 kilometers apart. The bandwidth of the network is 500 × 106 bits per second. The propagation speed of the media is 4 × 106 meters per second. It is needed to design a Go-Back-N sliding window protocol for this network. The average packet size is 107 bits. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then the minimum size in bits of the sequence number field has to be ___________.
8 | |
7 | |
6 | |
5 |
η=100% n = ?
∴a=Tp/Tn =2/0.02=100
Given Protocol, Go-back-N protocol, So η = w/(1+2a) where w = 2n-1
100/100 = w/(1+2a) ⇒ w = 1+2a
⇒ 2(n-1) = 1+2(100)
⇒ 2n – 1 = 201
⇒ 2n = 202 ⇒ 2n = 28
⇒ n = 8
Question 45 |
In the network 200.10.11.144/27, the fourth octet (in decimal) of the last IP address of the network which can be assigned to a host is ____________.
158 | |
157 | |
156 | |
155 |
Subnet mask is 255.255.255.224
Do AND with given IP and subnet mask then we get NID 200.10.11.128
In fourth octet first three bit will fixed for subnet and remaining 5 bits is for HID, so maximum value as 11111.
The address with all 1s in host part is broadcast address and can’t be assigned to a host.
So the maximum possible last octal in a host IP is 10011110 which is 158.
Question 46 |
Consider the following three statements about link state and distance vector routing protocols, for a large network with 500 network nodes and 4000 links.
-
[S1] The computational overhead in link state protocols is higher than in distance vector protocols.
[S2] A distance vector protocol (with split horizon) avoids persistent routing loops, but not a link state protocol.
[S3] After a topology change, a link state protocol will converge faster than a distance vector protocol.
Which one of the following is correct about S1, S2, and S3 ?
S1, S2, and S3 are all true. | |
S1, S2, and S3 are all false. | |
S1 and S2 are true, but S3 is false. | |
S1 and S3 are true, but S2 is false. |
S2: A distance vector protocol with split horizon avoid persistent routing loops is true, but not a link state protocol is false because link state protocols do not have count to infinity problem.
S3: As Distance vector protocol has count to infinity problem and converges slower. (True)
Question 47 |
Which of the following are used to generate a message digest by the network security protocols?
- (P) RSA
(Q) SHA-1
(R) DES
(S) MD5
P and R only | |
Q and R only | |
Q and S only | |
R and S only |
Question 48 |
Identify the correct order in which the following actions take place in an interaction between a web browser and a web server.
-
2. The web browser establishes a TCP connection with the web server.
3. The web server sends the requested webpage using HTTP.
4. The web browser resolves the domain name using DNS.
4,2,1,3 | |
1,2,3,4 | |
4,1,2,3 | |
2,4,1,3 |
Question 49 |
Consider a token ring network with a length of 2 km having 10 stations including a monitoring station. The propagation speed of the signal is 2×108 m/s and the token transmission time is ignored. If each station is allowed to hold the token for 2 µsec, the minimum time for which the monitoring station should wait (in µsec) before assuming that the token is lost is _______.
28μs to 30 μs | |
29μs to 31 μs | |
30μs to 32 μs | |
31μs to 33 μs |
No. of Stations (m) = 10
Propagation speed (v) = 2⨯108 m/s
THT = 2μs
So, Max, TRT = Tp in the ring + No. of Active Stations * THT
= 10 ⨯ 10-6 + 10 ⨯ 2 ⨯ 10-6
= 30 μs
Question 50 |
Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of the connection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.
1100 to 1300 | |
1101 to 1301 | |
1102 to 1302 | |
1103 to 1303 |
When Time Out occurs, for the next round of Slow Start, Threshold = (size of Cwnd) / 2
It means Threshold = 16KB
Slow Start
2KB
1RTT
4KB
2RTT
8KB
3RTT
16KB ———– Threshold reaches. So Additive Increase Starts
4RTT
18KB
5RTT
20KB
6RTT
22KB
7RTT
24KB
8RTT
26KB
9RTT
28KB
10RTT
30KB
11RTT
32KB
So, Total no. of RTTs = 11 → 11 * 100 = 1100