Which of the following protocol pairs can be used to send and retrieve e-mails (in that order)?
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)
⇒ 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
Consider three machines M, N, and P with IP addresses 22.214.171.124, 126.96.36.199, and 188.8.131.52 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.
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 Totient Function, then the prime factor of n which is greater than 50 is ______.
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
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.
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
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
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 _____.
Therefore the total required number of the switches = Ceil (15 /7) = 3
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 :
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 _________.
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)
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
Consider the following statements regarding the slow start phase of the TCP congestion control algorithm. Note that cwnd stands for the TCP congestion window and MSS denotes the Maximum Segment Size.
(i) The cwnd increase by 2 MSS on every successful acknowledgement.
(ii) The cwnd approximately doubles on every successful acknowledgedment.
(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.
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 ___________.
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.
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 __________.
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
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?
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:
A sender S sends a message m to receiver R, which is digitally signed by S with its private key. In this scenario, one or more of the following security violations can take place.
(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
(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.
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
So, the message 01011011 is transmitted as
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 _________.
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))
d×e = 1 mod ∅(n)
⇒ d×35 mod 192 = 1
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).
B = 1Mbps, L = 1980Bytes, Overhead = 20Bytes
TProc = 0.25ms, LAck = 20Bytes
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
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”.
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.
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).
The maximum number of IPv4 router addresses that can be listed in the record route (RR) option field of an IPv4 header is _________.
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.
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
P = L/B = 50000×8bits/ 106 bits/sec = 0.4sec = 400msec
q = d/v = 107m/ 2×108 m/s = 0.05sec = 50 msec
Which one of the following protocols is NOT used to resolve one form of address to another one?
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.
(i) and (ii) only
(ii) and (iii) only
(ii) and (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.
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.
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 _________.
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
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 _________.
S = C / (M - P)
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
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.
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
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
Anarkali digitally signs a message and sends it to Salim. Veriﬁcation of the signature by Salim requires
Anarkali’s public key.
Salim’s public key.
Salim’s private key.
Anarkali’s private key.
Receiver uses sender's public key to verify signature.
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.
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
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.
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
For the IEEE 802.11 MAC protocol for wireless communication, which of the following statements is/are TRUE?
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
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)
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 __________.
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
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.
If the estimated RTT at any given point of time is t second, then the value of the retransmission timeout is always set to greater than or equal to t sec.
The size of the advertized window may change during the course of the TCP connection depending on the processing capability at the receiver's side and the network traffic.
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.
Time to Live (TTL)
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.
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 _________.
The probability of sending a frame without any collision by any of these stations is
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 _________.
Tp = 20 ms
η ≥ 50%
listen, accept, bind recv
bind, listen, accept, recv
bind, accept, listen, recv
accept, listen, bind recv
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
In stop-and-wait, η=1/1+2a
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)
Offset of last fragment = (MTU-IP header ) *( number of fragments -1) / scaling factor = 1110 (scaling factor of 8 is used in offset field).
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
Only I is correct
Only I and III are correct
Only II and III are correct
All of I, II and III are correct
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
n = ?
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
Subnet mask is 255.255.255.224
Do AND with given IP and subnet mask then we get NID 184.108.40.206
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.
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)
P and R only
Q and R only
Q and S only
R and S only
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
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
16KB ----------- Threshold reaches. So Additive Increase Starts
So, Total no. of RTTs = 11 → 11 * 100 = 1100
Efficiency formula for SR protocol is
RIP uses distance vector routing and OSPF uses link state routing
OSPF uses distance vector routing and RIP uses link state routing
Both RIP and OSPF use link state routing
Both RIP and OSPF use distance vector routing
(b) The bind function assigns a local protocol address to a socket. With the Internet protocols, the protocol address is the combination of either a 32-bit IPv4 address or a 128-bit IPv6 address, along with a 16-bit TCP or UDP port number.
(c) The listen function converts an unconnected socket into a passive socket, indicating that the kernel should accept incoming connection requests directed to this socket.
(d) The accept function is called by a TCP server to return the next completed connection from the front of the completed connection queue. If the completed connection queue is empty, the process is put to sleep (assuming the default of a blocking socket).
Initially TTL value was 32, so at the Receiver it will become 32 – 6 = 26.
T1 < T2 < T3
T1 > T2 > T3
T2 = T3, T3 < T1
T1 = T3, T3 > T2
L = 103 byte
L = 1000 bytes
Header size = 100 bytes
Total Frame size = 1000 + 100 = 1100 bytes
∴ Tx = 1100 × 8/ 106×8 = 1100μs
L = 100 bytes
Header size = 100 bytes
Total Frame size = 100 + 100 = 200 bytes
∴ Tx = 200×8/ 106×8 = 200μs for 1 packet
For 10 packets ⇒ Tx = 2000μs
So, T2 = 2000+200+200 = 2400μs
L = 50 bytes
Header size = 100 bytes
Total Frame size = 50 + 100 = 150 bytes
∴ Tx = 150×8/ 106×8 = 150μs for 1 packet
For 20 packets ⇒ Tx = 3000μs
So, T3 = 3000+150+150 = 3300μs
∴ T1 = T3
T3 > T2
Only I1 and I2
Only I2 and I3
Only I3 and I4
An Intruder can learn [I2] through sniffing at R2 because Port Numbers are encapsulated in the payload field of IP Datagram.
An Intruder can learn [I3] through sniffing at R2 because IP Addresses and Routers are functioned at network layer of OSI Model.
An Intruder can’t learn [I4] through sniffing at R2 because it is related to Data Link Layer of OSI Model.
Network layer and Routing
Data Link Layer and Bit synchronization
Transport layer and End-to-end process communication
Medium Access Control sub-layer and Channel sharing
(b) Bit Synchronization is always handled by Physical Layer of OSI model but not Data Link Layer. So Option (b) is INCORRECT.
(c) End – to – End Process Communication is handled by Transport Layer. So Option (c) is CORRECT.
(d) MAC sub layer have 3 types of protocols (Random, Controlled and Channelized Access).
Output Bit string after stuffing is 01111100101.
(i) and (ii) only
(i) and (ii) only
(i), (ii) and (iii)
If we do AND operation between 255.254.0.0 and given IP 220.127.116.11, gives 18.104.22.168 which is matching with interface 1.
1000 unique Ids ⇒ 1Sec
218 unique Ids ⇒ 218/1000=28=256
MF bit: 0, Datagram Length: 1444; Offset: 370
MF bit: 1, Datagram Length: 1424; Offset: 185
MF bit: 1, Datagram Length: 1500; Offset: 370
MF bit: 0, Datagram Length: 1424; Offset: 2960
So Datagram with data 4404 byte fragmented into 3 fragments.
Only S1 is conflict-serializable.
Only S2 is conflict-serializable.
Both S1 and S2 are conflict-serializable.
Neither S1 nor S2 is conflict-serializable.
No cycle, so schedule S1 is conflict serializable.
There is a cycle, so S2 is not conflict serializable.
TCP, UDP, UDP and TCP
UDP, TCP, TCP and UDP
UDP, TCP, UDP and TCP
TCP, UDP, TCP and UDP
DNS runs over UDP protocol within port no-53.
Email needs, SMTP protocol which runs over TCP protocol within port no –25.
Encryption: X’s private key followed by Y’s private key; Decryption: X’s public key followed by Y’s public key
Encryption: X’s private key followed by Y’s public key; Decryption: X’s public key followed by Y’s private key
Encryption: X’s public key followed by Y’s private key; Decryption: Y’s public key followed by X’s private key
Encryption: X’s private key followed by Y’s public key; Decryption: Y’s private key followed by X’s public key
Encryption: Source has to encrypt with its private key for forming Digital signature for Authentication. Source has to encrypt the (M, σ) with Y’s public key to send it confidentially.
Decryption: Destination Y has to decrypt first with its private key, then decrypt using source public key.
Network layer – 4 times and Data link layer – 4 times
Network layer – 4 times and Data link layer – 3 times
Network layer – 4 times and Data link layer – 6 times
Network layer – 2 times and Data link layer – 6 times
From above given diagram, its early visible that packet will visit network layer 4 times, once at each node [S, R, R, D] and packet will visit Data Link layer 6 times. One time at S and one time at D, then two times for each intermediate router R as data link layer is used for link to link communication.
One at packet reaches R and goes up from physical –DL-Network and second time when packet coming out of router in order Network –DL-Physical.
∴104 bits -----5×108/104=104/5×108sec=1/5×104sec
1 sec------2×105 km
1/5×104-----2×105/5×104= 4 km
Maximum length of cable=4/2=2 km
Last fragment, 2400 and 2789
First fragment, 2400 and 2759
Last fragment, 2400 and 2759
Middle fragment, 300 and 689
HLEN = 10 - So header length is 4×10=40, as 4 is constant scale factor.
Total Length = 400 (40 Byte Header + 360 Byte Payload)
Fragment Offset = 300, that means 300×8 Byte = 2400 bytes are before this last fragment.
So the position of datagram is last fragment.
Sequence number of First Byte of Payload = 2400 (as 0 to 2399 Sequence no are used)
Sequence number of Last Byte of Payload = 2400+360-1=2759
245.248.136.0/21 and 245.248.128.0/22
245.248.128.0/21 and 245.248.128.0/22
245.248.132.0/22 and 245.248.132.0/21
245.248.136.0/24 and 245.248.132.0/21
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
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
block entire HTTP traffic during 9:00PM and 5:00AM
block all ICMP traffic
stop incoming traffic from a specific IP address but allow outgoing traffic to the same IP address
block TCP traffic from a specific user on a multi-user system during 9:00PM and 5:00AM
(B) Possible because it is network layer protocol.
(C) Possible because SP address is present in Network layer.
(D) Not possible, because to block specific user, we need user id which is present in Application layer.
(3, 2, 0, 2, 5)
(3, 2, 0, 2, 6)
(7, 2, 0, 2, 5)
(7, 2, 0, 2, 6)
N2 has made entry ∞
N4 has the distance of 8 to N1
N3 has the distance of 2 to N4
So 2 + 8 = 10
It can be used to prioritize packets
It can be used to reduce delays
It can be used to optimize throughput
It can be used to prevent packet looping
When we perform AND operation between IP address 10.105.1.113 and 255.255.255.224 result is 10.105.1.96 and when we perform AND operation between IP address 10.105.1.91 and 255.255.255.224 result is 10.105.1.64.
Therefore, 10.105.1.96 and 10.105.1.64 are different network, so D is correct answer.
Similarly, link R4-R6 will not be used, instead this link we can use R4-R5-R6 link which costs only 5 unit.
And only link that will be removed is R5-R6 link.
I and II
I and III
II and IV
III and IV
1. Generate randomly two “large” primes p and q.
2. Compute n=pq and ∅=(p-1)(q-1).
3. Choose a number e so that
4. Find the multiplicative inverse of e modulo ∅, i.e., find d so that
ed≡1 (mod ∅)
This can be done efficiently using Euclid’s Extended Algorithm.
The encryption public key is KE=(n,e) and the decryption private key is KD=(n,d).
The encryption function is
E(M)=Me mod n
The decryption function is
D(M)=Md mod n
The maximum packet lifetime is given is given 64s.
Maximum data rate possible(bandwidth) to avoid the wraparound = 232/64 = 226 Byte/sec.
The clock counter increments once per milliseconds = That means when then counter increments next possible sequence number is generated. The packet lifetime is 64 seconds and after this 64 seconds next sequence number is come. So that means in this 64 seconds only 1 sequence number is generated.
Hence the minimum rate is = 1/64 = 0.015/sec.
G(x) contains more than two terms
G(x) does not divide 1+xk, for any k not exceeding the frame length
1+x is a factor of G(x)
G(x) has an odd number of terms
I = 2
I = 3
I = 4
I = 5
Maximum number of frames that can be transmit to maximally pack them is=(Tt+2Tp)/Tx = (25+1)/1=26 which is window size
Minimum sequence numbers required = 26
Minimum number of bits required for sequence number is 5.
From above diagram we can say that Total RTT will be
1ms (transmitting time for first frame)
+ 25ms (propagation delay from S to R)
+ 1ms (transmitting time for piggybacked ACK)
+ 25ms (propagation delay from R to S)
= 52 ms
Also total time for sender needed to transmit 32 frames (22 = 25 = 32) is 32×1ms = 32ms
So sender has to wait for,
(52 - 32)ms
216 bytes-size of TCP header
Transport Layer - 65515 byte
Network layer - 65535 byte
Data link layer - 1500 byte
1) The client requests a connection by sending a SYN (synchronize) message to the server.
2) The server acknowledges this request by sending SYN-ACK back to the client.
3) The client responds with an ACK, and the connection is established.
does not increase
Number of bits assigned for host id is the number of zeros in subnet mask. Here 11 bits are used.
for host id so maximum possible hosts are= 211 -2=2046
A client process P needs to make a TCP connection to a server process S. Consider the following situation: the server process S executes a socket (), a bind () and a listen () system call in that order, following which it is preempted. Subsequently, the client process P executes a socket () system call followed by connect () system call to connect to the server process S. The server process has not executed any accept() system call. Which one of the following events could take place?
connect ( ) system call returns successfully
connect ( ) system call blocks
connect ( ) system call returns an error
connect ( ) system call results in a core dump
I-a, II-d, III-c, IV-b
I-b, II-d, III-c, IV-a
I-a, II-c, III-d, IV-b
I-b, II-c, III-d, IV-a
(ii) Kazaa and DC++ are P2P applications.
(iii) P2P slip is a processor of PPP.
(iv) DNS allows caching of entries at local server.
A 1Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 × 108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.
RTT = 4×Time to reach satellite (S1→S, S→S2, S2→S, S→S1)
∴ RTT = 0.48
None of these
Here, we need to change minimum 3-bits, and by doing it we get correct parity column wise and row wise (correction marked by boxed number).
S1, S2 and S4 only
S1, S3 and S4 only
S2 and S3 only
S1 and S4 only
→ In LSR shortest path is calculated at each every router. (B) is wrong.
→ In DVR also shortest path is calculated at each and every router. (C) is wrong.
→ Since DVR is based upon local knowledge whereas LSR is based upon global knowledge.
S1, S2 and S3 only
S1 and S3 only
S3 and S4 only
S1, S3 and S4 only
TCP handles both congestion and flow control ⇒ True.
UDP handles congestion but not flow control ⇒ False, because UDP neither handles congestion control nor flow control.
Fast retransmits deals with congestion but not flow control ⇒ True.
Slow start mechanism deals with both congestion and flow control ⇒ False, because it has nothing to do with flow control. Flow control is taken care by Advertisement window.
S2 and S3 only
S1 and S4
S1 and S3
S2 and S4
S2 → False, because if after ACK client immediately sends data then everything goes on without worry.
S3 → False.
S4 → True.
nC2 = n(n-1)/2
In case of public key, each sender has its own public key as well as private key. So, no. of keys are 2n.
XX.XX.XX.96, XX.XX.XX.64 and XX.XX.XX.128.
Subnet no. of host X is,
Now, the gateway must also have the same subnet number.
Let's take IP 192.168.1.110 of R1,
and hence this can be used by X.
Half the baud rate.
Twice the baud rate.
Same as the baud rate.
None of the above.
HTTP, Telnet and SMTP uses TCP.
500 metres of cable.
200 metres of cable.
20 metres of cable.
50 metres of cable.
Given Tp= 200 m / microsec
In, 1 microsec it covers 200m
Therefore, in 0.1 microsec it is 200 * 0.1 = 20 meters
62 subnets and 262142 hosts.
64 subnets and 262142 hosts.
62 subnets and 1022 hosts.
64 subnets and 1024 hosts.
From HID, we took 6-bits for subnetting.
Then total subnets possible = ( 26 ) - 2 = 64
Total hosts possible for each subnet = (210) - 2 = 1022
∴ Data transmitted is: 11001001011
N = (Tk + 2Tp)/Tk
Tp = l × l sec
Tk = K/R sec
So, N = (K/R+2Lt)/(K/R) = K+2LtR/K
Note that when asked for general sliding window protocol (not GBN nor SR) then we do not care about receiver's window size.
So, no. of bits required,
P – 2 Q – 1 R – 3 S – 5
P – 1 Q – 4 R – 2 S – 3
P – 1 Q – 4 R – 2 S – 5
P – 2 Q – 4 R – 1 S – 3
Q) BGP is network layer protocol that manages low packets are routed across the network.
R) TCP is a transport layer protocol.
S) PPP is a data link layer protocol.
(i) is false, but (ii) and (iii) are true
(i) and (iii) are false, but (ii) is title
(i) and (ii) are false, but (iii) is true
(i), (ii) and (iii) are false
The timeout value cannot be fixed for entire duration as it will turn timer to static timer, we need dynamic timer for timeout.
Statement-II: It is True.
Basic algorithm, Jacobson's algorithm, Karl's modification; these three algorithms are to be appropriate to RTT estimation algorithm used to set timeout value dynamically.
Statement-III: It is False.
Because timeout value is set to twice the propagation delay in data link layer where hop to hop distance is known, not in TCP layer.
Consider a TCP connection in a state where there are no outstanding ACKs. The sender sends two segments back to back. The sequence numbers of the first and second segments are 230 and 290 respectively. The first segment was lost, but the second segment was received correctly by the receiver. Let X be the amount of data carried in the first segment (in bytes), and Y be the ACK number sent by the receiver. The values of X and Y (in that order) are
60 and 290
230 and 291
60 and 231
60 and 230
Both are false
Statement (i) is true and the other is false
Statement (ii) is true and the other is false
Both are true
ii) It uses the P-Box permutation.
Statement-I is false, II is true.
A firewall is to be configured to allow hosts in a private network to freely open TCP connections and send packets on open connections. However, it will only allow external hosts to send packets on existing open TCP connections or connections that are being opened (by internal hosts) but not allow them to open TCP connections to hosts in the private network. To achieve this the minimum capability of the firewall should be that of
A combinational circuit
A finite automaton
A pushdown automaton with one stack
A pushdown automaton with two stacks
Turing machine can do everything as the normal computer can do, so firewall can be created by the TM.
The no. of bits that can be corrected is
Using distance vector routing protocol, F→D→B yeilds distance as 20 which eliminates option (B) and (D).
1000010111 and Differential Manchester respectively
0111101000 and Differential Manchester respectively
1000010111 and Integral Manchester respectively
0111101000 and Integral Manchester respectively
As per IEEE (B) is correct.
As per GE Thomas (A) is correct.
As we usually follow IEEE thus (B) is correct.
Let us consider a statistical time division multiplexing of packets. The number of sources is 10. In a time unit, a source transmits a packet of 1000 bits. The number of sources sending data for the first 20 time units is 6, 9, 3, 7, 2, 2, 2, 3, 4, 6, 1, 10, 7, 5, 8, 3, 6, 2, 9, 5 respectively. The output capacity of multiplexer is 5000 bits per time unit. Then the average number of backlogged of packets per time unit during the given period is
If the no. of packets transmitted is larger than 5 then the extra packets are backlogged. This means gets added to the next number and further backlog is calculated.
Average no. of backlogged packets = 89/20 = 4.45
A group of 15 routers are interconnected in a centralized complete binary tree with a router at each tree node. Router j communicates with router j by sending a message to the root of the tree. The root then sends the message back down to router j. The mean number of hops per message, assuming all possible router pairs are equally likely is
Message goes up from sender to root,
Average hops message goes to root
= (3×8) + (2×4) + (1×2) + (0×1)/15
Here, (3×8) represents 3 hops and 8 routers for bottom most level and so on.
Similarly, average hops when message comes down
= (3×8) + (2×4) + (1×2) + (0×1)/15
So, Total hops = 2×2.267 = 4.53
IMAP-(i), FTP-(ii), HTTP-(iii), DNS-(iv), POP3-(v)
FTP-(i), POP3-(ii), SMTP-(iii), HTTP-(iv), IMAP-(v)
POP3-(i), SMTP-(ii), DNS-(iii), IMAP-(iv), HTTP-(v)
SMTP-(i), HTTP-(ii), IMAP-(iii), DNS-(iv), FTP-(v)
(v) FTP needs two ports, 20 for data and 21 for control.
Hence, only (D) matches.
10100011 00110111 11101001 10101011
So, in total we have 32 bits. And for base 64we need 6 digits of binary no. to represent one digit of base 64 no.
So lets padd 4 bits on RHS, so that total digits will become 36 and we can separate then as group of 6 digits each.
Now, the longest substring will be from checking option is 'fpq'.
Ensure packets reach destination within that time
Discard packets that reach later than that time
Prevent packets from looping indefinitely
Limit the time for which a packet gets queued in intermediate routers
B1, B5, B3, B4, B2
B1, B3, B5, B2, B4
B1, B5, B2, B3, B4
B1, B3, B4, B5, B2
Port1 of B5 is connected to port1 of B2 and port2 of B3. B2 is connected with lower index port so B2 is traversed next.
Port 2 Both B3 and B4 is connected with port1 of B2, but B3 is Closer to root so B3 will be traversed next.
Depth First traversal is B1,B5,B2,B3,B4.
Frame sequence for 9 frame is shown below. Frame with bold sequence number gets lost.
1 2 3 4 [5 6 7] 5 6 [7 8 9] 7 8 9 9 = 16
Round trip delay = 2 * Tp = 80 ms (given)
Optimal window size is = (Tt + 2*Tp) / Tt = 82 / 2 = 41
Option is not given, closest option is 40.
C1 and C2 both assume they are on the same network
C2 assumes C1 is on same network, but C1 assumes C2 is on a different network
C1 assumes C2 is on same network, but C2 assumes C1 is on a different network
C1 and C2 both assume they are on different networks
Subnet mask for C1 is 255.255.128.0.
So it finds the Network ID as,
C1 → 22.214.171.124 AND 255.255.128.0 = 126.96.36.199
C2 → 188.8.131.52 AND 255.255.128.0 = 184.108.40.206
From C2 side,
Subnet mask for C2 is 255.255.192.0.
So it finds the network ID as,
C1 → 220.127.116.11 AND 255.255.192.0 = 18.104.22.168
C2 → 22.214.171.124 AND 255.255.192.0 = 126.96.36.199
Hence, option 'C' is correct.
FTP and HTTP
TELNET and POP3
HTTP and TELNET
SMTP and FTP
Both Ethernet frame and IP packet include checksum fields
Ethernet frame includes a checksum field and IP packet includes a CRC field
Ethernet frame includes a CRC field and IP packet includes a checksum field
Both Ethernet frame and IP packet include CRC fields
2 and 3 only
1 and 3 only
(2) Statement-2 is wrong, refer statement-1.
(3) Statement-3 is correct, for example hash function N%10, this will generate same values for 1 as well as 2!
188.8.131.52 = 144.16.68.01110101 AND 255.255.255.224 = 255.255.255.11100000
= 184.108.40.206(Not matching with destination)
Now, take 255.255.255.0
220.127.116.11 AND 255.255.255.0
= 18.104.22.168 (matched)
Hence, option (C) is correct.
Suppose that it takes 1 unit of time to transmit a packet (of fixed size) on a communication link. The link layer uses a window flow control protocol with a window size of N packets. Each packet causes an ack or a nak to be generated by the receiver, and ack/nak transmission times are negligible. Further, the round trip time on the link is equal to N units. Consider time i > N. If only acks have been received till time i(no naks), then the goodput evaluated at the transmitter at time i(in packets per unit time) is
1 – N/i
i/(N + i)
1 – e(i/N)
So, successful delivery of packet can be assured if ack has been received for it.
So till time 'i' we would have transmitted 'i' packets but only (i - N) can be acknowledged as minimum time for a packet to get acknowledged is N (since RTT is N which is equal to the window size, there is no waiting for the sender).
So, successfully delivered packets = (i - N)
Time for transmission = i
Goodput = Successfully delivered data/Time
= (i - N)/i
= 1 - N/i
In the 4B/5B encoding scheme, every 4 bits of data are encoded in a 5-bit codeword. It is required that the codewords have at most 1 leading and at most 1 trailing zero. How many such codewords are possible?
Codeword with first two bits '0'
= 0 0 x x x
Codeword with last two bits '0'
= x x x 0 0
= 23 = 8
Codeword with first two and last two bits '0'
= 0 0 x 0 0
Codeword with first or last two bits '0'
= 8 + 8 - 2
Therefore possible codewords
= 32 - 14
A router has two full-duplex Ethernet interfaces each operating at 100 Mb/s. Ethernet frames are at least 84 bytes long (including the Preamble and the Inter-Packet-Gap). The maximum packet processing time at the router for wirespeed forwarding to be possible is (in microseconds)
Tt = 84×8/10×106 = 6.72μs
But since a router has two full-duplex ethernet interfaces, so the maximum processing time should be,
6.72/2 μs = 3.36μs
A link of capacity 100 Mbps is carrying traffic from a number of sources. Each source generates an on-off traffic stream; when the source is on, the rate of traffic is 10 Mbps, and when the source is off, the rate of traffic is zero. The duty cycle, which is the ratio of on-time to off-time, is 1:2. When there is no buffer at the link, the minimum number of sources that can be multiplexed on the link so that link capacity is not wasted and no data loss occurs is S1. Assuming that all sources are synchronized and that the link is provided with a large buffer, the maximum number of sources that can be multiplexed so that no data loss occurs is S2. The values of S1 and S2 are, respectively,
10 and 30
12 and 25
5 and 33
15 and 22
Since there is no buffer and constraint given is there should not be any data lost, and no wastage of capacity as well.
Since data should not be lost, we calculate for the extreme case when all sources are ontime (that is transmitting).
10Mbps × n-station ≤ 100Mbps
n = 10 = S1
In the next part of question, it is given that the link is provided with large buffer and we are asked to find out large no. of stations.
For that we calculate expected value of bandwidth usage,
E = 1/3 × 10 + 1/3 × 10 + .......+ ....... n-station times ≤ 100Mbps
⇒ 1/3 × 10 × n-station ≤ 100Mbps ⇒ n-station = 30 = S2
So option (A) is answer.
On a wireless link, the probability of packet error is 0.2. A stop-and-wait protocol is used to transfer data across the link. The channel condition is assumed to be independent from transmission to transmission. What is the average number of transmission attempts required to transfer 100 packets?
So here it would be for one frame = 1/(1-0.2) = 1/0.8
So for 100 frames = 100/0.8 = 125
A program on machine X attempts to open a UDP connection to port 5376 on a machine Y, and a TCP connection to port 8632 on machine Z. However, there are no applications listening at the corresponding ports on Y and Z. An ICMP Port Unreachable error will be generated by
Y but not Z
Z but not Y
Neither Y nor Z
Both Y and Z
is necessarily 255.255.224.0
is necessarily 255.255.240.0
is necessarily 255.255.248.0
could be any one of 255.255.224.0, 255.255.240.0, 255.255.248.0
Broadcast address for subnet is
.95.255 or .01011111.11111111
(as in class B, 16 bits each are used for network and host)
So, we can take minimum 3 bits (from left) as subnet and make rest as host bits (as they are 1)
.224.0 → 11100000.00000000 (leftmost 3 bits for subnet)
.240.0 → 11110000.00000000 (leftmost 4 bits for subnet)
.248.0 → 11111000.00000000 (leftmost 5 bits for subnet)
R1 is one to many.
R2 is many to many.
→ E1 and E2 have separate table because they need to store multiple values.
→ R2 also have separate table by considering Primary keys E1 and E2 as foreign keys.
→ R1 is converted to many side table i.e., E2 as Primary key and E1 as Foreign key.
So, totally we need 3 tables to store the value.
Packet size have minimum header overhead.
Round trip propagation delay is RTT = 2*Tp
Minimum frame size of Ethernet can be found by using formula Tt = 2*Tp
Let L is minimum frame size. Then L / 10Mbps = 46.4 μs
It has nothing to do with jamming signal.
2n - 1
For Go-back N, the maximum window size can be 2n - 1.
In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges. Why is the spanning tree algorithm used for bridge-routing?
For shortest path routing between LANs
For avoiding loops in the routing paths
For fault tolerance
For minimizing collisions
TCP, but not UDP
TCP and UDP
UDP, but not TCP
Neither TCP nor UDP
Finding the IP address from the DNS
Finding the IP address of the default gateway
Finding the IP address that corresponds to a MAC address
Finding the MAC address that corresponds to an IP address
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
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.
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
For networks with high propagation delay this time becomes too long hence the minimum packet size required becomes too big to be feasible.
Bridge is a layer 2 device
Bridge reduces collision domain
Bridge is used to connect two or more LAN segments
Bridge reduces broadcast domain
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.
link state routing protocol.
distance vector routing protocol.
DNS while resolving host name.
TCP for congestion control.
The Bellman-Ford algorithm does not prevent routing loops from happening and suffers from the count-to-infinity problem.
A network with CSMA/CD protocol in the MAC layer is running at 1 Gbps over a 1 km cable with no repeaters. The signal speed in the cable is 2 × 108 m/sec. The minimum frame size for this network should be
L ≤ 2×Tp×B
L ≤ 2×(d/v)×B
d = 1Km = 1000m
v = 2×103 m/s
B = 109 bps
By solving the above equation we will set the value of L as,
A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. The transmission time of the acknowledgement frame is negligible. To get a channel efficiency of at least 50%, the minimum frame size should be
On a TCP connection, current congestion window size is Congestion Window=4 KB. The window size advertised by the receiver is Advertise Window=6 KB. The last byte sent by the sender is LastByteSent=10240 and the last byte acknowledged by the receiver is LastByteAcked=8192. The current window size at the sender is
= min (4KB, 6KB)
In a communication network, a packet of length L bits takes link L1 with a probability of p1 or link L2 with a probability of p2. Link L1 and L2 have bit error probability of b1and b2 respectively. The probability that the packet will be received without error via either L1 or L2 is
(1 – b1)L p1 + (1 – b2)Lp2
[1 – (b1 + b2)L]p1p2
(1 – b1)L (1 – b2)Lp1p2
1 – (b1 Lp1 + b2 Lp2)
Probability for no bit error for any single bit = (1 - b1)
Similarly, for link L2,
Probability of no bit error = (1 - b2)
Packet can go either through link L1 or L2, they are mutually exclusive events.
Probability packet will be received without any error = Probability of L1 being chosen and no errors in any of L bits + Probability of L2 being chosen and no error in any of the L bits
= (1 - b1)L p1 + (1 - b2)L p2
Hence, answer is option A.
In a TDM medium access control bus LAN, each station is assigned one time slot per cycle for transmission. Assume that the length of each time slot is the time to transmit 100 bits plus the end-to-end propagation delay. Assume a propagation speed of 2 × 108 m/sec. The length of the LAN is 1 km with a bandwidth of 10 Mbps. The maximum number of stations that can be allowed in the LAN so that the throughput of each station can be 2/3 Mbps is
A company has a class C network address of 22.214.171.124. It wishes to have three subnets, one with 100 hosts and two with 50 hosts each. Which one of the following options represents a feasible set of subnet address/subnet mask pairs?
10000000/128 (mask) - subnet id bit (1) (subnet 1)
01000000/192 (mask) - subnet id bit (01) (subnet 2)
0000000/192 (mask) - subnet id bit (00) (subnet 3)
Assume that “host1.mydomain.dom” has an IP address of 126.96.36.199. Which of the following options would be most appropriate as a subsequence of steps in performing the reverse lookup of 188.8.131.52? In the following options “NS” is an abbreviation of “nameserver”.
Query a NS for the root domain and then NS for the “dom” domains
Directly query a NS for “dom” and then a NS for “mydomain.dom” domains
Query a NS for in-addr.arpa and then a NS for 128.145.in-addr.arpa domains
Directly query a NS for 145.in-addr.arpa and then a NS for 128.145.in-addr.arpa domains
First we need to locate in-addr.apra, then perform reverse lookup of 184.108.40.206.in-addr.arpa which will point to host1.mydomain.com.
M = 1010001101
append 5 zeroes = M = 101000110100000
∴ CRC = 01110
Suppose that two parties A and B wish to setup a common secret key (D-H key) between themselves using the Diffie-Hellman key exchange technique. They agree on 7 as the modulus and 3 as the primitive root. Party A chooses 2 and party B chooses 5 as their respective secrets. Their D-H key is
where p is the primitive root and n is the modulus and 'a' and 'b' are the secret values selected by parity A & B.
So answer is,
32×5 mod 7 = 310 mod 7 = 4
P - 1, Q - 4, R - 3
P - 2, Q - 4, R - 1
P - 2, Q - 3, R - 1
P - 1, Q - 3, R - 2
Transport Layer :: Fourth layer of the OSI Model, Responsible for Service point addressing/Socket to socket connection or end to end connection with full reliability.
Network Layer :: Third layer of the OSI Model, Responsible for Host to Host.
Both bridge and router selectively forward data packets
A bridge uses IP addresses while a router uses MAC addresses
A bridge builds up its routing table by inspecting incoming packets
A router can connect between a LAN and a WAN
So bit rate is 9600 bps.
To send one char we need to send (1 + 8 + 2 +1) = 12
So total char send = 9600 / 12 = 800
A and B are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both A and B attempt to transmit a frame, collide, and A wins the first backoff race. At the end of this successful transmission by A, both A and B attempt to transmit and collide. The probability that A wins the second backoff race is
The probability that A wins the second back-off race = 5/8 = 0.625
More explanation in the video.
Eth1 and Eth2
Eth0 and Eth2
Eth0 and Eth3
Eth1 and Eth3
If results of ANDing subnet masks and IP address are same then subnet mask with higher number of 1s is preferred.
IP address 220.127.116.11 is AND with 255.255.255.0 results 18.104.22.168 Net ID which is similar to destination of this mask, but ANDing 22.214.171.124 with 255.255.255.128 also results same destination. So, here, mask with higher number of one is considered and router will forward packet to Eth1.
ANDing 126.96.36.199 with three subnet mask in table does not result in destination Net ID so router will forward this packet to default network via Eth2.
Data will be divided in three packets as:
First packet: 80 bytes + 20 byte of header
Second packet: 80 bytes + 20 byte of header
Third packet: 40 bytes + 20 byte of header
Note: Defragmentation (grouping of fragments) is done only at destination.
HC will receive total 260 bytes including header.
Application data is transferred at rate of (180/260) x 512 Kbps = 354.46 Kbps
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.
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
Sequence numbers can allow receivers to discard duplicate packets and properly sequence reordered packets.
If the congestion is deleted, the transmitter decreases the transmission rate by a multiplicative factor.
Acknowledgement allows the sender to determine when to retransmit lost packets.
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
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
= 26- 2
= 64 - 2
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
session and request only
request and response only
response and session only
session, request and response
l and II only
II and III only
l and III only
I, II and III
Then there are certain compliance rules which may be used for inheritance. So other statement (I) and (II) are True.
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
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).
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
M + N
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
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?
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?
0.16 = 0.2 × 0.8N-1
⇒ 0.8 = 0.8N-1
⇒ N = 2
Source IP address
Destination IP address
Destination port number
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?
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.
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
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?
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
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
Recovery from packet losses
Detection of duplicate packets
Packet delivery in the correct order
End to end connectivity
188.8.131.52 and 184.108.40.206
10.35.28.2 and 10.35.29.4
220.127.116.11 and 18.104.22.168
22.214.171.124 and 126.96.36.199
188.8.131.52 (Bitwise AND) 255.255.31.0 = 184.108.40.206
220.127.116.11 (Bitwise AND) 255.255.31.0 = 18.104.22.168
None of the above
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
L/107 = 2 x 10-5 = 200 bits = (200 / 8) bytes = 25 bytes
7.69 × 106 bps
11.11 × 106 bps
12.33 × 106 bps
15.00 × 106 bps
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.
010101 ⊕ 011001 = 001100
in this case receiver has to respond that receiver can be able to receive the data item.
Slowing down the communications