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Recent questions and answers in Computer Networks
0
votes
1
answer
Madeeasy Testseries CN Q4
Here why can’t the answer be p1(1-b1)^L (1-p2)(1-p3) + (1-p1)(p2)(1-b2)^L(1-p3) + (1-p1)(1-p2)(p3)(1-b3)^L ?
answered
6 days
ago
in
Computer Networks
by
Abhisheksmile94
(
193
points)
|
22
views
networks
0
votes
0
answers
Madeeasy Test series CN Q3
Which of the following is a valid multicast MAC Address ? 01:00:5E:00:00:00 01:00:5E:00:00:FF 01:00:5E:00:FF:FF 01:00:5E:FF:FF:FF
asked
6 days
ago
in
Computer Networks
by
Shivateja MST
(
45
points)
|
7
views
networks
0
votes
1
answer
Madeeasy Test series CN Q2
Which of the following is/are correct regarding IPV4 Datagram Header? The minimum length of HLEN field is 5 and maximum is 20. Total length field has a minimum value of 20 Bytes and maximum of 65516 Bytes Fragmentation offset maximum value is 65514 Bytes. None of the above
answered
6 days
ago
in
Computer Networks
by
Abhisheksmile94
(
193
points)
|
47
views
networks
0
votes
0
answers
Computer network (self doubt)
In datagrams fragmentation occurs at routers. What about in virtual circuits? As CPU bandwidth and buffers are reserved, is there a chance for fragmentation to occur in router?
asked
Jan 9
in
Computer Networks
by
Kishore7
(
5
points)
|
8
views
computernetwork
test-series
cidr
selfdoubt
0
votes
1
answer
Applied Gate Test Series Computer Network
answered
Jan 9
in
Computer Networks
by
Sahil91
(
645
points)
|
46
views
computernetwork
0
votes
0
answers
Ace test series
asked
Jan 2
in
Computer Networks
by
val_pro20
(
2
points)
|
22
views
test-series
0
votes
1
answer
#self-doubt #computer-networks
dhcp works at which layer?[ network layer or application layer] arp works at which layer?[data link layer or network layer]
answered
Dec 29, 2020
in
Computer Networks
by
wander
(
287
points)
|
16
views
computernetwork
0
votes
0
answers
#made-easy #computer-networks
consider the following figure of connection release.what will be B values ,if A is 21? in ans it is given 52.but it should be 51 bcz in connection release server sends FIN+ACK packet but ack packet does not consume any sequence number.Am i correct?
[closed]
asked
Dec 29, 2020
in
Computer Networks
by
404 found
(
31
points)
|
14
views
computernetwork
0
votes
1
answer
An IPv4 Packet in Hexadecimal digits.
Following is the Packet : 0 x 4 6 0 0 0 0 2 8 0 1 0 1 2 0 5 0 0 1 0 6 0 0 0 0 my question is that what is the size of options and how many bytes of data is being carried by this packet and what is the identification number of datagram fragment?
answered
Dec 29, 2020
in
Computer Networks
by
wander
(
287
points)
|
28
views
computernetwork
0
votes
0
answers
Made easy: IP data field can contain which of these?
asked
Dec 29, 2020
in
Computer Networks
by
ytheuyr
(
7
points)
|
16
views
selfdoubt
0
votes
1
answer
Applied Course test
An organisation have the network address as 129.50.0.0. The organisation wants to maximize the hosts and also wants 8 subnets for the organisation. To get the maximum hosts and 8 subnets what will be the subnet mask number.
answered
Dec 29, 2020
in
Computer Networks
by
saurabh111
(
11
points)
|
14
views
subnetting
ip-addressing
+1
vote
1
answer
GATE2006-IT-65 Video Solution
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 are such codewords possible? $14$ $16$ $18$ $20$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
10
views
gate2006-it
computer-networks
encoding
combinatory
normal
video-solution
0
votes
1
answer
GATE2007-IT-61 Video Solution
In the waveform (a) given below, a bit stream is encoded by Manchester encoding scheme. The same bit stream is encoded in a different coding scheme in wave form (b). The bit stream and the ... Manchester respectively $0111101000$ and Differential Manchester respectively $1000010111$ and Integral Manchester respectively $0111101000$ and Integral Manchester respectively
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
31
views
gate2007-it
computer-networks
communication
manchester-encoding
normal
video-solution
+1
vote
1
answer
GATE2004-IT-45 Video Solution
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 ... $136$ characters/sec $100$ characters/sec, $136$ characters/sec $80$ characters/sec, $153$ characters/sec
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
9
views
gate2004-it
computer-networks
serial-communication
normal
video-solution
+1
vote
1
answer
GATE2008-IT-18 Video Solution
How many bytes of data can be sent in $15$ seconds over a serial link with baud rate of $9600$ in asynchronous mode with odd parity and two stop bits in the frame? $10,000$ bytes $12,000$ bytes $15,000$ bytes $27,000$ bytes
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
6
views
gate2008-it
computer-networks
communication
serial-communication
normal
video-solution
+1
vote
1
answer
GATE2004-22 Video Solution
How many $8-bi$t characters can be transmitted per second over a $9600$ baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits and one parity bit? $600$ $800$ $876$ $1200$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
14
views
gate2004
computer-networks
serial-communication
normal
video-solution
+1
vote
1
answer
GATE2015-3-36 Video Solution
Two hosts are connected via a packet switch with $10^7$ bits per second links. Each link has a propagation delay of $20$ microseconds. The switch begins forwarding a packet $35$ microseconds after it receives the same. If $10000$ bits of data ... between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is ______.
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
15
views
gate2015-3
computer-networks
normal
numerical-answers
network-switching
video-solution
+1
vote
1
answer
GATE2007-IT-62 Video Solution
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 $20$ ... unit. Then the average number of backlogged of packets per time unit during the given period is $5$ $4.45$ $3.45$ $0$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
12
views
gate2007-it
computer-networks
communication
normal
video-solution
+1
vote
1
answer
GATE2008-IT-66 Video Solution
Data transmitted on a link uses the following $2D$ parity scheme for error detection: Each sequence of $28$ bits is arranged in a $4\times 7$ matrix (rows $r_0$ through $r_3$, and columns $d_7$ through $d_1$) and is padded with a column $d_0$ ... received by a receiver and has $n$ corrupted bits. What is the minimum possible value of $n$? $1$ $2$ $3$ $4$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
10
views
gate2008-it
computer-networks
normal
error-detection
video-solution
+1
vote
1
answer
GATE2005-IT-78 Video Solution
Consider the following message $M = 1010001101$. The cyclic redundancy check (CRC) for this message using the divisor polynomial $x^5+x^4+x^2+1$ is : $01110$ $01011$ $10101$ $10110$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
14
views
gate2005-it
computer-networks
crc-polynomial
normal
video-solution
+1
vote
1
answer
GATE2007-68, ISRO2016-73 Video Solution
The message $11001001$ is to be transmitted using the CRC polynomial $x^3 +1$ to protect it from errors. The message that should be transmitted is: $11001001000$ $11001001011$ $11001010$ $110010010011$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
11
views
gate2007
computer-networks
error-detection
crc-polynomial
normal
isro2016
video-solution
+1
vote
1
answer
GATE2006-IT-68 Video Solution
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 of transmission to transmission. What is the average number of transmission attempts required to transfer $100$ packets? $100$ $125$ $150$ $200$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
4
views
gate2006-it
computer-networks
sliding-window
stop-and-wait
normal
video-solution
+1
vote
1
answer
GATE2015-1-53 Video Solution
Suppose that the stop-and-wait protocol is used on a link with a bit rate of $64$ $\text{kilobits}$ per second and $20$ $\text{milliseconds}$ propagation delay. Assume that the transmission time for the acknowledgment and the processing time ... . Then the minimum frame size in bytes to achieve a link utilization of at least $50$ $\text{%}$ is_________________.
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
9
views
gate2015-1
computer-networks
stop-and-wait
normal
numerical-answers
video-solution
0
votes
1
answer
GATE2005-IT-72 Video Solution
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 acknowledgment frame is negligible. To get a channel efficiency of at least $50$\text{%}$, the minimum frame size should be $80$ $\text{bytes}$ $80$ $\text{bits}$ $160$ $\text{bytes}$ $160$ $\text{bits}$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
9
views
gate2005-it
computer-networks
network-flow
stop-and-wait
normal
video-solution
+1
vote
1
answer
GATE2015-2-8 Video Solution
A link has transmission speed of $10^6$ bits/sec. It uses data packets of size $1000$ $\text{bytes}$ each. Assume that the acknowledgment has negligible transmission delay and that its propagation delay is the same as the data propagation delay. Also ... $\text{%}$. The value of the one way propagation delay (in milliseconds) is_____.
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
10
views
gate2015-2
computer-networks
mac-protocol
stop-and-wait
normal
numerical-answers
video-solution
0
votes
1
answer
GATE2016-1-55 Video Solution
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 (1 Kbps = 1000 bits/second). Size of an acknowledgment is $100$ bytes ... . The one-way propagation delay is $100$ milliseconds. Assuming no frame is lost, the sender throughput is ________ bytes/ second.
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
11
views
gate2016-1
computer-networks
stop-and-wait
normal
numerical-answers
video-solution
+1
vote
1
answer
GATE2017-1-45 Video Solution
The values 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 ... (expressed in percentage) of the Stop-and-Wait ARQ protocol for the above parameters is _____________ (correct to $2$ decimal places).
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
15
views
gate2017-1
computer-networks
stop-and-wait
numerical-answers
normal
video-solution
+1
vote
1
answer
GATE2003-84 Video Solution
Host $A$ is sending data to host $B$ over a full duplex link. $A$ and $B$ are using the sliding window protocol for flow control. The send and receive window sizes are $5$ packets each. Data packets (sent only from $A$ to $B$) are all $1000$ bytes long and the transmission ... ? $7.69 \times 10^6$ Bps $11.11 \times 10^6$ Bps $12.33 \times 10^6$ Bps $15.00 \times 10^6$ Bps
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
10
views
gate2003
computer-networks
sliding-window
normal
video-solution
+1
vote
1
answer
GATE2006-44 Video Solution
Station $A$ uses $32$ $\text{byte}$ packets to transmit messages to Station $B$ using a sliding window protocol. The round trip delay between A and $B$ is $80$ $\text{milliseconds}$ and the bottleneck bandwidth on the path between $A$ and $B$ is $128$ $\text{kbps}$ . What is the optimal window size that $A$ should use? $20$ $40$ $160$ $320$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
9
views
gate2006
computer-networks
sliding-window
normal
video-solution
+1
vote
1
answer
GATE2007-69 Video Solution
The distance between two stations $M$ and $N$ is $L$ kilometers. All frames are $K$ bits long. The propagation delay per kilometer is $t$ seconds. Let $R$ bits/second be the channel capacity. Assuming that the processing delay is negligible, the $\text{minimum}$ number ... $\lceil \log_2 \frac{2LtR +K}{K} \rceil$ $\lceil \log_2 \frac{2LtR +2K}{2K} \rceil$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
9
views
gate2007
computer-networks
sliding-window
normal
video-solution
0
votes
1
answer
GATE2004-IT-81 Video Solution
In a sliding window $ARQ$ scheme, the transmitter's window size is $N$ and the receiver's window size is $M$. The minimum number of distinct sequence numbers required to ensure correct operation of the $ARQ$ scheme is $min (M, N)$ $max (M, N)$ $M + N$ $MN$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
6
views
gate2004-it
computer-networks
sliding-window
normal
video-solution
0
votes
1
answer
GATE2004-IT-83 Video Solution
A $20$ $\text{Kbps}$ satellite link has a propagation delay of $400$ $\text{ms}$. The transmitter employs the "go back $n$ $ARQ$" scheme with $n$ set to $10$. Assuming that each frame is $100$ $\text{byte}$ long, what is the maximum data rate possible? $5$ $\text{Kbps}$ $10$ $\text{Kbps}$ $15$ $\text{Kbps}$ $20$ $\text{Kbps}$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
10
views
gate2004-it
computer-networks
sliding-window
normal
video-solution
+1
vote
1
answer
GATE2008-IT-64 Video Solution
A $1$ $\text{Mbps}$ satellite link connects two ground stations. The altitude of the satellite is $36,504$ $\text{km}$ and speed of the signal is 3 108 m/s. What should be the packet size for a channel utilization of $25$ ... no errors during communication. $120$ $\text{bytes}$ $60$ $\text{bytes}$ $240$ $\text{bytes}$ $90$ $\text{bytes}$
answered
Dec 24, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
11
views
gate2008-it
computer-networks
sliding-window
normal
video-solution
0
votes
0
answers
Made Easy Test Series
asked
Dec 24, 2020
in
Computer Networks
by
Abhisheksmile94
(
193
points)
|
7
views
computernetwork
+1
vote
1
answer
GATE2006-46 Video Solution
Station $A$ needs to send a message consisting of $9$ packets to Station $B$ using a sliding window (window size $3$) and go-back-n error control strategy. All packets are ready and immediately available for transmission. If every 5th packet that $A$ transmits gets ... what is the number of packets that $A$ will transmit for sending the message to $B$? $12$ $14$ $16$ $18$
answered
Dec 23, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
23
views
gate2006
computer-networks
sliding-window
normal
video-solution
+1
vote
1
answer
GATE2015-3-28 Video Solution
Consider a network connecting two systems located $8000$ $\text{Km}$ apart. The bandwidth of the network is $500 \times 10^6$ $\text{bits}$ per second. The propagation speed of the media is $4 \times 10^6$ $\text{meters}$ per second ... . Assume that processing delays at nodes are negligible. Then, the minimum size in bits of the sequence number field has to be ______.
answered
Dec 23, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
15
views
gate2015-3
computer-networks
sliding-window
normal
numerical-answers
video-solution
+1
vote
1
answer
GATE2016-2-55 Video Solution
Consider a $128 \times 10^3$ 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. ... of acknowledgement. The minimum number of bits required for the sequence number field to achieve $100 \%$ utilization is ________.
answered
Dec 23, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
13
views
gate2016-2
computer-networks
sliding-window
normal
numerical-answers
video-solution
+1
vote
1
answer
GATE2012-44 Video Solution
Consider a source computer $(S)$ transmitting a file of size $10^{6}$ bits to a destination computer $(D)$ over a network of two routers $(R_{1}\text{ and }R_{2})$ and three links $(L_{1},L_{2},\text{ and } L_{3})$. $L_{1}$ connects $S$ to ... propagation delays in transmitting the file from $S$ to $D$? $\text{1005 ms}$ $\text{1010 ms}$ $\text{3000 ms}$ $\text{3003 ms}$
answered
Dec 23, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
11
views
gate2012
computer-networks
communication
normal
video-solution
+1
vote
1
answer
GATE2007-65 Video Solution
There are $n$ stations in slotted LAN. Each station attempts to transmit with a probability $p$ in each time slot. What is the probability that ONLY one station transmits in a given time slot? $np(1-p)^{n-1}$ $(1-p)^{n-1}$ $p(1-p)^{n-1}$ $1-(1-p)^{n-1}$
answered
Dec 23, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
9
views
gate2007
computer-networks
lan-technologies
probability
normal
video-solution
+1
vote
1
answer
GATE2015-1-29 Video Solution
Consider a LAN with four nodes $S_1, S_2, S_3,$ and $S_4$. 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 ... respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is__________________.
answered
Dec 23, 2020
in
Computer Networks
by
amitkhurana512
(
35
points)
|
10
views
gate2015-1
computer-networks
normal
numerical-answers
congestion-control
video-solution
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