Search results for sliding-window / GATE Overflow for GATE CSE

0 votes

2 answers

21

Sliding window, Go-Back-N protocol, All ack lost case
I understood that receiver will correctly discard the frames in GBN all ack lost case. But its given that its positive ack and hence receiver wont send ack for discarded frames.Wont the sender be sending the same frames again and again ... timeout as receiver will never send ack again? So how will the sender slide the window to the next frames?

I understood that receiver will correctly discard the frames in GBN all ack lost case. But its given that its positive ack and hence receiver wont send ack for discarded ...

fred20978

1.1k views

fred20978 asked Nov 8, 2022

5 votes

4 answers

22

ISRO2018-44
Station$A$ uses $32$ byte packets to transmit messages to Station $B$ using a sliding window protocol. The round trip delay between $A$ and $B$ is $80$ $ms$ and the bottleneck bandwidth on the path between $A$ and $B$ is $128$ $kbps$. What is the optimal window size that $A$ should use? $20$ $40$ $160$ $320$

Station$A$ uses $32$ byte packets to transmit messages to Station $B$ using a sliding window protocol. The round trip delay between $A$ and $B$ is $80$ $ms$ and the bottl...

Arjun

3.0k views

Arjun asked Apr 22, 2018

7 votes

6 answers

23

ISRO2008-8
The TCP sliding window can be used to control the flow of information always occurs when the field value is $0$ always occurs when the field value is $1$ occurs horizontally

The TCP sliding windowcan be used to control the flow of informationalways occurs when the field value is $0$always occurs when the field value is $1$occurs horizontally

go_editor

4.5k views

go_editor asked Jun 10, 2016

60 votes

5 answers

24

GATE IT 2006 | Question: 64
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 ... $1- \dfrac{ N}{i}$ $\dfrac{i}{(N + i)}$ $1$ $1 - e^{\left(\frac{i}{N}\right)}$

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...

Ishrat Jahan

12.9k views

Ishrat Jahan asked Nov 1, 2014

1 votes

1 answer

25

MADE EASY test series | Computer Networks
Can someone please suggest weather to go for 20.5 or 21 in these kind of problems in GATE? As logically 21 is correct, but mathematically 20.5 is correct. Both the answers can be reasoned correct and wrong, feels like “Umpire’s call” in DRS, cricket.

Can someone please suggest weather to go for 20.5 or 21 in these kind of problems in GATE? As logically 21 is correct, but mathematically 20.5 is correct.Both the answers...

Souvik33

435 views

Souvik33 asked Jan 3, 2023

2 votes

1 answer

26

Solution is 2 or 3?
Should the solution for the above question be 2 or 3?

Should the solution for the above question be 2 or 3?

h4kr

521 views

h4kr asked Dec 9, 2022

2 votes

2 answers

27

gateforum

h4kr

438 views

h4kr asked Nov 2, 2022

1 votes

3 answers

28

CN: Go Back N
Station A needs to send a message consisting of 15 packets to station B' using a sling window (window size 4) and go-back-N error control strategy. All packets are ready and immediately available for transmission. If every 6th packet that A' transmits gets lost (but no Acks from ... of packets that A' will transmit for sending the message to B' ? (a) 29 (b) 33 (c) 27 (d) 25

Station A needs to send a message consisting of 15 packets to station ‘B’ using a sling window (window size 4) and go-back-N error control strategy. All packets are r...

Prasanna

2.1k views

Prasanna asked Jan 4, 2016

0 votes

0 answers

29

Computer Networks
In the given solution , while calculating transmission time why 8 is not multiplied i.e. Length = 2000Byte = 2000 * 8 bits formula is not used.

In the given solution , while calculating transmission time why 8 is not multiplied i.e.Length = 2000Byte = 2000 * 8 bits formula is not used.

Overflow04

578 views

Overflow04 asked Aug 7, 2022

2 votes

0 answers

30

Testseries-doubt
“An SR sender’s window base cannot a have lower sequence number than the receiver window base” can some one explain this statement ? Why it is false

“An SR sender’s window base cannot a have lower sequence number than the receiver window base”can some one explain this statement ? Why it is false

Kabir5454

475 views

Kabir5454 asked Jul 13, 2022

1 votes

1 answer

31

MadeEasy Test Series: Computer Networks -Sliding Window
I am unable to understand their explanation,can anyone explain it in a better way?

I am unable to understand their explanation,can anyone explain it in a better way?

sripo

607 views

sripo asked Jan 5, 2019

43 votes

8 answers

32

GATE IT 2006 | Question: 68
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? $100$ $125$ $150$ $200$

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 in...

Ishrat Jahan

13.5k views

Ishrat Jahan asked Nov 1, 2014

0 votes

2 answers

33

Self doubt CN
In the connection termination phase, suppose the FIN segment from the client-side contains data ranging from sequence no 100 to 200, So will the ACK no from the server start from 201 or 202? Will the FIN flag consume 1 extra sequence number at the end, ... context of my question, will that sequence number be 99, before the beginning of the data segment or 201, after the data segment

In the connection termination phase, suppose the FIN segment from the client-side contains data ranging from sequence no 100 to 200,So will the ACK no from the server sta...

tusharb

1.1k views

tusharb asked Jan 24, 2022

0 votes

1 answer

34

slow start phase self doubt
when no of RTT is asked , i am confused when will start from 1 AND when will start from 2 rest part i known folow until thresold size and then increse with given MSS LET SUPPOSE THRESOLD 20 FOR EX ::: 1 2 4 8 16 20 22 24 26 OR 2 4 8 16 20 22 24 26 in ex 1 no of rtt is 9 while in ex 2 no of rtt is 8 can anyone tell me when we will start 1 and when we will start with 2

when no of RTT is asked ,i am confused when will start from 1 AND when will start from 2rest part i known folow until thresold size and then increse with given MSSLE...

jugnu1337

361 views

jugnu1337 asked Dec 28, 2021

30 votes

5 answers

35

GATE IT 2004 | Question: 81
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 $\min (M, N)$ $\max (M, N)$ $M + N$ $MN$

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 ...

Ishrat Jahan

10.6k views

Ishrat Jahan asked Nov 2, 2014

3 votes

6 answers

36

Testbook Test Series: Computer Networks - Sliding Window
packets are being transmitted using GB5 and here every 4th packet is lost.How many packets need to be transmitted to transmit 10 packets? my answr 18 given :20

packets are being transmitted using GB5 and here every 4th packet is lost.How many packets need to be transmitted to transmit 10 packets?my answr 18 given :20

Aboveallplayer

2.4k views

Aboveallplayer asked Jan 15, 2017

2 votes

10 answers

37

UGC NET CSE | January 2017 | Part 3 | Question: 26
Station $A$ uses $32$ byte packets t transmit messages to station $B$ using sliding window protocol. The round trip delay between $A$ and $B$ is $40$ milliseconds and the bottleneck bandwidth on the path between $A$ and $B$ is $64$ kbps. The optimal window size of $A$ is $20$ $10$ $30$ $40$

Station $A$ uses $32$ byte packets t transmit messages to station $B$ using sliding window protocol. The round trip delay between $A$ and $B$ is $40$ milliseconds and the...

go_editor

4.2k views

go_editor asked Mar 24, 2020

5 votes

3 answers

38

In Go-Back-N protocol, if the maximum window size is 127, what is the range of the sequence number?
In Go-Back-N protocol, if the maximum window size is 127, what is the range of the sequence number? A. 0 to 127 B. 0 to 128 C. 1 to 127 D. 1 to 128 Answer given is A Please explain :) Thanks.

In Go-Back-N protocol, if the maximum window size is 127, what is the range of the sequence number?A. 0 to 127 B. 0 to 128 C. 1 to 127 D. 1 to 128Answer given is APlease ...

sunil sarode

13.7k views

sunil sarode asked Oct 6, 2017

3 votes

4 answers

39

Go back N (arq) protocol
Q. If maximum sequence number in Go back N ( arq) protocol is 'n' then window size (sender) is ? If your answer is 'n' or 'n'+1' then please design GBN for sender window 5 and 11 .

Q. If maximum sequence number in Go back N ( arq) protocol is 'n' then window size (sender) is ?If your answer is 'n' or 'n'+1' then please design GBN for sender windo...

Abhi Som

12.9k views

Abhi Som asked Jul 19, 2017

0 votes

1 answer

40

Stop and wait protocol
R=transmission rate S=signal speed D=Distance between sender and receiver T=time to create one frame F=number of bits in a frame N=number of data bits in a frame A=number of bits in an acknowledgement F>N Q.1 Calculate the channel utilization My take: (F/R) / (F/R + 2*D/S + A/R ... .2 Calculate the effective data rate. My take: ( (F/R) / (F/R + 2*D/S + A/R + 2*T) ) * (N/F)

R=transmission rateS=signal speedD=Distance between sender and receiver T=time to create one frameF=number of bits in a frameN=number of data bits in a frameA=number of b...

Prashant Sharma 1

1.0k views

Prashant Sharma 1 asked Jun 22, 2016

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