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asked in Computer Networks by Boss (5.4k points) 3 5 29 | 73 views
2600 msec??
@Subhanshu:should be $1300$ msec , i think you have taken $MSS=1KB$
@sourav. it is given in the question PT not RTT.
oh yes, Actually i did not notice that . Thanks!
correct ans is 3000ms.

if the answer is 3000msec then question should be updated accordingly,

they should ask, Time taken (in msec) by TCP to get back to 38 KB congestion window and the data should be acknowledged.

yes you are right there should be 38 kB instead of 36 kB. but if 36 kB then ans will be 2800ms but you say 2600ms am i correct ??
it is 2600 ms or 2800 ms, it depends on wording of the question,

If it is written that, the data should be acknowledged, which means our task will complete when the data which we have sent will be acknowledged, which means we will get the ack, for that we require one more ack, then in that case answer will be 2800 ms.

And if it is written in the qeustion that "time required to sent window size of 36 KB" then in this case we have 2600 ms time, because in this case we are not worried about whether the data is recieved or not, they are just asking after what time you will send full window of 36 KB, so in that case answer should be 2600 ms,

According to given question, there intention is second case, which is 2600 ms.

@Shubhanshu i see many questions in which  after what time you will send full window  is mention but they consider last ack also .

In this @Bikram sir, said, both the answers are in the range in official key of gate.

okiii.. thank u

2 Answers

–1 vote


UGC NET (National Eligibility test) Exams is conducted by CBSE (Central Board of Secondary Education) for determining the eligibility of the Indian Nationals and aspirants for the post of the Assistant Professor for the Junior Research fellowship

The Exam covers Computer Science one of the important subject

There are various UGC Fake Universities you can check out

answered by (35 points)
–1 vote

UGC NET Computer science is a very important subject in UGC Exams

Paper III (A) will have 10 short essay type questions (300 words) carrying 16 marks each. In it there will be one question with internal choice from each unit (i.e. 10 questions from 10 units; Total marks will be 160).

UGC NET CS Paper III (B) will be compulsory and there will be one question from each of the Electives. The candidate will attempt only one question (one elective  in 800 words) carrying 40 marks. Total marks of UGC NET Computer Science  Paper III will be 200.

answered by (35 points)

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