The contest portal will be open tomorrow onwards, whoever has given 35 as answer please do contest. A difference of 1 mark can bring huge changes in rank/score. Even if it doesn't get accepted, at least we tried.
I think the answer would be 35, if they would have asked "minimum integer value" but they have asked "round to the nearest integer" which is 34.
If "Rounded to the nearest integer" is the main concern then check the language of this question. Same language but they've given a range for this one.
Please refer to this ... and experts give your valuable reviews..
Question no. 24
Wrap around time
BW- 1 Gbps = 10^9 bps
BW = 10^9/8= 125 *10^6 Bytes per second
Now we know that sequence number field in TCP header is 32 bits
Which implies 2^32 sequence numbers..
Now since TCP is byte stream protocol thus for each byte one sequence number will be dedicated ….
Now according to given bandwidth 125 *10^6 sequence numbers will be exhausted in one second…
Thus for 2^32 sequence numbers it will take…
(2^32)/(125*10^6)= 34.359 seconds
Which means at this particular instant we will be completely exhausted of the sequence numbers..
So only after this time previous sequence number can be used to avoid any collision..
Now in question it is written that the “minimum time before this (I.e 1234) sequence number can be used again..”
Actually the dilemma is if we just round off without any concept then answer will turn out to be 34, but before this we cannot use the same sequence number.
And if we go conceptually then we have to apply CEIL to the answer which will give us 35 as the accurate answer is 34.35, so before this time unit same sequence number can’t be reused.
One more thing if we go with the language of question then 35 is more appropriate as minimum time before which we can use..Thus 34.35 lies behind 35 thus rounding off to 35 is conceptually right and genuine but if we use 34 then it can’t be minimum as before 34 it 's not possible to use sequence again..
BEWARE!!!!! it's Question 25
Can we challenge this question?
We can count 1-ary + 2-ary + 3-ary + 4-ary + 5-ary + 6-ary min heaps. Is this right?