GATE CSE 2018 | Question: 25

Question

Consider a long-lived $\text{TCP}$ session with an end-to-end bandwidth of $1\; \text{Gbps}\; (=10^9$ bits-per-second$).$ The session starts with a sequence number of $1234$. The minimum time (in seconds, rounded to the closet integer) before this sequence number can be used again is _________.

Comments

but we have extra 10 bits in options, so by that logic the answer should be 2^42*2^3*10^-9, correct me if am wrong..??

Kindly let all know what would be the default to take n this case if this question reappears..??

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Total sequence numbers possible: $2^{32}$

Every byte is given a unique sequence number in TCP.

 

Consuming $2^{32}$ sequence numbers would take

$$\frac{2^{32}*8}{10^{9}}s$$

$$=> \frac{34359738368}{1000000000}$$

$$=> 34.36$$

$$=> 34$$

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So, will this remain same for any sequence number and 10^9 bandwidth?
Also, if we are asked to find wrap around time and sequence number and bandwidth are given,
 

Is the formula as below?
WAT = (2^32 *8) /bandwidth

Kindly help!

Answer 1

In question i think BW is 1Gbps... 1 Giga bits per second

and as we know TL is byte streaming protocol and it has 32 bits for sequence number ..

so ..its wrap around time = sequence number / BW

BW in GBps= 1Gbps/8   ( convert bits into bytes)

so wrap around time = (2^32)*8 /(10^9) seconds = 34.35 seconds ...

Comments

I have written 35. Should it be considered right ? Any Idea

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NO idea

Answer 2

total seq number =2^32

wrap around time =2^32/(10^9/8)=(approx)34