GATE CSE 2022 | Question: 50

Question

Consider the data transfer using $\text{TCP}$ over a $1 \; \text{Gbps}$ link. Assuming that the maximum segment lifetime $\text{(MSL)}$ is set to $60 \; \text{seconds},$ the minimum number of bits required for the sequence number field of the $\text{TCP}$ header, to prevent the sequence number space from wrapping around during the $\text{MSL}$ is ________________.

Comments

Answer 33 Ok But How can 33 Bits represent when TCP Header can Maximum can Hold 32 Bits of Sequence??? Am i Missing Something

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Manav_Sarang for this we use extra bit from option field..

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Answer 1

Answer : 33 bits

Given Bandwidth = $1Gbps$, MSL = $60 sec$.

Now, in $1$ sec, amount of data consumed = $1Gb$

Hence. in $60 sec$, amount of data consumed = $60Gb$ $\implies$ $(60Gb/8)$ $\implies$ $7.5GB$

Therefore, to prevent the sequence number space from wrapping around during the MSL, we need $7.5GB$ of data which can be represented with:-

$\lceil log(7.5 * 2^{30}) \rceil$ $\implies$ $\lceil log(7.5) + log(2^{30}) \rceil$ $\implies$ $\lceil 2.9+30 \rceil$  $\implies$ $33 bits$ for the sequence number field.

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The Answer will also be same if we use $1Gbps = 10^9bps$

Minimum no. of bits required in Sequence number field of the $\text{TCP}$ header to prevent the sequence number space from wrapping around during the $\text{MSL}$ is given by:-

$\lceil log(MSL*B_w) \rceil$ where $MSL$ (maximum segment lifetime) = $60 \; \text{seconds}$ [given]

and $B_w$ (Bandwidth in $Bps$) =  $10^9bps = 10^9/8 Bps$ [given]

Thus,

Min. no. of bits required= $\lceil log(60* 10^9/8) \rceil$

$=\lceil log(60*10^4) + log(10^5/8) \rceil$ [Breaking $10^9$ so that it becomes easy to use the scientific calculator in exam]

$=\lceil19.19460+  13.60964 \rceil$

$=\lceil 32.80424\rceil = 33bits$ for the sequence number field.

Comments

Perfect !

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good explanation

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In Bandwidth we take k=2^3,  M=2^6 G as 2^9 right ??

Answer 2

33 bits.

As in 60 seconds , 60Gb of data will get transferred which in turn is about 7.5GB and can be represented in 33 bits

Comments

to prevent the sequence number space from wrapping around during the MSL is=$\ln (BW*sement lifetime)$

but carefully see bandwidth is given in bit per sec.

so first convert bandwidth into byte because we have to find no of bit required for (2KB IS 11bit )

that means the unit of (  bw*time is byte if find no bit)

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($1$ Sequence Number $= 1$ Byte)

In $1$ sec you can transfer $→$ $1G\ bits = 2^{30}\ bits = 2^{27}\ Bytes = 2^{27}$ Sequence Numbers.

In $60$ sec you can transfer $→ 60*2^{27} Bytes = 60*2^{27}$ Sequence Numbers $\approx2^6*2^{27} = 2^{33}$ Sequence Numbers.

So you need $33\ bits$ for representing $2^{33}$ sequence numbers.

Note: The answer will be same if you assume $1G$ as $10^9$ .

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Why do we take upper bound?

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@psnt_rwt bcz wrap around time should be greater as compared to MSL

Answer 3

33 bits.

WAT = $2^{32}$/x (x is bandwidth which should be in Bytes/sec).
Value of x = $10^{9}$ bits/sec  = $10^{9}/8$  B/sec.

So, WAT = $2^{32} * 8 / 10^{9}$ =  34.35 sec.
And according to question the lifetime of TCP segment is set to 60sec.

It can easily be observed that : 2 * WAT ( 68.70 ) > TCP Lifetime ( 60 sec ).
For WAT already calculated using 32 bits and if we multiply it by 2 this will give 33 bits.

Comments

why * by 2?

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TCP Lifetime is given as 60sec. So we want to see that if we have 1 more bit then will WAT > TCP Lifetime. That’s why we multiplied it by 2. If * 2 don’t work then we will go for * 4 and so on.

Answer 4

let suppose SEQUNCE NO = $X$ BITS

WINDOW SIZE=$2^{X}$ B

WRAP AROUND TIME=  $(2^{X} * 2^{3}) / 2^{30}$ SEC

WRAP AROUND TIME > MSL

  $(2^{X} * 2^{3}) / 2^{30}$  $> 60$  {why taking greater bcz WAT should be greater than MSL bcz atleast after these many seconds it can repeat the sequence number}

= $33$

Comments

Best Answer by far. Straight forward and to the point. Thanks!

Answer 5

Just posting for points I know no one will read it

Comments

maine read kara bhai 😁

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Nice Explanation

Answer 6

 

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Bandwidth = 1 Gbps : 10^9 not 10^30

Maximum Segment Lifetime (MSL) = 60 seconds

Formula:
Minimum sequence number required to avoid wraparound within MSL =

Ceil ( Log (BW * MSL))

Calculations:

1. 1 second = 10910^9 sequence numbers (in Gb)
2. In 60 seconds: x=60×10^9 Gb

3. Convert x to GB by dividing by 8:

        x=60×10^9/8=75×10^8 GB

4. To calculate the minimum sequence bits:

Sequence bits=⌈log⁡(75×10^8)/log⁡2⌉

5. Simplify using logarithmic properties:

Sequence bits=⌈(log⁡(75)+8)/log⁡2⌉

6. Calculate:

Sequence bits=⌈(1.875+8)/0.301⌉=⌈32.8⌉=33

Final Answer:

Minimum sequence bits required = 33

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