GATE CSE 2023 | Question: 40

Question

Suppose you are asked to design a new reliable byte-stream transport protocol like $\text{TCP}.$ This protocol, named $\textsf{myTCP}$, runs over a $100 \mathrm{~Mbps}$ network with Round Trip Time of $150$ milliseconds and the maximum segment lifetime of $2$ minutes.

Which of the following is/are valid lengths of the Sequence Number field in the $\textsf{myTCP}$ header?

• A. $30$ bits
• B. $32$ bits
• C. $34$ bits
• D. $36$ bits

Comments

Since the sequence number field has a finite fixed length of $n$ bits, the sequence numbers will eventually be exhausted and "wrap around" back to zero. The total number of unique bytes that can be sequenced is $2^n$.

==> First, convert the network bandwidth $BW$ to the Byte transmission rate 

$$R = \frac{BW}{8}$$

==> Wrap-Around Time ($T_{wrap}$): The time to consume the entire sequence space of $2^n$ bytes.

$$T_{wrap} = \frac{2^n}{R}$$

The wrap-around time must be greater than or equal to the MSL.

$$T_{wrap} \ge \text{MSL}$$

By substituting 

$$\frac{2^n}{R} \ge \text{MSL}$$

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Here, RTT is given just for distraction or to waste our time. So avoid it when asked Sequence No.

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Answer : 32,34,36bits

Given Bandwidth = 1Mbps , MSL = 120sec.

1 sec, amount of data consumed = 100Mb

120sec, amount of data consumed = 120 * 100 Mbs⟹2^7 *2^7 * 2^20/2^3 = 2^31

# sequence number bits  ≥ ⌈log(2^31)⌉ ⟹31bits for the sequence number field.

Answer 1

Answer: B, C, D

Given bandwidth = $100 \text{ Mbps}$

$1 \text{ second} = 100*10^6 \text{ bits} = 125 * 10^5 \text{ bytes}$

$\text{In 120 seconds} = 15000 * 10^5 \text{ bytes}$  

Lifetime of a segment is $120$ seconds, and in that period $15*10^8 \text{ bytes}$ are generated,

So, $\text{# sequence number bits} \geq \lceil log_2(15*10^8) \rceil = \lceil (30.48) \rceil = 31 \text{ bits}$

Comments

Could you clarify @RygelKenoid. I don't understand what you said clearly....

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@RygelKenoid No it doesn't. The amount we can send has nothing to do with RTT. It only depends on the effective tcp window size and we have no reason to doubt that it is not adequate. So we don't have to wait for ACKS and transmit for however long we want. RTT in the question is a red herring to confuse the reader.

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@RygelKenoid

Can you please explain the significance of substracting it in 120 seconds....

lets say the packet reaches exactly at 2 minute so it will send a ACK then that will take 0.75 msec (as round trip is 150)

we should have a minimum sequence numbers so that we can keep of sending them so 2 min and 0.75 milisec at most (in worst case)

thats around 120+0.00075 sec = 120.00075 seconds 

IF I AM MISSING SOMETHING OR MY CONCEPTS ARE WRONG PLEASE FELL FREE TO CORRECT IT

Answer 2

The question asks about a hypothetical protocol called myTCP, which runs over a 100 Mbps network with a round-trip time of 150 milliseconds and a maximum segment lifetime of 2 minutes. You're asked to determine the valid lengths of the sequence number field in the myTCP header, with options being 30, 32, 34, or 36 bits​1​.

The calculation of the sequence number field size is based on the maximum amount of data that can be in flight in the network. Since the maximum segment lifetime is 2 minutes (or 120 seconds), we need to calculate how much data can be sent over the network in this time.

Given the bandwidth of 100 Mbps, 1 second translates to 100*10^6 bits, or 125 * 10^5 bytes. Hence, in 120 seconds, we can send 15000 * 10^5 bytes.

The sequence number field needs to be large enough to assign a unique number to each byte that could be in flight. This means we need a field size that can represent up to 15000 * 10^5 bytes. Taking the base-2 logarithm of this number (to get the number of bits required to represent this many distinct values) and rounding up to the nearest integer (since we can't have a fraction of a bit), we find that we need at least 31 bits.

Therefore, the valid lengths for the sequence number field in the myTCP header are 32 bits, 34 bits, and 36 bits, corresponding to options B, C, and D​

Answer 3

B, C, D are correct

Answer 4

Suppose you are asked to design a new reliable byte stream transport protocol like TCP.  

This protocol, named as myTCP, runs over a 100 mbps network with round trip time of  

150 milliseconds and max segment lifetime of 2 minutes.  

Which of following valid length of the sequence number field in the myTCP header?  

(a) 30 bit (b) 34 bit  

(c) 32 bit (d) 36 bit  

Ans. (b, c, d)  

 

Question 4.2.4

@goclasses

Comments

How b,c,d is coming?

Answer 5

lets assume sequence bit as s . now , window size is 2^x as with these sequence number we can differentiate every byte . MSL is 120 sec . now the time taken when BW is 100Mbps is 2^x / 100Mbps .

this value must be greater than 120seconds so that if an packet will get dealyed then it will be expired and cant be taken by host . 2^x / ( 100 x 10^6) > 120 . x >30 . Hence answer will be B,C,D .

Answer 6

Answer B, C, D

A. 30 bit

Maximum possible byte that we can address is 1GB(2^³°)

(In tcp we assign a sequence number 1st  byte of EACH segment)

Given that in question 100Mbps

 

So 1b= 1/100*10^6 sec

   1GB=(1/100*10^6) *(10^9) *8=80 sec

WAT (Wrap Around Time is the time taken to consume all the available sequence numbers)

LT(LIFE TIME)

WAT=80sec

We know that LT<WAT

IN question LT= 120 sec

So option A not possible

In this way you can check options B, C ,D

WHARE LT <WAT.