GATE CSE 2015 Set 2 | Question: 34

Question

Assume that the bandwidth for a $\text{TCP}$ connection  is $1048560$ bits/sec. Let $\alpha$ be the value of RTT in milliseconds (rounded off to the nearest integer) after which the $\text{TCP}$ window scale option is needed. Let $\beta$ be the maximum possible window size with window scale option. Then the values of $\alpha$ and $\beta$ are 

• A. $63$ milliseconds, $65535$ $\times $2$^{14}$
• B. $63$ milliseconds, $65535$ $\times $2$^{16}$
• C. $500$ milliseconds, $65535$ $\times $2$^{14}$
• D. $500$ milliseconds, $65535$ $\times $2$^{16}$

Comments

it is not sequence number....it is maximum window size

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Awesome question 😍

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An interesting read: https://lwn.net/Articles/92727/

Answer 1

Window size in TCP = 16 bits

 

 That means receiver Window can have at max 65536 sequence numbers.

If we want to increase this winow size, 14 bits in OPTIONS can be used to increase this.

So in totality we wil be having $30$ $bits$ of window $(16+14)$

$B$ =  1048560 bits/sec

 

which is equal to  

131070 Bps$

 

B*delay = 65535

delay = 65536/131070 = 0.5s = 500ms