Testbook Test Series: Computer Networks - Tcp

Question

TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence numbers differ by 12562.77 sec. Calculate the data rate(in KB) for the second party.

Comments

320 kBps?

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It should be 320KBps.

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for 1,

5 * 2^20 seq numbers in 1 sec

2^32 in 819.2 sec.

819.2 + 12562.77=13381.97sec

so

for 2, 2^32 seq in 13381.97 sec

so in 1 sec, 320,961 bytes = 320.961 KB

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5 * 2^20 seq numbers in 1 sec

you can't do like this. this is BW not data.

that will be 5*10^6.B

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why i cant do like this? kumar.dilip

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Utkarsh Joshi

5 MBps is Bandwith.As I know it should be power of 10.

 

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we are using sequence number for each byte 5MB in 1 sec so why not 5 * 2^20 Sequence numbers in 1 sec?

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@Utkarsh Joshi It has nothing to do with the initial sequence numbers right?

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@Utkarsh Joshi

Here intital sequence number doesn't matters ..for any TCP connection Wrap around time depends on BW and Sequence number bits ..right ??