Ace Test

Question

A TCP sender chooses and initial sequence number as 1093 and sends 1460 bytes of data to the receiver. The receiver has an initial sequence number 391 and acknowledgs the data to sender. The ACK number sent by receiver is:_____

My Answer is 2553. But the solution key says it's 2554. Please let me know your answer.

Answer 1

Your Question is actually incomplete as you have not specified the MSS(maximum segment size).

MSS $\Rightarrow$ is used to avoid fragmentation at Network layer, so that fragmentation takes place in the transport layer itself and the burden not moves to Network layer ,its default size is 536(https://tools.ietf.org/html/rfc879#section-1) though i am coming to your question directly.

total data send to destination =1460 B.

initial sequence number of source=1093 B

Acknowledgement number is actually the sequence number which to be expected next by destination .

Here (1460+1093)=2553 sequence number is consumed.the next sequence number demanded by destination to source (ACK)=2553+1=2554

Comments

As the initial sequence number is 1093, the address of the last byte received will be 2552 NOT 2553. Because, 2552 -  1093 + 1 = 1460 (required data)

Thus next ACK should be 2553 NOT 2554.

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@Sourav.

If I send 1 byte, the receiver send ACK 1094.  (i.e 1093 + 1)

If I send 2 bytes, the receiver send ACK 1095.  (i.e 1093 + 2)

So, If I send 1460 bytes, the receiver send ACK 2553.  (i.e 1093 + 1460)