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Even through B's browser has only viewed the first 80 bytes,
its stack has received up to byte 600 in sequence from A.
What will be the ACK and window in TCP header that B next
sends to A?
Note: If ACK is X and window in TCP header is Y then
write answer as X+Y.
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In TCP (Transmission Control Protocol), the ACK (Acknowledgment) field in the TCP header acknowledges the receipt of data. The window size represents the amount of data, in bytes, that the receiver is willing to accept.

Given the information provided:

  • B's browser has only viewed the first 80 bytes.
  • B's stack has received up to byte 600 in sequence from A.

This implies that B has received bytes up to 600 from A but has only processed/viewed up to 80 bytes.

So, the ACK field (X) would be 81 (the next byte expected from A), and the window size (Y) would be the remaining space in B's buffer, which is the difference between the last byte received (600) and the last byte processed/viewed (80), so 600 - 80 = 520.

Therefore, the answer is X + Y = 81 + 520 = 601.

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