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Unformatted text preview: y Suppose that the lowest-sequence number that the receiver is waiting for is packet m . In this case, its window is [ m , m + w-1] and it has received (and ACKed) packet m-1 and the w-1 packets before that, where w is the size of the window. y If none of those w ACKs have been yet received by the sender, then ACK messages with values of [ m-w , m-1] may still be propagating back. If no ACKs with these ACK numbers have been received by the sender, then the sender's window would be [ m-w , m-1]. y Thus, the lower edge of the sender's window is m-w, and the leading edge of the receiver window is m + w-1. In order for the leading edge of the receiver's window to not overlap with the trailing edge of the sender's window, the sequence number space must thus be big enough to accommodate 2 w sequence numbers. y That is, the sequence number space must be at least twice as large as the window size, k ≥ 2 w ....
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- Spring '11