4/30/2023 0 Comments Delay bandwidth productt=5, assuming the receiver can respond immediately and doesn't need any time at all to decide if even the 1st bit is received correctly, it immediately starts sending an ACK (say, a 1-bit ACK, like 1 for success, 0 for failure).t=5, the first bit starts arriving at the receiver, even as the next 4 bits are immediately following it and still in transit, as can be seen in the picture.t=0, sender starts transmitting, at 1 bit per second."The sender then waits for receiver acknowledgment for part of the burst before sending another burst." The key to understanding what the author meant, about "(2 × bandwidth × delay) bits", is in the last sentence. The sender then waits for receiver acknowledgment for part of the burst before sending another burst. The sender should send a burst of data of (2 × bandwidth × delay) bits. I don't understand why some people says the it takes 5s for receiver to receive the message, in this example, 5 bits is considered as a message, and the receiver receives the first bit after 5s, and it takes another 5s to receive the last bit, so it needs a total of 10s to receive the whole 5 bits, isn't it?īesides, we can only send another message before we get the ack from the receivers, so if we send 10 bits(two messages),isn't that against the meaning of acknowledge? Does it mean that we make each 5 bits data as a message, and each time we send two messages actually? According to what the textbook says, we need to send each burst of 10 bits. We also assume that the delay of the link is 5s. I'm a little bit confused here, let's say we have a bandwidth of 1 bps. To use the maximum capability of the link, we need to make the size of our burst 2 times the product of bandwidth and delay we need to fill up the full-duplex channel (two directions). If we need to send data in bursts and wait for the acknowledgment of each burst before sending the next one.
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