Consider a `b` bits per second link between two hosts that has a propagation delay of `t` seconds. Derive a formula that computes the time elapsed between the transmission of the first bit of a `d` bytes frame from a sending host and the reception of the last bit of this frame on the receiving host.
Transmission links have sometimes different upstream and downstream bandwidths. A typical example are access networks that use ADSL (Asymmetric Digital Subscriber Lines). Consider two hosts connected via an ADSL link having an upstream bandwidth of 1 Mbps and a downstream bandwidth of 50 Mbps. The propagation delay between the two hosts is 10 milliseconds. What is the maximum throughput, expressed in frames/second, that the alternating bit protocol can obtain on this link if each data frame has a length of 125 bytes and acknowledgments are 25 bytes long. Same question if the protocol is modified to support 1500 bytes long data frames.
To speedup the transmission of the frames, a student proposes to compute the CRC over the data part of the frame but not over the header. What do you think of this proposed solution ?
Derive a mathematical expression that provides the `goodput`, i.e. the amount of payload bytes that have been transmitted during a period of time, achieved by the Alternating Bit Protocol assuming that :
Consider a go-back-n sender and a go-back receiver that are directly connected with a 10 Mbps link that has a propagation delay of 100 milliseconds. Assume that the retransmission timer is set to three seconds. If the window has a length of 4 frames, draw a time-sequence diagram showing the transmission consisting of 10 data frames (each frame contains 10000 bits):