The distance between the two hosts is 10,000 km and the propagation speed along the link is 2 x 10 8 m/sec. At what data rate does the round trip delay equals the transmission delay for a 1 KB packet?Дў x Propagation delay = Transmission delayД«andwidth = (2 10 x 10 6 / 200 ) bytes per secД¬onsider two hosts X and Y connected by a single direct link of rate 10 6 bits/sec.

But here, channel capacity is actually given as bandwidth because its unit is b/sec.Д¬onsider a MAN with average source and destination 20 Km apart and one way delay of 100 Ојsec.

Generally, channel capacity is the total number of bits which a channel can hold.

Channel capacity = Bandwidth (here) = 10 9 b/sec.

If the packet size is 1 KB and propagation time is 15 msec, the channel capacity is 10 9 b/sec, then find the transmission time and utilization of sender in stop and wait protocol. Substituting the values in the above relation, we get.

Although it is incorrect, but we still ignore it for solving the question.

So, in the above formula of throughput, we have ignored the term T t from the denominator.

In the given question, we are not provided with the networkвЂ™s bandwidth.

What is the throughput achievable in stop and wait protocol by a maximum packet size of 1000 bytes and network span of 10 km.ДЄssume the speed of light in cable is 70% of the speed of light in vaccum.

L bits >= (20 x 10 -3 sec x 4 x 10 3 bits per sec) x 2ДЇrom here, frame size must be at least 160 bits. To get a channel efficiency of at least 50%, the minimum frame size should be-Д¬ondition For Efficiency To Be At least 50%-ДЇor efficiency to be at least 50%, we must have. The transmission time of the acknowledgement frame is negligible. = (2 10 x 8 bits) / (1.5 x 10 6 bits per sec)ДЄ channel has a bit rate of 4 Kbps and one way propagation delay of 20 msec.

C program for stop and wait protocol