# Transmission and propagation

Two stations are 400km apart from each other. You have access to a 100Mbit/s full duplex radio link which transmits packet of 4000 Bytes in each direction.

What is the transmission time?

I assume the full duplex link will take the 100Mbit/s and split it such that 50 are used for upload and 50 for download.

Then the transmission time is:

Transmission time = (4000*8) bit / (50*10^6) bit/s = 640 microseconds


What is the propagation delay?

delay = (400*10^3 m) / (3*10^8 m/s) = 3.88 milliseconds


What is the throughput if we use stop-and-wait ARQ with piggy backing?

throughput = 32000 bit / (3.88 * 10^-3) seconds = 8.4 Mbit/s


What should the buffer size be for the sliding window in order to fully make use of the link?

50*10^6 bit/s * 3,88 *10^(-3) seconds = 194000 bits
194000 bits / 32000 bits per packet = 6


so the window size should be 6.

Is my thinking correct here or did I make a mistake?

• To me, "100 Mbit/s full duplex" means 100Mbit/s in each direction. May 30, 2018 at 19:30
• You also need to research what a packet is. May 30, 2018 at 20:38
• Assuming that 4000 byte packets will use (4000 * 8) bit-times is unrealistically simplistic. May 30, 2018 at 21:50
• @brhans: This kind of simplification is typical of homework problems. Absent any other information, we have to assume that the 4000 bytes is the fully-encoded physical packet, ready to go over the air. May 31, 2018 at 12:14