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?

  • 3
    \$\begingroup\$ To me, "100 Mbit/s full duplex" means 100Mbit/s in each direction. \$\endgroup\$
    – brhans
    May 30, 2018 at 19:30
  • \$\begingroup\$ You also need to research what a packet is. \$\endgroup\$
    – Andy aka
    May 30, 2018 at 20:38
  • \$\begingroup\$ Assuming that 4000 byte packets will use (4000 * 8) bit-times is unrealistically simplistic. \$\endgroup\$
    – brhans
    May 30, 2018 at 21:50
  • \$\begingroup\$ @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. \$\endgroup\$
    – Dave Tweed
    May 31, 2018 at 12:14

1 Answer 1


You're generally thinking along the right lines, but you're oversimplifying in some areas.

For example, with stop-and-wait, the transmitter can't send a second packet until the the first packet has been fully received, and the ACK has propagated from the receiver back to the transmitter. Note also that "piggy backing" implies that the ACK itself is part of a full-size reverse channel packet that must be fully received by the transmitter.

This round-trip delay also factors into your sliding window calculation.

Also, double-check your math while you're at it.


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