If an ARQ algorithm is running over a 40-km point-to-point fiber optic link then:

a. Compute the propagation delay for this link, assuming that the speed of light is 2 x 10^8 meters per second in the fiber.
b. Suggest a suitable timeout value for the ARQ algorithm to use. Assume transmission time for data frames and returning ACKs is insignificant, compared to link propagation delay.
c. Why might it still be possible for the ARQ algorithm to time out and re-transmit a frame, given this timeout value?
d.

Answer and Explanation:

Given data:

Distance (D) = 40 KM

Speed of light in the fiber =Distance/ speed of light in the fiber

a) Delay (P) = Distance/ speed of light in the fiber

= (40,000 Meters/2×108 m/s)

=( 40×103 Meters/2×108 m/s)

Propagation delay (P) = 0.0002 seconds or 200 microseconds

b)

if propagation delay is 0.0002 Seconds roundup trip time (RTT) will be 0.0004 Seconds or 400 micro Seconds

Essentially since transmission times and returning ACKs are insignificant all we really need is a value slightly greater than 0.0004 seconds for our timeout value.

c)

The obvious reasons would be if the data frame was lost, or if the ACK was lost. Other possibilities include extremely slow processing on the receive side (late ACK).

Or Extremely Slow Processing of the ACK after it is received back at the send side.