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levacccp [35]
3 years ago
12

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

Computers and Technology
1 answer:
Dennis_Churaev [7]3 years ago
4 0

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.

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Answer:

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Explanation:

Solution

As regards to the above requirement where the application and database server in the production environment will need to exchange the data once ever day, the following access control account practices would be used in this situation:

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Azzam is reviewing a paper he just wrote about kinetic friction for his physics class. On the third page, he realizes that every
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