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.
Answer:
Following is attached the code that works accordingly as required. It reads in characters from standard input and outputs the number of times it sees an 'a' followed by the letter 'b'. All the description of program is given inside the code as comments.
I hope it will help you!
Explanation:
Answer:
B. Click the Next icon on the Reviewing toolbar to review and then accept or reject each edit.
Explanation:
Since Jack wants to keep some changes and reject others, he can't use a global solution (like presented in answers A and C).
He has to go through each and every change proposition and decide individually if he wants to keep the change or not. That's why it's answer B.
It's the only way to accept some, reject some.
At the end of this process, he'll have a clean document with Rob's recommendations and his original documents.
Answer:
See attached file.
Explanation:
See attached file for detailed code.