No thanks but Happy Mother’s Day!!!
Answer:
what s this to me I will tell him to do the science and I am rewatching to be there to be in the
Explanation:
go see the teacher that you will be able for me and u look like me I don't want you anymore to me it will not tell you that you send me a question about your course on this subject matter as we have a lot to experience with our class on Friday
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:
A.1.
Explanation:
CCS(Centum Call Seconds) is a unit to measure the network traffic in the telecommunication circuit network.
1 Traffic unit = 1 Erlang.
The relation between Erlang and CCS(Centum Call Seconds) is:-
36 CCS = 1 Erlang.
Hence the answer to this question is 1 Erlang.
Voice. voice uses the least bandwidth
