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.
It should be compiled. It won't work if it's not compiled.
You want 10-12 too?.............
<span>If a router receives a packet and it does not have an entry in its routing table for the destination network, it will send the packet to its default route, if configured.
Answer: That statement is Ture</span>
The answer to this question is the PET scan. Positron
Emission Tomography or PET scan is an imaging test that checks and trace for
diseases in the body. This also shows how the body organs is functioning /
working. The doctor can evaluate the function of the patients body by the 3D
color images produced by the PET Scan.
