A frame of size 5 x 106 bits enters a network at router A and exits at Router B.
ID: 3772430 • Letter: A
Question
A frame of size 5 x 106 bits enters a network at router A and exits at Router B. There are eleven routers in the path across the network, including router A and router B. The distance between a pair of routers is 200 km. It takes a router 2 µsec to queue the frame and 1 µ to process it. The link connecting a pair of routers has a bandwidth of 5 Mbps. Assuming the speed of light inside a link is 2 x 108 m, compute the total delay (latency) experienced by the frame traversing the network from router A to router B. Which component of the total delay is dominant? Which component is negligible?
Explanation / Answer
Given:
speed of light inside a link i.e propagation speed = 2 x 10^8 m/s
N= number of links =(Number of routers + 1)=11+1 =>12
Frame size= 5 x 10^6 bits
Bandwidth = 5Mbps =5*10^6 bps
Distance between pair of routers = 200Km => 200x10^3 m
Queuing delay at each router = 2 µsec
Frame Processing delay at each router =1 µsec
Propagation delay at each link =distance / speed
= 200x10^3 / 2 x 10^8
= 100 x 10^3 x 10^-8
= 10-3 sec
= 1ms
Transmission delay = Frame size / bandwidth
= 5 x 10^6 / 5*10^6
= 1 ms
Answer:
Formula to calculate a frame traversing source to destination:
End-to-end delay=N x ( transmission delay + propagation delay + processing delay + Queuing delay)
= 12 x (1s+1ms+1 µsec+ 2 µsec)
= 12 x (1ms +1ms+ 2*0.001)
= 12x(2.002)
=24.004 ms
The dominant component of the total delay: Transmission delay
The negligible components are: Frame processing delay and queuing delay
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