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if delays are recorded as 8 bit numbers in a 50 router network and delay vectors are excahngesd twice a second how much bandwidth per(full duplex)line is chewed up by the distributed routing algorithm? assume that each router has three lines to other routers.

a) 400 bps

b)800 bps

c)1200 bps

d)None of these

Answer is :- B

can any one provide explanation about this answer?

3 Answers

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A Network has 50 routers in which every router has a delay of 8bit .
Delay is actually a metric of router and has a proper field in routing table .
Hence total size of delay field is 50 * 8bits = 400bits
Also a router is connected to other routers which means it is a dedicated path , hence full utilization
of bandwidth will be there .

This field is updated twice a second onto each line .
Hence, 800 bps is needed in each direction (as it is full duplex)
for each line.
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(Terminology adopted from Kurose-Ross)

Consider the router as a toll plaza on a highway and the bits as cars that are going to be charged.. With this ,

each toll plaza will have 800 cars per 1/2 second (upside and downside combined) or 1600 cars per sec.

So, per router bandwidth consumed is 1600 bps.
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The routing table is 400 (50*8) bits.  Twice a second this table is written onto each line, so 800 bps are needed on each line in each direction.

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