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In a token bucket network transmission spreed  is $20^*(10^6)$ bps and maximum rate can only be sent for at most $10$ sec at a time, and at most $150$ Mb can be sent over any $15$ sec window, then the value for token input rate  is _________ mbps

we assume initial capacity is 150mb , read the question again and try to visualize please.
Refresh rate is Input rate here..

It is not clearly mentioned in the question that the capacity of the bucket is 150 Mb. I think that has caused problem for many people who gave this exam.

Consider  a token bucket with maximum rate R = 20*(10^6) bps

suppose we want to make sure that the maximum rate can only be sent for at-most 10 seconds at a time , and at- most ( maximum ) 150 Mb can be sent over any 15 second window.

Then the required value for which the new tokens are added at the rate of r Mbps which  we have to calculate in this question .

Capacity of the token bucket (b) = 150 Mbps
Maximum possible transmission rate (M) = 20*(106) bps
So the maximum burst time = b/(M-r) = 10

Duration = b/M-r, where b is the initial capacity, M is outgoing rate and r is incoming rate

now put b= 150  Mbps, M = 20 Mbps ,  we  need to find r

b= (M-r) 10

b = 10M - 10r

10r = (200 - 150 )

r = 50 /10
r = 5

The value for token input rate  is 5 Mbps

by

Here capacity is 150 Mb , but the bit rate is 20*10^6 bps but we have to convert right for bps

1Mbps = 10^6

but 150 Mb = 150*2^10

is this wrong ?
@Bikram sir...This might seem silly...but, how the window size could be in seconds? 🤔

“any”

@Arjun sir

Is token bucket in syllabus? I don't think so??

Yes, Token Bucket is in syllabus .

Token Ring is not in syllabus.
Please explain this answer how to put value in formula
@Nitish

plz see my answer now , hope it is clear .

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