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

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2 Answers

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Best answer

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

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4 Comments

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 ?
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@Bikram sir...This might seem silly...but, how the window size could be in seconds? πŸ€”
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β€œany”

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@Arjun sir

Is token bucket in syllabus? I don't think so??
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Yes, Token Bucket is in syllabus .

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

plz see my answer now , hope it is clear .
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Answer:

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