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Computer A has 19.5 MB to send on a network and transmits the data in burst 6 Mbps. The maximum transmission rate across router in the network is 4Mbps. If  Computer A transmission is shaped using a leaky bucket, how much capacity(Mb) must the queue in the bucket hold not to discard any data?
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+1
Since answer is asked in Megabit then it should be ( 19.5 × 8 ) - 104 = 52 Megabit
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Answer given as 109.04. But i am not getting how you got 52Mb. Please elobrate.
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Same type of question was given in made easy and according to the below mentioned approach the answer was correct.

+1 vote

In the above question leaky bucket is used.

So, consider the following diagram :

In the network of Computer A, the bandwidth acceptable is 6Mbps.

A is supposed to transmit a file of 19.5 MB which equals to 156 Mb

So, 6 Mb can be sent in 1 sec

156 Mb can be   sent in  156 / 6 = 26 secs

Now the problem is router network can accept the packet at a rate of 4Mbps

So, in 1 sec, the router network can accept a data of 4 Mb

In 26 secs, the router network can accept a   data of  26 * 4 = 104 Mb

So, the time at which A transmits a 156 Mb data, during that time the router network can accept 104 Mb of data.

Now what will happen to the remaining 156 – 104 = 52 Mb of data ?

So, the queue in the bucket has to take the responsibility for that else the remaining 52 Mb data will get discarded.

Hence, the capacity of the queue in the bucket should be 52 Mb.

Ans : 52 Mb.

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+1
Ohh.. Nice explaination. But answer given as 109.04 Mb. Your approach is really nice.
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Dnt knw why it is 109.04 Mb bt the made easy test series also followed the above approach.
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Ohh.. Then may be answer key have wrong solution.