leaky bucket

Question

Computer A has 19.5MB to be sent on a network and transmits the data in burst at 6mbps.The max transmission rate across routers in the network is 4mbps.If computer A's transmission is shaped using leaky bucket.Capacity that the queue in bucket must hold so that no data is discarded is ________________________________(in MB upto 1 decimal place)

Comments

6.5 MB?

Answer 1

6Mb can be transmitted by computer in => 1sec

1Mb => 1/6 sec

19.5 MB =>19.5*8/6 =26 sec.

Computer will transmit 19.5 MB in 26 sec.

The rate of network is 4Mbps means network can transmit 4Mb in one sec.

In 26 sec network can transmit => 4*26=102Mb.

The data that is still left is to be transmitted by network ,19.5MB-102Mb =156Mb-102Mb=52Mb.

Now this 52 Mb is the data which will be transmitted by computer but it cannot be transmitted by network because it is more than what network can transmit.

SO 52Mb or 6.5MB must be saved in bucket

 

Alternate:

Since computer is transmitting at 6Mbps and network at 4Mbps means in one sec the computer will transmit 6Mb and network can carry 4Mb.So the remaining 6-4=2Mb must be save in bucket.

So bucket will be filled at rate of 2Mbps

Now total data will be transmitted by computer in 26 second.

In 26 second bucket will size will be 2*26=52Mb or 6.5 MB

Comments

Question says Capacity that the queue in bucket must hold so that no data is discarded is ________

Leaky bucket discards packets if it becomes full

Now how do we know the full capacity??

we say 6.5MB...as the capacity

What if its the full capacity of the bucket...then everything will be discarded right

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What will be the longest burst time from the host if no data is lost from bucket?

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Very little mistake 4*26=104 may be ignored as explanation is superb