# Leaky bucket algorithm

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(30/6)*(6-4)=5*2=10 MBps

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Can you explain briefly how u got the result ?
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computer A has to send 30MB data to network and it sends data at 6MB rate to the network but routers data transmission rate is 4MB. So have to send data at min(4,6)=4MB to the network.But as computer is sending at 6 mb so each time 2 mb is left in the queue,now computer A has to send 5 times 6MB data so tatal (5*2)MB=10MB willbe there in the queue.
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Nice thnx
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30 MB is given, not Mb, so total, 30*8=240, now 240 Mb is sent input flow in 240/60=40s and output, 240/40=60s. So 60-40=20s left for which we require 20*6=120Mb require.

isnt it??

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