A leaky bucket with the capacity of bucket of 200 MB is at the host network interface. The data rate on the network is 2 Mbyte/s. If the host has 450 Mbytes to send onto the network and it sends the data in a burst then the maximum data speed from the host in order that no data is lost is ________ in Mbps.
(Upto 1 decimal place)
length of the packet = 450 M Bytes
The data rate on the network is 2 MByte/s
the data rate on the link from the host to the bucket is X MByte/s
Transmission time of packet of length 450MB put in the bucket = (450 /X) sec
therefore , actual data sent on network in (450 /X) sec = network rate * (450 /X) = (450 /X) * 2 = (900/x) MB
Capacity of bucket is = 200 MB
450 - 900/X = 200
450 X - 900 = 200 X
250 X = 900
X = 3.6 MByte /sec
= (3.6 /8 ) Mbps
= 0.45 mbps
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