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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)

asked in Computer Networks by (69 points) | 178 views

1 Answer

+5 votes

 

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

answered by Boss (13.6k points)
edited by
+1

@Magma How did you get this step,

450-900/X=200.

0
thnku boss nice explanation

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