Question

Consider a source that is sending on/off traffic (as in a square wave). In particular, for 0.5 seconds, it sends data traffic at a rate of 12 Mbps then it does not send any data traffic for 2 seconds. This process repeats. If this traffic is going through a router that has 10 Mbps capacity and a buffer of size B bits...

a) What is the average throughput achieved by this source? (your answer should be in Mbps)


b) How large B should be to prevent any data loss (that is to accommodate each square burst and buffer it)?


c) If we would like to remove this buffer, but still would like the source to send the same amount of data per burst, how would you choose the on period and the data rate?

Answer 1

throughput = 2.4 Mbps

buffer need = 1 sec

time = 0.6 second

Step-by-step explanation:

given data

sending on/off traffic = 0.5 sec

traffic rate = 12 Mbps

does not sen data = 2 sec

capacity = 10 Mbps

solution

we know here data send in 0.5 sec is

data send = 0.5 × 12 = 6

and total time is = 2 + 0.5 = 2.5 sec

so throughput is expressas

throughput = data send ÷ total time .............1

throughput = 6 ÷ 2.5

throughput = 2.4 Mbps

and

differenece in rate is

differenec in rate = 12 - 10 = 2 Mbps

and buffer need to data send in 0.5 sec is

buffer need = 0.5 × 2

buffer need = 1 sec

and

when we remove buffer we send same data rate i.e 10 Mbps

but in 0.5 sec we send 6 Mbps data

so for 6 Mbps data we no need buffer

so time will increase here

time = data ÷ speed ...............................2

time = 6 ÷ 10

time = 0.6 second