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A system uses Selective-Repeat protocol with a window size of 4.If each packet carries 5000 bits of data, then the time taken to transfer 5 million bits of data, if the distance between sender and receiver is 2500 Km, the propagation speed is 2*108m/s and data rate is 1Mbps. We assume no data or ACK is lost or damaged.

(Computer Networks - 2 marks)

A.   7.25s

B.   8.25s

C.   11.25s

D.   16.25s

I think answer should be around 7.5 to 7.8.. Can any one explain what the answer is ??

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d = 25x10^5
v = 2x10^8

tp=d/v;
tp=12.5ms

RTT=2*Tp=25ms

Window size=4
packet size=5000 bits.

Tt=5000/10^6 = 5ms

Therefore, we send 20,000 bits in 25+5ms =30ms.
(or)

efficiency = 4/(1+2a)

efficiency=4/(1+25/5) = 4/6=2/3

throughput = BW*efficiency
=10^6 *2/3 = 2/3Mbps

in 1 sec = (2/3)x10^6 bits
1 bit = 3/(2x10^6)
5 million bits = (3*5*10^6)/(2*10^6)=7.5 sec

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