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2 votes
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Consider two computers A and B are
connected through a network of 30 Mbps. Assume the distance between them is 3000km and the signal propagation speed is same as the speed of light and the packet size is 12 KB.
What is the minimum number of bits required for window to achieve 100% of utilization during GoBack-N and selective repeat protocol?
A . 5 and 6
B. 6 and 7
C. 6 and 6
D. 7 and 8
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1 Answer

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BW = 30Mbps, Distance = 3000Km, Speed = 3*10^8 m/s, Data Length = 12*10^3*8bits

Transmission Delay (Tt) = Data Length / BW

                                = (12 * 10^3* 8) / (30 * 10^6) = 3.2 ms

Propagation Delay (Tp) = Distance / Speed

                              = (3000 * 10^3) / (3 * 10^8) = 10 ms

Efficiency = N / (1 + 2a), where N = Window Size, and a = Tp / Tt = 10 / 3.2 = 3.125

            1  = N / (1 + 2 *3.125)

            N = ceil(7.25) = 8

For GB-N, N should be 8(Sender) + 1(Receiver) = 9, No. of bits required is 4

For Selective Repeat, N should be 8(Sender) +8(Receiver) = 16, No. of bits required is 4
Please Correct me if I am wrong

4 Comments

how do you know if it is megaByte or megabit
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Can you see big "B" after 12K??

Thats the identity of Byte. If it is small "b" then it is bit.
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edited by
Actually, i also think it should be 7 as 7.25 is the max possible packets which can be transmitted.
Maximum whole packets which can be transmitted in the duration of (Transmission Delay +2 * Propagation Delay) would be floor(7.25) i.e 7.

Min sequence number required for GBN will be log ( 7+1 ) i.e 3.

And for SR it should be ceil(log(7*2)) i.e 4.

Please correct me if i am wrong.
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