MadeEasy Test Series 2017: Computer Networks - Flow Control Methods

Question

Consider a network connecting two systems located 4000 kilometers apart. The bandwidth of the network is 64 Mbps. The propagation speed of the media is 2/3 of the speed of light in vacuum. It is needed to design selective repeat sliding window protocol for this network. The average packet size is of 8 Kb. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then, the minimum size in bits of the sequence number field has to be _________.

Comments

My answer is 10 but the solution says 9. I'm using  $sequence numbers = 2* (window size)$ for selective repeate. Is this wrong?

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i think u r rt...10 should be ans...

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i got 9 bits .

Tp=(4000*10³)/(2*10⁸)

Tp=0.02

Tt=(8*1024)/(64*10⁶)

Tt=1.28*10^-4

2*Wr<=ASN

312.5

ceil of log₂312.5=9

Answer 1

Selective Repeat Sliding Window Protocol.

Bandwidth = 64 Mbps, Packet size = 8 Kb, Distance = 4000 kilometers, Speed = 2/3*speed of light, No of frames in sender window is n.
The network is to be used to its full capacity.
so TT = 2*PT

(n*Packet size) / Bandwidth = 2* Distance/ Speed
(n* 8 Kb) / 64 Mbps = 2*4000 kilometers / (2/3*Speed of light)

n = 320

SIZE OF SENDER WINDOW = 1 + 320 = 321
Size of sender window needs 321 different numbers to identify one window. Same for reciever . so we need total 321+321 different numbers.

No of bits required to represent 640 different numbers is log₂(642) which is 10.

Comments

"The network is to be used to its full capacity. so TT = 2*PT".

How can we use this formula here?

Answer 2

your calculation is wrong.

check once again...

calculated window size= of 8 bits.

since it is selective repeat , so (1+8) = 9 bits

Comments

I think you should check once again because everybody else is also getting 10

Answer 3

Yes, The answer should be 10

Sender window size + receiver window size <= Available sequence numbers.