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3 votes
3 votes

The distance between A to B is 4000 km. The distance between B and C is 1000 km.The propogation delay is 5 micro sec/ km for both the links. The data rate between A and B is 100kbps. Both the links are full duplex. All data frames are 1000 bits long and ACK frames are negligible. Window size is 4.

What is the required transmission rate (in kbps) between B and C so that buffers of node B are not flooded?

A. 100

B.150

C.200

D.250

3 Answers

6 votes
6 votes

In order to ensure no flooding, data transfer rate between B-C must be same (or greater) as that between A-B. 

Lets consider the no. of bits transmitted by A until it receives ACK and let time for this be S. 

S = 2* Propagation delay  + Transmission delay for packet + Transmission delay for ACK 

= 2*20 ms + 10 ms + 0

= 50ms. 

Window size is 4. 

So, in 50ms, A could send up to 4 packets = 4000 bits. Data rate being 100 kbps, allows upto 100 *50 = 5000 bits in 50ms, so we are fine and we have effective data rate of 4kb/50ms = 80kbps.

Now, we know that in every 50ms, we are getting 4000 bits at B. (with a buffer of size 4000 bits we can assume no flooding for this)

Lets consider the no. of bits transmitted by B until it receives ACK and let time for this be T. 

T = 2* Propagation delay  + Transmission delay for packet + Transmission delay for ACK 

= 2 * 5ms + 1000/x ms, where x is the data rate in kbps 

In T ms, B sends 1000 bits. We want B to send at 80kbps which implies T = 1000/80  = 12.5ms

So, 12.5 = 10 + 1000/x

x = 1000/2.5 = 400 kbps. 

0 votes
0 votes
L=1000 bits

Tp = 5*1000 = 5000 $\mu$ sec or 5 ms

RTT = 5*2 = 10 ms

B * Delay = Bits flowing = L

B*10*10^-3=1000

B=100 kbps

Therefore the ans is A.
0 votes
0 votes

It should be 200 Kbps

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