MadeEasy Test Series: Computer Networks - Sliding Window

Question

Consider an error-free 64kbps satellite channel used to send 512 bytes data frames in one direction, with very short acknowledgements coming back the other way. assume the earth-satellite propagation time is 270 msec. What is the minimum window size so that the channel is fully utilized?

 

I am getting answer as 7. Answer given 10.

Answer 1

Tx  = (L/B) =  $\frac{512*8}{64*1000} = 64 ms$

Tp = 270ms

Throughput = $\frac{w}{1+2a}$ , where w is the window size

a = Tp/Tx = $\frac{270}{64}$ = 4.21875

 

for maximum throughput w > (1+2a) > 9.43

Hence w  = 10

Answer 2

I hope you have considered 512 bytes?

So,

1 + 2a

 = 1 + 2 * $\frac{Tp}{Tt}$

= $1 + 2*\frac{0.270 * 64000}{512*8}$

=541

window size = $\left \lceil log_{2}512 \right \rceil$

= 10