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4 votes
4 votes
Consider an error-free 64-kbps satellite channel used to send 512-byte data frames in
one direction, with very short acknowledgements coming back the other way. What is
the maximum throughput for window sizes of 1, 7, 15, and 127? The earth-satellite
propagation time is 270 msec.

4 Answers

6 votes
6 votes

Transmission time for 1 frame=512 B/(64/8)KBps = 64 ms

  • Propagation time = 270
  • Roundtrip time = 2x270= 540ms
  • Transmission time+ RTT =604 ms

a) Window size =1
means in 604 ms, only 1 frame is send 
Throughput= 512 B/604 msec = 678 bps

b) Window size =7
means in 604 ms, 7 frames are send 
Throughput= 7x512 B/604 msec = 7x678 bps = 4746 bps

c) We know maximum throughput is possible when Window size = [1+ 2Tpropagation /T transmision]

= 1+2 *270/64 =9.43 ~ 10
So Window size =15 will give throughput as maximum bandwidth available ie 64 Kbps

d) Window size =127
But maximum throughput is achieved when window size =10 onwards.
Hence 
Window size =127 also will give throughput as maximum bandwidth available ie 64 Kbps.
Therefore, Window size 15 and 127 have same throughput



 


 

edited by
4 votes
4 votes
Bandwidth = 64 Kbps

L = 512 Bytes

Tp =(for Satellites) 270 msec

Tx = L/B => (512 *8 bits)/64 * 10^-3 b/sec. => 64 msec

1 + 2a = 9.43

Throughput = Eff * Bandwidth

Eff = Window Size / (1 + 2a)

[ for Window Size = 1 ] => Eff = 1/ 9.43 => 0.1060

                                              Throughput = .1060 * 64 Kbps => 6.78 Kbps

[ for Window Size = 7] => Eff = 7/9.43 => .7423

                                             Throughput => .7423 * 64 Kbps => 47.50 Kbps

[ for Window Size = 15 and 127] => Since Window Size is > than (1 + 2a) Throughput will be 100% ie, 64Kbps

0 votes
0 votes

max window size = 1+2a = 1+2Tp/Tt

Tt= msg length/BW, Tp= 270 msec

2tp=270*2 msec

    =0.54 sec

Tt = (512*8)bit/(64 * 10^3)bps

     =64 * 10^-3

Ws = 1+(0.54/0.064) = 9.43(approx)

max throughput = efficiency * bw

efficiency=ws/1+2a

 ws=1,7,15,27

then we can calculate for each window size , it is max at window size = 127

0 votes
0 votes

Efficiency = W/1+2a where Ws= sender window size and a = Tp/Tt

and Transmission time = Length of pkt / B.W

Tt = 512*8/64*10sec

=  64 msec

and Tp = 270 msec given

i.e

applyting the formula for efficiency

efiiciency when ws = 1

(1 / (1+ (2*270)/64)) = .10 or 10% approx

now Throughtput = .10 * 64kbps  = 6.4kbps approx

and we can calculate for different window size just by putting the formula 

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