Flow Control

Question

Consider a satellite communication channel with 0% error rate and a link capacity of 64 kbps. The satellite is sending the images of size 512 bytes data in one direction. One Image can be accumulated in one frame. The acknowledgement packets are very small and can be negligible to calculate the delays. The propagation delay for this satellite channel is 270 msec, what is the maximum achievable throughput for window sizes of 1,7,15 and 90?

1.  6.7 Kbps, 47 Kbps, 101 Kbps, 61 Kbps
2.  6.7 Kbps, 47 Kbps, 64 Kbps, 64 Kbps
3.  6.7 Kbps, 47 Kbps, 101 Kbps, 0.6 Mbps
4.  6.7 Kbps, 47 Kbps, 61 Kbps, 0.6 Mbps

Answer 1

We know :

Throughput(effective data rate) = Link Utilisation * Bandwidth

So for window size of 1 , we have :

Link Utilisation = T.T. / [ T.T + 4 * P.T. ]

                      = [(512 * 8) / (64 * 10³)] / [ { (512 * 8) / (64 * 10³) } + 2 * 270 * 10^-3]

                      = 64 / (64 + 540)

                      = 64 / 604 

                      = 0.106  = 10.6 %

So throughput  = 0.106 * 64

                      = 6.7 Kbps in case window size

Now for 

b) Window size of 7 , efficiency = 0.106 * 7  =  70.42 %

So throughput  = 47 Kbps

c) Window size of 15 , throughput = 6.7 * 15  = 106 Kbps ..But it exceeds maximum link capacity(bandwidth) which is unrealistic.So it is limited to maximum bandwidth only..Hence throughput = 64 Kbps

d) Window size of 90 , throughput = 0.6 Mbps..Same here also..It will be 64 Kbps only.. 

Hence option 2) is correct..

Comments

It's $T_t +4T_p$ for satellite communication. rt??

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It is done when ground to ground satellite communication is mentioned..By default it is 2Tp..There was a debate in one earlier question on this topic ..:)