gate

Question

A 1Mbps satellite link connects two ground stations. The altitude of the satellite is 6000 km and speed of the signal is 3 × 10^8 m/s. What should be the packet size for a channel utilization of 50% for a satellite link using go-back-63 sliding window proto­col? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.

Answer 1

Here is my answer..

Comments

I think i got it wrong ..distance will be 12000..cause first it wo go the satellite and then the destination . someone pl s correct me

---

yes distance will be 2*6000 (Ground Station A -> Satellite, Satellite -> Ground Station B) so answer will be 640 bits

Answer 2

ans is 32*10^4
1/2=63/(1+2*(Tp/Tt))
l=320000 bits
or l=40000 Bytes

Answer 3

Answer is 79.3 byte's

Answer 4

Since we are talking about ground to ground station communication
So signal has to travel from ground to satellite and then satellite to ground, Total distance to be travel  = 12000 Km

Now $Propagation Delay = \frac{Total Distance}{Speed}$

i.e $Propagation Delay = \frac{12000 \times 10^{3} m}{3 \times 10^{8} m} sec$

i.e $Propagation Delay = 40,000 \times 10 ^{-6} sec$

$Efficiency (In Go Back N) = \frac{N \times TT}{TT + 2 \times PD}$

$\frac{50}{100} = \frac{63}{1+ 2 \times \frac{PD}{TT}}$

$\frac{PD}{TT} = \frac{125}{2}$

$TT =\frac{40000 \times 10^{-6} \times 2}{125} sec$

$\frac{Frame Size}{Bandwidth } = \frac{40000 \times 10^{-6} \times 2}{125} sec$

$Frame Size= \frac{40000 \times 10^{-6} \times 2 \times 10^{6}}{125} bits$

$Frame Size= 640 bits$

or $ Frame Size= 80 bytes$