GATE IT 2008 | Question: 64

Question

A $1\;\text{Mbps}$ satellite link connects two ground stations. The altitude of the satellite is $36,504\;\text{km}$ and speed of the signal is $3 \times 10^{8}\;\text{m/s}.$ What should be the packet size for a channel utilization of $25\%$ for a satellite link using go-back-$127$ sliding window proto­col? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.

• A. $120\;\text{bytes}$
• B. $60\;\text{bytes}$
• C. $240\;\text{bytes}$
• D. $90\;\text{bytes}$

Comments

I think one station is sending frames to other through satellite that’s why distance is doubled.

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η =  Useful time / Total Time,

where, Useful time = $Tt$

Total Time = $Tt + RTT$

See Bikram Sir’s comment under best answer.

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A $1\ Mbps$ satellite link connects two ground stations.

Since two ground stations are given, total propagation delay is $4\ T_p$.

If nothing is mentioned as in GATE2016-2-55, total propagation delay is $2\ T_p$.

Answer 1

Given altitude of the satellite is 36504 km. Let $A$ and $B$ be the two ground stations. Let $d$ be the distance between the two ground stations. 

By Trigonometry, we can say, $tan$ $45^{\circ}$ = $perpendicular / base$ $\implies$

$tan$ $45^{\circ}$ = $\frac{36504}{(d/2)}$ $\implies$ $d = 2*36504 = 73008 km$

Now, In $GBN$, Utilization = $\dfrac{n}{1+2a}$, where n is the window size and $\large a = \dfrac{\text{Propagation Time}}{\text{Transmission Time}}$

Given utilization = $25\%$ or $1/4$ and $n = 127$, $v = 3*10^8m/s$, $ packet size = L$, $Bandwidth = 1Mbps$

So, $1/4$ = $\frac{1}{1+ \frac{2*73008*10^3*10^6}{3*10^8*L}}$ $\implies$ $L = 960 bits$ = $120 B$ 

 

Comments

Very good explanation 👍

Answer 2

Distance between stations = L KM
Propogation delay per KM = t seconds
Total propagation delay = Lt seconds
 
Frame size = k bits
Channel capacity = R bits/second
Transmission Time = k/R

Let n be the window size.

UtiliZation = n/(1+2a) where a = Propagation time / transmission time
            = n/[1 + 2LtR/k]
            = nk/(2LtR+k) 
For maximum utilization: nk = 2LtR + k
Therefore, n = (2LtR+k)/k
Number of bits needed for n frames is Logn.