Question

A 1 Mbps satellite link connects two ground stations. The altitude of the satellite is 36.504 km and speed of the signal is $3\times {10}^{3}m/s$. 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 the there are no errors during communication.

• A. 120 bytes
• B. 90 bytes
• C. 240 bytes
• D. 60 bytes

Answer 1

The correct option is A 120 bytes



So propagation delay = Propagation from $S_1$ to S + Propagation from S + $S_2$
$=P.T+P.T$
$=2P.T$

Propagation delay
$\left(T_P\right)=\frac{Distance}{Speed}$
$=\frac{2\times 36504\times {10}^{3}m}{3\times {10}^{-3}m}sec$
$=24336\times {10}^{-5}sec$
$=243360\mu sec$
Transmission delay
$T_t=x\mu sec$
$\alpha =\frac{T_p}{T_t}=\frac{243360}{x}$
Efficiency
$\eta =\frac{1}{1+2a}$
$\Rightarrow 0.25=\frac{127x}{x+2\times 243360}$
$\Rightarrow x=960bits=120$Bytes