Problem 5 Consider a satellite connection with a oneway prop

Problem 5 Consider a satellite connection with a one-way propagation time of 0.27s, and a transmission rate of 1Mbit/s, with packets of 1000 bits. What is the minimum window size of a Go-Back-N protocol that results in an efficiency close to 100%?

Solution

The minimum window frame size = (Round Trip Propagation Delay) * (Transmission Speed)

we know that Round Trip Time (RTT )= 2*Transmission time + Propogation time

Transmission time = No. of bits per packet / Transmission speed

                                  1000/ 1000000 = 0.001 s

we know that prpopogation time is equal to 0.27s (given)

RTT= 2*0.27+ 0.001 = 0.541 s

Minimum frame size = RTT * Tansmission speed = 0.541* 1* 10^6 = 541Kbits