Question

The objective of this exercise is to realize high distance and high bandwidth affect the performance of stop and wait. The distance from Earth to a distant planet is approximately 9 x 10 km. Assume that the frame size is 32 KB and the speed of light is 3 108 m/s. Assume that the bit rate is 32 Mbps a) (5 points) What is the efficiency (channel utilization) if a stop-and-wait protocol is used? b) (15 points) Suppose we use a window protocol. What should be the window size in frames to achieve the maximal efficiency (channel utilization)? c) (20 points) Plot the efficiency versus the distance when the bit rate is set to 32 Mbps. Discuss this plot d) (20 points) Plot the efficiency versus the bit rate when distance is set to 9x 10'7 km. Discuss this plot

Answer 1

a). distance=*10^10m

data rate=64Mbps

size=32KByte= 256 Kbits

propagation speed= 3*10^8 m/s

Transmission delay= packet size/data rate

=256 Kb/64Mbps = 4ms= 0.004s

propagation delay= distance/propagation speed

=9*10^10/3*10^8

=300s

If no processing delay reciever and acknowledgement size is 0, then only one packet is sent in round trip time ie 600s

utilization= 0.004/(0.004+600)

=6*10^-6

b). to achieve efficiency of 100% window size should be equal to number of packets that can be sent in one round trip time

so RTT is 600s, and transmission time of one packet is 4ms

so we can send 1.5*10^5 packets in 1 RTT

so window size should be 150000.