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Answers to Problem 6.2

a. Slot Time = Propagation Time + Transmission Time = 103 m / (2 x 108 m/sec) + 10bits / 107 bps = 15x10-6 sec

Slot Rate = 1/Slot time = 6.67x104

Data Rate = 100bit/slot x Slot rate = 6.67x106 bps

Data Rate per station for N stations = (6.67106)/N

b. The end-to-end propagation delay is doubled compared to a). Hence

Slot Time = 20x10-6 sec

Slot Rate = 5x104

Data Rate =5 x106 bps

Data Rate per station = (5 x106)/N

c. The maximum end-to-end distance is 0.45+0.1 = 0.55 km. This must be doubled to compute end-to-end delay.

Slot Time = (1.1x103 m) / (2 x 108 m/sec) + 10bits / 107 bps=15.5x10-6 sec

Slot Rate = 6.45x104

Data Rate =6.45 x106 bps

Data Rate per station = (6.45 x106)/N

d. With round-robin, a slot time equals transmission time plus propagation time between stations plus one bit time

Slot Time = (103 m/N) / (2 x 108 m/sec) + 10bits / 107 bps + 10-7 =(10.1+5/N)x10-6 sec

Slot Rate = [N/(10.1xN+5)]x106

Data Rate =[100xN/(10.1xN+5)] x106 bps

Data Rate per station = [100/(10.1xN+5)] x106

e.

Slot Time = 102 m / (2 x 108 m/sec) + 10bits / 107 bps+ 10-7=10.6x10-6 sec

Slot Rate = 0.09x106

Data Rate =9.43 x106 bps

Data Rate per station = (9.43 x106)/N