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Telecom Assignment

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Assignment 5

1. A LAN has a data rate of 4 Mbps and a propagation delay between two stations at opposite ends of 20 μs. For what range of PDU sizes does stop-and-wait give an efficiency of at least 50%?

r = 4 d = 20 E > 0.5
The equation should look like: Range = 2 * 20 * 10^-6 * 4 * 10^-6 which is =

0.00000000016 or 160 bits

2. A disadvantage of the contention approach for LANs is the capacity wasted due to multiple stations attempting to access the channel at the same time. Suppose that time is divided into discrete slots, with each stations attempting to transmit with probability p during each slot. What fraction of slots are wasted due to multiple simultaneous transmission attempts?

The fraction of slots …show more content…

Therefore, we have:
Vulnerability Period = (transmission time + propagation delay) = 100/1 x 106 + 1 x 103/2 x 108 = 0.000105 sec

Probability of one or more packet arrivals at one of the stations = 1 – e-λ (vulnerability period) = 0.65

a. Slot Time = Transmission Time and Propagation Time
Transmission Time = (Number of bits / Rate)
Propagation Rate = 200,000,000 m / Second
Slot Rate = 1 / slot time = 100 bits per slot

Propagation Time = (10^4 / 2)(10^8) = 50 µs propagation delay
Transmission Time = (100 bps/10,000,000 bps) = 10 µs
Slot Time = 10µ seconds + 50 µ seconds = 60 µs
Slot Rate = (1 / .00006) = 16,666.7
Data Rate = (1,000 x 16,666.7) = 16666700Mbps

b. The duration of one bit delay = [(1 bit/ T seconds)=(10,000,000 bits / 1 second )], therefore, T= 0.1ms
So, one bit delay is imposed by (N+2) x 0.1 ms.
The propagation delay = 2d x (20ms/km) = (2x10) x (20ms/km) = 0.4 seconds or 400 ms.
But per the question, we need to use (2 x 10^8 m/s) for the propagation delay.
Transmission Time= [(100 bit/t seconds) = (10,000,000 bits / 1 second)] Therefore, t = 10 ms
Each 100-bit packets requires 10ms+(2 x 10^8)+(N+2) x 0.1

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