A baseband bus has 5 stations equally spaced at 500m from one another. The data

Question

A baseband bus has 5 stations equally spaced at 500m from one another. The data rate of the bus is 50 Mbps and the propagation speed is 10^5 km/s.

a. What will be the total time taken to send a packet of 2 KB from any station to its adjacent station, ,meeaured from the beginning of transmission to the end of reception?

b. What will be the total time taken to send the packet from a station at one end to the station at the other end, measured from the beginning of transmission to the end of reception, if the delay at each intermediate node is 10^-5 seconds?

c. If two adjacent stations begin to transmit at exactly the same time, when will their packets interfere with each other?

Explanation / Answer

a) Distance between adjacent stations is 500 m,
Total Time = transmission time + propagation time
= size of packet/data rate + distance/propagation speed
= (2 × 1024) / (50 × 106) + 500 / (100 × 105) = 90.96 µs.

b) Number of hops = 5-1 = 4.
Number of intermediate nodes = 3.
Time taken for traversing 1 hop = 90.96µs.
Delay at each hop = 10^-5 s = 10µs.
Total time = 4 × 90.96 + 3 × 10 = 393.84µs.

c) As both the packets have same traversing speed, they will collide at a distance of
500/2 = 250m from any end. As collision occurs when the first bits of the packets
meet, only propagation time is considered:
Time = distance /propagation speed = 250/(100 × 105) = 25µs.

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