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Objective Session

Suppose that ...

Question

Suppose that a transmitter operating at $10^{8}$ bps is connected to one end of a 230 km length of coaxial cable. The signal propagation speed in the cable is 230000 km/sec, If packet switching is used with a packet length of 2000 bits, how many packets have been transmitted and are propagating along the cable when the first bit reaches the other end?

A

50

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B

20

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C

80

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D

100

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Solution

The correct option is A 50
Propagation delay = $\frac{(230)}{230000}$ = 1 msec
Packet transmittion time = $\frac{2000}{10^{8}}$ = 0.02 msec
Number of packets in transit = $\frac{1}{0.02}$ = 50 packets.

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