Question

A 1000 byte packet is sent over a 50 kilo-bits-per-second (Kbps) point-to-point link whose propagation delay is 10 msec. The packet will reach the destination after ________ msec.

Answer 1

Answer:

The packet will reach the destination after 170 msec

Explanation:

1 byte = 8 bits

From the given problem, 1000 bytes are transferred

So, 1000 bytes = 8 x 1000 = 8000 bits or 8 KB

The given speed is 50 B / sec ( as per the given problem)

To calculate time,

Time (T) = 8 / 50 sec

= 8 x 20 msec

= 160 m sec.

with reference to the given problem, there is a propagation delay of 10 msec. Now let us add 10 msec to the calculated value.

160 + 10 msec = 170 msec.

Answer 2

Hello Dear,

◆ Answer -

170 msec

◆ Explaination -

# Given -

Packet size = 1000 byte = 8 kilobit

Transmission speed = 50 kilobit/sec

Propagation delay = 10 msec

# Solution -

Time required for transmission (except delay) is -

Transmission time = packet size / transmission speed

Transmission time = 8 / 50

Transmission time = 0.16 s

Transmission time = 160 msec

Total time taken for packet to reach destination is -

Total time = transmission time + propagation delay

Total time = 160 + 10

Total time = 170 msec

Thanks dear...