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. (assume 1k=1000)
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Answers & Comments
Explanation:
one byte = 8 bits
1000 bytes = 1000 × 8 = 8000 bits = 8 kilo bits
speed 50 kilobytes per second
time is = 8 / 50 seconds
= 8 × 20 msec
= 160 msec
delay of 10 seconds
the packet will reach the destination after 160 + 10 = 170 msec.
Hope it helps
The packet will reach the destination after 170msec.
given: size of packet = 100 byte
speed = 50 kbps
propagation delay = 10ms
to find : time required to reach the destination = ?
solution:
we know that , the number of bits in 1 byte is 8 bits
if, 1 byte = 8 bit
then 1000 byte = 1000×8 bits
= 8000 bits
= 8 kilo bit
if the speed is 50 kilo bit per second
then the time required for 8 kilo bit will be = 8/50 sec
= (8*100)/50ms
= 8 * 20ms
= 160ms
since we are dealing with a propagation delay of 10ms thus the packet will require 10ms more to reach the destination.
total time required = (160 + 10)ms
= 170ms
so we observe the effect of propagation delay here
hence, the packet will reach the destination after 170msec.
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