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.
Answers
Answered by
45
Answer:
170 msec
Explanation:
It can be solved like the answer for your question 10msec+(1000x8)/50K=170msec
Answered by
31
Answer:
170 msec
Explanation:
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.
1 Byte = 8 Bits
=> 1000 Bytes = 1000*8 = 8000 Bits = 8 Kilo Bytes
Speed = 50 kilo bytes per sec
Time = 8/50 Sec
= 8 * 20 msex
= 160 msec
Delay of 10 msec
The packet will reach the destination after 160 + 10 = 170 msec
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