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:

170 msec

Explanation:

It can be solved like the answer for your question 10msec+(1000x8)/50K=170msec

Answer 2

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