There are in all 132bytes in the message below, or 1056 bits, This is the first
ID: 3613373 • Letter: T
Question
There are in all 132bytes in the message below, or 1056 bits,
This is the first assignment from the Chem402 class. It is about packets and their
transmission over links with differentbandwidths.
Each letter of themessage above represents 1 byte (8 bits). That includes blanks andperiods.
(a) (6 points) Find thetime it will take to transmit this message on end to end linkswith
bandwidths of 56kbps(phone modem), 1.5Mbps and 4Mbps(DSL), assuming circuitswitching.
(b)(9 points) If youwere to break up the message into packets of size 20 bytes each,with 4 bytes
for header, how longwill it take to reach the destination if it must go through allthree links
Explanation / Answer
Given 132 bytes = 1056 bits When speed is 1) 56Kbps = 56000 bits per second transmit time = 1056/ 56000 seconds = 18.86milliseconds 2) 1.5 Mbps = 1.5 * 106 bits per second transmit time = 1056/ 1500 milliseconds = 0.704milliseconds = 704microseconds 3) 4 Mbps = 4 * 106 bits per second transmit time = 1056/ 4000 milliseconds = 0.264milliseconds = 264microseconds If it is divided into packets of 20 bytes each...and 4 bytes isheader then data sent per packet is 20-4 = 16 bytes so number of packets required to sent 132 bytes = 132/16 = 8.25 ==9 packets since number o fpackets is to be a natural number transmit time per packet 1) speed = 56Kbps time per packet = (20*8 /56) milliseconds = 2.857 milliseconds for 9 packets total time = 2.857 *9 =25.714milliseconds 2) speed = 1.5 Mbps time per packet = (20*8 /1500) milliseconds = 0.107 milliseconds =107 microseconds for 9 packets total time = 107 *9 = 963 microseconds 3) speed = 4 Mbps time per packet = (20*8 /4000) milliseconds = 0.04 milliseconds =40 milliseconds for 9 packets total time = 40 *9 = 360 microseconds
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