Calculate the minimum time required to download two million bytes of information using each of the following t

Question

Calculate the minimum time required to download two million bytes of information using each of the following technologies: [5x4=20]

a. V.32 modem

b. V.32bis modem

c. V.90 modem

d. Cable Modem

Answer 1

(a)   V.32 modem                  

V.32 modem operating bidirectional data transfer = 9.6 kbit/s = 9600 / 8 = 1200 bytes/sec

      Transfer 2 million bytes of information = 2 * 10^6 bytes

    Time required for down loading information  = 2* 10^6 / 1200 = 1666.67 sec = 27.78 min = 0.463 hrs



(b)   V.32bis modem                  

V.32bis modem operating bidirectional data transfer = 14.4 kbit/s = 14400 / 8 = 1800bytes/sec



    Transfer 2 million bytes of information = 2 * 10^6 bytes

    Time required for down loading information  = 2* 10^6 / 1800 = 1111.11 sec = 18.52 min = 0.31 hrs



(c) V.90 modem                  

V.90 modem operating bidirectional data transfer = 56 kbit/s = 56000 / 8 = 7000bytes/sec  

    Transfer 2 million bytes of information = 2 * 10^6 bytes

    Time required for down loading information  = 2* 10^6 / 7000 = 285.71sec = 4.76 min = 0.08 hrs



(d) Cable Modem                

Cable modem operating bidirectional data transfer = 6Mbits/s = 6*10^6 / 8 = 750000bytes/sec

  

    Transfer 2 million bytes of information = 2 * 10^6 bytes

  

    Time required for down loading information  = 2* 10^6 / 750000 = 2.67sec = 0.045 min = 7.41*10^-4 hrs

Answer 2

So, what's the issue?  You find the bitrate of each technology and divide into the download size.