Q1) Given a link with a maximum transmission rate of 99 Mbps. Only two computers

Question

Q1) Given a link with a maximum transmission rate of 99 Mbps. Only two computers, X and Y, wish to transmit starting at time t = 0 seconds. Computer X sends fileX (6 MiB) and computer Y sends fileY (299 KiB), both starting at time t = 0.

Computer X gets the transmission medium first, so Computer Y must wait.

For the following calculations, assume maximum transmission rate during transmission.

Suppose that entire files are sent as a stream (no packets, no multiplexing).

At what time (t = ?) would FileY finish transmitting?

Give answer in seconds, without units, and round to two decimal places (e.g. for an answer of 12.4567 seconds you would enter "12.46" without the quotes)

Q2)

Voice over IP (VoIP)

Given the attached image, and:h

Host A converts analog to digital at a = 51 Kbps

Link transmission rate R = 1.6 Mbps

Host A groups data into packets of length L = 58 bytes

Distance to travel d = 969 km

Propagation speed s = 2.5 x 108 m/s

Host A sends each packet to Host B as soon as it gathers a whole packet.

Host B converts back from digital to analog as soon as it receives a whole packet.

How much time elapses from when the first bit starts to be created until the conversion back to analog begins? Give answer in milliseconds (ms) to two decimal places, normal rounding, without units (e.g. 1.5623 ms would be entered as "1.56" without the quotes)

Explanation / Answer

Q1)

Ans:

It seems like a simple calculation.

With no other specified parameter such as the distance between the sending and the receiving end, the calculation should be:

Data : 5 MiB = 5*1024*1024*8 = 41943040

Speed : 19.7Mbps = 19.7*1000000 = 19700000

Time = Data / Speed ~ 2.13 seconds

Note: no payload size was considered since it was mentioned that no packetization is being done.

Q2)

Ans:

Given analog to digital conversion rate is = a = 42 Kbps.

         Link transmission rate is = R = 3.8 Mbps.

         Length of the packet is = L = 57 bytes.

         Distance to travel is = d = 846 km.

         Propagation speed is = s = 2.5 x 108 m/s

       Time elapsed from when the first bit starts to be created

       until the conversion back to analog begins is

                             =

              Analog to digital conversion time

                 +

      Transmission time

   +

Propagation time    

Here Analog to digital conversion time is = 57 B / 42Kbps

                                                            = 57 * 8 / 42K

                                                            = 10.86 / K

                                                            = 10.86 / 103

                                                           = 10.86 * 10-3 = 10.86 ms

      Hence Analog to digital conversion time is = 10.86 ms.

     Transmission time is = L / R

                                     = 57B / 3.8 * 106

                                                 = 57 * 8 / 3.8 * 106                                                 

                                     = 12 * 10-5 = 0.12 ms

      Hence Transmission time is = 0.12 ms.

      Propagation time is = d / v

                                     = 846 * 103 / 2.5 x 108

                                                 = 338.4 / 105                                                 

                                     = 338.4 * 10-5 = 3.38 ms

      Hence Propagation time is = 3.38 ms.

Hence Time elapsed is = 10.86 + 0.12 + 3.38 = 14.36 ms.

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