Time of File Transmission Given a link with a maximum transmission rate of 28.1

Question

Time of File Transmission

Given a link with a maximum transmission rate of 28.1 Mbps. Only two computers, X and Y, wish to transmit starting at time t = 0 seconds. Computer X sends fileX (10 MiB) and computer Y sends fileY (14 KiB), both starting at time t = 0.

Statistical multiplexing is used, with details as follows
Packet Payload Size = 1000 Bytes
Packet Header Size = 24 Bytes (overhead)
Ignore Processing and Queueing delays
Assume partial packets (packets consisting of less than 1000 Bytes of data) are padded so that they are the same size as full packets.
Assume continuous alternating-packet transmission.
Computer X gets the transmission medium first.
At what time (t = ?) would FileX 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)

Explanation / Answer

Size of each packet to be transmitted (along with the header) = (1000+24) bytes = 1024 bytes

Size of each packet in bits = 1024*8 bits = 8,192 bits        (1 Byte = 8 bits)

Transmission Rate = 28.1 Mbps, which means 28.1*10^6 bits are transmitted in one second.

Transmission time of each packet = size of packet/ transmission rate = (8,192)/ (28.1 * 10^6) = 291.53 * 10^-6 seconds

Size of FileX = 10MiB = 10,485,760 bytes    (1 Mib = 1024*1024 Bytes = 1048576 Bytes)

Since each packet has a payload size of 1000 bytes, one packet can contain 1000 bytes of data from the file.

Number of packets made to transfer whole FileX = 10,485,760/1000 = 10485.760 = 10486 packets (partial packets are padded)

Size of FileY = 14KiB = 14*1024 bytes = 14,336 bytes    (1 Kib = 1024 Bytes)

Since each packet has a payload size of 1000 bytes, one packet can contain 1000 bytes of data from the file.

Number of packets made to transfer whole FileY = 14,336/1000 = 14.336 = 15 packets (partial packets are padded)

Since computer X gets transmission medium first, packet from FileX will get transmitted first and then alternatively packets from FileY and FileX will be transmitted.

Since Statistical Multiplexing is followed, when file from any one of the resources is transmitted completely, the channel is no more shared with that resource and only the resource which has its file left to be transmitted is provided with the medium.

As it is clear from the sizes of the files, FileY will complete its transmission first since there are only 15 packets. First the medium is shared between Computer X and Y alternatively and after the transmission of 15 packets of FileY, the channel is dedicated to transmission of FileX only for its left over packets (10,486 – 15 = 10471). Therefore, time taken for FileX to be transmitted:

15 * (291.53 * 10^-6 seconds + 291.53 * 10^-6 seconds)    [one time slot each for FileX packet and FileY packet]

+

10471 * (291.53 * 10^-6 seconds)        [only packets of FileX left]

=

3,061,356.53 * 10^-6 seconds = 3.06 seconds

Therefore, at t=3.06 FileX would complete its transmission.

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