Data Communications and Networks. Data Transmission Calculations A file that has

Question

Data Communications and Networks. Data Transmission Calculations

A file that has an size of eleven megabytes is transferred over a lan network using ethernet. Calculate the number of frames needed. The frame structure of the protocol follows: the preamble is 7 bytes, start is 1 byte, Destination address is 6 bytes, source address is also 6 bytes, data length is 2 bytes, frame data is 1468 bytes, and the checksum is 4 bytes. Also find overhead of one frame in bytes and the overall overhead used for the data transmission.

Explanation / Answer

File size=11MB

=11*2^20

=11534336 bytes

and data transmission per frame =1468byte

the number of frames needed=11534336/1468

=7857.177

=~7858

overhead of one frame = (Preamble+ start of frame delimiter+inter-frame gap)

=7+1+inter-frame gap

you not written inter-frame gap.

overall overhead used for the data transmission=(7+1+inter-frame gap)*7858

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