The maximum allowed segment for Ethernet is 500 meters and the maximum number of

Question

The maximum allowed segment for Ethernet is 500 meters and the maximum number of segments that can be connected by repeaters is limited to five. The minimum length of the frame that can be transmitted is the sum of the round-trip delay and the repeater delays. Assume that the speed of transmission on the cable is 200 meters/microsecond and that the total round-trip delay in traversing all the repeaters is 25 microseconds. Show that the minimum frame size (number of bits per frame) of an Ethernet frame is 64 bytes. Note: The maximum frame size is 1,518 bytes.

Explanation / Answer

2500 total meters divided by 200 meters = 12.5 microseconds to get through the cables 1 way.

round trip through the cables would be 12.5 + 12.5 = 25 microsends round trip through the cable section

add 25 microsends for the repeaters = 50 microseconds

25 microseconds =round trip delay ( 500 per segment * 5 ) * 2 ( back and forth ) / 200

25 microseconds = round trip repeater delay ( given in question )

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50 seconds total delay time repeater and cable round trip

so theoritically you could have a 50 byte frame but they dont exist as any frame size has to work on binary 2 / 4 / 8 / 16 / 32 / 64

since 32 is less than 50 it wont work

Your next smallest one would be 64

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