Final answer:
The efficiency of transmission at the data link layer using Ethernet is calculated by dividing the data payload size by the total frame size. Only options (a) and (c) provide plausible efficiency ratios, with respective values of 0.222 and 0.285, as frame size typically cannot be smaller than the payload size.
Step-by-step explanation:
This question pertains to the calculation of the efficiency of transmission at the data link layer using Ethernet. The efficiency can be considered as the ratio of the actual data payload size to the total frame size. The options given in the question (a through d) present these ratios as simple arithmetic calculations.
- (a) For a 16-byte data payload and a 72-byte minimum frame size, the efficiency is calculated as 0.222 or 22.2%.
- (b) When the data payload is 72 bytes and the minimum frame size is 16 bytes, the ratio given is incorrect, as frame size usually cannot be smaller than the payload size.
- (c) With a 16-byte data payload and a 56-byte minimum frame size, the efficiency ratio is 0.285 or 28.5%.
- (d) For a 56-byte data payload and a 16-byte minimum frame size, the ratio again doesn't make sense as frame size typically includes the payload size plus additional overhead.
It's important to understand that the total frame size includes not just the data payload but also the overhead introduced by headers and footers, such as the Ethernet frame header, which is 18 bytes including the preamble, start frame delimiter (SFD), destination MAC, source MAC, type/length, and Frame Check Sequence (FCS).