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When 54 bytes of data is transferred using the UDP protocol, the efficiency is? (In per cent)

IN UDP, the maximum datagram size can be $2^{16}-1 \, bytes$

Now to this, 8 Bytes of UDP header is included and this together must fit within $65535\, bytes$ IPv4 packet limit(Including 20B IP Header size).

So, maximum data that we can send in 1 UDP datagram-> $65535-8-20=65507\,Bytes$

We could have sent this amount of data, but we are actually sending 54 Bytes, so the efficiency of the protocol is $\frac{54}{65507} \times 100(\,in\,\%)$

Am I correct?
in Computer Networks by Boss (30.6k points) | 167 views


why  20 is subtracted from the 65535-8?

max application layer data can be 65535-8 Bytes

given is 54 bytes

efficiency(%) should be (54/65527)*100


@Ayush Upadhyaya

why calculating effieicene against 65507.

65507 is the maximum size of data that UDP can send to the network layer.. but it's not fixed. It's not like, if UDP is not sending that much amount of data, then that much amount of space remain vacant..

efficiency (useful data sent)/(total data sent)

i don't think we need to find out the whole thing of 65507 here, because 65507 is total data that could have sent, but not sent.

simply 54B is useful data, and 54+8=62B is total data sent by UDP. Hence efficiency is 54/62

(also I think ip header overhead is not required to add in the total data, as efficiency is asked for udp not ip.)

I don't know why but I am feeling question is bogus

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