A station running TCP/IP needs to transfer a file to a host. The file contains 2048 bytes. Assume a payload size of 512 bytes would be sent per packet and both systems are running IPv4. Also assume that the three-way handshake and window size negotiation have been completed and that no errors occur during transmission. a. What is the protocol overhead

The percentage/protocol overheads is about 7.8125 %

Explanation:

Solution

Recall that the TCP header contains a  20 bytes and IP header also contains  20 bytes.

The total of the overhead now is 40 bytes

Now,

let 2048 bytes be  divided by 4 segments

Thus, it becomes  (40+512) *4 = 2208

Therefore  2208 byes are sent by only 4 segments.

Now what is the overhead protocol

(1)  Since 2048 = 100%

let 2208 be  = X%

Then,

X=2208 * 100/2048 = 107.8125, that is from 100 % increased the 7.8125

therefore. the percentage overheads is about 7.8125 %

hey there! :

mass = 5.75 g

volume = 7.5 cm³

therefore:

density = mass / volume

d = 5.75 / 7.5

d = 0.7 g/cm³

answer c

hope that !

i think its d if im not mistaken