Consider sending a 1,600-byte datagram into a link that has an mtu of 500 bytes. suppose the original datagram is stamped with the identification number 291. how many fragments are generated? what are the values in the various fields in the ip datagram(s) generated related to fragmentation?

Using the appropriate formula, the number of fragments which would be present in the datagram to be sent would be 4

The minimum length of IP header = 20 bytes

Maximum transmission unit (mtu) = 500 bytes

Hence, the payload would be calculated thus :

Hence, the maximum size of data field per Fragment = 480 bytes

The number of fragments required :

Hence, the number of fragments is 4

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Explanation:

Step one

The maximum size of data field in each fragment = 480

(because there are 20 bytes IP header) Thus the number of required

fragments  

thus the number of required fragment is 4

Step two

Each fragment will have identification number 291. each fragment except the last one will be of size 500 bytes (including IP header). the offset of the fragments will be 0, 60, 120, 180. each of the first 3 fragments will have

flag = 1; the last fragment will have flag =0

given the options, i would say your best bet is c. the modem supports up to 56 kbps, which is too low for video streaming. dial up modems use actual phone lines, which only can go up to 56 kbps as that's more than sufficient for traditional phone calls. this is also the reason why when a dial-up modem connects, you hear all those sounds. it's literally using the phone line to establish a connection.