Suppose an IP packet is fragmented into 10 fragments, each with a 1% (independent) probability of loss. To a reasonable approximation, this means there is a 10% chance of losing the whole packet due to loss of a fragment. What is the probability of net loss of the whole packet if the packet is transmitted twice? a. assuming all fragments received must have been part of the same transmission?
b. assuming any given fragment may have been part of either transmission?
c. explain how use of the IDENT field might be applicable here.

a. 0.01

b. 0.001

c. The identification field of the packet fragment can be used to uniquely identify and collate the fragments lost in transmission.

Explanation:

The probability of losing a packet is 10% or 0.1, so the probability of losing the packet twice during transmission;

= 0.1 x 0.1 = 0.01

When any fragments have been part of the transmission, the probability of the packet is dependent on the fragments;

= 0.01 x 0.1 = 0.001

The identification field is a unique 16-bit value assigned to an IPv4 packet, when a packet is fragmented for transmission, its field is used to collate the unique fragments in the packet.

i don't think so. although it would be nice. i am pretty sure that was an option an update or two ago. i guess they want you to stay 14 forever. haha.

i don't know to be honest but you on your own

step-by-step explanation:

ignitia made me retarted fool im stoopid laugh like shi

explanation: