Answer:
Application C has 281.3 Mbps.
Explanation:
The TCP A, B and C connections all share a 3Gbps link.
Where 3Gbps is the bandwidth ( the number of bits or bytes the link can transmit in one second) and is equivalent to 3,000,000,000 bytes.
Connection on TCP A = 25,
Connection on TCP B = 4,
Connection on TCP C = 3,
The total number of connections are = 25 + 4 + 3 = 32.
When all connections are using the link,
The TCP C bandwidth is = (3/32) x 3,000,000,000 bytes.
= 0.09375 x 3,000,000,000
= 281250000 bytes.
= 281.25 Mbps or 281.3 Mbps.