Case 1: Assume that there is no segmentation needed.
IP header = 20 bytes
TCP header = 20 bytes
TCP payload can be 1500 bytes.
Ethernet header = 8 + 6 + 6 + 2 bytes
Ethernet checksum = 4 bytes
Therefore, the size of first frame is 1566 bytes.
The time to live = 120 sec.
The length of sequence number = 32 bits.
Therefore, the maximum number of bytes that a host can send out without wrapping around the sequence number is 2 ^32/ 1500.
Since Ethernet frames may be sent continuously, the maximum line speed, V can be:
V <= 1566 * 8 * (2^32 / 1500) (bits)/120 (sec)
= 298929723.8016 bits/sec
= 298.9 Mbps
The maximum line speed is 298.9 Mbps.
Case 2: Since the maximum size of Ethernet payload is 1500 bytes, the TCP payload needs to be segmented as followings.
Frame 1:
IP header = 20 bytes
TCP header = 20 bytes
Since the maximum Ethernet Payload is 1500 bytes including IP header and TCP header, TCP payload can be 1460 bytes.
Ethernet header = 8 + 6 + 6 + 2 bytes
Ethernet checksum = 4 bytes
Therefore, the size of first frame is 1526 bytes
Frame 2:
IP header = 20 bytes
TCP header = 20 bytes
TCP payload remained after segmentation = 40 bytes (= 1500 – 1460 bytes)
Ethernet header = 8 + 6 + 6 + 2 bytes
Ethernet checksum = 4 bytes
Therefore, the size of second frame is 106 bytes
The time to live = 120 sec.
The length of sequence number = 32 bits.
Therefore, the maximum number of bytes that a host can send out without wrapping around the sequence number is 2 ^32/ 1500.
Since Ethernet frames may be sent continuously, the maximum line speed, V can be:
V <= (1526 + 106) * 8 * (2^32 / 1500) (bits)/120 (sec)
= 1632 * 8 * (2^32 /1500) (bits)/ 120 (sec)
= 311528294.5365 bits/sec
= 311.5 Mbps
The maximum line speed is 311.5 Mbps.
