A device is sending a PDU of 5000 B on a link with an MTU of 1500 B.
If the PDU includes 20 B of IP header, which statement is true considering the most efficient way to transmit this PDU?
Click on the arrows to vote for the correct answerA. B. C. D.
In this scenario, a device is sending a Protocol Data Unit (PDU) of 5000 bytes on a link with an MTU (Maximum Transmission Unit) of 1500 bytes. Additionally, the PDU includes 20 bytes of IP header.
To transmit this PDU efficiently, the device will need to break it down into smaller packets that can be transmitted over the link. The most efficient way to do this is to use the Maximum Segment Size (MSS) value in the TCP (Transmission Control Protocol) header.
The MSS is the largest amount of data that a device can send in a single TCP segment, and it is determined by subtracting the TCP and IP header sizes from the MTU. In this case, the MSS would be 1480 bytes, calculated as follows:
MTU - TCP header size - IP header size = MSS 1500 - 20 - 20 = 1460
However, since the MSS should always be a multiple of 8, it will be rounded up to 1480 bytes.
Using this MSS value, the device can break down the PDU into smaller TCP segments that can be transmitted over the link. The first three packets will have a packet payload size of 1480 bytes, and the last packet will have a payload size of 560 bytes, calculated as follows:
PDU size - IP header size = payload size per packet 5000 - 20 = 4980
Number of packets needed = PDU size / MSS 4980 / 1480 = 3.37
The first three packets will have a payload size of 1480 bytes, and the last packet will have a payload size of 560 bytes, calculated as follows:
First three packets: 1480 bytes each Last packet: 560 bytes
Therefore, the correct answer is option A: The first three packets will have a packet payload size of 1400.