PC A has a 1500 byte MTU size and 1460 byte MSS size. PC A needs to send 9000 byte of data to PC B. PC B has a 1400 byte MTU(I am not 100%, but I heard MTU size can be changed) size and obviously 1360 byte MSS size. PC A will segment 9000 byte of data into 6 1460 byte and 1 240 byte, then send all 7 segments to PC B. Since first 6 packets' size is bigger than PC B's MTU size, PC B would discard them. But since the last packet size is less than PC B's MTU size, PC B will take that last packet. If my understanding or explanation is wrong, please please let me know.
P.S. If you know how to simulate that senario in the packet tracer, please let me know :)
Your conclusion is missing an important information which is the SYN packet MSS negotiation. The packet exchange will do by the lowest MSS capacity.
Well, if must exist negotiations then presumably we can say that there's no MSS for UDP, only TCP.
This is not a problem at all. Fragmentation is performed on the sender following the link´s MTU. UDP rely on Fragment Offset, and More Fragments (MF) present on IP header to organized the stream.
PC A will segment the 9000 byte according to link MTU and sent this to PC B so that PC B can receive all the information. There´s no way PC sent 9000 byte cause there´s no MTU with that size.
We are certainly talking about Ethernet here. Ethernet has MTU of 1500 right? Then, using your scenario, PC A needs 6 packets of 1500 byte plus one more packet with 240 byte.
Also, we are talking about L4 only here. TCP/UDP has no idea what L3 is about. They are only a payload inside L3 packet.
Is that making sense?