totally ordered multicast with Lamport timestamp Watch
>"It is sufficient to multicast any other type of message, as long as that message has a timestamp larger than the received message. The condition for delivering a message m to the application, is that another message has been received from each other process with a large timestamp. This guarantees that there are no more messages underway with a lower timestamp."
This is a definition from a book. I tried to apply this definition to an example but I guess that something is wrong.
>There are 4 processes and they multicast the following messages (second number in parentheses is timestamp) :
P1 multi-casts (m11, 5); (m12, 12); (m13, 14);
P2 multi-casts (m21, 6); (m22, 14);
P3 multi-casts (m31, 5); (m32, 7); (m33, 11);
P4 multi-casts (m41, 8); (m42, 15); (m43, 19).
Supposing that there are no acknoledgments, can I guess which messages can be delivered and which not? Based on definition, my guess is that only m11 and m31 can be delivered to the application, because all the other messages received will have a timestamp greater, but this seems very strange, and I think I didn't understand the delivery condition very well.
While you're waiting for an answer, did you know we have 300,000 study resources that could answer your question in TSR's Learn together section?
We have everything from Teacher Marked Essays to Mindmaps and Quizzes to help you with your work. Take a look around.
If you're stuck on how to get started, try creating some resources. It's free to do and can help breakdown tough topics into manageable chunks. Get creating now.
Not sure what all of this is about? Head here to find out more.