Comment by olafA

please bear with me on this, running through the example even when the wait kicked in trxA's timestamp (14ms nodeA time) is still greater than trxB's timestamp (9ms nodeB time) which violate gcp's definition of external consistency?

> To be externally consistent, a transaction must see the effects of all the transactions that complete before it and none of the effects of transactions that complete after it, in the global serial order

Example when both node waited

    real-world time = 7ms
    node A local time = 7ms
    node B local time = 0ms


    trxA arrive nodeA at (real-world time 7ms, A time 7ms, B time 0ms)
    wait 7ms before commit
    trxA commit by nodeA at (real-world time 14ms, A time 14ms, B time 7ms)

    trxB arrive nodeB at (real-world time 9ms, A time 9ms, B time 2ms)
    wait 7ms before commit
    trxB is commit by nodeB at (real-world time 16ms, A time 16ms, B time 9ms)

result, localtime(trxA) = 14ms , localtime(trxB) = 9ms

No wait example

    real-world time = 7ms
    node A local time = 7ms
    node B local time = 0ms


    trxA arrive nodeA at (real-world time 7ms, A time 7ms, B time 0ms)
    trxA take 1ms to commit
    trxA commit by nodeA at (real-world time 8ms, A time 8ms, B time 1ms)

    trxB arrive nodeB at (real-world time 9ms, A time 9ms, B time 2ms)
    trxB take 1ms to commit
    trxB is commit by nodeB at (real-world time 10ms, A time 10ms, B time 3ms)

result, localtime(trxA) = 8ms , localtime(trxB) = 3ms

in both cases localtime(trxA) > localtime(trxB) , external consistency is not achieved?

and in the no wait example the transactions are not concurrent, only when it waited it became concurrent?

I must be missing something here?