4 ms·

If you're looking at the first graph, the explanation is right above - the overhead of synchronization.

• aidenn0 7y ago
  That's nonsensical. There is surely at least as much synchronization overhead with 4 threads as with 2.
  • jonesetc 7y ago
    If the overhead is X, and the amount of work that can be done ignoring it is Y, you have Y - X. If you double the amount of work that can be down, but overhead stays the same you have 2Y - X, which is more than 2(Y - X) = 2Y - 2X. You're over performing doubling by not having to double the synchronization cost.
  • paulddraper 7y ago
    Not necessarily. There can be overhead introduced between the "1 thread" and ">1 thread" cases.
  • ricardobeat 7y ago
    From 1 -> 2 threads the overhead is introduced, if the cost is somewhat fixed, the gains from 2 -> 4 can be proportionally larger.