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cmuratori's profile
Casey Muratori
Casey Muratori
Casey Muratori
@cmuratori

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Casey Muratori

@cmuratori

I'm worried that the baby thinks people can't change.

Seattle
caseymuratori.com
Joined March 2009

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    1. Casey Muratori‏ @cmuratori Mar 15

      Is there any standard equation for the cost to process n items through a pipeline of fixed cost stages? Maybe from queuing theory? Like t_total(n) = t_single + t_worst*(n - 1) where t_single is the sum of the stage costs and t_worst is cost of the most expensive stage?

      4 replies 0 retweets 10 likes
      Show this thread
    2. Rik‏ @IsntTrivial Mar 15
      Replying to @cmuratori

      max(t_i) * (n-1) + sum(t_i) is a lower bound and tight in case max(t_i) = t_0. Upper bound is sum(t_i)*n which is tight in case there's only one stage. Neither *has* to be tight though. Don't think a general formula exists.

      1 reply 0 retweets 0 likes
    3. Casey Muratori‏ @cmuratori Mar 15
      Replying to @IsntTrivial

      Where are you getting those bounds from, though? Why would it matter if max(t_i) is on t_0 or anywhere else?

      1 reply 0 retweets 0 likes
    4. Rik‏ @IsntTrivial Mar 15
      Replying to @cmuratori

      The upper bound is simply the total amount of work if the pipeline is executed on a single processor. The lower bound is (I realize now) incorrect, but the idea is that all tasks have to get to the bottleneck and be processed there, before you can process the final task fully.

      2 replies 0 retweets 0 likes
    5. Casey Muratori‏ @cmuratori Mar 15
      Replying to @IsntTrivial

      But I'm pretty sure the equation I proposed already accounts for that. Can you show an example where it fails?

      1 reply 0 retweets 0 likes
    6. Rik‏ @IsntTrivial Mar 15
      Replying to @cmuratori

      t = [1, 3, 1, 3], n=2. The formula gives 7 + 3 = 10. However, this will take 12 steps. Diagram in a pastebinhttps://pastebin.com/N0JfzxP6 

      1 reply 0 retweets 0 likes
    7. Casey Muratori‏ @cmuratori Mar 15
      Replying to @IsntTrivial

      You cheated by counting two different ways :) Here is the same diagram you drew, but for a single input. Using your counting, the "t single" is clearly 9, not 7. So in your example, the formula gives 9 + 3 = 12, which is exactly what your diagram showed.https://pastebin.com/2RicjZqX 

      1 reply 0 retweets 0 likes
    8. Rik‏ @IsntTrivial Mar 15
      Replying to @cmuratori

      Oof, caught red-handed! I think t_single is 8 though, since the dot is already out at time 9 ;)

      1 reply 0 retweets 0 likes
    9. Casey Muratori‏ @cmuratori Mar 15
      Replying to @IsntTrivial

      It doesn't matter how you count, you just have to count the same. So if you want t_single to be 8, your example has to be 11. Either way, my formula works correctly for your example as far as I can tell, so it is not a case where the formula fails?

      1 reply 0 retweets 0 likes
    10. Rik‏ @IsntTrivial Mar 15
      Replying to @cmuratori

      I think you are right and I am wrong. I also tried with 3 items and various combinations of stage lengths, but I'm not able to make it last any longer or shorter. Seems like it doesn't matter that there could be various "other bottlenecks". Far from a proof though. :/

      1 reply 0 retweets 0 likes
      Casey Muratori‏ @cmuratori Mar 15
      Replying to @IsntTrivial

      I think a proof is not too hard. We may try it and see. But it still doesn't change the fact that I feel like this should have been proved before, and it should probably have a name, like "Milton's Conjecture" or "Bonham's Law" or something.

      3:30 PM - 15 Mar 2021
      • 2 Likes
      • Ivan Braidi Daniel Bokser
      1 reply 0 retweets 2 likes
        1. New conversation
        2. Rik‏ @IsntTrivial Mar 16
          Replying to @cmuratori

          Pinedo's "Scheduling Theory, Algorithms, and Systems" has your formula (Theorem 6.1.8, proof as exercise), but it's for when each task takes a fixed amount of time for any stage, rather than each stage taking a fixed amount of time for any task.

          1 reply 0 retweets 0 likes
        3. Rik‏ @IsntTrivial Mar 16
          Replying to @IsntTrivial @cmuratori

          I should get to work. If you want to look for yourself (it's probably easier to just find a proof), the problem is mathematically described as a permutation flow shop, you're optimizing the makespan. In notation Fm|prmu|C_max.

          0 replies 0 retweets 0 likes
        4. End of conversation

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