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I have a single worker that can work on N different tasks, but his total time T is limited. Let's assume time is in steps of 10 minutes (1 step = 1 mn). The payoff from a single job is 0 for the first step (setup cost) and then $\log_{10}(t)$, eg payoff is 0.778 after 60 minutes (6 steps). If a worker goes back to an old job and continues it, there is another setup cost of 1, so working 2 times 60 minutes on the same job, payoff is log(12-1).

We can then show that working on each job for 30 minutes (3 steps) and then not going back to it, is better than working on fewer jobs for 60 minutes each. It maximises the total payoff. In fact working on many jobs 11 minutes (1.1 steps) is kind of the optimal thing to do.

How can I make this more realistic? Imagine the worker is a human, and the jobs are meetings. Obviously you would not switch meetings all the time.

I appreciate this is not 100% well defined. But grateful for any ideas.

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    $\begingroup$ $\log_{10}(1)=0$, so working for 10 minutes on a job has no value (even without setup cost). Is this intentional? $\endgroup$ – prubin Jul 11 at 22:45
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    $\begingroup$ Your criterion function pays for work done but not for job completion. In most contexts, a partially done job is not all that valuable. $\endgroup$ – prubin Jul 11 at 22:47
  • $\begingroup$ Yes no value for first step is intentional, this is the setup cost. Yes I realise there is no benefit for getting the job done, it's just continuous $\endgroup$ – Dirk Nachbar Jul 13 at 8:28
  • $\begingroup$ If the first step is always setup, then shouldn't two 60 minute (six step) stints on a job be worth $\log_{10}(12-2)$ rather than $\log_{10}(12-1)$? $\endgroup$ – prubin Jul 14 at 15:57
  • $\begingroup$ Sounds like a work model for my teenager. Get everything set up, do a minute of work and then leave it for later....repeat on other tasks. $\endgroup$ – AirSquid Jul 24 at 6:03

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