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In 1962 Herbert Simon wrote a paper in which he presents his watchmaker parable about how piecemeal construction is more robust than one-shot construction. I won't repeat the full argument here - see the link - though the gist is below. My question is about the 'straightforward calculation'. While the math itself is simple as can be, I can't seem to fully pinpoint why it makes sense to use the ratios as he does.

To summarize the givens: To complete a watch...

  • ... Tempus has to construct 1 assembly (of 1000 parts), has a $.99^{1000}$ chance to achieve that, and loses (the time of making) 100 parts on average when he fails;
  • ... Hora has to construct 111 assemblies (of 10 parts each), has a $.99^{10}$ chance to achieve each assembly, and loses (the time of making) 5 parts on average when he fails.

And the question is: How much longer does it take Tempus to complete a watch?

(Note that Simon does quite a bit of rounding to get to his ~4000 figure -- from 3774.542... -- but that is not the issue here.)

So, Hora has to build 111 assemblies, against 1 for Tempus:

$$\frac{111}{1}$$

But building one assembly is easier for Hora than for Tempus, so it makes sense to want to throw in the achievement probabilities. Now it has to be easier for Hora to make an assembly (bringing down the 111 figure, relatively), so the multiplier needs to be small, relatively, which you get by taking the inverses of the probabilities:

$$\frac{111}{1} × \frac{\frac{1}{.99^{10}}}{\frac{1}{.99^{1000}}} = \frac{111}{1} × \frac{.99^{1000}}{.99^{10}}$$

But how to understand - in words - this inverse-taking? I would be inclined to want to multiply e.g. 111 with $.99^{10}$ (rather than $.99^{1000}$) from the idea that Hora is accumulating probabilities as he goes along building assemblies. Except that doesn't fit with the reasoning in the previous paragraph, and furthermore 111 times $.99^{10}$ equals 100.386... which is not a probability of any kind... I'm lost: what are we actually calculating here? And then when the average losses are added in to obtain

$$\frac{111}{1} × \frac{.99^{1000}}{.99^{10}} × \frac{5}{100}$$

I get still more lost as to its interpretation.

Can anyone clarify? Much obliged.

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The following model seems to be used: Each watchmaker picks up one piece per second, say, until they have the required 1000 or 10 pieces together. And it is assume that a phone call occurs with $p=0.01$ within any such second (or calls occur once every $100$ seconds on average).

Tempus' construction process "survives" with probability $(1-p)^{1000}$, so for one watch he needs $\frac1{(1-p)^{1000}}$ attempts, which means that until success there were $\frac1{(1-p)^{1000}}-1$ phone calls, followed by $1000$ seconds of uninterrupted silence. In round figures, this means that it takes about $\frac1{(1-p)^{1000}}$ times the typical between-calls interval of $100$ seconds to complete a watch. That is $$\frac{ 100}{\left(1-\frac1{100}\right)^{1000}}\approx \frac{100}{e^{-10}}\approx 220.000\quad\text{seconds}$$

One of Hora's (sub-)constructions "survives" with the much larger probability $(1-p)^{10}$. And because now a phone call is a rather "rare" event, we may assume for simplicity that it happens at a random moment of the $10$ second process, thus costing us $5$ seconds (instead of $100$ seconds) on average. Instead of making an exact calculation, just note that with some luck, it would take Hora only $111\times 10=1110$ seconds to assemble a watch. But within these $1110$ seconds, we must expect about $11.1$ phone calls, costing us a total of $55.5$ work-seconds, so that the expected construction time increases to $1165.5$ seconds. But within these extra $55.5$ seconds, we'd expect another $0.555$ phone calls, adding another $2.775$ seconds as penalty. We could contuinue this reasoning, but the penalties decrease very quickly, so we see that Hora completes a watch about every $1170$ seconds, so about $$ 200$$ times as fast (instead of $4000$ times as fast).

Actually, in the original text, it is unclear (to me) why there should be a $111:1$ penalty for Hora due to the subconstructions made. This would only be valid if the full construction step of Tempus took as long as ach of the sub-steps of Hora, but that would invaliudate the est of the argument about the avearge number of steps lost.

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  • $\begingroup$ $100 e^{10} \approx 2.2 \times 10^6$, not $2.2\times 10^5$, for a final ratio of $2000:1$ (which I think is approximately correct; by my calculations, Simon's estimate is off by a factor of about two). $\endgroup$ – David K Feb 22 '16 at 15:07
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Under Simon's assumptions, the chance of completing any step of a particular assembly process without interruption is $1 - 0.01 = 0.99$. If an interruption occurs at any step, the watchmaker has to start the entire assembly from the beginning.

Since Tempus has to complete $1000$ steps in order to make one assembly, the chance of completing all of them without interruption is $0.99^{1000} \approx 4.32\times 10^{-5}.$ As only about $43$ out of a million attempted assemblies are successful, it takes Tempus on average $1/(0.99^{1000}) \approx 23,000$ attempts in order to complete one assembly (at which point they have a complete watch).

Hora, on the other hand, has a $0.99^{10} \approx 0.9$ chance to complete each assembly of ten parts, so on average $1.1$ attempts in order to complete one assembly.

In other words, the figure $1/(0.99^{1000})$ is the number of attempted assemblies Tempus must make per watch, and $111/(0.99^{10})$ is the number of attempted assemblies Hora must make. Intuitively, "more assemblies required" is bad, and "less chance to succeed" is bad. If we multiplied the number of assemblies by the chance of completion, and we then double the number of assemblies and cut the chance of completion in half, the resulting product would be unchanged; that is, the two "bad" effects would have canceled, while really they should have compounded each other. Dividing the number of assemblies by the chance of completion is a better way to combine the two effects.

For Tempus, since almost all attempted assemblies fail, the number of steps to complete one watch is dominated by the number of failed assemblies and the number of steps lost during each failure, which is $100$ steps on average. The fact that the last attempt (which is successful) takes $1000$ steps instead of $100$ a mere rounding error in the estimated number of steps, which is about $23,000 \times 100 = 2.3 \times 10^6$.

However, I believe Simon's final ratio is off by a factor of almost two due to an underestimation of Hora's time. Simon took the $111$ assemblies Hora must complete, multiplied by $1.1$ to account for the extra attempts required due to interruptions, and then multiplied by $5$ for the amount of time per attempt. The error here is that almost all attempts are successful, and therefore take ten steps rather than five. That is, Simon in effect gives Hora $111 \times 1.1 \times 5 \approx 600$ steps to complete one watch, whereas the actual average number of steps is $$ 111 \times 10 + 111 \times 5 \times \left(\frac{1}{0.99^{10}} - 1\right) \approx 1170. $$ That is, in round numbers, $111\times 1.1\times 10 \approx 1200$ would have been a better estimate of the number of steps for Hora.

The final ratio should be (in round numbers) $$ \frac{2.3 \times 10^6}{1200} \approx 2000. $$

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