Maximum Error

See Normal Completion for background.

For a given task, the difference between the quality Q1 achieved over time with efficiency E1 and the quality Q2 achieved over time with expected efficiency E2 is the "error":

Error = Q2 - Q1 = (1 - E1)^t - (1 - E2)^t

Here, t is in units of best-case time to perform the task. Note that this is the negative of the traditional definition of error, Q1 - Q2, so that the value is positive.

The maximum measurable quality Qmax is defined by the resolution, or Units composing the task:

Qmax = 1 - 1 / Units

Setting E2 = Qmax ensures that Qmax will be achieved at t = 1 with the expected efficiency.

The following graph shows how Error varies over time for an average person (E1 = 0.5) and Units equal to the world's population in 2013 (P₂₀₁₃  = 7.1 E+09):

This pattern is common to all values of E1 and E2, with an early peak, an error equal to E2 - E1 at t = 1, and a decay towards zero.

The maximum error occurs at the following time, with the value as calculated by Q2 - Q1 above, and where LN is the natural logarithm:

t_{Max. Error} = LN[LN(1 - E2)/LN(1 - E1)] / LN[(1 - E1) / (1 - E2)]

This time and the maximum error are graphed below for different values of E1 and E2 is the same as above (1 - 1 / P₂₀₁₃). Note that the red dots indicate specific values of E1 (from right to left: 0.1, 0.2, ... 0.9).

The following graph shows what happens to the peak when we just vary E2. In this case, E1 = 0.5, and Units are doubled 52 times (starting at 4, from right to left, with each doubling indicated by a red dot). A curve fit is calculated for reference. Note that if E1 was less than 0.5, the peak time could exceed 1.