= Solution
Assume the usual one-lead-biased measurement, $V_s=V_{sd}$ with drain grounded and the gate referred to that ground; the printed map does not separately specify the wiring. This assumption is needed to assign individual source/drain capacitances. In this <Coulomb diamond capacitance extraction>, the smaller zero-bias gate intervals are 5 mV wide. In the <constant interaction model> with no orbital gap between those additions,
$$
C_g=\frac e{\Delta V_g}=\frac{1.602\times10^{-19}}{5\times10^{-3}}
=32.04\ \mathrm{aF}.
$$
An upper diamond tip lies at gate 123.75 mV between the zero-bias vertices 120 and 125 mV and at bias 2.50 mV. Its positive and negative edge slopes are therefore $s_+=2.50/3.75=2/3$ and $s_-=-2.50/1.25=-2$. Reservoir alignment gives
$$
s_-=-\frac{C_g}{C_s},\qquad
s_+=\frac{C_g}{C_\Sigma-C_s}=\frac{C_g}{C_g+C_d}.
$$
The <gate capacitance> is not negligible in the second denominator. Consequently
$$
\boxed{C_s\simeq16.0\ \mathrm{aF},\quad C_d\simeq16.0\ \mathrm{aF},\quad
C_g\simeq32.0\ \mathrm{aF},\quad C_\Sigma\simeq64.1\ \mathrm{aF}.}
$$
The consistency check is $e^2/C_\Sigma=2.50\ \mathrm{meV}$, matching the smaller tip's bias energy. The conventional electrostatic <charging energy> $E_C=e^2/(2C_\Sigma)$ is 1.25 meV; it is half the <charging energy> contribution to the <addition energy>. A different applied-bias convention changes the slope formulas, so the convention must accompany the extraction.
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