= Solution
Since $n_0=qn_k$ and the $k$ experimental arms each contain $n_k$ patients,
$$
n_k=\frac{n_{\mathrm{tot}}}{k+q}.
$$
Consequently
$$
V=\frac1{n_0}+\frac1{n_k}
=\frac{(q+1)(q+k)}{qn_{\mathrm{tot}}}
=\frac{q+k+1+k/q}{n_{\mathrm{tot}}}.
$$
Differentiation gives $V'(q)=(1-k/q^2)/n_{\mathrm{tot}}$, so the stationary point is $q=\sqrt k$. Since
$$
V''(q)=\frac{2k}{q^3n_{\mathrm{tot}}}>0,
$$
this is the unique minimum. Thus a shared standard arm should be $\sqrt k$ times the size of each new-treatment arm.
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