Since

n_{0} = q n_{k}

and the

k

experimental arms each contain

n_{k}

patients,

n_{k} = \frac{n _{tot}}{k + q} .

Consequently

V = \frac{1}{n _{0}} + \frac{1}{n _{k}} = \frac{( q + 1 ) ( q + k )}{q n _{tot}} = \frac{q + k + 1 + k / q}{n _{tot}} .

Differentiation gives

V^{'} (q) = (1 - k / q^{2}) / n_{tot}

q = k

. Since

V^{''} (q) = \frac{2 k}{q ^{3} n _{tot}} > 0,

this is the unique minimum. Thus a shared standard arm should be

k

times the size of each new-treatment arm.

ii

Solution

For

k = 4

, the optimum is

q = 2

. The supplied two-sided sample size formula gives

n_{k} = \frac{1 + 2}{2} (\frac{2}{2})^{2} (0.84 + 1.96)^{2} = \frac{3}{2} (7.84) = 11.76,

so round upward to

12

patients in each new-treatment arm and take

n_{0} = 24

. The analyzed total is therefore

n_{tot} = 4 (12) + 24 = 72.

Allowing for ten percent attrition requires

⌈ \frac{72}{0.9} ⌉ = 80

patients to be recruited.

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