Assume we have a set of age nodes \(0=x_0,x_1,\ldots,x_n\) and survival fractions \(1=l(x_0),l(x_1),\ldots,l(x_n)\). For each age interval \(I_i\equiv [x_i,x_{i+1})\), \(i=0,\ldots,n-1\) calculate \(\tilde{\mu}_i\) using the Constant Force method $$ \begin{eqnarray*} \tilde{\mu}_i & = & \frac{-1}{dx_i} \log\left(\frac{l(x_{i+1})}{l(x_i)}\right),\quad i=0,\ldots,n-1\\ & = & \frac{1}{dx_i}\int_{x_i}^{x_{i+1}} \mu(a) da \end{eqnarray*} $$ where \(dx_i = x_{i+1}-x_i\). From the above expression we see that \(\tilde{\mu}_i\) is the average of \(\mu(x)\) over the interval \(I_i\). Let \(\mu_i\) be the force at age \(x_i\). For the interior nodes \(i=1,\ldots,n-1\) calculate \(\mu_i\) by linearly interpolating \(\tilde{\mu}_i\) and \(\tilde{\mu}_{i+1}\) from the mid points of \(I_{i-1}\) and \(I_i\): $$ \mu_i=\frac{dx_i}{x_{i+1}-x_{i-1}} \tilde{\mu}_{i-1} + \frac{dx_{i-1}}{x_{i+1}-x_{i-1}} \tilde{\mu}_i ,\quad i=1,\ldots,n-1 . $$ In the interior intervals \(I_i\) for \(i=1,\ldots,n-2\) make \(\mu(x)\) a quadratic function of age $$ \mu(x) = \frac{x_{i+1} - x}{dx_i}\mu_i + \frac{x-x_i}{dx_i}\mu_{i+1} + \frac{C_i}{2}(x_{i+1}-x)(x-x_i),\quad x \in I_i $$ Since the average of \(\mu(x)\) over \(I_i\) is \(\tilde{\mu}_i\) we must have $$ C_i=\frac{12}{dx_i^2}\left(-\tilde{\mu}_i + \frac{\mu_i+\mu_{i+1}}{2}\right),\quad i=1,\ldots,n-2 $$ For the first interval \(I_0\) make the force linear in age. This implies $$ \begin{eqnarray*} \mu_0 &=& 2 \tilde{\mu}_0 - \mu_1\\ C_0 & = & 0 \end{eqnarray*} $$ If \(T_n\), total person-years lived after \(x_n\), is not specified then we impose the condition that \(\mu(x)\) is linear over \(I_{n-1}\) which implies $$ \begin{eqnarray*} \mu_n &=& 2 \tilde{\mu}_{n-1} - \mu_{n-1}\\ C_{n-1} & = & 0 \end{eqnarray*} $$ If \(T_n\) is specified then we impose the condition that \(\mu(x)\) is linear after \(x_n\) with both level and slope continuous at \(x_n\). Since \(mu(x)\) is quadratic on \(I_{n-1}\) there is a linear relationship between level and slope: $$ \begin{equation} \mu_n = A + B \mu'_n \end{equation} $$ where $$ \begin{eqnarray*} A&=&\frac{1}{2}\left(3 \tilde{\mu}_{n-1}-\mu_{n-1}\right)\\ B&=&\frac{dx_{n-1}}{4} \end{eqnarray*} $$ Also, to match \(T_n\) we must have $$ \frac{1}{m}= \sqrt{\frac{\pi}{2 \mu'_n}} e^{\sigma^2} \mathrm{erfc}(\sigma) $$ where \(m = l(x_n)/T_n\) is the open-intervl death rate, \(\sigma = \mu_n/\sqrt{2 \mu'_n}\) and \(\mathrm{erfc}(x)\) is the complementary Error Function. Using these two equations we can solve for \(\mu_n\). \(C_{n-1}\) is then given by the same equation as for the interior intervals.
Although this build method as it stands does not suffer as much as Quadratic Force from wild swings in the interpolated mortality rate it is still possible for the interpolated force to be negative even when all the \(\tilde{\mu}_i\) are positive. To ensure that interpolated forces are always positive we modify the \(\mu_i\) on the internal nodes using $$ \mu_i=\max\left(0,\min\left(\mu_i,2\min\left(\tilde{\mu}_{i-1},\tilde{\mu}_i\right)\right)\right) $$