Formula 1 ADbackoff (u)=1n∑i=1ni2[ln(bi+1,i)−ln(bi,i)]−2∑i=1ni(bi+1,i−bi,i)+n2∑i=1n(bi+1,i2−bi,i2)\mathrm{AD}_{\text {backoff }}(\mathbf{u})=\frac{1}{\mathbf{n}} \sum_{\mathbf{i}=1}^{\mathbf{n}} \mathbf{i}^{2}\left[\ln \left(\mathbf{b}_{\mathbf{i}+\mathbf{1}, \mathbf{i}}\right)-\ln \left(\mathbf{b}_{\mathbf{i}, \mathbf{i}}\right)\right]-2 \sum_{\mathbf{i}=1}^{\mathbf{n}} \mathbf{i}\left(\mathbf{b}_{\mathbf{i}+\mathbf{1}, \mathbf{i}}-\mathbf{b}_{\mathbf{i}, \mathbf{i}}\right)+\frac{\mathbf{n}}{2} \sum_{\mathbf{i}=1}^{\mathbf{n}}\left(\mathbf{b}_{\mathbf{i}+\mathbf{1}, \mathbf{i}}^{2}-\mathbf{b}_{\mathbf{i}, \mathbf{i}}^{2}\right)ADbackoff (u)=n1i=1∑ni2[ln(bi+1,i)−ln(bi,i)]−2i=1∑ni(bi+1,i−bi,i)+2ni=1∑n(bi+1,i2−bi,i2)