Years of depositing/Years of calculation Number of months necessary for degradation -m- Number of deposited months in accordance with calendaristic years depositing(12n) The equation of every calculus year -AT after the first depositing year: $\left(3+\left(8n\right)\right)t+7=\left(\left(12n\right)+13\right)-t$ , Equation(2) 1(2000) 0 $\left(12+1\right)$ $3t+7=13-t$ (Equation (1)) 2(2001)1 9 $\left(\left(1×12\right)+13\right)$ $11t+7=25-t$ (Equation (2-1)) 3(2002)2 7 $\left(\left(2×12\right)+13\right)$ $19t+7=37-t$ (Equation (2-2)) 4(2003)3 14 $\left(\left(3×12\right)+13\right)$ $27t+7=49-t$ (Equation (2-3)) 5(2004)4 13 $\left(\left(4×12\right)+13\right)$ $35t+7=61-t$ (Equation (2-4)) 6(2005)5 12 $\left(\left(5×12\right)+13\right)$ $43t+7=73-t$ (Equation (2-5)) 7(2006)6 11 $\left(\left(6×12\right)+13\right)$ $51t+7=85-t$ (Equation (2-6)) 8(2007)7 9 $\left(\left(7×12\right)+13\right)$ $59t+7=97-t$ (Equation (2-7)) 9(2008)8 13 $\left(\left(8×12\right)+13\right)$ $67t+7=109-t$ (Equation (2-8)) 10 (2009)9 11 $\left(\left(9×12\right)+13\right)$ $75t+7=121-t$ (Equation (2-9)) 11(2010)10 10 $\left(\left(10×12\right)+13\right)$ $83t+7=133-t$ (Equation (2-10)) 12(2011)11 7 $\left(\left(11×12\right)+13\right)$ $91t+7=145-t$ (Equation (2-11)) 13(2012)12 9 $\left(\left(12×12\right)+13\right)$ $99t+7=157-t$ (Equation (2-12))