Pembahasan Ujian PAI: A60 – No. 30 – November 2016

\(\begin{array}{*{20}{c}}{\begin{array}{*{20}{c}}{{A_{40}} = 0,22}\\{{A_{45}} = 0,24}\\{{A_{50}} = 0,26}\\{{A_{60}} = 0,30}\end{array}}&{\begin{array}{*{20}{c}}{{}_5{E_{40}} = 0,8}\\{{}_{10}{E_{40}} = 0,64}\\{{}_{20}{E_{40}} = 0,4}\\{d = 0,04}\end{array}}\end{array}\)

• \(v = 1 – d = 0.96\)
• \({}_kV = A_{x:\left. {\overline {\, n \,}}\! \right| }^1 – P \cdot {\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }}\)
• \(P = \frac{{A_{x:\left. {\overline {\, n \,}}\! \right| }^1}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}\)
• \({A_x} = A_{x:\left. {\overline {\, n \,}}\! \right| }^1 + {}_n{E_x} \cdot {A_{x + n}}\)
• \({A_{x:\left. {\overline {\, n \,}}\! \right| }} = A_{x:\left. {\overline {\, n \,}}\! \right| }^1 + {}_n{E_x}\)
• \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{1 – {A_{x:\left. {\overline {\, n \,}}\! \right| }}}}{d}\)
• \({}_n{E_x} = {v^n} \cdot {}_n{p_x}\)
• \({}_n{p_x} = \frac{{{l_{x + n}}}}{{{l_x}}}\)

Dari persamaan (1) dan (2) diperoleh
\(\frac{{{{\left( {0.96} \right)}^5}}}{{0.8}}{l_{45}} = \frac{{{{\left( {0.96} \right)}^{10}}}}{{0.64}}{l_{50}}\) \(\frac{{{l_{50}}}}{{{l_{45}}}} = {}_5{p_{45}} = \frac{{0.8}}{{{{\left( {0.96} \right)}^5}}}\)

Dari persamaan (1) dan (3) diperoleh
\(\frac{{{{\left( {0.96} \right)}^5}}}{{0.8}}{l_{45}} = \frac{{{{\left( {0.96} \right)}^{20}}}}{{0.4}}{l_{60}}\) \(\frac{{{l_{60}}}}{{{l_{45}}}} = {}_{15}{p_{45}} = \frac{{0.5}}{{{{\left( {0.96} \right)}^{15}}}}\)

Menghitung “net premium reserve”
\(A_{45:\left. {\overline {\, {15} \,}}\! \right| }^1 = {A_{45}} – {v^{15}} \cdot {}_{15}{p_{45}} \cdot {A_{60}} = 0.24 – {\left( {0.96} \right)^{15}}\frac{{\left( {0.5} \right)}}{{{{\left( {0.96} \right)}^{15}}}}\left( {0.3} \right) = 0.09\) \({A_{45:\left. {\overline {\, 5 \,}}\! \right| }} = {A_{45}} – {v^5} \cdot {}_5{p_{45}} \cdot {A_{50}} + {v^5} \cdot {}_5{p_{45}}\) \(= 0.24 – {\left( {0.96} \right)^5}\frac{{\left( {0.8} \right)}}{{{{\left( {0.96} \right)}^5}}}\left( {0.26} \right) + {\left( {0.96} \right)^5}\frac{{\left( {0.8} \right)}}{{{{\left( {0.96} \right)}^5}}} = 0.832\) \({\ddot a_{45:\left. {\overline {\, 5 \,}}\! \right| }} = \frac{{1 – {A_{45:\left. {\overline {\, 5 \,}}\! \right| }}}}{d} = \frac{{1 – 0.832}}{{0.04}} = 4.2\) \({}_5V = A_{45:\left. {\overline {\, {15} \,}}\! \right| }^1 – P \cdot {\ddot a_{45:\left. {\overline {\, 5 \,}}\! \right| }} = 0.09 – \frac{5}{{383}}\left( {4.2} \right) = 0.03517\)