Pembahasan-Soal-Ujian-Profesi-Aktuaris-1024x481 (3) pai

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

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A60 - Matematika AktuariaAktuariaEdukasiPembahasan Ujian Profesi Aktuaris (Persatuan Aktuaris Indonesia / PAI)

Pembahasan Soal Ujian Profesi Aktuaris

Institusi : Persatuan Aktuaris Indonesia (PAI)
Mata Ujian : Matematika Aktuaria
Periode Ujian : November 2016
Nomor Soal : 30

SOAL

Untuk suatu asuransi “fully discrete 10-pay 20 year term life” dengan manfaat 1 pada (40) diberikan:

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

Hitunglah “net premium reserve” pada akhir tahun ke-5

  1. 0,008
  2. 0,014
  3. 0,028
  4. 0,035
  5. 0,042
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Diketahui Untuk suatu asuransi “fully discrete 10-pay 20 year term life” dengan manfaat 1 pada (40) diberikan:

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

Rumus yang digunakan
  • \(v = 1 – d = 0.96\)
  • \({}_kV = A_{x:\left. {\overline {\, n \,}}\! \right| }^1 – P \cdot {\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }}\)
  • \(P = \frac{{A_{x:\left. {\overline {\, n \,}}\! \right| }^1}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}\)
  • \({A_x} = A_{x:\left. {\overline {\, n \,}}\! \right| }^1 + {}_n{E_x} \cdot {A_{x + n}}\)
  • \({A_{x:\left. {\overline {\, n \,}}\! \right| }} = A_{x:\left. {\overline {\, n \,}}\! \right| }^1 + {}_n{E_x}\)
  • \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{1 – {A_{x:\left. {\overline {\, n \,}}\! \right| }}}}{d}\)
  • \({}_n{E_x} = {v^n} \cdot {}_n{p_x}\)
  • \({}_n{p_x} = \frac{{{l_{x + n}}}}{{{l_x}}}\)
Proses pengerjaan Menghitung “net premium”
\(A_{40:\left. {\overline {\, {20} \,}}\! \right| }^1 = {A_{40}} – {}_{20}{E_{40}} \cdot {A_{60}} = 0.22 – \left( {0.4} \right)\left( {0.3} \right) = 0.1\) \({A_{40:\left. {\overline {\, {10} \,}}\! \right| }} = {A_{40}} – {}_{10}{E_{40}} \cdot {A_{50}} + {}_{10}{E_{40}} = 0.22 – \left( {0.64} \right)\left( {0.26} \right) – 0.64 = 0.6936\) \({\ddot a_{40:\left. {\overline {\, {10} \,}}\! \right| }} = \frac{{1 – {A_{40:\left. {\overline {\, {10} \,}}\! \right| }}}}{d} = \frac{{1 – 0.6936}}{{0.04}} = 7.66\) Maka \(P = \frac{{A_{40:\left. {\overline {\, {20} \,}}\! \right| }^1}}{{{{\ddot a}_{40:\left. {\overline {\, {10} \,}}\! \right| }}}} = \frac{{0.1}}{{7.66}} = \frac{5}{{383}}\) \({}_5{E_{40}} = {v^5} \cdot {}_5{p_{40}} = {v^5} \cdot \frac{{{l_{45}}}}{{{l_{40}}}}\) \(0.8 = {\left( {0.96} \right)^5} \cdot \frac{{{l_{45}}}}{{{l_{40}}}}\) \({l_{40}} = \frac{{{{\left( {0.96} \right)}^5}}}{{0.8}}{l_{45}}\) ……………………………… (1)

\({}_{10}{E_{40}} = {v^{10}} \cdot {}_{10}{p_{40}} = {v^{10}} \cdot \frac{{{l_{50}}}}{{{l_{40}}}}\) \(0.64 = {\left( {0.96} \right)^{10}} \cdot \frac{{{l_{50}}}}{{{l_{40}}}}\) \({l_{40}} = \frac{{{{\left( {0.96} \right)}^{10}}}}{{0.64}}{l_{50}}\)  ………………………………… (2)

\({}_{20}{E_{40}} = {v^{20}} \cdot {}_{20}{p_{40}} = {v^{20}} \cdot \frac{{{l_{60}}}}{{{l_{40}}}}\) \(0.4 = {\left( {0.96} \right)^{20}} \cdot \frac{{{l_{60}}}}{{{l_{40}}}}\) \({l_{40}} = \frac{{{{\left( {0.96} \right)}^{20}}}}{{0.4}}{l_{60}}\)  …………………………….. (3)

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”
\(A_{45:\left. {\overline {\, {15} \,}}\! \right| }^1 = {A_{45}} – {v^{15}} \cdot {}_{15}{p_{45}} \cdot {A_{60}} = 0.24 – {\left( {0.96} \right)^{15}}\frac{{\left( {0.5} \right)}}{{{{\left( {0.96} \right)}^{15}}}}\left( {0.3} \right) = 0.09\) \({A_{45:\left. {\overline {\, 5 \,}}\! \right| }} = {A_{45}} – {v^5} \cdot {}_5{p_{45}} \cdot {A_{50}} + {v^5} \cdot {}_5{p_{45}}\) \(= 0.24 – {\left( {0.96} \right)^5}\frac{{\left( {0.8} \right)}}{{{{\left( {0.96} \right)}^5}}}\left( {0.26} \right) + {\left( {0.96} \right)^5}\frac{{\left( {0.8} \right)}}{{{{\left( {0.96} \right)}^5}}} = 0.832\) \({\ddot a_{45:\left. {\overline {\, 5 \,}}\! \right| }} = \frac{{1 – {A_{45:\left. {\overline {\, 5 \,}}\! \right| }}}}{d} = \frac{{1 – 0.832}}{{0.04}} = 4.2\) \({}_5V = A_{45:\left. {\overline {\, {15} \,}}\! \right| }^1 – P \cdot {\ddot a_{45:\left. {\overline {\, 5 \,}}\! \right| }} = 0.09 – \frac{5}{{383}}\left( {4.2} \right) = 0.03517\)
Jawaban D. 0,035
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