Pembahasan Ujian PAI: A60 – No. 28 – Juni 2016

1. Manfaat meninggal dibayarkan pada akhir tahun kematian
2. Level premium dibayarkan setiap awal kuartal
3. Mortality mengikuti select and ultimate life table dengan 2 tahun periode seleksi:
   

   \(x\)	\({l_{\left[ x\right]}}\)	\({l_{\left[ x \right] + 1}}\)	\({l_{x + 2}}\)	\(x + 2\)
   75	15.930	15.668	15.286	77
   76	15.508	15.224	14.816	78
   77	15.050	14.744	14.310	79

4. Kematian berdistribusi seragam setiap tahun
5. \(i = 0,06\)

Untuk membuatnya kuartal dengan syarat seragam
\({}_3{E_{\left[ {75} \right]}} = {v^3} \cdot {}_3{p_{\left[ {75} \right]}} = {\left( {\frac{1}{{1.06}}} \right)^3}\left( {\frac{{14,816}}{{15,930}}} \right) = 0.780904\) \({i^{\left( 4 \right)}} = 4\left[ {{{\left( {1.06} \right)}^{\frac{1}{4}}} – 1} \right] = 0.058695\) dan \({d^{\left( 4 \right)}} = 4\left[ {1 – {{\left( {1.06} \right)}^{ – \frac{1}{4}}}} \right] = 0.057847\) \(\alpha \left( 4 \right) = \frac{{id}}{{{i^{\left( 4 \right)}}{d^{\left( 4 \right)}}}} = \frac{{\frac{{\left( {0.06} \right)\left( {0.06} \right)}}{{1.06}}}}{{\left( {0.058695} \right)\left( {0.057847} \right)}} = 1.000264\) \(\beta \left( 4 \right) = \frac{{i – {i^{\left( 4 \right)}}}}{{{i^{\left( 4 \right)}} \cdot {d^{\left( 4 \right)}}}} = \frac{{0.06 – 0.058695}}{{0.058695 – 0.057847}} = 0.384351\) \(\ddot a_{\left. {\overline {\, {[75]:3} \,}}\! \right| }^{(4)} = \alpha (4){{\ddot a}_{\left. {\overline {\, {[75]:3} \,}}\! \right| }} – \beta (4){A_{\left[ {75} \right]}} = \alpha (4){{\ddot a}_{\left. {\overline {\, {[75]:3} \,}}\! \right| }} – \beta (4)\left( {1 – {}_3{E_{\left[ {75} \right]}}} \right)\) \(\ddot a_{\left[ {75} \right]:\left. {\overline {\, 3 \,}}\! \right| }^{\left( 4 \right)} = \left( {1.000264} \right)\left( {2.7819} \right) – \left( {0.384351} \right)\left( {1 – 0.780904} \right) = 2.698425\)

Nilai sekarang asuransi
\({d_{\left[ {75} \right]}} = {l_{\left[ {75} \right]}} – {l_{\left[ {75} \right] + 1}} = 15,930 – 15,668 = 262\) \({d_{\left[ {75} \right] + 1}} = {l_{\left[ {75} \right] + 1}} – {l_{77}} = 15,668 – 15,286 = 382\) \({d_{77}} = {l_{77}} – {l_{78}} = 15,286 – 14,816 = 470\) \(A_{\left[ {75} \right]:\left. {\overline {\, 3 \,}}\! \right| }^1 = \sum\nolimits_{k = 0}^2 {{v^{k + 1}} \cdot {}_{\left. k \right|}{q_{\left[ {75} \right]}}} \) \(A_{\left[ {75} \right]:\left. {\overline {\, 3 \,}}\! \right| }^1 = v \cdot {}_{\left. 0 \right|}{q_{\left[ {75} \right]}} + {v^2} \cdot {}_{\left. 1 \right|}{q_{\left[ {75} \right]}} + {v^3} \cdot {}_{\left. 2 \right|}{q_{\left[ {75} \right]}}\) \(A_{\left[ {75} \right]:\left. {\overline {\, 3 \,}}\! \right| }^1 = \left( {\frac{1}{{1.06}}} \right)\left( {\frac{{262}}{{15,930}}} \right) + {\left( {\frac{1}{{1.06}}} \right)^2}\left( {\frac{{382}}{{15,930}}} \right) + {\left( {\frac{1}{{1.06}}} \right)^3}\left( {\frac{{470}}{{15,930}}} \right) = 0.0616302\)

nilai dari premi kuarteran: \(\frac{{1000A_{\left[ {75} \right]:\left. {\overline {\, 3 \,}}\! \right| }^1}}{{4\ddot a_{\left[ {75} \right]:\left. {\overline {\, 3 \,}}\! \right| }^{\left( 4 \right)}}} = \frac{{1000\left( {0.0616302} \right)}}{{4\left( {2.698425} \right)}} = 5.71\)