Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Suatu polis asuransi biasanya memuat klausa bahwa bila usia tertanggung diketahui tidak tepat pada saat diterbitkan, maka manfaat dari polis tersebut dapat disesuaikan sebesar selisih kalau polis tersebut dibeli dengan usia yang tepat. Suatu polis asuransi berjangka diskrit selama 3 tahun sebesar 1.000 dijual kepada seseorang yang menyatakan berusia 30 pada saat penerbitan polis. Akan tetapi, pada tahun ke tiga, diketahui sesungguhnya orang tersebut berusia 31 tahun pada saat penerbitan
Bila diketahui: \(i = 0,04\), \({q_{30}} = 0,01\), \({q_{31}} = 0,02\), \({q_{32}} = 0,03\), dan \({q_{33}} = 0,04\)
Hitunglah berapa besar manfaat yang harus disesuaikan (besar manfaat yang dikurangkan)
- 264,10
- 664,10
- 864,10
- 335,90
- 135,90
| Diketahui |
|
| Rumus yang digunakan | \(P = \frac{{{b_t}A_{x:\left. {\overline {\, n \,}}\! \right| }^1}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}\), \(d = 1 – v = 1 – \frac{1}{{1 + i}}\) \(A_{x:\left. {\overline {\, n \,}}\! \right| }^1 = {A_{x:\left. {\overline {\, n \,}}\! \right| }} – {A_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^1 }} = {A_{x:\left. {\overline {\, n \,}}\! \right| }} – {}_n{E_x} = \left( {1 – d{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}} \right) – {v^n} \cdot {}_n{p_x}\) \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \sum\limits_{k = 0}^{n – 1} {{v^k} \cdot {}_k{p_x}} \), \({}_t{p_x} = \prod\limits_{k = 0}^{t – 1} {{p_{x + k}}} \) |
| Proses pengerjaan | Jika dibeli pada usia 30, maka premi yg harus dibayar \({{\ddot a}_{30:\left. {\overline {\, 3 \,}}\! \right| }} = \sum\limits_{k = 0}^2 {{v^k} \cdot {}_k{p_{30}}} \) \({{\ddot a}_{30:\left. {\overline {\, 3 \,}}\! \right| }} = 1 + v \cdot {p_{30}} + {v^2} \cdot {}_2{p_{30}}\) \({{\ddot a}_{30:\left. {\overline {\, 3 \,}}\! \right| }} = 1 + \frac{{0.99}}{{1.04}} + \frac{{\left( {0.99} \right)\left( {0.98} \right)}}{{{{1.04}^2}}} = 2.848928\) \(A_{30:\left. {\overline {\, 3 \,}}\! \right| }^1 = {A_{30:\left. {\overline {\, 3 \,}}\! \right| }} – {}_3{E_{10}}\) \(A_{30:\left. {\overline {\, 3 \,}}\! \right| }^1 = {A_{30:\left. {\overline {\, 3 \,}}\! \right| }} – {}_3{E_{10}}\) \(A_{30:\left. {\overline {\, 3 \,}}\! \right| }^1 = 1 – \frac{{0.04}}{{1.04}}\left( {2.848928} \right) – \frac{{\left( {0.99} \right)\left( {0.98} \right)\left( {0.97} \right)}}{{{{1.04}^3}}} = 0.053797\) \(P = \frac{{{b_1}A_{30:\left. {\overline {\, 3 \,}}\! \right| }^1}}{{{{\ddot a}_{30:\left. {\overline {\, 3 \,}}\! \right| }}}} = \frac{{1,000\left( {0.053797} \right)}}{{2.848928}}\) \(P = 18.883243\) |
| Karena usia ternyata 31 tahun maka \({{\ddot a}_{31:\left. {\overline {\, 3 \,}}\! \right| }} = \sum\limits_{k = 0}^2 {{v^k} \cdot {}_k{p_{31}}} \) \({{\ddot a}_{31:\left. {\overline {\, 3 \,}}\! \right| }} = 1 + v \cdot {p_{31}} + {v^2} \cdot {}_2{p_{31}}\) \({{\ddot a}_{31:\left. {\overline {\, 3 \,}}\! \right| }} = 1 + \frac{{0.98}}{{1.04}} + \frac{{\left( {0.98} \right)\left( {0.97} \right)}}{{{{1.04}^2}}} = 2.821191\) \(A_{30:\left. {\overline {\, 3 \,}}\! \right| }^1 = {A_{30:\left. {\overline {\, 3 \,}}\! \right| }} – {}_3{E_{10}}\) \(A_{30:\left. {\overline {\, 3 \,}}\! \right| }^1 = \left( {1 – d{{\ddot a}_{30:\left. {\overline {\, 3 \,}}\! \right| }}} \right) – {v^3} \cdot {}_3{p_{30}}\) \(A_{30:\left. {\overline {\, 3 \,}}\! \right| }^1 = 1 – \frac{{0.04}}{{1.04}}\left( {2.821191} \right) – \frac{{\left( {0.98} \right)\left( {0.97} \right)\left( {0.96} \right)}}{{{{1.04}^3}}} = 0.080216\) \(P = \frac{{{b_1}A_{30:\left. {\overline {\, 3 \,}}\! \right| }^1}}{{{{\ddot a}_{30:\left. {\overline {\, 3 \,}}\! \right| }}}}\) \(18.883243 = \frac{{{b_1}\left( {0.080216} \right)}}{{2.821191}}\) \({b_1} = 664.12231\) | |
| Penyesuaiannya adalah: \(1,000 – 664.12231 = 335.87769\) | |
| Jawaban | d. 335,90 |