Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | November 2014 |
| Nomor Soal | : | 13 |
SOAL
Tentukan nilai dari \(1000\left( {{}_2{V_{x:\left. {\overline {\, 3 \,}}\! \right| }} – {}_1{V_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right)\), bila menggunakan tingkat bunga tahunan 6% dan nilai sebagai berikut: \({l_x} = 100\), \({l_{x + 1}} = 90\), dan \({P_{x:\left. {\overline {\, 3 \,}}\! \right| }} = 0,3251\)
- 330,38
- 230,83
- 130,83
- 133,38
- Tidak ada jawaban yang benar
| Diketahui | Tingkat bunga tahunan 6% dan nilai sebagai berikut: \({l_x} = 100\), \({l_{x + 1}} = 90\) dan \({P_{x:\left. {\overline {\, 3 \,}}\! \right| }} = 0,3251\) |
| Rumus yang digunakan | Rumus rekursif untuk reserves \({}_{k + s}{V_{x:\left. {\overline {\, n \,}}\! \right| }} \cdot {}_s{p_{x + k}} + {b_{k + 1}} \cdot {}_s{q_{x + k}} = \left( {{}_k{V_{x:\left. {\overline {\, n \,}}\! \right| }} + {b_{k + 1}} \cdot {P_{x:\left. {\overline {\, n \,}}\! \right| }}} \right){\left( {1 + i} \right)^s}\) \({P_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{{A_{x:\left. {\overline {\, n \,}}\! \right| }}}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}\) dengan \({A_{x:\left. {\overline {\, n \,}}\! \right| }} = 1 – d{\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }}\) dan \(d = 1 – v = 1 – \frac{1}{{1 + i}}\) \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \sum\limits_{k = 0}^{n – 1} {{v^k}{}_k{p_x}} \) dengan \({}_t{p_x} = \prod\limits_{k = 0}^{t – 1} {{p_{x + k}}} \) dan \({}_t{p_x} = \frac{{{l_{x + t}}}}{{{l_x}}}\) |
| Proses pengerjaan | \({P_{x:\left. {\overline {\, 3 \,}}\! \right| }} = \frac{{{A_{x:\left. {\overline {\, 3 \,}}\! \right| }}}}{{{{\ddot a}_{x:\left. {\overline {\, 3 \,}}\! \right| }}}} = \frac{{1 – d{{\ddot a}_{x:\left. {\overline {\, 3 \,}}\! \right| }}}}{{{{\ddot a}_{x:\left. {\overline {\, 3 \,}}\! \right| }}}}\) \(0.3251 = \frac{{1 – \left( {\frac{{0.06}}{{1.06}}} \right){{\ddot a}_{x:\left. {\overline {\, 3 \,}}\! \right| }}}}{{{{\ddot a}_{x:\left. {\overline {\, 3 \,}}\! \right| }}}}\) \(0.3251{{\ddot a}_{x:\left. {\overline {\, 3 \,}}\! \right| }} = 1 – 0.056604{{\ddot a}_{x:\left. {\overline {\, 3 \,}}\! \right| }}\) \({{\ddot a}_{x:\left. {\overline {\, 3 \,}}\! \right| }} = \frac{1}{{0.3251 + 0.056604}} = 2.619833\) |
| \({p_x} = \frac{{{l_{x + 1}}}}{{{l_x}}} = \frac{{90}}{{100}} = 0.9\) \({{\ddot a}_{x:\left. {\overline {\, 3 \,}}\! \right| }} = \sum\limits_{k = 0}^2 {{v^k}{}_k{p_x}} = {v^0}{}_0{p_x} + v{p_x} + {v^2}{}_2{p_x}\) \(2.619833 = 1 + \frac{{0.9}}{{1.06}} + \frac{{0.9{p_{x + 1}}}}{{{{1.06}^2}}}\) \({p_{x + 1}} = \frac{{{{1.06}^2}}}{{0.9}}\left( {2.619833 – 1 – \frac{{0.9}}{{1.06}}} \right)\) \({p_{x + 1}} = 0.962272\) | |
| \({V_{x:\left. {\overline {\, 3 \,}}\! \right| }} \cdot {p_x} + {b_1} \cdot {q_x} = \left( {{}_0{V_{x:\left. {\overline {\, 3 \,}}\! \right| }} + {b_1} \cdot {P_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right)\left( {1 + i} \right)\) \(0.9{V_{x:\left. {\overline {\, 3 \,}}\! \right| }} + 1,000\left( {0.1} \right) = \left( {325.1} \right)\left( {1.06} \right)\) \({V_{x:\left. {\overline {\, 3 \,}}\! \right| }} = \frac{{\left( {325.1} \right)\left( {1.06} \right) – 100}}{{0.9}} = 271.784444\) | |
| \({}_2{V_{x:\left. {\overline {\, 3 \,}}\! \right| }} \cdot {p_{x + 1}} + {b_2} \cdot {q_{x + 1}} = \left( {{}_1{V_{x:\left. {\overline {\, 3 \,}}\! \right| }} + {b_2} \cdot {P_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right)\left( {1 + i} \right)\) \(0.962272{}_2{V_{x:\left. {\overline {\, 3 \,}}\! \right| }} + 1,000\left( {1 – 0.962272} \right) = \left( {271.784444 + 325.1} \right)\left( {1.06} \right)\) \({}_2{V_{x:\left. {\overline {\, 3 \,}}\! \right| }} = \frac{{\left( {271.784444 + 325.1} \right)\left( {1.06} \right) – 1,000\left( {1 – 0.962272} \right)}}{{0.962272}} = 618.296605\) | |
| Maka, \(1000\left( {{}_2{V_{x:\left. {\overline {\, 3 \,}}\! \right| }} – {}_1{V_{x:\left. {\overline {\, 3 \,}}\! \right| }}} \right) = 618.296605 – 271.784444 = 346.512161\) |
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| Jawaban | e. Tidak ada jawaban yang benar |
