Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Misalkan Anda melakukan smoothing deret waktu \({y_t}\) menggunakan metode exponential smoothing 2-parameter dari Holt:
| \(t\) | \({y_t}\) | \({\tilde y_t}\) | \({r_t}\) |
| 1995 | 120,50 | 117,50 | 12,00 |
| 1996 | 135,00 | 130,88 | 12,96 |
| 1997 | 147,70 | 144,80 | 13,64 |
| 1998 | 146,60 | \({\tilde y_{1998}}\) | \({r_{1998}}\) |
Hitunglah forecast 2-periode \({\hat y_{2000}}\) dengan terlebih dahulu melengkapi table di atas dengan deret exponential 2-parameter dari Holt
- Lebih kecil dari 166
- Paling sedikit 166, tetapi lebih kecil dari 172
- Paling sedikit 172, tetapi lebih kecil dari 176
- Paling sedikit 176, tetapi lebih kecil dari 180
- Paling sedikit 180
| Diketahui | smoothing deret waktu \({y_t}\) menggunakan metode exponential smoothing 2-parameter dari Holt:
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| Rumus yang digunakan | \({{\tilde y}_t} = \alpha {y_t} + \left( {1 – \alpha } \right)\left( {{{\tilde y}_{t – 1}} + {r_{t – 1}}} \right)\) \({r_t} = \gamma \left( {{{\tilde y}_t} – {{\tilde y}_{t – 1}}} \right) + \left( {1 – \gamma } \right){r_{t – 1}}\) \({{\hat y}_{T + l}} = {{\tilde y}_T} + l{r_T}\) | ||||||||||||||||||||
| Proses pengerjaan | Mencari \(\alpha \) dan \(\gamma \) \({{\tilde y}_{1996}} = \alpha {y_{1996}} + \left( {1 – \alpha } \right)\left( {{{\tilde y}_{1995}} + {r_{1995}}} \right)\) \(130.88 = 135\alpha + \left( {1 – \alpha } \right)\left( {117.5 + 12} \right)\) \(130.88 = 135\alpha + 129.5 – 129.5\alpha \) \(5.5\alpha = 1.38\) \(\alpha = 0.2509091 \approx 0.25\) \({r_{1996}} = \gamma \left( {{{\tilde y}_{1996}} – {{\tilde y}_{1995}}} \right) + \left( {1 – \gamma } \right){r_{1995}}\) \(12.96 = \gamma \left( {130.88 – 117.5} \right) + \left( {1 – \gamma } \right)12\) \(12.96 = 13.38\gamma + 12 – 12\gamma \) \(1.38\gamma = 0.96\) \(\gamma = 0.6956522 \approx 0.7\) | ||||||||||||||||||||
| Melengkapi tabel \({\tilde y_{1998}}\) dan \({r_{1998}}\) \({{\tilde y}_{1998}} = \alpha {y_{1998}} + \left( {1 – \alpha } \right)\left( {{{\tilde y}_{1997}} + {r_{1997}}} \right)\) \({{\tilde y}_{1998}} = 0.25\left( {146.6} \right) + 0.75\left( {144.8 + 13.64} \right)\) \({{\tilde y}_{1998}} = 155.48\) \({r_{1998}} = \gamma \left( {{{\tilde y}_{1998}} – {{\tilde y}_{1997}}} \right) + \left( {1 – \gamma } \right){r_{1997}}\) \({r_{1998}} = 0.7\left( {155.48 – 144.8} \right) + 0.3\left( {13.64} \right)\) \({r_{1998}} = 11.568\) | |||||||||||||||||||||
| \({{\hat y}_{1998 + 2}} = {{\tilde y}_{1998}} + 2{r_{1998}}\) \({{\hat y}_{1998 + 2}} = 155.48 + 2 \cdot 11.568\) \({{\hat y}_{1998 + 2}} = 178.616\) | |||||||||||||||||||||
| Jawaban | d. Paling sedikit 176, tetapi lebih kecil dari 180 |