ACTSC221 Lecture 1: chap2solstudent_notes.pdf
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Part a: s = 100(1. 055)5 = . 70, = . 65: s = 50(1. 005)48 = . 52, 13: s = 500(1 + 0. 04, s = 500(1 + 0. 08, s = 500(1 + 0. 12. 365 )365 1 = 0. 127474614 = 12. 75% 3. a) (1 + i)4 = (1. 04)2 b) (1 + i)2 = (1. 05)4. I = (1. 04)1/2 1 j4 = 4[(1. 04)1/2 1] = 7. 92% I = (1. 015)2 1 j2 = 2[(1. 015)2 1] = 6. 05% c) d) e) f) g) h) (1 + i)4 = (1 + 0. 18. 12 )2 i = (1. 015)3 1 j4 = 4[(1. 015)3 1] = 18. 27% (1 + i)12 = (1 + 0. 1. 6 )6 i = (1 + 0. 1. I = (1. 02)2 1 j2 = 2[(1. 02)2 1] = 8. 08% (1 + i)2 = (1. 02)4 (1 + i)2 = (1 + 0. 04. 52 )52 i = (1 + 0. 04. 52 )26 1] = 4. 04% (1 + i)12 = (1 + 0. 0525. )1/6 1 j12 = 12[(1 + 0. 0525.