Solutions to Extra Problems 3 a

# Solutions to Extra Problems 3 a - ACTSC 231 Solutions Extra...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ACTSC 231 Solutions Extra Problems 3 a 1. (a) 300a6j = 300 1 (b) 300a6j = 300 1 (c) 300a6j = 300 1 ln 1 + i(m) m m v6 v6 = 300 1 = 300 1 e :04(6) :04 (1:04) ln 1:04 1 6 = 1600: 291 = 1603: 888 3 Note 4(6) = ln (1 + i) Note = v6 = 300 (1+ :04 ) 4 4 ln(1+ :04 ) 4 = 1601: 207 1 (d) 300a6j = 300 1 (e) a 1 v6 = 300 1 (1 :04)6 ln(1 :04) 1 2 = 1596: 508 5 Note = =e 1 2 [ln(1+:01t) ln (1 ln(1)] d) = (t) = e 1 2 eln(1+:01t) = (1 + :01t) Rt :005 ds 0 1+:01s =e 1 2 ln (1+:01s)jt ds 0 300a6j = = = 300 300 300 Z 6 a 6 1 (t) dt 1 2 1 2 (1 + :01t) 2 :01 0 1 2 = 300 (1:06) 2 1 :01 = 1773: 780 8 Z 0 (1 + :01t) dt 6 0 2. a 1 (t) = e 0 s ds = e 0 ::005=(1+:0025s)ds = e : 2 ln(1+:0025s)j0 = e 2 (1 + :0025t) R 10 R 10 2 (400 + t) (1 + :0025t) dt = 400 0 (1 + :0025t) (1 + :0025t) 0R 10 10 1 400 400 0 (1 + :0025t) dt = :0025 ln (1 + :0025t) 0 = 160000 ln (1:025) = 3950:82 3. Let A denote the P.V. Rt Rt t 2 ln(1+:0025) = 2 dt = 1 A= = = 5v + 5v 2 + 5 (1:02) v 3 + 5 (1:02) v 4 + 5 (1:02) v 5 + 5 (1:02) v 6 + 5 (1:02) v 7 + 5 (1:02) v 8 + 5 v + v 2 + 5 (1:02) v 2 v + v 2 + 5 (1:02) v 4 v + v 2 + 5 (1:02) v 6 v + v 2 + 0 1 !2 !3 1:02 1:02 1:02 A 5 v + v 2 @1 + + + 2+ 2 2 (1:05) (1:05) (1:05) • 5 v + v 2 a1jj , where vj = 1 2 3 2 2 3 3 = = = = 5 (1:05) 5 (1:05) + (1:05) + (1:05) 2 1 2 1 1:02 (1:05) = )j= 2= 1+j 1:02 (1:05) 1 :1 vj 1 1 02 1 (1::05)2 2 1 = 8:0882353% 124: 242 42 2 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

Solutions to Extra Problems 3 a - ACTSC 231 Solutions Extra...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online