chapter3A - Additional Solutions: Chapter Three: Cash Flow...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Additional Solutions: Chapter Three: Cash Flow Analysis 3S.1 Mr Dashwood must find the least value of N for which 1 500 < 100 (P/A,5%,N) which is equivalent to finding N such that (P/A,5%,N) > 15 Consulting Appendix A, this first occurs when N = 29 years. If the interest rate is 10%, we come to the end of the table in Appendix A before finding a value of N for which (P/A,10%,N) > 15. We therefore go on to look at the capitalized value of the annuity, which is 100/0.1, or £1 000. Thus, if he can invest his money at 10%, Mr Dashwood can afford to support his widowed half-sister indefinitely for a cost equivalent to a single present payment of £1 000. *3S.2 There are several ways of tackling this. One way is to convert the bi-annuity to an equivalent annuity: A1 = A(A/F,i,2) where ¥ A1 is the annual payment equivalent to getting a final payment of ¥ A at the end of two years. Then we can convert ¥ A1 to its present worth using a formula we already know: P=A1(P/A,i,2N)=A(A/F,i,2)(P/A,i,2N) (Don’t forget that it’s now 2 N rather than N .) Another solution is to calculate the effective biennial interest rate, j , from the equation j = (1 + i ) 2 -1 = 2 i + i 2 and then to use a conversion factor based on j : P = A(P/A, j, N) A third alternative is to construct a formula based on the one at the back of the book, (P/A, j, N) = ((1 + j ) N -1)/( j (1 + j ) N ) and substitute in the value of j , the effective biennial interest rate, as calculated above. This is a correct solution, but it’s more work. Also, I would advise against using these algebraic formulas, except when you’re constructing a spreadsheet – evaluating them is more work than looking a number up in the tables, and there are more opportunities for making a slip.
Background image of page 1

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

View Full DocumentRight Arrow Icon
3S.3 Having drawn a cash-flow diagram, we write down the PW equivalent of each of the costs: PW = 32 000 -16 000 (P/F, 15%,8 ) + 20 000 (P/ A , 15%,8 ) + 600 + 550 (P/A, 15%,7 ) – 50 (P/G, 15%,7) = 32 000 -16 000(0.3269) + 20 000 (4.6587) + 600 + 550(4.1604) – 50(10.192) = 122 322. The P/ A denotes a continuous cash flow, continuously compounded – this is the kind that you look up in Appendix C. We represent the decreasing insurance costs as an annuity plus a negative arithmetic gradient. There isn’t a P/G column in Appendix A, but we can make one by multiplying the P/A entry by the A/G entry. We have to treat the first insurance payment of 600 separately in order for the remaining payments to fit the standard pattern of an annuity, which always starts at the end of Year 1. Another possible interpretation of the question is to treat the labour costs as occurring at year’s end
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/16/2009 for the course ENSC 201 taught by Professor Dr.johnjones during the Fall '08 term at Simon Fraser.

Page1 / 5

chapter3A - Additional Solutions: Chapter Three: Cash Flow...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online