HW2 Soln 10

# HW2 Soln 10 - Eng 106 Homework #2 Solution Winter 2010 2.40...

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Eng 106 Homework #2 Solution Winter 2010 2.40 Given: Cash deposits and withdrawals (negative amounts) below, i=12%/yr. EOY: 0 1 2 3 4 5 6 7 8 9 10 Flow: 1,000 800 600 400 200 0 -C -2C -3C -4C -5C Find: the one of four values of C given in the book that makes the deposit series equivalent to the withdrawal series. Use linear gradient relationships for both the deposits and withdrawals. There are several correct ways of addressing this problem, but all will give the same result. Here is one possibility: bring all amounts to EOY 4. 1000(F/A, 12%, 5) – 200(P/G, 12%, 5)(F/P, 12%, 5) – C(P/G, 12%, 6) = 0 Using table values: 1000*6.3528 – 200*6.3970*1.7623 – C*8.9302 = 0 Solve for C = 458.90, which is answer (d) given in the book 2.43 What is the amount A of each of 10 equal annual deposits, the first being @EOY 1, that can provide five annual withdrawals, where the first withdrawal of \$3,000 is made @EOY 11 and subsequent withdrawals increase at the rate of 6% per year over the previous year’s, if a) i=8%/yr There are multiple ways of solving this. One is to bring the two series to EOY 10. A(F/A, 8%, 10) – \$3k(P/A 1 , g, i, N) = 0 where g=6%, i=8%, N=5 A*14.4866 - \$3k{[1-(1.06) 5 (1.08) -5 ]/(0.08-0.06)} = 0 A*14.4866 - \$13,383.92 = 0 Solve for A = \$923.88 b) i=6%/yr Bringing the two series to EOY 10: A(F/A, 6%, 10) – \$3k(P/A 1 , g, i, N) = 0 where g=i=6%, N=5 A*13.1808 - \$3k{5/1.06} = 0 A*13.1808 - \$14,150.94 = 0 Solve for A = \$1,073.60 2.53 Given: Deposits and interest rates shown below, current balance = \$1,000. Year: -4 -3 -2 -1 0 Deposit @EOY, \$: 200 X 0 300 0 i during the year, %/yr 6 8 12 15 Find: Amount X of deposit @EOY 3. there are many possible approaches, although the easiest is to apply the relevant interest rates to each deposit separately to bring them to the present balance. This isolates the unknown X.

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## HW2 Soln 10 - Eng 106 Homework #2 Solution Winter 2010 2.40...

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